Concept Of Physics Author : Pradeep Concepts of Physics is a calculus-based physics textbook by H. C. Verma. The book was first published in 1992. It is published as a two-volume set, with each volume roughly covering the physics syllabus of class XI and class XII respectively. The book is extensively popular amongst students preparing for competitive exams, especially the Joint Entrance Examination. The first volume covers mechanics, waves and optics, while the second volume covers thermodynamics, electromagnetism, modern physics and Theory of relativity. 1. Find the dimensions of B. frequency and A. linear momentum, C. pressure. Solutions : B. Frequency 1 t =[M 0 L 0 T -1 ] A. mv =[MLT -1 ] C. F orce Area = [MLT -2 ] L 2 =[ML -1 T -2 ] 2. Find the dimensions of D. moment of interia I A. angular speed ω ω = θ 2 - θ 1 t 2 - t 1 ,α = ω 2 - ω 1 t 2 - t 1 , Γ= F.r and I = mr 2 The symbols have standard meanings. C. torque Γ and Some of the equations involving these quantities are B. angular acceleration α Solutions : D. moment of interia I = Mr 2 =[M][L 2 ]=[ML 2 T 0 ] A. angular speed ω = θ t =[M 0 L 0 T -1 ] C. torque Γ = Fr =[MLT -2 ][L]=[ML 2 T -2 ] B. angular acceleration α = ω t = M 0 L 0 T -2 T =[M 0 L 0 T -3 ] 3. Find the dimensions of The relevant equations are lt;brgt; F = qE,F = qvB, and B = μ 0 I 2πα A. electric field E, C. magnetic permeability μ 0 where F is force, q is charge,v is speed, I is current, and a is distance. B. magnetic field B and Solutions : A. Electric field E = F q = MLT -2 [IT ] =[MLT -3 -1] C. Magnetic permeability μ 0 = B2πa I = MT -2 I -1 [L] [I ] =[MLT -2 I -2 ] B. Magnetic field B = F qv = MLT -2 [IT ][LT -1 ] =[MT -2 I -1 ] 4. Find the dimensions of A. electric dipole moment p and The defining equations are p = q.d and M = IA; where d is distance, A is area, q is charge and I is current. B. electric dipole moment M Solutions : A. Electric dipole moment P = qI =[IT ][L]=[LT I ] B. Magnetic dipole moment M = IA =[I ][L 2 ][L 2 I ] 5. Find the dimensions of Planck’s constant h from the equation E = hv where E is the energy and v is the frequency. Solutions : E = hv where E = energy and v = frequency lt;brgt; h = E v = [ML 2 T -2 ] T -1 =[ML 2 T -1 ] 6. Find the dimensions of Some of the equations involving these quantities are A. the specific heat capacity c, C. the gas constant R. Q = mc(T 2 - T 1 ),lt = l 0 [1 + a(T 2 - T 3 )] and PV = nRT.
Transcript
Concept Of Physics
Author : PradeepConcepts of Physics is a calculus-based physics textbook by H. C. Verma. The book was first published in 1992. It is published
as a two-volume set, with each volume roughly covering the physics syllabus of class XI and class XII respectively. The book isextensively popular amongst students preparing for competitive exams, especially the Joint Entrance Examination. The first volumecovers mechanics, waves and optics, while the second volume covers thermodynamics, electromagnetism, modern physics and Theoryof relativity.
1. Find the dimensions of
B. frequency and
A. linear momentum,
C. pressure.
Solutions :B. Frequency
1
t= [M0L0T−1]
A. mv = [MLT−1]
C.Force
Area=
[MLT−2]
L2= [ML−1T−2]
2. Find the dimensions of
D. moment of interia I
A. angular speed ω
ω =θ2 − θ1t2 − t1
, α =ω2 − ω1
t2 − t1,Γ = F.r and I = mr2 The symbols have standard meanings.
C. torque Γ and
Some of the equations involving these quantities are
B. angular acceleration α
Solutions :D. moment of interia I = Mr2 = [M ][L2] = [ML2T 0]
A. angular speed ω =θ
t= [M0L0T−1]
C. torque Γ = Fr = [MLT−2][L] = [ML2T−2]
B. angular acceleration α =ω
t=M0L0T−2
T= [M0L0T−3]
3. Find the dimensions of
The relevant equations are lt;brgt; F = qE, F = qvB, and B =µ0I
2παA. electric field E,
C. magnetic permeability µ0
where F is force, q is charge,v is speed, I is current, and a is distance.
B. magnetic field B and
Solutions :
A. Electric field E =F
q=MLT−2
[IT ]= [MLT−3−1]
C. Magnetic permeability µ0 =B2πa
I=MT−2I−1[L]
[I]= [MLT−2I−2]
B. Magnetic field B =F
qv=
MLT−2
[IT ][LT−1]= [MT−2I−1]
4. Find the dimensions of
A. electric dipole moment p and
The defining equations are p = q.d and M = IA; where d is distance, A is area, q is charge and I is current.
B. electric dipole moment M
Solutions :A. Electric dipole moment P = qI = [IT ][L] = [LTI]
B. Magnetic dipole moment M = IA = [I][L2][L2I]
5. Find the dimensions of Planck’s constant h from the equation E = hv where E is the energy and v is the frequency.
Solutions :E = hv where E = energy and v = frequency lt;brgt;
h =E
v=
[ML2T−2]
T−1= [ML2T−1]
6. Find the dimensions of
Some of the equations involving these quantities are
Taking force, length and time as fundamental quantityF
LT−2[LT−1]2 =
F
LT−2[L2T−2] = [FL]
B. Pressure =F
A=
F
L2= [FL−2]
D. Energy =1
2mv2 =
Force
acceleration(velocity)2
A. Density =m
V=
(force/acceleration)
V olume=
[F/LT−2]
[L2]=
F
L4T−2= [FL−4T 2]
8. Suppose the acceleration due to gravity at a place is 10 m/s2 . Find its value in cm/(minute)2
Solutions :g = 10
meter
sec2= 36× 105cm/min2
9. If the average speed of a snail is 0.02 mile/hr and the average speed of leopard is 70 miles/hr. Convert these speeds in S.I Units?
Solutions :The average speed of snail = 0.02 mile/hr converting to S.I units, 0.02X1.6X1000
3600m/sec [1mile = 1.6km = 1600m] = 0.0089ms−1
the average speed of leopard = 70 miles/hr In S.I Units = 70 miles/hour = 70X1.6X10003600
= 31 m/s
10. The height of mercury column in a barometer in a Calcutta laboratory was recorded to be 75 cm. Calculate this pressure in SIand CGS units using the following data : Specific gravity of mercury =13.6, Density of water = 1033 kg/m3, g = 9.8 in/s2 atCalcutta. Pressure =hpg in usual symbols.
Solutions :Height h = 75cm, Density of mercury = 13600kg/m3, g = 9.8ms−2 then, Pressure = hfg = 10X104 N/m2 (approximately) In C.G.SUnits, P = 10X105 dyne/cm
11. Express the power of a 100 watt bulb in CGS unit.
12. The normal duration of I.Sc. Physics practical period in Indian colleges is 100 minutes. Express this period in micro centuries.1 micro century = 10−6 x 100 years. How many micro centuries did you sleep yesterday ?
Solutions :In 1 micro century = 104 X 100 years = 104 X 365 X 24 X 60 min . So 100 min = 105/5256 = 1.9 micro century.
13. The surface tension of water is 72 dyne/cm. Convert it in SI unit.
Solutions :Surface tension of water = 72 dyne/cm. In S.I Unit, 72 dyne/cm = 0.072 N/m
14. The kinetic energy K of a rotating body depends on its moment of inertia I and its angular speed ω. Assuming the relation tobe K = kIaωb where k is a dimensionless constant, find a and b. Moment of inertia of a sphere about its diameter is 2
5Mr2.
Solutions :According to principle to homogeneity of dimensions, [ML2T−2] = [ML2T−2][T−1]b
K = KIaωb where k = kinetic energy of rotating body and k = dimensionless constant
Dimensions of right side are, Ia = [ML2]a , ω = [T−1]b
Equating the dimensions of both sides, 2 = 2a and −2 = −b ⇒ a = 1 and b = 2
Dimensions of left side are, K = [ML2T−2]
15. Theory of relativity reveals that mass can be converted into energy. The energy E so obtained is proportional to certain powersof mass m and the speed c of light. Guess a relation among the quantities using the method of dimensions.
Solutions :Dimensions of right side, Ma = [M ]a , [C]b = [LT−1]b
Let energy E ∞ MaCb, where M = Mass, C = Speed of light
⇒ a = 1 ; b = 2
Dimensions of the left side, E = [ML2T−2]
Therefore, [ML2T−2] = [M ]a[LT−1]b
⇒ E = KMaCb K = ( Proportionality constant)
So, the relation is E = KMC2
Page 2
24Tutors.com
16. Let I = current through a conductor, R = its resistance and V = potential difference across its ends. According to Ohm’s law,product of two of these quantities equals the third. Obtain Ohm’s law from dimensional analysis. Dimensional formulae for Rand V are ML2I−2T−3 and ML2T 3I−1 respectively.
17. The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass perunit length m. Guess the expression for its frequency from dimensional analysis.
Solutions :La = [La], F b = [MLT−2]b, Mc = [ML−1]c
Frequency f = KLaF bMcM = Mass/unit length, L = length, F = tension (force)
a = −1, b = 12
and c = −12
M0L0T−1 = KMb+cLa+b−cT−2b
−c+ a+ b = 0
Dimensions of right side,
Now, solving the equation we get,
Therefore, [T−1] = K[L]a [MLT−2]b [ML−1]c
Therefore, b+ c = 0
Dimensions of f = [T−1]
Therefore, So frequency f = KL−1F12M
−12 = K
LF
12M
−12 = K
L=√
FM
Equating the dimensions of both sides,
−2b = −1
18. Test if the following equations are dimensionally correct :
d) ν = 12π
√mglI
a) h = 2SCosθprg
c) V = πPr4t8ηι
where h = height, S = surface tension, ρ = density, P = pressure, V = volume, η = coefficient of viscosity, ν = frequency and I = momentof inertia.
b) v =√pρ
Solutions :Density = ρ = M
V= [ML−3T 0]
So relation is correct.
RHS =√
(mgl/I) =
√[M ][LT−2]
[ML2]= [T−1]
a) h = 2SCosθprg
Dimension of p = [ML−1T−2], r4 = [L4], t = [T ]
Dimension of p = FA
= [ML−1T−2]
So, the relation is correct.
RHS = 2SCosθprg
=[MT−2]
[ML−3T0][L][LT−2]= [M0L1T 0] = [L]
So, the relation is correct
b) v =√Pρ
where v = velocity
RHS = πpr4t8ηI
=[ML−1T−2][L4][T ]
[ML−1T−2][L]
Surface tension =S = F/I = MLT−2
L= [MT−2]
LHS = Dimesion of v = [T−1]
RHS =√Pρ
=
√[ML−1T−2]
[ML−3]= [L2T−2]
12 = [LT−1]
LHS = Dimension of V = [L3]
LHS = dimension of V = [LT−1]
Radius = r = [L], g = [LT−2]
So, the relation is correct.
LHS = RHS
LHS = [L]
Coefficient of viscosity = [ML−1T−1]
Dimension of ρ = mV
= [ML−3]
d) v = 12π
√(mgl/l)
LHS = RHS
c) V =(πpr4t)(8ηI)
LHS = dimension of V = [Lt−1]
Page 3
24Tutors.com
19. Let x and a stand for distance. Is∫
dx√a2−x2
= 1asin−1 a
xdimensionally correct?
Solutions :So, the equation is dimensionally incorrect.
Dimensions of the left side =∫
dx√(a2−x2)
=∫
L√(L2−L2)
= [L0]
So, dimensions of∫
dx√(a2−x2)
6= 1aSin−1
(ax
)
Dimensions of the right side = 1aSin−1
(ax
) = [L−1]
20. A vector ~A makes an angle of 20 and ~B makes an angle of 110 with the X-axis . The magnitudes of these vectors are 3m and4m respectively. Find the resultant.
Solutions :Resultant R =
√A2 +B2 + 2ABcosθ = 5m
As shown in the figure ,
β = tan−1(
4sin90o
3+4cos90o) = tan−1
(43
) = 53o
| ~A| = 3m and | ~B| = 4m
Let β be the angle between ~R and ~A
The angle between ~A and ~B = 110o - 20o = 90o
therefore, Resultant vector makes angle (53o + 20o) = 73o with x-axis
21. Let ~A and ~B be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles 30 and 60 respectively,find the resultant.
Solutions :β be the angle between ~R and ~A
Angle between ~A and ~B is θ = 60o - 30o = 30o
Therefore, Resultant makes 15o + 30o = 45o angle with x-axis
R =√
10o + 10o + 2.10.10.cos30o = 19.3
β = tan−1(
10sin30o
10+10cos30o) = tan−1
(1
2+√
3) = tan−1(0.26795) = 15o
| ~A| and | ~B| = 10 unit
22. Add vectors ~A, ~B and ~C each having magnitude of 100 unit and inclined to the X-axis at angles 45, 135 and 315 respectively.
Solutions :x component of ~A = 100cos45o = 100√
2unit
The resultant is 100 unit at 45o with x-axis
y component of ~B = 100cos135o = 100√2
Resultant = 100
x component of ~C = 100cos315o = 100√2
⇒ α = tan−1(1) = 45o
y component of ~A = 100cos45o = 100√2
unit
Resultant Y component = 100√2
+ 100√2
- 100√2
= 100√2
x component of ~B = 100cos135o = 100√2
y component of ~C = 100cos315o = 100√2
Tan α = y componentx component
= 1
Resultant x component = 100√2
- 100√2
+ 100√2
= 100√2
23. Let ~a = ~4i + ~3j and ~b = ~3i + ~4j.
(a) Find the magnitudes of
(a) ~a , (b) ~b, (c) ~a+~b and (d) ~a−~b .
Solutions :c) |~a+~b| = |~7i+ ~7j| = 7
√2
~a = ~4i + ~3j , ~b = ~3i + ~4j
a) |~a+~b| =√
12 + (−1)2 =√
2
b) |~b| =√
9 + 16 = 5
a) ~a−~b = (−3 + 4) i + (−4 + 3)j = i− ja) |~a| =
√42 + 3 ∗ 2 = 5
24. Refer to figure (2 − E1). Find (a) the magnitude, (b) x and y components and (c) the angle with the X-axis of the resultantof ~OA, ~BC and ~DE.
Solutions :y components of ~OA = 2sin30o = 1
x components of ~OA = 2cos30o =√
3
⇒ α = tan−11.32
y components of ~DE = 1sin270o = −1
So, R = Resultant = 1.6m
x components of ~DE = 1cos270o = 0
Page 4
24Tutors.com
Tan α = y componentx component
= 1.32
y components of ~BC = 1.5sin120o = 1.3
Ry = resultant y component = 1 + 1.3 - 1 = 1.3m
x components of ~BC = 1.5cos120o = −0.75
Rx = x component of resultant =√
3 - 0.75 + 0 = 0.98m
If it makes and angle α with positve x-axis
25. Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them if the magnitude of theresultant is (a) 1 unit, (b) 5 unit and (c) 7 unit.
Solutions :b)√
32 + 42 + 2.3.4.cosθ = 5
|~a| = 3m |~b| = 4m
c)√
32 + 42 + 2.3.4.cosθ = 7
θ = 180o
θ = 90o
Angle between them is 0o
a) If R = 1 Unit ⇒√
32 + 42 + 2.3.4.cosθ = 1
θ = 0o
26. A spy report about a suspected car reads as follows. ”The car moved 2.00 km towards east, made a perpendicular left turn,ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped”. Find the displacement of the car.
Solutions :θ = tan−1
(112
)
~AD = 2i + 0.5j+ 4K = 6i + 0.5j
T anθ = DEAE
= 112
The displacement of the car is 6.02 km along the distance tan−1 with positive x-axis
AD =√AE2 +DE2 = 6.02 KM
27. A carrom board (4 ft x 4 ft square) has the queen at the centre. The queen, hit by the striker moves to the front edge, reboundsand goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the centre to the frontedge, (b) from the front edge to the hole and (c) from the centre to the hole.
Solutions :a) In ∆ABC, AC =
√AB2 +BC2 = 2
3
√10 ft
In ∆ABC, tanθ = x2
and in ∆DCE, tanθ =(2−x)
4tanθ = x
2=
(2−x)4
= 4x
⇒ 6x = 4 ⇒ x = 23
ft
⇒ 4 = 2x = 4x
28. A mosquito net over a 7 ft x 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through which a mosquito entersthe net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of themosquito. (b) Taking the hole as the origin, the length of the bed as the X-axis, its width as the Y-axis, and vertically up asthe Z-axis, write the components of the displacement vector.
Solutions :Here is the displacement vector ~r = 7i + 4j + 3k
b) the components of displacement vector are 7 ft, 4 ft and 3 ft
a) magnitude of displacement =√
74 ft
29. Suppose ~a is a vector of magnitude 4.5 unit due north. What is the vector (a) 3~a, (b)−4~a?
Solutions :−4|~a| = −4 x 1.5 = −6unit
~a is a vector of magnitude 4.5 unit due north.
3~a is along north having magnitude 13.5 units
−4~a is a vector of magnitude 6 unit due south.
a) 3|~a| = 3 x 4.5 = 13.5
30. Two vectors have magnitudes 2 m and 3 m. The angle between them is 60. Find (a) the scalar product of the two vectors, (b)the magnitude of their vector product.
Solutions :|~a x ~b| = |~a| . |~b|sin60o = 2 x 3 x
√32
= 3√
3 m2
|~a| = 2m , |~b| = 3m
~a . ~b = |~a| . |~b|cos60o = 2 x 3 x 12
= 3m2
angle between them θ = 60o
31. Let A1 A2 A3 A4 A5 A6 A1 be a regular hexagon. Write the x-components of the vectors represented by the six sides taken inorder. Use the fact that the resultant of these six vectors is zero, to prove that cos0 + cosπ
3+ cos 2π
3+ cos 3π
3+ cos 4π
3+ cos 5π
3
= 0. Use the known cosine values to verify the result.
Solutions :[As a resultant is zero. X component of resultant Rx = 0]
Page 5
24Tutors.com
We know that according to polygon law addition, the resultant of these six vectors is zero.
Note: similarly it can be proved that,
So, Rx = A cosθ + A cosπ3
+ A cos 2π3
+ A cos 3π3
+ A cos 4π4
+ A cos 5π5
= 0
= cosθ +cosπ3
+cos 2π3
+ cos 3π3
+ cos 4π3
+ cos 5π3
= 0
Here A = B = C = D = E = F (magnitude)
sinθ + sinπ3
+ sin 2π3
+ sin 3π3
+ sin 4π3
+ sin 5π3
= 0
32. Let ~a = 2~i + 3~j + 4~k and ~b = 3~i + 4~j + 5~k . Find the angle between them.
a)
33. Prove that ~A . ( ~A x ~B ) = O.
Solutions :Thus, ~A . ( ~A X ~B) = 0
~A . ( ~A X ~B) = 0 (claim)
AB sinθ n is a vector which is perpendicular to the plane containing ~A and ~B , that implies that it is also perpendicular to ~A . As dotproduct of two perpendicular vector is zero.
As ~A x ~B = AB sinθ n
34. If ~A = 2~t +3~j + 4~k and ~B = 4~t + 3~j + 2~k , Find ~A x ~B.
Solutions :~A = 2t +3j + 4k , ~B = 4t + 3j + 2k
~A x ~B =
∣∣∣∣∣∣∣∣∣i amp; j amp; k2 amp; 3 amp; 4
4 amp; 372If ~A, ~B, ~C are mutually perpendicular, show that ~C x ( ~A x ~B ) = O. Is the converse true ?Solutions :Therefore, angle between ~C and ~A x ~B is 0o or 180o (fig.1) Given that ~A, ~B and ~C are mutually perpendicular Also ~C is perpendicular to ~A and ~B ~A X ~B is vector which direction is perpendicular to the plane containing ~A and ~BA particle moves on a given straight line with a constant speed v. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that ~OP x ~v is independent of the position P .Solutions :It can be seen from the figure, OQ = OP sin θ = OP ′ sinθ the particle moves on the straight line PP ′atspeedv.Therefore,XvisindependentofthepositionP.xv=(OP)vsinθn = v(OP ) sinθn = v(OQ) n So, whatever may be the position of the particle, the magnitude and direction of ~OP x v remain constant. From the figure,The force on a charged particle due to electric and magnetic fields is given by ~F = q ~E + q ~v x ~B. Suppose ~E is along the X-axis and ~B along the Y -axis. In what direction and with what minimum speed v should a positively charged particle be sent so that the net force on it is zero ?Solutions :Again, E = vB sin θ ⇒ v = EBsinθ
Give ~F = q ~E + q(~v x ~B) = 0 So, the practicle must be projected at a minimum speed of EB
along +ve z-axis (θ = 90o) as shown in the figure, so that the force is zero. So, direction of ~v x ~B should be opposite to the direction of ~E . Hence, ~v should be in the positive yz-plane. For v to be minimum, θ = 90o and so vmin = FB⇒ ~E = -(~v x ~B) = 0Give an example for which ~A . ~T = ~C but ~A 6= ~C.Solutions : ~C along north For example, as shown in the figure, Therefore, ~A · ~B = ~B · ~C But ~B 6= ~C ~B ⊥ ~C . ~A along south ~A · ~B = 0 ~B · ~C = 0 ~A ⊥ ~B . ~B along west EQAEDraw a graph from the following data. Draw tangents at x= 2, 4, 6 and 8. Find the slopes of these tangents. Verify that the curve drawn is y = 2x2 and the slope of tangent is tanθ = dy
dx= 4x.x 1 2 3 4 5 6 7 8 9 10 y 2 8 18 32 50 72 98 128 162 200 Solutions :Slope = tan θ = dy
dx= d
dx(2x2) = 4x The graph y = 2x2 should be drawn by the student on a graph for exact results. It can be checked that, To find slope at any point, draw a tangent at the point and extend the line to meet x-axis . Then find tan θ as shown in the figure.A curve is represented by y = sin x. If x is changed from π
3to π
3+ π
100. Find approximately the change in y.Solutions :=
(π3
+ π100
) - sin π3
= 0.0157 . y = sin x ∆y = sin (x+ ∆x) - sin x So, y + ∆y = Sin (x+ ∆x)The electric current in a charging R-C circuit is given by i = i0 , e−tRC where io , R and C are constant parameters of the circuit and t is time. Find the rate of change of current at (a) t = 0, (b) t = RC , (c) t = 10 RC.Solutions :b) When t = RC , di
dt= −i
RCeGiven that, i = i0e
−tRC
∴ Rate of change of current = didt
= ddti0e−iRC = I0
ddte−tRC = −i0
RCx e−tRC a) t= 0 , di
dt= −i
RCc) When t = 10 RC , di
dt= −i0
RCe10when,The electric current in a discharging R - C circuit is given by i = io , e
−tRC where io , R and C are constant parameters and t is time. Let io = 2 · 00 A, R = 6 · 00 x 105 Ω and C = 0 · 500 µF (a) Find the current at t = 0 · 3 s. (b) Find the rate of change of current at t = 0 · 3 s. (c) Find approximately the current at t = 0 · 31 s.6.67x10−11 x 106
6.64x10−17
Solutions :Equation i = i0e
−tRC
I0 = 2A, R = 6 x 10−5 Ω , C = 0.0500 x 10−8 F = 5 x 10−7 Fa) i = 2 X e −0.3
6×03×5×10−7 = 2 x e−0.30.3
= 2eamp .
b) didt
= −i0RC
e −tRC
when, t = 0.3 sec ⇒ didt
= - 20.30
e(−0.30.3
) = −203e
Amp/sec
c) At t = 0.31 sec, i = 2e(−0.30.3
) = 5.83e
Amp
Find the area bounded under the curve y = 3x62 + 6x + 7 and the X-axis with the ordinates at x = 5 and x = 10.
Solutions :y = 3x2 + 6x + 7
Area =∫ y0 dy =
∫ 105 (3x2 + 6x + 7)dx = 3x
3
3
]10
5+ 5x
2
3
]10
5+ 7x
]10
5= 1135 sq.units
∴ Area bounded by the curve, x axis with coordinates with x = 5 and x = 10 is given by,
Find the area enclosed by the curve y = sin x and the X-axis between x = 0 and x = n .
Solutions :Area =
∫ y0 dy =
∫ x0 sinxdx = - [cosx]π0 = 2
Find the area bounded by the curve y = e−x, the X-axis and the Y -axis.
Solutions :So, the required area can be found out by integrating the function from 0 to ∞ .The given function is y = e−x
X increases , Y value decreases and only at x = ∞ , y = 0So , Area =
∫∞0 e−xdx = -[e−x]∞0 = 1
When x = 0 , y = e−0 = 1
A rod of length L is placed along the X-axis between x = 0 and x = L. The linear density (mass/length) ρ of the rod varieswith the distance x from the origin as ρ = a + bx. (a) Find the S.I units of a and b. (b) Find the mass of the rod in terms of a, band L
Solutions :∴ dm = mass of the element = ρ dx = ( a +bx ) dxρ = mass
length= a + bx
b) Let us consider a small element of length ’dx’ at a distance x fro, the orgin as shown in the figure.
So, mass of the rod = m=∫dm = |intL0 ( a + bx ) dx =
[ax+ bx2
2
]L0
a0 S.I. unit of ’a’ = kgm
and S.I. unit of ’b’ = kgm2
The momentum p of a particle changes with time t according to the relation dpdt
= (10N) + (2Ns
)t . If the momentum is zeroat t = 0, what will the momentum be at t = 10 s
Solutions :dp = [(10N) + 2Nst]dtdpdt
= (10 N) + (2 NS
) t∴ momentum at t = 10 sec will be∫ p0 dp =
∫ 100 (10dt) +
∫ 100 (2tdt) = 10t]10
0 + 2 t2
2
]10
0= 200 kg m
smomentum is zero at t = 0
The changes in a function y and the independent variable x are related as dydx
= x2 . Find y as a function of x .
Solutions :∫dy =
∫x2 dx ⇒ y = x3
3+ c
The change in a function of y and the independent variable x are related as dydx
= x2 .Taking integration of both sides,
∴ y as a function of x represented by y = x3
3+ c
Page 6
24Tutors.com
⇒ dy = x2 dx
Write the number of significant digits in (a) 1001, (b) 100 · 1, (c) 100 · 10, (d) 0 · 001001.
Solutions :b) 100.1The number significant digitsc) 100.10No. of significant digits = 4No. of significant digits = 4d) 0.001001 No. of significant digits = 4a) 1001No. of significant digits = 5
A metre scale is graduated at every millimetre. How many significant digits will be there in a length measurement with thisscale ?
Solutions :So, the no. of significant digits may be 1, 2, 3, or 4.The metre scale is graduated at every millimeter.The maximum no. of significant digit may be 1 ( e.g. for measurements like 5 mm, 7 mm etc) and the maximum no. of significant digits
may be 4 (e.g. 1000 mm)1 m = 100 mm
Round the following numbers to 2 significant digits. (a) 3472, (b) 84 · 16, (c) 2 · 55 and (d) 28 · 5.
Solutions :b) value = 84
c) 2.6d) value is 28.
In the value 3472, after the digit 4, 7 is present. Its value is greater than 5.∴ value becomes 3500So, the next two digits are neglected and the value of 4 is increased by 1
The length and the radius of a cylinder measured with a slide callipers are found to be 4 · 54 cm and 1 · 75 cm respectively.Calculate the volume of the cylinder.
Solutions :Since, the minimum mo. of significant digits on a particular term is 3, the result should have 3 significant digits and others rounded off.Given that, for the cylinderSince, it is to be rounded off to 3 significant digits, V = 43.7 cm3
Volume = πr2I = π x (4.54) x (1.75)2
So, volume V = πr2I = (3.14) x (1.75) x (1.75) x (4, 54) = 43.6577 cm3
Length = I = 4.54 cm, radius = r = 1.75 cm
The thickness of a glass plate is measured to be 2 · 17 mm, 2 · 17 mm and 2 · 18 mm at three different places. Find the averagethickness of the plate from this data.
The length of the string of a simple pendulum is measured with a metre scale to be 90 · 0 cm. The radius of the bob plusthe length of the hook is calculated to be 2 · 13 cm using measurements with a slide callipers. What is the effective length of thependulum ? (The effective length is defined as the distance between the point of suspension and the centre of the bob.)
Solutions :So, the addition must be done by considering only 2 significant digits of each measurement.As shown in the figure,But, in the measurement 90.0 cm, the no. of significant digits is only 2.So, effective length = 90.0 + 2.1 = 92.1 cmActual effective length = (90.0 + 2.23) cm
The thickness of a glass plate is measured to be 2.17 mm, 2.17 mm and 2.18 mm at three different places. Find the averagethickness of the plate from this data.
Solutions :We know that,Rounding off to 3 significant digits, average thickness = 2.17 mmAverage thickness = 2.17+2.17+2.18
3= 2.1733 mm
The length of the string of a simple pendulum is measured with a metre scale to be 90.0 cm. The radius of the bob plusthe length of the hook is calculated to be 2.13 cm using measurements with a slide callipers. What is the effective length of thependulum ? (The effective length is defined as the distance between the point of suspension and the centre of the bob.)
DHTDFJHGA Ωω~A = 22i+Rthy = x2 + 2x+ 1 ... (1)The left side of the equation describes acceleration, and may be composed of time-dependent and convective components (also theeffects of non-inertial coordinates if present). ~A = 22i+RthThe right side of the equation is in effect a summation of hydrostatic effects, the divergence of deviatoric stress and body forces(such as gravity).
JGFJH
Solutions :zczcdsccs
1. A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the accelerationof the particle at t = 4 s
Page 7
24Tutors.com
Solutions :so, w = 2π
T= 2π
6= π
3sec−1
Given, r = 10cm.
Acceleration a = -w2x = -(π2
9
)x(-10) = 10.9 ≈ 0.11 cm/sec
so, 5 = 10 sin(w x 0 x φ) = 10 sin φ[y = r sin wt]
(ii) At t = 4 secondT = 6 sec= 10 sin
(3π2
)= 10 sin
(π + π
2
)= -10 sin
(π2
)= -10
At, t = 0,x = 5cmtherefore Equation of displacement x = (10cm) sin
(π3
)At t = 0, x = 5cmsin φ = 1/2 ⇒ φ = π
6
x = 10 sin[π3
x 4 + π6
]= 10 sin
[8π+π
6
]A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field. (a) What distance he has to walk to reach
the field ? (b) What is his displacement from his house to the field ?
Solutions :a) Distance travelled = 50 + 40 + 20 = 110 mb) AF = AB - BF = AB - DC = 50 - 20 = 30 M
His displacement is AD
AD =√AF 2 −DF 2 =
√302 + 402 = 50 M
In ∆AED tanθ = DEAE
= 3040
= 34
⇒ θ = tan−1 ( 34
)His displacement from his house to the field is 50 m ,
tan−1( 34
) north to east
A particle starts from the origin, goes along the X-axis to the point (20m, 0) and then returns along the same line to the point(−20m, 0). Find the distance and displacement of the particle during the trip.
Solutions :
O → Starting point origin.i ) Distance travelled = 20 + 20 + 20 = 60 mii ) Displacement is only OB = 20 m in the negative direction.Displacement → Distance between final and initial position.
It is 260 km from Patna to Ranchi by air and 320 km by road. An aeroplane takes 30 minutes to go from Patna to Ranchiwhereas a delux bus takes 8 hours.(a) Find the average speed of the plane.(b) Find the average speed of the bus.(c) Find the average velocity of the plane.(d) Find the average velocity of the bus.
Solutions :d) Straight path distance between plane to ranchi is equal to the displacement of bus.
∴ velocity = ~Vavg = 2608
= 32.5 kmhr
a) Vavg of plane (DistanceTime
) = 2600.5
= 520 kmhr
c) Plane goes in straight path
velocity = ~Vavg = 2600.5
= 520 kmhr
b) Vavg of bus = 3208
= 40 kmhr
When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns home after two hoursthe reading is 12416 km.(a) What is the average speed of the car during this period ?(b) What is the average velocity ?
Solutions :a ) Total distance covered 12416 - 12352 = 64 km in 2 hours.
speed = 642
= 32 kmhr
b) As he returns to his house , the displacement is zero.
Velocity = displacementtime
= 0(zero)
An athelete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average acceleration ?
Solutions :
Page 8
24Tutors.com
Initial velocity u = 0 ( ∴ starts from rest)final velocity v = 18 km
hr= 5 sec.
(i.e max velocity)The interval t = 2 sec∴ acceleration = aavg = v−u
t= 5
2= 2.5m
s2
The speed of a car as a function of time is shown in figure (3-E1). Find the distance travelled by the car in 8 seconds and itsacceleration.
Solutions :
In the interval 8 sec the velocity changes from 0 to 20 ms
average acceleration = 208
= 2.5 ms2
( changeinvelocitytime
)
Distance Travelled s = ut+ 12at2
⇒ 0 + 12
(2.5)82 = 80m.
Distance Travelled s = ut+1
2at2
=⇒ 0 +1
2(2.5)82 = 80m
The acceleration of a cart started at t = 0, varies with time as shown in figure (3-E2). Find the distance travelled in 30 secondsand draw the position-time graph.
Solutions :
In 1st 10 sec S1 = ut + 12
at2 ⇒ 0 + ( 12× 5× 102) = 250 ft.
At 10 sec v = u + at = 0 + 5 × 10 = 50 ftsec
.
∴ From 10 to 20 sec (∆t = 20 - 10 = 10 sec) it moves with uniform velocity 50 ftsec
.Distance S2 = 50× 10 = 500ft.Between 20 sec to 30 sec Acceleration is constant i.e -5 ft
s2. At 20 sec velocity is 50 ft
sec.
t = 30 - 20 = 10 sec.S3 = ut+ 1
2at2 = 50× 10 + ( 1
2)(−5)(10)2 = 250 m
Total distance travelled in 30 sec = S1 + S2 + S3 = 250 + 500 + 250 = 1000 ft.
Figure (3− E3) shows the graph of velocity versus time for a particle going along the X-axis. Find(a) the acceleration,(b) the distance travelled in 0 to 10 s and(c) the displacement in 0 to 10 s.
Solutions :
a) Initial velocity u = 2 ms
.Final velocity v = 8 m
s.
time = 10 sec,acceleration = v−u
at= 8−2
10= 0.6 m
s2
b) v2 - u2 = 2aS
⇒ Distance S = v2−u2
2a= 82−22
2×0.6= 50 m.
c) Displacement is same as distance travelled .Displacement = 50 m.
Figure (3-E4) shows the graph of the x-coordinate of a particle going along the X-axis as a function of time. Find(a) the average velocity during 0 to 10 s,(b) instantaneous velocity at 2, 5, 8 and 12s.
Solutions :b) At 2 sec it is moving with uniform velocity 50
2.5= 20 m
s.
a) Displacement in 0 to 10 sec is 1000 m.time = 10 sec.vavg = s
t= 100
10= 10 m
s
Page 9
24Tutors.com
b) At 2 sec it is moving with uniform velocity 502.5
= 20 ms
At 2 sec . Vinst = 20 ms.
At 5 sec it is at rest.Vinst = zeroAt 8 sec it is moving with uniform velocity 20 m
sVinst = 20 m
sAt 12 sec velocity is negative as it move toward initial position . Vinst = -20 m
s
From the velocitytime plot shown in figure (3-E5), find the distance travelled by the particle during the first 40 seconds. Alsofind the average velocity during this period.
Solutions :
Distance in first 40 sec is , ∆OAB + ∆BCD= 1
2× 5× 20 + 1
2× 5× 20 = 100 m.
Average velocity is 0 as the displacement is zero.
Figure (3-E6) shows x− t graph of a particle. Find the time t such that the average velocity of the particle during the period 0to t is zero.
Solutions :
Consider the point B , at t = 12 secAt t = 0 ; s = 20 mand t = 12 sec ; s = 20mSo for time interval 0 to 12 sec.Change in displacement is zero.So average velocity = displacement
time= 0
∴ The time is 12 sec.
A particle starts from a point A and travels along the solid curve shown in figure (3−E7). Find approximately the position Bof the particle such that the average velocity between the positions A and B has the same direction as the instantaneous velocity atB.
Solutions :
We can see that ~AB is along ~BC i.e they are in same direction.
Vavg = displacementtime
=( ~AB)t
t= timeThe point is B (5m , 3m).
At position B instantaneous velocity has direction along ~BC. For average velocity between A and B.
An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s2 for 5.0 s. Find the distance travelled during the periodof acceleration.
Solutions :u = 4 m
s, a = 1.2 m
s2, t = 5 sec
Distance = s = ut + 12at2
= 4(5) + 12
(1.2)52 = 35 m.
A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s2 to his scooter. How far will it travel beforestopping ?
Solutions :Initial velocity u = 43.2 km
hr= 12 m
su = 12 m
s, v = 0
a = −6 ms2
(deceleration)
Distance S = v2−u2
2(−6)= 12 m
Page 10
24Tutors.com
A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied andthe train comes to rest in one minute. Find(a) the total distance moved by the train,(b) the maximum speed attained by the train and(c) the position(s) of the train at half the maximum speed
Solutions :v′ = 0, t = 60 sec ( 1 min)Initial velocity u = 0
Acceleration a = 2 ms2
. Let final velocity be v ( before applying breaks )t = 30 secv = u+ at⇒ 0 + 2× 30 = 60 m
sb) The maximum speed attained by train v = 60 m
sWhen it accelerates the distance travelled is 900 m. Then again declarates and attain 30 m
s.
when breaks are applied u’ = 60ms
Distance = v2−u2
2a= 302−602
2(−1)= 1350 m
Position is 900 + 1350 = 2250 = 2.25 km from starting point.
Declaration a′ =(v−u)t
==(0−60)
60= -1 m
s2.
S2 =(v′)2−(u′)2
2a′ = 1800 mTotal S = S1 + S2 = 1800 + 900 = 2700 m = 2.7 km.
Distance S = v2−u2
2a= 302−02
2×2= 225 m from starting point
a) S1 = ut+ 12at2 = 900m
c) Half the maximum speed = 602
= 30 ms
∴ u = 60 ms
, v = 30 ms
, a = -1 ms2
A bullet travelling with a velocity of 16 m/s penetrates a tree trunk and comes to rest in 0.4 m. Find the time taken duringthe retardation.
Solutions :u = 16m
s(initial) , v = 0, s = 0.4m
Time = t = v−ua
= 0−16−320
= 0.05 sec.
Deceleration a = v2−u2
2s= −320 m
s2.
A bullet going with speed 350 m/s enters a concrete wall and penetrates a distance of 5.0 cm before coming to rest. Find thedeceleration.
Solutions :u = 350 m
s, s = 5 cm = 0.05m , v = 0
Deceleration is 12.2× 105 ms2.
Deceleration = a = v2−u2
2a=
0−(350)2
2×0.05= −12.2× 105 m
s2.
A particle starting from rest moves with constant acceleration. If it takes 5.0 s to reach the speed 18.0 km/h find(a) the average velocity during this period, and(b) the distance travelled by the particle during this period.
Solutions :u = 0 , v = 18 km
hr= 5 m
s, t = 5 sec
a) Average velocity Vavg =(12.5)
5= 2.5m
s.
b) distance travelled is 12.5ma = v−u
t= 5−0
5= 1 m
s2.
s = ut+ 12at2 = 12.5m
A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is drivinga car at a speed of 54 km/h and the brakes cause a deceleration of 6.0 m/s2, find the distance travelled by the car after he sees theneed to put the brakes on
Solutions :In reaction time the body moves with the speed 54 km
hr= 15 m
sec(constant speed)
Distance travelled in this time S1 = 15× 0.2 = 3m.When brakes are applied ,u = 15 m
s, v = 0, a = −6m
s2(deceleration)
S2 = v2−u2
2a= 0−152
2(−6)= 18.75m
Total distance s = s1 + s2 = 3 + 18.75 = 21.75 = 22m.
A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72 km/h. The jeep followsit at a speed of 90 km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, howfar from the turning will the jeep catch up with the bike ?
Solutions :
VP = 90 kmh
= 25ms.
VC = 72 kmh
= 20ms.
In 10 sec culprit reaches at point B from A.Distance converted by culprit S = vt = 20× 10 = 200mAt time t = 10sec the police jeep is 200m behind the culprit.Time = s
v= 200
5= 40s. (Relative velocity is considered).
In 40s the police jeep will move from A to a distance S, where S = vt = 25× 40 = 1000m = 1.0km away.∴ The jeep will catch up the bike , 1km far from the turning.
Page 11
24Tutors.com
A car travelling at 60 km/h overtakes another car travelling at 42 km/h. Assuming each car to be 5.0 m long, find the timetaken during the overtake and the total road distance used for the overtake.
Solutions :In 2 sec the 1st car moved = 16.6× 2 = 33.2
H also covered its own length 5 m∴ Total road distance used for the overtake = 33.2 + 5 = 38m
V1 = 60 kmhr
= 16.6ms.
V2 = 42 kmhr
= 11.6ms.
Relative velocity between the cars = (16.6− 11.6) = 5ms.
Distance to be travelled by first car is 5 + t = 10m.T ime = t = s
v= 0
5= 2sec to cross the 2nd car.
A ball is projected vertically upward with a speed of 50 m/s. Find(a) the maximum height,(b) the time to reach the maximum height,(c) the speed at half the maximum height. Take g = 10 m/s2
Solutions :⇒v =
√(u2 + 2as) =
√(502 + 2(−10)(62.5)) = 35m
s.
u = 50ms, g = −10m
s2When moving upward , v = 0 (at highest point)
b) t =(v−u)a
=(0−50)−10
= 5sec
c) s′ = 1252
= 62.5m, u = 50ms, a = −10m
s2, v2 − u2 = 2as
a) S = v2−u2
2a= 0−502
2(−10)= 125m
maximum height reached = 125 m
A ball is projected vertically upward with a speed of 50 m/s. Find(a) the maximum height,(b) the time to reach the maximum height,(c) the speed at half the maximum height. Take g = 10 m/s2
A ball is projected vertically upward with a speed of 50 m/s. Find(a) the maximum height,(b) the time to reach the maximum height,(c) the speed at half the maximum height. Take g = 10 m/s2
A ball is projected vertically upward with a speed of 50 m/s. Find(a) the maximum height,(b) the time to reach the maximum height,(c) the speed at half the maximum height. Take g = 10 m/s2
A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping theball, how long will the ball take in reaching the ground ?
Solutions :taking positive sign t = 7+35.34
10= 4.2sec ( ∴ t 6= -ve)
Initially the ball is going upwardu = −7m
s, s = 60m , a = g = 10m
s2
s = ut+ 12at2 ⇒ 60 = −7t+ 1
210t2
t=7±√
49−4.5(−60)
2×5= 7±35.34
10Therefore , the ball will take 4.2 sec to reach the ground.⇒ 5t2 − 7t− 60 = 0
A stone is thrown vertically upward with a speed of 28 m/s.(a) Find the maximum height reached by the stone.(b) Find its velocity one second before it reaches the maximum height.(c) Does the answer of part (b) change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s ?
Solutions :t′ = 2.85− 1 = 1.85u = 28m
s, v = 0 , a = g = −9.8m
s2
∴ The velocity is 9.87 ms.
b) time t = v−ua
= 0−28−9.8
= 2.85
v′ = u+ at′ = 28− (9.8)(1.85) = 9.87ms
a) S = v2−u2
2a= 0−282
2((9.8)= 40m
c) No it will not change. As after one second velocity becomes zero for any initial velocity and deceleration is g = 9.8ms2
remains same. forinitial velocity more than 28m
smax height increases.
A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd,4th and 5th ball when the 6th ball is being dropped.
Solutions :S3 = ut+ 1
2at2 = 0 + 1
2(9.8)32 = 4.9m below the top.
S2 = 0 + 12gt2 = 1
2(9.8)22 = 19.6m below the top (u=0).
For 3rd ball t = 3sec
S3 = ut+ 12at2 = 0 + 1
2(9.8)12 = 4.9m below the top.
Page 12
24Tutors.com
For 4th ball t = 2sec
For every ball , u = 0, a = g = 9.8ms2
∴ 4th ball move for 2 sec, 5th ball 1sec and 3rd ball 3sec when 6th ball is being dropped.For 5th ball t = 1sec
A healthy youngman standing at a distance of 7 m from a 11.8 m high building sees a kid slipping from the top floor. Withwhat speed (assumed uniform) should he run to catch the kid at the arms height (1.8m) ?
Solutions :At point B( i.e. over 1.8m from ground) the kid should be catched.
For kid initial velocity u=0Acceleration = 9.8m
s2
Distance S = 11.8− 1.8 = 10mS = ut+ 1
2at2 ⇒ 10 = 0 + 1
2(9.8)t2
⇒ t2 = 2.04⇒ t = 1.42.In this time the man has to reach at the bottom of the building.
Velocity st
= 71.42
= 4.9ms.
An NCC parade is going at a uniform speed of 6 km/h through a place under a berry tree on which a bird is sitting at a heightof 12.1 m. At a particular instant the bird drops a berry. Which cadet (give the distance from the tree at the instant) will receivethe berry on his uniform ?
Solutions :my namename
Let the time of fall be ’t’ . initial velocity u=0Acceleration a = 9.8m
s2Distance S = 12
1m
∴ S = ut+ 12at2
⇒ 12.1 = 0 + 12
(9.8)× t2
For cadet velocity = 6 kmhr
= 1.66 msec
Distance = vt = 1.57× 1.66 = 2.6m.The cadet , 2.6 m away from tree will receive the berry on his uniform.⇒ t2 = 12.1
4.9= 2.46⇒ t = 1.57sec
A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from whichit was dropped. Take g = 10 m/s2.
Solutions :For last 6 m distance travelled s = 6m ,u =?
t = 0.2 sec, a = g = 9.8 ms2
S = v2−u2
2a= 292−02
2×9.8= 42.05 m.
For last 6 m distance travelled s = 6m
Page 13
24Tutors.com
S = ut+ 12at2 ⇒ 6 = u(0.2) + 4.9× 0.04
⇒ u = 5.80.2
= 29 ms.
For distance x , u = 0, v = 29 ms, a = g = 9.8 m
s2
Total distance = 42.05 + 6 = 48.05 = 48 m.
A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find theretardation of the ball in sand assuming it to be uniform
Solutions :Consider the motion of ball from A to B.
B → just above the sand (just to penetrate)u = 0, a = 9.8 m
s2, s = 5 m.
S = ut+ 12at2
⇒ 5 = 0 + 12
(9.8)t2
⇒ t2 = 54.9
= 1.02⇒ t = 1.01.∴ velocity at B, v = u+ at = 9.8× 1.01 (u = 0) = 9.89 m
s.
F rom motion of ball in sandu1 = 9.89 m
s, v1 = 0,
a =?, s = 10 cm = 0.1m.
a =v21−u
21
2s=
0−(9.89)2
2×0.1= −490 m
s2
The retardation in sand is 490 ms2.
An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at themoment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that thecoin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator
Solutions :For elevator and coin u = 0
As the elevator descends downward with acceleration a′ (say)The coin has to move more distance than 1.8m to strike the floor. T ime taken t = 1 sec.Sc = ut+ 1
2a′t2 = 0 + 1
2g(1)2 = 1
2g
Se = ut+ 12at2 = u+ 1
2a(1)2 = 1
2a
Total distance covered by coin is given by = 1.8 + 12a = 1
2g
⇒ 1.8 + a2
= 9.82
= 4.9
⇒= 6.2ms2
= 6.2× 3.28 = 20.34 fts2.
A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find(a) the time it takes to reach the ground,(b) the horizontal distance it travels before reaching the ground,(c) the velocity (direction and magnitude) with which it strikes the ground.
Solutions :=√
2×1009.8
= 4.51 sec.
b) Horizontal range x = ut = 20× 4.5 = 90 m.c) Horizontal velocity remains constant through out the motionAt A, V = 20 m
sAVy = u+ at = 0 + 9.8× 4.5 = 44.1 m
s.
Resultant velocity Vr =√
(44.1)2 + 202 = 48.42 ms.
tanβ =VyVx
= 44.120
= 2.205
The ball strikes the ground with a velocity 48.42 msat an angle 66 with horizontal.
It is a case of projectile fired horizontally from a heighth = 100 m, g = 9.8 m
s2
a) T ime taken to reach the ground t =√
( 2hg
)
⇒ β = tan−1(2.205) = 60
Page 14
24Tutors.com
A ball is thrown at a speed of 40 m/s at an angle of 60 with the horizontal. Find(a) the maximum height reached and(b) the range of the ball.Take g = 10 m/s2
Solutions :u = 40 m
s, a = g = 9.8 m
s2, θ = 60 Angle of projection.
a) Maximum height h = u2sin2θ2g
=402(sin60)2
2×10= 60 m
b) Horizontal range X =(u2sin2θ)
g=
(402sin2(60))10
= 80√
3m
In a soccer practice session the football is kept at the centre of the field 40 yards from the 10 ft high goalposts. A goal isattempted by kicking the football at a speed of 64 ft/s at an angle of 45 to the horizontal. Will the ball reach the goal post ?
Solutions :y =usinθ(t)− 1
2gt2 = 64 1√
2(2.65)− 1
2(32.2)(2.65)2
In time 2.65, the ball travels horizontal distance 120 ft (40 yd) and vertical height 7.08 ft which is less 10 ft. The ball will reach the goalpost.⇒= t = x
ucosθ= 120
64cos45 = 2.65 sec= 7.08 ft which is less than the height of goal post.g = 9.8 m
s2, 32.2 ft
s2; 40 yd = 120 ft
horizontal range x = 120 ft, u = 64 fts, θ = 45
We know that horizontal range X = u cosθt
A popular game in Indian villages is goli which is played with small glass balls called golis. The goli of one player is situated ata distance of 2.0 m from the goli of the second player. This second player has to project his goli by keeping the thumb of the lefthand at the place of his goli, holding the goli between his two middle fingers and making the throw. If the projected goli hits thegoli of the first player, the second player wins. If the height from which the goli is projected is 19.6 cm from the ground and thegoli is to be projected horizontally, with what speed should it be projected so that it directly hits the stationary goli without fallingon the ground earlier ?
Solutions :The goli moves like a projectile
Here h = 0.196 m.Horizontal distance X = 2 m.Acceleration g = 9.8 m
s2.
T imetoreachthegroundi.e.
t =√
2hg
=√
2×0.1969.8
= 0.2 sec.
Horizontal velocity with which it is projected be u.∴ x = ut⇒ u = x
t= 2
0.2= 10 m
s.
Figure (3-E8) shows a 11.7 ft wide ditch with the approach roads at an angle of 15 with the horizontal. With what minimumspeed should a motorbike be moving on the road so that it safely crosses the ditch ? Assume that the length of the bike is 5 ft,and it leaves the road when the front part runs out of the approach road.
Solutions :Horizontal range X = 11.7 + 5 = 16.7 ft covered by te bike.
g = 9.8 ms2
= 32.2 fts2.
y = xtanθ − gx2sec2θ2u2 ⇒ u2 = gx2sec2θ
2xtanθ= gx
2sinθ(cosθ)= gx
sin2θ
⇒ u =
√(32.2)(16.7)
12
(beacuse sin30 = 12
)
⇒ u = 32.79 fts
= 32 fts.
Page 15
24Tutors.com
A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontallyaway. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall ?
Solutions :tanθ = 171
228⇒ θ = tan−1( 171
228)
The motion of projectile (i.e. the packed) is from A. Taken references axis at A.∴ θ = −37 as u is below x− axis.u = 15 ft
s, g = 32.2 ft/s2, y = −171 ft
y = xtanθ − x2gsec2θ2u2
∴ −171 = −x(0.7536)− x2g(1.568)2(225)
⇒ 0.1125x2 + 0.7536x− 171 = 0x = 35.78 ft (can be calculated)Horizontal range covered by the packet is 35.78 ft.So, the packet will fall 228− 35.78 = 192 ft short of his friend.
A ball is projected from a point on the floor with a speed of 15 m/s at an angle of 60 with the horizontal. Will it hit a verticalwall 5 m away from the point of projection and perpendicular to the plane of projection without hitting the floor ? Will the answerdiffer if the wall is 22 m away ?
Solutions :Here u = 15 m
s, θ = 60, g = 9.8 m
s2
Horizontal range X = u2sin2θg
=152sin(2×60)
9.8= 19.88 m.
In first case the wall is 5 m away from projection point, so it is in the horizontal range of projectile . So the ball will hit the wall. In secondcase (22 m away) wall is not within the horizontal range. So the ball would not hit the wall.
Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speedu at an angle 0 with the horizontal.
Solutions :Total of flight T = 2usinθ
g
Average velocity = change in displacementtime
from the figure , it can be said AB is horizontal . So there is no effect of vertical component of the velocity during this displacement. Sobecause the body moves at a constant speed of u′cosθ′ in horizontal direction. The average velocity during this displacement will be ucosθ inthe horizontal direction.
A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode vertically below theplane. Is the statement true if the plane flies with uniform speed but not horizontally ?
Solutions :During the motion of bomb its horizontal velocity u remains constant and is same as that of aeroplane at every point of its path. Suppose
the bomb explode i.e. reach the ground in time t. Distance travelled in horizontal direction by bomb = ut = the distance travelled by aeroplane.so bomb explode vertically below the aeroplane. Suppose the aeroplane move making angle θ with horizontal. For both bomb and aeroplane ,horizontal distance is ucosθt . t is time for bomb to reach the ground. so in this case also , the bomb will explode vertically below aeroplane.
A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with anacceleration of 1 m/s2 and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall onthe car ?
Solutions :Let the velocity of car be u when the all is thrown. Initial velocity of car is = Horizontal velocity of ball. Distance travelled by ball B
Sb = ut (in horizontal direction) And by car Sc = ut+ 12at2 where t→ time of flight of ball in air.
∴ Car has travelled extra distance Sc − Sb = 12at2.
Ball can be considered as a projectile having θ = 90. ∴ t = 2usinθg
= 2×9.89.8
= 2 sec.
Page 16
24Tutors.com
∴ Sc − Sb = 12at2 = 2 m.
∴ The ball will drop 2m behind the body.
A staircase contains three steps each 10 cm high and 20 cm wide (figure 3-E9). What should be the minimum horizontal velocityof a ball rolling off the uppermost plane so as to hit directly the lowest plane ?
Solutions :At minimum velocity it will move just touching point E reaching the ground. A is origin of reference coordinate. it u is the minimum
speed. X = 40, Y = −20, θ = 0
∴ Y = xtanθ − g x2sec2θ2u2 (because g = 10 m
s2= 1000 cm
s2)
⇒ −20 = xtanθ − 1000×402×12u2
⇒ u = 200 cms
= 2 m/s∴ Theminimumhorizontalvelocityis2 m/s.
A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball insuch a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection(a) as seen from the truck,(b) as seen from the road
Solutions :a) As seen from the truck the ball moves vertically upwards comes back. Time taken = time taken by truck to cover 58.8 m. ∴ time =
ucos53 = 24.42 m/s.The speed of ball is 24.42 m/s at an angle 53 with horizontal as seen from the road.
b) From road it seems to be projectile motion . Total time of flight = 4 sec In this time horizontal range covered 58.8 m = x∴ X = ucosθt⇒ ucosθ = 14.7..(1)
The benches of a gallery in a cricket stadium are 1 m wide and 1 m high. A batsman strikes the ball at a level one metreabove the ground and hits a mammoth sixer. The ball starts at 35 m/s at an angle of 53 with the horizontal. The benches areperpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit ?
Solutions :From the equation , y can be calculated.
∴ y = 5⇒ n− 1 = 5⇒ n = 6.The ball will drop in sixth bench.
θ = 53, socos53 = 35
Sec2θ = 259and tanθ = 4
3Suppose the ball lands on nth benchSo, y = (n− 1)1 ...(1) [ball starting point 1 m above ground]
Again y = xtanθ − gx2sec2θ2u2 [x = 110 + n− 1 = 110 + y]
⇒ y = (110 + y)( 43
)− 10(110+y)2( 259
)
2×352
⇒= 4403
+ 43y − 250(110+y)2
18×352
Page 17
24Tutors.com
A man is sitting on the shore of a river. He is in the line of a 1.0 m long boat and is 5.5 m away from the centre of the boat. Hewishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/s, find the minimum and maximumangles of projection for successful shot. Assume that the point of projection and the edge of the boat are in the same horizontallevel.
Solutions :When the apple just touches the end B of the boat.
x = 5 m, u = 10 m/s, g = 10 m/s2, θ =?
x = u2sin2θg
⇒ 5 = 102sin2θ10
⇒ 5 = 10 sin 2θ
⇒ sin2θ = 12⇒ sin 30 or sin150
⇒ θ = 15 or 75
Similarly for end C, x = 6 mThen 2θ1 = sin−1( gx
u2 ) = sin−1(0.6) = 182 or 71.So, for a successful shot, θ may vary from 15 to 18 or 71 to 75 .
A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/s with respect to the water, in adirection perpendicular to the river.(a) Find the time taken by the boat to reach the opposite bank.(b) How far from the point directly opposite to the starting point does the boat reach the opposite bank ?
Solutions :b) The boat will reach at point C.
In ∆ABC, tan θ = BCAB
= BC400
= 15
⇒ BC = 4005
= 80 m.a) Here the boat moves with the resultant velocity R. But the vertical component 10 m/s takes him to the opposite shore.
tan θ = 210
= 15
V elocity = 10 m/sdistance = 400 mTime = 400
10= 40 sec.
A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to water is 3 km/h.(a) If he heads in a direction making an angle 0 with the flow, find the time he takes to cross the river.(b) Find the shortest possible time to cross the river
Solutions :b) Here vertical component of velocity i.e. 3 km/hr takes him to opposite side.
TIme = DistanceV elocity
= 0.53
= 0.16 hr
∴ 0.16 hr = 60× 0.16 = 9.6 = 10 minute.
a) The vertical component 3sinθ takes him to opposite side.Distance = 0.5 km, V elocity = 3sinθ km/hT ime = Distance
V elocity= 0.5
3sinθhr
= 10sinθ
min.
Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to hisstarting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance thathe has to walk.
Solutions :Velocity of man ~Vm = 3 km/hr
BD horizontal distance for resultant velocity R.X-component of resultant Rx = 5 + 3 cosθt = 0.5
3sinθwhich is same for horizontal component of velocity.
An aeroplane has to go from a point A to another point B, 500 km away due 30 east of north. A wind is blowing due north ata speed of 20 m/s. The air-speed of the plane is 150 m/s.(a) Find the direction in which the pilot should head the plane to reach the point B.(b) Find the time taken by the plane to go from A to B.
Solutions :∴ 20sin A
= 150sin 30 ⇒ sinA = 20
150sin30 = 20
150× 1
2= 1
15
⇒ A = sin−1( 115
)
b) sin−1( 115
) = 348′
⇒ 30 + 348′ = 3348′
R =√
1502 + 202 + 2(150)20cos3348′ = 167 m/s.T ime = s
v= 500000
167= 2994 sec = 49 = 50 min.
In resultant direction ~R the plane rech the point B.V elocity of wind ~Vw = 20 m/s
V elocity of aeroplane ~Va = 150 m/sIn ∆ACD according to sine formula
a) The direction is sin−1( 115
) east of the line AB
1. The gravitational force acting on a particle of 1 g due to a similar particle is equal to 6.67 x 10−17N. Calculate the separationbetween the particles.
Solutions :
m = 1gm = 11000
kg
F = 6.67 x 10−17 N ⇒ F = Gm1m2r2
∴ 6.67 x 20−17 =6.67×10−11×( 1
1000)×( 1
1000)
r2
⇒ r2 = 6.67×10−11×10−6
6.64×10−17 = 10−17
10−17= 1
⇒ r =√
1 = 1 metre.So. The separation between the particle is 1m.
2. Calculate the force with which you attract the earth.
Solutions :A man is standing on the surface of earth.
The force acting on the man = mg.....(I)Assuming that, m = mass of the man = 50 Kgandg = accelaration due to gravity on the surface of the earth = 10 m
s2
W = mg = 50 x 10 = 500 N = force acting on the manSo, the man is also attracting the earth with a force of 500 N.
3. At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight ?
Solutions :The force of attraction between the two charges = 1
4πε0
q1q2r2
= 9 x 109 1r2
The force of attraction is equal to the weight.
Mg = 9×109
r2
⇒ r2 = 9×109
m×10= 9×108
m[taking g = 10 m/s2]
⇒ r =√
9×108
m= 3×104
√m
mt
For example,Assuming m = 64 kg.
r = 3×104√
64= 3
8104 = 3750 m.
4. Two spherical bodies, each of mass 50 kg, are placed at a separation of 20 cm. Equal charges are placed on the bodies andit is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude. Find the magnitude of the chargeplaced on either body.
Page 19
24Tutors.com
Solutions :mass = 50kg.
r = 20 cm = 0.2m
FG = Gm1m2r2
= 6.67×10−11×25000.04
Coulomb’s force Fc = 14πε0
q1q2r2
= 9×109×q20.04
Since, Fg = Fc = 6.7×10−11×25000.04
= 9×109×q20.04
⇒ q2 = 6.7×10−11×25000.04
= 6.67×10−9
9×109 x 25
= 18.07 x 10−18
q =√
18.07× 10−18 = 4.3 x 10−9c
5. A monkey is sitting on a tree limb. The limb exerts a normal force of 48N and a frictional force of 20N . Find the magnitudeof the total force exerted by the limb on the monkey.
Solutions :
The limb exerts a normal force 48N and frictional force of 20N .Resultant magnitude of the force.R =
√(48)2 + (20)2
=√
2304 + 400=√
2704= 52N.
6. A body builder exerts a force of 150N against a bullworker and compresses it by 20 cm. Calculate the spring constant of thespring in the bullworker.
Solutions :
The body builder exerts a force = 150N.Compression x = 20CM = 0.2m∴ Total force exerted by the man = f = kx.⇒ Kx = 150⇒ K= 150
0.2= 1500
2= 750N
m
7. A satellite is projected vertically upwards from an earth station. At what height above the earth’s surface will the force onthe satellite due to the earth be reduced to half its value at the earth station ? (Radius of the earth is 6400 km.)
Solutions :Suppose the height is h.
At earth station F = GMmR2
M = mass of earth.m=mass of satelite.R = radius of earth.F = GMm
8. Two charged particles placed at a separation of 20cm exert 420N of Coulomb force on each other. What will be the force ifthe separation is increased to 25cm ?
Solutions :Two charged particle placed at a sehortion 2m. exert a force of 20m.
F1 = 20 N.F2 = ?r1 = 20 cm.
Page 20
24Tutors.com
r2 = 25 cm.Since, F = 1
4πε0
q1q2r2
,
F = 1r2
f1f2
=r22r21⇒ F2 = F1 x
( r1r2
)2= 20 x
(2025
)2= 20 x 16
25= 64
5= 12.8 N = 13 N.
Two charged particle placed at a sehortion 2m. exert a force of 20m.F1 = 20 N.F2 = ?r1 = 20 cm.r2 = 25 cm.Since, F = 1
4πε0
q1q2r2
,
F = 1r2
f1f2
=r22r21⇒ F2 = F1 x
( r1r2
)2= 20 x
(2025
)2= 20 x 16
25= 64
5= 12.8 N = 13 N.
9. The force with which the earth attracts an object is called the weight of the object. Calculate the weight of the moon fromthe following data : The universal constant of gravitation G = 6.67 x 10− 11N −m2/kg2, mass of the moon = 7.36 x 1022kg, massof the earth = 6 x 1024kg and the distance between the earth and the moon = 3.8 x 105km.
Solutions :The force between the earth and the moon ,
F = G mmmcr2
F = 6.67×10−11×7.36×1022×6×1024
(3.8×108)2
= 6.67×7.36××1035(3.8)2×1016
= 20.3× 1019 = 2.03× 1020N = 2× 1020N .
10. Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.
Solutions :Charge on portion = 1.6× 10−19
∴ Felectrical = 14πε0
x q1q2r2
=9×109×(1.6)2×10−38
r2
mass of proton = 1.732× 10−27 Kg
FGravity = G m1m2r2
=6.67×10−11×(1.732)×10−54
r2
FeFg
=9×(1.6)2×10−29
6.67(1.732)210−65
= 1.24× 1036
11. The average separation between the proton and the electron in a hydrogen atom in ground state is 5.3 x 10 − 11i n. (a)Calculate the Coulomb force between them at this separation. (b) When the atom goes into its first excited state the averageseparation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force inthis state ?
Solutions :The average separation between proton and electron of hydrogen atom is = 5.3 x 10−11m.
a) Coulomb’s force F = 9 x 109 x q1q2r2
=9×109×(1.0×10−19)2
(5.3×10−11)2= 8.2 x 10−8N .
b) When the average distance between proton and electron becomes 4times that of its ground state.
Coulomb’s force F = 14πε0
x q1q2(4r)2
=9×109×(1.6×10−19)2
16×(5.3)2×10−22
=9×(1.6)2
16×(5.3)2x 10−7
=0.0512× 10−7 = 5.1× 10−9N.
12. The geostationary orbit of the earth is at a distance of about 36000 km from the earth’s surface. Find the weight of a120− kg equipment placed in a geostationary satellite. The radius of the earth is 6400km.
Solutions :The geostationary orbit of earth is at distance of about 36000km.
We know that, g1 = GM(R+h)2
At h = 36000km.g1 = GM(36000+6400)2
∴ g1
g= 6400×6400
42400×42400= 256
106×106= 0.0227
⇒g1 = 0.0227 x 9.8 = 0.223[ taking g = 9.8 m
s2at the surface of the earth]
A120kg equipment placed in a geostationary satellite will have weightMg = 0.233 x 120 = 26.79 = 27N
Thanks
Solutions :Two charged particle placed at a sehortion 2m. exert a force of 20m.
F1 = 20 N.F2 = ?r1 = 20 cm.r2 = 25 cm.Since, F = 1
4πε0
q1q2r2
,
F = 1r2
f1f2
=r22r21⇒ F2 = F1 x
( r1r2
)2= 20 x
(2025
)2= 20 x 16
25= 64
5= 12.8 N = 13 N.
Thanks
Page 21
24Tutors.com
Solutions := 20.3× 1019 = 2.03× 1020N = 2× 1020N .The force between the earth and the moon ,
F = G mmmcr2
= 6.67×7.36××1035(3.8)2×1016
F = 6.67×10−11×7.36×1022×6×1024
(3.8×108)2
Thanks
Solutions :Charge on portion = 1.6× 10−19
∴ Felectrical = 14πε0
x q1q2r2
=9×109×(1.6)2×10−38
r2
mass of proton = 1.732× 10−27 Kg
FGravity = G m1m2r2
=6.67×10−11×(1.732)×10−54
r2
FeFg
=9×(1.6)2×10−29
6.67(1.732)210−65
= 1.24× 1036
Thanks
Solutions :a) Coulomb’s force = F =9 x 109 x q1q2
r2
The average separation between proton and electron of hydrogen atom is = 5.3 x 10−11m.
a) Coulomb’s force F = 9 x 109 x q1q2r2
=9×109×(1.0×10−19)2
(5.3×10−11)2= 8.2 x 10−8N .
b) When the average distance between proton and electron becomes 4times that of its ground state.
Coulomb’s force F = 14πε0
x q1q2(4r)2
=9×109×(1.6×10−19)2
16×(5.3)2×10−22
=9×(1.6)2
16×(5.3)2x 10−7
=0.0512× 10−7 = 5.1× 10−9N.
A block of mass 2kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F . It is found tomove 10m in the first two seconds. Find the magnitude of F .
A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average forcemust be applied on it ?
Solutions :u = 40km
hr= 40000
3600= 11.11 m
sm = 2000 kg : v = 0 : s = 4m
acceleration ′a′ = v2−u2
2s=
02−(11.11)2
2x4= - 123.43
8= −15.42 m
s2(deceleration)
So, braking force = F = ma = 2000 x 15.42= 30840 = 3.08 104 N (Ans)
A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure (5-E1). Find the forceacting on the particle at t = 2, 4 and 6 seconds.
Solutions :
m= 50g = 5 x 10−2 kgSlope of OA = Tanθ AD
OD= 15
3= 5m
s2
So, at t = 2 sec acceleration is 5ms2
Force = ma = 5 x 10−2 x 5 = 0.25N along the motionAt t = 4 secslope of AB = 0, acceleration = 0 [tanθo = 0]Force = 0At t = 6 sec, acceleration = slope of BCIn ∆BEC = tanθ = BE
EC= 15
3= 5
Slope of BC = tan (180o - θ) = -tanθ = -5 ms2
(deceleration)
Force = ma = 5 x 10−2 5 = 0.25 N . Opposite to the motion.
In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5 x 106m/s in travellingone centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is 9.1 x 10- 31kg
Solutions :Initial Velocity u = 0 (negligible)
v = 5 x 106 ms
s = 1 cm = 1 x 10−2 m
acceleration a = v2−u2
2s=
(5×62)−0
2×1×10−2 = 25×102
2×10−2 = 12.5 x 1014 ms−2
F = ma = 9.1 x 10−31 x 12.5 x 1014 = 113.75 x 10−17 = 1.1 x 10−15 N
Page 22
24Tutors.com
Two blocks A and B of mass mA, and mB respectively are kept in contact on a frictionless table. The experimenter pushes theblock A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by theexperimenter on A ?
Solutions :
Let F → contact force between mA amp; mB .And, f → force exerted by experimenter.F + mAa - f = 0 , mBa - f = 0⇒ F = f - mAa ......(i)⇒ F = mBa......(ii)From eqn (i) and eqn (ii)⇒ f - mAa = mBa ⇒ f = mBa + maA ⇒ f = a (mA +mB)f = F
mB(mB + mA) = F
(1 + mA
mB
)[because a = F
mB]
∴ The force exerted by the experimenter is F(1 + mA
mB
)A block of mass 0.2kg is suspended from the ceiling by a light string. A second block of mass 0.3kg is suspended from the first
block through another string. Find the tensions in the two strings. Take g = 10 ms2
Solutions :
g = 10 ms2
T - 0.3g = 0 ⇒ T = 0.3 g = 0.3 x 10 = 3 NT1 = (0.2g + T ) = 0 ⇒ T1 = 0.2g + T = 0.2 x 10 + 3 = 5 N
Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining themwith a constant force F . Find the tension in the string joining the blocks.
Solutions :T + ma - F = 0
(T - ma = 0 ⇒ T = ma - (Eqn.1)⇒ F = T + ma ⇒ F = T + T ....(fromEqn.1)⇒ 2T = F ⇒ T = F
2
Raindrops of radius 1mm and mass 4mg are falling with a speed of 30 ms
on the head of a bald person. The drops splash on thehead and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the forceexerted by each drop on the head.
Solutions :r = 1 mm = 10−3
′m′ = 4 mg = 4 x 10−6kg8 = 10−3m.v = 0u = 30m
s
So, a = v2−u2
2s= −30×30
2×10−3 = −4.5 x 105 ms2
(decelerating)
Taking magnitude only deceleration is 4.5 x 106 ms2
So, force F = 4 x 10−6 x 4.5 X 106 = 1.8 N
A particle of mass 0.3 kg is subjected to a force F = −kx with k = 15 Nm
. What will be its initial acceleration if it is releasedfrom a point = 20 cm ?
Solutions :x = 20 cm = 0.2 m, k = 15N
m, m = 0.3 kg
Acceleration a = Fm
= −kxx
=−15(0.2)
0.3= 3
0.3= −10m
s2deceleration)
So, the acceleration is 10ms2
opposite to the direction of motion.
Both the springs shown in figure (5-E2) are un stretched. If the block is displaced by a distance x and released, what will bethe initial acceleration?
Solutions :
Let the block m towards left through displacement xF1 = k1 x (compressed)F2 = k2 x (expanded)They are in same direction.Resultant F = F1 + F2 ⇒ F = k1x + k2x ⇒ F = x(k1 + k2)
So, a = acceleration = Fm
=x(k1+K2)
mopposite to the displacement.
Page 23
24Tutors.com
A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleysas shown in figure (5-E4). Show that the force (assumed equal for both the friends) exerted by each friend on the road increases asthe man moves up. Find the force when the man is at a depth h.
Solutions :
a) at any depth let the ropes make angle θ with the vertical From the free body diagramF cosθ+ F cosθ - mg = 0⇒ 2F cosθ = mg ⇒ F = mg
2cosθAs the man moves up, θ increases i.e, cos θ decreases. Thus F increases.
b) When the man is at depth hcos θ = h√
d22+h2
Force = mgh
= mg4h
√d2 + 42√
d2
4+ h2
The elevator shown in figure (5-E5) is descending with an acceleration of 2ms2
. The mass of the block A is 0.5kg. What force isexerted by the block A on the block B ?
Solutions :
From the free diagram∴ R + 0.5 x 2 - w = 0⇒ = w - 0.5 x 2= 0.5 (10− 2) = 4NSo, the force exerted by the block A on the block B is 4N
A pendulum bob of mass 50g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goesup with acceleration 1.2m
s2, (b) goes up with deceleration 1.2m
s2, (c) goes up with uniform velocity, (d) goes down with acceleration
1.2ms2
, (e) goes down with deceleration 1.2ms2
and (f) goes down with uniform velocity.
Solutions :
a)The tension in the string is found out for the different conditions from the body diagram shown below.T - (W + 0.06× 1.2) = 0⇒ = 0.05 x 9.8 + 0.05 x 1.2= 0.56 N .
b) ∴ T + 0.05 x 1.2 - 0.05 x 9.8 = 0⇒ T = 0.05 x 9.8 - 0.5 x 1.2
Page 24
24Tutors.com
= 0.43 N .
c) When the elevator makes uniform motionT - W = 0⇒ T = W - 0.05 x 9.8= 0.49 N
d) T + 0.05 x 1.2 - W = 0⇒ T = W - 0.05 x 1.2= 0.43 N
e) T - (W + 0.05×) = 0 ⇒ T = W + 0.05 x 1.2= 0.55 N
f) When the elevator goes down with the uniform velocity acceleration = 0T -W=0⇒T=W = 0.05 x 9.8=0.49 N
A small block B is placed on another block A of mass 5kg and length 20cm. Initially the block B is near the right end of blockA (figure 5-E3). A constant horizontal force of 10N is applied to the block A. All the surfaces are assumed friction less. Find thetime elapsed before the block B separates from A.
Solutions :
m = 5kg of mass Ama = 10 N⇒ a 10
5= 2m
s2
As there is no friction between A and B, when the block A moves, Block B remains at rest in its motioninitial velocity of A = u = 0Distance to cover so that B seperate out s = 0.2mAcceleration a = 2m
s2
∴ s = ut + 12at2
⇒ 0.2 = 0 + ( 12
) x 2 x t2 ⇒ t2 = 0.2 ⇒ t = 0.44 sec ⇒ t = 0.45 sec
A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration,moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72kgand 60kg. Assuming that the magnitudes of the acceleration and the deceleration are the same, find (a) the true weight of theperson and (b) the magnitude of the acceleration. Take g = 9.9m
s2.
Solutions :
Page 25
24Tutors.com
When the elevator is accelerating upwards, maximum weight will be recorded.R - (W + ma) = 0⇒ R = W +ma = m(g + a) max.wtWhen decelerating upwards, maximum weight will be recorded.R + ma - W = 0⇒ R = W - ma = m(g − a)So, m(g + a) = 72 x 9.9.....(1)m(g − a) = 60 x 9.9.....(2)Now, mg + ma = 72 x 9.9 ⇒ mg - ma = 60 x 9.9⇒ 2mg = 1306.8⇒ m = 1306.8
2×9.9= 66 Kg
So the true weight of the man is 66 kgAgain, to find the acceleration, mg + ma = 72 x 9.9⇒ a = 72×9.9−66×99
66= 9.9
11= 0.9 m
s2.
Find the reading of the spring balance shown in figure (5-E6). The elevator is going up with an acceleration of g10
, the pulleyand the string are light and the pulley is smooth.
Solutions :
Let the acceleration of the 3 kg mass relative to the elevator is ′a′ in the downward direction.As, shown in the free body diagramT = 1.5 g = 1.5( g
10) = 1.5 a = 0.....from figure (1)
and, T - 3g - 3( g10
) + 3a = 0.....from figure (2)⇒ T = 1.5g + 1.5( g
10) + 1.5a .....(i)
And T = 3g + 3( g10
) - 3a.....(ii)Equation (i) x 2 ⇒ 3g +3( g
10) + 3a = 2T
Equation (ii) x 1 ⇒ 3g +3( g10
) - 3a = TSubtracting the above two equations we get, T = 6aSubtracting T = 6a in equation (ii)6a = 3g + 3( g
10) - 3a
⇒ 9a = 33g10⇒ a =
(9.8)3310
= 32.34 ⇒ a = 3.59∴ T = 6a = 6 x 3.59 = 21.55T 1 = 2T = 2 x 21.55 = 43.1 N cut is T1 shown in spring.Mass = wt
g= 43.1
9.8= 4.39 = 4.4 kg
A block of 2kg is suspended from the ceiling through a massless spring of spring constant k = 100Nm
What is the elongation ofthe spring ? If another 1kg is added to the block, what would be the further elongation ?
Solutions :
Given m = 2kg, k = 100Nm
From the free body diagram, kl - 2g = 0 ⇒ kl = 2g⇒ l = 2g
k= 2×9.8
100= 19.6
100= 0.196 = 0.2 m
Suppose further elongation when 1 kg block is added to be x,Then k(1 + x) = 3g⇒ kx = 3g - 2g = g = 9.8 N⇒ x = 9.8
100= 0.098 = 0.1 m
Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of 2.0ms2
. Find theelongations.
Solutions :
a = 2ms2
kl - (2g + 2a) = 0⇒ kl = 2g + 2a= 2 x 9.8 + 2 x 2 = 19.6 + 4⇒ l = 23.6
100= 0.0236 = 0.24
Page 26
24Tutors.com
When 1kg body is added total mass (2 + 1)kg = 3kg.elongation be l1,kl1 = 3g + 3a = 3 x 9.8 + 6⇒ l1 = 33.4
100= 0.0334 = 0.36
Further elongation = l1 - 1 = 0.36 - 0.12 m
The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force ofair resistance on the balloon acts opposite to the direction of velocity and is proportional to it. The balloon carries a mass M andis found to fall down near the earth’s surface with a constant velocity V . How much mass should be removed from the balloon sothat it may rise with a constant velocity V ?
Let, the air resistance force is F and Buoyant force is BGiven thatFa ∞ v where v → velocity⇒ Fa = kv, where k → proportionality constant.When the balloon is moving downward,B + kv = mg .....(i)
⇒ M = B+kvg
For the balloon to rise with a constant velocity v, (upward) let the mass be mHere, B - (mg + kv) = 0 .....(ii)⇒ B = mg + kv⇒ m = B−kw
g
So, amount of mass that should be removed = M - m
=B+kvg
- B−kvg
= B+kv−B+kvg
= 2kvg
=2(Mg−B)
G= 2(M − (B
g)
Solutions :
Let, the air resistance force is F and Buoyant force is BGiven thatFa ∞ v where v → velocity⇒ Fa = kv, where k → proportionality constant.When the balloon is moving downward,B + kv = mg .....(i)
⇒ M = B+kvg
Page 27
24Tutors.com
For the balloon to rise with a constant velocity v, (upward) let the mass be mHere, B - (mg + kv) = 0 .....(ii)⇒ B = mg + kv⇒ m = B−kw
g
So, amount of mass that should be removed = M - m
=B+kvg
- B−kvg
= B+kv−B+kvg
= 2kvg
=2(Mg−B)
G= 2(M − (B
g)
An empty plastic box of mass m is found to accelerate up at the rate of g6
when placed deep inside water. How much sandshould be put inside the box so that it may accelerate down at the rate of g
6?
Solutions :
When the box is accelerating upward,U - mg - m( g
6= 0
⇒ U= mg + mg6
= m(g + ( g6
)) = 7 mg7
.....(i)
⇒ m = 6U7g
Whenit is accelerating downward, let the required mass be M .U - Mg + Mg
6= 0
⇒ U = 6Mg−mg6
= 5Mg6⇒ M = 6U
5g
Mass to be added = M - m = 6U5g
- 6U7g
= 6Ug
(15
- 17
)= 6U
g
(235
)= 12
35
(Ug
)= 12
35
( 7mg6× 1
g
)from (i)
= 25m.
∴ The mass to be added is 2m5
A force−→F = −→v x
−→A is exerted on a particle in addition to the force of gravity, where −→v is the velocity of the particle and
−→A is
a constant vector in the horizontal direction. With what minimum speed a particle of mass m be projected so that it continues tomove undeflected with a constant velocity ?
Solutions :
Given that−→F = −→u x
−→A and −→mg act on the particle.
For the particle to move undeflected with constant velocity, net force should be zero.
∴ (−→u x−→A ) + (−→mg) = 0
∴ (−→u x−→A ) = −−→mg
Because, (−→u x−→A ) is perpendicular to the plane containing −→u and
−→A , −→u should be in the x-z plane
Again ,u A sinθ = mg∴ u = mg
Asinθu will be minimum , when sin θ = 1 ⇒ θ = 90o
∴ umin = mgA
along Z-axis
In a simple Atwood machine, two unequal masses m1, and m2 are connected by a string going over a clamped light smoothpulley. In a typical arrangement (figure 5-E7) m, = 300g and m2 = 600g. The system is released from rest. (a) Find the distancetravelled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp onthe pulley
Initial velocity u = 0So, distance travelled by the body is,S = ut + 1
2at2 ⇒ 0 + 1
2(3.266) 22 = 6.5 m.
b) From (i) T = m1(g + a) = 0.3 (9.8 + 3.26) = 3.9 N
c) The force exerted by the clamp on the pully is giiven by F - 2T = 0F = 2T = 2 x 3.9 = 7.8 N
Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment 2.0s after the system is setinto motion. Find the time elapsed before the string is tight again.
Solutions :
a = 3.26 ms2
T = 3.9NAfter 2 sec mass m1 the velocity V = u + at = 0 + 3.26 x 2 = 6.52m
supward.
At this time m2 is moving 6.52ms
downward.
At time 2 sec, m2 stops for a moment, But m1 is moving upward with velocity 6.52ms
Itwill continue to move till final velocity (at highest point) because zeroHere , v = 0 ; u = 6.52A = −g = −9.8m
s2[moving upward m1]
V = u + at ⇒ 0 = 6.52 + (−9.8)t⇒ t = 6.52
9.8= 0.66 = 2
3sec
During the period 23
, m2 mass also starts moving downward. So the string becomes tight after a time of 23
sec.
Figure (5-E8) shows a uniform rod of length 30cm having a mass of 3.0kg. The strings shown in the figure are pulled by constantforces of 20N and 32N . Find the force exerted by the 20cm part of the rod on the 10cm part. All the surfaces are smooth and thestrings and the pulleys are light.
Solutions :
Mass per unit length 330
kgcm
= 0.10 kgcm
Mass of 10 cm part = m1 = 1 kgMass of 20 cm part = m2 = 2 kgLet, F = contact force between them,From the free body diagramF - 20 - 10 = 0 .....(i)And, 32 - F - 2a = 0 .....(ii) From eqn (i) and (ii)3a - 12 = 0 ⇒ a = 12
3= 4m
s2
Contact force F = 20 + 1a = 20 + 1 x 4 = 24 N .
Consider the situation shown in figure (5-E9). All the surfaces are frictionless and the string and the pulley are light. Find themagnitude of the acceleration of the two blocks.
Solutions :
Page 29
24Tutors.com
Sinθ1 = 45
gSinθ1 − (a+ T ) = 0T - g Sinθ2-a = 0Sinθ2 = 3
5⇒ gSinθ1 = a +T .......(i)
⇒ T = g sinθ2 + a .....(ii)⇒ T + a - g sinθ2 = 0From eqn (i) and (ii) , g sinθ2 + a + a - g sinθ1 = 0⇒ 2a = g sinθ1 - g sinθ2 = g
(45
- 35
)= g
5
⇒ a = g5
x 12
= g10
A constant force F = m2g
is applied on the block of E mass m, as shown in figure (5-E10). The string and the pulley are lightand the surface of the table is smooth. Find the acceleration of m.
Solutions :
From the above free body diagramM1A + F - T = 0 ⇒ T = m1a + F .....(i)From the above free diagramm2a + T - m2g = 0 .....(ii)⇒ m2a +m1a + F - m2g = 0 (from (i)⇒a(m1 +m2) +m2
g2−m2g = 0 because f = m2 g
2⇒ a(m1 +m2) -m2g = 0⇒ a(m1 +m2) =m2g
2⇒ a = m2
2(m1=m2g)
In figure (5-E11) m1 = 5kg, m2 = 2kg and F = 1N . Find the acceleration of either block. Describe the motion of m1, if thestring breaks but F continues to act.
Solutions :
From the above free body diagramT + m1a−m(m1g + F )= 0From the above free body diagramT − (m2g + F +m2a) = 0 ⇒ T = m1g + F −m1a⇒ T = 5g + 1− 5a.....(i)⇒ T = m2g + F +m2a⇒ T = 2g + 1 + 2a.....(ii)From the eqn (i) and (ii)5g + 1− 5a = 2g + 1 + 2a⇒ 3g − 7a = 0⇒ 7a = 3g⇒ a = 3g
7= 29.4
7= 4.2m
s2[g = 9.8m
s2]
a) acceleration of block is 4.2ms2
b) After the string breaks m1 move downward with force F acting downward.ma = F +m2g = (1 + 5g) = 5(g + 0.2)Force = 1N , acceleration = 1
5= 0.2m
s
So, acceleration = ForceMass
=5(g+0.2)
5=
(g+0.2m)
s2
Let m1 = 1kg, m2 = 2kg and m3 = 3kg in figure (5-E12). Find the accelerations of m1 , m2 and m3. The string from the upperpulley to m, is 20cm when the system is released from rest. How long will it take before m, strikes the pulley ?
Let the block m1 + 1 moves upward with acceleration a, and the two blocks m2 and m3 have relative acceleration a2 due to the differenceof weight between them. So, the actual acceleration at the blocks m1,m2,m3 will be a1.(a1 − a2) and (a1 + a2) as shown
Page 30
24Tutors.com
T = 1g − 1a2 = 0....(i) from fig(2)T2− 2g − 2(a1 − a2) = 0....(ii) from fig(3)
T2− 3g − 3(a1 + a2) = 0....(iii) from fig(4)
From eqn (i) and (ii), eliminating T we get, 1g + 1a2 = 4g + 4(a1 + a2)⇒ 5a2 − 4a1 = 3g (iv)From eqn (ii) and (iii), we get 2g + (a1 − a2) = 3g − 3(a1 − a2)⇒ 5a1 + a2 = (v)
Solving (iv) and (v), a1 = 2g29
and a2 = g − 5a1 = g − 10g29
= 19g29
So, a1 - a2 = 2g29− 19g
29= − 17g
29
a1 + a2 = 2g29
+ 19g29
= 21g29
. So acceleration of m1,m2,m3 are 19g29
(up)17g29
(down)21g29
(down) respectively
Again, m1, u = 0,s = 20cm = 0.2m and a2 = 1929g [g = 10m
s2]
∴ S = ut+ 12at2 = 0.2 = 1
2× 19
29gt2 ⇒ t = 0.25sec.
Solutions :
Let the block m1 + 1 moves upward with acceleration a, and the two blocks m2 and m3 have relative acceleration a2 due to the differenceof weight between them. So, the actual acceleration at the blocks m1,m2,m3 will be a1.(a1 − a2) and (a1 + a2) as shownT = 1g − 1a2 = 0....(i) from fig(2)T2− 2g − 2(a1 − a2) = 0....(ii) from fig(3)
T2− 3g − 3(a1 + a2) = 0....(iii) from fig(4)
From eqn (i) and (ii), eliminating T we get, 1g + 1a2 = 4g + 4(a1 + a2)⇒ 5a2 − 4a1 = 3g (iv)From eqn (ii) and (iii), we get 2g + (a1 − a2) = 3g − 3(a1 − a2)⇒ 5a1 + a2 = (v)
Solving (iv) and (v), a1 = 2g29
and a2 = g − 5a1 = g − 10g29
= 19g29
So, a1 - a2 = 2g29− 19g
29= − 17g
29
a1 + a2 = 2g29
+ 19g29
= 21g29
. So acceleration of m1,m2,m3 are 19g29
(up)17g29
(down)21g29
(down) respectively
Again, m1, u = 0,s = 20cm = 0.2m and a2 = 1929g [g = 10m
s2]
∴ S = ut+ 12at2 = 0.2 = 1
2× 19
29gt2 ⇒ t = 0.25sec.
In the previous problem, suppose m2 = 2.0 kg and m3 = 3.0 kg. What should be the mass m so that it remains at rest ?
Solutions :
m1 should be at restT −m1g = 0⇒ T = m1g.....(i)
T2− 2g − 2a1 = 0⇒ T − 4g − 4a1 = 0.....(ii) T
2− 3g − 3a1 = 0⇒ T = 6g − 6a1.....(iii)
From eqn(ii) amp; (iii) we get 3T − 12g = 12g − 2T ⇒ T = 24g5
= 408gPutting the value of T eqn(i) we get m1 = 4.8kg.
Calculate the tension in the string shown in figure (5-E13). The pulley and the string are light and all surfaces are frictionless.Take g = 10m
s2.
Solutions :
T + 1a = 1g....(i)T − 1a = 0⇒ T = 1a....(ii)From eqn (i) and (ii),we get1a+ 1a = 1g ⇒ 2a = g ⇒ a = g
2= 10
2= 5m
s2
From (ii) T = 1a = 5N
Consider the situation shown in figure (5-E14). Both the pulleys and the string are light and all the surfaces are friction less.(a) Find the acceleration of the mass M . (b) Find the tension in the string. (c) Calculate the force exerted by the clamp on thepulley A in the figure.
Solutions :
Ma− 2T = 0⇒Ma = 2T ⇒ T = Ma2
.
T +Ma−Mg = 0⇒ Ma2
+ma = Mg (because T = Ma2
)
⇒ 3Ma = 2Mg ⇒ a = 2g3
a) acceleration of mass M is 2g3b) Tension T = Ma
2= M
2= 2g
3=Mg
3
c) Let, R1 = resultant of tensions = force exerted by the clamp on the pulleyR1 =
√T 2 + T 2 =
√2T
∴=√
2T =√
2Mg3
=√
2Mg3
Again, Tanθ = TT
= 1⇒ θ = 45o
Page 31
24Tutors.com
So, it is√
2Mg3
at an angle of 45o with horizontal.
Find the acceleration of the block of mass M in the situation shown in figure (5-E15). All the surfaces are frictionless and thepulleys and the string are light.
Find the mass M of the hanging block in figure (5-E16) which will prevent the smaller block from slipping over the triangularblock. All the surfaces are friction less and the strings and the pulleys are light.
Solutions :
As the block m does not slinover M l, ct will have acceleration as that of M l
From the freebody diagrams.T +Ma−Mg = 0....(i)T −M ′a−R sin θ = 0....(ii)R sin θ −ma = 0....(iii)R cos θ −mg = 0....(iv)
Eliminating T , R and a from the above equation,we get M =Ml+mcotθ−1
Find the acceleration of the blocks A and B in the three situations shown in figure (5-E17).
Solutions :
a) 5a+ T − 5g = 0⇒ T = 5G− 5a.....(i)Again ( 1
2− 4g − 8a = 0⇒ T = 8g − 16a.....(ii)
From eqn(i) and (ii), we get 5g − 5a = 8g + 16a⇒ 21a = −3g ⇒ a = − 17g
So, the acceleration of 5kg mass is g7
upward and that of 4kg mass is 2a = 2g7
(downward).
b) 4a− V 2 = 0⇒ 8a− T = 0⇒ T = 8a.....(ii)Again, T + 5a− 5g = 0⇒ 8a+ 5a+ 5g = 0⇒ 13a− 5g = 0⇒ a = 5g
13downward.
Acceleration of mass(A) kg is 2a = 1013
(g) amp; 5kg (B) is 5g13
c) T + 1a− 1g = 0⇒ T = 1g − 1a.....(i)Again, T
2− 2g − 4a = 0⇒ T − 4g − 8a = 0.....(ii)
⇒ 1g − 1a− 4g − 8a = 0[From(i)]⇒ a = −( g
3)downward
Acceleration of mass 1kg(b) is g3
(up)
Acceleration of mass 2kg(A) is 2g3
(downward)
Find the acceleration of the 500g block in figure (5-E18).
A monkey of mass 15 kg is climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of1ms2
, how much force should it apply to the rope ? If the rope is 5 m long and the monkey starts from rest, how much time will ittake to reach the ceiling ?
Solutions :
m = 15kg of monkey.a = 1m
s2
From the free body diagram∴ T − [15g + 15(1)] = 0⇒ T = 15(10 + 1)⇒ T = 15× 11⇒ T = 165NThe monkey should apply 165N force to the rope.Initial velocity u=0; acceleration a = 1m
s2; s = 5m.
∴ s = ut+ 12at2
5 = 0 + ( 12
)1t2
⇒ t2 = 5× 2⇒ t =
√10sec
TIme required is√
10sec
A block A can slide on a frictionless incline of angle θ and length 1, kept inside an elevator going up with uniform velocity v(figure 5-E22). Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline.
Solutions :
The driving force on the block which n the body to move down the plane is F =mg sin θ .So , acceleration = g sin θInitial velocity of block u = 0s = l, a =mg sin θNow, S = ut+ 1
2at2
→ l=0+ 12l
(g sin θ) t2 ⇒ g2= 2lgsinθ
⇒ t =√
2lgsinθ
Time taken is√
2lgsinθ
A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure5-E19). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equalacceleration. If initially both were at rest, their separation will not change as time passes.
Solutions :
Suppose the monkey accelerates upward with acceleration ′a′ amp; the block, accelerate downward with acceleration a1. let the forceexerted by the monkey is equal to ′T ′
From the free body diagram of the monkey∴ −mg −ma = 0 ....(i)⇒ T = mg +maAgain from the FBD of the block,T = ma1 −mg = 0⇒ mg +ma+ma1 −mg = 0[From(i)]⇒ ma−ma1 ⇒ a = a1
Acceleration ′ − a′ downward i.e ′a′ upward. ∴ The block amp; the monkey move in the same direction with equal acceleration.If initially they are rest (no force is exerted by monkey)no motion of monkey of block occurs as they have same weight (same mass). theirseperation will not change as time passes.
The monkey B shown in figure (5-E20) is holding on to the tail of the monkey A which is climbing up a rope. The masses ofthe monkeys A and B are 5kg and 2kg respectively. If A can tolerate a tension of 30N in its tail, what force should it apply on therope in order to carry the monkey B with it ? Take g = 10m
s2.
Solutions :
Page 33
24Tutors.com
Suppose ′A′ move upward with acceleration a, such that in the tail of A maximum tension 30N produced.T − 5g − 30− 5a = 0.....(i)30− 2g − 2a = 0......(ii)⇒ T = 5 + 30 + (5× 5) = 105N (max)⇒ 30− 20− 2a = 0⇒ a = 5m
s2
So. A can apply a maximum force of 105n in the rope to carry the monkey B with it.For minimum force there is no acceleration of mnkey ′A′ and B ⇒ a = 0Now equation (ii) is T l1 − 2g = 0⇒ T l1 = 20N (wt of monkey B)Equation (ii) is T − 5g − 20 = 0 [AsT l1 = 20N ]⇒ T = 5g + 20 = 50 + 20 = 70N.∴ the monkey A should apply force between 70N and 105N to carry the monkey B with it.
A car is speeding up on a horizontal road with an acceleration a. Consider the following situations in the car. (i) A ball issuspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle. (ii) A block iskept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.
Solutions :From the free body diagram T cos θ-mg = 0
⇒ T cos θ = mgT = mg
cosθ.....(i)
ma-Tsin θ=0⇒ t = ma
sinθ.....(ii)
From (i) amp; (ii) mgcosθ
= masinθ
⇒ tan θ = ag⇒ θ=tan−t a
g
The angle is Tan−t ag
with vertical
(ii)m⇒ mass of blockSuppose the angle of incline is θFrom the diagramma cos θ-mg sin θ = 0⇒ ma cos θ =mg sin θ⇒ sinθ
cosθ= a
g
⇒ tan θ = ag⇒ θ = tan−1(a
g)
Suppose pendulum makes θ angle with the vertical. let, m = mass of the pendulum.From the free body diagram T cos θ-mg = 0⇒ T cos θ = mgT = mg
cosθ.....(i)
ma-Tsin θ=0⇒ t = ma
sinθ.....(ii)
From (i) amp; (ii) mgcosθ
= masinθ
⇒ tan θ = ag⇒ θ=tan−t a
g
The angle is Tan−t ag
with vertical
(ii)m⇒ mass of blockSuppose the angle of incline is θFrom the diagramma cos θ-mg sin θ = 0⇒ ma cos θ =mg sin θ⇒ sinθ
cosθ= a
g
⇒ tan θ = ag⇒ θ = tan−1(a
g)
6. A 3cm tall object is placed at a distance of 7.5cm from a convex mirror of focal length 6cm. Find the location, size andnature of the image.
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12ms2
. Find thedisplacement of the block during the first 0.2s after the start. Take g = 10m
s2.
Solutions :
Because, the elevator is moving downward with an acceleration 12ms2
(gt; g), the body gets seperated. So, body moves with an acceleration
g = 10ms2
[freely falling body] and the elevator move with acceleration 12ms2
Now, the block has acceleration g= 10ms2
u = 0t = 0.2secSo, the distance traveled by the block is given by,∴ s = ut+ 1
2at2
= 0 + ( 12
)10(0.2)2 = 5× 0.04 = 0.2m = 20cm.The displacement of body is 20cm during first 0.2sec
Page 34
24Tutors.com
Figure (5-E21) shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hangingfrom a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages tokeep the box at rest, what is the weight shown by the machine ? What force should he exert on the rope to get his correct weighton the machine?
Solutions :
(i) Given , Mass of man = 60kgLet Rl = apparent weight of man in this case .Now, Rl + T − 60g = 0 [FromFBDofman]⇒ T = 60g −Rl.....(i)T −Rl − 30g = 0.....(ii) [FromFBDofbox]⇒ 60g −Rl −Rl − 30g = 0 [From(i)]⇒ Rl = 15g The weight shown by the machine is 15kg.
(ii) To get his correct weight suppose the applied force is T and so, accelerates upward with ′a′.In this case, given that correct weight= R = 60g, where g = accn due to gravityFrom the FBD of the man T 1 +R− 60g − 60a = 0T 1 − 60a = 0[∴ R = 60g] T 1 = 60a.....(i)From the FBD of the boxT 1 −R− 30g − 30a = 0⇒ T 1 − 60g − 30g − 30a = 0⇒ T 1 − 30a = 90g = 900⇒ T 1 = 30a− 900.....(ii)From the eq (i) and(ii) we get T 1 = 2T 1 − 1800⇒ T 1 = 1800N∴ So, he should exert 1800N force on the rope to get correct reading.
1- A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s 2. What is the coefficient of kinetic frictionbetween the block and the plane ?
Solutions :Let m = mass of the block from the freebody diagram,
R - mg = 0 ⇒ R = mg ...(1)Again ma - µ R = 0 ⇒ ma = µ mg (from (1))⇒ a = µg ⇒ 4 = µg ⇒ µ = 4
g= 4
10= 0.4
The co-efficient of kinetic friction between the block and the plane is 0.4//undefined
2- A block is projected along a rough horizontal road with a speed of 10 m/s. If the coefficient of kinetic friction is 0.10, howfar will it travel before coming to rest ?
Solutions :Due to friction the body will decelerate
Let the deceleration be ’a’R - mg = 0 ⇒ R = mg ...(1)ma - µ R = 0 ⇒ ma = µ R = µ mg (from(1))⇒ a = µg = 0.1×10 = 1m/sec2
Initial velocity u = 10m/sFinal velocity v = 0m/sa = -1m/s2(deceleration)
S = v2−u2
2a= 0−102
2(−1)= 100
2=50m
It will travel 50m before coming to rest.
3- A block of mass m is kept on a horizontal table. If the static friction coefficient is 1.t, find the frictional force acting on theblock.
Solutions :Body is kept on the horizontal table. If no force is applied, no frictional force will be there
f→ frictional forceF→ Applied forceFrom grap it can be seen that when applied force is zero, frictional force is zero.
4- A block slides down an inclined surface of inclination 30 with the horizontal. Starting from rest it covers 8 m in the first twoseconds. Find the coefficient of kinetic friction between the two.
Solutions :From the free body diagram,
R - mg cos θ = 0 ⇒ R = mg cos θ ...(1)For the blockU = 0, s = 8m, t = 2sec.∴s = ut + 1
2at2 ⇒ 8 = 0 + 1
2a 22 ⇒ a = 4m
s2
Again, µR + ma - mg sin θ = 0⇒ µ mg cos θ + ma - mg sin θ = 0 [from(1)]m(µg cos θ + a - g sinθ) = 0
Page 35
24Tutors.com
⇒ µ × 10 × cos 30 = g sin 30 - a
⇒ µ × 10 ×√
( 33
) = 10 × ( 12
) - 4
⇒ ( 5√3
)µ = 1 ⇒ µ = 1(5/√
3)= 0.11
∴ Co-efficient of kinetic friction between the two is 0.11.
5- Suppose the block of the previous problem is pushed down the incline with a force of 4 N. How far will the block move in thefirst two seconds after starting from rest ? The mass of the block is 4 kg.
Solutions :From the free body diagram
4 - 4a - µR + 4g sin 30 = 0 ...(1)R - 4g cos 30 = 0 ...(2)⇒ R = 4g cos 30Putting the values of R is amp; in equn. (1)4 - 4a - 0.11 × 4g cos 30 + 4g sin 30 = 0⇒ 4 - 4a - 0.11 4 10 (
√3/2 ) + 4 10 (1/2) = 0
⇒ 4 - 4a - 3.81 + 20 = 0 ⇒ a ≈ 5 ms2
For the block u =0, t = 2sec, a = 5ms2
Distance s = ut + 1/2 at2 ⇒ s = 0 + (1/2) 5 22 = 10mThe block will move 10m.
6- A body of mass 2 kg is lying on a rough inclined plane of inclination 30. Find the magnitude of the force parallel to theincline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0.2.
Solutions :To make the block move up the incline, the force should be equal and opposite to the net force acting down the incline = µ R + 2 g sin 30
= 0.2 (9.8)√
3 + 2 × 9.8 × ( 12
) [from(1)]= 3.39 + 9.8 = 13NWith this minimum force the body move up the incline with a constant velocity as net force on it is zero.b) Net force acting down the incline is given by,F = 2 g sin 30 - µR= 2 9.8 (1/2) - 3.39 = 6.41NDue to F = 6.41N the body will move down the incline with acceleration.No external force is required.∴ Force required is zero.
7- Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.
Solutions :From the free body diagram
g = 10ms2
, m = 2kg, θ = 30, µ = 0.2R - mg cos θ - F sin θ = 0⇒ R = mg cosθ + F sin θ ...(1)And mg sin θ + µR - F cos θ = 0⇒ mg sin θ + µ(mg cos θ + F sin θ) - F cos θ = 0⇒ mg sin θ + µ mg cos θ + µ F sin θ - F cos θ = 0
⇒ F =(mgsinθ−µmgcosθ)
(µsinθ−cosθ)
⇒ F =2×10×
(12
)+0.2×2×10×
(√3
2
)0.2×( 1
2)−(√
32
)= 13.464
0.76= 17.7N ≈ 17.5N
8- In a children-park an inclined plane is constructed with an angle of incline 45 in the middle part (figure 6-E1). Find theacceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g =10m
s2.
Solutions :m → mass of child
R - mg cos 45 = 0⇒ R = mg cos 45 = mg
v2...(1)
Net force acting on the boy due to which it slides down is mg sin 45 - µR= mg sin 45 - µ mg cos 45= m 10( 1√
2) - 0.6 m 10 ( 1√
2)
= m( 2√2
)
acceleration = Forcemass
=m(2√
2)m
= 2√
2ms2
9- A body starts slipping down an incline and moves half meter in half second. How long will it take to move the next halfmeter ?
Solutions :Suppose, the body is accelerating down with acceleration a.
From the free body diagramR mg cos θ = 0⇒ R = mg cosθ ...(1)ma + mg sinθ - µ R = 0
⇒ a =mg(sinθ−µcosθ)
m= g (sinθ − µcosθ)
For the first half mt. u = 0, s = 0.5m, t = 0.5 sec.So, v = u + at = 0 + (0.5)4 = 2m
s
S = ut + 12
at2 ⇒ 0.5 = 0 + 12
a 0.52 ⇒ a = 4ms2
...(2)For the next half metreu‘ = 2m
s, a = 4m
s2, s = 0.5
⇒ 0.5 = 2t + ( 12
) 4 t2 ⇒ 2 t2 + 2 t - 0.5 = 0
⇒ 4 t2 + 4 t - 1 = 0∴ = 1.656
8= 0.207sec
Time taken to cover next half meter is 0.21sec.
Page 36
24Tutors.com
10- The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle offriction. Show that, if X be the angle of friction and la the coefficient of static friction, λ ≤ tan−1µ
Solutions :f → applied force
F‘ → contact forceF → frictional forceR → normal reactionµ = tan λ = F
RWhen F = µR, F is the limiting friction (max friction). When applied force increase, force of friction increase upto limiting friction (µR)Before reaching limiting frictionF lt; µR∴ tan λ = F
R≤ µR
R⇒ tan λ ≤ tan−1 µ
11- Consider the situation shown in figure (6-E2). Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension in thestring connecting the 1.0 kg blocks and (c) the tension in the string attached to 0.50 kg.
Solutions :From the free body diagram
T + 0.5a 0.5 g = 0 ...(1)µR + + 1a +T1 - T = 0 ...(2)µR + 1a - T1 = 0µR + 1a = T1 ...(3)From (2) and (3) ⇒ µR + a = T - T1
∴ T - T1 = T1
⇒ T = 2T1
Equation (2) becomes µR + a + T1 - 2T1 = 0⇒ µR + a - T1 = 0⇒ T1 = µR + a = 0.2g + a ...(4)Equation (1) becomes 2T1 + 0/5a - 0.5g = 0
⇒ T1 = 0.5g−0.5a2
= 0.25g - 0.25a ...(5)From (4) amp; (5) 0.2g + a = 0.25g 0.25a⇒ a = 0.05
1.25× 10 = 0.04×10 = 0.4m
s2
a) Accln of 1kg blocks each is 0.4ms2
b) Tension T1 = 0.2g + a + 0.4 = 2.4Nc) T = 0.5g 0.5a = 0.5 10 0.5 0.4 = 4.8N
12- If the tension in the string in figure (6-E3) is 16 N and the acceleration of each block is 0.5 ms2
, find the friction coefficientsat the two contacts with the blocks.
Solutions :From the free body diagram
µ1 R + 1 - 16 = 0⇒ µ1 (2g) + (-15) = 0⇒ µ1 = 15
20= 0.75
µ2 R1 + 4 × 0.5 + 16 - 4g sin 30 = 0= 0⇒ µ2 (20
√3) + 2 + 16 - 20 = 0
⇒ µ2 = 220√
3= 1
17.32= 0.057 ≈ 0.06
∴Coefficient of friction µ1 = 0.75 and µ2 = 0.06.
13- The friction coefficient between the table and the block shown in figure (6-E4) is 0.2. Find the tensions in the two strings.
Equation (ii) ⇒ T = 5 10 + 10 3.6 + 0.2 5 10⇒ 96N in the left stringEquation (iii) T1 = 5g + 5a = 5 10 + 5 3.6 =68N in the right string.
14- The friction coefficient between a road and the tyre of a vehicle is 4/3. Find the maximum incline the road may have so thatonce hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m.
Solutions :s = 5m, µ = 4
3, g = 10m
s2
u = 36km/h = 10m/s, v = 0 ,
a = v2−u2
2s= 0−102
2×5= -10m
s2
From the freebody diagrams,R mg cos θ = 0 ; g = 10m
s2
⇒ = mg cosθ...(i) ; µ = 43
.Again, ma + mg sin θ- µ R = 0⇒ a+ g sin θ - mg cos θ = 0⇒30 + 30 sinθ - 40cosθ = 0⇒ 4 cosθ - 3 sinθ = 3⇒ 4√
15- The friction coefficient between an athelete’s shoes and the ground is 0.90. Suppose a superman wears these shoes and racesfor 50 m. There is no upper limit on his capacity of running at high speeds. (a) Find the minimum time that he will have to takein completing the 50 m starting from rest. (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimumtime will he take to stop ?
Solutions :to reach in minimum time, he has to move with maximum possible acceleration.
Let, the maximum acceleration is a∴ ma µR = 0 ⇒ ma = µ mg⇒ a = µ g = 0.9 × 10 = 9m
s2
a) Initial velocity u = 0, t = ?a = 9m
s2, s = 50m
s = ut + 12
at2 ⇒ 50 = 0 + ( 12
) 9 t2 ⇒ t =√
1009
= 103
b) After overing 50m, velocity of the athelete isV = u + at = 0 + 9 10
3= 30m
sHe has to stop in minimum time. So deceleration ia a = 9m
s2(max)
[R = mama = µR((max frictional force)⇒ a = µg = 9m
s2( Deceleration )]
u1 = 30ms
, v1 = 0
t = v1−u1
a= 0−30−a = −30
−a = 103
sec.
16- A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30 (figure 6-E5). The frictioncoefficient between the road and the tyre is 1/2
√3. Show that no matter how hard the driver applies the brakes, the car will reach
the bottom with a speed greater than 36 kmhr
. Take g = 10 ms
.
Solutions :Hardest brake means maximum force of friction is developed between cars type amp; road.
Max frictional force = µRFrom the free body diagramR - mg cos θ = 0⇒ R = mg cos θ ...(i)and µR + ma - mg sin θ = 0 ...(ii)⇒ µmg cos θ + ma - mg sinθ = 0
⇒ a = 5 - 1 - (2√
3) × 10 (√
32
) = 2.5 ms2
When, hardest brake is applied the car move with acceleration 2.5ms2
S = 12.8m, u = 6ms
SO, velocity at the end of inclineV =
√u2 + 2as =
√62 + 2(2.5)(12.8) =
√36 + 64 = 36 km
h
Hence how hard the driver applies the brakes, that car reaches the bottom with least velocity 36 kmh
.
17- A car starts from rest on a half kilometer long bridge. The coefficient of friction between the tyre and the road is 1.0. Showthat one cannot drive through the bridge in less than 10 s.
Solutions :Let, a maximum acceleration produced in car.
∴ ma = µ [For more acceleration, the tyres will slip]⇒ ma = µ mg ⇒ a = µg = 1 10 = 10m
s2
For crossing the bridge in minimum time, it has to travel with maximum acceleration u = 0,s = 500m, a = 10ms2
s = ut + 12at2
⇒ 500 = 0 + ( 12
) 10t2 ⇒ t = 10sec.If acceleration is less than 10m
s2, time will be more than 10sec. So one cant drive through the bridge in less than 10sec.
18- Figure (6-E6) shows two blocks in contact sliding down an inclined surface of inclination 30. The friction coefficient betweenthe block of mass 2.0 kg and the incline is µ1 and that between the block of mass 4.0 kg and the incline is µ2. Calculate theacceleration of the 2.0 kg block if (a) µ1 = 0.20 and µ2, = 0.30, (b) µ1 = 0.30 and µ2 = 0.20. Take g = 10 m
s2.
Solutions :From the free body diagram
R = 4g cos 30 = 4 × 10 ×√
32
= 20√
3 ...(i)µ2 R + 4a - P - 4g sin 30 = 0 ⇒ 0.3 (40) cos 30 + 4a - P - 40 sin 20 = 0 ...(ii)
P + 2a + µ1 R1 2g sin 30 = 0 ...(iii)
R1 = 2g cos 30 = 2 × 10 ×√
32
= 10√
3...(iv)Equn. (ii) 6
√3 + 4a - P - 20 = 0
Equn (iv) P + 2a + 2√
3 - 10 = 0From Equn (ii) amp; (iv) 6
√3 + 6a - 30 + 2
√3 = 0
⇒ 6a = 30 - 8√
3 = 30 - 13.85 = 16.15⇒ a = 16.15
6= 2.69 = 2.7m
s2
b) can be solved. In this case, the 4 kg block will travel with more acceleration because, coefficient of friction is less than that of 2kg. So, theywill move separately. Drawing the free body diagram of 2kg mass only, it can be found that, a = 2.4m
s2.
19- Two masses M1 and M2 are connected by a light rod and the system is slipping down a rough incline of angle θ with thehorizontal. The friction coefficient at both the contacts is pt. Find the acceleration of the system and the force by the rod on oneof the blocks.
Solutions :
Page 38
24Tutors.com
R1 = M1 g cos θ ...(i)R2 = M2 g cos θ ...(ii)T + M1 g sin θ - m1 a - µ R1 = 0 ...(iii)T - M2 - M2 a + µR2 = 0 ...(iv)Equn (iii) ⇒ T + M1g sinθ - M1 a - µ M1g cosθ = 0Equn (iv) ⇒T - M2 g sin θ + M2 a + µ M2 g cos θ= 0 ...(v)
R1 = M1 g cos θ ...(i)R2 = M2 g cos θ ...(ii)T + M1 g sin θ - m1 a - µ R1 = 0 ...(iii)T - M2 - M2 a + µR2 = 0 ...(iv)Equn (iii) ⇒ T + M1g sinθ - M1 a - µ M1g cosθ = 0Equn (iv) ⇒T - M2 g sin θ + M2 a + µ M2 g cos θ= 0 ...(v)Equn (iv) amp; (v) ⇒ g sin θ (M1 + M2 ) - a(M1 + M2 ) - µ g cos θ (M1 + M2 ) = 0⇒ a (M1 + M2) = g sin θ (M1 + M2) - µ g cos θ (M1 + M2)⇒ a = g(sinθ - µ cosθ)∴ The blocks (system has acceleration g(sin θ µ cos θ)The force exerted by the rod on one of the blocks is tension.Tension T = - M1 g sin θ + M1 a + µ M1 g sin θ⇒ T = -M1g sinθ + M1(g sin θ - µ g cos θ) + µ M1g cosθ⇒ T = 0.
20- A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surfaceis µ. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block ? In which directionshould this force act ?
Solutions :Let p be the force applied to at an angle θ
From the free body diagramR + P sin θ - mg = 0⇒ R = - P sinθ + mg ...(i)µR - p cos θ ...(ii)Equn. (i) is µ(mg - P sin θ) - P cos θ = 0⇒ µ mg = µ P sinθ - P cos θ⇒ P = µmg
µsinθ+cosθ
Applied force P should be minimum, when µ sin θ + cos θ is maximum.Again, µ sin θ + cos θ is maximum when its derivative is zero.∴ ddθ
(µ sin θ + cos θ) = 0
⇒ µ cosθ - sinθ = 0 ⇒ θ = tan1 µSo, P = µmg
µsinθ+cosθ= µmgsecθ
1+µtanθ= µmgsecθ
1+tan2θ
= µmgsecθ
= µmg√(1+tan2θ)
= µmg√1+µ2
Minimum force is µmg√1+µ2
at an angle θ = tan1µ.
21- The friction coefficient between the board and the floor shown in figure (6-E7) is µ. Find the maximum force that the mancan exert on the rope so that the board does not slip on the floor.
Solutions :Let, the max force exerted by the man is T.
From the free body diagramR + T - Mg = 0⇒ R = Mg - T ...(i)R1 = R + mg ...(iii)And T - µ R1 = 0⇒ T - µ(R+ mg) = 0[From equn.(ii)]⇒ T - µ R - µ mg = 0⇒ T - µ (Mg + T) - µ mg = 0 [from (i)]
⇒ T =µ(M+m)g
1+µ
Maximum force exerted by man isµ(M+m)g
1+µ
22- A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of frictionbetween the blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a) the upper block,(b) the lower block. Take g= 10 m
2kg block has acceleration 4m/s amp; that of 4 kg is 1ms2
(ii) R1 = 2g = 20Ma - µ R1 = 0⇒ 2a = 0.2 (20) = 4⇒ a = 2m
s2
4a + 0.2 × 2 × 10 - 12 = 0⇒ 4a + 4 = 12
Page 39
24Tutors.com
⇒ 4a = 8⇒ a = 2 m
s2.
23- Find the accelerations a1 , a2 ,a3 of the three blocks shown in figure (6-E8) if a horizontal force of 10 N is applied on (a) 2kg block, (b) 3 kg block, (c) 7 kg block. Take g= 10 m
s2.
Solutions :a) When the 10N force applied on 2kg block, it experiences maximum frictional force
µR1 = mu × 2kg = (0.2) × 20 = 4N from the 3kg block.So, the 2kg block experiences a net force of 10 - 4 = 6NSo, a1 = 6
2= 3 m
s2
But for the 3kg block, (fig-3) the frictional force from 2kg block (4N) becomes the driving force and the maximum frictional force between 3kgand 7 kg block isµ2R2 = (0.3) × 5kg = 15NSo, the 3kg block cannot move relative to the 7kg block. The 3kg block and 7kg block both will have same acceleration (a2 = a3) which willbe due to the 4N force because there is no friction from the floor.∴a2 = a3 = 4
10= 0.4m
s2
b) When the 10N force is applied to the 3kg block, it can experience maximum frictional force of 15 + 4 = 19N from the 2kg block amp; 7kgblock.So, it can not move with respect to them.As the floor is frictionless, all the three bodies will move together∴ a1 = a2 = a3 = 10
12= ( 5
6)ms2
c) Similarly, it can be proved that when the 10N force is applied to the 7kg block, all the three blocks will move together.Again a1 = a2 = a3 = ( 5
6)ms2
24- The friction coefficient between the two blocks shown in figure (6-E9) is µ but the floor is smooth. (a) What maximumhorizontal force F can be applied without disturbing the equilibrium of the system ? (b) Suppose the horizontal force applied isdouble of that found in part (a). Find the accelerations of the two masses.
Solutions :Both upper block amp; lower block will have acceleration 2 m
s2
R1 = mg ...(i)F - µR1 - T = 0 ⇒ F - µmg - T = 0 ...(ii)T - µR1 = 0⇒ T = µmg
∴ F = µ mg + µ mg = 2 µ mg [putting T = µmg]b) 2F - T - µ mg - ma = 0 ...(i)
T - Ma - µ mg = 0⇒ T = Ma + µ mgPutting value of T in (i)2f - Ma - µ mg - µ mg - ma = 0⇒ 2(2µmg) - 2 µ mg = a(M + m) [Putting F = 2 µmg]⇒ 4µ mg - 2 µ mg = a (M + m)
⇒ a = 2µmgM+m
Both blocks move with this acceleration a in opposite direction.
25- Suppose the entire system of the previous question is kept inside an elevator which is coming down with an acceleration alt; g. Repeat parts (a) and (b).
Solutions :R1 + ma - mg =0
⇒ R1 = m(g-a) = mg - ma ...(i)T - µ R1 = 0 ⇒ T = m (mg - ma) ...(ii)Again, F - T - µ R1 = 0⇒ F - µ(mg - ma) - µ(mg - ma) = 0⇒ F - µ mg + µ ma - µmg + µ ma = 0F - µ mg + µ ma - µ mg + µma = 0⇒ F = 2 µ mg - 2µ ma ⇒ F = 2µ m(g-a)b) Acceleration of the block be a1
R1 = mg ma ...(i)2F - T - µR1 - ma1 = 0 ...(ii)T1 - µR1 - M a1 = 0⇒ T = µR1 + M a1
⇒ T = µ mg - µ ma + M a1
Subtracting values of F amp; T, we get2(2µm(g a)) 2(µmg µma + Ma1 ) µmg + µ ma µ a1 = 0⇒ 4µ mg 4 µ ma 2 µ mg + 2µ ma = ma1 + M a1
⇒ a1 =2µm(g−a)M+m
Both blocks move with this acceleration but in opposite directions.
26- Consider the situation shown in figure (6-E9). Suppose a small electric field E exists in the space in the vertically upwarddirection and the upper block carries a positive charge Q on its top surface. The friction coefficient between the two blocks is g butthe floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium ? [Hint : The force on acharge Q by the electric field E is F = QE in the direction of E.]
Solutions :R1 + QE - mg = 0
R1 = mg - QE ...(i)F - T - µR1 = 0⇒ F - T µ(mg QE) = 0⇒ F - T - µ mg + µQE = 0 ...(2)
Page 40
24Tutors.com
T - µ R 1 = 0⇒ T = µ R1 = µ (mg - QE) = µ mg - µQENow equation (ii) is F mg + µ QE µ mg + µ QE = 0⇒ F - 2 µ mg + 2µQE = 0⇒ F = 2µmg - 2µ QEF = 2µ(mg - QE)Maximum horizontal force that can be applied is 2µ(mg QE).
27- A block of mass m slips on a rough horizontal table under the action of a horizontal force applied to it. The coefficient offriction between the block and the table is g. The table does not move on the floor. Find the total frictional force applied by thefloor on the legs of the table. Do you need the friction coefficient between the table and the floor or the mass of the table ?
Solutions :Because the block slips on the table, maximum frictional force acts on it.
From the free body diagramR = mg∴ F - µ R = 0 ⇒ F = µR = µmgBut the table is at rest. So, frictional force at the legs of the table is not µ R1 . Let be f, so form the free body diagram.f0 - µ R1 = 0 ⇒ f0 = µR = µ mgTotal frictional force on table by floor is µ mg.
28- Find the acceleration of the block of mass M in the situation of figure (6-E10). The coefficient of friction between the twoblocks is µ1, and that between the bigger block and the ground is µ2.
Solutions :Let the acceleration of block M is a towards right. So, the block m must go down with an acceleration 2a.
As the block m is in contact with the block M, it will also have acceleration a towards right. So, it will experience two inertia forces as shownin the free body diagram-1.From free body diagram -1R1 - ma = 0 ⇒ R1 = ma ...(i)Again, 2ma + T - mg + µ1 R1 = 0⇒ T = mg - (2 - µ1)ma ...(ii)From free body diagram-2T + µ1 R1 + mg - R2 = 0⇒ R2 = T + µ1 ma + Mg [Putting the value of R 1 from (i)]= (mg 2ma µ1 ma) + µ1 ma + Mg [Putting the value of T from (ii)]∴ R2 = Mg + mg - 2ma ...(iii)Again, form the free body diagram -2T + T - R - Ma -µ2R2 = 0⇒ 2T - MA - mA - µ2(Mg + mg - 2ma) = 0 [Putting the values of R 1 and R 2 from (i) and (iii)]⇒ 2T = (M + m) + µ2 (Mg + mg - 2ma) ...(iv)From equation (ii) and (iv)2T = 2 mg - 2(2+ µ1)mg = (M + m)a + µ2(Mg + mg - 2ma)⇒ 2mg - µ2(M + m)g = a (M + m - 2µ2m + 4m + 2µ1m)
⇒ a =[2m−µ2(M+m)]gM+m[5+2(µ1−µ2)]
29- A block of mass 2 kg is pushed against a rough vertical all with a force of 40 N, coefficient of static friction being 0.5.Another horizontal force of 15 N, is applied on the block in a direction parallel to the wall. Will the block move ? If yes, in whichdirection ? If no, find the frictional force exerted by the wall on the block.
So, the block will move at an angle 53 with an 15N force.
30- A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and theother wall B by his back (figure 6-E11). Assume that the friction coefficient between his body and the walls is 0.8 and that limitingfriction acts at all the contacts. (a) Show that the person pushes the two walls with equal force. (b) Find the normal force exertedby either wall on the person. Take g = 10m
s2.
Solutions :a) Mass of man = 50kg. g =10m
s2
Frictional force developed between hands, legs amp; back side with the wall the wt of man. So he remains in equilibrium.He gives equal force on both the walls so gets equal reaction R from both the walls. If he applies unequal forces R should be different he cantrest between the walls.Frictional force 2µR balance his wt.From the free body diagramµR + µR = 40g ⇒ 2 µR = 40 × 10 ⇒ R = 40×10
2×0.8= 250N
b) The normal force is 250 N.
31- Figure (6-E12) shows a small block of mass m kept at the left end of a larger block of mass M and length l.The system canslide on a horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the roadand the bigger block is g and that between the block is µ
2Find the time elapsed before the smaller blocks separates from the bigger
block.
Solutions :Let a 1 and a 2 be the accelerations of ma and M respectively.
Here, a1 gt; a2 so that m moves on MSuppose, after time t m separate from M.In this time, m covers vt + 1
2a1t2 and SM = vt + 1
2a2t2
For m to m to m separate from M. vt + 12a1 t2 = vt + 1
2a2t2 + l
Again from free body diagramMa1 + µ
2R = 0
⇒ ma1 = - (µ2
) mg = - (µ2
)m 10 ⇒ a1 = -5µ
Page 41
24Tutors.com
Again,Ma2 + µ (M + m)g - (µ
2)mg = 0⇒ 2M a2 + µ mg - 2µMg - 2 µmg
⇒ a2 = −µmg−2µMg2M
Putting values of a1 amp; a2 in equation (1) we can find that
T =√
4ml(M+m)µg
11. Find the acceleration of the moon with respect to the earth from the following data : Distance between the earth and themoon = 3.85 x 10 5 km and the time taken by the moon to complete one revolution around the earth = 27.3 days.
Solutions :Distance between Earth and Moon
T = 27.3 days =24x3600x(27.3)sec= 2.36x106sec
v = 2πfT
= 2×3.14×3.85×108
2.36×106 = 1025.42m/sec
a = v2
r=
(1025.42)2
64×105 = 0.00273m/sec2=2.73× 10−3m/sec2
2. Find the acceleration of a particle placed on the surface of the earth at the equator due to earth’s rotation. The diameter ofearth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.
Solutions :Diameter of earth = 12800km
Radius R = 6400km = 64 x 105
V =2 πRT
= 2×3.14×64×105
24×3600m/sec=465.185
a = V 2
R=
(46.5185)2
64×105 = 0.0338m/sec2
3. A particle moves in a circle of radius 1.0 cm at a speed given by v = 2.0 t where v is in cm/s and t in seconds. (a) Findthe radial acceleration of the particle at t = 1 s. (b) Find the tangential acceleration at t = 1 s. (c) Find the magnitude of theacceleration at t = 1 s.el.
Solutions :V = 2t, r= 1cm
a) Radial acceleration at t= 1 sec
a= V 2
R= 22
1= 4cm/sec2
b) Tangential acceleration at t =1 seca= dv
dt= d
dt(2t) = 2cm/sec2
c)Magnitude of acceleration at t=1 sec
a=√
42
+√
22
=√
20cm/sec2
4. A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal forceon the scooter is needed to make the turn possible ?
Solutions :Given that m = 150kg.
v=36km/hr = 10 m/sec, r=30m
Horizontal force needed is mv2
r=
150×(10)2
30= 150×100
30= 500N
5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, whatshould be the proper angle of banking ?
Solutions :In the diagram Rcosθ = mg..(i)
Rsinθ = mv2
r
Dividing equation (i) by equation (ii) Tanθ = mv2
rmg= v2
rg
v = 36km/hr = 10m/sec, r = 30m
Tanθ = v2
rg= 100
30×10= (1/3)
⇒ θ = tan−1(1/3)
6. A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle ofbanking ?
Solutions :Radius of Park = r = 10m
speed of vehicle = 18km/hr =5m/sec
Angleofbankingtanθ = v2
rg
⇒ θ = tan−1 v2
rg= tan−1 25
100= tan−1(1/4)
7. If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that ascooter going at 18 km/hr does not skid ?
Solutions :The road is horizontal (no banking)
mv2
R= µN
mv2
R= µmg v = 5m/sec,R = 10m
⇒ 2510
= µg ⇒ µ = 25100
= 0.25
8. A circular road of radius 50 m has the angle of banking equal to 30. At what speed should a vehicle go on this road so thatthe friction is not used ?
Solutions :Angle of banking = θ = 30
Radius =r = 50mtanθ = v2
rg⇒ tan30 = v2
rg
Page 42
24Tutors.com
⇒ 1√3
= v2
rg⇒= v2 = rg√
3= 50×10√
3
⇒ v =√
500√3
= 17m/sec.
9. In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. Theproton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coulomb attraction. In the groundstate, the electron goes round the proton in a circle of radius 5.3 x 10 -11 m. Find the speed of the electron in the ground state.Mass of the electron = 9.1 x 10−3kgandchargeoftheelectron = 1.6x10− 19C.
Solutions :Electron revolves around the proton in a circle having proton at the centre.
Centripetal force is provided by coulomb attraction.r = 5.3→ t10−11m m = mass of electron = 9.1× 10−3kgcharge of electron = 1.6× 10−19cmv2
r= k q
2
r2⇒ v2 = kq2
rm= 9×109×1.6×1.6×10−38
5.3×10−11×9.1×10−31= 23.04
48.23× 1013
v2 = 0.477× 1013 = 4.7× 1012
⇒ v =√
4.7× 1012 = 2.2× 106m/sec .
10. A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the minimum speed the stonecan have at the highest point of the circle.
Solutions :At the highest point of a vertical circle
mv2
R= mg
⇒ v2 = Rg ⇒ v =√Rg
11. A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed.Consider a particle of mass 1 g sticking at the outer end of a blade. How much force does it experience when the fan runs at fullspeed ? Who exerts this force on the particle ? How much force does the particle exert on the blade along its surface ?
Solutions :Diameter of the ceiling fan= 120cm
∴ Radius = r = 60cm = 0/6mMass of particle on the outer end of the blade is 10.n = 1500rev/min = 25rev/secω = 2πn = 2π × 25 = 157.14Force of the particle on the blade = Mrω2 = (0.001)× 0.6× (157.14) = 14.8NThe fan runs at full speed in a circular path. This exerts the force on the particle (inertia).The particle also exerts a force of 14.8N on the blade along the surface.
12. A mosquito is sitting on an L.P. record disc rotating on a turn table at 33 13
per minute. The distance 3 of the mosquitofrom the centre of the turn table is 10 cm. Show that the friction coefficient between the record and the mosquito is greater thanIt 2/81. Take g =10m/s2.
Solutions :A pendulum is suspended from the ceiling of a car taking a turn
r = 10m, v = 36km/hr = 10m/sec g = 10m/sec2
From the figure Tsinθ = mv2r..(i)Tcosθ = mg ..(ii)
⇒ sinθcosθ
= mv2
mg⇒ tanθ = v2
rg⇒ θ = tan−1 v
2
rg
= tan−1 10010×10
= tan−1(1)⇒ θ = 45
13. A simple pendulum is suspended from the ceiling of a car taking a turn of radius 10 m at a speed of 36 km/h. Find theangle made by the string of the pendulum with the vertical if this angle does not change during the turn. Take g= 10m/sec2.
Solutions :A pendulum is suspended from the ceiling of a car taking a turn
r = 10m, v = 36km/hr = 10m/sec g = 10m/sec2
From the figure Tsinθ = mv2r..(i)Tcosθ = mg ..(ii)
⇒ sinθcosθ
= mv2
mg⇒ tanθ = v2
rg⇒ θ = tan−1 v
2
rg
= tan−1 10010×10
= tan−1(1)⇒ θ = 45
14. The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Findthe tension in the string at this instant.
Solutions :Bob has a velocity of 1.4m/sec,when the string makes an angle of 0.2 radian.
15. Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with thevertical. Find the tension at this instant. You can use cosθ = 1- θ2/2 and sinθ = 0 for small θ.
Solutions :
Page 43
24Tutors.com
Bob has a velocity of 1.4m/sec,when the string makes an angle of 0.2 radian.m=100g=0.1kg r=1m, v=1.4m/secFrom the diagram,
16. Suppose the amplitude of a simple pendulum having a bob of mass m is θ . Find the tension in the string when the bob isat its extreme position.
Solutions :At the extreme position, velocity of the pendulum is zero.
So there is no centrifugal force.So T=mgcosθ
17. A person stands on a spring balance at the equator. (a) By what fraction is the balance reading less than his true weight? (b) If the speed of earth’s rotation is increased by such an amount that the balance reading is half the true weight, what will bethe length of the day in this case ?
Solutions :a)Net force on the spring balance.
R=mg -mω2rSo the fraction less than the true weight (3mg) is
=mg−(mg−mω2r)
mg= ω2
g= [ 2π
24×3600
2]× 6400×103
10= 3.5× 10−3
b) When the balance reading is half the true weight,mg−(mg−mω2r)
mg= 1/2
ω2r = g/2⇒ ω =√
g2r
=√
102×6400×103 rad/sec
∴ Duration of thedayis
T = 2πω
= 2π ×√
2×6400×103
9.8sec = 2π
√64×106
49sec = 2π×8000
7×3600hr = 2hr
18. A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between theroad and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up ?
Solutions :Given v=36km/hr =10m/s r=20m, µ = 0.4
The road is banked with an angle ,
tan−1[ v2
rg] = tan−1[ 100
20×10] = tan−1 1
2ortanθ = 0.5
When the car travels at max speed such that it slips upwards, µR1acts downwards as shown in Fig.1
So R1−mgcosθ − mv12
rsinθ = 0..(i)
And µR1 +mgsinθ − mv12
rcosθ = 0..(ii)
Solving the equations we get,
V1=√rg tanθ−µ
1+µtanθ=√
20× 10× 0.11.2
= 4.082m/s = 14.7km/hr
So the possible speeds lie between 14.7 km/hr and 54km/hr
19. A motorcycle has to move with a constant speed on an overbridge which is in the form of a circular arc of radius R andhas a total length L. Suppose the motorcycle starts from the highest point. (a) What can its maximum velocity be for which thecontact with the road is not broken at the highest point ? (b) If the motorcycle goes at speed 1/42 times the maximum found inpart (a), where will it lose the contact with the road ? (c) What maximum uniform speed can it maintain on the bridge if it doesnot lose contact anywhere on the bridge ?
Solutions :So the possible speeds are between 15.7 km/hr and 54 km/hr
R= radius of the bridgeL=total length of the overbridgea) At the highest point
mg=mv2
R⇒ v2 = Rg ⇒ v =
√Rg
b) Given,v = 1√2
√Rg
suppose it loses contact at . So, at B, mgcosθ = mv2
R
⇒ v2 = Rg2cosθ
[√Rv2
]2 = Rgcosθ ⇒ Rg2
= Rgcosθ ⇒ cosθ = 12⇒ 60 = π
3
θ = lr→ l = rθ = πR
3
So it will lose contact at distance πR3
from the highest pointc) Let the uniform speed on the bridge be v.The chance of losing contact is maximum at the end of the bridge for which a= L
2R
So, mv2
R= mgcosα⇒ v =
√gRcos[ l
2R]
20. A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate dv/dt = a. The frictioncoefficient between the road and the tyre is µ. Find the speed at which the car will skid.
Solutions :
Page 44
24Tutors.com
Since the motion is non uniform, the acceleration has both radial and tangential components.
ar = V 2
r
at = dvdt
= a
Resultant magnitude =√
[ v2
r]2 + a2
Now µN =√
[ v2
r]2 + a2 ⇒ µmg = m
√[ v
2
r]2 + a2 ⇒ µ2g2 = V 4
r2+ a2
⇒ v4 = (µ2g2 − a2)r2 ⇒ v = [(µ2g2 − a2)r2]1/4
21. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is g. The ruler isfixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal planethrough the fixed end. (a) What can the maximum angular speed be for which the block does not slip ? (b) If the angular speedof the ruler is uniformly increased from zero at an angular acceleration a, at what angular speed will the block slip ?
21. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is g. The ruler isfixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal planethrough the fixed end. (a) What can the maximum angular speed be for which the block does not slip ? (b) If the angular speedof the ruler is uniformly increased from zero at an angular acceleration a, at what angular speed will the block slip ?
Solutions :a) When the ruler makes uniform circular motion in the horizontal plane
µ = mg = mω21L
ω1 =√µgL
b) When the ruler makes uniformly accelerated circular motion
µmg =√
(mω22L)2 + (mLα)2 ⇒ ω4
2 + α2 = µ2g2
L2 ⇒ ω2 = [(µgL
)2 − α2]1/4
22. A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure (7-E1).Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18km/h on the track. (a) Find the normal contact force by the road on the cycle when it is at B and at D. (b) Find the force offriction exerted by the track on the tyres when the cycle is at B, C and D. (c) Find the normal force between the road and the cyclejust before and just after the cycle crosses C. (d) What should be the minimum friction coefficient between the road and the tyre,which will ensure that the cyclist can move with constant speed ? Take g= 10 m
s2
Solutions :Radius of the curves = 100m
Weight= 100KgVelocity = 18km/hr=5m/sec
a) at B mgmv2
R= N ⇒ N = (100× 10)− 100×25
100= 1000− 25 = 975N
At d, N= mg+mv2
R= 1000 + 25 = 1025N
b)At B amp; D the cycle has no tendency to slide. So at B amp; D,frictional force is zero.At ’C’, mgsinθ = F ⇒ F = 1000× 1√
2= 707N
c) (i) Before ’C’ mgcosθ −N = mv2
R⇒ N = mgcosθ − mv2
R= 707− 25 = 683N
(ii) N-mgcosθ = mv2
R+mgcosθ = 25 + 707 = 732N
d) To find out the minimum desired coeff. of friction,we have to consider a point just before C.( Where N is minimum)Now, µN = mgsinθ ⇒ µ× 682 = 707So,µ = 1.037
23. In a children’s park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod alwaysremains horizontal. Two kids hold the rod near the ends and thus rotate with the rod (figure 7-E2). Let the mass of each kid be15 kg, the distance between the points of the rod where the two kids hold it be 3.0 m and suppose that the rod rotates at the rateof 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids.
Solutions :d=3m⇒ R = 1.5m
R=distance from the centre to one of the kidsN=20 rev per min=20/60=1/3 rev per secω = 2πr = 2π/3m = 15kg
∴ FrictionalforceF = mrω2 = 15× (1.5)× (2π)2
9= 5× (0.5)× 4π2 = 10π2
∴ Frictional force for one of the kids is 10π2
24. A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in thebowl at a position where the radius makes an angle 0 with the vertical. The block rotates with the bowl without any slipping. Thefriction coefficient between the block and the bowl surface is u. Find the range of the angular speed for which the block will notslip.
Solutions :If the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force acts downwards. Here, r= Rsinθ
1(Rsinθ)cosθ = 0 ..(ii)Substituting the value of R1 from Eq (i) in Eq(ii), it can be found out that
ω1 = [g(sinθ+µcosθ
Rsinθ(cosθ−µsinθ) ]1/2
Again, for minimum speed, the frictional forceµR2acts upward from FBD-2, it can be proved that
ω2 = [g(sinθ−µcosθ)
Rsinθ(cosθ+µsinθ)]1/2
∴ the range of speed is between ω1 andω2
Page 45
24Tutors.com
25. A particle is projected with a speed u at an angle 0 with the horizontal. Consider a small part of its path near the highestposition and take it approximately to be a circular arc. What is the radius of this circle ? This radius is called the radius ofcurvature of the curve at the point.
Solutions :Particle is projected with speed ’u’ at an angle θ. At the highest point, the vertical component of velocity is ’0’.
So, at the point, velocity =ucosθ
centripetal force = mu2cos2( θr
)At the highest pt.
mg=mv2
r⇒ r = u2cos2θ
g
26. What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where theparticle velocity makes an angle 0/2 with the horizontal ?
Solutions :Let ’u’ be the velocity at the pt where it makes an angle θ/2 with the horizontal. The horizontal component remains unchanged
So, v cosθ/2 = ωcosθ ⇒ v = ucosθ
cos θ2
...(ii)
From the figure
mgcos(θ/2) = mv2
r⇒ r = V 2
gcos(θ/2)
putting the value of ’v’ from equn(i)
r = u2cos2θgcos3(θ/2)
27. A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room onwhich the block moves is smooth but the friction coefficient between the wall and the block is g. The block is given an initial speedv0. As a function of the speed v write (a) the normal force by the wall on the block, (b) the frictional force by the wall and (c) thetangential acceleration of the block. (d) Integrate the tangential acceleration ( dv
dt= v dv
dx) to obtain the speed
of the block after one revolution.
Solutions :A block of mass ’m’ moves on a horizontall circle against the wall of a cylindrical room of radius ’R’.
Friction coefficient between wall amp; the block is µ
a)Normal reaction by the wall on the block is mv2
R
b)∴ Frictional force by the wall =µmv2
R
c) µmv2
R= ma⇒ a = −µv2
R⇒ ds = −R
µdvv
d)Now, dvdt
= v dvds
= −µv2R⇒ ds = −R
µdvv
⇒ s = −RµlnV + CAt s=0,v=v∴, c = R
µlnV
So ,s = −Rµln VV⇒ V
V= e−µs/R
For one rotation s = 2πR, so V=Ve−2piµ
28. A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity co in a circular pathof radius R (figure 7-E3). A smooth groove AB of length L( R) is made on the surface of the table. The groove makes an angle 0with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released tomove along AB. Find the time taken by the particle to reach the point B.
Solutions :The cabin rotates with angular velocity ω amp; radius R.
∴The particle experiences a force mRω2
. The component of mRω2 provides the required force to the particle to move along AB.∴ mRω2cosθ = ma⇒ a = Rω2cosθlength of groove = LL=ut+1/2at2 ⇒ L = 1/2Rω2cosθt2
⇒ t2 = 2LRω2cosθ
⇒ t = 1√
2LRω2cosθ
29. A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small wooden plate is kept onthe seat with its plane perpendicular to the radius of the circular road (figure 7-E4). A small block of mass 100 g is kept on the seatwhich rests against the plate. The friction coefficient between the block and the plate is 1.1 = 0.58. (a) Find the normal contactforce exerted by the plate on the block. (b) The plate is slowly turned so that the angle between the normal to the plate and theradius of the road slowly increases. Find the angle at which the block will just start sliding on the plate.
Solutions :v= Velocity of car= 36 km/hr =10m/s
r=Radius of circular path = 50m m= mass of small body = 100g =0.1kgµ= Friction coefficient between plates amp; body = 0.58a) The normal contact force exerted by the plate on the block
N=mv2
r= 0.1×100
50= 0.2N
b) The plate is turned so the angle between the normal to the plate amp; the radius of the road slowly increases
N=mv2
rcosθ ..(ii)
µN = mv2
rsinθ..(ii)
Putting the value of N from (i)
Page 46
24Tutors.com
µmv2
rcosθ = mv2
rsinθ ⇒ µ = tanθ ⇒ θ = tan−1µ = tan−1(0.58) = 30
30. A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R . A smooth pulleyof small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string going over thepulley. Initially the masses are held by a person with the strings along the outward radius and then the system is released from rest(with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension inthe string.
Solutions :Let the bigger mass accelerate towards the right with acceleration ’a’
T-ma=mω2R = 0 ..(i)
T+2ma-2mω2R = 0 ..(ii) ⇒ a = mω2R3
Substituting the value of a in Equation (i), we get T= 4/3mω2R.
The mass of cyclist together with the bike is 90 kg. Calculate the increase in kinetic energy if the speed increases from 6′0 km/hto 12 km/h.
Solutions :M = mc + mb = 90kg
u = 6km/h = 1.666m/secv = 12km/h = 3.333 m/secincrease in K.E. = 1
A block of mass 2.00 kg moving at a speed of 10.0 m/s accelerates at 3.00 m/s for 5.00 s. Compute its final kinetic energy.
Solutions :mb = 2kg
u = 10m/seca = 3m/aec2
t = 5secv = u + at = 10 + 3I5 = 25m/sec∴ F.K.E = 1
2mv2 = 1
2× 2 × 625 = 625J
A box is pushed through 4.0 m across a floor offering 100 N resistance. How much work is done by the resisting force ?
Solutions :F = 100N
S = 4m, θ = 0o
ω = F.S = 100 × 4 = 400 J
A block of mass 5.0 kg slides down an incline of inclination 30 and length 10 m. Find the work done by the force of gravity.
Solutions :m = 5kg
θ = 30o
S = 10 m F = mgSo work done by the force of gravityω = mgh = 5 × 9.8 × 5 = 245 J
A constant force of 2.50 N accelerates a stationary particle of mass 15 g through a displacement of 2.50 m. Find the work doneand the average power delivered.
Solutions :F = 2.50N , S = 2.5m, m =15g = 0.015kg
So, w= F × S ⇒ a = Fm
= 2.50.015
= 5003
m/s2
= F times S cos 0o ( acting alone the same line)2.5 × 2.5 = 6.25 JLet the velocity of the body at b = U. Applying work energy principle 1
2mv2 - 0 = 6.25
⇒ V =√
6.25×20.015
= 28.86 m/sec
So time taken to travel from A to B.⇒ t = v−u
a= 28.86×3
500
∴ Average Power = Wt
= 6.25×500(28.86)×3
= 36.1
A particle moves from a point ~r1 = (2 m)~i + (3 m)~j to another point ~r2 = (3 m)~i + (2 m)~j during which a certain force ~F =(5 N)~i + (5 N)~j acts on it. Find the work done by the force on the particle during the displacement.
A man moves on a straight horizontal road with a block of mass 2 kg in his hand. If he covers a distance of 40 in with anacceleration of 0.5 m/s2, find the work done by the man on the block during the motion.
Solutions :mb = 2kg, s = 40m, a = 0.5m/sec2
So, forceappliedbythemanontheboxF = mba = 2× (0.5) = 1Nω = FS = 1× 40 = 40J
A force F = a+ bx acts on a particle in the x-direction, where a and b are constants. Find the work done by this force duringa displacement from x = 0 to x = d.
Page 47
24Tutors.com
Solutions :Given That F = a+ bx
Where a and b are constants.So, work done by this force during this force during the displacement x = 0 and x = d is given by
W =∫ d0 F dx =
∫ d0 (a+ bx) dx = ax+ (bx2/2) = [a+ 1
2bd]d
A block of mass 250 g slides down an incline of inclination 37 with a uniform speed. Find the work done against the friction asthe block slides through 1.0 m.
Solutions :mb = 250g = .250kg
θ = 37o, S = 1m.Frictional force f = µ R
mg sin θ = µ R ...(1)mg cos θ ...(2)So, work done against µR = µRS cos 0o = mg sin θ S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
A block of mass m is kept over another block of mass M and the system rests on a horizontal surface (figure 8-E1). A constanthorizontal force F acting on the lower block produces an acceleration F
2(m+M)in the system, the two blocks always move together.
(a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface. (b) Find the frictional force actingon the smaller block. (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacementd of the system.
Solutions :a = F
2(m+M)(given)
a) from fig (1)ma = µk R1 and R1 = mg⇒ µ = ma
R1= F
2(m+M)g
b) Frictional force acting on the smaller block f = µ R = F2(m+M)g
×mg = m×F2(M+m)
c) work done w = fs s = dw = mF
2(M+m)× d = mFd
2(M+m)
A box weighing 2000 N is to be slowly slid through 20 m on a straight track having friction coefficient 0.2 with the box. (a) Findthe work done by the person pulling the box with a chain at an angle θ with the horizontal. (b) Find the work when the personhas chosen a value of θ which ensures him the minimum magnitUde of the force.
Solutions :Weight = 2000 N, S = 20m, µ = 0.2
a) R + P sin θ - 2000 = 0 ..(1)P cos θ - 0.2 R = 0 ...(2)From (1) and (2) P cos θ - 0.2 (2000 - P sin θ) = 0P = 400
cosθ+0.2sinθ...(3)
So, work done by the person , W = PS cos θ = 8000cosθcosθ+0.2sinθ
= 80001+0.2sinθ
= 400005+tanθ
b) for minimum magnitude of force from equn (1)ddθ
(cos θ + 0.2 sin θ) = 0 ⇒ tan θ = 0.2putting the value in equn (3)W = 40000
5+tanθ= 40000
(5.2)= 7690J
A block of weight 100 N is slowly slid up on a smooth incline of inclination 37 by a person. Calculate the work done by theperson in moving the block through a distance of 2.0 m, if the driving force is (a) parallel to the incline and (b) in the horizontaldirection.
Solutions :w = 100 N, θ = 37o, s = 2m
Force F = mg sin 37o = 100 × 0.60 = 60 NSo, work done, when the force is parallel to incline.w = Fs cos θ = 60 × 2 × cos θ = 120 JIn ∆ABC AB = 2mCB = 37o
So, h = c = 1m∴ work done when the force in horizontal directionW = mgh = 100 × 1.2 = 120 J
Find the average frictional force needed to stop a car weighing 500 kg in a distance of 25 m if the initial speed is 72 km/h.
Solutions :m = 500 kg , s = 25m, u = 72 km/h = 20 m/s,
(-a) = v2−u2
2S⇒ a = 400
50= 8m/sec2
Frictional force f = ma = 500 × 8 = 4000 N
Find the average force needed to accelerate a car weighing 500 kg from rest to 72 km/h in a distance of 25 m.
Solutions :m = 500 kg, u = 0, v = 72 km/h = 20 m/s
a = v2−u2
2s= 400
50= 8m/sec2
force needed to accelerate the car F = ma = 500 × 8 = 4000 N
A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equationv = a
√x, where a is a constant. Find the total work done by all the forces during a displacement from x = 0 to x = d
Solutions :Given v = a
√x, (uniformly accelerated motion)
displacement s = d - 0 = dputting x = 0, v1 = 0
Page 48
24Tutors.com
putting x = d, v2 = a√d
a =v22−u
22
2s= a2d
2d= a2
2
force f = ma = ma2
2
work done w = FS cos θ = ma2
2× d = ma2d
2
A block of mass 2.0 kg kept at rest on an inclined plane of inclination 37 is pulled up the plane by applying a constant force of20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J.(b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kineticenergy of the block at the instant the force ceases to act. Take g = 10m/s2.
Solutions :m = 2 kg, θ = 37o, F = 20 N
From the free body diagramF = (2g sin θ) + ma ⇒ a = (20 - 20 sin θ )/s = 4 m/sec2
S = ut + 12at2 (u=0, t=1s, a = 1.66) = 2m
So, work done w = Fs = 20 × 2 = 40 Jb) If W = 40 JS = W
F= 40
20h = 2 sin 37o = 1.2 mSo, work done W = -mgh = -20 × 1.2 = -24 Jc) v = u + at = 4 × 10 = 40 m/secSo, K.E. = 1
2mv2 = 1
2× 2× 16 = 16J
A block of mass 2.0 kg is pushed down an inclined plane of inclination 37 with a force of 20 N acting parallel to the incline. Itis found that the block moves on the incline with an acceleration of 10 m/s2. If the block started from rest, find the work done (a)by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting onthe block in the first second. Take g = 10 m/s2.
Solutions :m = 2 kg, θ = 37o, F = 20 N, a = 10 m/sec2
a) t = 1secSo, s = ut + 1
2at2 = 5m
work done by the applied force w = FS cos 0o = 20 × 5 = 100 Jb) BC (h) 5 sin 37o = 3mSo, work done by the weight W = mgh = 2 × 10 × 3 = 60 Jc) So, frictional force f = mg sin θwork done by the frictional forces w = fs cos 0o = (mg sin θ) s = 20 ×0.60 × 5 = 60 J
A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if itis initially moving at a speed of 40 cm/s. If the friction coefficient between the table and the block is 0.1, how far does the blockmove before coming to rest ?
Solutions :Given m = 250 g = 0.250 kg,
u = 40 cm/sec = 0.4 m/secµ = 0.1 , v = 0Here, µ R = ma where a = decelerationa = µR
m= µmg
m= µg = 0.1× 9.8 = 0.98m/sec2
S = v2−u2
2a= 0.082m = 8.2cm
Again, work done against friction is given by- w = µ RS cos θ= 0.1 × 2.5 × 0.082 × 1(θ = 0o ) = 0.02 J⇒ W = -0.02 J
Water falling from a 50 m high fall is to be used for generating electric energy. If 1.8× 105 kg of water falls per hour and halfthe gravitational potential energy can be converted into electric energy, how many 100 W lamps can be lit ?
Solutions :h = 50 m, m = 1.8× 105 kg/hr, P = 100 watt,
P.E. = mgh = 1.8× 105 × 9.8× 50 = 882× 105J/hrBecause, half the potential energy is converted into electricity,Electrical energy 1
2P.E. = 441 ×105J/hr
So, power in watt (J/sec) is given by = 441×105
3600∴ number of 100 W lamps, that can be lit 441×105
3600×100= 122.5 ≈ 122
A person is painting his house walls. He stands on a ladder with a bucket containing paint in one hand and a brush in other.Suddenly the bucket slips from his hand and falls down on the floor. If the bucket with the paint had a mass of 6.0 kg and was ata height of 2.0 m at the time it slipped, how much gravitational potential energy is lost together with the paint ?
Solutions :m = 6kg, h = 2m
P.E. at a height ’2m’ = mgh 6 × 9.8 × 2 = 117.6 JP.E. at floor = 0Loss in P.E. = 117.6 - 0 = 117.6 J ≈ 118 J
A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed whenit hits the ground.
Solutions :h = 40m, u = 50 m/sec
Let the speed be ’v’ when it strikes the ground.Applying law of conservation of energymgh + 1
The 200 m free style women’s swimming gold medal at Seol Olympic 1988 went to Heike Friendrich of East Germany when sheset a new Olympic record of 1 minute and 57.56 seconds. Assume that she covered most of the distance with a uniform speed andhad to exert 460 W to maintain her speed. Calculate the average force of resistance offered by the water during the swim.
Solutions :t = 1 min 57.56 sec = 11.56 sec, p = 400 w, s = 200m
p = wt
, work w = pt = 460× 117.56J
Again W = FS = 460×117.56200
= 270.3 ≈ 270N
The US athlete Florence Griffith-Joyner won the 100 m sprint gold medal at Seol Olympic 1988 setting a new Olympic record of10.54 s. Assume that she achieved her maximum speed in a very short-time and then ran the race with that speed till she crossedthe line. Take her mass to be 50 kg. (a) Calculate the kinetic energy of Griffith-Joyner at her full speed. (b) Assuming that thetrack, the wind etc. offered an average resistance of one tenth of her weight, calculate the work done by the resistance during therun. (c) What power Griffith- Joyner had to exert to maintain uniform speed ?
Solutions :S = 100 m, t = 10.54 sec, m = 50 kg
The motion can be assumed to be uniform because the time take for acceleration is minimum.a) Speed v = S
t= 9.487 e/s
So, K.E. = 12mv2 = 2250J
b) Weight = mg = 490 Jgiven R = mg /10 = 49 JSo, work done against resistance Wf = −RS = −49× 100 = −4900Jc) To maintain her uniform speed , she has to expert 4900 j of energy to overcome frictionP = W
t= 4900/10.54 = 465W
A water pump lifts water from a level 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligiblevelocity. Calculate the minimum horsepower the engine should have to do this.
Solutions :h = 10 m
flow rate = (mt
) = 30kg/min = 0.5kg/sec
power P = mght
= (0.5)× 9.8× 10 = 49W
So, horse power (h.p.) P/746 = 49/746 = 6.6× 10−2hp
An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection,the stone is 150 cm above the ground and has a speed of 3.00 m/s. Calculate the work done by the demonstrator during the process.If it takes one second for the demonstrator to lift the stone and throw, what horsepower does he use ?
Solutions :m = 200 g = 0.2kg , h = 150 cm = 1.5 m, v = 3 m/sec, t = 1 sec
In a factory it is desired to lift 2000 kg of metal through a distance of 12 m in 1 minute. Find the minimum horsepower of theengine to be used.
Solutions :m = 200 kg, s = 12 m, t = 1 min = 60 sec
So, work W = F cos θ = mgs cos 0o [θ = 0o for minimum work]= 2000× 10× 12 = 240000JSo, power p = W
t= 240000
60= 4000watt
h.p. = 4000746
= 5.3hp.
A scooter company gives the following specifications about its product.Weight of the scooter - 95 kgMaximum speed - 60 km/hMaximum engine power - 3.5 hpPick up time to get the maximum speed - 5 sCheck the validity of these specifications
Solutions :The specification given by the company are
U = 0, m = 95 kg, pm = 3.5 hpVm = 60km/h = 50/3 m/sec tm = 5secSo, the maximum acceleration that can be produced is given by,
a =(50/3)−0
5= 10
3So, the driving force is given byF = ma = 95× 10
3= 950
3N
So, the velocity that can be attained by maximum h.p. while supplying 9503
will be
v = pF⇒ v = 3.5×746×5
950= 8.2m/sec.
Because the scooter can reach a maximum of 8.2 m/sec while producing a force of 950/3 N, the specification given are some what over claimed.
A block of mass 30.0 kg is being brought down by a chain. If the block acquires a speed of 40.0 cm/s in dropping down 2.00 m,find the work done by the chain during the process.
Solutions :Given m = 30 kg, v = 40 cm/sec = 0.4 m/sec, s = 2m
From the free body diagram, the force given by the chain is,F = (ma−mg) = m(a− g) [where a = acceleration of the block]
The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is 16.0 N when thesystem is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is releasedfrom rest
Solutions :Given T = 19 N
From the free body diagrams,T − 2mg + 2ma = 0 ...(1)T −mg −ma = 0 ...(2)From equation (1) & (2) T = 4ma ⇒ a = T
4m⇒ A = 16
4m= 4
mm/s2
Now, S = ut+ 12at2
⇒ S = 12× 4m× 1⇒ S = 2
mm [because u= 0]
Net mass = 2m−m = mDecrease in P.E. = mgh ⇒ P.E. = m× g × 2
m⇒ P.E. = 9.8× 2⇒ P.E. = 19.6J
The two blocks in an Atwood machine have masses 2.0 kg and 3.0 kg. Find the work done by gravity during the fourth secondafter the system is released from rest.
Solutions :Given m1 = 3kg,m2 = 2kg t = during 4th second
From the free body diagramT − 3g + 3a = 0 ...(1)T − 2g − 2a = 0 ...(2)Equation (1) & (2) we get 3g − 3a = 2g + 2a⇒ a = g
5m/sec2
Distance traveled in 4th sec is given by
S4th = a2
(2n− 1) =g5s
(2× 4− 1) = 7g10
= 7×9.810
mNet mass ’m’ = m1 −m2 = 3− 2 = 1kgSo, decrease in P.E.= mgh = 1× 9.8× 7
10× 9.8 = 67.2 = 67J
Consider the situation shown in figure (8-E2). The system is released from rest and the block of mass 1.0 kg is found to have aspeed 0.3 m/s after it has descended through a distance of 1 m. Find the coefficient of kinetic friction between the block and thetable.
(v1 = 2x2 m this system)h = 1m = height decent by 1 kg blocks = 2× 1 = 2m distance traveled by 4 kg blocku = 0Applying change in K.E. = work done (for the system)[(1/2)m1v2
1 + (1/2)m2v2m]− 0 = (−µR)S +m2g
[R = 4g = 40 N]⇒ 1
2× 4× (0.36)× 1
2× 1× (0.09) = −µ× 40× 2 + 1× 40× 1
⇒ 0.72 + 0.045 = −80µ+ 10⇒ µ 9.235
80= 0.12
A block of mass 100 g is moved with a speed of 5.0 m/s at the highest point in a closed circular tube of radius 10 cm kept ina vertical plane. The cross-section of the tube is such that the block just fits in it. The block makes several oscillations inside thetube and finally stops at the lowest point. Find the work done by the tube on the block during the process.
Solutions :Given, m = 100g = 0.1 kg, v = 5m/sec, r = 10 cm.
work done by the block = total energy at A - total energy at B(1/2mv2 +mgh)− 0⇒W = 1
2mv2 +mgh− 0 = 1
2× (0.1)× 25 + (0.1)× 10× (0.2) [h = 2r = 0.2 m]
⇒W = 1.25− 0.2⇒W = 1.45JSo, the work done by the tube on the body isWt = −1.45J
A car weighing 1400 kg is moving at a speed of 54 km/h up a hill when the motor stops. If it is just able to reach the destinationwhich is at a height of 10 m above the point, calculate the work done against friction (negative of the work done by the friction).
Solutions :m = 1400 kg , v = 54 km/h = 15 m/sec, h = 10 m
Work done = (total K.E.) - total P.E.= 0 + 1
2mv2 −mgh = 1
2× 1400× (15)2 − 1400× 9.8× 10
= 157500− 137200 = 20300So, work done against friction, Wt = 20300J
A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3.2 m high. How much work wasrequired (a) to lift the block from the ground and put it at the top, (b) to slide the block up the incline ? What will be the speed ofthe block when it reaches the ground, if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline ?Take g = 10 m/s2
Solutions :m = 200 g = 0.2 kg , s = 10 m, h = 3.2 m, g = 10 m/sec2
a) work done W = mgh = 0.2× 10× 3.2 = 6.4Jb) work done to slide the block up the inclinew = (mg sinθ) = (0.2)× 10× 3.2
10× 10 = 6.4J
c) Let, the velocity be v when falls on the ground vertically,12mv2 − 0 = 6.4J ⇒ v = 8m/s
Page 51
24Tutors.com
d) Let V be the velocity when reaches the ground by liding12mV 2 − 0 = 6.4J ⇒ v = 8m/sec
In a children’s park, there is a slide which has a total length of 10 m and a height of 8.0 m (figure 8-E3). Vertical ladder areprovided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. Theaverage friction offered by the slide is three tenth of his weight. Find (a) the work done by the ladder on the boy as he goes up, (b)the work done by the slide on the boy as he comes down. Neglect any work done by forces inside the body of the boy.
Solutions :l = 10m, h = 8m, mg = 200 N
f = 200× 310
= 60Na) work done by the ladder on the boy is zero when the boy is going up because the work is done by the boy himself.b) work done against frictional force, W = µRS = fl = (−60)× 10 = −600Jc) work done by the force inside the boy isWb = (mg sinθ)× 10 = 200× 8
10× 10 = 1600J
36. Figure (8-E4) shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particlestarts slipping from the point A, how far away from the track will the particle hit the ground ?
Solutions :H = 1m, h = 0.5m
Applying law of conservation of Energy for point A & BmgH = 1
2mv2 +mgh⇒ g = (1/2)v2 + 0.5g
⇒ v22(g − 0.59) = g ⇒ v =√g = 3.1m/s
After point B the body exhibits projectile motion for whichθ = 0o, v = −0.5So, −0.5 = (u sinθ)t− (1/2)gt2 ⇒ 0.5 = 4.9t2 ⇒ t = 0.31secSo, x = (4 cosθ)t = 3.1× 3.1 = 1m.So, the particle will hit the ground at a horizontal distance in from B.
A block weighing 10 N travels down a smooth curved track AB joined to a rough horizontal surface (figure 8-E5). The roughsurface has a friction coefficient of 0.20 with the block. If the block starts slipping on the track from a point 1.0 m above thehorizontal surface, how far will it move on the rough surface ?
Solutions :mg = 10N, µ = 0.2, H = 1m, u=v=0
change in P.E. = work done.Increase in K.E.⇒ w = mgh = 10× 1 = 10JAgain on the horizontal surface the frictional forceF = µR = µmg = 0.2× 10 = 2NSo, the K.E. is used to overcome friction⇒ S = W
F= 10J
2N= 5m
A uniform chain of mass m in and length l overhangs a table with its two third part on the table. Find the work to be done bya person to put the hanging part back on the table.
Solutions :Let ’dx’ be the length of an element at a distance x from the table
mass of ’dx’ length = (m/l)dxWork done to put dx part back on the tableW = (m/l)dx g(x)So, total work done to put l/3 part back on the table
W =∫ 1
30 (m/l)gx dx⇒ w = (m/l)g
[x2
2
] l3
0
= mgl2
18l= mgl
18
A uniform chain of length L and mass M overhangs a horizontal table with its two third part on the table. The friction coefficientbetween the table and the chain is 11. Find the work done by the friction during the period the chain slips off the table.
Solutions :Let, x length of chain is on the table at a particular instant .
So, work done by frictional force on a small element ’dx’
dWf = µRx = µ
(MLdx
)gx [where dx = M
Ldx]
Total work done by friction,Wf = ∫ 0
2L/3µM
Lgx dx
∴Wf = µMLg
[x2
2
]0
2L/3
= µML
[4L2
18
]= 2µMg L/9
A block of mass 1 kg is placed at the point A of a rough track shown in figure (8-E6). If slightly pushed towards right, it stopsat the point B of the track. Calculate the work done by the frictional force on the block during its transit from A to B.
Solutions :m = 5kg, x = 10cm = 0.1m, v = 2m/sec,
h = ? G = 10m/sec2
So, k = mgx
= 500.1
= 500N/m
Total energy just after the blow E = 12mv2 + 1
2kx2 ...(1)
Total energy a a height h = 12k(h− x)2 +mgh ...(2)
12mv2 + 1
2kx2 = 1
2k(h− x)2 +mgh
Page 52
24Tutors.com
on solving we can get ,H = 0.2 m = 20 cm
A block of mass 5.0 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block.The block is given a sharp impulse from below so that it acquires an upward speed of 2.0 m/s. How high will it rise ? Take g = 10m/s2
Solutions :m = 5kg, x = 10cm = 0.1m, v = 2m/sec,
h = ? G = 10m/sec2
So, k = mgx
= 500.1
= 500N/m
Total energy just after the blow E = 12mv2 + 1
2kx2 ...(1)
Total energy a a height h = 12k(h− x)2 +mgh ...(2)
12mv2 + 1
2kx2 = 1
2k(h− x)2 +mgh
on solving we can get ,H = 0.2 m = 20 cm
A block of mass 250 g is kept on a vertical spring of spring constant 100 N/m fixed from below. The spring is now compressedto have a length 10 cm shorter than its natural length and the system is released from this position. How high does the block rise? Take g =10 m/s2.
Solutions :m = 250 g = 0.250 kg,
k = 100 N/m, m = 10 cm = 0.1 mg = 10m/sec2
Applying law of conservation of energy
12kx2 = mgh⇒ h = 1
2
(kx2
mg
)=
100×(0.1)2
2×0.25×10= 0.2m = 20cm
Figure (8-E7) shows a spring fixed at the bottom end of an incline of inclination 37. A small block of mass 2 kg starts slippingdown the incline from a point 4.8 m away from the spring. The block compresses the spring by 20 cm, stops momentarily and thenrebounds through a distance of 1 m up the incline. Find (a) the friction coefficient between the plane and the block and (b) thespring constant of the spring. Take g = 10 m/s2.
Solutions :m = 2kg, s1 = 4.8m, x = 20cm = 0.2 m, s2 = 1m,
sin 37 = 0.60 = 3/5, θ = 37, cos 37 = .79 = 0.8 = 4/5g = 10m/sec2
Applying work-Energy principle for downward motion of the body0− 0 = mg sin 37× 5− µR× 5− 1
⇒ −12− 16µ+ 0.02K = 0 ...(2)Adding equation (1) & equation (2), We get 96µ = 48⇒ µ = 0.5Now putting the value of µ in equation (1) K = 1000 N/m
A block of mass m moving at a speed v compresses a spring through a distance x before its speed is halved. Find the springconstant of the spring.
Solutions :Let the velocity of the body at A be v
So,the velocity of the body at B is v/2Energy at point A = Energy at point BSo, 1
2mv2
A = 12mv2
B + 12kx2+
⇒ 12kx2 = 1
2mv2
A −12mv2
B ⇒ kx2 = m(V 2+−A V 2
B)
⇒ kx2 = m
(v2 − v2
4
)⇒ k = 3mv2
3x2
Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assumingno friction in the pulley, find the maximum elongation of the spring.
Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assumingno friction in the pulley, find the maximum elongation of the spring.
Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assumingno friction in the pulley, find the maximum elongation of the spring.
Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assumingno friction in the pulley, find the maximum elongation of the spring.
Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assumingno friction in the pulley, find the maximum elongation of the spring.
Solutions :Mass of the body = m
Let the elongation be xSo, 1
2kx2 = mgx
⇒ x = 2mg/k
A block of mass m. is attached to two unstretched springs of spring constants k1 and k2 as shown in figure (8-E9). The blockis displaced towards right through a distance x and is released. Find the speed of the block as it passes through the mean position
Page 53
24Tutors.com
shown.
Solutions :The body is displaced x towards right
Let the velocity of the body be v at its mean positionApplying law of conservation of energy12mv2 = 1
2k1x2 ⇒ mv2 = x2(k1 + k2)⇒ v2 =
x2(k1+k2)m
⇒ v = x√k1+k2m
A block of mass m, sliding on a smooth horizontal surface with a velocity ~v meets a long horizontal spring fixed at one end andhaving spring constant k as shown in figure (8-E10). Find the maximum compression of tin spring. Will the velocity of the blockbe the same as ~v when it comes back to the original position shown ?
Solutions :Let the compression be x
According to the law of conservation of the energy12mv2 = 1
2kx2 ⇒ x2 = mv2/k ⇒ x = v
√(m/k)
b) No. it will be in the opposite direction and magnitude will be less due to loss in spring.
A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5.0 cm (figure8-E11). The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit theground 2 m below the spring ?
Solutions :m = 100g = 0.1 kg, x = 5cm = 0.05m, k = 100 N/m
when the body leaves the spring, let the velocity be v12mv2 = 1
2kx2 ⇒ v = x
√(k/m) = 0.05×
√1000.1
= 1.58m/sec
For the projectile motion, θ = 0, Y = -2Now, y = (u sin θ)t - 1
2gt2
⇒ −2 = (−1/2)× 9.8× t2 ⇒ t = 0.63sec.So, x = (u cos θ)t ⇒ 1.58× 0.63 = 1m
A small heavy block is attached to the lower end of a light rod of length l which can be rotated about its clamped upper end.What minimum horizontal velocity should the block be given so that it moves in a complete vertical circle ?
Solutions :Let the velocity of the body at A is ’V’ for minimum velocity given at A velocity of the body at point B is zero.
Applying law of conservation of energy at A & B12mv2 = mg (2l)⇒ v =
√(4gl) = 2
√gl
Figure (8-E12) shows two blocks A and B, each having a mass of 320 g connected by a light string passing over a smooth lightpulley. The horizontal surface on which the block A can slide is smooth. The block A is attached to a spring of spring constant 40N/m whose other end is fixed to a support 40 cm above the horizontal surface. Initially, the spring is vertical and unstretched whenthe system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take g = 10 m/s2
Solutions :m = 320g = 0.32kg
k = 40N/mh = 40cm = 0.4mg = 10 m/s2
From the free body diagram,kx cosθ = mg(when the block breaks off R = 0)
⇒ cosθ = mg/kxSo, 0.4
0.4+x= 3.2
40×x ⇒ 16x = 3.2x+ 1.28⇒ x = 0.1m
So, s = AB =√
(h+ x)2 − h2 =√
(0.5)2 − (0.4)2 = 0.3mLet the velocity of the body at B be vCharge in K.E. = work done (for the system)12mv2 + 1
2mv2 = − 1
2kx2 +mgs
⇒ (0.32)× v2 = − 12× 40× (0.1)2 + 0.32× 10× (0.3)
⇒ v = 1.5m/s.
One end of a spring of natural length h and spring constant k is fixed at the ground and the other is fitted with a smooth ringof mass m which is allowed to slide on a horizontal rod fixed at a height h (figure 8-E13). Initially, the spring makes an angle of 37with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical.
Solutions :θ = 37 ; l = h = natural length
Let the velocity when the spring is vertical be ’v’.Cos 37 = BC/AC = 0.8 = 4/5AC = (h + x) = 5h/4 (because BC = h)So, x = (5h/4) - h = h/4Applying work energy principle 1
2kx2 = 1
2mv2
⇒ v = x√
(k/m) = h4
√km
Figure (8-E14) shows a light rod of length l rigidly attached to a small heavy block at one end and a hook at the other end.The system is released from rest with the rod in a horizontal position. There is a fixed smooth ring at a depth h below the initialposition of the hook and the hook gets into the ring as it reaches there. What should be the minimum value of h so that the blockmoves in a complete circle about the ring ?
Page 54
24Tutors.com
Solutions :The minimum velocity required to cross the height point c =
√2gl
Let the rod release from a height h.Total energy at A = total energy at Bmgh = 1
2mv2;mgh = 1
2m (2gl)
[ Because v = required velocity at B such that the block makes a complete circle.]So, h = l.
The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity.√
10 gl, where 1 is the length of the pendulum.Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angleof 60 with the upward vertical.
Solutions :a) Let the velocity at B be v2
12mv2
1 = 12mv2
2 +mgl
⇒ 1/2 m (10 gl) = 12mv2
2 +mgl
v22 = 8 gl
So, the tension in the string at horizontal position
T = mv2
R= m8gl
l= 8 mg
b) Let the velocity at C be V312mv2
1 = 12mv2
3 +mg (2l)
⇒ 12m (logl) = 1
2mv2
3 + 2mgl
⇒ v23 = 6 mgl
So, the tension in the string is given by
Tc = mv2
l−mg = 6glm
lmg = 5 mg
c) Let the velocity at point D be v4
Again, 12mv2
1 = 12mv2
4 +mgh12×m× (10/gl) = 1.2 mv2
4 +mgl(1 + cos60)
⇒ v24 = 7 gl
So, the tension in the string isTD = (mv2/l)−mg cos60
A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makesan angle of 37 with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.
Solutions :From the figure, cosθ = AC/AB
⇒ AC = AB cosθ ⇒ (0.5)× (0.8) = 0.4.So, CD = (0.5) - (0.4) = (0.1) mEnergy at D = energy at B12mv2 = mg(CD)
v2 = 2× 10× (0.1) = 2So, the tension is given by,
T = mv2
r+mg = (0.1)
(2
0.5+ 10
)= 1.4 N.
Figure (8-E15) shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring ofspring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses thetrack with a force mg when it reaches the point P, where the radius of the track is horizontal.
Solutions :Given, N = mg
As shown in the figure, mv2/R = mg⇒ v2 = gR ...(1)Total energy at point A = energy at P12kx2 = mgR+2mgR
2[because v2 = gR]
⇒ x2 = 3mgR/k ⇒ x =√
(3mgR)/k
The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of√
3 gl . Find the angle rotated by thestring before it becomes slack.
Solutions :V =
√3gl
12mv2 − 1
2mu2 = −mgh
v2 = u2 − 2gl(l + lcosθ)⇒ v2 = 3gl − 2gl(1 + cosθ) ...(1)Again,mv2/l = mg cosθv2 = lg cosθFrom equation (1) and (2) we get3gl − 2gl − 2gl/cosθ = gl cosθ3 cosθ = 1⇒ cosθ = 1/3θ = cos−1(1/3)So, angle rotated before the string becomes slack= 180 − cos−1(1/3) = cos−1(−1/3)
A heavy particle is suspended by a 1.5 m long string. It is given a horizontal velocity of√
57 m/s. (a) Find the angle made bythe string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximumheight reached by the particle over the point of suspension. Take g =10 m/s2
Solutions :
Page 55
24Tutors.com
l = 1.5 m; u =√
57m/sec.a) mg cosθ = mv2/l v2 = lg cosθ ...(1)change in K. E. = work done12mv2 − 1
2mu2 = mgh
⇒ v2 − 57 = −2× 1.5g(1 + cosθ) ...(2)⇒ v2 = 57− 3g(1 + cosθ)Putting the value of v from equation (1)15 cosθ = 57− 3g(1 + cosθ)⇒ 15 cosθ = 57− 30− 30 cosθ⇒ 45 cosθ = 27⇒ cosθ = 3/5.⇒ θ = cos−1(3/5) = 53
b) v =√
57− 3g(1 + cosθ) from equation (2)
=√
9 = 3 m/sec.c)As the string becomes slack at point B, the particle will start making projectile motion.
H = OE +DC = 1.5 cosθ + u2sin2θ2g
= (1.5× (3/5) +9×(0.8)2
2×10= 1.2 m.
A simple pendulum of length L having a bob of mass m is deflected from its rest position by an angle θ and released (figure8-E16). The string hits a peg which is fixed at a distance x below the point of suspension and the bob starts going in a circle centredat the peg. (a) Assuming that initially the bob has a height less than the peg, show that the maximum height reached by the bobequals its initial height. (b) If the pendulum is released with θ = 90 and x = L/2 find the maximum height reached by the bobabove its lowest position before the string becomes slack. (c) Find the minimum value of x/L for which the bob goes in a completecircle about the peg when the pendulum is released from θ = 90.
Solutions :a) When the bob has an initial height less that the peg and then released from rest (figure 1),
let body travels from A to B.Since, Total energy at A = Total energy at B∴ (K.E.)A = (PE)A = (KE)B + (PE)B(PE)A = (PE)B [because, (KE)A = (KE)B = 0]So, the maximum height reached by the bob is equal to initial height.
b) When the pendulum is released with θ = 90 and x = L2
(figure 2) the path of the particleis shown in the figure 2.At point C, the string will become slack and so the particle will start making projectile motion.12mv2
c − 0 = mg (L/2) (1− cos α)because, distance between A and C in the verticle direction is L/2(1− cos α)⇒ v2
c = gL(1− cos θ) ...(1)Again, from the free body diagrammv2cL/2
= mg cos αbecauseTc = 0
So, V 2
c = gL2cos α ...(2)
From Eqn. (1) and equn (2),
gL(1− cos α) = gL2cos α
⇒ 1− cos α = 12cos α
⇒ 32cos α = 1⇒ cos α = 2/3 ...(3)
To find highest position C, before the string becomes slack
BF = L2
+ L2cos θ = L
2+ L
2× 2
3= L
(12
+ 13
)So, BF = (5L/6)c) If the particle has to comlpete a vertical circle, at the point C.mv2c
(L−x)= mg
⇒ V 2c = g (L− x) ...(1)
Again applying energy principle between A and C,12mv2
c − 0 = mg (OC)
⇒ 12v2c = mg[L− 2(L− x)] = mg (2x− L)
⇒ v2c = 2g(2x− L) ...(2)
From equn (1) and equn (2)g(L− x) = 2g (2x− L)⇒ L− x = 4x− 2L⇒ 5x = 3L∴ xL
= 35
= 0.6So, the rates (x/L) should be 0.6
A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radiusthrough the particle, when it leaves contact with the sphere.
Solutions :Let the velocity be v when the body leaves the surface.
From the free body diagram
mv2
R= mg cos θ [Because normal reaction]
v2 = Rg cos θ ...(1)Again, from work-energy principle,Change in K.E. = work done⇒ 1
2mv2 − 0 = mg(R−R cos θ)
v2 = 2gR (1− cos θ) ...(2)From (1) and (2)
Page 56
24Tutors.com
Rg cos θ = 2gR(1− cos θ)3gR cos θ = 2 gRcos θ = 2/3θ = cos−1(2/3)
A particle of mass m is kept on a fixed, smooth sphere of radius R at a position, where the radius through the particle makes anangle of 30 with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particlejust after the release ? (b) Find the distance travelled by the particle before it leaves contact with the sphere.
Solutions :a) When the particle is released from rest, the centrifugal force is zero.
N force is zero = mg cos θ
= mg cos 30 =√
3mg2
b) When the prticle is leaves contact with the surface , N = 0.
So, mv2
Rmg cos θ
⇒ v2 = Rg cos θ ...(1)Again, 1
2mv2 = mgR(cos 30 − cos θ)
⇒ v2 = 2Rg
(√
32− cos θ
)...(2)
From equn (1) and equn (2)Rgcos θ =
√3 Rg − 2Rg cos θ
⇒ 3 cos θ =√
3
⇒ cos θ = 1√3⇒ θ = cos−1
(1√3
)So, the distance travelled by the particle before leaving contact,l = R(θ − π/6) [Because 30 = π/6]putting the value of θ we get l = 0.43R
A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontalspeed v. (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimumvalue of v for which the particle does not slip on the sphere ? (c) Assuming the velocity v to be half the minimum calculated inpart, (d) find the angle made by the radius through the particle with the vertical when it leaves the sphere.
Solutions :a) Radius = R
horizontal speed = vfrom the free body diagram,
N = Normal force = mg − mv2
Rb) when the particle is given maximum velocity so that the centrifugal force balance the weight, the particle does not slip on the sphere.mv2
R= mg ⇒ v =
√gR
c) If the body is given velocity v1
v1 =√gR/2
v21 − gR/4
Let the velocity be v2 when it leaves contact with the surface,
So, mv2
R= mg cos θ
⇒ v22 = Rg cos θ ...(1)
Again, 12mv2
2 −12mv2
1 = mgR (1− cos θ)⇒ v2
2 = v21 + 2gR (1− cos θ) ...(2)
From equn (1) and equn (2)Rg cos θ = (Rg/4) + 2gR (1− cos θ)⇒ cos θ = (1/4) + 2− 2 cos θ⇒ 3 cos θ = 9/4⇒ cos θ = 3/4⇒ θ = cos−1(3/4)
Figure (8-E17) shows a smooth track which consists of a straight inclined part of length 1 joining smoothly with the circularpart. A particle of mass m is projected up the incline from its bottom. (a) Find the minimum projection-speed vo for which theparticle reaches the top of the track. (b) Assuming that the projection-speed is 2vo and that the block does not lose contact withthe track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection-speed is onlyslightly greater than vo, where will the block lose contact with the track ?
Solutions :a) Net force on the particle between A and B, F = mg sin θ
work done to reach B, W = FS = mg sin θ lAgain work done to reach B to C = mgh = mg R (1− cos θ)So, total work done = mg[lsin θ +R(1− cos θ)]Now, change in K.E. = work done⇒ 1
2mv2
o = mg [l sin θ +R (1− cosθ)]⇒ vo =
√2g(R(1− cos θ) + l sin θ)
b) When the block is projected at a speed 2vo.Let the velocity at C will be Vc.Applying energy principle,12mv2
c − 12m (2vo)2 = −mg [l sin θ +R(1− cos θ)]
= v2c = 4vo − 2g [l sin θ +R(1− cos θ)]
4.2g [l sin θ +R(1− cos θ)]− 2g [l sin θ +R(1− cos θ)]= 6g [lsin θ +R(1− cos θ)]So, force acting on the body,
Page 57
24Tutors.com
⇒ N =mv2cR
= 6mg [(l/R) sin θ + 1− cos θ]c) Let the block loose contact after making an angle θmv2
A chain of length l and mass m lies on the surface of a smooth sphere of radius Rgt; l with one end tied to the top of thesphere. (a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere. (b) Suppose thechain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle θ. (c) Find thetangential acceleration dv
dtof the chain when the chain starts sliding down.
Solutions :Let us consider a small element which makes angle ’dθ’ at the centre.
∴ dm = (m/l)Rd θa) Gravitational potential energy of ’dm’ with respect to centre of the sphere= (dm)g R cos θ(mg/l) R cos θ dθSo, Total, G.P.E.
=
∫ l/r
0
mgR2
lcos θ dθ [ α = (l/R)]
(angle subtended by the chain at the centre)......
= mR2gl
[sin θ](l/R) = mRglsin (l/R)
b) When the chain is released from rest and slides down through an angle θ, the K.E. of the chain is givenK.E. = Change in potential energy.
= mR2gl
sin(l/R)−m ∫gR2
lcos θ d θ......?
= mR2gl
[sin(l/R) + sin θ − sin
θ + (l/R)
]c) Since, K.E. = 1
2mv2 = mR2g
l
[sin(l/R) + sin θ − sin
θ + (l/R)
]Taking derivative of both sides with respect to ’t’
(1/2)× 2v × dvdt
= R2gl
[cos θ × dθ
dt− cos(θ + l/R) dθ
dt
]∴ (R dθ
dt) dvdt
= R2gl× dθ
dt
[cos θ − cos(θ + (l/R))
]When the chain start sliding down, θ = 0.So, dv
dt= Rg
l[1− cos (l/R)]
A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the topof the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to thesphere as a function of the angle θ it slides
Solutions :Let the sphere move towards left with an acceleration ’a’
Let m = mass of the particleThe particle ’m’ will also experience the inertia due to acceleration ’a’ as it is on the sphere. It will also experience the tangential inertia force(m (dv/dt)) and centrifugal force (mv2/R).m dvdt
= ma cos θ +mg sin θ ⇒ mv dvdt
= ma cos θ(R dθdt
)+mg sin θ(
R dθdt
)Because, v = R dθ
dt⇒ vd v = a R cos θ dθ + gR sin θ dθIntegrating both sides we get,v2
2= a R sin θ − gR cos θ + C
Given that, at θ = 0, v = 0, So, C = gR
So, v2
2= a R sin θ − gR cos θ + gR
∴ v2 = 2R (a sin θ + g − g cos θ)⇒ v = [2R (a sin θ) + g − g cos θ)]1/2
Three particles of masses 1.0 kg, 2.0 kg and 3.0 kg are placed at the corners A, B and C respectively of an equilateral triangleABC of edge 1 m. Locate the center of mass of the system.
The structure of a water molecule is shown in figure (9−E1). Find the distance of the center of mass of the molecule from thecenter of the oxygen atom.
Solutions :
Let θ be the origin of the systemIn the above figurem1 = 1kg, x1= -
(0.96* 10−10
)sin 52o, y1 = 0
m2 = 1kg, x2= -(
0.96* 10−10)sin 52o, y2 = 0
x3 = 0, y3= -(
0.96* 10−10)cos 52o,
The position of center of mass
(m1x1+m2x2+m3x3
m1+m2+m3, m1y1+m2y2+m3y3
m1+m2+m3
)
=
(−(0.96×10−10
)sin52o+
(0.96×10−10
)sin52o+
(16×0
)1+1+16
, 0+0+16y318
)=(
0, 89
(0.96× 10−10
)cos52o
)Seven homogeneous bricks, each of length L, are arranged as shown in figure (9− E2). Each brick is displaced with respect to
the one in contact by L10
. Find the x-coordinate of the center of mass relative to the origin is shown.
Solutions :
Let ’O’ (0,0) be the origin of the system,Each brick is mass ’M’ amp; length ’L’Each brick is displaced w.r.t. one in contact by L
10∴ The X coordinate of the center of mass
Xcm =m(L2
)+m(L2
+ L10
)+m(L2
+ 2L10
)+m(L2
+ 3L10
)+m(L2
+ 3L10− L
10
)+m(L2
+ L10
)+m(L2
)7m
=L2
+L2
+ L10
+L2
+L5
+L2
+ 3L10
+L2
+L5
+L2
+ L10
+L2
7
=7L2
+ 5L10
+ 2L5
7= 35L+5L+4L
10×7= 44L
70= 11
35L
A uniform disc of radius R is put over another uniform disc of radius 2R of the same thickness and density. The peripheries ofthe two discs touch each other. Locate the center of mass of the system
Solutions :
Let the center of the bigger disc be the origin2R = Radius of the bigger discR = Radius of the smaller discm1 = πR2 × T × ρm2 = π
(2R)
2 | T × ρwhere T = Thickness of the two discρ = Density of the two disc
Page 59
24Tutors.com
∴ the position of the center of the mass
(m1x1+m2x2m1+m2
, m1y1+m2y2m1+m2
)x1 = R, y1 = 0x2 = 0, y2 = 0(
πR2TρR+0
πR2Tρ+π(2R)2Tρ
, 0m1+m2
)=
(πR2TρR5πR2Tρ
, 0
)=
(R5
,0
)A disc of radius R is cut out from a larger disc of radius 2R in such a way that the edge of the hole touches the edge of the disc.
Locate the center of mass of the residual disc.
Solutions :
Let ’O’ be the origin of the system.R = Radius of the smaller disc2R = Radius of the bigger discThe smaller disc is cut out from bigger discAs from the figurem1= πR2Tρ, x1=R, y1=0m2= π
(2R)
2 T ρ, x2=0, y2=0
The position of the C.M. =
(−πR2TρR+0
−πR2Tρ+π(2R)2Tρ
, 0m1+m2
)
=
(−πR2TρR3πR2Tρ
, 0
)=
(−R3
,0
)A square plate of edge d and a circular disc of diameter d are placed touching each other at the midpoint of an edge of the plate
as shown in figure (9−Q2). Locate the center of mass of the combination, assuming same mass per unit area for the two plates.
Solutions :
Let m be the mass per unit area.∴ Mass of the square plate = M1 = d2m
Mass of the circular disc = M2 = πd2
4m
Let the center of the circular disc be the origin of the system.
∴ Position of the center of mass =
(d2md+π
(d2/4
)m×0
d2md+π(d2/4
)m
, 0+0M1+M2
)=
(d3m
d2m(1+π
4
) , 0
)=
(4dπ+4
,0
)
The new center of mass is
(4dπ+4
)right of the center of the circular disc.
Calculate the velocity of the centre of mass of the system of particles shown in figure (9− E3).
Two blocks of mares 10kg and 20 kg are placed on the X-axis. The first mass is moved on the axis by a distance of 2 cm. Bywhat distance should the second mass be moved to keep the position of the centre of mass unchanged?
Solutions :Two masses m1 amp; m2 are placed on X-axis,
m1 = 10kg, m2 = 20 kgThe first mass is displaced by a distance of 2cm∴ ~Xcm = m1x1+m2x2
m1+m2= 10×2+20x2
30
⇒ 0 ⇒ 20+20x230
⇒ 20 + 20x2 = 0
Page 60
24Tutors.com
⇒ 20 =−20x2 ⇒ x2 =−1∴ The 2nd mass should be displaced by 1cm towards left so as to kept the position of center of mass unchanged.
Two blocks of masses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. Bywhat distance should the second mass be moved to raise the centre of mass by 1 cm ?
Solutions :Two masses m1 amp; m2 are kept in the vertical line,
m1 = 10kg, m2 = 30 kgThe first block is rised through a height of 7cmThe center of mass is raised by 1 cm,∴ 1 = m1y1+m2y2
m1+m2= 10×7+30y2
40
⇒ 1 ⇒ 10×7+30y240
⇒ 10× 7 + 30y2 = 40 ⇒ 30y2 =−30 ⇒ y2 =−1The mass of 30 kg body should be displaced 1 cm downword inorder to raised cener of mass through 1cm.
Consider a gravity-free hall in which a tray of mass M, carrying a cubical block of ice of mass m and edge L, is at rest in themiddle (figure9− E4). If the ice melts, by what distance does the centre of mass of ”the tray plus the ice” system descend ?
Solutions :
As the hall is gravity-free, after the ice melts, it would tend to acquire the spherical shape. But, there is no external force acting on thesystem.So, the center of mass of the system would not move.
Mr. Verma (50kg) and Mr. Mathur (60kg) are sitting at the two extremes of a 4 m long boat (40kg) standing still in water. Todiscuss a mechanics problem, they come to the middle of the boat. Neglecting friction with water, how far does the boat move onthe water during the process ?
Solutions :
m1 = 60kg, m2 = 40kg, m3 = 50kg,Let A bethe origin of the system.Initially Mr. Verma amp; Mr. Mathur are at extreme position of the boat.∴ The center of mass will be at a distance = 60×0+40×2+50×4
150= 280
150= 1.87m from ′A′.
When they come to the mid point of the boat the CM lies at 2m from ′A′
∴ The shift in CM = 2− 1.87 = 0.13m towards right.But as there is no external force in longitudinal direction their CM would not shift.So, the boat moves 0.13m or 13cm towards right.
A cart of mass M is at rest on a frictionless horizontal surface and a pendulum bob of mass m hangs from the roof of the cart(figure9 − E6). The string breaks, the bob falls on the floor, makes several collisions on the floor and finally lands up in a smallslot made in the floor. The horizontal distance between the string and the slot is L. Find the displacement of the cart during thisprocess.
Solutions :
Let the bob fall at A. The mass of bob = mThe mass of cart = M
Initially, their center of mass will be at m×L+M×0M+m
=
(m
M+m
)L
Distance from P, when the bob falls in the slot the CM is at distance ’O’ from P.Shift in CM = 0 - mL
M+m= − mL
M+mtowards left.
= mLM+m
towards right.
But there is no external force in horiontal direction.So the cart displaces a distance mL
M+mtowards right.
The balloon, the light rope and the monkey shown in figure (9−E7) are at rest in the air. If the monkey reaches the top of therope, by what distance does the balloon descend ? Mass of the balloon = M , mass of the monkey = m and the length of the ropeascended by the monkey = L.
Solutions :
Page 61
24Tutors.com
Initially, the monkey y amp; balloon are at rest.So the CM is at ’P’When the monkey descends through a distance ’L’The CM will shift t0 = m×L+M×0
M+m= mL
M+m from P.
So, the balloon descends through a distance mLM+m
Find the ratio of the linear momenta of two particles of masses 1.0 kg and 4.0 kg if their kinetic energies are equal.
Solutions :Let the mass of the two particles be m1 amp; m2 respectively
m1 = 1kg, m1 = 4kg,∴ According to question 1
2m1v1
2 = 12m2v2
2
⇒ m1m2
= v22
v12 ⇒ v2v1
=√m1m2
= v1v2
=√m2m1
Now, m1v1m2v2
= m1m2×√m2m1
=√m1√m2
=√
1√4
= 12
⇒ m1v1m2v2
= 1:2.
A uranium-238 nucleus, initially at rest, emits an alpha particle with a speed of 1.4× 107 m/s. Calculate the recoil speed of theresidual nucleus thorium-234. Assume that the mass of a nucleus is proportional to the mass number.
Solutions :As the uranium-238 nucleus emits a α-particle with a speed of 1.4× 107m/sec. Let v2 be the speed of the residual nucleus thorium-234.
∴ m1v1 = m2v2
⇒ 4× 1.4× 107 = 234× v2
⇒ v2 = 4×1.4×107
234= 2.4× 105 m/sec.
A man of mass 50 kg starts moving on the earth and acquires a speed of 1.8m/s. With what speed does the earth recoil ? Massof earth = 6× 1024 kg.
Solutions :m1v1 = m2v2
⇒ 50× 1.8 = 6× 1024 × v2
⇒ v2 = 50×1.86×1024
= 1.5× 10−23 m/sec.
So, the earth will recoil at a speed of 1.5× 10−23 m/sec.
A neutron initially at rest, decays into a proton, an electron and an antineutrino. The ejected electron has a momentumof 1.4 × 10−26 kg-m/s and the antineutrino 6′4 × 10−27kg-m/s. Find the recoil speed of the proton (a) if the electron and theantineutrino are ejected along the same direction and (b) if they are ejected along perpendicular directions. Mass of the proton =1.67× 10−27 kg.
Solutions :
Mass of proton = 1.67× 10−27
Let ’Vp’ be the velocity of protonGiven momentum of electron = 1.4× 10−26kg m/secGiven momentum of antineutrino = 6.4× 10−27kg m/seca) The electron amp; the antineutrino are ejected in the same direction. As the total momentum is considered the proton should be ejected inhte opposite direction.1.67× 10−27 × Vp = 1.4× 10−26 + 6.4× 10−27 = 20.4× 10−27
⇒ Vp = ( 20.41.67
) = 12.2 m/sec in the opposite direction.
Page 62
24Tutors.com
b) The electron amp; the antineutrino are ejected ⊥ to each other.Total momentum of electron and antineutrino,=√
(142) + (6.42)× 10−27kg m/s = 15.4× 10−27kg m/secSince, 1.67× 10−27Vp = 15.4× 10−27 kg m/sec.So, Vp = 9.2 m/sec.
A man of mass M having a bag of mass m slips from the roof of a tall building of height H and starts falling vertically(figure9− E8). When at a height h from the ground, he notices that the ground below him is pretty hard, but there is a pond ata horizontal distance x from the line of fall. In order to save himself he throws the bag horizontally (with respect to himself) in thedirection opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water.If the man just succeeds to avoid the hard ground, where will the bag land ?
Solutions :
Mass of man = M, Initially velocity = 0, Mass of bad = mLet the man throws the bag towards left with a velocity v towards left. So, there is no external force in the horizontal direction.The momentum will be conserved. Let he goes right with the velocity m = MV ⇒ mv
M⇒ v = MV
m
Let the total time he will take to reach ground =√
2H/g = t1Let the total time he will take to reach the height h = t2 =
√2(H − h)/g
Then the time of his flying = t1 − t2 =√
2H/g −√
2(H − h)/g =√
2/g(√H −
√H − h)
Within in this time he reaches the ground in the pond covering a horizontal distance x ⇒ x = V × t ⇒ V = x/t
∴ v = Mmxt
= Mm×
√g√
2(√H−√H−h)
As there is no external force in horizontal direcion, the x-coordinate of CM will remain at that position.
0 =M×(x)+m×x
M+m⇒ x1 = −M
mx
∴ The bag will reach the bottom at a distance (M/m) × towards left of the line it falls.
A ball of mass 50 g moving at a speed of 2.0 m/s strikes a plane surface at an angle of incidence 45. The ball is reflected by theplane at equal angle of reflection with the same speed. Calculate (a) the magnitude of the change in momentum of the ball (b) thechange in the magnitude of the momentum of the ball.
Solutions :
Mass = 50g = 0.05Kg // v = 2cos45o~i− 2sin45o~j
v1 = 2cos45o~i− 2sin45o~ja) change in momentum = m~v −m~v1
c) The change in magnitude of the momentum of the ball − | ~Pi | − | ~Pf |= 2× 0.5− 2× 0.5 = 0.
Find the centre of mass of a uniform plate having semicircular inner and outer boundaries of radii R, and R2 (figure9− E5)
Solutions :
Page 63
24Tutors.com
The center of mass of the plate will be on the symmetrical axis.
⇒ ~ycm =
(πR2
2
2
)(4R23π
)−(πR1
2
2
)(4R13π
)πR2
2
2−πR1
2
2
=
(2/3)R2
3−(2/3)R1
3
π/2(R2
2−R12) = 4
3π
(R2−R1
)(R2
2+R12+R1R2
)(R2−R1
)(R2+R1
)= 4
3π
(R2
2+R12+R1R2
)R!+R2
above the center.
Light in certain cases may be considered as a stream of particles called photons. Each photon has a linear momentum hIX whereh is the Planck’s constant and is the wavelength of the light. A beam of light of wavelength A. is incident on a plane mirror at anangle of incidence O. Calculate the change in the linear momentum of a photon as the beam is reflected by the mirror.
Solutions :
PIncidence = (h/λ)cosθ~i− (h/λ)sinθ~j
PReflected = (h/λ)cosθ~i− (h/λ)sinθ~jThe change in momentum will be only in the x-axis direction i.e.| ∆P |= (h/λ)cosθ − ((h/λ)cosθ) = (2h/λ)cosθ
A block at rest explodes into three equal parts. Two parts start moving along X and Y axes respectively with equal speeds of10 m/s. Find the initial velocity of the third part
Solutions :
As the block is exploded only due to its internal energy. So net external force during this process is 0. So the center mass will not change.Let the body while exploded was at the origin of the coordinate system.If the two bodies of equal mass is moving at a speed of 10 m/s in +xamp; +y axis directions respectively,√
102 + 102 + 210.10cos90o = 10√
2m/s 45ow.r.t.+ xaxisIf the center mass is at rest, then the third mass which has equal mass with other two, will move in the opposite direction (i.e. 1135ow.r.t.+xaxis)of the resultant at the same velocity.
Two fat astronauts each of mass 120 kg are travelling in a closed spaceship moving at a speed of 15 km/s in the outer space farremoved from all other material objects. The total mass of the spaceship and its contents including the astronauts is 660 kg. If theastronauts do slimming exercise and thereby reduce their masses to 90 kg each, with what velocity will the spaceship move ?
Solutions :Since the spaceship is removed from any material object amp; totally isolated from the surrounding, the missions by astronauts couldn’t
slip away from the spaceship. So the total mass of the spaceship remain unchanged and also it’s velocity.
During a heavy rain, hailstones of average size 1.0 cm in diameter fall with an average speed of 20 m/s. Suppose 2000 hailstonesstrike every square meter of a 10m × 10 m roof perpendicularly in one second and assume that the hailstones do not rebound.Calculate the average force exerted by the falling hailstones on the roof. Density of a hailstone is 900kg/m3.
Solutions :d = 1cm v = 20m/s u = 0 ρ = 900kg/m3 = 0.9gm/cm3
volume = (4/3)πr3 = (4/3)π(0.5)3 = 0.5238cm3
∴ mass = vρ = 0.5238× 0.9 = 0.4714258gm∴ mass of 2000 hailstone = 2000× 0.4714 = 947.857∴ Rate of change in momentum per unit area = 947.857× 2000 = 19N/m3
∴ Total force exerted = 19× 100 = 1900N
A ball of mass m is dropped onto a floor from a certain height. The collision is perfectly elastic and the ball rebounds to thesame height and again falls. Find the average force exerted by the ball on the floor during a long time interval.
Solutions :A ball of mass m is dropped onto a floor from a certain height let ’h’.
∴ v1 =√
2gh, v1 = 0, v2 = −√
2gh amp; v2 = 0∴ Rate of change of velocity :-
F = m×2√
2ght
∴ v =√
2gh, s = h, v = 0,⇒ v = u+ at
Page 64
24Tutors.com
⇒√
2gh = gt⇒ t =√
2hg
∴ Total time 2√
2ht
∴ F = m×2√
2gh
2√
2hg
= mg
A railroad car of mass M is at rest on frictionless rails when a man of mass m starts moving on the car towards the engine. Ifthe car recoils with a speed v backward on the rails, with what velocity is the man approaching the engine ?
Solutions :A railroad car of mass M is at rest on frictionless rails when a man of mass m starts moving on the car towards the engine. The car recoils
with a speed v backward on the rails.Let the mass is moving with a velocity x w.r.t. the engine.∴ The velocity of the mass w.r.t. earth is (x− v) towards right.Vcm = 0 ( Initially at rest)∴ 0 = −Mv +m(x− v)
⇒Mv = m(x− v)⇒ mx = Mv +mv ⇒ x =
(M+mm
)v ⇒ x =
(1 + M
m
)v
A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50 m where m is the mass ofone shell. If the velocity of the shell with respect to the gun (in its state before firing) is 200 m/s, what is the recoil speed of thecar after the second shot ? Neglect friction.
Solutions :A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50m where m is the mass of the one shell.
The muzzle velocity of the shells is 200m/s.Initial, Vcm = 0∴ 0 = 49m× V +m× 200⇒ V = −200
49m/s
∴ 20049m/s towards left
When another shel is fired, then the velocity of the car, with respect to the platform is,⇒ V = 200
49m/s towards left
When another shel is fired, then the velocity of the car, with respect to the platform is,⇒ V = 200
48m/s towards left
∴ Velocity of the car w.r.t. the earth is
(20049
+ 20048
)m/s towards left
Two persons each of mass m are standing at the two extremes of a railroad car of mass M resting on a smooth track (figure9−E10). The person on left jumps to the left with a horizontal speed u with respect to the state of the car before the jump. Thereafter,the other person jumps to the right, again with the same horizontal speed u with respect to the state of the car before his jump.Find the velocity of the car after both the persons have jumped off
Solutions :Two persons each of mass m are standing at the two extremes of a railroad car of mass m resting on a smooth track.
Case - ILet the velocity of the railroad car w.r.t. the earth is V after the jump of the man.∴ 0 = −mu+ (M +m)V⇒ V = mu
M+mtowards right
Case - IIWhen the man on the right jumps, the velocity of it w.r.t. the car is u.∴ 0 = mu = Mv′
⇒ v′ = muM
(V’ is the change in velocity of the platform when platform itself is taken as reference assuming the car to be at rest)
∴ So, net velocity towards left (i.e. the velocity of the car w.r.t. the earth) = mvM− mvM+m
= mMu+m2v−MmuM(M+m)
= m2vM(M+m)
Figure(9 − E11) shows a small block of mass m which is started with a speed v on the horizontal part of the bigger block ofmass M placed on a horizontal floor. The curved part of the surface shown is semicircular. All the surfaces are frictionless. Findthe speed of the bigger block when the smaller block reaches the point A of the surface.
Solutions :
A small block of mass m which is started with a velocity V on the horizontal part of the bigger block of mass M placed on a horizontalfloor.Since the small body of mass m is started with a velocity V in the horizontal direction, so the total initial momentum at the initial position inthe horizontal direction will remain same as the total final momentum at the point A on the bigger block in the horizontal direction.From L.C. K. m : mv +M ×O = (m+M)v ⇒ v′ = mv
M+m
In a typical Indian Bugghi (a luxury cart drawn by horses), a wooden plate is fixed on the rear on which one person can sit. Abugghi of mass 200 kg is moving at a speed of 10 km/h. As it overtakes a school boy walking at a speed of 4 km/h, the boy sits onthe wooden plate. If the mass of the boy is 25 kg, what will be the new velocity of the bugghi ?
Solutions :Mass of the bugghi = 200kg, VB = 10km/hour
∴ Mass of the body = 2.5kg amp; VBoy = 4km/hourIf we take the boy and bugghi as a system then total momentum before the process of sitting will remain constant after the proces of sitting.∴ mbVb = mboyVboy = (mb +mboy)v
Page 65
24Tutors.com
⇒ 200× 10 + 25× 4 = (200 + 25)× v⇒ v = 2100
225= 28
3= 9.3m/sec
A ball of mass 0.50 kg moving at a speed of 5.0 m/s collides with another ball of mass 1.0 kg. After the collision the balls sticktogether and remain motionless. What was the velocity of the 1.0 kg block before the collision ?
Solutions :Mass of the ball = m1 = 0.5kg, the velocity of the ball = 5 m/s
Mass of the another ball m2 = 1kg.Let it’s velocity = v’ m/s. Using the law of conservation of momentum,0.5× 5 + 1× v′ = 0⇒ v′ = −2.5∴ Velocity of second ball is 2.5m/s opposite to the direction of motion of 1st ball.
A 60 kg man skating with a speed of 10 m/s collides with a 40 kg skater at rest and they cling to each other. Find the loss ofkinetic energy during the collision.
Solutions :Mass of the man =m1 = 60kg
Speed of the man = v1 = 10kgMass of the skater =m2 = 40kgLet the velocity = v’∴ 60× 10 + 0 = 100× v′ ⇒ v′ = 6m/sLoss in K.E. = (1/2)60× (10)2 = (1/2)× 100× 36 = 1200 J.
Consider a head-on collision between two particles of masses m1 and m2. The initial speeds of the particles are u1 and u2 inthe same direction. The collision starts at t = 0 and the particles interact for a time interval At. During the collision, the speed ofthe first particle varies asv(t) = u1 + t
∆t(v1 − u1)
Find the speed of the second particle as a function of the time during the collision.
Solutions :Using law of conservation of momentum
m1u1 +m2u2 = m1v(t) +m2v′
Where v’ = speed of 2nd paricle during collision.⇒ m1u1 +m2u2 = m1u1 +m1 + (t/∆t)(v1 − u1) +m2v′
⇒ m2u2m2 − m1
m2
t∆t
(v1 − u1)v′
∴ v′ = u2 − m1m2
t∆t
(v1 − u)
A bullet of mass m moving at a speed v hits a ball of mass M kept at rest. A small part having mass m’ breaks from the balland sticks to the bullet. The remaining ball is found to move at a speed v1 in the direction of the bullet. Find the velocity of thebullet after the collision
Solutions :Mass of the Bullet = m and speed = v
Mass of the ball = Mm’ = frictional mass from the ballUsing law of conservation of momentum,mv + 0 = (m′ +m)v′ + (M −m′)v1
where v’ = final velocity of the bullet + frictional mass
⇒ v′ =mv−(M+m′)V1
m+m′
A ball of mass m moving at a speed v makes a head-on collision with an identical ball at rest. The kinetic energy of the ballsafter the collision is three fourths of the original. Find the coefficient of restitution.
Solutions :Mass of the 1st ball = m and speed = v
Mass of the 2nd ball = mLet final velocities of 1st and 2nd ball are v1 and v2 respectively.Using law of conservation of momentum,m1(v1 + v2) = mv⇒ v1 + v2 = v ........... (1)Also, v1 − v2 = ev ............ (2)Given that final K.E. = 3/4 Initial K.E.⇒ 1/2mv2
1 + 1/2mv22 = 3/4× 1/2mv2
⇒ v21 + v2
2 = 3/4v2
⇒ (v1+v2)2+(v1−v2)2
2= 3
4v2
⇒ (1+e2)v2
2= 3
4v2 ⇒ 1 + e2 = 3
2⇒ e2 = 1
2⇒ e = 1√
2
A block of mass 2.0 kg moving at 2.0 m/s collides head on with another block of equal mass kept at rest. (a) Find the maximumpossible loss in kinetic energy due to the collision. (b) If the actual loss in kinetic energy is half of this maximum, find the coefficientof restitution.
Solutions :Mass of the block = 2kg and speed = 2m/s
Mass of the 2nd block = 2kgLet final velocities of 2nd block =vUsing law of conservation of momentum,2× 2 = (2 + 2)v ⇒ v = 1m/s∴ Loss in K.E. in inelastic collision−( 1
2)× 2× (2)2v − 1
2(2 + 2)× (1)2 − 4− 2− 2J
b) Actual Loss = maximum loss2
= 1J
−( 12
)× 2× (2)2 − ( 12
)2× v21 + ( 1
2)× 2× v2
2 = 1
⇒ 4− (v21 + v2
2) = 1
Page 66
24Tutors.com
⇒ 4− (1+e2)×42
= 1
2(1 + e2) = 3⇒ 1 + e2 = 32⇒ e2 = 1
2⇒ e = 1√
2
A particle of mass 100 g moving at an initial speed u collides with another particle of same mass kept initially at rest. If thetotal kinetic energy becomes 0.2 J after the collision, what could be the minimum and the maximum value of u.
Solutions :
Final K.E. = 0.2JInitial K.E. = 1/2mv2
1 + 0 = 1/2× 0.1u2 = 0.05u2
mv1 = mv2 = muWhere v1 and v2 are final velocities of 1st and 2ndblock respectively.⇒ v1 + v2 = u ...........(1)(v1 − v2) + l(11 − u2) = 0⇒ la = v2 − v1 ............(2)u2 = 0, u1 = u.Adding Eq.(1) and Eq.(2)2v2 = (1 + l)u⇒ v2 = (u/2)(1 + l)∴ v1 = u− u
2− u
2l
v1 = u2
(1− l)Given(1/2)mv2
1 + (1/2)mv22 = 0.2
⇒ v21 + v2
2 = 4
⇒ u2
4(1− l)2 + u2
4(1 + l)2 = 4 ⇒ u2
2(1− l2) = 4 u2 = 8
1−l2For maximum value of u, denominator should be minimum,⇒ l = 0⇒ u2 = 8⇒ u = 2
√2m/s
For maximum value of u, denominator should be maximum,⇒ l = 1⇒ u2 = 4⇒ u = 2m/s
Two friends A and B (each weighing 40 kg) are sitting on a frictionless platform some distance d apart. A rolls a ball of mass 4kg on the platform towards B which B catches. Then B rolls the ball towards A and A catches it. The ball keeps on moving backand forth between A and B. The ball has a fixed speed of 5 m/s on the platform. (a) Find the speed of A after he rolls the ball forthe first time. (b) Find the speed of A after he catches the ball for the first time. (c) Find the speeds of A and B after the ball hasmade 5 round trips and is held by A. (d) How many times can A roll the ball ? (e) Where is the centre of mass of the system ”A+ B + ball” at the end of the nth trip ?
Solutions :
Two friends A amp; B (each 40kg) are sitting on a frictionless platform some distance d apart A rols a ball of mass 4 kg on the platformtowards B, which B catches. Then B rolls the ball towards A and A catches it. The ball keeps on moving back amp; forth between A and B.The ball has a fixed velocity 5 m/s.a) Case -I :- Total momentum of the man A amp; the ball will remain constant.∴ 0 = 4× 5− 40× v ⇒ v = 0.5m/s towards left.b) Case -II :- When B catches the ball , the momentum between the B amp; the ball will remain constant.⇒ 4× 5 = 44v ⇒ v = (20/44)m/sCase-III :- When B throws the ball, then L.C.L.M.,⇒ 44× (20/44) = −4× 5 + 40× v ⇒ v = 1m/s(towards right)Cases-IV :- When A catches the ball , then applying L.C.L.M.⇒ −4× 5 + (−0.5)× 40 = −44v ⇒ v = 10
11m/s towards left
c) Case-V :- When A throws the ball, by applying L.C.L.M.44× (10/11) = 4× 5− 40× V ⇒ V = 60/40 = 3/2m/s towards leftCase -VI:- When B receives the ball then by applying L.C.L.M.40× 1 + 4× 5 = 44× v ⇒ V = 60/44 m/s towards rightCase-VII:- When B throws the ball, then applying L.C.L.M.44× (66/44) = −4× 5 + 40× V ⇒ V = 80/40 = 2m/s towards right.Case-VIII:- When A catches the ball then applying L.C.L.M.⇒ −4]times5− 40× (3/2) = −44v ⇒ v = (80/44) = (20/11)m/s towards leftSimilarly after 5 round tripsThe velocity of A will be (50/110) amp; velocity of B will be 5 m/s.
Page 67
24Tutors.com
d) Since after 6 round trip, the velocity of A is 60/11 i.e. gt;5m/s So, it can’t catch the ball. So it can only roll the ball six.e)Let the ball amp; the body A at the initial position be at origin.
∴ Xc = 40×0+40×0+40×d40+40+4
= 1011d
A ball falls on the ground from a height of 2.0 m and rebounds up to a height of 1.5 m. Find the coefficient of restitution.
Solutions :
u =√
2gh = velocity on the ground when ball approaches the ground.⇒ u =
√2× 9.8× 2
v = velocity of ball when it separates from the ground.~v + l~u = 0
⇒ l~u = −~v ⇒ l =√
2×9.8×1.5√2×9.8×2
=√
34
=√
32
In a gamma decay process, the internal energy of a nucleus of mass M decreases, a gamma photon of energy E and linearmomentum E I c is emitted and the nucleus recoils. Find the decrease in internal energy.
Solutions :
K.E. of nucleus =(1/2)mv2 = (1/2)m
(Emc
)2
= E2
2mc2
Energy limited by Gamma photon = EDecrease in internal energy = E + E2
2mc2
A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s (figure 9-E12) towards anotherblock of equal mass kept at rest. The spring constant of the spring fixed at one end is 100 N/m. Find the maximum compressionof the spring.
Solutions :
Mass of each block MA and MB = 2kg.Initial velocity of the 1st block (V) = 1m/s.VA = 1m/s VB = 0m/sSpring constant of the spring = 100N/mThe block A strikes the spring with a velocity 1m/sAfter the collision it’s velocity decreases continuously and at a instant the whole system (Block A + the compund spring + Block B) movetogether with a common velocity.Let the velocity be V.Using Conservation of energy, (1/2)MAV
2A + (1/2)MBV
2B = (1/2)MAV 2 + (1/2)MBV 2 = (1/2)kx2.
(1/2)× 2(1)2 + 0 = (1/2)× 2× v2 + (1/2)× 2× v2 + (1/2)x2 × 100(Where x = max compression of the spring )⇒ 1 = 2v2 + 50x2 .......(1)As there is no external force in the horizontal direction, the momentum should be conserved.⇒MAVA +MBVB = (MA +MB)V⇒ 2× 1 = 4× v⇒ V = (1/2)m/s ........(2)Putting in eq.(1)1 = 2× (1/4) + 50x+ 2(1/2) = 50x2
x2 = 1/100m2
Page 68
24Tutors.com
x = (1/10)m = 0.1m = 10cm.
A bullet of mass 20 g travelling horizontally with a speed of 500 m/s passes through a wooden block of mass 10.0 kg initiallyat rest on a level surface. The bullet emerges with a speed of 100 m/s and the block slides 20 cm on the surface before coming torest. Find the friction coefficient between the block and the surface (figure 9-E13).
Solutions :
Mass of bullet m = 0.02kgInitial velocity of bullet V1 = 500m/sMass of block M = 10kg.Initial velocity of block u2 = 0Final velocity of bullet = 100m/s = vLet the final velocity of block when the bullet emerges out, if block =v’.mv1 +Mu2 = mv +Mv′
⇒ 0.02× 500 = 0.02× 100 + 10× v′⇒ v′ = 0.8m/sAfter moving a distance 0.2m it stops.⇒ change in K.E. = Work done0− (1/2)× 10× (0.8)2 = −µ× 10× 10× 0.2⇒ µ = 0.16.
A projectile is fired with a speed u at an angle 0 above a horizontal field. The coefficient of restitution of collision between theprojectile and the field is e. How far from the starting point, does the projectile makes its second collision with the field ?
Solutions :
The projected velocity = u.The angle of projection = θWhen the projectile hits the ground for the 1st time, the velocity would be the same i.e. u.Here the Component of velocity parallel to ground, ucosθ should remain constant.But the vertical component of the projectile undergoes achange after the collision.⇒ e = usinθ
v⇒ v = eusinθ
Now for the 2nd projectile motion,
U = velocity of projection =√
(ucosθ)2 + (eusinθ)2 and Angle of projetion = α = tan−1
(eusinθacosθ
)= tan−1(etanθ)
or tanalpha = etanθ ......(2)
Because, y = xtanα− gx2sec2α2u2 ......(3)
Here, y = 0, tanα = etanθ, sec2α = 1 + e2tan2θAnd u2 = u2cosθ + e2sin2θPutting the above values in the equation (3)
xetanθ =gx2(1+e2tan2θ)
2u2(cos2θ+e2sin2θ)
⇒ x =2eu2tanθ(cos2θ+e2sin2θ
g(1+e2tan2θ)
⇒ x = 2eu2tanθ−cos2θg
= eu2sin2θg
⇒ So, From the starting point O, it will fall at a distance = u2sin2θg
+ eu2sin2θg
= u2sin2θg
(1 + e)
A ball falls on an inclined plane of inclination 0 from a height h above the point of impact and makes a perfectly elastic collision.Where will it hit the plane again ?
Solutions :
Angle inclination of the plane = θM the body falls through a height of h,The Striking velocity of the projectile with the inclined plane v =
√2gh
Now, the projectile makes on angle (90o − 2θ)Velocity of projection = u =
√2gh
Let AB = LSo, x = lcosθ, y = −lsinθFrom equation of trajectory, y = xtanα− gx2sec2α
Solve the previous problem if the coefficient of restitution is e. Use 0 = 45, e = 34
and h = 5 m.
Solutions :
h = 5m, θ = 45o, e = ( 34
)
Here the velocity with ehih it would strike = v =√
2g × 5 = 10m/secAfter collision, let it make anangle β with hoeizontal . The horizontal component of velocity 10cos45o will remain unchanged and the velocityin the perpendicular direction to the plane after willsine.⇒ Vy = e× 10sin45o
let the distance where it falls = L⇒ x = Lcosθ, y = −LsinθAngle of projection (α) = −8o
Using equation of trajectory, y = xtanα− gx2sec2α2u2
⇒ −lsinθ = lcosθ × tan8o − g2× lcos2θsec28o
u2
⇒ −sin45o = cos45o − tan8o − 10cos245osec8o
(8.83)2(l)
Solving the above equation we get l = 18.5m
A block of mass 200 g is suspended through a vertical spring. The spring is stretched by 1.0 cm when the block is in equilibrium.A particle of mass 120 g is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact. Find themaximum extension of the spring. Take g = 10 m/s 2
Solutions :
Block of the particle = m = 120gm = 0.12kgIn the equillibrium condition, the spring is streched by a distance x = 1.00 cm = 0.01m.⇒ 0.2× g = K.x.⇒ 2 = K]times0.01⇒ K = 200N/mThe velocity with the particle m will strike M is given by u=√
2× 10× 0.45 =√
9 = 3m/secSo after the collision, the velocity of the particle and the block isv = 0.12×3
0.32= 9
8m/sec
Let the spring be streched through an extra deflection of δ.0− (1/2)× 0.32× (81/64) = 0.32× 10× δ − (1/2× 200× (δ + 0.1)2 − (1/2)× 200× (0.01)2)Solving the above equation we getδ = 0.045 = 4.5cm
A bullet of mass 25 g is fired horizontally into a ballistic pendulum of mass 5.0 kg and gets embedded in it (figure9−E14). Ifthe centre of the pendulum rises by a distance of 10 cm, find the speed of the bullet.
Page 70
24Tutors.com
Solutions :Mass of the bullet = 25g = 0.025kg.
Mass of pendulum = 5kg.The vertical displacement h =10cm = 0.1mLet it strike the pendulum with a velocity u.Let the final velocity be v,⇒ mu = (M +m)v⇒ v = m
M+mu = 0.025
5.025× u = u
201Using conservation of energy,
0− (1/2)(M +m)V 2 = −(M +m)g × h ⇒ u2
(201)2= 2× 10× 0.1 = 2
⇒ u = 201×√
2 = 280m/sec.
A bullet of mass 20 g moving horizontally at a speed of 300 m/s is fired into a wooden block of mass 500 g suspended by a longstring. The bullet crosses the block and emerges on the other side. If the centre of mass of the block rises through a height of 20.0cm, find the speed of the bullet as it emerges from the block.
Solutions :Mass of the bullet = M = 20 gm = 0.02kg.
Mass of wooden block M = 500gm = 0.5KgVelocity of the bullet with ehich it strikes u = 300 m/secLet the bullet emerges out with velocity V and vlocity of block = V’As per law of conservation of momentum.mu = Mv′ +mv ...........(1)Again applying work-enrgy principle for the block after the collision,0− (1/2)M × V ′2 = −Mgh (where h =0.2m)⇒ V 2 = 2ghV ′ =
√2gh =
√20× 0.2 = 2m/sec
Substituting the value of V’ in equation (1), we get0.02× 300 = 0.5× 2 + 0.2× V⇒ V = 6.1
0.02= 250m/sec.
Two masses m, and m2 are connected by a spring of spring constant k and are placed on a frictionless horizontal surface. Initiallythe spring is stretched through a distance xo when the system is released from rest. Find the distance moved by the two massesbefore they again come to rest.
Solutions :
Mass of the two blocks are m1,m2.Initially the spring is stretched by xoSpring constant K.For the blocks to come to rest again,Be x1 and x2 towards right and left respectively,As o external force acts in horizontal direction,m1x1 = m2x2 ..........(1)Again , the energy would be conserved in the spring.⇒ (1/2)k × x2 = (1/2)k(x1 + x2 − x0)2
⇒ x0 = x1 + x2 − x0
⇒ x1 + x2 = 2x0 ................(2)
⇒ x1 = 2x0 − x2 similarly x1 =
(2m2
m1+m2
)x0
⇒ m1(2x0 − x2) = m2x2 ⇒ 2m1x0 −m1x2 = m2x2 ⇒ x2 =
(2m1
m1+m2
)x0
Two blocks of masses m, and m2 are connected by a spring of spring constant k (figure 9-E15). The block of mass m2 is givena sharp impulse so that it acquires a velocity vo towards right. Find (a) the velocity of the centre of mass, (b) the maximumelongation that the spring will suffer.
Solutions :
a) ∴ velocity of cw=enter of mass = m2×v0+m1×0m1+m2
= m2v0m1+m2
b) The spring will attain maximum elongation when both velocity of two blocks will attain the velocity of center of mass.d) x → maximum elongation of the spring.Change of Kinetic energy = Potential stored in spring.
⇒ (1/2)m2v20 − (1/2)(m1 +m2)
(m2v0m1+m2
)2
= (1/2)kx2
⇒ m2v20
(1− m2
m1+m2
)= kx2 ⇒ x =
(m2
m1+m2
)1/2
× v0
Consider the situation of the previous problem. Suppose each of the blocks is pulled by a constant force F instead of any impulse.Find the maximum elongation that the spring will suffer and the distances moved by the two blocks in the process.
Solutions :
If both the blocks are pulled by some force, they suddenly move with some acceleration and instantaneously stop at same position wherethe elongation of spring is maximum∴ Let x1, x2 → extension by block m1 and m2
Total work done =Fx1 + Fx2 ......(1)∴ Increase the potential energy of spring = (1/2)K(x1 + x2)2 .........(2)
KSince the net external force of the two blocks is zero thus same force act on opposite direction.∴ m1x1 = m2x2 ..........(3)And (x1 + x2) = 2F
K∴ x2 = m1
m2× 1
Substituting m1m2× 1 + x1 = 2F
K
⇒ x1
(1 + m1
m2
)= 2F
K⇒ x1 = 2F
Km2
m1+m2
Similarly, x2 =⇒ x1 = 2FK
m2m1+m2
Consider the situation of the previous problem. Suppose the block of mass m, is pulled by a constant force F, and the otherblock is pulled by a constant force F2. Find the maximum elongation that the spring will suffer.
Solutions :
Acceleration of mass m1 = F1−F2m1+m2
Similarly Acceleration of mass m2 = F1−F2m1+m2
Due to F1 and F2 block of mass m1 and m2 will experience different acceleration and experience inertia force.∴ Net force on m1 = F1 −m1a= F1 −m1 × F1−F2
m1+m2= m1F1+m2F2−m1F1+F2m1
m1+m2= m2F1+m1F2
m1+m2
Similarly Net force on m2 = F2 −m2a= F2 −m2 × F2−F1
m1+m2= m1F2+m2F2−m2F2+F1m2
m1+m2= m1F2+m2F2
m1+m2
If m1 displaced by a distance x1 and x2 by m2 the maximum extension of the spring is x1 +m2
Consider a gravity-free hall in which an experimenter of mass 50 kg is resting on a 5 kg pillow, 8 ft above the floor of the hall.He pushes the pillow down so that it starts falling at a speed of 8 ft/s. The pillow makes a perfectly elastic collision with the floor,rebounds and reaches the experimenter’s head. Find the time elapsed in the process.
Solutions :
Mass of the man (Mm) is 50 kg.Mass of the pillow (Mp) is 5 kg. When the pillow is pushed by the man, the pillow will go down while the man goes up. It because of externalforce on the system which is zero.⇒ acceleration of center of mass is zero.⇒ velocity of center of mass is zero∴ As the initial velocity of the system is zero.∴Mm × Vm = Mp × Vp ......(1)Given the velocity of pillow is 80ft/sWhich is relative velocity of pillow w.r.t. man.~Vp/m = ~Vp − ~Vm = ~Vp − (−~Vm) = ~Vp + ~Vm ⇒ ~Vp = ~Vp/m − ~VmPutting in equation (1)Mm × Vm = Mp(Vp/m − Vm)
⇒ 50× Vm =⇒ 10× Vm = 8− Vm ⇒ Vm = 811
= 0.721m/s∴ Absolute velocity of pillow = 8− 0.727 = 7.2 ft/sec∴ Time taken to reach the floor = S
V=8
7.2=1.1secAs the mass of ball gt;gt;gt; then pillowThe velocity of block before the collision = velocity after the collision⇒ Time of ascent = 1.11 sec∴ Total time taken = 1.11 + 1.11 = 2.22sec.
The track shown in figure (9-E16) is frictionless. The block B of mass 2m is lying at rest and the block A of mass m is pushedalong the track with some speed. The collision between A and B is perfectly elastic. With what velocity should the block A bestarted to get the sleeping man awakened ?
Solutions :
Let the velocity of A =u1
Let the final velocity when reaching B becomes collision = v1
∴ (1/2)mv21 − (1/2)mu2
1 = mgh
v21 − u2
1 = 2gh ⇒ v1 =√
2gh− u21 .....(1)
When the block B reached at the upper man’s head, the velocity of B is just zero.∴ (1/2)× 2m× 02 − (1/2)× 2m× v2 = mgh ⇒ v =
√2gh
Page 72
24Tutors.com
∴ Before collision velocity of uA = v1 ub = 0After collision velocity of vA = v (say) vB =
√2gh
Since it is an elastic collision the momentum and K.E. should be conserved∴ m× v1 + 2m× 0 = m× v + 2m×
√2gh
⇒ v1 − v = 2√
2ghAlso, (1/2)×m× v2
1 + (1/2)2m× 02 = (1/2)×m× v2 + (1/2)× 2m× (√
2gh)2
⇒ v21 − v2 = 2
√2gh×
√2gh ......(2)
Dividing (1) by (2),(v1+v)(v1−v)
v1+v= 2×
√2gh×
√2gh
2×√
2gh→ v1 + v =
√2gh ......(3)
Adding (1) and (3), 2v1 = 3√
2gh⇒ v1 = (3/2)√
2gh
But v1 =√
2gh+ u2 =
(32
)√
2gh
⇒ 2gh+ u2 = 94× 2gh
u = 2.5√
2ghSo the block will travel with the velocity greater than 2.5
√2gh so awake the man by B.
A bullet of mass 10 g moving horizontally at a speed of 5017 m/s strikes a block of mass 490 g kept on a frictionless track asshown in figure (9-E17). The bullet remains inside the block and the system proceeds towards the semicircular track of radius 0.2m. Where will the block strike the horizontal part after leaving the semicircular track ?
Solutions :
Mass of the block = 490 gm.Mass of the bullet = 10 gm.Since the bulet embedded inside the block, it is an plastic collision.Initially velocity of bullet v1 = 50
√7 m/sec
Velocity of the block is v2 = 0Let the final velocity of both = v∴ 10× 10−3 × 50×
√7 + 10−3 × 190|0 = (490 + 10)× 10−3 × VA
VA =√
7m/sWhen the block lossess the contact at ’D’ the component mg will act on it.m(VB)2
r= mgsinθ ⇒ (VB)2 = grsinθ ....(1)
Prutting work energy principle(1/2)m× (VB)2 − (1/2)×m× (VA)2 = −mg(0.2 + 0.2sinθ)⇒ (1/2)× grsinθ − (1/2)× (
∴ Distance travelled in horizontal direction.s = V cosθ × t =
√grsinθ × t =
√9.8× 2× (1/2)× 0.247 = 0.196m.
Total distance = (0.2− 0.2cos30o) + 0.196 = 0.22m.
Two balls having masses m and 2m are fastened to two light strings of same length 1 (figure 9-E18). The other ends of thestrings are fixed at 0. The strings are kept in the same horizontal line and the system is released from rest. The collision betweenthe balls is elastic. (a) Find the velocities of the balls just after their collision. (b) How high will the balls rise after the collision ?
Solutions :
Let the velocity of m reaching at lower end = v1
from work energy principle,∴ (1/2)×m× (1/2)×m× 02 = mgl⇒ v1 =
√2gl
Similarly velocity of heavy block will be v2 =√
2ghv1 = v2 = u(say)Let the final velocity of m and 2m v1 and v2 respectively,According to law of conservation of momentum.m× x1 + 2m× V2 = mv1 + 2mv2
b) Putting the work energy principle,(1/2)× 2m× 02 − (1/2)× 2m× (v2)2 = −2m× g × h
Page 73
24Tutors.com
[h → height gone by heavy ball]
⇒ (1/2) 2g9
= l × h h = l9
Similarly, (1/2)×m× 02 − (1/2)×m× v21 = m× g × h2
[ height reached by small ball ]
⇒ (1/2)× 50gl9
= g × h2 ⇒ h2 = 25l9
Some h2 is more than 2l, the velocity at height point will not be zero and the ’m’ will rise by a distance 2l.
A uniform chain of mass M and length L is held vertically in such a way that its lower end just touches the horizontal floor.The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming that the chain does notform a heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor.
Solutions :
Let us consider a small element at a distance ’x’ from the floor of length ’dy’.So, dm = M
Ldx
So, the velocity with which the element will strike the floor is v =√
2gx∴ So, the momentum transferred to the floor is,M = (dm)v = M
L× dx×
√2gx [because the element comes to rest]
S0, the force exerted on the floor change in momentum is given by, F1 = dMdt
= ML× dx
dt×√
2gx
Because, v = dxdt
=√
2gx( for the chain element)
F1 = ML×√
2gx×√
2gx = ML× 2gx = 2Mgx
L
Again, the force exerted due to ’x’ length of chain on the floor due to its own weight is given by,W = M
L(x)× g = Mgx
L
So, the total force exerted is given by, F = F1 +W = 2MgxL
+ MgxL
= 3MgxL
The blocks shown in figure (9-E19) have equal masses. The surface of A is smooth but that of B has a friction coefficient of 0.10with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. Find the distance travelled by B if (a) thecollision is perfectly elastic and (b) the collision is perfectly inelastic. Take g = 10 mls 2.
Solutions :
V1 = 10m/s V2 = 0V1, v2 → velocity of ACB after collision.a) If the edlision is perfectly elastic,mV1 +mV2 = mv1 +mv2
⇒ 10 + 0 = v1 + v2
⇒ v1 + v2 = 10 .......(1)Again, v1 − v2 = −(u1 − v2) = −(10− 0) = −10 .......(2)Subtracting (2) from (1)2v2 = 20 ⇒ v2 = 10m/sThe deacceleration of B = µgPutting work energy principle,∴ (1/2)×m× 02 − (1/2)×m× v2
2 = −m× a× h⇒ −(1/2)× 102 = −µg × h ⇒ h = 100
2×0.1×10= 50m
b) If the collision perfectly elastic.m× u1 +m× u2 = (m+m)× v⇒ m× 10 +m× 0 = 2m× v ⇒ v = 10
2= 5m/s
The two blocks will move together sticking to each otherPutting work energy princilple,(1/2)× 2m× 02 − (1/2)× v2 = 2m× µg × s⇒ 52
0.1×10×2= s ⇒ s = 12.5m
The friction coefficient between the horizontal surface and each of the blocks shown in figure (9-E20) is 0.20. The collision
Page 74
24Tutors.com
between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take g = 10m/s2
Solutions :
Let the velocity of 2kg block on reaching the 4 kg block before collision = u1.Given, V2 = 0(velocity of 4kg block).∴ From work energy principle,(1/2)m× u2
1 − (1/2)m× 1 = −m× ug × s
⇒ u21−1
2= −2× 5 ⇒ −16 =
u21−1
4
64× 10−2 = u21 − 1 ⇒ u1 = 6m/s
Since it is a perfectly elastic collision,Let V1, V2 → velocity of 2kg amp; 4 kg block after collision,m1V1 +m2V2 = m1v1 +m2v2
⇒ 2× 0.6 + 4× 0 = 2v1 + 4v2 ⇒ v1 + 2v2 = 0.6 ...(1)Again, V1 − V2 = −(u1 − u2) = −(0.6− 0) = −0.6Subtracting (2) from (1)3v2 = 1.2 ⇒ v2 = 0.4m/s∴ v1 = 0.6 + 0.4 = −0.2m/s∴ Putting work energy principle for 1st 2kg block when come to rest.(1/2)× 4× 02 − (1/2)× 4× (0.4)2 = −4× 0.2× 10× s2× 0.4× 0.4 = 4× 0.2× 10× s ⇒ S2 = 4cmDistance between 2kg amp; 4 kg block = S1 + S2 = 1 + 4 = 5cm
A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown in figure(9-E21). Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end
Solutions :The block ’m’ will slide down the inclined plane of mass M with acceleration a1gsinα (relative) to the inclined plane.
The horizontal component of a1 will be, ax = gsinαcosα, for which the block M will accelerate towards left. Let, the acceleration be a2.According to the concept of center of mass, (in horizontal direction external force is zero).max = (M +m)a2
⇒ a2 = maxM+m
= mgsinαcosαM+m
....(1)
So, the absolute (Resultant) acceleration of ’m’ on the block ’M’ along the direction of incline will be
a = gsinα− a2cosα = gsinα− mgsinαcos2αM+m
= gsinα
(1− mcos2α
M+m
)
= gsinα
(M+m−mcos2α
M+m
)
So, a = gsinα
(M+msin2αM+m
)
Let the time taken by the block ’m’ to reach the bottom end br ’t’,Now, S = ut+ (1/2)at2
⇒ hsinα
= (1/2)at2 ⇒ t =√
2asinα
So, the velocity of the bigger block after time ’t’ will be.
Vm = u+ a2t = mgsinαcosαM+m
√2h
asinα=
√2m2g2hsin2αcos2α
(M+m)2asinα
Now. Subtracting the value of a from equation (2) we get,
VM =
(2m2g2hsin2αcos2α
(M+m)2asinα× (M+m)
gsinα(M+msin2α)
)1/2
or,
VM =
(2m2g2hcos2α
(M+m)(M+msin2/alpha)
)1/2
Figure (9-E22) shows a small body of mass m placed over a larger mass M whose surface is horizontal near the smaller massand gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed v and the system is left to itself.Assume that all the surfaces are frictionless. (a) Find the speed of the larger block when the smaller block is sliding on the verticalpart. (b) Find the speed of the smaller mass when it breaks off the larger mass at height h. (c) Find the maximum height (fromthe ground) that the smaller mass ascends. (d) Show that the smaller mass will again land on the bigger one. Find the distancetraversed by the bigger block during the time when the smaller block was in its flight under gravity.
Page 75
24Tutors.com
Solutions :
The mass ’m’ is given a velocity ’v’ over the largest mass M.a) When the smaller block is travelling on the vertical part, let the velocity of the bigger block be v1 towards left.From law of conservation of momentum, (in the horizontal direction)mv = (M +m)v1
⇒ v1 = mvM+m
b) When the smaller block breaks off, let its resultant velocity is v2.From law of conservation of energy,(1/2)mv2 = (1/2)Mv2
1 + (1/2)mv22 +mgh
⇒ v22 = v2 − M
mv21 − 2gh .....(1)
⇒ v22 = V 2
(1− M
m× m2
(M+m)2
)− 2gh
⇒=
((m2+Mm+m2)
(M+m)2v2 − 2gh
)1/2
c) Now the vertical component of the velocity v2 of mass ’m’ is given by, v2y = v2
2 = v21
= M2+Mm+m2
(M+m)2v2 − 2gh− m2v2
(M+m)2[∴ v1 = mv
M+v]
⇒ v2y = M2+Mm+m2−m2
(M+m)2v2 − 2gh
⇒ v2y = Mv2
(M+m)v2 − 2gh .......(2)
To find the maximum height (from the ground), let us assume the body rises to the height ’h’ over and above ’h’,
now, (1/2)mv2y = mgh1 ⇒ h1 =
V 2y
2g.....(3)
So, Total height = h+ h1 = h+V 2y
2g= h+ Mv2
(M+m)2g− h
[From equation (2) and (3)]
⇒ H = Mv2
(M+m)2g
d) Because, the smaller mass has also got a horizontal component of velocity v1 at the time it breaks off from ’M’ (which has a velocityv1), theblock′m′willagainlandontheblock′M ′(biggerone).Letusfindoutthetimeofflightofblock′m′afteritbreaksoff.Duringtheupwardsmotion(BC),0 = vy − gt1
⇒ t1 =vyg
= 1g
(Mv2
(M+m)− 2g
)1/2
...(4) [from the equation (2)]
So, the time for which the smaller blocks was in its flight is given by,
T = 2t1 = 2g
(Mv2−2(M+m)gh
(M+m)
)1/2
So, the distance travelled by the bigger block during this time is,
S = v1T = mvM+m
× 2g
(Mv2−2(M+m)gh)1/2
(M+m)1/2
or, S =2mv(Mv2−2(M+m)gh)1/2
g(M+m)3/2
A small block of superdense material has a mass of3× 1024 kg. It is situated at a height h (much smaller than the earth’s radius) from where it falls on the earth’s surface. Find
its speed when its height from the earth’s surface has reduced to h /2. The mass of the earth is 6× 1024 kg.
Solutions :
Given h lt;lt;lt; RGmass = 6|1024kg.Mb = 3× 1024kg.Let Va → velocity of earth.Vb → velocity of thr blockThe two block are attracted by gravitational force of attraction. The gravitational potential energy stored will be the K.E. of two blocks.
~Gplm
(1
R+(h/2)− 1R+h
)= (1/2)me × v2
e + (1/2)mb × v2b
Again as the an internal forces acts
MeVe = mbVb Ve = mbVbMe
...........(2)
Putting in equation (1)
Gme ×mb
(2
2R+h− 1R+h
)= (1/2)×Me ×
m2bv
2b
M2e× v2
e + (1/2)Mb × V 2b
= (1/2)×mb × V 2b
(MbMe
+ 1
)
Page 76
24Tutors.com
⇒ GM
(2R+2h−2R−h(2R+h)(R+h)
)= (1/2)× V 2
b ×(
3×1024
6×1024 + 1
)⇒[
GM×h2R2+3Rh+h2
]= (1/2)× V 2
b × (3/2)
As h lt;lt;lt; R, if can be neglected,GM×h
2R2 = (1/2)× V 2b × (3/2) ⇒ Vb =
√2gh3
A body of mass m makes an elastic collision with another identical body at rest. Show that if the collision is not head-on, thebodies go at right angle to each other after the collision.
Solutions :
Since it is not an head on collision, th two bodies move in different dimensions. Let V1, V2 → velocities of the bodies vector collision. Since,the collision is elastic. Applying law of conservation of momentum on X-direction.mu1 +mx = mv1cosα+mv2cosβ⇒ v1cosa+ v2cosb = u1 ......(1)Putting law of conservation of momentum in y direction.0 = mv1sinα−mv2sinβ⇒ v1sinα = v2sinβ ......(2)Again ( 1
A small particle travelling with a velocity u collides elastically with a spherical body of equal mass and of radius r initially keptat rest. The centre of this spherical body is located a distance p(lt; r) away from the direction of motion of the particle (figure9-E23). Find the final velocities of the two particles.[Hint : The force acts along the normal to the sphere through the contact. Treat the collision as one- dimensional for this direction.In the tangential direction no force acts and the velocities do not change].
Solutions :
Let the mass of both the particle and the spherical body be ’m’. The particle velocity ’v’ has two components, vcosα normal to the sphereand vsinα tangential to sphere.After the collision, they will exchange their velocities. So, the spherical body will have a velocity vcosα and the particle will not have anycomponent of velocities in this direction[The collision will due to the component vcosα in the normal direction. But, the tangential velocity, of the particle vsinα will be unaffected ].
So, the velocity of the sphere = vcosα = vr
√r2 − ρ2[from (fig-2)]
And the velocity of the particle = vsinα = vρr
Two spherical balls of mass 10 kg each are placed 10 cm apart. Find the gravitational force of attraction between them.
Solutions :Gravitational force of attraction
F = GMmr2
6.67×10−11×10×10(0.1)2
= 6.67 ×10−7 N
Four particles having masses m, 2 m, 3 m and 4 m are placed at the four corners of a square of edge a. Find the gravitationalforce acting on a particle of mass m placed at the centre.
Solutions :
15. A pole of length 1′00m stands half dipped in a swimming pool with water level 50′0cm higher than the bed. The refractiveindex of water is 1′33 and sunlight is coming at an angle of 45 with the vertical. Find the length of the shadow of the pole on thebed.
Solutions :Shadow length =BA1 = BD +A1D = 0.5 + 0.5 tan r
Now. 1.33 = sin450
sinr⇒ sin r = 0.53.
Page 77
24Tutors.com
⇒ cos r =√
1− sin2r =√
1− (0.53)2 = 0.85So, tan r = 0.6235So, shadow length = (0.5) (1 + 0.6235) = 81.2 cm
The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, 1m/s and 10 m/s 2 at a certain instant. Find the amplitude and the time period of the motion.
Solutions :A = 10msec−2
Given that, at a particular instance,⇒ r = 0.0489 ≈0.049m
⇒ ω =√ax
=√
100.02
=√
500 = 10√
5ωv = 1m/sec1 = 500 (r2-0.0004)We Know that a = ω2xAgain, amplitude r is given by v = ω (
√r2 − x2)
x =2cm = 0.02m∴ r = 4.9 cmT = 2π
ω= 2
10√
5= 2x3.14
10x2.236= 0.28seconds.
⇒ v2 = ω2(r2-x2)
A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6s. At t = 0 it is at position x = 5 cmgoing towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration ofthe particle at t = 4s.
Solutions :At t = 0,x = 5cmGiven r = 10cmsin φ = 1/2 ⇒ φ = π
6
x = 10 sin[π3
x4 + π6
]= 10 sin
[8π+π
6
]so, w = 2π
T= 2π
6= π
3sec−1
Acceleration a = - w2 x = -(π2
9
)x (-10) = 10.9 ≈ 0.11cm/sec
so, 5 = 10 sin (wx0xφ) = 10 sin φ[y = r sin wt]
(ii) At t = 4 secondsAt t = 0,x =5cmtherefore Equation of displacement x = (10cm) sin
(π3
)= 10 sin
(3π2
)= 10 sin
(π + π
2
)= -10 sin
(π2
)= -10
A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from -the mean position are thekinetic and potential energies equal ?
Solutions :r2 - y2 = y2 ⇒ 2y2 = r2 ⇒ y = r√
2= 10√
2= 5√
2 cm form the mean position
r = 10cmso (1/2) m ω2 (r2-y2) = (1/2) m ω2y2
Bacause, K.E. = P.E. so (1/2) m ω2 (r2- y2) = (1/2) m ω2y2
The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s2. Find theposition(s) of the particle when the speed is 8 cm/s.
⇒ (4 - y2) = 6425⇒ y2 = 1.44 ⇒ y = ± 1.2 cm from mean position.
⇒ ω2 = 50y
= 50r
......(2)
Again to find out the position where the speed is 8m/sec.⇒ r ω = 10
∴ ω =√
100r2
= 5 sec2
⇒ 64 = 25 (4 - y2)
A particle having mass 10 g oscillates according to the equation x = (2.0cm) sin[(100 s−1 )t + π/6]. Find (a) the amplitude,the time period and the spring constant (b) the position, the velocity and the acceleration at t = 0.
Solutions :ω = 100s−1
x = (2.0cm) sin [(100s−1) t + (π/6)c) a = -ω2 x =1002 x 1 = 100 m/s2
We know that T = 2π√mk⇒ T 2 = 4π2 x m
k⇒ k = 4π2
T2 m
x = 2 cm sin(π6
)= 2 x (1/2) = 1 cm.from the mean position.
a) Amplitude = 2cm
v = A cos (0 + π6
) = 200 x√
32
= 100√
3 sec−1 = 1.73m/s
∴ T = 2π100
= π50sec = 0.063 sec
b) At t = 0m =10g= 105 dyne/cm = 100 N/M .
[because ω = 2πT
= 100sec−1]
Page 78
24Tutors.com
we know that x = A sin (ωt + φ)
The equation of motion of a particle started at t = 0 is given by x = 5 sin (20t+n/3) w here x is in centim etre and t in second.When does the particle(a) first come to rest(b) first have zero acceleration(c) first have maximum speed ?
Solutions :∴ x = 5 = Amplitude.x = 5 sin (20t+ π/3)⇒ t = π/30 sec.⇒ 20t = π - π/3 = 2π/3sin (20t+ π/3) = 1 = sin (π/2)b) a = ω2 x = ω2 [5 sin (20t+ π/3)]At the extream position, the velocity becomes ’0’.when, v is maximum i.e cos (20t+ π/3) = −1 = cos π⇒ 20t = π - π/3 = 2π/3∴ 5 = 5 sin (20t+ π/3)⇒ t = π/120 sec it first comes to rest.a) Max. displacement from the mean position = Amplitude of the particle.c) v = A ω cos (ωt+ π/3)⇒ 20t + π/3 = π/2⇒ t = π/30 sec.For a = 0,5 sin (20t+ π/3) = 0 ⇒ sin (20t+ π/3) = sin (π)
Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 π t + tan−1 - 0.75) where x isin centimetre and t in second. The motion is started at t = 0.(a) When does the particle come to rest for the first time ?(b) When does the acceleration have its maximum magnitude for the first time ?(c) When does the particle come to rest for the second time ?
Solutions :⇒ sin (50πt + 0.643) = 0a) x = 2.0 cos (50πt = tan−1 0.75)50 π t + 0.643 = 2π⇒ 50πt + 0.643 = πFor Maximum acceleration cos (50 π t + 0.643) = −1 cos π (max)(so a is max)v = dv
dt= - 100 sin (50 π t + tan−1 - 0.75) = 2.0 cos (50πt + 0.643)
c) When the particle comes to rest for second time,As the particle comes to rest for the 1st timeb) Acceleration a = dv
dt= - 100π x 50 π cos (50 π t + 0.643)
a) x = 2.0 cos (50πt = tan−1 0.75) = 2.0 cos (50πt + 0.643)⇒ t = 3.4 x 10−2 s.t = 1.6 x 10−2 sec⇒ t = 1.6 x 10−2 sec
Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change value from halfthe amplitude to the amplitude.
Solutions :Again, y2 = r sin ωt2y1 = r
2, y2 = r (for the two given position)
so, t2 - t1 = t4
- t12
= t6
⇒ r2
= r sin ωt1 ⇒ sin ωt1 = 12⇒ ωt1 = π
2⇒ 2π
tx t1 = π
6⇒ t1 - t
12
⇒ r = r sin ωt2 ⇒ sin ωt2 = 1 ⇒ ωt2 = π2⇒(
2πt
)t2 = π
2⇒ t2 - t
4Now, y1 = r sin ωt1The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant 0’1 N/m. What mass
should be attached to the spring ?
Solutions :k = 0.1 N/M T = 2π
√mk
= 2 sec [Time period of pendulum of a clock = 2 sec]
∴ m = kπ2 = 0.1
10= 0.01kg ≈ 10 gm
So, 4π2 +(mk
)= 4
A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of anequivalent simple pendulum i.e., a pendulum having frequency same as that of the block
Solutions :
Page 79
24Tutors.com
Time period of simple pendulum = 2π√
1g
Time period of spring is 2π√mk
⇒√
1g
=√mk⇒ 1
g= m
k
⇒ 1 = x (proved)Tp = Ts [Frequency is same]
⇒ 1 = mgk
= FK
= x. (Because , restoring force = weight = F =msg)
A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period0’314 s. Find the maximum force exerted by .the spring on the block.
Solutions :
x = r = 0.1 mT = 0.314 secm = 0.5 kg.Total force exerted on the block = weight of the block + spring force.f = Kx = 201.1 x 0.1 = 20N∴ Force exerted by the spring on the block is
T = 2π√mk⇒ 0.314 = 2π
√0.5k⇒ k = 200 N/m
∴ Maximum force = f + weight = 20 = 5 = 25N
A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations arestopped and the body hangs in equilibrium, find the potential energy stored in the spring.
Solutions :⇒b4 = 2π
√2k⇒ 2 = π
√2k
m = 2kg.But, we know that f = mg = kx
T = 2π√mk⇒ 4 = 2π
√2k⇒ 2 = π
√2k
⇒ 4 = π2(
2k
)⇒ k = 2π2
4⇒ k = π2
2= 5 N/M
∴ Potential Energy = (1/2) k x2 = (1/2) x 5x16 = 5 x 8 = 40 JT =4sec.⇒ x = mg
k= 2×10
5= 4
A spring stores 5 J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A block fastened to its otherend is made to undergo small oscillations. If the block makes 5 oscillations each second, what is the mass of the block ?
Solutions :So, T = 1/5sec.x = 25 cm = 0.25m⇒(1/2)kx2 = 5 ⇒ (1/2) k (0.25)2 = 5 =⇒ K = 160 N/m.f = 5Now P.E. = (1/2) Kx2
E = 5j
Again, T = 2π√mk⇒ 1
5= 2π
√m
160⇒ m = 0.16kg
A small block of mass m is kept on a bigger block of mass M which is attached to a vertical spring of spring constant k as shownin the figure. The system oscillates vertically.
Page 80
24Tutors.com
(a) Find the resultant force on the smaller block when it is displaced through a distance x above its equilibrium position.(b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude ?(c) What can be the maximum amplitude with which the two blocks may oscillate together ?
Solutions :
a) From the free body diagram,∴ R + mω2x - mg =0 ......(1)
Resultant force mω2x = mg - R⇒ mω2x = m
(k
M+m
)⇒ x = mkx
M+m
[ ω =√K/(M +m) for spring mass system ]
b) R = mg - mω2x = mg - m kM+m
x = mg - mkxM+m
For R to be smallest mω2x should be max i.e. x is maximum.The particle shoild be at the high point.
c) We have R = mg - mω2xThe two blocks may oscillates together is such a way that R is grater than 0.At limiting condition,
R = 0, mg = mω2x X = mgmω2 =
mg(M+m)mk
So, the maximum amplitude is =g(M+m)
k
The block of mass in, shown in figure (12− E2) is fastened to the spring and the block of mass m2 is placed against it.(a) Find the compression of the spring in the equilibrium position.(b) The blocks are pushed a further distance (2/h)(m1 + m2)g sinθ against the spring and released. Find the position where thetwo blocks separate.(c) What is the common speed of blocks at the time of separation ?
Solutions :
a) At the equilibrium condition,kx = (m1 +m2)g sin θ (Given)
⇒ x =(m1+m2)gsinθ
k
b) x1 = 2k
(m1 +m2) g sinθ (Given)when the system is released, it will start to make SHM
where ω =√
km1+m2
When the block lose contact, P = 0So m2 g sin θ = m2 x2 ω2 = m2x2
(k
m1+m2
)⇒ x2 = x k(m1 +m2)g sin θSo the block will lose contact with each other when the springs attain its natural length.c) Let the common speed attained by both the blocks be v.1/2 (m1 +m2)v2 - 0 = 1/2 k(x1 + x2)2 - (m1 +m2) g sin θ (x+ x1) [x+ x1 = total compression]⇒ (1/2) (m1 +m2) v2 = [(1/2)k(3/k) ((m1 +m2)g sin θ - (m1 +m2)g sin θ] (x+ x1)⇒ (1/2) (m1 +m2) v2 = [(1/2) (m1 +m2)g sin θ x (3/k) (m1 +m2)g sin θ
⇒ v =√
3k(m1+m2)
g sinθ
A particle of mass in is attatched to three springs A, B and C of equal force constants k as shown in figure (12 − E6). If theparticle is pushed slightly against the spring C and released, find the time period of oscillation.
Solutions :
P.E = (1/2)k (x+ δ)2 = (1/2) x 100(0.1 + 0.2)2 = 50 x0.09 = 4.5 Jd) Let the amplitude be’x’ which means the distance between the mean position and the extream position.Since, in SHM,the total enegy remaining constant.e) Potential Energy at the left extreme is given by.
The different value in (b) (e) and (f) do not violate law of conservation of energy as the work is done bythe external force 10N.
e) Potential Energy at the left extreme is given by.Give, k = 100 N/m,
m = 1kg and F = 10 NP.E = (1/2)k (x+ δ)2 - F (2x)
[2x = distance between two extremes]f) Potential Energy at the right extream is given by,So, 50(x+ 0.1)2 = 2.5 +10x
c) Time period = 2π√Mk
= 2π√
1100
= π5sec
So, in the extream position, compression of the spring is (x+ δ).∴ 50x2 + 0.5 + 10x = 2.5 + 10x
∴ 50x2 = 2 ⇒ x2 = 250
= 4100⇒ x = 2
10m = 20cm.
b) The below imparts a speed of 2m/s to block towards left.Since, in SHM,the total energy remaining constant.= 4.5 - 10(0.4) = 0.5J∴ 50x2 = 2 ⇒ x2 = 2
50= 4
100⇒ x = 2
10m = 20cm.
f) Potential Energy at the right extream is given by,P.E = (1/2)k (x+ δ)2 = (1/2) x 100(0.1 + 0.2)2 = 50 x0.09 = 4.5 JSo, in the extreme position, compression of the spring is (x+ δ).So, 50(x+ 0.1)2 = 2.5 +10x= (1/2) x 100 x (0.1)2 + (1/2) x 1 x 4 = 0.5 + 2 = 2.5 Ja) In the euilibrium position,
compression δ = F/k = 10/100 = 0.1m = 10 cmSo, in the extreme position, compression of the spring is (x+ δ).∴ 50x2 + 0.5 + 10x = 2.5 + 10xP.E = (1/2)k (x+ δ)2 - F (2x)
[2x = distance between two extremes]
Find the time period of the oscillation of mass m in figures (12−E4 a, b, c) What is the equivalent spring constant of the pairof springs in each case ?
Solutions :Resultant force F = F1 + F2 = (K1 +K2)xa) Equivalent spring constant = k = k1 + k2 (parallel)The equivalent spring constant k = k1 + k2
b) Let us, displace the block m towards left through displacement ’x’
Time period T = 2π√displacementAcceleration
= 2π√
xm(K1+K2)
m
=2π√
MK1+K2
T = 2π Mk
= 2πm(k1+k2)k1k2
T = 2π MK
= 2π√
MK1+K2
c) In series conn equivalent spring constant be k. So, 1k
= 1k1
+ 1k2
= k2+k1k1k2
⇒ k = k1k2k2+k1
The spring shown in figure (12−E5) is unstretched when a man starts pulling on the cord. The mass of the block is M . If theman exerts a constant force F , find(a) the amplitude and the time period of the motion of the block,(b) the energy stored in the spring when the block passes through the equilibrium position and(c) the kinetic energy of the block at this position.
Solutions :
a) We have F = kx ⇒ x = Fk
Acceleration = Fm
Time period T = 2π√displacementAcceleration
= 2π√
F/kF/m
=2π√mk
Amplitude = max displacement = F/kb) The energy stored in the spring when the block passes through the equilibrium position(1/2)kx2 = (1/2)k(F/k)2 = (1/2)k(F 2/k2) = (1/2)(F 2/k)c) At the mean position, P.E. is 0.K.E. is (1/2)kx2 = (1/2)(F 2/x)
A particle of mass in is attatched to three springs A, B and C of equal force constants k as shown in figure (12 − E6). If theparticle is pushed slightly against the spring C and released, find the time period of oscillation.
Solutions :Acceleration = 2kx
m
Page 82
24Tutors.com
Suppose the particle is pushed slightly against the spring ’C’ through displacement ’x’.[Cause : − when the body pushed again ’C’ the spring C , tries to pull the block towards XL.At that moment the spring A and B tries to
pull the block with force kx√2
and
∴ Total Resultant force = kx +√
( kx√2
) + ( kx√2
) = kx + kx = 2kx.
Tme period T = 2π displacementAcceleration
= 2π√
x2kxm
= 2π√
m2k
Total resultant force on the particle is kx due to spring C and kx√2
due to spring A B.
kx√2
respectively towards xy and xz respectively. So the total force on the block is due to the spring force ’C′ as well as the component of
two spring force A and B]
Repeat the previous exercise if the angle between each pair of springs is 120 initially.
Solutions :
In this case , if the particle ’m’ is pushed against ’C′ a by distance ’x’.Total resultant force acting on man ’m’ is given by,
F = kx + kx2
= 3kx2
[ Because net force A amp; B =√
( kx2
)2 + ( kx2
)2 + 2( kx2
)( kx2
)cos1200 = kx2
∴ a = Fm
= 3km2m
⇒ ax
= 3k2m
= ω2√
3k2m
∴ Time period T 2πω
= 2π√
2m3k
The springs shown in the figure (12− E7) are all unstretched in the beginning when a man starts pulling the block. The manexerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.
Solutions :
k2 and k3 are in series.Let equivalent spring constant be k4
∴ 1k4
= 1k2
+ 1k3
= k2+k3k2k3
⇒ k4 = k2k3k2+k3
Now k4 and k1 are in parallel.So equivalent spring constant k = k3+k4 = k2k3
k2+k3+k1 = k2k3+k1k2+k1k3
k2+k3
∴ T = 2π =√Mk
= 2π√
M(k2+k3)k2k3+k1k2+k1k3
b) frequency = 1T
= 12π
√k2k3+k1k2+k1k3
M(k2+k3)
c) Amplitude x = Fk
=F (k2+k3)
k1k2+k2k3+k1k3
Find the elastic potential energy stored in each spring shown in figure (12 − E8), when the block is in equilibrium. Also findthe time period of vertical oscillation of the block.
Solutions :
Page 83
24Tutors.com
k1k2k3 are in series.1k
= 1k1
+ 1k2
+ 1k3
⇒ k = k1k2k3k1k2+k2k3+k1k3
Time period T = 2π√mk
2π√
k1k2k3k1k2+k2k3+k1k3
= 2π√m( 1
k1+ 1k2
+ 1k3
)
Now, Force = weight = mg.∴ At k1 spring, x1 = mg
k1Similarly x2 = mg
k2and x3 = mg
k3∴ PE1 =(1/2)
k1x21
= 12k1(Mg
k1)2 = 1
2k1m2g2
k21= m2g2
2k1
Similarly PE2 = m2g2
2k2and PE3 = m2g2
2k3
The string, the spring and the pulley shown in figure (12− E9) are light. Find the time period of the mass m.
Solutions :
When only ’m’ is hanging, let the extension in the spring be ’l’So T1 = kl = mg.When a force F is applied, let the further extension be’x’∴ T2 = k(x+ l)∴ Driving force = T2 - T1 = k(x+ l) - kl = kx∴ Acceleration = kl
m
T = 2π√displacementAcceleration
= 2π√
xKxm
= 2πmk
Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it
Solutions :
Page 84
24Tutors.com
Let us solve the problem by ’energy method’.initial extension of the spring in the mean position.δ = mg
kDuring osillation, at any position ’x’ below the equilibrium position, let the velocity of ’m’ be v and
angular velocity of the pulley be ’ω’. if r is the radius of the pulley, then v = rω.At any instant, Total Energy = constant (for SHM)
(δ = mg/k)Taking derived of both sides either respect to ’t’.mv dv
dt+ I
r2V dvdt
+kx dvdt
= 0
⇒ a(m+ I
r2
)= kx
(∴ x = dxdt
anda= dxdt
)
⇒ ax
= k
m+ Ir2
= ω2 ⇒ = T = 2π
√m+ I
r2
k
Consider the situation shown in figure (12 − E10). Show that if the blocks are displaced slightly in opposite directions andreleased, they will execute simple harmonic motion. Calculate the time period
Solutions :
The center of mass of the system should not change during the motion. So,if the block ’m’ on the leftmoves towards right a distance ’x’,the block on the right moves left a distance ’x’.So,totalcompression of the spring is 2x.
By energy method, 12k(2x)2+ 1
2mv2 + 1
2mv2 + 1
2mv2 = C ⇒ mv2 + 2kx2 = C.
Taking derivation of both sides with respect to ’t’.m x 2v dv
dt+ 2k x 2x dx
dt= 0
∴ ma + 2kx = 0[because v = dx/dt and a = dv/dt]
⇒ ax
= − 2km
= ω2 ω =√
2km
⇒ Time period T = 2π√
m2k
A rectangular plate of sides a and b is suspended from a ceiling by two parallel strings of length L each (figure12−E11). Theseparation between the strings is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will executesimple harmonic motion. Find the time period.
Solutions :For small angle θ, sin θ = θ.Here we have to consider oscillation of center of mass∴ a = x1
Acceleration = a = Fm
= g sin θ∴ a = g θ = g( x
L)
[where g and L are constant]
Time period T = 2π√DisplacementAcceleration
= 2π√
x( gxL
)= 2π
√Lg
Driving force F = mg sin θ.So the motion is simple Harmonic
A 1 kg block is executing simple harmonic motion of amplitude 0’1 m on a smooth horizontal surface under the restoring forceof a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the meanposition. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.
Solutions :
Page 85
24Tutors.com
Total mass = 3 + 1 = 4kg (when both the blocks are moving together)∴ Initial momentum = Final momentum∴ (1/2) x (1 x v2) = (1/2) x 100 (0.1)2
∴ Frequency = 52π
Hz.
∴ 1/4 = 100 δ2 ⇒ δ =√
1400
= 0.05m = 5cm.
Now the two block have velocity 1/4 m.s. at its mean poison.Amplitude = 0.1 mAfter the 3kg block is gentely placed on the 1kg, then let, 1kg + 3kg = 4kg block and the spring be one
system.For this mass spring system, there is do external force. (when oscillation takes place). theThe momentum should be conserved. Let, 4kg block has velocity v|.
KE. = (1/2)mv2 = (1/2)mx2
Where x→ Amplitude = 0.1 m.When the block are going to the extreme position, there will be only potential energy.
∴ T = 2π√Mk
= 2π√
4100
= 2π5sec.
∴ PE = (1/2)kδ2 = (1/2) x (1/4) where δ → new amplitude.∴ 1 x v = 4 x v| ⇒ v| = 1/4 m/s
(as v = 1m/s from equation (1))⇒ v = 1m/sec ..........(1)So amplitude = 5cm.Again at the mean position, let 1kg block has velocity v.KEmass = (1/2)m|v2 = (1/2) 4 x (1/4)2 = (1/2) x (1/4).
The left block in figure (12−E13) moves at a speed v towards the right block placed in equilibrium. All collisions to take placeare elastic and the surfaces are frictionless. Show that the motions of the two blocks are periodic. Find the time period of theseperiodic motions. Neglect the widths of the blocks.
Solutions :
is LV
+ LV
= 2( LV
)When the block A moves with velocity ’V ’ and collides withe the block B, it transfers all energy to the
block B.(Because it is a elastic colosion).The block A will move a distance ′x’ against the spring, again the block B will return to the originalpoint and completes half of the oscillation.
The block B collides with the block A and comes to rest at the point.The block A again moves a further distance ’L’ to return to its original position.∴ Time taken by the block to move from M → N and N → M
∴ So time period of the periodic motion is 2( LV
) + π√mk
So, the time period of B is2π√mk
2= π
√mk
Find the time period of the motion of the particle shown in figure (12 − E14). Neglect the small effect of the bend near thebottom.
Solutions :
Let the time taken to travel AB and BC be t1 and t2 respectivelyFor part AB,a1 = g sin 450. s1 = 0.1
sin450 = 2mLet v = velocity at B
∴ v2 - u2 = 2a1s1⇒ v2 = 2 x g sin 450. s1 = 0.1
sin450 = 2
⇒ v =√
2m/s
∴ t1 = v−ua1
=√
2−0g√2
= 2g
= 210
= 0.2 sec
Again for part BC , a2 = −g sin 600 , u =√
2 , v = 0
∴ t2 = 0−√
2
−g(√
32
)= 2√
2√3g
=2×(1.414)(1.732)×10
= 0.165sec.
So, time period = 2(t1 + t2) = 2(0.2+0.1555) = 0.71 sec
All the surfaces shown in figure (12−E15) are frictionless. The mass of the car is M , that of the block is m and the spring hasspring constant k. Initially, the car and the block are at rest and the spring is stretched through a length x0, when the system isreleased.
Page 86
24Tutors.com
(a) Find the amplitudes of the simple harmonic motion of the block and of the car as seen from the road.(b) Find the time period(s) of the two simple harmonic motions.
Solutions :Let the amplitude of oscillation of ’m’ and ’M ’ be x1 and x2 respectively.
a) From law of conservation of momentum,mX1 = mX2 .....(1) [ because only internal force are present ]Again, (1/2) kx2
g = (1/2) k (x1 + x2)2
∴ x0 = x1 + x2 .....(2)[ Block and mass oscillation in opposite direction. But x → stretched part ]From equation (1) and (2)∴ x0 = x1 + m
Mx1 = (M+m
M) x1
∴ x1mx0M+m
So , x2 = x0 - x1 = x0 [1− MM+m
] = mx0M+m
respectively.
b) At any position , let the velocity be v1 and v2 respectively.By energy method
Total Energy = Constant(1/2) Mv2 + (1/2) m(V1 − V2)2 + (1/2)k(X1 +X2)2 = Constant ...(i)[ v1 - v2 = Absolute velocity of mass ’m’ as seen from the road.]Again , from law of conservation of momentum.12Mv2
2 + 12mM2
m2 v22 + 1
2kx2
2 (1 + Mm
)2 = constant
⇒ a2X2
= -k(M+m)Mm
= ω2
Here, v1 = velocity of ’m’ with respect to M .Taking derivation of both sides,
So, Time period, T = 2π√
Mmk(M+m)
⇒ mv22 + k(1 + M
m)2 x2
2 = constant
⇒ ma2 + k(M+m)m
x2 = 0 [ because , v2 = dx2dt
]putting the above value in equation (1), we get
∴ ω =√k(M+m)Mm
mv2 = m(v1 − v2) ⇒ (v1 − v2) = Mmv2 ....(2)
∴ M(1 + Mm
)v2 + k(1 + Mm
)2 x22 = constant.
M x 2v2dv2dt
+ k(M+m)m
-ex22dv2dt
= 0
A uniform plate of mass M stays horizontally and symmetrically on two wheels rotating in opposite directions (figure12−E16).The separation between the wheels is L. The friction coefficient between each wheel and the plate is N . Find the time period ofoscillation of the plate if it is slightly displayed along its length and released
Solutions :Taking moment about G, we getLet ’x’ be the displacement of the plank towards left.Now the center of gravity is also displaced through ’x’
R1 + R2 = mgSince, F1 gt; F2 ⇒ F1 - F2 = ma = 2µmg
ιx
⇒ R1 =mg(2x+ι)
2ι....(2)
R1( ι2
) - R1x = mg ι2
- R1x + mgx - R1ι2
⇒ R1ι2
+ R1ι2
= mg(x+ ι2
)In displaced position
Similarly F2 = µR2 =µmg(ι−2x)
2ι⇒ R1
ι2
-R1x = mg ι2
- R1x + mgx - R1ι2
∴ Time period = 2π√
ι2rg
⇒ R1ι =mg(2x+ι)
2
A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of thistype. Find, the length of a seconds pendulum at a place where g = π2m/s2.
Solutions :T = 2 sec.⇒ 2 = 2π
√ιg⇒ ι
g= 1
π2 ⇒ ι = 1 cm ( ∴ π2 ≈ 10 )
T = 2π√
ιg
The angle made by the string of a simple pendulum with the vertical depends on time as θ = π90sin [(πs−1)t].Find the length
of the pendulum if g = π2m/s2.
Page 87
24Tutors.com
Solutions :⇒ 2π
T= π ⇒ T = 2 sec.
From the equation.∴ Length of the pendulum is 1m.∴ ω = πsec−1 ( comparing with the equation of S.H.M)
We know that T = 2π√
ιg⇒ 2 = 2
√ιg⇒ 1 =
√ιg⇒ ι = 1m.
θ = π sin [(πsec−1)t]
The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours ?
The pendulum of the clock has time period 2.04sec.So in one day it is slower by,Given that, T1 =2sec ,g1 = 9.8m/s2
But , in each oscillation it is slower by (2.04− 2.00) = 0.04secSo, the clock runs 28.8 minutes slower in one day.
A pendulum clock giving correct time at a place where g = 9.800 m/s2 is taken to another place where it loses 24 seconds during24 hours. Find the value of g at this new place.
Solutions :For the pendulum, T1
T2=√g2g1
Given that, T1 =2sec ,g1 = 9.8m/s2
T2 = 24×3600
( 24×3600−24 )= 2 x 3600
3599
Now , g2g1
= (T1T2
)2
∴ g2 = (9.8) ( 35993600
)2 = 9.795m/s2
A simple pendulum is constructed by hanging a heavy ball by a 5.0 m long string. It undergoes small oscillations.(a) How many oscillations does it make per second ?(b) What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is 1.67 m/s 2.
Solutions := 2π 5
1.76L = 5m.b) When it is taken to the moon T = 2π ι
g′
where g′ → Acceleration in the moon.
∴ f = 1T
= 12π
√1.67
5= 1
2π(0.577) = 1
2π√
3times.
a) T = 2π√
ιg
= 2π√
0.5 = 2π(0.7)
∴ In 2π(0.7)sec, the body completes 1 oscillation,In 1 second, the body will complete 1
2π(0.7)oscillation
∴ f = 12π(0.7)
= 1014π
= 0.70π
times
The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude.
Solutions :The tension in the pendulum is maximum at the mean position and minimum on the extreme positionHere (1/2) mv2 - 0 = mg ι(1 - cosθ)
A small block oscillates back and forth on a smooth concave surface of radius R (figure12 − E17). Find the time period ofsmall oscillation.
Solutions :
Given that, R = radius.Let N = normal reaction.Driving force F = mg sinθ.Acceleration = a = mg sinθ
Page 88
24Tutors.com
As, sinθ is very small, sinθ → θLet x be the displacement from the mean position of the body,∴ θ = x
R⇒ a = gθ = g(x/R) ⇒ (a/x) = (g/R)So the body makes S.H.M.
∴ T = 2π√DisplacementAcceleration
= 2π√
xgx/R
= 2π√Rg
therefore Acceleration a = gθ
A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes smalloscillations about the lowest point, Find the time period.
Solutions :
Let the angular velocity of the system about the point os suspension at any time be ’ω’So, vc = (R− r)ωAgain vc = rω1 [where ,ω1 = rotational velocity of the sphere]
ω1 = vcr
= (R+(−r)
r)ω ....(1)
By Energy method, Total energy in SHM is constant.So, mg(R− r)(1− cosθ) + (1/2)mv2
A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep.Calculate the time period of the pendulum there. Radius of the earth = 6400 km
Solutions := 2π
√0.47.35
= 2π√
0.054 = 2π x 0.023 = 2 x 3.14 x 0.23 = 1.465 ≈ 1.47sec.
Length of the pendulum = 40cm = 0.4.Let acceleration due to gravity be g at the depth of 1600km.
∴ Time period T′
= 2πsqrt lgδ
∴ gd = g(1− d/R) = 9.8 (1− 16006400
) = 9.8 (1− 14
) =9.8 x 34
= 7.35m/s2
Assume that a tunnel is dug across the earth (radius = R) passing through its centre. Find the time a particle takes to coverthe length of the tunnel if(a) it is projected into the tunnel with a speed of
√gR
(b) it is released from a height R above the tunnel(c) it is thrown vertically upward along the length of tunnel with a speed of
√gR.
Solutions :
Page 89
24Tutors.com
Let M be the total mass of the earth.At any position x,
∴ M′
M=
ρ×( 43
)π×x3
ρ×( 43
)π×R3 = x3
R3 ⇒ M′
= Mx3
R3
So force on the particle is given by,
∴ FX = GM′m
x2= GMm
R3 ......(1)So, acceleration of the mass ’M ’ at the position is given by,ax = GM
R2 x ⇒ axx
= w2 = GMR3 = g
R(∴ Gm
R2 )
So, T = 2π√Rg
= Time period of oscillation.
a) Now, using velocity - displacement equation.
V = ω√
(A2 −R2) [Where, A = amplitude]
Given when, y = R, v =√gR , ω =
√gR
⇒√gR =
√gR
√(A2 −R2) [because ω =
√gR
]
⇒ R2 A2 - R2 ⇒ A =√
2R[Now, the phase of the particle at the point P is greater than π/2 but less than π and at Q is greater than π but less than 3π/2. Let the
times taken by the particle to reach the positions P and Q be t1 amp; t2 respectively, then using displacement time equation]y = r sin ωtWe have ,R =
√2 R sin ωt1
⇒ωt1 = 3π/4amp; −R =
√2 R sin ωt2
⇒ωt2 = 5π/4So , ω (t1 − t2) = π/2 ⇒ t1 − t2 = π
2ω= π
2√
(R/g)
Time taken by the particle to travel from P to Q is t2t1 π
2√
(R/g)sec.
b) When the body is dropped from a height R, then applying conservation of energy, change in P.E. = gain in K.E.⇒ GMm
R- GMm
2R= 1
2mv2 ⇒ v =
√gR
Since, the velocity is same at P , as in part (a) the body will take same time to travel PQ.c) When the body is projected vertically upward from P with a velocity
√gR, its velocity will be Zero at the highest point.
The velocity of the body, when reaches P , again will be v =√gR , hence, the body will take same
time π
2√
(R/g)to travel PQ.
Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance R/2 from the earth’s centre where R is theradius of the earth. The wall of the tunnel is frictionless.(a) Find the gravitational force exerted by the earth on a particle of mass m placed in the tunnel at a distance x from the centreof the tunnel.(b) Find the component of this force along the tunnel and perpendicular to the tunnel.(c) Find the normal force exerted by the wall on the particle.(d) Find the resultant force on the particle.(e) Show that the motion of the particle in the tunnel is simple harmonic and find the time period.
Solutions :
M = 4/3 πR3ρ.M1 = 4/3 πx3
1ρ.
m1 = ( MR3 )x3
1a) F = Gravitational force exerted by the earth on the particle of mass x is,
F = Gm1mx21
= GMmR3
x31x21
= GMmR3 x1 = GMm
R3
√x2 + (R
2
4)
b) Fy = F cos θ = GMmR3
xX1
= GMmR3
Fx = F sin θ = GMmx1R3
R2X1
= GMm2R2
c) Fx = GMm2R2 since Normal force exerted by the wall N = Fx]
d) Resultant force = GMmxR3
Page 90
24Tutors.com
e) Acceleration = Drivingforcemass
= GMmxR3m
GmxR3
So, a α x (The body makes SHM)
∴ ax
= W 2 = GMR3 ⇒ w =
√GMR3 ⇒ T = 2π
√R3
GM
A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if theelevator(a) is going up with an acceleration a0
(b) is going down with an acceleration a0, and(c) is moving with a uniform velocity.
Solutions :
Here driving force F = m(g + a0) sin θ ....(1)
Acceleration a = Fm
= (g + a0)sinθ =(g+a0)x
l(Because when θ is small sin θ → θ = x
l
∴ a =(g+a0)x
l.
∴ acceleration is proportional to displacement.So, the motion is SHM.
Now ω2 =(g+a0)
l.
∴ T = 2π√
lg+a0
b) When the elevator is going downwards with acceleration a0
Driving force = F = m (g − a0) sinθ.
Acceleration = (g − a0) sinθ =(g−a0)x
l= ω2x
T = 2πω
= 2π√
lg−a0
c) When moving with uniform velocity a0 = 0.For, the simple pendulum, driving force = mgx
l
⇒ a = gxl
= ⇒ xa
= lg
T = 2π√displacementacceleration
2π√
lg
A simple pendulum of length 1 feet suspended from the ceiling of an elevator takes π/3 seconds to complete one oscillation.Find the acceleration of the elevator.
Solutions :
Let the elevator be moving upward accelerating a0
Here driving force F = m(g + a0) sin θAcceleration = (g + a0) sin θ
= (g + a0) θ (sintheta→ θ)(g+a0)x
l= ω2x
T = 2π√
lg+a0
Given that, T = π/3 Sec,l = 1ft and g = 32 ft/sec2
π3
= 2π√
132+a0
19
= 4( 132+a
)⇒ 32 + a = 36
⇒ a = 36 - 32 = 4ft/sec2
A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. Whenthe accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of thecar.
Solutions :⇒ 3.99 = 2π
√l
g2+a20When the car moving with uniform velocitySolving for ’a0’ we can get a0 = g/10 ms−2
When the car makes accelerated motion, let the acceleration be a0
T = 2π√
lg2+a20
Page 91
24Tutors.com
Now T
T′ = 4
3.99=
(g2+a20)1/4√g
T = 2π√
lg⇒ 4 = 2π
√lg
...(1)
A simple pendulum of length l is suspended from the ceiling of a car moving with a speed v on a circular horizontal road ofradius r.(a) Find the tension in the string when it is at rest with respect to the car.(b) Find the time period of small oscillation.
Solutions :
From the freebody diagram,
T =√
(mg2 + (mv2
r2)
=M√g2 + v4
r2= ma , where a = acceleration = (g2 + v4
r2)1/2
The time period of small accellations is given by,
T = 2π sqrt lg
= 2π√
l
(g2+ v4
r2)1/2
The ear-ring of a lady shown in figure (12− E18) has a 3 cm long light suspension wire.(a)Find the time period of small oscillations if the lady is standing on the ground.(b) The lady now sits in a merry-go-round moving at 4 m/s in a circle of radius 2 m.Find the time period of small oscillations of the ear-ring.
Solutions :
a) l = 3cm = 0.03m
T = 2π√
lg
= 2π√
0.039.8
= 0.34 second.
b) When the lady sets on the Merry-go-round the ear rings also experience centrepetal acceleration
a = v2
r= 42
2= 8 m/s2
Resultant Acceleration A =√g2 + a2 =
√100 + 64 = 12.8 m/s2
Time period T = 2π√
lA
= 2π√
0.0312.8
= 0.30 second.
Find the timik period of small oscillations of the following systems.(a) A metre stick suspended through the 20 cm mark. .(b) A ring of mass in and radius r suspended through a point on its periphery. .(c) A uniform square plate of edge a suspended through a corner. .(d) A unifrom disc of mass m and radius r suspended through a point r/2 away from the centre.
Solutions :
a) M.I about the pt A = l = lC.G + Mh2
= Ml2
12+ MH2 = Ml2
12+ m(0.3)2 = M(
112
0.09) = M ( 1+1.08
12) = M ( 2.08
12)
∴ T = 2π√
l
mgl′ = 2π 2.08m
m×9.8× 0.3(l′
= dis. between C.G and pt. of suspension)
≈ 1.52sec.
Page 92
24Tutors.com
b) Moment of in isertia about Al = lC.G + mr2 = mr2 + mr2 = 2mr2
∴ Time period = 2π√
lmgl
= 2π√
2mr2
mgr= 2π
√2rg
c) lzz (comes) = m (a2+a2
3) = 2ma2
3
In the ∆ABC, l2+l2 = a2
∴ l = a√2
∴ T = 2π√
lmgl
= 2π√
2ma2
3mgl= 2π
√2a2
3ga√
2= 2π
√√8a
3g
d) h = r/2 , l = r/2 = Dist. Between C.G and suspension point.
M.L. about A, l = lC.G. + Mh2 = mc2
2+ n( r
2)2 = mr2 ( 1
2+ 1
4) = 3
4mr2
∴ T = 2π√
lmgl
= 2π√
3Mr2
4mgl= 2π
√3R2
4g( r2
)= 2π
√3r2g
A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simplependulum having the time period equal to that of the rod.
Solutions :Let, the time period T is equal to the time period of simple pendulum of length x.Let A → suspension of point.
B → Center of Gravity.
l′
= l/2, h = l/2Moment of inertia about A is
⇒ T = 2π√
l
mg l2
= 2π√
2ml2
3mgl= 2π
√2l3g
∴ T = 2π√xg
. So, 2l3g
= xg⇒ x = 2l
3
∴ Length of the simple pendulum = 2l3
l = lC.G. + mh2 = ml2
12+ ml2
4= ml2
3
A uniform disc of radius r is to be suspended through a small hole made in the disc. Find the minimum possible time period ofthe disc for small oscillations. What should be the distance of the hole from the centre for it to have minimum time period ?
Solutions :For T is minimum dt2
dx= 0
Suppose that the point is x distance from C.G.Let m = mass of the disc., Radius = rHere l = x⇒ 2π2r2
g
(− 1x2
)+ 4π2
g= 0
So putting the value of equation (1)
T 2π√
lmgl
= 2π
√m(r2
2+X2
)mgx
= 2πm(r2/2+x2)
2mgx= 2π
√r2+2x2
2gx....(1)
∴ ddxT 2 = d
dx
(4π2r2
2gx+ 4π22x2
2gx
)⇒ −π
2r2
gx2+ 2π2
g⇒ 2x2 = r2 ⇒ x = r√
2
M.I. about A = lC.G. + mx2 = mr2/2 +mx2 = m(r2/2 + x2)
⇒ −π2r2
gx2+ 2π2
g= 0
T = 2π
√r2+(r2
2
)2gx
= 2π√
2r2
2gx= 2π
√r2
g( r√2
)2π√√
2r2
gr= 2π
√√2rg
A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation ofthis pendulum. How does it differ from the time period calculated using the formula for a simple pendulum ?
Solutions :
Page 93
24Tutors.com
According to Energy equation,mgl(1− cosθ) + (1/2) lω2 = const.mg(0.2)(1− cosθ) + (1/2) lω2 = C.
Again, l = 2/3 m(0.2)2 + m(0.2)2
= m[
0.0083
+ 0.04]
= m(
0.12083
)m. Where I → Moment of Inertia about the pt of suspension A
From equationDifferenting and putting the value of I and 1 is
ddt
[mg(0.2)(1− cosθ) + 1
20.1208
3mω2
]= d
dt(C)
⇒ mg(0.2)sinθ + dθdt
+ 12
(0.1208
3
)m2ω dω
dt= 0
For simple pendulum T = 2π√
0.1910
= 0.86sec.
⇒ 2 sinθ = 0.12083
α[because, g = 10m/s2
]⇒ α
θ= 6
0.1208= ω2 = 58.36
⇒ ω = 7.3. So T = 2πω
= 0.89sec.∴ it is about 0.3
A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 2and time period 2 s. Find(a) the radius of the circular wire,(b) the speed of the particle farthest away from the point of suspension as it goes through its mean position,(c) the acceleration of this particle as it goes through its mean position and(d) the acceleration of this particle when it is at an extreme position.Take g = π2m/s2.
Solutions :
(For a compound pendulum)
a) T = 2π√
lmgl
= 2π√
lmgr
The MI of the circular wire about the point of suspension is given by∴ I = mr2 + mr2 = 2 mr2 is Moment of inertia about A.
∴ 2 = 2π√
2mr2mgr = 2π√
2rg
⇒ 2rg
= 1π2 ⇒ r = g
2π2 = 0.5π = 50cm.(Ans)
b) (1/2) ω2 - 0 = mgr (1− cosθ)⇒ (1/2) 2mr2 - ω2 = mgr (1− cos20)⇒ ω2 = g/r (1− cos20)⇒ ω = 0.11 rad/sec [ putting the value of g and r ]⇒ v = ω x 1r = 11 cm/sec.c) Acceleration at the end position will be centripetal.
= an = ω2 (2r) = (0.11)2 100 = 1.2 cm/s2
The direction of an is towards the point of suspension.
So, tangential acceleration = α(2r) = 2π3
180x 100 = 34 cm/s2.
α = ω2θ = π2 x 2π180
= 2π3
180[10 = π
180radious]
d) At the extreme position the centrepetal acceleration will be zero. But, the particle will still haveacceleration due to the SHM.Because, T = 2 sec.Angular frequency ω = 2π
T(π = 3.14)
So, angular acceleration at the extreme position,
A uniform disc of mass m and radius r is suspended through a wire attached to its centre. If the time period of the torsionaloscillations be T , what is the torsional constant of the wire.
Solutions :
M.I. of the centre of the disc. = mr2/2
Page 94
24Tutors.com
T = 2π√
lk
= 2π√mr2
2k[where K = Torsional constant]
T 2 = 4π2 mr2
2k= 2π2 mr2
k
⇒ 2π2 mr2 = KT 2 ⇒ K = 2mr2π2
T2
∴ Torsional constant K = 2mr2π2
T2
Two small balls, each of mass m are connected by a light rigid rod of length L. The system is suspended from its centre by athin wire of torsional constant k. The rod is rotated about the wire through an angle θ0, and released. Find the tension in the rod’as the system passes through the mean position.
Solutions :
The M.I of the two ball systeml = 2m(L/2)2 = m L2/2At any position θ during the oscillation, [fig-2]
Torque = kθ So, work done during the displacement 0 to θ0,
W =∫ ba kθdθ = k θ2
0/2By work energy method,
(1/2) l ω2 - 0 = Work done = k θ20/2
∴ ω2 =kθ202l
=kθ20mL
Now, from the freebody diagram of the rod,T 2 =
√(mω2L2) + (mg)2
=
√(m
kθ20mL2 × L
)2+m2g2 =
k2θ40L2 +m2g2
A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the firstmotion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a)0, (b) 60, (c) 90.
Solutions :b) When φ = 600,
R =√
(32 + 42 + 2× 3× 4cos600) = 6.1cmc) When φ = 900,
R =√
(32 + 42 + 2× 3× 4cos900) = 5cmThe particle is subjected to two SHMs of same time period in the same direction/ Given, r1 = 3cm, r2 = 4cm and φ = phase difference.a) When φ = 00,
R =√
(32 + 42 + 2× 3× 4cos00) = 7cm
Resultant amplitude = R =√r21 + r2
2 + 2r1r2cosφ
Three simple harmonic motions of equal amplitudes A and equal time periods in the same direction combine. The phase ofthe second motion is 60 ahead of the first and the phase of the third motion is 60 ahead of the second. Find the amplitude of theresultant motion.
Solutions :
Three SHMs of equal amplitudes A and equal time periods in the same dirction combine.The vectors representing the three SHMs are shown it the figure.Using vector method,Resultant amplitude = Vector sum of the three vectors= A +A cos 600 + A cos 600 = A + A/2 + A/2 = 2ASo the amplitude of the resultant motion is 2A.
A particle is subjected to two simple harmonic motions given byx1 = 2.0 sin(100t) and x2 = 2.0 sin(120πt+ π/3) where x is in centimeter and t in second. Find the displacement of the particleat(a) t = 0.0125,(b) t = 0.025.
Solutions :x1 = 2 sin 100 πt
x2 = W sin (120πt+ π/3)So, resultant displacement is given by,x = x1 + x2 = 2 [sin (100πt) + sin (120πt+ π/3)]
b) At t = 0.025s,x = 2 [sin (100π × 0.025) + sin (120π × 0.025 + π/3)]= 2 [ sin 5π/2 + sin (3π + π/3)]= 2 [1 + (−0.8666)] = 0.27cm.
a) At t = 0.0125s,x = 2 [sin (100π × 0.0125) + sin (120π × 0.0125 + π/3)]
Page 95
24Tutors.com
= 2 [ sin 5π/4 + sin (3π/2 + π/3)]= 2 [(−0.707) + (−0.5)] = −2.41cm.
A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of 45with the X-axis, The two motions are given by x = x0 sinωt and s = s0 sinωt Find the amplitude of the resultant motion.
Solutions :The particle is subjected to two simple harmonic motions represented by,
x = x0 sin wts = s0 sin wtand, angle between two motions = θ = 45∴ Resultant motion will be given by,
R =√
(x2 + s2 + 2xscos450)
=√x2
0sin2wt+ s20sin
2wt+ 2x0s)0sin2wtx(1/√
2)
= [x+ 02 + s20 =√
2x0s0]1/2 sin wt
∴ Resultant amplitude = [x+ 02 + s20 =√
2x0s0]1/2
A load of 10 kg is suspended by a metal wire 3 m long and having a cross-sectional area 4 mm2. Find (a) the stress (b) thestrain and (c) the elongation. Young’s modulus of the metal is 2 · 0 x 1011 N/m2.
Solutions :Y = FL
A∆L⇒ ∆L
L= F
Y AF = mgStrain ∆L
L
Stress = FA
A vertical metal cylinder of radius 2 cm and length 2 m is fixed at the lower end and a load of 100 kg is put on it. Find (a) thestress (b) the strain and (c) the compression of the cylinder. Young’s modulus of the metal = 2 x 1011 N/m2.
Solutions :ρ = stress = mg
ACompression ∆L = eLe = strain = ρ
Y
The elastic limit of steel is 8 x 108 N/m2 and its Young’s modulus 2 x 1011 N/m2. Find the maximum elongation of a half-metersteel wire that can be given without exceeding the elastic limit.
Solutions :Y = F
A. L
∆L⇒ ∆L = FL
AY
A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination issubjected to a tension. Find the ratio of (a) the stresses developed in the two wires and (b) the strains developed. Y of steel = 2 x1011 N/m2. Y of copper = 1 · 3 X 1011 N/m2.
In figure (14− E1) the upper wire is made of steel and the lower of copper. The wires have equal cross-section. Find the ratioof the longitudinal strains developed in the two wires.
Solutions :(∵ Acu = Ast) = Ycu
Yst(∆LL
)st
= FAYst
Strain steel wireStrain om copper wire
= FAYst
x AYcuF(
∆LL
)cu
= FAYcu
The two wires shown in figure (14−E2) are made of the same material which has a breaking stress of 8 x 108 N/m2. The areaof cross-section of the upper wire is 0 · 006 cm2 and that of the lower wire is 0 · 003 cm2. The mass m1 = 10 kg, m2 = 20 kg andthe hanger is light. (a) Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break firstif the load is increased ? (b) Repeat the above part if m1 = 10 kg and m2 = 36 kg.
Solutions :T1A1
= m1g + ωgA1
= 8 x 108 ⇒ w = 14 kg
Stress in the lower rod = T1A1⇒ m1g + ωg
A1⇒ w = 14 kg
The maximum load that can be put is 2kg. Upper wire will break first if load is increasedFor same stress, the max load that can be put bis 14 kg. If the load is increased the lower wire will break first.T2Au⇒ m2g + m1g + ωg
Au= 8 x 108 ⇒ ω0 = 2 kg
Stress in upo rod = T2Au⇒ m2g + m1g + wg
Au⇒ w = 0.18 kg
Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find the Young’s modulus ofthe material of the rope if it extends in length by 1 cm. Original length of the rope = 2 m and the area of cross-section = 2 cm2.
Solutions :Y = F
A. L
∆L
A steel rod of cross-sectional area 4 cm2 and length 2 m shrinks by 0 · 1 cm as the temperature decreases in night. If the rod isclamped at both ends during the day hours, find the tension developed in it during night hours. Young’s modulus of steel = 1 · 9 x1011 N/m2.
Solutions :Y = F
AL
∆L⇒ F = Y A∆L
L
Consider the situation shown in figure (14− E3). The force F is equal to the m2 g/2. If the area of cross-section of the stringis A and its Young’s modulus Y , find the strain developed in it. The string is light and there is no friction anywhere.
Page 96
24Tutors.com
Solutions :From equation (1) and (2) , we get m2g
2(m1 + m2)
m2g - T = m2a ....(1)Now, Y = FL
A∆L⇒ ∆L
L= F
AY
⇒ a = m2g − Fm1 + m2
Again , T = F + m1aand T - F = m1a ...(2)
⇒ ∆LL
=(m2
2 + 2m1m2)g
2(m1 + m2)AY=
m2g(m2 + 2m1)2AY (m1 + m2)
A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium, thereis a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ withthe vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young’s modulus of the metalof the wire is 2 · 0 x 1011 N/m2. Make appropriate approximations.
Solutions :The cahneg in tension is due to centrifugal force ∆T = mv2
r.......(1)
At equilibrium ⇒ T = mg⇒ 1
2mv2 - 0 = mgr(1− cosθ)
⇒ F = ∆T⇒ T ′ = mg + mv2
r⇒ Again, by work energy principle,
So, ∆T =m [ 2gr(1−cosθ) ]
r= 2mg(1− cosθ)
When it moves to angle θ, and released, the tension the T at lowest point is⇒ v2 = 2gr(1− cosθ) ......(2)⇒ F =Y A∆L
L= 2mg - 2mg cosθ ⇒ 2mg cosθ = 2mg - Y A∆L
L= cosθ = 1 - Y A∆L
L(2mg)
A steel wire of original length 1 m and cross-sectional area 4 · 00 mm2 is clamped at the two ends so that it lies horizontallyand without tension. If a load of 2 · 16 kg is suspended from the middle point of the wire, what would be its vertical depression ?Y of the steel = 2.0 x 1011 N/m2. Take g = 10 m/s2
Solutions :So, ∆L = 2(I2 + x2)
12 - 100 .....(2)
Y= FA
I∆I
....(3)
Here, AC = (I2 + x2)12
From equation (1), (2) and (3) and the frebody diagram,2I cosθ = mg.
Increase in length ∆L = (AC + CB) - AB
A copper wire of cross-sectional area 0 ·01 cm2 is under a tension of 20 N . Find the decrease in the cross-sectional area. Young’smodulus of copper = 1 · 1 x 10114N/m2 and Poisson’s ratio = 0 · 32. [Hint : ∆A
A= 2 ∆r
r]
Solutions :⇒ ∆A = 2∆r
r
y = FLA∆L
⇒ ∆LL
= FAy
Again, ∆AA
= 2∆rr
σ =∆D/D∆L/L
⇒ ∆DD
= ∆LL
Find the increase in pressure required to decrease the volume of a water sample by 0 · 01water = 2 · 1 x 109 N/m2.
Solutions :B = Pv
∆v⇒ P = B
(∆vv
)Estimate the change in the density of water in ocean at a depth of 400 m below the surface. The density of water at the surface
= 1030 kg/m3 and the bulk modulus of water = 2 x 109 N/m2.
Solutions :B = ρ0gh
(V0 − Vd)⇒ 1 - Vd
v0= ρ0gh
B
ρ0 = mv0
= mvd
Vol.strain = V0 − VdV0
⇒ vDv0
=(1−ρ0gh
B)
So, ρdρ0
= V0Vd
.......(1)
A steel plate of face-area 4 cm2 and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is appliedon the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel= 8 · 4 X 1010 N/m2.
Solutions :η = F
AθLateral displacement = Iθ
A 5.0 cm long straight piece of thread is kept on the surface of water. Find the force with which the surface on one side of thethread pulls it. Surface tension of water = 0 · 076 N/m.
Solutions :F = T I
Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and (c) an air bubbleof radius 4 mm formed inside a tank of water. Surface tension of mercury, soap solution and water are 0 · 465 N/m, 0 · 03 N/m and0 · 076 N/m respectively.
Solutions :a) P =
2THgr
c) P =2Tgr
Page 97
24Tutors.com
b) P =42Tgr
Consider a small surface area of 1 mm2 at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area (a)by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure = 1 · 0 x105 Pa and surface tension of mercury = 0 · 465 N/m. Neglect the effect of gravity. Assume all numbers to be exact.
Solutions :P = 2T
ra) F = P0AF = P ′A = (P0 + 2T
r)A
F = PA = 2TrA
b) pressure = P0 + ( 2Tr
)
The capillaries shown in figure (14−E4) have inner radii 0 · 5 mm, 1 · 0 mm and 1.5 mm respectively. The liquid in the beakeris water. Find the heights of water level in the capillaries. The surface tension of water is 7 · 5 x 10−2 N/m.
Solutions :a) hA = 2Tcosθ
rA − ρg
c) hc = 2Tcosθrcρg
b) hB = 2TcosθrBρg
The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outerlevel. If the same tube is immersed in water, upto what height will the water rise in the capillary ?
Solutions :h−Hg =
2THgCosθHgrρHgg
hωhHg
= TωTHg
xρHgρω
x CosθωCosθHg
hω = 2tωCosθωrρωg
where, the symbols have their usual meanbings.
A barometer is constructed with its tube having radius 1 · 0 mm. Assume that the surface of mercury in the tube is spherical inshape. If the atmospheric pressure is equal to 76 cm of mercury, what will be the height raised in the barometer tube. The contactangle of mercury with glass = 135 and surface tension of mercury = 0 · 465 N/m. Density of mercury = 13600 kg/m3.
Solutions :h = 2Tcosθ
rρg
A capillary tube of radius 0 · 50 mm is dipped vertically in a pot of water. Find the difference between the pressure of the waterin the tube 5 · 0 cm below the surface and the atmospheric pressure. Surface tension of water = 0 · 075 N/m.
Solutions :P = 2T
r
P = Fr
Find the surface energy of water kept in a cylindrical vessel of radius 6 · 0 cm. Surface tension of water = 0 · 075 J/m2.
Solutions :A = πr2
A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension ofmercury = 0 · 465 J/m2.
Solutions :43πR3 = 4
3πr3 × 8
Increase in surface energy = TA’ - TA⇒ r = R
2= 2
A capillary tube of radius 1 mm is kept vertical with the lower end in water. (a) Find the height of water raised in the capillary.(b) If the length of the capillary tube is half the answer of part (a), find the angle θ made by the water surface in the capillary withthe wall.
Solutions :h = 2Tcosθ
rρg, h′ = 2Tcosθ
rρg
So, θ = cos−1 12
= 60o
⇒ cosθ =h′rρg
2T
The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury. (a) Find the depression of mercury columnin the capillary. (b) If the length dipped inside is half the answer of part (a), find the angle made by the mercury surface at the endof the capillary with the vertical. Surface tension of mercury = 0 · 465 N/m and the contact angle of mercury with glass = 135.
Solutions :a) h = 2Tcosθ
rρg
∴ cosθ =hrρg
2T
b) T × 2πrcosθ = πr2h× ρ× gTwo large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates
equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = 0 · 075 N/m.
Solutions :T (2I) = [1× (10−3)× h]ρg
Consider an ice cube of edge 1 · 0 cm kept in a gravity free hall. Find the surface area of the water when the ice melts. Neglectthe difference in densities of ice and water.
Solutions :Surface area = 4πr2
A wire forming a loop is dipped into soap solution and taken out so that a film of soap solution is formed. A loop of 6 · 28 cmlong thread is gently put on the film and the film is pricked with a needle inside the loop. The thread loop takes the shape of acircle. Find the tension in the thread. Surface tension of soap solution = 0 · 030 N/m.
Page 98
24Tutors.com
Solutions :dFy = 2T rdθ, sinθ [dFx = 0]The length of small element = r d θTension ⇒ 2Tt = T × 2r ⇒ T1 = TrConsidering symmetric elements,
So, F = 2Tr∫ π/20 sinθdθ = 2Tr[cosθ]
π/20 = T × 2r
dFy = T × r d θA metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine
on the sphere when the speed of the sphere is 1 cm/s, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) theterminal velocity with which the sphere will move down without acceleration. Density of glycerine = 1260 kg/m3 and its coefficientof viscosity at room temperature = 8 · 0 poise
Solutions :v = 2
9r2(ρ − σ)g
η⇒ 2
9r2
( m(4/3)πr3
− σ)g
n
a) Viscous force = 6πηrvc) 6πη rv +
(43
)πr3σg = mg
b) Hydrostatic force = B =(
43
)πr3σg
Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = 0.02 cm, viscosity of air = 1 · 8x 10−4 poise, g = 9.9 m/s2 and density of water = 1000 kg/m3.
Solutions :Because, σ of air is very small, the force of buoyancy may be neglected.To find terminal velocity of rain drops, the forces acting on the drop are,6πη rv +
(43
)πr3σg
ii) Force of buoyancy (4/3)πr3 σg upward.Thus,
v = 2r2ρg9η
i) The weight (4/3)πr3 ρg downward.or
Water flows at a speed of 6 cm/s through a tube of radius 41 cm. Coefficient of viscosity of water at room temperature is 0 · 01poise. Calculate the Reynolds number. Is it a steady flow ?
Solutions :v = Rη
ρD⇒ R = vρD
η
A steel tube of length 1.00 m is struck at one end. A person with his ear close to the other end hears the sound of the blowtwice, one travelling through the body of the tube and the other through the air in the tube. Find the time gap between the twohearings. Use the table in the text for speeds of sound in various substances.
Solutions :Vair = 230 m/s.V5 = 5200 m/s. Here s = 7 mSo, t = t1 − t2 =
(1
330− 1
5200
)= 2.75× 10−1 sec = 2.75 ms
At a prayer meeting, the disciples sing JAI − RAM JAI − RAM . The sound amplified by a loudspeaker comes back afterreflection from a building at a distance of 80 m from the meeting. What maximum time interval can be kept between one JAI−RAMand the next JAI −RAM so that the echo does not disturb a listener sitting in the meeting. Speed of sound in air is 320 m/s.
Solutions :Here given S = 80m× 2 = 160 mSo minimum time interval will be
t = 5v
= 160320
= 0.5 seconds.v = 320m/s
A man stands before a large wall at a distance of 50 · 0 m and claps his hands at regular intervals. Initially, the interval is large.He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10times during every 3 seconds, find the velocity of sound in air.
Solutions :He ha to clap 10 times in 3 seconds.
So time interval between two clap = ( 310
second).
So the taken go the wall =(
32× 10
)= 3
20seconds.
= 333 m/s.
A person can hear sound waves in the frequency range 20 Hz to 20 kHz. Find the minimum and the maximum wavelengths ofsound that is audible to the person. The speed of sound is 360 m/s.
Solutions :a) For minimum wavelength n = 20 KHz
as(η ∞ 1
λ
)b) For minimum wavelength, n = 20KHz
∴ λ = 360(20×103)
= 18× 10−3 m = 18 mm
⇒ x = ( vn
) = 36020
= 18 m
Find the minimum and maximum wavelengths of sound in water that is in the audible range (20 − 20000 Hz) for an averagehuman ear. Speed of sound in water = 1450 m/s.
Solutions :a) For minimum wavelength n = 20 KHz
⇒ v = nλ⇒ λ =(
145020×103
)= 7.25 cm
b) for minimum wavelength n should be minimum⇒ v = nλ⇒ λ = v
n⇒ 1450
20= 72.5 m
Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger thanthe diameter of the loudspeaker. (a) Calculate the frequency for which the wavelength of sound in air is ten times the diameter of
Page 99
24Tutors.com
the speaker if the diameter is 20 cm. (b) Sound is essentially transmitted in the forward direction if the wavelength is much shorterthan the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one tenth of the diameter of thespeaker described above. Take the speed of sound to be 340 m/s.
Solutions :According to the question,
a) λ = 20 cm× 10 = 200 cm = 2 mv = 340 m/sSo, n = 340
Ultrasonic waves of frequency 4.5 MHz are used to detect tumour in soft tissues. The speed of sound in tissue is 1 · 5 km/s andthat in air is 340 m/s. Find the wavelength of this ultrasonic wave in air and in tissue.
Solutions :a) Given Vair = 340 m/s, n = 4.5× 106 Hz
⇒ λair =(
3404.5
)× 10−6 = 7.36× 10−5 m.
b) Vtissue = 1500 m/s⇒ λ =(
15004.5
)× 10−6 = 3.3× 10−4 m.
The equation of a travelling sound wave is y = 6 · 0 sin (600 t − 1 · 8x) where y is measured in 10−5 m, t in second and x inmetre. (a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave. (b) Find the ratio of thevelocity amplitude of the particles to the wave speed.
Solutions :Here Vy = 3600× 10−5 m/sHere given ry = 6.0× 10−5 m
a) Given 2πλ
= 1.8⇒ λ = ( 2π1.8
)
So the ratio of(Vyv
)= 3600×3×10−5
1000.
b) Let, velocity amplitude = Vy
V = dydt
= 3600 cos (600t− 1.8)× 10−5 m/s
Again, λ = 2π1.8
and T = 2π600⇒ wave speed = v = λ
T= 600
1.8= 1000
3m/s
So,ryλ
=6.0×(1.8)×10−5 m/s
2π= 1.7× 10−5 m/s
A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is 350 m/s. (a) By how much is the phasechanged at a given point in 2 ·5 ms ? (b) What is the phase difference at a given instant between two points separated by a distanceof 10 · 0 cm along the direction of propagation ?
Solutions :Here given n = 100, v = 350 m/s.
⇒=v
n=
350
100= 3.5 m
In 2.5 ms, the distance travelled by the particle is given by,∆x = 350× 2.5× 103
So, phase difference φ =2π
λ×∆x⇒
2π
350/100× 350× 2.5× 10−3 =
(π2
)b) In the second case, Given ∆η = 10 cm = 101 m
So, φ =2π
x∆x
2π × 10−1
(350/100)=
2π
35Two point sources of sound are kept at a separation of 10 cm. They vibrate in phase to produce waves of wavelength 5 · 0 cm.
What would be the phase difference between the two waves arriving at a point 20 cm from one source (a) on the line joining thesources and (b) on the perpendicular bisector of the line joining the sources ?
Solutions :a) Given ∆x = 10 cm, λ = 5.0 cm
⇒ φ =2π
λ×∆η = 2π5× 10 = 4π
So phase difference is zero.b) Zero, as the particle is in same phase because of having same path.
Calculate the speed of sound in oxygen from the following data. The mass of 22 · 4 litre of oxygen at STP (T = 273 K and p =1 · 0 × 105N/m2) is 32 g, the molar heat capacity of oxygen at constant volume is Cv = 2 · 5 R and that at constant pressure isCp = 3 · 5 R.
Solutions :Given that p = 1.0× 105N/m2, T = 273K, M = 32g = 32× 103kg
V = 22.4litre = 22.4× 103m3
C
Cv= r =
3.5R
2.5R= 1.4
⇒ V =
√rp
f=
√1.2× 1.0× 10−5
32/22.4= 310m/s (because ρ = m/v)
The speed of sound as measured by a student in the laboratory on a winter day is 340 m/s when the room temperature is 17C.What speed will be measured by another student repeating the experiment on a day when the room temperature is 32C ?
Solutions := 340×
√305
290= 349 m/s
V1 = 330 m/s, V 2 =?We know v ∞
√T√
V1√V2
=
√T1√T2⇒ V2 =
V1 ×√T2√
T1T1 = 273 + 17 = 290K, T2 = 272 + 32 = 305K
Page 100
24Tutors.com
At what temperature will the speed of sound be double of its value at 0C ?
Solutions :T1 = 273 V2 = 2V1
V1 = v T2 =?So temperature will be (4× 273) 273 = 819c.
We know that V ∞√T ⇒
T2
T1=V 2
2
V 21
⇒ T2 = 273× 22 = 4× 273K
The absolute temperature of air in a region linearly increases from T1 to T2 in a space of width d. Find the time takenby a sound wave to go through the region in terms of T1, Tvd and the speed v of sound at 273 K. Evaluate this time forT1 = 280 K, T2 = 310 K, d = 33 m and v = 330 m/s.
Solutions :The variation of temperature is given by,
T = T1 +(T2− T2)
dx (1)
We know that V ∞√T ⇒
Vt
V=
√T
273⇒ V T = v
√T
273
⇒ dt =dx
VT= dfracduV ×
√273
T
⇒ t =273
V
∫ d0
dx
[T1 + (T2 − T1/dx)]1
2
=
√273
V×
2d
T2 − T1[T1 +
T2 − T1
dx]d0 =
(2d
V
)( √273
T2 − T1
)×√T2√T1
= T =2d
V
√273
√T2 +
√T1
Putting the given value we get
=2× 33
330
√273
√280 +
√310
= 96 ms
Find the change in the volume of PO litre kerosene when it is subjected to an extra pressure of 2.0×105 N/m2 from the followingdata. Density of kerosene = 800 kg/m 3 and speed of sound in kerosene = 1330 m/s.
Solutions :We know that v =
√K
ρWhere K = bulk modulus of elasticity⇒ K = v2 ρ = (1330)2 × 800 N/m2
We know K =( F/A
∆V/V
)⇒ ∆V =
Pressures
K=
2× 105
1330× 1330× 800So, ∆V = 0.15 cm3
Calculate the bulk modulus of air from the following data about a sound wave of wavelength 35 cm travelling in air. Thepressure at a point varies between (1 · 0 × 10514) Pa and the particles of the air vibrate in simple harmonic motion of amplitude5 · 5× 10−6 m.
Solutions :We know that,
Bulk modulus B =∆p
(∆V/V )=
P0λ
2πS0
Where P0 = pressure amplitude ⇒ P0 = 1.0× 105
S0 = displacement amplitude ⇒ S0 = 5.5× 106 m
⇒14× 35× 10−2 m
2π(5.5)× 10−6 m= 1.4× 105N/m2
A source of sound operates at 2 · 0 kHz, 20 W emitting sound uniformly in all directions. The speed of sound in air is 340 m/sand the density of air is 1 · 2 kg/m3. (a) What is the intensity at a distance of 6 · 0 m from the source ? (b) What will be thepressure amplitude at this point ? (c) What will be the displacement amplitude at this point ?
Solutions :a) Here given Vair = 340 m/s., Power =
E
t= 20 W
f = 2, 000Hz, ρ = 1.2kg/m3
So, intensity I =E
t.A
=20
3πr2=
20
4× π × 62= 44 mw/m2 (because r = 6 m)
b) We know that I =p2
0
2ρVair⇒ P0 =
√1× 2ρVair
=√
2× 1.2× 340× 44× 10−3 = 6.0N/m2
c) We know that I = 2π2S20v
2ρV where S0 = displacement amplitude
⇒ S0 =
√I
π2ρ2ρVair
Putting the value we get Sg = 1.2× 106 m.
The intensity of sound from a point source is 1 · 0 × 10−6 W/m2 at a distance of 5 · 0 m from the source. What will be theintensity at a distance of 25 m from the source ?
Solutions :
Page 101
24Tutors.com
Here I1 = 1.0× 108 W1/m2; I2 =?r1 = 5.0m, r2 = 25 m.
We know that I ∞1
r2
⇒ I1r21 = I2r2
2 ⇒I1r2
1
r22
=1.0× 10−8 × 25
625= 4.0× 10−10 W/m2
The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from thesource ?
Solutions :IA
IB=r2B
r2A
=(50
5
)2 ⇒ 10(βAβB) = 102
We know that β = 10 log10
( II0
)⇒ βB = 40− 20 = 20 dβIA
I0= 10(βA/10) ⇒
IB
Io= 10(βB/10)
⇒βA − βB
10= 2⇒ βA − βB = 102
βA = 10 logIA
I0, βB = 10 log
IA
I0If the intensity of sound is doubled, by how many decibels does the sound level increase ?
Solutions :We know that, β = 10 log10
J
I0According to the questions,
βA = 10 log10
(2I
I0
)⇒ βB − βA = 10 log
(2I
I
)= 10× 0.3010 = 3 dB
Sound with intensity larger than 120 dB appears painful to a person. A small speaker delivers 2 · 0 W of audio output. Howclose can the person get to the speaker without hurting his ears ?
Solutions :If sound level = 120 dB, then I = intensity = 1 W/m2
Given that, audio output = 2WLet the closest distance be x.
So, intensity =( 2
4πx2
)= 1⇒ x2 =
( 2
2π
)⇒ x = 0.4 m = 40 cm
If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased ?
Solutions :β1 = 50 dB, β2 = 60 dB
∴ I1 = 107W/m2, I2 = 106 W/m2
(because β = 10 log10( II0
), where I0 = 1012W/m2)
Again,I2
I1=(p2
p1
)2=(106
107
)= 10 (where p = pressure amplitude)
∴(p2
p1
)=√
10.
The noise level in a class-room in absence of the teacher is 50 dB when 50 students are present. Assuming that on the averageeach student outputs same sound energy per second, what will be the noise level if the number of students is increased to 100 ?
Solutions :So, βA = 50 + 3 = 53dB.Let the intensity of each student be I.
According to the question
βA = 10 log1050I
I0;βB = 10log10
(100I
I0
)= 10log
(100I
50I
)= 10log102 = 3
βB βA = 10log1050I
I0− 10log10
(100I
I0
)In a Quincke’s experiment the sound detected is changed from a maximum to a minimum when the sliding tube is moved through
a distance of 2 · 50 cm. Find the frequency of sound if the speed of sound in air is 340 m/s.
Solutions :Distance between tow maximum to a minimum is given by, λ/4 = 2.50 cm
⇒ λ = 10cm = 10−1 mWe know, V = nx
⇒=V
λ=
340
10−1= 3400 Hz = 3.4 kHz
In a Quincke’s experiment, the sound intensity has a minimum value I at a particular position. As the sliding tube is pulled outby a distance of 16 · 5 mm, the intensity increases to a maximum of 9I. Take the speed of sound in air to be 330 m/s. (a) Find thefrequency of the soundsource. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it doesnot change much between the positions of minimum intensity and maximum intensity.
Solutions :
Page 102
24Tutors.com
a) According to the data λ/4 = 16.5 mm⇒ λ = 66mm = 66× 10−6=3 m.
⇒ n =V
λ=
330
66× 10−3= 5kHz
b) Iminimum = K(A1A2)2 = I ⇒ A1A2 = 11Imaximum = K(A1 +A2)2 = 9⇒ A1 +A2 = 31
So, the ratio amplitudes is 2.
So,A1 +A2
A1 +A2=
3
4⇒
A1
A2=
2
1Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place
6 · 0 m from one of the speakers and 6 · 4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz,what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air = 320 m/s.
Solutions :The path difference of the two sound waves is given by,
∆L = 6.46.0 = 0.4 m
The wavelength of either wave = λ =V
ρ=
320
ρ(m/s)
For destructive interference ∆L =(2n+ 1)λ
2where n is an integers.
or 0.4 m =2n+ 1
2×
320
ρ⇒ ρ = n = = 8000 Hz = (2n+ 1)400 Hz
Thus the frequency within the specified range which cause destructive interference are 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz.
A source of sound S and a detector D are placed at some distance from one another. A big cardboard is placed near the detectorand perpendicular to the line SD as shown in figure (16-E1). It is gradually moved away and it is found that the intensity changesfrom a maximum to a minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted.Velocity of sound in air is 336 m/s.
Solutions :According to the given data
V = 336 m/s,λ/4 = distance between maximum and minimum intensity.= (20cm)⇒= 80 cm
⇒ n = frequency =V
λ=
336
80× 10−2= 420 Hz
A source S and a detector D are placed at a distance d apart. A big cardboard is placed at a distance√
2d from the source andthe detector as shown in figure (16 − E2). The source emits a wave of wavelength = d/2 which is received by the detector afterreflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distanceshould the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave ?
Solutions :Here given λ =
d
2
Initial path difference is given by = 2
√(d2
)+ 2d2 − d
If it is now shifted a distance x then path difference will be
= 2√
+ (√
2d+ x)2 − d =d
4
(2d+
d
4
)→(d
2
)2=
169d2
64⇒=
153
64d2
⇒√
2d+ x = 1.54 d⇒ x = 1.54d− 1.414d = 0.13d.
Two stereo speakers are separated by a distance of 2 · 40 m. A person stands at a distance of 3 · 20 m directly in front of one ofthe speakers as shown in figure (16−E3). Find the frequencies in the audible range (20− 2000Hz) for which the listener will heara minimum sound intensity. Speed of sound in air = 320 m/s.
Solutions :As shown in the figure the path differences 2.4 = ∆x =
√(3.2)2 + (2.4)2 − 3.2
Again, the wavelength of the either sound waves =320
ρWe know, destructive interference will be occur
If ∆x =(2n+ 1)λ
2
⇒√
(3.2)2 + (2.4)2 − (3.2) =(2n+ 1)
2
320
ρSolving we get,
⇒ V =(2n+ 1)400
2= 200(2n+ 1)
where n = 1, 2, 3, ..........49. (audile region)
Two sources of sound, S1 and S2 , emitting waves of equal wavelength 20 · 0 cm, are placed with a separation of 20 · 0 cmbetween them. A detector can be moved on a line parallel to S1 S2 and at a distance of 20 · 0 cm from it. Initially, the detector isequidistant from the two sources. Assuming that the waves emitted by the sources are in phase, find the minimum distance throughwhich the detector should be shifted to detect a minimum of sound.
Solutions :According to the data,
λ = 20 cm, S1S2 = 20 cm, BD = 20 cmLet the detector is shifted to left for a distance x for hearing the minimum sound. So path difference AI = BCAB=√
(20)2 + (10 + x)2 −√
(20)2 + (10− x)2
So the minimum distances hearing for minimum
Page 103
24Tutors.com
=(2n+ 1)λ
2=λ
2=
20
2= 10 cm
⇒√
(20)2 + (10 + x)2 −√
(20)2 + (10− x)2 = 10 solving we get x = 12.0 cm
Two speakers S1 and S2 , driven by the same amplifier, are placed at y = 1 · 0 m and y = −1 · 0 m (figure16 − E4). Thespeakers vibrate in phase at 600 Hz. A man stands at a point on the X-axis at a very large distance from the origin and startsmoving parallel to the Y -axis. The speed of sound in air is 330 m/s. (a) At what angle θ will the intensity of sound drop to aminimum for the first time ? (b) At what angle will he hear a maximum of sound intensity for the first time ? (c) If he continuesto walk along the line, how many more maxima can he hear ?
Solutions :b) For minimum intensity, x = 2n(
λ
2
)yd
D= λ⇒
y
D= θ =
λ
D=
0.55
2= 0.275 rad
∴ θ = 16
Given, F = 600 Hz, and v = 330 m/s⇒v
f=
330
600= 0.55 mm
a) For minimum intensity, x = (2n+ 1)(λ/2)yd
D=λ
2[for minimum y, x =
λ
2]
∴y
D= θ =
λ
2=
0.55
4= 0.1375 rad = 0.1375× (57.1) = 7.9
c) For more maxima,yd
D= 2λ, 3λ, 4λ, ..
⇒y
D= θ = 32, 64, 128
But since, the maximum value of θ can be 90, he will hear two more maximum i.e. at 32 and 64.
Let OP = D, PQ = y ⇒ θ =y
R(1)
Now path difference is given by, x = S2QS1Q =yd
DWhere d = 2 m
[The proof of x =yd
Dis discussed in interference of light waves]
Three sources of sound S1 S2 and S3 of equal intensity are placed in a straight line with S1S2 = S2S3, (figure 16− E5). At apoint P , far away from the sources, the wave coming from S2 is 120 ahead in phase of that from S1. Also, the wave coming fromS3 is 120 ahead of that from S2. What would be the resultant intensity of sound at P?
Solutions :Because the 3 sources have equal intensity, amplitude are equal
So, A1 = A2 = A3
As shown in the figure, amplitude of the resultant = 0 (vector method)So, the resultant, intensity at B is zero.
Two coherent narrow slits emitting sound of wavelength λ in the same phase are placed parallel to each other at a smallseparation of 2λ. The sound is detected by moving a detector on the screen
∑at a distance D( λ) from the slit S1 as shown in
figure (16− E6). Find the distance x such that the intensity at P is equal to the intensity at O.
Solutions :The two sources of sound S1 and S2 vibrate at same phase and frequency.
Resultant intensity at P = I0a) Let the amplitude of the waves at S1 and S2 be r.When θ = 45, path difference = S1PS2P = 0 (because S1P = S2P )
So, when source is switched off, intensity of sound at P isI0
4
. b) When θ = 60, path difference is also 0. Similarly it can be proved that, the intensity at P isI0
4when one is switched off.
Figure(16 − E7) shows two coherent sources S1 and S2 which emit sound of wavelength λ in phase. The separation betweenthe sources is 3λ. A circular wire of large radius is placed in such a way that S1S2 lies in its plane and the middle point of S1S2 isat the centre of the wire. Find the angular positions θ on the wire for which constructive interference takes place.
Solutions :If V = 340 m/s, I = 20 cm = 20× 102 m
Fundamental frequency =V
21=
340
2× 20× 10−2= 850 Hz
We know first over tone =2V
21=
2× 340
2× 20× 10−2(for open pipe) = 1750 Hz
Second over tone = 3( V
21
)= 3× 850 = 2500 Hz
Two sources of sound S1 and S2 vibrate at same frequency and are in phase (figure16− E8). The intensity of sound detectedat a point P as shown in the figure is I0. (a) If θ equals 45, what will be the intensity of sound detected at this point if one of thesources is switched off ? (b) What will be the answer of the previous part if θ = 60 ?
Solutions :According to the questions V = 340 m/s, n = 500 Hz
We know thatV
4I(for closed pipe)
⇒ I =340
4× 500m = 17 cm
Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. Speed of sound inair is 340 m/s.
Solutions :
Page 104
24Tutors.com
Here given distance between two nodes is = 4.0 cm,⇒ λ = 2× 4.0 = 8 cmWe know that v = nλ
⇒ η =328
8× 10−2= 4.1 Hz
A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. Speed of sound in air =340 m/s.
Solutions :V = 340 m/s
Distances between two nodes or antinodes
⇒λ
4= 25 cm
⇒ λ = 100 cm = 1 m⇒ n =
v
λ= 340 Hz.
In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4 · 0 cm. If the speed of sound in air is328 m/s, what is the frequency of 73 the source ?
Solutions :Here given that 1 = 50 cm, v = 340 m/s
As it is an open organ pipe, the fundamental frequency f1 = (v
21)
=340
2× 50× 10−2= 340 Hz
So, the harmonies aref3 = 3× 340 = 1020Hzf5 = 5× 340 = 1700, f6 = 6× 340 = 2040 Hzso, the possible frequencies are between 1000 Hz and 2000 Hz are 1020, 1360, 1700.
The separation between a node and the next antinode in a vibrating air column is 25 cm. If the speed of sound in air is 340 m/s,find the frequency of vibration of the air column.
Solutions :Here given I2 = 0.67 m, l1 = 0.2 m, f = 400 Hz
We know thatλ = 2(l2l1)⇒ λ = 2(6220) = 84 cm = 0.84 m.So, v = nλ = 0.84× 400 = 336 m/sWe know from above that,
l1 + d =λ
4⇒ d =
λ
4l1 = 2120 = 1 cm.
A cylindrical metal tube has a length of 50 cm and is open at both ends. Find the frequencies between 1000 Hz and 2000 Hzat which the air column in the tube can resonate. Speed of sound in air is 340 m/s.
Solutions :According to the questions
f1 first overtone of a closed organ pipe P1 =3v
4l=
3× V4× 30
f2 fundamental frequency of a open organ pipe P2 =V
2I2
Here given3V
4× 30=
V
2I2⇒ I2 = 20 cm
∴ length of the pipe P2 will be 20 cm
In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the aircolumn has a length of 20.0 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find thespeed of sound in air. (b) How much distance above the open end does the pressure node form ?
Solutions :Length of the wire = 1.0 m
For fundamental frequency λ/2 = l⇒ λ = 2l = 2× 1 = 2 mHere given n = 3.8 km/s = 3800 m/s
We know ⇒ v = nλ⇒ n =3800
2= 1.9 kH.
So standing frequency between 20 Hz and 20 kHz which will be heard are= n× 1.9 kHz where n = 0, 1, 2, 3, 10
The first overtone frequency of a closed organ pipe P1 is equal to the fundamental frequency of an open organ pipe P2 . If thelength of the pipe P1 is 30 cm, what will be the length of P2 ?
Solutions :Let the length will be l.
Here given that V = 340 m/s and n = 20 HzHere λ/2 = l⇒ λ = 2l
We know V = nλ⇒ l =V
n=
340
2× 20=
34
4= 8.5 cm (for maximum wavelength, the frequency is minimum)
A copper rod of length 1 · 0 m is clamped at its middle point. Find the frequencies between 20 Hz − 20, 000 Hz at whichstanding longitudinal waves can be set up in the rod. The speed of sound in copper is 3.8 km/s.
Solutions :a) Here given l = 5 cm = 5× 102m, v = 340 m/s
⇒ n =V
2I=
340
2× 5× 10−2= 3.4 kHz
b) If the fundamental frequency = 3.4 KHz
Page 105
24Tutors.com
⇒ then the highest harmonic in the audible range (20 Hz20 KHz)
=20000
3400= 5.8 = 5 (integral multiple of 3.4KHz).
Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearingrange (20− 20, 000 Hz). Speed of sound in air = 340 m/s.
Solutions :The resonance column apparatus is equivalent to a closed organ pipe.
Here I = 80 cm = 10× 102 m; v = 320 m/s
⇒ n0 =v
4I=
320
4× 50× 10−2= 100 Hz
So the frequency of the other harmonics are odd multiple of n0 = (2n+ 1) 100 HzAccording to the question, the harmonic should be between 20 Hz and 2 KHz.
An open organ pipe has a length of 5 cm. (a) Find the fundamental frequency of vibration of this pipe. (b) What is the highestharmonic of such a tube that is in the audible range ? Speed of sound in air is 340 m/s and the audible range is 20− 20, 000 Hz.
Solutions :Let the length of the resonating column will be = 1
Here V = 320 m/s
Then the two successive resonance frequencies are(n+ 1)v
4Iand
nv
4I
Here given(n+ 1)v
4I= 2592; λ =
nv
4I= 1944
⇒(n+ 1)v
4I−nv
4I= 2592− 1944 = 548 cm = 25 cm.
An electronically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of air columnin the tube is 80 cm. The frequency of the loudspeaker can be varied between 20 Hz − 2 kHz. Find the frequencies at which thecolumn will resonate. Speed of sound in air = 320 m/s.
Solutions :Let, the piston resonates at length l1 and l2
Here, l = 32 cm; v =?, n = 512 HzNow ⇒ 512 = v/λ⇒ v = 512× 0.64 = 328 m/s.
Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speedof sound in air is 324 m/s.
Solutions :Let the length of the longer tube be L2 and smaller will be L1.
According to the data 440 =3× 330
4× L2.............(1) (first over tone)
And 440 =330
4× L1..............(2) (fundamental)
solving equation we get L2 = 56.3 cm and L1 = 18.8 cm.
A piston is fitted in a cylindrical tube of small cross-section with the other end of the tube open. The tube resonates with atuning fork of frequency 512 Hz. The piston is gradually pulled out of the tube and it is found that a second resonance occurs whenthe piston is pulled out through a distance of 32 · 0 cm. Calculate the speed of sound in the air of the tube.
Solutions :Let n0 = frequency of the turning fork, T = tension of the string
L = 40cm = 0.4 m, m = 4g = 4× 103 kgSo, m = Mass/Unit length = 102 kg/m
n0 =1
21
√T
m
So, 2nd harmonic 2n0 =( 2
2I
)√ T
mAs it is unison with fundamental frequency of vibration in the air column
⇒ 2n0 =340
4× 1= 85 Hz
⇒ 85 =2
2× 0.4
√T
14⇒ T = 852 × (0.4)2 × 10−2 = 11.6 Newton.
A U-tube having unequal arm-lengths has water in it. A tuning fork of frequency 440 Hz can set up the air in the shorter armin its fundamental mode of vibration and the same tuning fork can set up the air in the longer arm in its first overtone vibration.Find the length of the air columns. Neglect any end effect and assume that the speed of sound in air = 330 m/s.
Solutions :Given, m = 10 g = 10× 103kg, l = 30 cm = 0.3 m
Let the tension in the string will be = Tµ = mass / unit length = 33× 103 kg
The fundamental frequency ⇒ n0 =1
2I
√T
µ.......(1)
The fundamental frequency of closed pipe
⇒ n0 =( v
4I
) 340
4× 50× 102= 170 Hz .....(2)
According to equations (1) ×(2)weget,
170 =1
2× 30× 10−2×
√T
33× 10−3
⇒ T = 347 Newton
Consider the situation shown in figure (16-E9). The wire which has a mass of 4 · 00 g oscillates in its second harmonic and setsthe air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is 340 m/s, find the
Page 106
24Tutors.com
tension in the wire.
Solutions :We know that f ∞
√T
According to the question f + ∆f ∞√
∆T + T
⇒f + ∆f
f=
√∆t+ T
T⇒ 1 +
∆f
f=(
1 +∆T
T
)1
2=1+
1
2
∆T
T+........
(neglecting other terms)
⇒∆f
f=(1
2
)∆T
T.
A 30 ·0−cm-long wire having a mass of 10 ·0 g is fixed at the two ends and is vibrated in its fundamental mode. A 50 ·0−cm-longclosed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire.Find the tension in the wire. Speed of sound in air = 340 m/s.
Solutions :We know that the frequency = f, T = temperatures
f ∞√T
So,f1
f2=
√T1√T2⇒
293
f2=
√293√
295
⇒ f2 =293×
√295
√293
= 294
Show that if the room temperature changes by a small amount from T to T + ∆T , the fundamental frequency of an organ pipechanges from v to v + ∆v, where
∆v
v=
1
2
∆T
T
Solutions :We know that f ∞
√T
According to the question f + ∆f ∞√
∆T + T
⇒f + ∆f
f=
√∆t+ T
T⇒ 1 +
∆f
f=(
1 +∆T
T
)1
2=1+
1
2
∆T
T+........
(neglecting other terms)
⇒∆f
f=(1
2
)∆T
T.
The fundamental frequency of a closed pipe is 293 Hz when the air in it is at a temperature of 20C. What will be its fundamentalfrequency when the temperature changes to 22C ?
A Kundt’s tube apparatus has a copper rod of length 1 · 0 m clamped at 25 cm from one of the ends. The tube contains air inwhich the speed of sound is 340 m/s. The powder collects in heaps separated by a distance of 5 · 0 cm. Find the speed of soundwaves in copper.
A Kundt’s tube apparatus has a steel rod of length 1 · 0 m clamped at the centre. It is vibrated in its fundamental mode at afrequency of 2600 Hz. The lycopodium powder dispersed in the tube collects into heaps separated by 6.5 cm. Calculate the speedof sound in steel and in air.
A source of sound with adjustable frequency produces 2 beats per second with a tuning fork when its frequency is either 476 Hzor 480 Hz. What is the frequency of the tuning fork ?
A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with alittle wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork ?
Calculate the frequency of beats produced in air when two sources of sound are activated, one emitting a wavelength of 32 cmand the other of 32 · 2 cm. The speed of sound in air is 350 m/s.
A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe oflength 40 cm to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded withwax. Find its original frequency. The speed of sound in air is 320 m/s.
A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire B produces 6 beats per second with itwhen the tension in A is slightly increased. Find the ratio of the tension in A to the tension in B.
A tuning fork of frequency 256 Hz produces 4 beats per second with a wire of length 25 cm vibrating in its fundamental mode.The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire beshortened so that it produces no beats with the tuning fork ?
A traffic policeman standing on a road sounds a whistle emitting the main frequency of 2 ·00 kHz. What could be the appparentfrequency heard by a scooter-driver approaching the policeman at a speed of 36 · 0 km/h ? Speed of sound in air = 340 m/s.
The horn of a car emits sound with a dominant frequency of 2400 Hz. What will be the apparent dominant frequency heard bya person standing on the road in front of the car if the car is approaching at 18 · 0 km/h ? Speed of sound in air = 340 m/s.
A person riding a car moving at 72 km/h sounds a whistle emitting a wave of frequency 1250 Hz. What frequency will be heardby another person standing on the road (a) in front of the car (b) behind the car ? Speed of sound in air = 340 m/s.
A train approaching a platform at a speed of 54 km/h sounds a whistle. An observer on the platform finds its frequency to be1620 Hz. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hearafter the train has crossed the platfrom ? The speed of sound in air = 332 m/s.
A bat emitting an ultrasonic wave of frequency 4.5× 104 Hz flies at a speed of 6 m/s between two parallel walls. Find the twofrequencies heard by the bat and the beat frequency between the two. The speed of sound is 330 m/s.
A bullet passes past a person at a speed of 220 m/s. Find the fractional change in the frequency of the whistling sound heardby the person as the bullet crosses the person. Speed of sound in air = 330 m/s.
Two electric trains run at the same speed of 72 km/h along the same track and in the same direction with a separation of 2.4 kmbetween them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of 500 m fromthe track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at 500 Hz and the speedof sound in air is 340 m/s, find the frequencies heard by the person
Page 107
24Tutors.com
A violin player riding on a slow train plays a 440 Hz note. Another violin player standing near the track plays the same note.When the two are close by and the train approaches the person on the ground, he hears 4 · 0 beats per second. The speed of soundin air = 340 m/s. (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train ?
Two identical tuning forks vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runsbetween the forks at a speed of 3 · 0 m/s so that he approaches one tuning fork and recedes from the other (figure 16-E10). Findthe beat frequency observed by the listener. Speed of sound in air = 332 m/s.
Figure (16-E11) shows a person standing somewhere in between two identical tuning forks, each vibrating at 512 Hz. If boththe tuning fbrks move towards right at a speed of 5 · 5 m/s, find the number of beats heard by the listener. Speed of sound in air= 330 m/s.
A small source of sound vibrating at frequency 500 Hz is rotated in a circle of radius 100/π cm at a constant angular speed of5 · 0 revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and the maximum frequencyof the sound observed. Speed of sound in air = 332 m/s.
Two trains are travelling towards each other both at a speed of 90 km/h. If one of the trains sounds a whistle at 500 Hz, whatwill be the apparent frequency heard in the other train ? Speed of sound in air = 350 m/s.
A traffic policeman sounds a whistle to stop a car-driver approaching towards him. The car-driver does not stop and takes theplea in court that because of the Doppler shift, the frequency of the whistle reaching him might have gone beyond the audible limitof 20 kHz and he did not hear it. Experiments showed that the whistle emits a sound with frequency close to 16 kHz. Assumingthat the claim of the driver is true, how fast was he driving the car ? Take the speed of sound in air to be 330 m/s. Is this speedpractical with today’s technology ?
A car moving at 108 km/h finds another car in front of it going in the same direction at 72 km/h. The first car sounds a hornthat has a dominant frequency of 800 Hz. What will be the apparent frequency heard by the driver in the front car ? Speed ofsound in air = 330 m/s.
Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of 36 km/h and the other at54 km/h relative to the water. The first submarine sends a sound signal (sound waves in water are also called sonar) at a frequencyof 2000 Hz. (a) At what frequency is this signal received by the second submarine ? (b) The signal is reflected from the secondsubmarine. At what frequency is this signal received by the first submarine. Take the speed of the sound wave in water to be1500 m/s.
A small source of sound oscillates in simple harmonic motion with an amplitude of 17 cm. A detector is placed along the lineof motion of the source. The source emits a sound of frequency 800 Hz which travels at a speed of 340 m/s. If the width of thefrequency band detected by the detector is 8 Hz, find the time period of the source.
A boy riding on his bike is going towards east at a speed. of 4√
2 m/s. At a certain point he produces a sound pulse offrequency 1650 Hz that travels in air at a speed of 334 m/s. A second boy stands on the ground 45 south of east from him. Findthe frequency of the pulse as received by the second boy.
A sound source, fixed at the origin, is continuously emitting sound at a frequency of 660 Hz. The sound travels in air at aspeed of 330 m/s. A listener is moving along the line x = 336 m at a constant speed of 26 m/s. Find the frequency of the soundas observed by the listener when he is (a) at y = 140 m, (b) at y = 0 and (c) at y = 140 m.
1. A person looks at different trees in an open space with the following details. Arrange the trees in decreasing order of theirapparent sizes.Tree
Solutions :
The visual angles made by the trees with the eyes can be calculated be below.θ = Height of the tree
Distance from the eye= AB
OB⇒ θA = 2
50= 0.04
simillerly, θB = 2.580
= 0.03125
θC = 1.870
= 0.02571
θD = 2.8100
= 0.028Since, θAgt; θBgt; θDgt; θC , the arrangement in decreasing order is given by A, B, D and C.
An object is to be seen through a simple microscope of focal length 12 cm. Where should the object be placed so as to producemaximum angular magnification ? The least distance for clear vision is 25 cm.
Solutions :
For the given simple microscope,f = 12 cm and D = 25 cmFor maximum angular magnification, the image should be produced at least distance of clear vision.So, v = −D = −25 cmNow, 1
v− 1u
= 1f
⇒ 1u
= 1v− 1f
= 1−25− 1
12= − 37
300⇒ u = −8.1 cmSo, the object should be placed 8.1 cm away from the lens.
Page 108
24Tutors.com
A simple microscope has a magnifying power of 3.0 when the image is formed at the near point (25cm) of a normal eye. (a)What is its focal length ? (b) What will be its magnifying power if the image is formed at infinity ?
Solutions :The simple microscope has, m = 3, when image is formed at D = 25 cm
a) m = 1 + Df⇒ 3 = 1 + 25
f
⇒ f = 25/2 = 12.5 cmb) when the image is formed at infinity (normal adjustment )Magnifying power = D
f= 25
12.5= 2.0
A child has near point at 10 cm. What is the maximum angular magnification the child can have with a convex lens of focallength 10 cm ?
Solutions :The child has D = 10 cm and f = 10 cm
The maximum angular magnification is obtained when the image is formed at near point.m = 1 + D
f= 1 + 10
10= 1 + 1 = 2
A simple microscope is rated 5 X for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whosenear point is 40 cm ?
Solutions :The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm).
Because, the eye is relaxed the image is formed at infinity (normal adjustment)So, m = 5 = D
f= 25
f⇒ f = 5cm
For the relaxed farsighted eye, D = 40 cmSo, m = D
f= 40
5= 8
So, its magnifying power is 8X
Find the maximum magnifying power of a compound Microscope having a 25 diopter lens as the objective, a 5 diopter lens asthe eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is 25 cm.
Solutions :
For the given compound microscopef0 = 1
25 diopter= 0.04 m = 4 cm, fe = 1
5 diopter= 0.2 m = 20 cm
D = 25cm, separation between objective and eyepiece = 30 cmThe magnifying power is maximum when the image is form bythe eye piece at least distance of clear vision i.e. D = 25 cmfor the eye piece, ve = −25 cm, f = 20 cmFor lens formula, 1
ve− 1ue
= 1fe
⇒ 1ue
= 1ve− 1fe⇒ 1−25− 1
20⇒ ue = 11.11cm
So, for the objective lens, the image distance should bevD = 30− (11.11) = 18.89 cmNow , for the objective lens,v0 = +18.89 cm (Because real image is produced)fD = 4 cm
So, 1u0
= 1v0− 1f0⇒ 1
18.89− 1
4= 0.053− 0.25 = −0.197
⇒ u0 = −5.07cmSo, the maximum magnificent power is given by
m = −u0v0
[1 + D
fe
]= − 18.89
−5.07
[1 + 25
20
]= 3.7225× 2.25 = 8.376
The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm.If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if theimage is always needed at 24 cm from the eye
Solutions :For the given compound microscope
fo = 1 cm, fe = 6 cm, D = 24 cmFor the eye piece, ve = −24 cm, fe = 6 cmNow, 1
ve− 1ue
= 1fe
⇒ 1ue
= 1ve− 1fe⇒ −
[124
+ 16
]= − 5
24
⇒ ue = −4.8 cm
Page 109
24Tutors.com
a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective lens must be (9.8)-(4.8) = 5.0 cmNow, 1
v0− 1u0
= 1f0
⇒ 1u0
= 1v0− 1f0
= 15− 1
1= − 4
5
⇒= u0 = − 54
= −1.25 cmSo, the magnifying power is given by,
m = v0u0
[1 + D
f
]= −5−1.25
[1 + 24
6
]= 4× 5 = 20
b) When the separation is 11.8 cm,v0 = 11.8− 4.8 = 7.0cm, f0 = 1 cm⇒ 1
u0= 1
v0− 1f0
= 17− 1
1= − 6
7
So, m = − v0u0
[1 + D
f
]= −7−( 7
6)
[1 + 24
6
]= 6× 5 = 30
So, the range of magnifying power will be 20 to 30.
An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placedat 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated bya distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of theobject which can now be distinguished ?
Solutions :For the given compound microscope.
f0 = 120D
= 0.05 m = 5 cm, fe = 110D
= 0.1 m = 10 cm.
D = 25 cm, separation between objective & eyepiece= 20 cm For the minimum separation between two points which can be distin-guished by eye using the microscope, the magnifying power should be maximum.
For the eyepiece, v0 = 25 cm, fe = 10 cm
So, 1ue
= 1ve− 1fe
= 1−25− 1
10= −
[2+550
]⇒ ue = − 50
7cm
So, the image distance for the objective lens should be,v0 = 20− 50
7= 90
7cm
Now, for the objective lens,
1u0
= 1v0− 1f0
= 790− 1
5= − 11
90
⇒ u0 = − 9011cm
So, the maximum magnifying power is given by,
m = −v0u0
[1 + D
fe
]
=
(907
)(− 90
11
)[1 + 2510
]
= 117× 3.5 = 5.5
Thus, minimum separation eye can distinguish = 0.225.5
mm = 0.04 mm
A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal lengthof 0.5 cm and the tube length is 6.5 cm. Find the focal length of the eyepiece.
Solutions :
For the give compound microscope,f0 = 0.5cm, tube length = 6.5 cm
Page 110
24Tutors.com
magnifying power = 100 (normal adjustment)Since, the image is formed at infinity, the real image produced by theobjective lens should lie on thefocus of the eye piece.
So, v0 + fe = 6.5 cm ...(1)Again, magnifying power = v0
u0× Dfe
[for normal adjustment]
⇒ m = −[
1− v0f0
]Dfe
[∵ v0u0
= 1− v0f0
]
⇒ 100 = −[
1− v00.5
]× 25fe
[Taking D = 25 cm]
⇒ 100 fe = −(1− 2v0)× 25⇒ 2v0 − 4fe = 1 ...(2)Solving equation (1) and (2) we can get,
V0 = 4.5 cm and fe = 2 cmSo, the focal length of the eye piece is 2cm.
A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placedat a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects aninverted real image of the object on a screen 30 cm behind the eyepiece ?
Solutions :
Given that,
fo = = 1 cm, fe = 5 cm, u0 = 0.5 cm, ve = 30 cmFor the objective lens, u0 = 0.5 cm, f0 = 1 cm.
From lens formula,
1v0− 1u0
= 1f0
⇒ 1v0
= 1u0
+ 1f0
= 1−0.5
+ 11
= −1⇒ v0 = −1 cmSo, a virtual image is formed by the objective on the same side as that of the object at a distance of 1 cm from the objective lens. This imageacts as a virtual object for the eyepiece.For the eyepiece,1v0− 1u0
= 1f0
⇒ 1u0
= 1v0− 1f0
= 130− 1
5= −5
30= −1
6⇒ u0 = −6 cmSo, as shown in figure,Separation between the lenses = u0v0 = 6 1 = 5 cm
An optical instrument used for angular magnification has a 25 D objective and a 20 D eyepiece. The tube length is 25 cm whenthe eye is least strained.(a) Whether it is a microscope or a telescope ? (b) What is the angular magnification produced ?
Solutions :
The optical instrument has
f0 = 125D
= 0.04 m = 4 cm
fe = 120D
= 0.05 m = 5 cm
tube length = 25 cm (normal adjustment)
(a) The instrument must be a microscope as f0lt; fe(b) Since the final image is formed at infinity, the image produced by the objective should lie on the focal plane of the eye piece.
So, image distance for objective = v0 = 25 5 = 20 cmNow, using lens formula.1v0− 1u0
= 1f0
⇒ 1u0
= 1v0− 1f0
= 120− 1
4= −4
20= −1
5⇒ u0 = −5 cm
So, angular magnification = m = v0u0× Dfe
[Taking D = 25 cm]
= − 20−5× 25
5= 20
An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tubeis 102 cm, find the powers of the objective and the eyepiece.
Solutions :
Page 111
24Tutors.com
For the astronomical telescope in normal adjustment.Magnifying power = m = 50, length of the tube = L = 102 cmLet f0 and fe be the focal length of objective and eye piece respectively.m = f0
fe= 50⇒ f0 = 50 fe ...(1)
and, L = f0 + fe = 102 cm ...(2)
Putting the value of f0 from equation (1) in (2), we get,
f0 + fe = 102⇒ 51fe = 102⇒ fe = 2 cm = 0.02 mSo, f0 = 100 cm = 1 m
∵ Power of the objective lens = 1f0
= 1D
And Power of the eye piece lens = 1fe
= 10.02
= 50D
The eyepiece of an astronomical telescope has a focal length of 10 cm. The telescope is focused for normal vision of distantobjects when the tube length is 1.0 m. Find the focal length of the objective and the magnifying power of the telescope
Solutions :For the given astronomical telescope in normal adjustment,
Fe = 10 cm, L = 1 m = 100cmS0, f0 = L fe = 100 10 = 90 cmand, magnifying power = f0
fe= 90
10= 9
A Galilean telescope is 27 cm long when focused to form an image at infinity. If the objective has a focal length of 30 cm, whatis the focal length of the eyepiece ?
Solutions :For the given Galilean telescope, (When the image is formed at infinity)
f0 = 30 cm, L = 27 cmSince L = f0 | fe |[Since, concave eyepiece lens is used in Galilean Telescope]⇒ fe = f0 − L = 30 27 = 3 cm
A farsighted person cannot see objects placed closer to 50 cm. Find the power of the lens needed to see the objects at 20 cm.
Solutions :For the far sighted person,
u = 20 cm, v = 50 cmfrom lens formula 1
v− 1u
= 1f
1f
= 1−50− 1−20
= 120− 1
50= 3
100
⇒ f = 1003
cm = 13m
So, power of the lens = 1f
= 3 Diopter
A nearsighted person cannot clearly see beyond 200 cm. Find the power of the lens needed to see objects at large distances.
Solutions :For the near sighted person,
u = ∞ and v = 200 cm = 2mSo, 1
f= 1
v− 1u
= 1−2− 1∞ = − 1
2= −0.5
So, power of the lens is 0.5D
A person wears glasses of power −2.5 D. Is the person farsighted or nearsighted ? What is the far point of the person withoutthe glasses ?
Solutions :The person wears glasses of power 2.5D
So, the person must be near sighted.u =∞ v = far point, f = 1
−2.5= −0.4m = 40 cm
Now, 1v− 1u
= 1f
⇒ 1v
= 1u
+ 1f
= 0 + 1−40
⇒ v = 40 cm
So, the far point of the person is 40 cm
A professor reads a greeting card received on his 50th birthday with + 2.5 D glasses keeping the card 25 cm away. Ten yearslater, he reads his farewell letter with the same glasses but he has ’to keep the letter 50 cm away. What power of lens should henow use ?
Solutions :On the 50th birthday, he reads the card at a distance 25cm using a glass of +2.5D.
Ten years later, his near point must have changed.So after ten years,u = 50 cm, f = 1
2.5D= 0.4m = 40 cm v = near point
Now, 1v− 1u
= 1f⇒ 1
v= 1
u+ 1f
= 1−50
+ 140
= 1200
So, near point = v = 200cmTo read the farewell letter at a distance of 25 cm,U = 25 cm
Page 112
24Tutors.com
For lens formula,
1v− 1u
= 1f⇒ 1
f= 1
200− 1−25
= 1200
+ 125
= 9200
⇒ f = 2009cm = 2
9cm
⇒ Power of the lens = 1f
= 92
= 4.5D
∴ He has to use a lens of power +4.5D.
A normal eye has retina 2 cm behind the eye-lens. What is the power of the eye-lens when the eye is (a) fully relaxed, (b) moststrained ?
Solutions :
Since, the retina is 2 cm behind the eye-lensv = 2cm(a) When the eye-lens is fully relaxedu = ∞ , v = 2cm = 0.02 m⇒ 1
f= 1
v− 1u
= 10.02− 1∞ = 50D
So, in this condition power of the eye-lens is 50D(b) When the eye-lens is most strained,u = 25 cm = 0.25 m, v = +2 cm = +0.02 m
⇒ 1f
= 1v− 1u
= 10.02− 1−0.25
= 50 + 4 = 54D
In this condition power of the eye lens is 54D.
The near point and the far point of a child are at 10 cm and 100 cm. If the retina is 2.0 cm behind the eye-lens, what is therange of the power of the eye-lens ?
Solutions :The child has near point and far point 10 cm and 100 cm respectively.
Since, the retina is 2 cm behind the eye-lens, v = 2cmFor near point u = 10 cm = 0.1 m, v = 2 cm = 0.02 m
So, 1fnear
= 1v− 1u
= 10.02− 1−0.1
= 50 + 10 = 60D
For far point, u = 100 cm = 1 m, v = 2 cm = 0.02 m
So, 1ffar
= 1v− 1u
= 10.02− 1−1
= 50 + 1 = 51D
So, the rage of power of the eye-lens is +60D to +51D
A nearsighted person cannot see beyond 25 cm. Assuming that the separation of the glass from the eye is 1 cm, find the powerof lens needed to see distant objects.
Solutions :For the near sighted person,
v = distance of image from glass= distance of image from eye separation between glass and eye= 25 cm 1cm = 24 cm = 0.24m
So, for the glass, u = ∞ and v = 24 cm = 0.24mSo, 1
f= 1
v− 1u
= 1−0.24
− 1∞ = −4.2D
A person has near point at 100 cm. What power of lens is needed to read at 20 cm if he/she uses (a) contact lens, (b) spectacleshaving glasses 2.0 cm separated from the eyes?
Solutions :The person has near point 100 cm. It is needed to read at a distance of 20cm.
(a) When contact lens is used,u = 20 cm = 0.2m, v = 100 cm = 1 m
So, 1f
= 1v− 1u
= 1−1− 1−0.2
= −1 + 5 = +4D
(b) When spectacles are used,
u = (20 2) = 18 cm = 0.18m, v = 100 cm = 1 mSo, 1
f= 1
v− 1u
= 1−1− 1−0.18
= 1 + 5.55 = + 4.5D
A lady uses + 1.5 D glasses to have normal vision from 25 cm onwards. She uses a 20 D lens as a simple microscope to see anobject. Find the maximum magnifying power if she uses the microscope (a) together with her glass (b) without the glass. Do theanswers suggest that an object can be more clearly seen through a microscope without using the correcting glasses ?
Solutions :The lady uses +1.5D glasses to have normal vision at 25 cm.
So, with the glasses, her least distance of clear vision = D = 25 cmFocal length of the glasses = 1
1.5m = 100
1.5cm
So, without the glasses her least distance of distinct vision should be more
Page 113
24Tutors.com
If, u = 25cm, f = 1001.5
cm
Now, 1v− 1u
= 1f
= 1.5100− 1
25= 1.5−4
100= −2.5
100
⇒ v = 40cm = near point without glasses.Focal length of magnifying glass = 1
20m = 0.05m = 5 cm = f
(a) The maximum magnifying power with glassesm = 1 + D
f= 1 + 25
5= 6 [ ∵ D = 25cm]
(b) Without the glasses, D = 40cmSo, m = 1 + D
f= 1 + 40
5= 9
A lady cannot see objects closer than 40 cm from the left eye and closer than 100 cm from the right eye. While on a mountaineeringtrip, she is lost from her team. She tries to make an astronomical telescope from her reading glasses to look for her teammates. (a)Which glass should she use as the eyepiece ? (b) What magnification can she get with relaxed eye ?
Solutions :The lady can not see objects closer than 40 cm from the left eye and 100 cm from the right eye.
For the left glass lens,v = 40 cm, u = 25 cm
∴ 1f
= 1v− 1u
= 1−40− 1−25
= 125− 1
40= 3
200
⇒ f = 2003cm
For the right glass lens,v = 100 cm, u = 25 cm
1f
= 1v− 1u
= 1−100
− 1−25
= 125− 1
100= 3
100
⇒ f = 1003cm
(a) For an astronomical telescope, the eye piece lens should have smaller focal length. So, she should use the right lens (f = 1003cm) as the
eye piece lens.
(b) With relaxed eye, (normal adjustment)f0 = 200
3cm, fe = 100
3cm
magnification = m = f0fe
=(200/3)(100/3)
= 2
1. A flint glass prism and a crown glass prism are to be combined in such a way that the deviation of the mean ray is zero. Therefractive index of flint and crown glasses for the mean ray are 1’620 and 1’518 respectively. If the refracting angle of the flint prismis 6.0, what would be the refracting angle of the crown prism ?
Solutions :Given that
Refractive index of flint glass = µf = 1.620Refractive index of crown glass= µc = 1.528Refracting angle of flint prism = Af=6.0
For zero net deviation of mean ray(µ1 − 1)Af = (µc − 1)Ac
⇒ Ac = µc−1µc−1
Af = 1.620−11.518−1
(6.0) = 7.2
2. A certain material has refractive indices 1’56, 1’60 and 1’68 for red, yellow and violet light respectively. (a) Calculate thedispersive power. (b) Find the angular dispersion produced by a thin prism of angle 6 made of this material.
3. The focal lengths of a convex lens for red, yellow and violet rays are 100 cm, 98 cm and 96 cm respectively. Find the dispersivepower of the material of the lens.
Solutions :The focal length of a lens is given by
1f
= (µ− 1)( 1R1− 1R2
)
⇒ (µ− 1) = 1f× 1
( 1R1− 1R2
)= K
f..(1)..(1)
So µr − 1 = K100
..(2)
µy − 1 = k98..(3)
And µv − 1 = K96
(4)
So dispersive power = ω = µv−µrµy−1
=(µv−1)−(µr−1)
µy−1=
k96− k
100k98
=k96− k
100k98
= 98×49600
= 0.0408
5. A thin prism is made of a material having refractive indices 1’61 and 1’65 for red and violet light. The dispersive power of thematerial is 0’07. It is found that a beam of yellow light passing through the prism suffers a minimum deviation of 4’0 in favourableconditions. Calculate the angle of the prism.
Solutions :Given that µr = 1.61, µv = 1.65, ω = 0.07 and δy = 4
Now, ω = µv−µrµy−1
⇒ 0.07 = 1.65−1.61µy−1
⇒ µy − 1 = 0.040.07
= 47
Page 114
24Tutors.com
Again, δ = (µ− 1)A
⇒ A =δy
µy−1= 4
(4/7)= 7
4. The refractive index of a material changes by 0’014 as the colour of the light changes from red to violet. A rectangular slabof height 2’00 cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1’32cm below the top surface of the slab. Calculate the dispersive power of the material.
Solutions :
Given that µv − µr = 0.014Again, µy = RealDepth
Apparentdepth= 2.00
1.300= 1.515
So, dispersive power =µv−µrµy−1
= 0.0141.515−1
= 0.027
6. The minimum deviations suffered by red, yellow and violet beams passing through an equilateral transparent prism are 38’4,38’7 and 39’2 respectively. Calculate the dispersive power of the medium.
7. Two prisms of identical geometrical shape are combined with their refracting angles oppositely directed. The materials ofthe prisms have refractive indices 1’52 and 1’62 for violet light. A violet ray is deviated by 1’0 when passes symmetrically throughthis combination. What is the angle of the prisms ?
Solutions :
Two prisms of identical geometrical shape are combined.Let A = Angle of the prismsµ′v = 1.52andµv = 1.62, δv = 1
δv = (µv − 1)A− (µ′v − 1)A[SinceA = A′]
⇒ A = δvµv−µ′v
= 11.62−1.52
= 10
7. Two prisms of identical geometrical shape are combined with their refracting angles oppositely directed. The materials ofthe prisms have refractive indices 1’52 and 1’62 for violet light. A violet ray is deviated by 1’0 when passes symmetrically throughthis combination. What is the angle of the prisms ?
8. Three thin prisms are combined as shown in figure . The refractive indices of the crown glass for red, yellow and violet raysare µr , µs and µy, respectively and those for the flint glass are µ′r,µ
′yand µ′s, respectively. Find the ratio A’/A for which (a) there
is no net angular dispersion, and (b) there is no net deviation in the yellow ray.
Solutions :
Total deviation for yellow ray produced by the prism combination isδy = δcy − δfy + δcy = 2δcy − δfy = 2(µcy − 1)A− (µcy − 1)A′
Similarly, the angular dispersion produced by the combination isδv − δr = [(µvc − 1)A− (µvf − 1)A′ + (µvc − 1)A]− [(µrc − 1)A− (µd − 1)A′ + (µr − 1)A)]= 2(µvc− 1)A− (µV f − 1)A′
⇒ A′
A=
2(µcv−µrc)(µvf−µrf )
=2(µV −µr)(µ′v−µ′r)
9. A thin prism of crown glass (µr, = 1.515, µs, = 1’525) and a thin prism of flint glass (µr, = 1’612,. µs = 1’632) are placed incontact with each other. Their refracting angles are 5’0 each and are similarly directed. Calculate the angular dispersion producedby the combination.
Solutions :
Given that µcr=1.515,µcv=1.525 and µfr=1.612,µfv=1.632 and A=5.Since, they are similarly detected, the total deviation produced is given by,δ = δc + δr = (µc − 1)A+ (µr − 1)A = (µc + µr − 2)ASo, the angular dispersion of the combination is given by,
10. A thin prism of angle 6.0,ω = 0.07 and µy = 1.50 is combined with another thin prism having ω= 0.08 and µy = 1.60. Thecombination produces no deviation in the mean ray. (a) Find the angle of the second prism. (b) Find the net angular dispersionproduced by the combination when a beam of white light passes through it. (c) If the prisms are similarly directed, what will bethe deviation in the mean ray ? (d) Find the angular dispersion in the situation described in (c).
Solutions :Given thatA = 6, ω′ = 0.07 µ′y = 1.50
A=? ω = 0.08 µy = 1.60The combination produces no deviation in the mean ray
(c) If the prisms are similarly directed, δy = (µy − 1)A+ (µ′y − 1)A=(1.60-1)5+(1.50-1)6=3 + 3 = 6
(d) Similarly, if the prisms are similarly directed , the net angular dispersion is given by,δv − δr = (µy − 1)ωA− (µ′y − 1)ω′A′ = 0.24 + 0.21 = 0.45
11. The refractive index of a material M1 changes by 0’014 and that of another material M2 changes by 0.024 as the colour ofthe light is changed from red to violet. Two thin prisms one made of M1(A = 5.3) and other made of M2(A = 3.7) are combinedwith their refracting angles oppositely directed. (a) Find the angular dispersion produced by the combination. (b) The prisms arenow combined with their refracting angles similarly directed. Find the angular dispersion produced by the combination.
(a) When the prisms are oppositely directedangular dispersion = (µv − µr)A− (µ′v − µ′r)A′=0.024×3.7 + 0.014× 5.3 = 0.163
1. In an experiment to measure the speed of light by Fizeau’s apparatus, following data are used : Distance between the mirrors= 12.0 km, Number of teeth in the wheel = 180. Find the minimum angular speed of the wheel for which the image is not seen.
Solutions :In the given Fizeau apparatus,
D=12km=12×103kmn=180c=3×108m/s.We know, c= 2Dnω
π
⇒ ω = πc2Dn
rad/sec = πc2Dn
× 180πdeg/sec
⇒ ω = 180×3×108
24×103×180= 1.25× 104deg/sec
2. In an experiment with Foucault’s apparatus, the various distances used are as follows : Distance between the rotating andthe fixed mirror = 16 m Distance between the lens and the rotating mirror = 6 m,
Solutions :In the given Focault expirement
, R=Distance between fixed and rotating mirror=16mω = Angularspeed = 356rev/′ = 356× 2rad/secb=distance between lens and rotating mirror = 6ma= distance between source and lens=2ms=shift in image = 0.7cm= 0.7× 10−3mSo, speed of light is given by,
C= 4R2ωas(R+b)
= 4×162×356×2π×20.7×10−3(16+6)
= 2.975× 108m/s
3. In a Michelson experiment for measuring speed of light, the distance traveled by light between two reflections from therotating mirror is 4.8 km. The rotating mirror has a shape of a regular octagon. At what minimum angular speed of the mirror(other than zero) the image is formed at the position where a non-rotating mirror forms it ?
Solutions :
Page 116
24Tutors.com
In the given Michelson experiment,D=4.8km=4.8×103mN=8We know, c=DωN2π
⇒ ω = 2πcDN
rad/sec = cDN
rev/sec = 3×108
4.8×103×8= 7.8× 103rev/sec
A source emits 45 joules of energy in 15s. What is the radiant flux of the source ?
Solutions :Radiant Flux = Tota energy emitted
Time= 45
15s= 3w
A photographic plate records sufficiently intense lines when it is exposed for 12s to a source of 10W . How long should it beexposed to a 12W source radiating the light of same colour to get equally intense lines ?
Solutions :To get equally intense lines on the photographic plate, the radiant flux (energy) should be same. So, 10W × 12sec = 12W × t
⇒ t = 10W×12sec12W
= 10sec.
Using figure (22.1), find the relative luminosity of wavelength (a) 480nm, (b) 520nm (c) 580nm and (d) 600nm.
Solutions :It can be found out from the graph by the student.
The relative luminosity of wavelength 600nm is 0.6. Find the radiant flux of 600nm needed to produce the same brightnesssensation as produced by 120W of radiant flux at 555nm.
Solutions :Relative luminousity = Luminous flux of a source of given wave length
Luminous flux of asource of 555nm of same power
Let the radiant flux needed be P watt.Ao, 0.6 = Luminous flux of source ′P ′ watt
685 p
∴ Luminous flux of the source = (685P )× 0.6 = 120× 685⇒= 120
0.6= 200W
The luminous flux of a monochromatic source of 1W is 450lumen/watt. Find the relative luminosity at the wavelength emitted.
Solutions :The luminous flux of the given source of 1W is 450 lumen/watt
⇒ Relative luminosity = Luminous flux of the source of given wavelengthLuminous flux of 555nm source of same power
= 450685
= 66%
[∴ Since, luminous flux of 555nm source of 1w = 685 lumen]
A source emits light of wavelengths 555nm and 600nm. The radiant flux of the 555nm part is 40 W and of the 600nm partis 30W . The relative luminosity at 600nm is 0.6. Find (a) the total radiant flux, (b) the total luminous flux, (c) the luminousefficiency.
A source emits light of wavelengths 555nm and 600nm. The radiant flux of the 555nm part is 40 W and of the 600nm partis 30W . The relative luminosity at 600nm is 0.6. Find (a) the total radiant flux, (b) the total luminous flux, (c) the luminousefficiency.
Solutions :The radiant flux of 555nm part is 40W and of the 600nm part is 30W
a) Total radiant flux = 40W + 30W = 70Wb) Luminous flux = (L.F lux)555nm + (L.F llux)600nm
A light source emits monochromatic light of wavelength 555nm. The source consumes 100W of electric power and emits 35Wof radiant flux. Calculate the overall luminous efficiency.
Solutions :Overall luminous efficiency = Total luminous flux
Power input= 35×685
100= 239.75 lumen/W
A source emits 31.4W of radiant flux distributed uniformly in all directions. The luminous efficiency is 60lumen/watt. What isthe luminous intensity of the source ?
A point source emitting 628 lumen of luminous flux uniformly in all directions is placed at the origin. Calculate the illuminanceon a small area placed at (1′0m, 0, 0) in such a way that the normal to the area makes an angle of 37 with the X-axis
The illuminance of a small area changes from 900lumen/m2 to 400lumen/m2 when it is shifted along its normal by 10cm.Assuming that it is illuminated by a point source placed on the normal, find the distance between the source and the area in theoriginal position.
Solutions :
Page 117
24Tutors.com
Let, l = Luminous intensity of sourceEA = 900lumen/m2
EB = 400lumen/m2
Now, Ea = lcosθX2 and EB = lcosθ
(X+10)2)
So, l = EAX2
cosθ=
EB(X+10)2
cosθ
⇒ 900x2 = 400(x+ 10)2 ⇒ XX+10
= 23⇒ 3X = 2X + 20⇒ X = 20cm
A point source emitting light uniformly in all directions is placed 60cm above a table-top. The illuminance at a point on thetable-top, directly below the source, is 15lux. Find the illuminance at a point on the table-top 80cm away from the first point.
Solutions :
Given that, Ea = 15 lux l0602
⇒ l0 = 15× (0.6)2 = 5.4 candela
So, EB = l0cosθ(OB)2
=5.4×( 3
5)
12 = 3.24 lux
Light from a point source falls on a small area placed perpendicular to the incident light. If the area is rotated about the incidentlight by an angle of 60, by what fraction will the illuminance change ?
Light from a point source falls on a small area placed perpendicular to the incident light. If the area is rotated about the incidentlight by an angle of 60, by what fraction will the illuminance change ?
Solutions :The illuminance will not change.
A student is studying a book placed near the edge of a circular table of radius R. A point source of light is suspended directlyabove the centre of the table. What should be the height of the source above the table so as to produce maximum illuminance atthe position of the book ?
Solutions :
Let the height of source is ’h’ and the luminous intensity in the normal direction is l0,So, illuminance at the book is given by,E = l0cosθ
r2= l0h
r3= l0h
(r2+h2)32
For maximum E, dEdh
= 0⇒ l0[(R2+h2)32− 3
2h×(R2+h2)
t2×2h]
(R2+h2)3
⇒ (R2 + h2)12 [R2 + h2 − 3h2] = 0
⇒ R2 − 2h2 = 0⇒ h = R√2
Figure (22 − E1) shows a small diffused plane source S placed over a horizontal table-top at a distance of 2.4m with its planeparallel to the table-top. The illuminance at the point A directly below the source is 25lux. Find the illuminance at a point B ofthe table at a distance of 1.8m from A.
An electric lamp and a candle produce equal illuminance at a photometer screen when they are placed at 80cm and 20cm fromthe screen respectively. The lamp is now covered with a thin paper which transmits 49luminous flux. By what distance should thelamp be moved to balance the intensities at the screen again ?
