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Solution to Subjective Assignments
LEVEL – I
1. Comparing the general expression 0v = (v0 cos θ0) î + (v0 sin θ0) ĵ with thegiven equation v0 = 20 î + 10 ĵ , we obtain
v0 cos θ0 = 20 and v0 sin θ0 = 10
!he x"component o# velocit$ v a#ter a time interval t = 1 sec, v x = v0 cos θ0 = 20m%sec because the hori&ontal component o# velocit$ o# a pro'ectile remains
constant and the corresponding $"component o# velocit$ v is given as v$ = (v0 sin
θ0) gt putting v0 sin θ0 = 10 m%sec, g = 10 m%sec2 and t = 1 sec, we obtainv$ = 10 10 * 1 = 0
⇒ !he required velocit$ = v = vx î + v$ ĵ = 20 î
2. !he velocit$ o# the bod$ at an$ time t #rom the instant o# pro'ection is given as
= 0 gt
e#erring to the derived #ormulae, time average velocit$ is given as
∫ = 0t
0dt+
t
1+ -utting = (0 gt) we obtain
∫
−=−⊥= 0
0
t
0
000
t 2
tg+dt)gt+(+
.here t0 = total time o# ascent /ince = 0 at t = t0 ⇒ t0 =g
+0
-uttingg
+t 00 = , we obtain
2
++ 0=
3. !he velocit$ o# the bomb 'ust a#ter its release is equal to the velocit$ o# the
aeroplane !he displacement r can be directl$ written b$ re#erring the derived
ormula as r = $ 1g$
v2 20 +
$ putting
v0 = 100 m%sec, g = 10, $ = 1000 m, we obtain
r = 132 m (approximatel$)
4nd φ = tan"1
0v2
g$ = tan"1
×100
2
100010
⇒ φ = tan"1 00
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8/18/2019 Kinemat-Hints and Solution
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4. 5et the initial velocit$ has the
magnitude v0, inclined at an angle
θ0 with hori&ontal !he time ta6enb$ the #ootball to cover 70 m is
equal to 7 seconds
⇒ t =00 cosv
70
θ
⇒ v0 cos θ0 = 70%t = 70%7 = 10 m%s
!he 6inematical equation #or
vertical motion is
$ = (v0 sin θ0)t"2gt
2
1
50 mt ' 5 (
)50, - *.5+
θ0
v0
y
t ' 0
y ' *.5 m)0,0+
⇒ (v0sin θ0) (t) =2gt
2
1" $
⇒ (v0sin θ0) (7) =2)7)(10(
2
1 " 17
⇒ v0sin θ0 = 287
7123= m%s
'9)sinv(i9)cosv(v 00000 θ+θ=
) '928i910(v0 += m%s
⇒ v0 = 2::8 m%s ; θ0 = tan−1 10
28= 67.960.
5. !he distance travelled b$ the roc6et in the #irst 1 minute in which resultant
acceleration is verticall$ upwards and 10 m%s2 will be
h1 = 0 × :0 +2
1 × 10 × :02 = 1
8/18/2019 Kinemat-Hints and Solution
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6. 5et the$ meet a#ter time t #rom the instant o# release at point - as shown in the
#igure /ince /1 is downwards,
/1 = /g + /i = @gt2 + 1t (a)
; /2 is upwards ⇒ /2 = /i " /g
⇒ /2 = 2t " @ gt2 (b)
(a) + (b) ⇒ /1 + /2 = (1 + 2)t
⇒++
h t
+=
(/1 + /2 = h ; 1 = 2 = )
⇒2V
) t =
t = t
p
t = 0
v1 = v
t = 0
v2 = v
1s
2s
7. &!'5et the total distance covered = h m, and the totaltime ta6en = ! seconds
4ccording to the question,
the path covered during !th second = h%2
⇒ Aistance covered during (!" 1) seconds = h%2⇒ (1% 2)g(!"1)2 = h%2,where h =(1%2) g!2
⇒ (1%2)g (! 1)2 = (1% 8)g!2
⇒ (!"1)%! = ± 1%√2⇒ ± ! = √2! " √2⇒ !(√2 1) = √2
t '0
A
t 'T-*
t 'TC
B
⇒ ! =12
2
±s
/ince ! B 1, we accept the other value
⇒ ! =12
2
−= √2(√2+1)= &2* 2' s.
&b' !he height o# #all = h = (1%2) g!2 = (1% 2)(
8/18/2019 Kinemat-Hints and Solution
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θ=θ−=∆2
sin22r ) cos 1(2r r 222
⇒2 sinr 2 r
θ=∆
ow the magnitude o# average velocit$
= v =t
r
∆∆
t
2 sinr 2
v
θ
=
(i)
we 6now that
r
v ;
t=ωω=θ (when w =
constant)
⇒v
r t θ= (ii)
r ∆
v
Ft=t
t=0
-θ v
1r
2r
G
-utting t #rom (ii) in (i) we obtain
θ
θ
=θ
θ= 2
sinv2
v
r
)2%(sin r 2v ⇒
-
.2'& sin2
v
v θ=
10. .e 6now that velocit$ o# rain wrt man v rm is given as mr rm v v v −=
5et21 mm
v ; v are the initial ; #inal velocities o# the man, then
2mr 2rm1mr 1rmv"vv ;v"vv ==
⇒ 2m2rm1m1rmr vvv v v +=+=
e#erring the vector diagram we obtain
21
21 rmmr
vvv +=1m
v
1rmv
θ2rm
v
2mv
21rm
2r vvv += (a)
-utting C = ∆v = nvwe obtain
θ= cot nvv1rm (b)
Hsing (a) ; (b) we #ind22
r ) cot v(vv θη+=-cot/1v v 22" +=
θα
1rmv 2rm
v
1mv
2mv
A
C 4
v nv
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LEVEL – II
1. 5et a#ter a time t the particles attain the ground level !he particles acquire equal
vertical velocit$ a#ter time t, that is equal to gt
!he hori&ontal components o# the velocities remain uncharged !here#ore the
total velocities o# the particle a#ter time t are given as
'9 gti9vv 11 +−=′ ; '9 gti9vv 22 +=′
/ince the$ move perpendicularl$ 'ust be#ore touching the ground, there#ore
0v v 21 =
⇒ 0 ) '9 gt i9(v ) '9 gti9v( 21 =++−
⇒ g2t2 = v1v2 ⇒ g
vvt
21=
!he vertical displacement o# each particle =2
1h = gt2 where h = height o# the
pole
;g
vvt
21=
⇒g2
vv 21=
=
2
21
g
vv g
2
1 h
+1 +2 =0 =0
=t =t +2 +1
2v ′ gt
1v′
gt
2. 4t the point o# collision, x1 = x2 and $1 = $2
⇒ vo cosθ1t1 = vo cosθ2t2 or t2 =
θ
θ
2
1
cos
cos
t1
.here θ1 and θ2 are the angles o# pro'ection and t1, t2 are the time o# #light
/imilarl$ #or equal vertical displacement ($1 = $2)
vosinθ1t1 − 2
1g 21t = vosinθ1t1 −
2
1g 212t
vosinθ1t1 − vosinθ2
( )222112
1 tt2
gt
cos
cos−=
θθ
vot1( )
θ
θ−θ=
θθ−θ
2
2
12
22
21
2
21
cos
coscost
2
g
cos
sin
⇒ t1 =( )
12
22
221o
coscos
cossinv2
θ−θ
θθ−θ
t2 = t1 =2
1
cos
cos
θθ
∴ !ime interval ∆t = t1 − t2 =( )
21
21o
coscosg
sinv2
θ+θθ−θ
/ubstituting the given values,
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∆t = 10 s
3. orward Iourne$
ector C represents wind velocit$
4 represents velocit$ plane relative to
ground 4C represents velocit$ o# plane in still air
800cos i9 + (800 sinθ + 70) '9 = 4
tan87° = θ+θ
cos800
70sin800
4+
0
1 E
S
2
3
A
⇒ tan θ +<
secθ = 1
⇒ tan2θ − 2tanθ + 1 =:8
1tan2 +θ
⇒ :3 tan2θ − 12< tanθ + :3 = 0
⇒ tanθ =12:
7180
12:
70
8/18/2019 Kinemat-Hints and Solution
7/15
∴ s = s1 + s2 = u1t + u2t = (u1 + u2)g
uu 21
/ubstituting the values, the distance between the particles =
8/18/2019 Kinemat-Hints and Solution
8/15
⇒ i9)20v(v x −=ωµ + v$ '9
/ince the wind appears to blow #rom north east, the x ; $ components o# mv ω
must be equal
⇒ (vωm)$ = (vωm)x and both are negativehence vx " 20 = 10 " 20 = "10 m%s
⇒ wind speed = 10√2 m%s and the wind is blowing #rom north west
9 xdx
+d+a ==
⇒ dxx+d+ = ⇒ ∫ = dxx+d+
⇒ 2%320
2
x3
2
2
++=
− ⇒3%2
20
2
8
)++(3x
−=
-utting = 2 0 we obtain , x =
3%8
0
3%2
2
02
+3+
8
C
=
10. 5et us consider that velocit$ o# train with respect to ground
=v! and velocit$ o# car in the #irst part o# 'ourne$
= ve and velocit$ o# car in the last part o# 'ourne$
=v′e
4lso distance o# the point where the ob'ect #ell and where the car turns bac6
=x6m
?enceM ve − v! =:
16m%min (1)
v′e + v! = :
1
6m%min (2)
1v
x
e
= (3)
ev
N
′ = 2 (8)
/olving these (1), (2), (3) and (8)
ve = 1338 6m%hr, v′e = :: 6m%hr v! = 338 6m%hr x = 1338 6m
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Solutions to 5bjective Assignments
LEVEL – I
1. !he stone is sub'ected to earthPs gravitational #ield a#ter losing contact with theelevator !here#ore its acceleration will be equal to g = < m%s 2 pointed verticall$
downwards
2. 5et this pro'ectiles be pro'ected with velocities
'9)sinv(i9)cosv( '9$i9vv1111$x1 11
θ+θ=+=
and '9)sinv(i9)cosv( '9$i9vv 2222$x2 22 θ+θ=+=
!he velocities o# the pro'ectiles a#ter a time t are given as
'9)gtsinv(i9)cosv(v11111
−θ+θ=′
'9)gtsinv(i9)cosv(v22222
−θ+θ=′
'9)sinvsinv(i9)cosvcosv(vv2211221121 θ−θ+θ−θ=′−′
/ince the relative velocit$ 21 vv ′−′ is a constant and so does not var$ with time,the locus o# one wrt the other is a straight line
3. v = a1t1 = a2t2
∴ t1 = (a2%a1) t2 = (8%2)t2 = 2t2t1 + t2 = 3 or 2t2 + t2 = 3t2
t2 = 1 sec and t1∴ v = 2 × 2 = 8 m%s
4. N =ct2 ; $ = bt2
⇒ ct2dt
dx= ; bt2
dt
d$=
/ince v =22
dt
d$
dt
dx
dt
ds
+
=
⇒ v = 22 )bt2()ct2( +
⇒ v = 2t 22 cb +
+. 5et G- = r 4ngular speed about the origin = ω =tp0
t0p v where,r
v = !he
component o# velocit$ o# - wrt G perpendicular to G-
⇒r
sinv
θ=ω where r = b cosec θ
⇒b
sinv
2θ=ω
⇒ (C) is correct
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!he angular speed o# an$ point - wrt another point F = ωpq = E(!he componento# relative velocit$ between them is perpendicular to pq) % (!he distance o#
separation pq)D
!here#ore $ou can neither writeb
v
sinop
v nor
cosecb
v
op
v=
θθ=
!he last choice (A) is dimensionall$ wrong
6. 5et the particle be pro'ected verticall$ up"ward with velocit$ u #urther, let t bethe time when pro'ectile be at height h !hen
h = ut "2
1gt2 or
2
1gt2 ut + h = 0
∴ t2 " 0g
h2t
g
u2=+ (1)
#rom equation (1)
sum o# roots, t1 + t2 = " g
u2
a
b= (2)
product o# roots, t1t2 = gh2
a
c = (3)
Qiven that3
1
t
t
2
1 = or t2 = 3t1 (8)
#rom equation (2) and (8), we get
8t1 = g2
uort
g
u21 = (7)
#rom equation (3) and (8) we get
32
1t =
g
h2
g2
u3or
g
h22
=
or3h8
g2u2 = (:)
4lso, maximum height, ? =g2
u2
()
#rom eq (:) and () we obtain,
? =8
?3orh
3
h8=
7. $ geometr$ o# the #igure the length o# the string is = 8x 4 + x
⇒dt
dx
dt
dx8
dt
d 4 +=
/ince the length o# the string is constant,dt
d = 0
-utting
4
4 vdt
dx ;v
dt
dx==
8
v v 4 = numericall$
4 = 4v"v = v 4 + v =
8
v + v = 127v = 127v
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/ince the displacement o# is #our times that o# 4, their speeds can not be equal
,. !ime ta6en =mv
d
= 2w
2mw vv
d
− = 22
8/18/2019 Kinemat-Hints and Solution
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1+. !he height at which stone was released = 80melocit$ o# balloon in the upward direction at the time o# release o# stone
= 127 ×
8/18/2019 Kinemat-Hints and Solution
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1. = 2dx
tdt
S(i)
31 t$
2 3=
2d$ t
dt 2= S(ii)
t = 1, vx = 1, v$ =*
19 9v i '2
= +r
2
2
d x2t
dt= S(iii)
2
2
d $t
dt= S(iv)
at t = 1 s ax = 2 and a$ = 1
ˆ ˆa i j∴ = +r
2. Hsing constrained equation, / *c0(v vθ =
Gn di##erentiation / *c0(a aθ =
3. 2 2($ h) x h− + + = l
2 2
d$ x dx0
dt dtx h+ =
+
2 2
d$ x dx
dt dtx h= −
+
Ady vdt 5
= −
, ' - c m
h
'
1
c
m
y
4
3J u J v
7= S(i)
2 2
42 2 2
d $ hv
3dt (x h )2
=+
4 3
1:a v
(7)=
4
1:a v
127= S(ii)
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4. i#
2 2 2 2u cos u sin
g 2g
θ θ>
*tan /−θ <
i#
2 2 2 2u cos u sin
g 2g
θ θ
<1tan 2−θ >
+. inal velocit$ o# each ball
v )/gh+=!his is independent o# path
urther,1 2h
tsin g
= ÷θ
or *
t(in
µθ
#or same h
1
2
t sin
t sin
β∴ =α
7. 4s velocit$ o# the particle increases #rom&ero and #inall$ it decreases to &ero,hence acceleration can not remainpositive #or all the time o# motion>inimum magnitude o# acceleration canbe #ound out b$ the velocit$" time curveAisplacement = 4rea bounded b$ the graph and time axis
>agnitude o# minimum acceleration = 8 m%s2∴ (4) ; (C)
COMPREHENSIONS
1$
dv2t
dt=
∴ 2$d$
v tdt
= =
and $ =t
2 Ari#t = 0 × time o# crossing
3 0x v t= S(i)3t
$3
= S(ii)
#rom (i) and (ii)
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8/18/2019 Kinemat-Hints and Solution
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3
3
0
x$
3v=
8 . 0 2 - 0 2
* 0 m / (
5 m / (
α
v ,
v y5 - m / (
v
!ime a#ter which velocit$ vector becomes perpendicular to initial velocit$ vector is
u 10 2t
gsin 10 sin :0 3= = =
θ seconds
5et v$ be the vertical component o# velocit$ at that instant thenv$ = u + at
v$ =10 2
7 33
×−
v '
v ' 5,
α
v ',
5
-
* 0
-
$
7v
3= −
2
2 7v 73
∴ = + ÷
10
v3
= m%s2v
gcos,
α =
2
vgcos
=α
= 203 3
= m
7 4cceleration o# particle is equal to acceleration due to gravit$
: at = g sin α73
1010
3
= × = 7 m%s2
%A( (E 5LL5I8
1 jo o 99u :0cos30 :0sin30→
= i +
( )$ 9a g '= −a 0=
x
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