PHY10 DIAGNOSTIC

Phys101 Term: 122 Online HW-Ch05-Lec01

KFUPM-Physics Department 1

Q1: An object moving at constant velocity in an inertial frame must:

A. have a net force on it B. eventually stop due to gravity C. not have any force of gravity on it D. have no frictional force on it E. have zero net force on it

Ans:

Q2:

Two students are dragging a box of mass m across a horizontal frozen lake (frictionless surface). The first student pulls with force F1= 50.0 N, while the second pulls with force F2

= 30 N. The direction of both forces is shown in the figure. Find the angle (in degrees) between the resultant force and the positive x-axis. (Give your answer in three significant figures form)

Ans: The resultant force F F = F 1 + F 2 F = (F1x + F2x) + F1y + F2y Where F1x = F1cos60 = 50 cos60 = 25 N F1y = F1sin60 = 50 sin60 = 43.3 N F2x = F2cos30 = 30 sin30 26 N F2y = F2sin30 = 30 cos30 = 15 N F = (25 + 26) + (43.3 15) N = (51 + 28.3 ) N & = tan1 28.351 = 29.0

Phys101 Term: 122 Online HW-Ch05-Lec01


Q3:

Only two forces, F1 and F2 act upon a 10 kg box. One of the forces is F1 = (3 i -4 j) N. If the box moves at a constant velocity v = (12 i +20 j) m/s, what is the magnitude (in N) of the second force F2

Ans: ? (Give your answer in three significant figures form)

Since the velocity is constant then the acceleration is zero.

Newtons second law:

F 1 + F 2 = 0 F 2 = F 1 and F 2 = F 1 = 32 + 42 = 5.00 N

F net = ma 0

Phys101 Term:123 Online HW-Ch06-Lec02


Q1: A small car with mass 1.60 kg move at constant speed of v = 12 m/s in a vertical circle with radius 5.00 m (Figure). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point A (at the bottom of the vertical circle) and point B (at the top of the vertical circle)?

A) = 61.8 N, = 30.4 N B) = 23.8 N, = 10.2 N C) = 76.8 N, = 82.1 N D) = 22.1 N, = 10.4 N E) = 53.8 N, = 64.4 N

Ans: A FA = m g + v2R = 61.8 N FB = mg v2R = 30.4 N Q2: A flat (unbanked) curve on a highway has a radius of 220.0 m. A car rounds the

curve at a speed of 25 m/s? What is the minimum coefficient of friction that will prevent sliding?

Ans: fs = m v2R s = v2gR = 0.290 Q3:

At what angle should the roadway on a curve with a 50 m radius be banked to allow cars to negotiate the curve at 12 m/s even if the roadway is icy (and the frictional force is zero)?

Ans: Tan = v2Rg = tan1 v2Rg = 16.3

Chap6

A boy pulls a wooden box along a rough horizontal floor at constant speed by means of a force as shown. In the

diagram f is the magnitude of the force of friction, N is the magnitude of the normal force, and Fgis the magnitude of the

force of gravity. Which of the following must be true?

1. Test Bank, Question 14

P = f and N = Fg

P = f and N > Fg

P > f and N < Fg

P > f and N = Fg

none of these

A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If the coefficients of friction

are s = 0.5 and k = 0.4, the magnitude of the frictional force on the crate is:


8 N

12 N

16 N

20 N

40 Nb

A 24-N horizontal force is applied to a 40-N block initially at rest on a rough horizontal surface. Ifthe coefficients of

friction are s = 0.5 and k = 0.4, the magnitude of the frictional force on the block is:


8 N

12 N

16 N

1 7 Print Assignment: Chap6 ::false

2011/10/3http://edugen.wileyplus.com/edugen/shared/assignment/test/agprint.uni?numberQuest=true&titleT...

20 N

400 N

A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static friction is 0.40. To start the crate moving with the weakest possible applied force, in what direction should the force be applied?


Horizontal

24 below the horizontal

22 above the horizontal

24 above the horizontal

66 below the horizontal

Body A in Fig. 6-33 weighs 91 N, and body B weighs 84 N. The coefficients of friction between A and the incline are s = 0.57 and k = 0.26. Angle is 44. Let the positive direction of an x axis be down the slope. What is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

Fig. 6-33 Problems 27 and 28.

5. *Chapter 6, Problem 27

(a) Number 0 Units m/s^2

(b) Number -0.210924482284 Units m/s^2

(c) Number -2.117117405998 Units m/s^2

In Fig. 6-46, a box of ant aunts (total mass m1 = 1.76 kg) and a box of ant uncles (total mass m2 = 4.13 kg) slide down

an inclined plane while attached by a massless rod parallel to the plane. The angle of incline is = 21. The coefficient of kinetic friction between the aunt box and the incline is 1 = 0.295; that between the uncle box and the incline is 2 = 0.135. Compute (a) the tension in the rod and (b) the common acceleration of the two boxes.




Fig. 6-46 Problem 60.

(a) Number 1.806530177532 Units N

(b) Number 1.839462493411 Units m/s^2

7. Chapter 6, Concept Question 5

The following questions are in regard to the situation where you press an apple crate against a wall so hard that the crate cannot slide down the wall.

What is the direction of the static frictional force on the crate from the wall?

Horizontal, away from you

Horizontal, towards you

Downward

Upward

What is the direction of the normal force on the crate from the wall?

Horizontal, away from you

Downward

Upward

Horizontal, towards you

If you increase your push, what happens to fs?

Increase

Decrease

No change

If you increase your push, what happens to FN?



Increase

Decrease

Remain the same

If you increase your push, what happens to fs, max?

Increase

Decrease

Remain the same

The two blocks (m = 14 kg and M = 100 kg) in Fig. 6-38 are not attached to each other. The coefficient of static friction between the blocks is s = 0.41, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force required to keep the smaller block from slipping down the larger block?


Significant digits are disabled; the tolerance is +/-5%


Number 381.482926829268 Units N

A block of mass mt = 5.00 kg is put on top of a block of mass mb = 6.00 kg. To cause the top block to slip on the

bottom one while the bottom one is held fixed, a horizontal force of at least 16.0 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table (Fig. 6-47). Find the magnitudes of (a) the maximum

horizontal force that can be applied to the lower block so that the blocks will move together and (b) the resulting

acceleration of the blocks.





(a) Number 35.2 Units N

(b) Number 3.2 Units m/s^2

Fig. 6-53 shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.) The bob has a mass of 0.034 kg, the string has length L = 0.63 m and negligible mass, and the bob follows a circular path of circumference 0.65 m. What are (a) the tension in the string and (b) the period of the motion?



(a) Number 0.337785155677 Units N

(b) Number 1.582228931849 Units s

When a small 4.00 g coin is placed at a radius of 5.30 cm on a horizontal turntable that makes three full revolutions in 3.44 s, the coin does not slip. What are (a) the coin's speed, the (b) magnitude and (c) direction (radially inward-denote 0 or outward-denote 1) of the coin's acceleration, and the (d) magnitude and (e) direction (inward-denote 0 or outward-denote 1) of the frictional force on the coin? The coin is on the verge of slipping if it is placed at a radius of 14.0 cm. (f) What is the coefficient of static friction between coin and turntable?

11. *Chapter 6, Problem X9

(a) Number 29.041466972138 Units cm/s

(b) Number 159.133359225244 Units cm/s^2

(c) 0

(d) Number 0.006365334369 Units N

(e) 0

(f) Number 0.428493648982 Units This answer has no units



A certain string can withstand a maximum tension of 31 N without breaking. A child ties a 0.36 kg stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.93 m, slowly increasing the speed until the string breaks. (a) Where is the stone on its path when the string breaks? 1 - Top of the circle; 2 - Middle of the circle; 3 - Bottom of the circle (Give the number of the correct answer.) (b) What is the speed of the stone as the string breaks?

12. *Chapter 6, Problem X7

(a) 3

(b) Number 8.424329844761 Units m/s

A circular curve of highway is designed for traffic moving at 94 km/h. Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is 300 m, what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at 94 km/h?


(a) Number 13.056190281952 Units (degrees)

(b) Number 0.231901402536 Units This answer has no units

An 800-N passenger in a car, presses against the car door with a 200 N force when the car makes a left turn at 13 m/s. The (faulty) door will pop open under a force of 800 N. Of the following, the least speed for which the man is thrown out of the car is:


14 m/s

19 m/s

20 m/s

26 m/s

54 m/s

A ball is thrown downward from the edge of a cliff with an initial speed that is three times the terminal speed. Initially its acceleration is


upward and greater than g

upward and less than g

downward and greater than g

downward and less than g



downward and equal to g

A 909 kg boat is traveling at 86 km/h when its engine is shut off. The magnitude of the frictional force k between boat

and water is proportional to the speed v of the boat: fk = 83v, where v is in meters per second and fk is in newtons. Find

the time required for the boat to slow to 44 km/h.

Significant digits are disabled; the tolerance is +/-5%


Number 7.339437530876 Units s



-2-N force is applied horizontally to the free end, the force of the string on the toy, atthe other end, is

Y. KubotaNov. 14, 1997

Key: 1: d, 2: b, 3: e, 4: d, 5: c, 6: a, 7: b 8: a, 9: a, 10: c,(a) 0.15 N11: c, 12: b,13: e,14: e,15: e,16: d,17: b,18: c, 19: c, 20: d, 21: d.(b) 6.0 10-3 N

GOOD LUCK! (c) 2.5 10-2 N---------------------------------------------------------------------------------------------- (d) 3.0 10-2 N1 mile = 1.6 Km, 1 ft = 30.5 cm, 1 yd = 91.4 cm, 1 inch = 2.54 cm, (e) 3.5 10-2 N1 gallon = 3.8 liters = 3800 cm3 = 0.14 ft3, k(kilo) = 103, c(centi) = 10-2, m(mili) = 10-3,A B=AB cos = AxBx+AyBy+AzBz, 6. The velocity of a 0.5-kg hockey puck, sliding across a level ice surface, decreases at the rate

of 0.6 m/s2. The coefficient of kinetic friction between the puck and ice is|A B|=AB sin , A B= (AyBz- AzBy, AzBx- AxBz, AxBy- AyBx)

(a) 0.06x(t) - x(0) = vot + 12at

2 and v(t) = vo+ at when a is a constant.(b) 0.12

ax2 + bx + c = 0 fi x = (- b b2 - 4ac )/2a g = 9.8 m/s2. (c) 0.5(d) 2.0

SF i = ma, W(weight) = mg (downward) a =v2/r. (e) 16f sN, f = kN. 7. A block rests on a rough horizontal surface (ms = 0.50, mk = 0.40). A constant horizontal

force, just sufficient to start the block in motion, is applied. The acceleration of the block afterit starts moving, in m/s2, is

DK=W, K=12mv2, W=SFi D r.

----------------------------------------------------------------------------------------------1. A circus performer of weight W is walking

along a "high wire" as shown. The tension inthe wire is

(a) 0(b) 0.98

(a) approximately W(c) 3.3(d) 4.5

(b) approximately W/2 (e) 8.9(c) much less than W(d) much more than W 8. A boy pulls a wooden box along a rough horizontal floor at constant speed by means of a

force P as shown. Which of the following must be true (f is the magnitude of the force offriction, N is the magnitude of the normal force, and W is the magnitude of the weight)

(e) depends on whether he stands onone or two feet

(a) P=f and N=W2. A 25-N crate slides down a frictionless incline that is 25 above the horizontal. The magnitudeof the normal force of the incline on the crate is (b) P=f and N>W

(a) 11 N(c) P>f and Nf and N=W

(b) 23 N (e) none of these(c) 25 N(d) 100 N(e) 220 N

9. A block of mass m is pulled at constantvelocity along a rough horizontal floor by anapplied force T as shown. The frictional forceis

3. Two blocks (60 lb and 100 lb) are connected by a string thatpasses over a massless pulley as shown. The tension in thestring is

(a) 40 lb (a) T cos(b) 60 lb

(b) T sin(c) 100 lb(c) zero(d) 160 lb(d) mg(e) none of these(e) mg cos

4. Three books (X, Y, and Z) rest on a table. The weightof each book is indicated. The net force acting on bookY is

10. A horizontal force of 5.5 N pushes a 0.50-kg block against a vertical wall. The block isinitially at rest. If ms = 0.6 and mk = 0.80, the frictional force is

(a) 4 N down (a) 0(b) 5 N up (b) 3.9 N(c) 9 N down (c) 4.4 N(d) zero (d) 4.9 N(e) none of these (e) 5.5 N

11. A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.49 and thecoefficient of kinetic friction is 0.40. After one end of the plank is raised so the plank makesan angle of 25 with the horizontal, the force of friction is

17. A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the track, causes the object to speed up as it goes around. Thework done by the external force as the mass makes one revolution is

(a) 0 N(a) 24 J(b) 18 N

(c) 21 N (b) 47 J(d) 22 N (c) 59 J(e) 44 N (d) 94 J

(e) 120 J12. A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.49 and the

coefficient of kinetic friction is 0.40. After one end of the plank is raised so the plank makesan angle of 30 with the horizontal, the force of friction is

18. A 700-kg elevator accelerates downward at 3.0 m/s2. The force exerted by the cable on theelevator is

(a) 0 N (a) 2.1 kN, up(b) 18 N (b) 2.1 kN, down(c) 21 N (c) 4.8 kN, up(d) 22 N (d) 4.8 kN, down(e) 44 N (e) 9.0 kN, up

13. The driver of a 1000-kg car tries to turn through a circle of radius 100 m on an unbankedcurve at a speed of 10 m/s. The actual frictional force between the tires and slippery road is900 N. The car will

19. A particle moves with constant speed around the circle shown tothe right. When it is at point A its coordinates are x = 0, y = 2mand its velocity is (4m/s)i. When it is at point B its velocity andacceleration are (i and j are unit vectors pointing in the x and ydirections.)(a) slide into the inside of the curve

(b) make the turn(a) -(4m/s)j and (8m/s2)i, respectively(c) slow down due to the frictional force

(d) make the turn only if it went faster (b) (4m/s)j and -(8m/s2)i, respectively(e) slide off to the outside of the curve (c) (4m/s)j and (8m/s2)i, respectively

(d) (4m/s)i and (2m/s2)j, respectively14. One end of a 1.0-m string is fixed, the other end is attached to a 2.0-kg stone. The stone

swings in a vertical circle, passing the bottom point at 4.0 m/s. The string tension (innewtons) at this point is about

(e) (4m/s)j and 0, respectively

20. Which of the curves on the graph to the right bestrepresents vy vs. t for a projectile fired at an angle of45 above the horizontal?

vy

t

A

O

D

B

F

E

C

(a) 0(b) 12

(a) OC(c) 20(d) 32 (b) DE(e) 52 (c) AB

(d) AE15. A crate moves to

the right on ahorizontalsurface as awoman pulls onit with a 10-N

(e) AF

21. An object moving at constant velocity in an inertialframe must

(a) have a net force acting on itforce. Rank the situations shown below according to the work done by the 10-N force, leastto greatest. The displacement is the same for all cases.

(b) eventually stop due to gravity(c) not have any force of gravity acting on it

(a) 1,2,3(d) have zero net force acting on it(e) have no frictional force acting on it

(b) 2,1,3(c) 2,3,1(d) 1,3,2(e) 3,2,1

16. A baseball is hit high into the upper bleachers of left field. Over its entire flight the work doneby gravity and the work done by air resistance, respectively, are

(a) positive; positive(b) positive; negative(c) negative; positive(d) negative; negative(e) unknown since vital information is lacking

Must knows!!Multiple Choice

Problems

Physics IExam 2 Review

Christopher Lane1,2 Julia Bielaski1,2

1Department Physics, Clarkson University2Department Mathematics, Clarkson University

October 10, 2010

Clarkson University Physics Club Physics I Exam 2 Review


Problems

Outline

1 Must knows!!

2 Multiple ChoiceChapter 5: FORCE AND MOTION IChapter 6: FORCE AND MOTION II

3 ProblemsProblem 1Problem 2



Problems

Must Knows!!

Constants:g = 9.81 m

s2Volumes:

VSphere =43pir 3

VCylinder = pir2h Surface Area:

ASphere = 4pir2



Problems

Chapter 5: FORCE AND MOTION IChapter 6: FORCE AND MOTION II

Multiple Choice



Problems


Chapter 5: FORCE AND MOTION I



Problems


Question 1

An object moving at constant velocity in an inertial frame must:

A have a net force on it

B eventually stop due to gravity

C not have any force of gravity on it

D have zero net force on it

E have no frictional force on it

Answer: D



Problems


Question 2

Acceleration is always in the direction:

A of the displacement

B of the initial velocity

C of the final velocity

D of the net force

E opposite to the frictional force

Answer: D



Problems


Question 3

The inertia of a body tends to cause the body to:

A speed up

B slow down

C resist any change in its motion

D fall toward earth

E decelerates due to friction

Answer: C



Problems


Question 4

Equal forces F act on isolated bodies A and B. The mass of B is three timesthat of A. The magnitude of the acceleration of A is:

A three times that of B

B 1/3 that of B

C the same as B

D nine times that of B

E 1/9 that of B

Answer: A



Problems


Chapter 6: FORCE AND MOTION II



Problems


Question 1

A brick slides on a horizontal surface. Which of the following will increase themagnitude of the frictional force on it?

A putting a second brick on top

B decreasing the surface area of contact

C increasing the surface area of contact

D decreasing the mass of the brick

E none of the above

Answer: A



Problems


Question 2

Why do raindrops fall with constant speed during the later stages of theirdecent?

A The gravitational force is the same for all drops

B Air resistance just balances the force of gravity

C The drops all fall from the same height

D The force of gravity is negligible for objects as small as raindrops

E Gravity cannot increase the speed of a falling object to more than 9.8 m/s

Answer: B



Problems


Question 3

A ball is thrown upward into the air with a speed that is greater than terminalspeed. On the way up it slows down and, after its speed equals the terminalspeed but before it gets to the top of the trajectory:

A its speed is constant

B it speeds up

C it continues to slow down

D its motion becomes jerky

E none of the above

Answer:C



Problems


Question 4

An object moves around a circle. If the radius is doubled keeping the speed thesame then the magnitude of the centripetal force must be:

A twice as great

B half as great

C four times as great

D one-fourth as great

E the same

Answer: B



Problems

Problem 1Problem 2

Problems



Problems

Problem 1Problem 2

Problem 1

In the following system m1 accelerates downward in the negative direction. Consider

all pulleys to be massless and frictionless and consider both cords to be massless.

m1

m2

m3

m4

m1=6kg =/8 m2=1kg m3=3kg m4=5kg

A Draw and label all forces acting on each block.

B Find a simplified mathematical model describing the magnitude of theacceleration of the system.

C Find the magnitude of the tension in each cord for the values given.

D Find how long it takes m3 to reach the bottom the pit, given the depth, d ,of the pit is 3m, and the system starts from rest.

E Find a value for, , to keep the system from accelerating when released.



Problems

Problem 1Problem 2

Problem 2


Multiple choice Problem 1

A 50.0-N box is sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the box is sliding to the right at 1.75 m/s and that it stops in 2.25 s with uniform acceleration. The force that friction exerts on this box is closest to:

a) 3.97 N b) 490 N c) 50.0 N d) 8.93 N e) 38.9N Mass of box ism = 50N 9.8m / s2 = 5.1kg fk v0x = 1.75 m/s ax

Problem 2 A professor holds an eraser against a vertical chalkboard by pushing horizontally on it. She pushes with a force that is much greater than is required to hold the eraser. The force of friction exerted by the board on the eraser increases if she: WARNING: The correct answer may surprise you. Think about the amount of static friction in each case. A) pushes with slightly greater force B) pushes with slightly less force C) pushes so her force is slightly downward but has the same magnitude D) stops pushing E) pushes so her force is slightly downward but has the same magnitude The three diagrams below shows the possible scenario. In all cases the normal force is FN = Mgcos! (! = 0 for A and B). The maximum static friction is fsmax = FNs . We will assume that in all cases, the upward (+ y, as indicated below) vertical force cancels the downward forces so that the eraser do not fall. A and B C E

fs +y fs fs Static Friction +x Fcos

!F

!F

Applied Fsin Fsin Force

!F Fcos

Mg Mg Mg If the eraser is to remain on the wall (not fall) the y-component of the net force must be zero: Fynet = 0 . The static friction fs < fsmax is calculated for all scenarios below. ANSWER: Clearly E!

m

Final velocity (x) vx = 0 vx = v0x ! axt,vx = 0ax = v0x / t = 1.75m / s 2.25s = 0.778m / s

2nd law taking left as positive Fxnet = fk = max = 5.1kg( ) .778m / s2( ) = 3.97N answer a)

A and B y-component Fynet = fs ! Mg = 0fs = Mg

C, y-component +y (shown above) Fynet = fs ! Mg + F sin" = 0fs = Mg ! F sin"

E, y-component +y (shown above) Fynet = fs ! Mg ! F sin" = 0fs = Mg + F sin!

Problem 3

A series of weights connected by massless cords are given an upward acceleration of 4 m/s2 by a pull P as shown below. A, B and C are the tensions in the connecting cords. The smallest of the three tensions A,B, and C is closest to: a. 483 N pull P b. 621 N c. 196 N d. 276 N 5.00 kg e. 80.0 N

C 10.0 kg B 15.0 kg A

20.0kg

It is obvious that Tension A, TA, between 15.0kg and 20.0 kg mass, is the smallest tension. To solve see free body diagram on 20.0 kg (m = 20.0kg) below: ANSWER: D TA a = 4.0 m/s2 Use 2nd law Fynet = TA ! mg = ma TA = m(g + a) = 20kg 13.8m / s2( ) mg TA = 276N Kinematics with Calculus Problem 4 Kinematics with Calculus The position of a particle moving in an xy plane is given by

!r = 2t4 ! 3( ) i + t5 ! 2t( ) j ,

with !r in meters and t in seconds. A) find the average velocity and acceleration for the

time interval between t = 1s and t = 3s. B) find the velocity and acceleration at t = 1s in unit-vector notation. C) What is the angle between the positive direction of the +x axis and a line tangent (i.e.

!v ) to the particle's path at t = 1 s? Give your answer in the range of (-180o; 180o).

A) !r t( ) = xi + yj = 2t4 ! 3( ) i + t5 ! 2t( ) j

at t = 1s, !r 1s( ) = 2 1s( )4 ! 3( ) i + 1s( )5 ! 2 1s( )( ) j = !1mi !1mj

at t = 3s, !r 3s( ) = 2 3s( )4 ! 3( ) i + 3s( )5 ! 2 3s( )( ) j = 159mi + 237mj

!vavg =!x!t i +

!y!t j =

x 3s( ) " x 1s( )3s "1s i +

y 3s( ) " y 1s( )3s "1s j = 80

ms i +119

ms j

!v t( ) = vxi + vy j =dxdt i +

dydt j =

d 2t 4 ! 3( )dt i +

d t 5 ! 2t( )dt j = 8t

3( ) i + 5t 4 ! 2( ) j at t = 1s,

!v 1s( ) = 8 1s( )3( ) i + 5 1s( )4 ! 2( ) j = 8 ms i + 3ms j

at t = 3s,

!v 1s( ) = 8 3s( )3( ) i + 5 3s( )4 ! 2( ) j = 216 ms i + 403ms j

!aavg =!vx!t i +

!vy!t j =

vx 3s( ) " vx 1s( )3s "1s i +

vy 3s( ) " vy 1s( )3s "1s j = 104

ms2 i + 200

ms2 j

b)

!v 1s( ) = 8 1s( )3( ) i + 5 1s( )4 ! 2( ) j = 8 ms i + 3ms j

!a t( ) = axi + ay j =dvxdt i +

dvydt j =

d 8t 3( )dt i +

d 5t 4 ! 2( )dt j = 24t

2( ) i + 20t 3( ) j at t = 1s,

!a 3s( ) = 24 1s( )2 i + 20 1s( )3 j = 24 ms2 i + 20ms2 j

C)

!v 1s( ) = 8 1s( )3( ) i + 5 1s( )4 ! 2( ) j = 8 ms i + 3ms j

vx > 0 and vy > 0 ,1st quadrant ( !900 < " < 0! ). ! = tan"1 3m8m = 20.5

! , ! = 20.5!

Work by Graphical Integration Problem 5

The figure gives the acceleration of a 3.0 kg particle as an applied force moves it from rest along an x axis from x = 0 to x = 9.0 m. The scale of the figure's vertical axis is set

by as = 5.0ms2

.

A) Calculate the work done after the particle move from xi = 0m to xf = 1.0m. Repeat for xi = 0.5 m to xf = 1.0m. Repeat for the interval xi = 0 to xf = 5.0m. B) If at x = 1m, the particle is at rest (at v = 0), calculate the speed of the particle when it is at xf = 5.0m.

W = Fdxxix f! = m adxxi

x f! = m area under the graph[ ] For xi = 0 to xf = 1.0m:

adx01m

! = =12 base( ) ! height( ) =

12 1m( ) as( ) =

12 1m( ) 5

ms2

"#$

%&'= 2.5 m

2

s2

HINT: For part a) Use F = ma , and equation on work in the back. For part b) use the equation on power in the back.

W = m adxxix f! = 3kg " 2.5

m2s2

= 7.5J For xi = 0.5m to xf = 1.0m: Here I will enlarge the figure for clarity as as/2

adx0.5m1m

! = = minus 0m 1m 0.5m 1 m 0.5m 1.0m

adx0.5m1m! =

12 base( ) height( )big "

12 base( ) height( )small

= 12 1m( ) 5ms2

#$%

&'("

12 0.5m( ) 2.5

ms2

#$%

&'(= 1.875 m

2

s2

W = m adx0.5m1.0m! = 3kg "1.875

m2s2

= 5.625J For xi = 0 to xf = 5.0m:

adx05m

! = + +

=12 1m( ) 5

ms2

!"#

$%&+ 3m( ) 5 m

s2!"#

$%&+12 1m( ) 5

ms2

!"#

$%&= 20 m

2

s2

W = m adxxix f! = 3kg " 20

m2s2

= 60J B) For xi = 1m to xf = 5.0m:

adx05m

! = +

= 3m( ) 5 ms2

!"#

$%&+12 1m( ) 5

ms2

!"#

$%&= 17.5 m

2

s2

W = m adxxix f! = 3kg "17.5

m2s2

= 52.5J

Use the work-energy theorem W = !K = 12 mvf2 "

12 mvi

2 , vi = 0 at x = 1m. The

speed at x = 5m is W = 52.5J = 12 mvf2 ! vf =

2 " 52.5J3kg = 5.9

ms

Angular Momentum Problem 6 In diagram below a 2kg rock is at point P traveling horizontally with a speed of 12 m/s. At this instant what is the magnitude and direction of the angular momentum? If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of the angular momentum?

Angular Momentum

!L = !r ! !p = m!r ! !v valid for point particle w.r.t. point O

!r , r = 8m

36.9 143.1

!v For the magnitude

L =

!L = mvr sin143.1" = 2kg !12m / s ! 8m ! .6

L= 115.2kg-m2/s. Torque due to gravity on particle w.r.t. point O.

!! = !r "

!Fg ,Fg = mg = 2kg " 9.8m / s2 = 19.6N

!r , r = 8m

53.1 36.9

!Fg

Since the rate of change of angular momentum !! = d

!L / dt has opposite direction (+z)

compared to the direction of the current angular momentum !L = !r ! !p = m!r ! !v (-z), the

angular momentum is decreasing. Can you see the similarity with our much earlier discussion on linear kinetics? FINAL COMMENT AND ADVICE: Angular momentum

!L and Torque

!! depend on

the origin (O). For example, !L = 0 and

!! = 0may be zero for origin O, but nonzero

!L ! 0 and

!! " 0 in another origin O/. Study angular momentum problems and static

equilibrium problems.

Direction perpendicular to x-y plane ! indicates +z out of the page ! indicates z into page Also +x right +y up

Using the right hand rule on

!L = !r ! !p = m!r ! !v it is easy to see that the direction of

!L is z or into the page

Using the right hand rule on !! = !r "

!Fg it is easy to see that the

direction of !! is +z or out the page.

! = rFg sin53.1! = 8m "19.6N " .8 = 125.4N im

This can be expressed in a different unit ! = 125.4kg m2 / s2 Using second law for rotation in terms of angular momentum

!! = d

!Ldt . Hence the net torque is the rate of change of angular momentum

Solutions to 1st Major 111

Q1. Consider a cube of iron of mass 8.0 kg and side 4.0 inches. What is its density in kg/m3? (1 inch = 2.54 cm)

A) 7.6103B) 6.9103C) 9.8103 D) 4.3103E) 10103 ( )

3

33 3 3 3

3 33

2.54 14.0 1.02 101 100

1.02 10 1.05 10

8.0 7.6 10 /1.05 10

cm ml inch minch cm

V l m m

m kg mV

= =

= = =

= = =

Q2. Consider the following physical relation: M = C arb, where M is mass, is density, r is distance and a and b are exponents. C is a dimensionless constant. What are the values of a and b so that the equation is dimensionally correct?

A) a = 1 and b = 3 B) a = 1 and b = 2 C) a = 2 and b = 2 D) a = 2 and b = 1 E) a = 3 and b = 1 [ ] [ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

a b

a b a b

aa b b

3

3 3

M C r

M C r r C 1

M r L

/1& 3 3

a a b b a a

ML

M M L L ML M La b a

=

= = =

= = = =

= = =

Q3. A hot air balloon carrying a 10.0 kg block is descending vertically at a constant speed of 10.0 m/s. When the balloon is 100 m above the ground, the block is released. How long does it take the block to reach the ground? (Neglect air resistance)

A) 3.61 s B) 2.53 s C) 1.64 s D) 5.43 s E) 9.12 s

( ) ( )

2 20

22

2

10.0 2 9.80 10045.4 /

45.4 10.0 9.803.61

v v ay

vv m sv v at

tt s

= +

=

=

= +

=

=

Q4. Figure 1 shows the velocity-time graph of a particle moving along the x-axis. What is the average acceleration of the particle during the time interval t = 1.0 s to t = 8.0 s?

25 10 2.1 /8 1

f iavg

f i

v vva m st t t

= = = =

Q5. Which of the graphs shown in Figure 2 represents an object moving with a negative constant velocity?

A) (3) B) (2) C) (1) D) (4) E) (5)

Q6. A car moving along the positive x-axis with constant acceleration covered the distance between two points 60 m apart in 6.0 s. Its velocity as it passes the second point was 15 m/s. What was its velocity at the first point?

A) 5.0 m/s B) 10 m/s C) 2.0 m/s D) 4.0 m/s E) 15 m/s

0

00 0

constant. , 15 /60 , 6.0 , ?

22

a is v m sx m t s v

v v xx t v v find vt

= = = =

+ = =

Q7. A car travels 30 km due south and then D km in an unknown direction. The magnitude of the resultant displacement is 50 km and its direction is 53o west of south. Find the magnitude and direction of the unknown displacement D.

A) 40 km due west B) 40 km due east C) 45 km due west D) 45 km due east E) 54 km due west

1

1 2 2 1

30 , 50sin 53 50cos53 40 30

40 30 30 40

d j R i j i j

R d d d R d i j j i

= = =

= + = = + =

45

45 60 105

4cos105 4sin105 1.04 3.86

o

o o o

o o

A makes with ve the x axis

B makes with ve the x axis

B i j i j

+

+ = +

= + = +

Q9. Which one of the following statements concerning vectors and scalars is FALSE?

A) A vector that has zero magnitude may have components other that zero. B) A vector that has a negative component, has a positive magnitude. C) A scalar component may be either positive or negative. D) Two vectors are equal only if they have the same magnitude and same direction. E) In calculations, the vector components of a vector may be used in place of the vector itself.

Q10. An airplane makes a gradual 90.0o turn while flying at a constant speed of 200 m/s. The process takes 20.0 seconds to complete. For this turn the magnitude of the average acceleration of the plane is:

A) 14.1 m/s2B) zero C) 40.4 m/s2D) 20.8 m/s2 E) 10.3 m/s2

2 2 2

200 200

200 200 10.0 10.020.0

10.0 10.0 10.0 2 14.1 /

i f

f iavg

avg

assume v i and v j

v vv j ia i jt t

a m s

= =

= = = = +

= + = =

Q11. Identical guns fire identical bullets horizontally at the same speed from the same height above level planes, one on the Earth and one on the Moon. Which of the following three statements is/are TRUE? (gmoon = 1/6 gearth) I. The horizontal distance traveled by the bullet is greater on the Moon. II. The flight time is less for the bullet on the Earth. III. The velocities of the bullets at impact are the same.

A) I and II only B) I only C) I and III only D) II and III only E) I, II and III

Q12. A ball is thrown horizontally from the top of a 20-m high hill as shown in Figure 3. It strikes the ground at an angle of 45o. With what speed was it thrown?

A) 20 m/s B) 14 m/s C) 28 m/s D) 32 m/s E) 40 m/s

2

:0, 20 , 2 2 9.8 20 400

20 /

(45 ) 20 / 20 /

yo y

y

ox y o x

on the y axisv y m a g v ayv m s

on the ground angle v v m s v v m s

= = = = = =

=

= = = =

Q14. A boy wishes to swim across a river to a point directly opposite as shown in Figure 5. He can swim at 2.0 m/s in still water. The river is flowing at 1.0 m/s toward the west. At what angle with respect to the line joining the starting and finishing points should he swim?

( )

2.0 sin 1.0 sin 0.5 30x r

o

his v V speed of river to v in the x direction

east of north

=

= = =

1 2

2

2

2

(2 3 ) 2(2 )

2(2 ) (2 3 )4 2 2 32 5

netF ma

F F a

i j k F i j

F i j i j ki j i j ki j k

=

+ =

+ + =

= + = += +

2 2

12

10 5(0) 2.0 / & (5) 1 /5.0 5.0

. 2.0 5 (7 5) 1 10 1 9

9 2 9 2 7 /f i f f

Fa a m s a m sm

v adt area under the acc curve

v v v v v m s

= = = = =

= = = + = =

= = = + = =

1 2

2

30.0, k k kk

take the system to the two blocks togetherM m m kgN Mg f N Mg

T f Ma find a

= + =

= = =

=

1 1 1

1 1 1

, k k kk

take the system to block one onlyN m g f N m g

T f m a find T = = =

=

( )

2

2

,max

2

,

2.0 , 0.50

s

s s s s

ss s

mvf mg Nr

mvslipping f f N mg mgr

gr grv v r m

= =

= = = =

= = = =

Q19. A 0.10 kg stone is tied to the end of a 1.0-m long rope. The stone is moved in a circle in the vertical plane. What is the tension in the rope when the stone is at its lowest position and has a speed of 5.0 m/s?

A) 3.5 N B) 0.98 N C) 0 N D) 0.49 N E) 1.5 N

2 225.0 25 /

1.0

( ) 0.10(9.8 25) 3.5net

va m sr

F T mg maT m g a N

= = =

= =

= + = + =

Q20. An object is being accelerated in the absence of friction by a 100-N force. A second force of 100-N is then applied to the object in a direction opposite to the direction of motion. The object with these two forces acting on it will

A) Move at a constant velocity B) Slow down C) Move in a circle D) Stop rapidly E) Move backward

GENERAL PHYSICS PH 221-3A (Dr. S. Mirov) Test 1 (09/14/09)

STUDENT NAME: __________KEY_________ STUDENT id #: ___________________________ ---------------------------------------------------------------------------------------------------------------------------------------

ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS. NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)

Important Formulas: 1. Motion along a straight line with a constant acceleration v d = [dist taken]/[time trav ]=S/t;vaver. speed = [dist. taken]/[time trav.]=S/t; vaver.vel. = x/t; vins =dx/t; aaver.= vaver. vel./t; a = dv/t; v = vo + at; x= 1/2(vo+v)t; x = vot + 1/2 at2; v2 = vo2 + 2ax (if xo=0 at to=0) 2. Free fall motion (with positive direction )2. Free fall motion (with positive direction )g = 9.80 m/s2; y = vaver. t vaver.= (v+vo)/2; v = vo - gt; y = vo t - 1/2 g t2; v2 = vo2 2gy (if yo=0 at to=0) 3. Integration in Motion Analysis (non-constant acceleration)

1t

v v a d t= + 1o

ot

v v a d t= + 1

1

o

t

ot

x x v d t= + 4. Motion in a plane vx = vo cos; vy = vo sin;vy vo sin; x = vox t+ 1/2 ax t2; y = voy t + 1/2 ay t2; vx = vox + at; vy = voy + at; 5. Projectile motion (with positive direction ) vx = vox = vo cos; x = vox t; xmax = (2 vo2 sin cos)/g = (vo2 sin2)/g for yin = yfin; vy = voy - gt = vo sin - gt;

2y = voy t - 1/2 gt2; 6. Uniform circular Motion a=v2/r, T=2r/v

7. Relative motion PA PB BA

PA PB

v v va a

= +=

r r r

r r

8. Component method of vector addition

A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2; A A Ax y= +2 2 ; = tan-1 Ay /Ax; The scalar product A = cosa b ab rr

( ) ( )x y z x y za b a i a j a k b i b j b k = + + + +rr

= x x y y z za b a b a b a b + +rr

y y

The vector product ( ) ( )x y z x y za b a i a j a k b i b j b k = + + + +rr

y z x yx z

i j ka a a aa a

a b b a a a a i j k = = = + =r rr r

( ) ( ) ( )

x y zy z x yx z

x y z

y z y z z x z x x y x y

a b b a a a a i j kb b b bb b

b b b

a b b a i a b b a j a b b a k

+

= + +

1. Starting at time t=0, an object moves along a straight line with velocity in m/s given by v(t)=98-2t2, where t is in seconds. When it momentarily stops what will be its acceleration?

2

1) Time at which the object will stop corresponds to 098 2 0 7

v =

298 2 0; t=7s

3) 4

4) f t 7 28

tdva tdt

m

= = =

24) for t=7s 28ma s =

2. At time t=0 a car has a velocity of 16 m/s. It slows down with an acceleration given by -0.50t, in m/s2 for t in seconds. At what instant of time it will stop?

1 1 2 22

1 11

0

1

016 ( 0.5 ) 16 ( ) 16 04 4 4

8.0o

t t

ot

t tv v adt t dt

t s

= + = + = + + = ==

1 8.0t s

3. At a stop light, a truck traveling at 15 m/s passes a car as it starts from rest. The truck travelsat constant velocity and the car accelerates at 3 m/s2. How much time does the car take tocatch up to the truck?

When the car catches up with the truck they both will have the same displacement

with respect to the origin point (stop light), and it will take them the same time to reach this point.

xt pFor the truck

2

moving with constant speed 15 / ;

For the car moving with constant acceleration from rest 02

v m s x vtata x

= == +

2

2

2 2 (15 / )Hence, and 102 (3 / )

at v m svt t sa m s

= = = =

4. A projectile is shot vertically upward with a given initial velocity. It reaches a maximumheight of 100m. If, on a second shot, the initial velocity is doubled then what will be the new maximum height?maximum height?

21 1

Given: a) 100 ; 0; 9.8 /fy m v g m s= = =

22 2 02 01

2

b) ?; 0; 9.8 / ; 2

Solution

f

fy v g m s v v

v

= = = =

2 0101 1 1

22 0101 2 2

0 2 ; 1002

40 4 2 ; 4002

vv gy y mg

vv gy y mg

= = =

= = =2g

5. At time t=0 s, a puck is sliding on a horizontal table with a velocity 3.00 m/s, 65.0 above the +x axis. As the puck slides, a constant acceleration acts on it that has the following components: ax=-0.460 m/s2 and ay=-0.980 m/s2. What is the velocity of the puck at time t=1.50 s?

y

ay

x

vo

ax

x direction

x vox vx ax t ? 3.00cos65.0=1.268 m/s ? -0.460 m/s2 1.5 s

21.268 / (0.460 / ) (1.5 ) 0.578 /x ox xv v a t m s m s s m s= + = =

y direction

y voy vy ax t 2? 3.00sin65.0=2.720 m/s ? -0.980 m/s2 1.5 s

22.720 / (0.980 / ) (1.5 ) 1.250 /y oy yv v a t m s m s s m s= + = =

2 2 2 22 2 2 20.578 1.250 1.38 /x yv v v m s= + = + =

1.250arctan arctan 65.20.578

y

x

vv

= = =

6. In the diagram, Aur

has magnitude 12 m and Bur

has magnitude 8 m.

(b)

rr

(b)

Find the resultant of the vectors A

ur and B

ur. Express in

(a) component notation, (b) graphical form,( ) g p ,(c) magnitude-angle form

(a) Component notation

co s 4 5 8 .5 ; s in 4 5 8 .5 ; 8 .5 8 .5

co s 6 0 4 ; s in 6 0 7 ; B 4 7x yA A m A A m A i j

B B m B B m i j

= = = = = +ur r rur r r

1

co s 6 0 4 ; s in 6 0 7 ; B 4 7

( ) ( ) 1 2 .5 1 .5 1 1 0 2

(c) M ag n itu d e-an g le fo rm

x y

x x y y

B B m B B m i j

r A B i A B j i j i j

= = = = = = + + + = + = +r r r r r r r

2 2 11 2 .5 1 .5 1 2 .6 1 1 0

arctan arcyx

r mrr

= + = = = = 1 .5tan 7

1 2 .5=

8. A Ferris wheel with a radius of 8.0 m makes 1 revolution every 10 s. When a passenger is atthe top, essentially a diameter above the ground, he releases a ball. How far from the point onthe ground directly under the release point does the ball land?

y vo

x o

2 2

G iven: 10 ; 8.0 ; y 16 ; 9.8 / ; 0 /

Find: =?y xT s R m m a m s a m s

x= = = = =

Find: ?1) M agnitude of initial velocity of the ball equal to velocity of the radial point of the Ferris wheel

2 2 (8.0 ) 5.0 /(10 )

2) Directi

o

x

R mv m sT s = = =

on of the initial velocity is tangent to the path. At the top of the wheel it is directed ) y g p palong positive direction of x axis. 0.

3) Consider motion of a ball along vertical y direction. Find time oyv =

2

2

of flight

2 2( 16 ); 1.82 ( 9 8 / )oy

t

at y my v t t sa m s

= + = = =2 ( 9.8 / )

4) Consider motion of a ball along x direction.(5.0 / ) (1.8 ) 9.0

ox o

a m s

x v t v t m s s m

= = = =

GENERAL PHYSICS PH 221-3A (Dr. S. Mirov) Test 1 (09/17/07)

STUDENT NAME: ________________________ STUDENT id #: ___________________________ -------------------------------------------------------------------------------------------------------------------------------------------

ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS. NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)

Important Formulas: 1. Motion along a straight line with a constant acceleration vaver. speed = [dist. taken]/[time trav.]=S/t; vaver.vel. = x/t; vins =dx/t; aaver.= vaver. vel./t; a = dv/t; v = vo + at; x= 1/2(vo+v)t; x = vot + 1/2 at2; v2 = vo2 + 2ax (if xo=0 at to=0) 2. Free fall motion (with positive direction ) g = 9.80 m/s2; y = vaver. t vaver.= (v+vo)/2; v = vo - gt; y = vo t - 1/2 g t2; v2 = vo2 2gy (if yo=0 at to=0) 3. Motion in a plane vx = vo cos; vy = vo sin; x = vox t+ 1/2 ax t2; y = voy t + 1/2 ay t2; vx = vox + at; vy = voy + at; 4. Projectile motion (with positive direction ) vx = vox = vo cos; x = vox t; xmax = (2 vo2 sin cos)/g = (vo2 sin2)/g for yin = yfin; vy = voy - gt = vo sin - gt; y = voy t - 1/2 gt2; 5. Uniform circular Motion a=v2/r, T=2r/v 6. Relative motion

P A P B B A

P A P B

v v va a

= +

=

G G GG G

5. Component method of vector addition

Key

1

Test 1 sample

A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2; A A Ax y= +2 2 ; = tan-1 Ay /Ax;

The scalar product A = cosa b ab GG

( ) ( )x y z x y za b a i a j a k b i b j b k = + + + +

GG

= x x y y z za b a b a b a b + +GG

The vector product ( ) ( )x y z x y za b a i a j a k b i b j b k = + + + +GG

( ) ( ) ( )

y z x yx zx y z

y z x yx zx y z


i j ka a a aa a

a b b a a a a i j kb b b bb b

b b b


= = = + =

= + +

G GG G

-------------------------------------------------------------------------------------------------------------------------------------------

2

1.Startingattimet=0,anobjectmovesalongastraightline.Itscoordinateinmetersisgivenbyx(t)=75t1.0t3,wheretisinseconds.Whatisitsaccelerationwhenitmomentarilystops?

2

2

2

1) 75 3.0

2) Time at which the object will stop corresponds to 075 3.0 0; t=5s

3) 6.0

4) for t=5s 30

dxv tdt

vt

dva tdt

ma s

= =

=

=

= =

=

3

5. Let and is the angle between and when they are drawn with their tails at the same point. Which of the following is not true?

A. sin It is true by definition.

B. it

R S T S T

R S T

R T S

=

=

=

G GG G G

GG GGG G

is

( ) ( ) ( )

true sin

( ) ( )

ce

y z x yx zx y z

y z x yx zx y z


y z x yx zx y z

y z x yx zx y z

y z y z z x z x

i j ka a a aa a

a b a a a i j kb b b bb b

b b b


i j kb b b bb b

b a b b b i j ka a a aa a

a a a

a b b a i a b b a

= = + =

= + +

= = + =

=

GG

G G

C. 0 . It is true since c

os90 0

D. 0 It is true since cos90 0

E. 0 It is not true sin

( ) (

ce co

)

s 0

x y x y

R S R S RSR T R T RT

S

j a b b a k b a

T

= = =

=

=

= =

=

G GG GG G

G G

G GG G

7

6. According to an ancient Greek source, a stone throwing machine on one occasion achieved a range of 730 m. If this is true,(a) What must have been the minimal initial speed of the stone as it was ejected from the engine?(b) When ejected with this speed, how long would the stone have taken to reach its target?

2 22

2

Given: 730

=45Find (a) ?(b)

:sin 2( ) (730 ) (9.8 / ) 84.6 /

( ) Consider y motion. Given: 1) Time of flight corresponds to y=0;2) a=-9.8 m/s 3) sin 45 84.

o

f

o oo

oy o

x m

vt

Solutionv va x v xg m m s m s

g g

bv v

=

=

= = = = =

= =

D

D

22

6sin 45 59.8 /

Use equation (3) to find time of flight21 2(59.8 / )0= 12.2

2 ( 9.8 / )oy

oy f f f

m s

v m sv t at t sa m s

=

+ = = =

8

7. A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45. With what speed was it thrown?

a) Consider motion along y direction. Choose the origin point at the top of the hill and yaxis directed vertically upwards. The ball is in a free fall from the height of 20 mGiven: 1) 0; 2) 20oyv y m= =

2

2 2 2

; 3) 9.8 / . Find =?

Use equation 4. = 2 ; 2 2( 9.8 / )( 20 ) 19.8 /

b) Since a ball strikes the ground at 45 final x component of velocity equal final y component of velocity

y

y oy y

a m s v

v v ay v ay m s m m s

v

=

+ = = =

=

D

20 /ox xv v m s= =

v

y

9

8. A girl wishes to swim across a river to a point directly opposite as shown. She can swim at 2m/s in still waterand the river is flowing at 1m/s. At what angle with respect to the line joining the starting and finishing points should she swim?

velocity of a girl with respect to the bank of the river

velocity of a water with respect to the bank of the river velocity of a girl with respect to the still water

According to the rel

gb

wb

gw

vvv

GGG

1

ative motion rule

1arctan( ) arctan( ) 3 10 2

gb wb gw

wb

gw

v v vvv

= +

= = = D

G G G

10

9. A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of25m along the circumference of the circle, in 5.0 s. What is the magnitude of her acceleration?

( ) Find speed of the uniform circular motion.since qurter of a full revolution travel corresponds to 25 m, and time it takes equal to 5.0 s the speed of the motion is 25/5.0=5.0m/s

(b) Find radius

a

2 22

of a circular trajectory. 2 25 , 15.9

4

(c) Find centripital acceleration.

(5 / ) 1.6 /(15.9 )

rr m r m

v m sa m sr m

= =

= = =

11

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