usapho problems (2007-2014)

United States Physics Team

Contest Papers with Solutions

(2007-2014)


F= ma Contest

2007

2007 F=ma Contest 1. An object moves in two dimensions according to

2 ( ) (4.0 9.0) (2.0 5.0) ,r t t i t j= + G where r is in meters and t in seconds. When does the object cross the x-axis?

(a) 0.0 s (b) 0.4 s (c) 0.6 s (d) 1.5 s (e) 2.5 s

2. The graph shows velocity as a

function of time for a car. What was the acceleration at time = 90 seconds?

(a) 0.22 m/s2 (b) 0.33 m/s2 (c) 1.0 m/s2 (d) 9.8 m/s2 (e) 30 m/s2

3. The coordinate of an object is given as a function of time by x = 8t - 3t2, where x is in

meters and t is in seconds. Its average velocity over the interval from t = 1 to t = 2s is

(a) -2 m/s (b) -1 m/s (c) -0.5 m/s (d) 0.5 m/s (e) 1 m/s

4. An object is released from rest and falls a distance h during the first second of time.

How far will it fall during the next second of time?

(a) h (b) 2h (c) 3h (d) 4h (e) h2

1

5. A crate of toys remains at rest on a sleigh as the sleigh is pulled up a hill with an increasing speed. The crate is not fastened down to the sleigh. What force is responsible for the crates increase in speed up the hill?

(a) the force of static friction of the sleigh on the crate (b) the contact force (normal force) of the ground on the sleigh (c) the contact force (normal force) of the sleigh on the crate (d) the gravitational force acting on the sleigh (e) no force is needed

6. At time t = 0 a drag racer starts from rest at the origin and moves along a straight line

with velocity given by v = 5t2, where v is in m/s and t in s. The expression for the displacement of the car from t = 0 to time t is

(a) 5t3 (b) 5t3/3 (c) 10t (d) 15t2 (e) 5t/2

7. The chemical potential energy stored in a battery is converted into kinetic energy in a toy

car that increases its speed first from 0 mph to 2 mph and then from 2 mph up to 4 mph. Ignore the energy transferred to thermal energy due to friction and air resistance. Compared to the energy required to go from 0 to 2 mph, the energy required to go from 2 to 4 mph is

(a) half the amount. (b) the same amount. (c) twice the amount. (d) three times the amount. (e) four times the amount.

8. When two stars are very far apart their gravitational potential energy is zero; when they

are separated by a distance d the gravitational potential energy of the system is U. If the stars are separated by a distance 2d the gravitational potential energy of the system is

(a) U/4 (b) U/2 (c) U (d) 2U (e) 4U

2

9. A large wedge rests on a horizontal frictionless surface, as shown. A block starts from rest and slides down the inclined surface of the wedge, which is rough. During the motion of the block, the center of mass of the block and wedge

(a) does not move (b) moves horizontally with constant speed (c) moves horizontally with increasing speed (d) moves vertically with increasing speed (e) moves both horizontally and vertically

10. Two wheels with fixed hubs, each having

a mass of 1 kg, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertia is I = mR2. In order to impart identical angular accelerations about their respective hubs, how large must F2 be?

(a) 0.25 N (b) 0.5 N (c) 1 N (d) 2 N (e) 4 N

11. A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same

outer radius, are each free to rotate about a fixed axis through its center. Assume the hoop is connected to the rotation axis by light spokes. With the objects starting from rest, identical forces are simultaneously applied to the rims, as shown. Rank the objects according to their kinetic energies after a given time t, from least to greatest.

(a) disk, hoop, sphere (b) sphere, disk, hoop (c) hoop, sphere, disk (d) disk, sphere, hoop (e) hoop, disk, sphere

3

12. A 2-kg rock is suspended by a

massless string from one end of a uniform 1-meter measuring stick. What is the mass of the measuring stick if it is balanced by a support force at the 0.20-meter mark?

(a) 0.20 kg (b) 1.00 kg (c) 1.33 kg (d) 2.00 kg (e) 3.00 kg

13. A particle moves along the x-axis. It collides elastically head-on with an identical

particle initially at rest. Which of the following graphs could illustrate the momentum of each particle as a function of time?

14. When the speed of a rear-drive car is increasing on a horizontal road, the direction of

the frictional force on the tires is

(a) backward on the front tires and forward on the rear tires. (b) forward on the front tires and backward on the rear tires. (c) forward on all tires. (d) backward on all tires. (e) zero.

4

15. A uniform disk (I = MR2) of mass 8.0 kg can rotate without friction on a fixed axis. A string is wrapped around its circumference and is attached to a 6.0 kg mass. The string does not slip. What is the tension in the cord while the mass is falling?

(a) 20.0 N (b) 24.0 N (c) 34.3 N (d) 60.0 N (e) 80.0 N

16. A baseball is dropped on top of a basketball. The basketball hits the

ground, rebounds with a speed of 4.0 m/s, and collides with the baseball as it is moving downward at 4.0 m/s. After the collision, the baseball moves upward as shown in the figure and the basketball is instantaneously at rest right after the collision. The mass of the baseball is 0.2 kg and the mass of the basketball is 0.5 kg. Ignore air resistance and ignore any changes in velocities due to gravity during the very short collision times. The speed of the baseball right after colliding with the upward moving basketball is

(a) 4.0 m/s (b) 6.0 m/s (c) 8.0 m/s (d) 12.0 m/s (e) 16.0 m/s

17. A small point-like object is thrown horizontally off of a 50.0-m high building with an

initial speed of 10.0 m/s. At any point along the trajectory there is an acceleration component tangential to the trajectory and an acceleration component perpendicular to the trajectory. How many seconds after the object is thrown is the tangential component of the acceleration of the object equal to twice the perpendicular component of the acceleration of the object? Ignore air resistance.

(a) 2.00 s (b) 1.50 s (c) 1.00 s (d) 0.50 s (e) The building is not high enough for this to occur.

5

18. A small chunk of ice falls from rest down a frictionless parabolic ice sheet shown in the figure. At the point labeled A in the diagram, the ice sheet becomes a steady, rough incline of angle with respect to the horizontal and friction coefficient

030k .

This incline is of length h23 and ends

at a cliff. The chunk of ice comes to rest precisely at the end of the incline. What is the coefficient of friction k ?

300

(a) 0.866 (b) 0.770 (c) 0.667 (d) 0.385 (e) 0.333

19. A non-Hookian spring has force

where k is the spring constant and x is the displacement from its unstretched position. For the system shown of a mass m connected to an unstretched spring initially at rest, how far does the spring extend before the system momentarily comes to rest? Assume that all surfaces are frictionless and that the pulley is frictionless as well.

2kxF =

(a) 2

1

23

kmg

(b) 2

1

k

mg

(c) 2

12

kmg

(d) 3

1

3

kmg

(e) 3

1

233

kmg

6

20. A point-like mass moves horizontally between two walls on a frictionless surface with initial kinetic energy E. With every collision with the walls, the mass loses its kinetic energy to thermal energy. How many collisions with the walls are necessary before the speed of the mass is reduced by a factor of 8?

(a) 3 (b) 4 (c) 6 (d) 8 (e) 16

21. If the rotational inertia of a sphere about an axis through the center of the sphere is I, what is the rotational inertia of another sphere that has the same density, but has twice the radius?

(a) 2I (b) 4I (c) 8I (d) 16I (e) 32I

22. Two rockets are in space in a negligible gravitational field. All observations are made by an observer in a reference frame in which both rockets are initially at rest. The masses of the rockets are m and 9m. A constant force F acts on the rocket of mass m for a distance d. As a result, the rocket acquires a momentum p. If the same constant force F acts on the rocket of mass 9m for the same distance d, how much momentum does the rocket of mass 9m acquire?

(a) p/9 (b) p/3 (c) p (d) 3p (e) 9p

7

23. If a planet of radius R spins with an angular velocity about an axis through the North Pole, what is the ratio of the normal force experienced by a person at the equator to that experienced by a person at the North Pole? Assume a constant gravitational field g and that both people are stationary relative to the planet and are at sea level.

(a) 2Rg (b) gR 2 (c) gR 21 (d) 21 Rg+ (e) gR 21 +

24. A ball of mass m is launched into the air. Ignore air resistance, but assume that there is a wind that exerts a constant force Fo in the x direction. In terms of F0 and the acceleration due to gravity g, at what angle above the positive x-axis must the ball be launched in order to come back to the point from which it was launched?

(a) )(tan 01 mgF

(b) )/(tan 01 Fmg

(c) )(sin 01 mgF

(d) the angle depends on the launch speed (e) no such angle is possible

25. Find the period of small oscillations of a water pogo, which is a stick of mass m in the shape of a box (a rectangular parallelepiped.) The stick has a length L, a width w and a height h and is bobbing up and down in water of density . Assume that the water pogo is oriented such that the length L and width w are horizontal at all times. Hint: The buoyant force on an object is given by ,VgFbuoy = where V is the volume of the medium displaced by the object and is the density of the medium. Assume that at equilibrium, the pogo is floating.

(a) 2 Lg

(b) 2 2

2

gw Lmh

(c) 2

2 22mhL w g

(d) 2g

mwL

(e) g

mwL

8

Questions 26 38: Be sure to show all of your work on the corresponding Free Response Answer Form as well as to record your answer on the optical mark answer sheet.

26. A sled loaded with children starts from rest and slides down a snowy 250 (with respect

to the horizontal) incline traveling 85 meters in 17 seconds. Ignore air resistance. What is the coefficient of kinetic friction between the sled and the slope? (5 pts.)

(a) 0.36 (b) 0.40 (c) 0.43 (d) 1.00 (e) 2.01

27. A space station consists of two living modules attached to a central hub on opposite sides of the hub by long corridors of equal length. Each living module contains N astronauts of equal mass. The mass of the space station is negligible compared to the mass of the astronauts, and the size of the central hub and living modules is negligible compared to the length of the corridors. At the beginning of the day, the space station is rotating so that the astronauts feel as if they are in a gravitational field of strength g. Two astronauts, one from each module, climb into the central hub, and the remaining astronauts now feel a gravitational field of strength g . What is the ratio gg in terms of N? (5 pts)

(a) 2 /( 1)N N (b) )1( NN (c) NN )1( (d) )1( NN (e) none of the above

9


Questions 28-30

A simplified model of a bicycle of mass M has two tires that each comes into contact with the ground at a point. The wheelbase of this bicycle (the distance between the points of contact with the ground) is w, and the center of mass of the bicycle is located midway between the tires and a height h above the ground. The bicycle is moving to the right, but slowing down at a constant rate. The acceleration has a magnitude a. Air resistance may be ignored.

Case 1 (Questions 28 29): Assume that the coefficient of sliding friction between each tire and the ground is , and that both tires are skidding: sliding without rotating. Express your answers in terms of w, h, M, and g.

28. What is the maximum value of so that both tires remain in contact with the ground? (5 pts)

(a) h

w2

(b) wh

2

(c) wh2

(d) hw

(e) none of the above

10


29. What is the maximum value of a so that both tires remain in contact with the ground?

(5 pts)

(a) h

wg

(b) h

wg2

(c) w

hg2

(d) wgh

2


Case 2 (Question 30): Assume, instead, that the coefficient of sliding friction between each tire and the ground is different: 1 for the front tire and 2 for the rear tire. Let

1 22 = 30. Assume that both tires are skidding: sliding without rotating. What is the maximum

value of a so that both tires remain in contact with the ground? (5 pts)

(a) wgh

(b) 3wg

h

(c) 23wgh

(d) 2hgw


11


Questions 31 33

A thin, uniform rod has mass m and length L. Let the acceleration due to gravity be g. Let the rotational inertia of the rod about its center be . 2md

31. Find the ratio L / d. (3 pts)

(a) 23 (b) 3 (c) 12 (d) 32 (e) none of the above

The rod is suspended from a point a distance kd from the center, and undergoes small

oscillations with an angular frequency gd

. 32. Find an expression for in terms of k. (7 pts)

(a) 21 k+(b) 21 k+ (c)

kk+1

(d) k

k+1

2

(e) none of the above 33. Find the maximum value of . (5 pts)

(a) 1 (b) 2 (c) 1/ 2 (d) does not attain a maximum value (e) none of the above

12


Questions 34 36

A point object of mass m is connected to a cylinder of radius R via a massless rope. At time t = 0 the object is moving with an initial velocity v0 perpendicular to the rope, the rope has a length L0, and the rope has a non-zero tension. All motion occurs on a horizontal frictionless surface. The cylinder remains stationary on the surface and does not rotate. The object moves in such a way that the rope slowly winds up around the cylinder. The rope will break when the tension exceeds Tmax. Express your answers in terms of Tmax, m, L0, R, and v0.

34. What is the angular momentum of the object with respect to the axis of the cylinder at

the instant that the rope breaks? (6 pts)

(a) Rmv0

(b) 32

0

maxTm v

(c) 00 Lmv

(d) 2

max

0

T Rv


13


35.What is the kinetic energy of the object at the instant that the rope breaks? (4 pts)

(a) 2

20mv

(b) 0

20

2LRmv

(c) 20

220

2LRmv

(d) 2 2

0 022

mv LR


36. What is the length (not yet wound) of the rope? (6 pts)

(a) RL 0 (b) RL 20 (c) 0 18L R (d)

20

max

mvT


14


Questions 37 38

A massless elastic cord (that obeys Hookes Law) will break if the tension in the cord exceeds Tmax. One end of the cord is attached to a fixed point, the other is attached to an object of mass 3m. If a second, smaller object of mass m moving at an initial speed v0 strikes the larger mass and the two stick together, the cord will stretch and break, but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one-dimensional collision, the cord will still break, and the larger mass will move off with a final speed of vf. All motion occurs on a horizontal, frictionless surface.

37. Find vf/v0. (7 pts)

(a) 1 12 (b) 21 (c) 61 (d) 1 3 (e) none of the above

38. Find the ratio of the total kinetic energy of the system of two masses after the perfectly elastic collision and the cord has broken to the initial kinetic energy of the smaller mass prior to the collision. (7 pts)

(a) 1 4 (b) 1/ 3(c) 1/ 2(d) 3 / 4(e) none of the above

15


Multiple Choice Answers

1. e 2. b 3. b 4. c 5. a 6. b 7. d 8. b 9. d 10. d 11. e 12. c 13. d 14. a 15. b 16. b 17. a 18. b 19. a 20. c 21. e 22. d 23. c 24. b 25. d 26. b 27. e 28. a 29. b 30. e 31. d 32. e 33. c 34. b 35. a 36. d 37. c 38. d

16

Copyright 2007, American Association of Physics Teachers Page 1

2007 F=ma Contest SOLUTIONS

Multiple Choice Answers

1. e 2. b 3. b 4. c 5. a 6. b 7. d 8. b 9. d 10. d 11. e 12. c 13. d 14. a 15. b 16. b 17. a 18. b 19. a 20. c 21. e 22. d 23. c 24. b 25. d 26. b 27. e 28. a 29. b 30. e 31. d 32. e 33. c 34. b 35. a 36. d 37. c 38. d


Solutions to Free Response

26. Since the acceleration is constant and the sled starts from rest,

212

x at = (26-1) so a = 0.588 m/s2. With the y-axis perpendicular to the incline, ay = 0, so the normal force is

cosN mg = (26-2)

Applying Newtons second law parallel to the incline with f = force of kinetic friction

sinmg f ma = (26-3)

Using f N= along with equations (26-2) and (26-3) we find that

tancosa

g =

2

2

0.588 /tan 25 0.40(10 / ) cos 25

m sm s

= =

27. An accelerated reference frame is equivalent to a gravitational field. We will denote all quantities that change when the astronauts move with a primed superscript after the move. Due to circular motion and the fact that the radius does not change and that v r= , we find that

2 2

2 2

g vg v

= = (27-1)

Angular momentum is conserved since there is no external torque acting on the system. Therefore,

I I = (27-2)

Since the corridors are long, we can consider the astronauts to be point masses. So, with r = the distance from the central hub to the living modules, m = the mass of one astronaut, and with two


living modules each with N astronauts originally, we find that the rotational inertia before the astronauts move is

22I Nmr= (27-3)

After the two astronauts climb into the central hub,

22( 1)I N mr = (27-4)

When we substitute (27-3) and (27-4) into (27-2) we obtain

2 22 2( 1)Nmr N mr = (27-5)

1N

N = (27-6)

Finally, substituting (27-6) into (27-1), we find

2 2

1g Ng N

= = (27-7)

28. For static equilibrium in an accelerated reference frame, we need to calculate torques about the center of mass. Let N1 be the normal force on the front tire and N2 the normal force on the rear tire. Let f1 be the force of friction on the front tire and f2 the force of friction on the rear tire. If the front tire just barely remains in contact with the ground then N1 = f1 = 0. Then setting the counter-clockwise torque due to friction on the rear tire = the clockwise torque due to the normal force on the rear tire, we have

2 2 2wf h N= (28-1)

Substituting 2 2f N= into (28-1),

2 2 2wN h N = (28-2)


2wh

= (28-3)

29 30. Applying Newtons Second Law to the horizontal direction,

1 2f f Ma+ = (29-1)

Setting clockwise torque = counterclockwise torques:

2 11 2( )2 2

N w N w f f h= + + (29-2)

Substituting (28-4) into (28-5) and solving for a,

2 1( )2wMah N N= (29-3)

2 1( )

2N Nwa

Mh= (29-4)

The maximum acceleration will clearly occur when N1=0. In that case, N2 = Mg, and

2w ga

h= (29-5)

(This is the answer to question 29. Note that this answer did not depend at all upon whether the coefficient of sliding friction for each tire and the ground is the same or different. Therefore, this is the answer to question 30 also.) 31. The rotational inertia of a thin, uniform rod about its center is

2112cm

I mL= (31-1)

We are given that the rotational inertia of the rod about its center is md2. Setting this expression equal to (31-1), we obtain

2 2112

md mL= (31-2)


Therefore,

2 212L d= (31-3)

and

12 2 3Ld= = (31-4)

32. The torque due to gravity is the same as if the entire mass were located at the center of mass. Therefore, the gravitational torque on the rod about an axis through the suspension point a distance kd from the center when the rod making an angle to the vertical is

sinp mgkd = (32-1)

where the subscript p denotes the pivot point. We now need the parallel axis theorem to find the rotational inertia about the pivot point.

2 2 2( )p cmI I mh md m kd= + = + (32-2)

2 2(1 )pI md k= + (32-3)

Now, apply the rotational analogue of Newtons Second Law to the axis through the pivot point. Noting that the force of the pivot does not exert a torque about an axis through the pivot and using equations (32-1) and (32-3), we find

p pI = (32-4)

22 2

2sin (1 )dmgkd md kdt = + (32-5)

For small oscillations, sin . Therefore,

2

2 2(1 )d gkdt d k = + (32-6)


Since an object oscillates with angular frequency when the objects motion is governed by the differential equation

22

2

ddt = (32-7)

we find that

2 2(1 ) 1gk k g

d k k d = =+ + (32-8)

gd

= (32-9)

where

21kk

= + (32-10) 33. To find the value of k that gives the maximum value of , square equation (32-10) and then differentiate both sides with respect to k.

2 2

2 2 2 2

(1 ) (2 ) 12(1 ) (1 )

d k k k kdk k k + = =+ + (33-1)

0ddk = when k = 1 (33-2)

Substituting k = 1 into (32-10), we find that

12

= (33-3)

34 36. Since the velocity is perpendicular to the rope, the rope does not do any work on the object. Since the object is moving on a horizontal frictionless surface, the net work done on the object is zero and therefore the change in kinetic energy of the object is zero.


Thus, the kinetic energy of the object at the instant that the rope breaks is the same as the initial kinetic energy of the object:

20

2mvK = (34-1)

(This is the answer to #35.) Therefore, the speed of the object is always v0. The angular momentum of the object with respect to the axis of the cylinder is

0L mv r= (34-2) where r is the radius of the circular orbit (which is the length of the not yet wound rope.) At the time that the rope breaks, the tension is

20

maxmvT

r= (34-3)

Solve equation (34-3) for r

20

max

mvrT

= (34-4) (This is the answer to #36.) Substitute (34-4) into (34-2).

2 30

max

m vLT

= (34-4)

(This is the answer to #34.)


37 38. The cord breaks when it has exceeded a certain tension, which happens when it exceeds a certain potential energy, U0. For the inelastic collision, when the cord is slack, we use conservation of momentum

0 4mv mv= (37-1)

0

4vv = (37-2)

The kinetic energy of the two masses immediately after the collision is

20

1 8mvK = (37-3)

All of this kinetic energy gets transferred into potential energy so we know that the cord breaks when

20

0 1 8mvU K= = (37-4)

Now for the elastic collision: First, find the velocities of each mass immediately after the collision while the cord is slack. The easy way to do this is to find that the velocity of the center of mass is

0

4cmvv = (37-5)

Then in the center of mass reference frame, before the collision the velocity of m is 034v and the

velocity of 3m is 04v . In a one-dimensional elastic collision, in the center of mass reference frame,

each blocks velocity after the collision is the same magnitude, but in the opposite direction, of its velocity before the collision. So the velocity of the 3m object right after the collision in the center of

mass reference frame is 04v+ . Using (37-5) to transform back to the lab reference frame, we find that

the velocity of the 3m object immediately after the collision is 02v and therefore its kinetic energy

immediately after the collision (while the cord is still slack) is


20

23

8mvK = (37-6)

But since we know that the cord breaks, we know that U0 of the kinetic energy of the 3m block will be consumed by the cord. Therefore, the final kinetic energy of the 3m block, using conservation of energy along with equations (37-4) and (37-6) is

2 2 20 0 0

3 2 13

8 8 4mv mv mvK K K= = = (37-7)

Now we can find that

2 20

3

32 4

fmv mvK = = (37-8)

0

16

fvv

= (37-9)

The velocity of the object of mass m in the center of mass reference frame immediately after the

collision was 034v . Transforming back to the lab reference frame, we find that the mass m has a

velocity after the collision of 02v . Therefore, the kinetic energy of m after the elastic collision is

20

4 8mvK = (37-10)

The total kinetic energy of the system after the elastic collision and the cord is broken, using (37-7) and (37-10) is

2 2 20 0 0

3 43

8 4 8mv mv mvK K+ = + = (37-11)

So, the ratio of the total kinetic energy of the system after the elastic collision and the cord is broken to the initial kinetic energy of the smaller mass prior to the collision is


20

20

3384

2

mv

mv= (37-12)


Semi Final Contest

2007

2007 Semi-Final Exam

INSTRUCTIONS

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Work Part A first. You have 90 minutes to complete all four problems. After you have completed Part A, you may take a break. Then work Part B. You have 90 minutes to complete both problems. Show all your work. Partial credit will be given. Start each question on a new sheet of paper. Be sure to put your name in the upper right-

hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie A1 1/3 A hand-held calculator may be used. Its memory must be cleared of data and programs. You

may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDAs, or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.

Questions with the same point value are not necessarily of the same difficulty. Do not discuss the contents of this exam with anyone until after March 27th. Good luck!

Possibly Useful Information - (Use for both part A and for part B) Gravitational field at the Earths surface g = 9.8 N/kg Newtons gravitational constant G = 6.67 x 10-11 Nm2/kg2Coulombs constant k = 1/4 = 8.99 x 109 Nm2/C2Biot-Savart constant km = /4 = 10-7 Tm/A Speed of light in a vacuum c = 3.00 x 108 m/s Boltzmanns constant kB = 1.38 x 10-23 J/K Avogadros number NA = 6.02 x 1023 (mol)-1Ideal gas constant R = NAkB = 8.31 J/(molK) Stefan-Boltzmann constant = 5.67 x 10-8 J/(sm2K4) Elementary charge e = 1.602 x 10-19 C 1 electron volt 1 eV = 1.602 x 10-19 J Plancks constant h = 6.63 x 10-34 Js = 4.14 x 10-15 eVs Electron mass m = 9.109 x 10-31 kg = 0.511 MeV/c2Binomial expansion (1 + x)n 1 + nx for |x|

Copyright 2007, American Association of Physics Teachers

Semi-Final Exam Part A

A1. A group of 12 resistors is arranged along the edges of a cube as shown in the diagram below. The vertices of the cube are labeled a-h.

a b

c d

f

g h

e

a. (13 pts) The resistance between each pair of vertices is as follows:

Rab = Rac = Rae = 3.0 Rcg = Ref = Rbd = 8.0 Rcd = Rbf = Reg = 12.0 Rdh = Rfh = Rgh = 1.0 What is the equivalent resistance between points a and h?

b. (12 pts) The three 12.0 resistors are replaced by identical capacitors. Ccd = Cbf = Ceg =

15.0 F. A 12.0 V battery is attached across points a and h and the circuit is allowed to operate for a long period of time. What is the charge (Qcd, Qbf, Qeg) on each capacitor after this long period of time?


,

A2. A simple gun can be made from a uniform cylinder of length L0 and inside radius rc. One end of the cylinder is sealed with a moveable plunger and the other end is plugged with a cylindrical cork bullet. The bullet is held in place by friction with the walls of the cylinder. The pressure outside the cylinder is atmospheric pressure, . The bullet will just start to slide out of the cylinder if the pressure inside the cylinder exceeds .

0P

crP a. There are two ways to launch the bullet: either by heating the gas inside the cylinder and keeping the plunger fixed, or by suddenly pushing the plunger into the cylinder. In either case, assume that an ideal monatomic gas is inside the cylinder, and that originally the gas is at temperature , the pressure inside the cylinder is , and the length of the cylinder is 0T 0P 0.L

(8 pts) i. Assume that we launch the bullet by heating the gas without moving the plunger.

Find the minimum temperature of the gas necessary to launch the bullet. Express your answer in terms of any or all of the variables: and . 0 0 0, , , ,cr T L P crP

(8 pts) ii. Assume, instead that we launch the bullet by pushing in the plunger, and that we

do so quickly enough so that no heat is transferred into or out of the gas. Find the length of the gas column inside the cylinder when the bullet just starts to move. Express your answer in terms of any or all of the variables: and . 0 0 0, , ,cr T L P crP

b. (9 pts) It is necessary to squeeze the bullet to get it into the cylinder in the first place. The bullet normally has a radius that is slightly larger than the inside radius of the cylinder;

, is small compared to . The bullet has a length The walls of the cylinder apply a pressure to the cork bullet. When a pressure is applied to the bullet along a given direction, the bullets dimensions in that direction change by

br

b cr r r = cr 0.h LP

x Px E =

for a constant E known as Youngs modulus. You may assume that compression along one direction does not cause expansion in any other direction. (This is true if the so-called Poisson ratio is close to zero, which is the case for cork.) If the coefficient of static friction between the cork and the cylinder is , find an expression for . Express your answer in terms of any or all of the variables: crP 0 , , , , ,P h E r and . cr


A3. A volume fV of fluid with uniform charge density is sprayed into a room, forming spherical drops. As they float around the room, the drops may break apart into smaller drops or coalesce into larger ones. Suppose that all of the drops have radius R. Ignore inter-drop forces and assume that 3.fV R

(10 pts) a. Calculate the electrostatic potential energy of a single drop. (Hint: suppose the

sphere has radius r. How much work is required to increase the radius by dr?). (4 pts) b. What is the total electrostatic energy of the drops?

Your answer to (b) should indicate that the total energy increases with R. In the absence of surface tension, then, the fluid would break apart into infinitesimally small drops. Suppose, however, that the fluid has a surface tension . (This value is the potential energy per unit surface area, and is positive.)

(4 pts) c. What is the total energy of the drops due to surface tension? (7 pts) d. What is the equilibrium radius of the drops?


A4. A nonlinear circuit element can be made out of a parallel plate capacitor and small balls, each of mass m, that can move between the plates. The balls collide inelastically with the plates, dissipate all kinetic energy as thermal energy, and immediately release the charge they are carrying to the plate. Almost instantaneously, the balls then pick up a small charge of magnitude q from the plate; the balls are then repelled directly toward the other plate under electrostatic forces only. Another collision happens, kinetic energy is dissipated, the balls give up the charge, collect a new charge, and the cycle repeats. There are n0 balls per unit surface area of the plate. The capacitor has a capacitance C. The separation d between the plates is much larger than the radius r of the balls. A battery is connected to the plates in order to maintain a constant potential difference V. Neglect edge effects and assume that magnetic forces and gravitational forces may be ignored.

(5) a. Determine the time it takes for one ball to travel between the plates in terms of any or

all of the following variables: m, q, d, and V. (5) b. Calculate the kinetic energy dissipated as thermal energy when one ball collides

inelastically with a plate surface in terms of any or all of the following variables: m, q, d, and V.

(5) c. Derive an expression for the current between the plates in terms of the permittivity of free space, 0 , and any or all of the following variables: m, q, n0, C, and V.

(5) d. Derive an expression for the effective resistance of the device in terms of 0 , and any or all of the following variables: m, q, n0, C, and V.

(5) e. Calculate the rate at which the kinetic energy of the balls is converted into thermal

energy in terms of 0 , and any or all of the following variables: m, q, n0, C, and V.


Semi-Final Exam

Part B

B1. A certain mechanical oscillator can be modeled as an ideal massless spring connected to a moveable plate on an incline. The spring has spring constant k, the plate has mass m, and the incline makes an angle with the horizontal. When the system is operating correctly, the plate oscillates between points A and B in the figure, located a distance L apart. When the plate reaches point A it has zero kinetic energy, but then trips a small lever that instantaneously loads a block of mass M onto the plate. The block and plate then move down the incline to point B, where the force from the spring stops the plate. At this point, the block falls through a hole in the incline, allowing the plate to move back up under the force of the spring. Upon returning to point A it collects another block, and the cycle repeats. Both the plate and the block have a coefficient of friction with the incline for both kinetic and static friction. It is reasonable that the motion in either direction is simple harmonic in nature.

(10 pts) a. Let c be the critical value of the coefficient of friction where the block will just

start to slide under the force of gravity on an incline (without the spring acting on

it). Then let 2

c = . Find in terms of g, the acceleration of free fall, and any or all of the following variables: and M.


(14 pts) b. In order for this system to work correctly, it is necessary to have the correct ratio

between the mass of the block and the mass of the plate. These masses are chosen so that the downward moving block and plate just stop at point B while the

upward moving plate just stops at point A. Find the ratio .MRm

=

(13 pts) c. The system delivers blocks to point B with period , until the blocks run out.

After that, the plate alone oscillates with a period T

0T

. Find the ratio 0TT .

(13 pts) d. The plate only oscillates a few times after delivering the last block. At what

distance up the incline, measured from point B, does the plate come to a permanent stop?


e

B2. A model of the magnetic properties of materials is based upon small magnetic moments generated by each atom in the material. One source of this magnetic moment is the magnetic field generated by the electron in its orbit around the nucleus. For simplicity, we will assume that each atom consists of a single electron of charge e and mass , a single proton of charge +e and mass , and that the electron orbits in a circular orbit of radius R about the proton.

em

pm m

a. Magnetic Moments. Assume that the electron orbits in the x-y plane. (3 pts) i. Calculate the net electrostatic force on the electron from the proton. Express your

answer in terms of any or all of the following parameters: e, , R, and the permittivity of free space,

em ,pm

0 , where 0

14 k

= . (k is the Coulombs Law constant).

(5 pts) ii. Determine the angular velocity 0 of the electron around the proton in terms of any or all of the following parameters: e, , R, and em 0 .

(8 pts) iii. Derive an expression for the magnitude of the magnetic field eB due to the orbital

motion of the electron at a distance from the x-y plane along the axis of orbital rotation of the electron. Express your answer in terms of any or all of the following parameters: e, , R,

z R

em 0 , and the permeability of free space ,z 0 .

(4 pts) iv. A small bar magnet has a magnetic field far from the magnet given by 0

3 ,2mBz

=

where z is the distance from the magnet on the axis connecting the north and south poles, m is the magnetic dipole moment, and 0 is the permeability of free space. Assuming that an electron orbiting a proton acts like a small bar magnet, find the dipole moment m for an electron orbiting an atom in terms of any or all of the following parameters: e, , R, andem 0 .


,

b. Diamagnetism. We model a diamagnetic substance to have all atoms oriented so that the electron orbits are in the x-y plane, exactly half of which are clockwise and half counterclockwise when viewed from the positive z axis looking toward the origin. Some substances are predominantly diamagnetic.

(3 pts) i. Calculate the total magnetic moment of a diamagnetic substance with N atoms.

Write your answer in terms of any or all of the following parameters: and , , ,ee m R N 0.

(6 pts) ii. An external magnetic field 0 0 B B z=

Gis applied to the substance. Assume that the

introduction of the external field doesnt change the fact that the electron moves in a circular orbit of radius R. Determine , the change in angular velocity of the electron, for both the clockwise and counterclockwise orbits. Throughout this entire problem you can assume that 0 . Write your answer in terms of

and, ee m , 0B only.

(6 pts) iii. Assume that the external field is turned on at a constant rate in a time interval t . That is to say, when 0t = the external field is zero and when the external field is

t = t0.BG

Determine the induced emf E experienced by the electron. Write your answer in terms of any or all of the following parameters: , , , ,ee m R N 0B , 0 , and

0.

(6 pts) iv. Verify that the change in the kinetic energy of the electron satisfies E. This justifies our assumption in (ii) that R does not change.

K e =

(6 pts) v. Determine the change in the total magnetic moment m for the N atoms when the

external field is applied, writing your answer in terms of 0, , , ,ee m R N and 0.B

(3 pts) vi. Suppose that the uniform magnetic field used in the previous parts of this problem is replaced with a bar magnet. Would the diamagnetic substance be attracted or repelled by the bar magnet? How does your answer show this?

2007 Semifinal Exam Solutions 1

AAPT UNITED STATES PHYSICS TEAMAIP 2007

Solutions to Problems

Part A

Question 1

a. There is a high degree of symmetry present. Points b, c, and e are at the same potential;similarly, points d, f , and g are at the same potential. The circuit then reduces to a seriesconnection of three parallel resistor clusters.

The three parallel clusters have effective resistances of 1 , 8/5 , and 1/3 . The effectiveresistance of the circuit is then 44/15 .

b. After a long time no current will flow through the branches of the circuits containing capaci-tors. The circuit then reduces to a parallel connection of three series resistor clusters.

The effective resistance of the circuit is 4 , the current through the circuit is then

(12 V)/(4 ) = 3 A. (A1-1)

The three branches are identical; each then carries 1 A.

The potential drop across each capacitor is the same as the potential drop across the 8 resistors, so

VC = (1 A)(8 ) = 8 V. (A1-2)

Finally, the charge on each capacitor is

Q = (15 F)(8 V) = 120 C. (A1-3)

Question 2

a. Two parts, solved individually.

i. Isochoric compression:PfPi

=TfTi, (A2-1)

soT =

PcrP0

T0. (A2-2)

Copyright c2007 American Association of Physics Teachers


ii. Adiabatic compression:PV = const, (A2-3)

where = CP /CV = (CV + 1)/CV = 5/3 for a monatomic gas. Consequently

P0L0 = PcrL

, (A2-4)

and then

L = L0(P0Pcr

)5/3. (A2-5)

b. The normal pressure on the bullet comes from

P =r

rcE. (A2-6)

Therefore, the normal force on the bullet is

FN =r

rcE (2pirch), (A2-7)

and finally the force of friction is FN . The force due to the pressure difference between theinside of the barrel and the outside must equal the normal force, so

(pir2c )(Pcr P0) = 2pih r E, (A2-8)

and thenPcr = P0 +

2Ehr2c

r. (A2-9)

Question 3

a. If the sphere has radius r, it has charge

q =43pir3 (A3-1)

and thus its surface is at electrostatic potential

V =q

4pi0r=

r2

30(A3-2)

To increase the radius by dr, an additional charge dq = 4pir2dr must be brought in frominfinity, requiring work

dU = V dq =4pir42

30dr (A3-3)

Thus to grow the sphere from r = 0 to r = R requires

U = R0

4pir42

30dr =

4piR52

150(A3-4)



b. Each drop has volume Vd = 43piR3, so the number of drops is

n =VfVd

=Vf

43piR

3(A3-5)

Since we are ignoring inter-drop forces, the total energy of the drops is simply the sum of theenergies of each individual drop:

Ue,tot = nU =Vf

43piR

3

4piR52

150=

R22

50Vf (A3-6)

c. Each drop has surface area 4piR2 and thus surface tension energy 4piR2. As before, the totalenergy due to surface tension is just the sum of the energies of the individual drops:

Us,tot = 4piR2n = 4piR2Vf

43piR

3=

3RVf (A3-7)

d. The total potential energy from both sources is

Utot =

(R22

50+

3R

)Vf (A3-8)

Equilibrium is reached when the total energy is a minimum; since U at both R 0and R, it must have an interior minimum.

d

dRUtot =

(2R2

50 3R2

)Vf (A3-9)

Setting this equal to zero,2R2

50=

3R2

(A3-10)

R3 =15022

(A3-11)

R =(15022

) 13

(A3-12)

Question 4

a. The electric field between the plates is given by E = V/d. The force on the charged ball isthen F = Eq = V q/d. The acceleration of the ball is a = V q/md.

Kinematics gives us d = at2/2 for the time of flight. So

t =2d/a =

2md2/qV . (A4-1)

b. The kinetic energy collected by a ball will be K = qV as it moves between the plates. Thatswhat will be dissipated.



c. The current is given by I = Q/t. The total number of balls is N = n0A, where A is thesurface area of a plate. The charge Q is then Q = n0qA, so the current is

I =Qt

=n0qA2md2/qV

. (A4-2)

We cant stop here, since this is not in terms of the allowed variables. The problem is A andd, but since C = 0A/d, we have

I =n0qA2md2/qV

, (A4-3)

=A

dn0q

qV

2m, (A4-4)

=C

0n0q

qV

2m. (A4-5)

d. R = V/I, so

R =V

I=

0V

Cn0q

2mqV

. (A4-6)

We can simplify, slightly, with

R =0

Cn0q

2mVq

. (A4-7)

e. P = V I, so

P = VC

0n0q

qV

2m=

02n02C2q3V 3

2m. (A4-8)

Part B

Question 1

a. To not slip, from a free-body diagram, we must have

mg cos mg sin (B1-1)

so tan . (B1-2)

Therefore c = tan and hence

=tan 2

. (B1-3)

b. In one cycle the energy input into the system is

MgL sin , (B1-4)

the energy of the block dropping.



The energy loss on the way up isLmg cos (B1-5)

and the energy loss on the way down is

L(m+M)g cos (B1-6)

ThusMgL sin = Lmg cos + L(m+M)g cos (B1-7)

and since 2 cos = sin ,

M =m

2+m+M

2, (B1-8)

= 2m, (B1-9)R = M/m = 2. (B1-10)

c. The period of a mass m oscillating on a spring of spring constant k is

T = 2pim

k. (B1-11)

In this case, the friction force is constant on both the up and down trips, and so each trip issimple harmonic (with different equilibrium points). Hence

T0 = pim

k+ pi

3mk, (B1-12)

T = 2pim

k, (B1-13)

T0/T =

1 +3

2. (B1-14)

d. As mentioned in part (c), both the up and down trips are simple harmonic, this time witha mass of m both ways. The equilibrium points for the two trips are different, however. Onthe up trip, the equilibrium point is clearly at a distance L/2 from B, since the plate stopsat both B and A and hence those are the endpoints of the oscillation and the equilibrium ishalfway between. For the trip down, the equilibrium point will shift by a distance y such that

ky = 2mg cos = mg sin (B1-15)

because 2mg cos is the difference between the friction forces on the trip up and the tripdown.

The place where the plate finally comes to a stop is the first place that is at the end of anoscillation (either up or down) and where the total force being exerted by gravity and thespring is less than the maximal force of friction. For that to happen, the plate needs to nothave gone past the other equilibrium point during that oscillation.

So we start by determining where the endpoints of the oscillations are. For the first trip upthese are B and A. For the following trip down, the plate stops at a distance of (2mg sin )/kfrom B (because the equilibrium shifts up by (mg sin )/k. For the following trip up, the



plate stops a distance L (2mg sin )/k from B, since the equilibrium point is again in themiddle of the incline. And so forth.

Thus the stopping points are located at

n(2mg sin )/k and L n(2mg sin )/k (B1-16)for integer n. The plate will stop permanently once either

n(2mgsin)/k > L/2 (B1-17)

orL n(2mg sin )/k < L/2 + (mg sin )/k, (B1-18)

whichever happens first. (The first condition corresponds to going down and ending up abovethe midpoint at the end of the down trip, the second condition corresponds to going up andstopping below the upper equilibrium.) The second condition can be rewritten as(

n+12

)(2mg sin )/k > L/2. (B1-19)

Question 2

a. Magnetic Moments

i. From Coulombs Law,

F =e2

4pi0R2(B2-1)

ii. For circular motion,

F =mev

2

R= meR20, (B2-2)

The force is provided by the Coulomb force, so

meR20 =

e2

4pi0R2, (B2-3)

0 =

e2

4pi0meR3(B2-4)

iii. From the law of Biot and Savart,

~Be =0i

4pi

d~s ~rr3

, (B2-5)

Be =0i

4pi2piR

R

(z2 + r2)3/2, (B2-6)

0iR2

2z3. (B2-7)

For the current, i, we can writei =

q

t=

e02pi

. (B2-8)

Then

Be =0e0R

2

4piz3. (B2-9)



iv. By substitution,

m =e0R

2. (B2-10)

b. Diamagnetism

i. If half go one way and half go the other, M = 0.

ii. Additional force from magnetism,

FB = qvB0 = eRB0 (B2-11)

modifies previous central force problem to give

meR2 =

e2

4pi0R2 eR0B0, (B2-12)

where the positive sign corresponds to anticlockwise motion, the negative to clockwisemotion.A little math,

meR(2 20) = eRB0, (B2-13)me( 0)( + 0) = eB0, (B2-14)

me()(20) = e0B0, (B2-15)

where in the last line we have used the approximation 0. Then

= eB02me

. (B2-16)

iii. The emf is given by

E = nt

=nt

, (B2-17)

but n/t is a measure of the number of turns made by the electron in a time intervalt, so

nt

=0R

2piR=

02pi

. (B2-18)

ThenE = 0

2piB0piR

2 =120b0R

2. (B2-19)

iv. The change in kinetic energy is given by

K = (12me

2R2), (B2-20)

= meR2 , (B2-21) meR20 , (B2-22)= me0R2

( eB02me

), (B2-23)

= eE . (B2-24)



v. M = Nm, where N is the number of atoms, and m the change in magnetic momentin each. The change is

m = (e0R

2

). (B2-25)

=eR

2, (B2-26)

=e2R2B04me

, (B2-27)

so

M = Ne2R2B04me

. (B2-28)

vi. Repelled, by Lenzs law.



F = ma Contest

2008

2008 F = ma Exam 1


2008 F = ma Contest

25 QUESTIONS - 75 MINUTES

INSTRUCTIONS

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Use g = 10 N/kg throughout this contest. You may write in this booklet of questions. However, you will not receive any credit foranything written in this booklet.

Your answer to each question must be marked on the optical mark answer sheet. Select the single answer that provides the best response to each question. Please be sure touse a No. 2 pencil and completely fill the box corresponding to your choice. If you change ananswer, the previous mark must be completely erased.

Correct answers will be awarded one point; incorrect answers will result in a deduction of 14point. There is no penalty for leaving an answer blank.

A hand-held calculator may be used. Its memory must be cleared of data and programs. Youmay use only the basic functions found on a simple scientific calculator. Calculators maynot be shared. Cell phones may not be used during the exam or while the exam papers arepresent. You may not use any tables, books, or collections of formulas.

This test contains 25 multiple choice questions. Your answer to each question must be markedon the optical mark answer sheet that accompanies the test. Only the boxes preceded bynumbers 1 through 25 are to be used on the answer sheet.

All questions are equally weighted, but are not necessarily the same level of difficulty. In order to maintain exam security, do not communicate any information about

the questions (or their answers or solutions) on this contest until after February20, 2008.

The question booklet and answer sheet will be collected at the end of this exam. You maynot use scratch paper.



2008 F = ma Exam 2

1. A bird flying in a straight line, initially at 10 m/s, uniformly increases its speed to 18 m/s whilecovering a distance of 40 m. What is the magnitude of the acceleration of the bird?

(a) 0.1 m/s2

(b) 0.2 m/s2

(c) 2.0 m/s2

(d) 2.8 m/s2

(e) 5.6 m/s2

2. A cockroach is crawling along the walls inside a cubical room that has an edge length of 3 m. If thecockroach starts from the back lower left hand corner of the cube and finishes at the front upperright hand corner, what is the magnitude of the displacement of the cockroach?

(a) 32 m

(b) 3 32 m

(c) 33 m

(d) 3 m

(e) 9 m

3. The position vs. time graph for an object moving in a straight line is shown below. What is theinstantaneous velocity at t = 2 s?

0

2

4

3212

Posit

ion

(m)

Time (s)

(a) 2 m/s(b) 12 m/s(c) 0 m/s

(d) 2 m/s

(e) 4 m/s


2008 F = ma Exam 3

The information below is for the next two problems

Shown below is the velocity vs. time graph for a toy car moving along a straight line.

0

2

4

3212

Vel

ocity

(m/s)

Time (s)

4. What is the maximum displacement from start for the toy car?

(a) 3 m

(b) 5 m

(c) 6.5 m

(d) 7 m

(e) 7.5 m

5. Which of the following acceleration vs. time graphs most closely represents the acceleration of thetoy car?

Acc

eler

atio

n

0321

Time (s)

Acc

eler

atio

n

0321

Time (s)

(a) (b)

Acc

eler

atio

n

0321

Time (s)

Acc

eler

atio

n

0321

Time (s)

(c) (d)

Acc

eler

atio

n

0321

Time (s)

(e)


2008 F = ma Exam 4

6. A cannon fires projectiles on a flat range at a fixed speed but with variable angle. The maximumrange of the cannon is L. What is the range of the cannon when it fires at an angle pi6 above thehorizontal? Ignore air resistance.

(a)

32 L

(b) 12L

(c) 13L

(d) 12L

(e) 13L

7. A toboggan sled is traveling at 2.0 m/s across the snow. The sled and its riders have a combinedmass of 120 kg. Another child (mchild = 40 kg) headed in the opposite direction jumps on the sledfrom the front. She has a speed of 5.0 m/s immediately before she lands on the sled. What is thenew speed of the sled? Neglect any effects of friction.

(a) 0.25 m/s

(b) 0.33 m/s

(c) 2.75 m/s

(d) 3.04 m/s

(e) 3.67 m/s

8. Riders in a carnival ride stand with their backs against the wall of a circular room of diameter8.0 m. The room is spinning horizontally about an axis through its center at a rate of 45 rev/minwhen the floor drops so that it no longer provides any support for the riders. What is the minimumcoefficient of static friction between the wall and the rider required so that the rider does not slidedown the wall?

(a) 0.0012

(b) 0.056

(c) 0.11

(d) 0.53

(e) 8.9


2008 F = ma Exam 5

9. A ball of mass m1 travels along the x-axis in the positive direction with an initial speed of v0. Itcollides with a ball of mass m2 that is originally at rest. After the collision, the ball of mass m1has velocity v1xx+ v1y y and the ball of mass m2 has velocity v2xx+ v2y y.

Consider the following five statements:

I) 0 = m1v1x +m1v2xII) m1v0 = m1v1y +m2v2yIII) 0 = m1v1y +m2v2yIV) m1v0 = m1v1x +m1v1yV) m1v0 = m1v1x +m2v2x

Of these five statements, the system must satisfy

(a) I and II

(b) III and V

(c) II and V

(d) III and IV

(e) I and III

The following information applies to the next two problems

An experiment consists of pulling a heavy wooden block across a level surface with a spring forcemeter. The constant force for each try is recorded, as is the acceleration of the block. The dataare shown below.

Force F in Newtons 3.05 3.45 4.05 4.45 5.05

acceleration a in meters/second2 0.095 0.205 0.295 0.405 0.495

10. Which is the best value for the mass of the block?

(a) 3 kg

(b) 5 kg

(c) 10 kg

(d) 20 kg

(e) 30 kg

11. Which is the best value for the coefficient of friction between the block and the surface?

(a) 0.05

(b) 0.07

(c) 0.09

(d) 0.5

(e) 0.6


2008 F = ma Exam 6

12. A uniform disk rotates at a fixed angular velocity on an axis through its center normal to the planeof the disk, and has kinetic energy E. If the same disk rotates at the same angular velocity aboutan axis on the edge of the disk (still normal to the plane of the disk), what is its kinetic energy?

(a) 12E

(b) 32E

(c) 2E

(d) 3E

(e) 4E

13. A mass is attached to the wall by a spring of constant k. When the spring is at its natural length,the mass is given a certain initial velocity, resulting in oscillations of amplitude A. If the springis replaced by a spring of constant 2k, and the mass is given the same initial velocity, what is theamplitude of the resulting oscillation?

(a) 12A

(b) 12A

(c)2A

(d) 2A

(e) 4A

14. A spaceborne energy storage device consists of two equal masses connected by a tether and rotatingabout their center of mass. Additional energy is stored by reeling in the tether; no external forcesare applied. Initially the device has kinetic energy E and rotates at angular velocity . Energy isadded until the device rotates at angular velocity 2. What is the new kinetic energy of the device?

(a)2E

(b) 2E

(c) 22E

(d) 4E

(e) 8E


2008 F = ma Exam 7

15. A uniform round tabletop of diameter 4.0 m and mass 50.0 kg rests on massless, evenly spaced legsof length 1.0 m and spacing 3.0 m. A carpenter sits on the edge of the table. What is the maximummass of the carpenter such that the table remains upright? Assume that the force exerted by thecarpenter on the table is vertical and at the edge of the table.

3.0 m

4.0 m

1.0 m

(a) 67 kg

(b) 75 kg

(c) 81 kg

(d) 150 kg

(e) 350 kg

16. A massless spring with spring constant k is vertically mounted so that bottom end is firmly attachedto the ground, and the top end free. A ball with mass m falls vertically down on the top end ofthe spring, becoming attached, so that the ball oscillates vertically on the spring. What equationdescribes the acceleration a of the ball when it is at a height y above the original position of thetop end of the spring? Let down be negative, and neglect air resistance; g is the magnitude of theacceleration of free fall.

(a) a = mv2/y + g

(b) a = mv2/k g(c) a = (k/m)y g(d) a = (k/m)y + g(e) a = (k/m)y g


2008 F = ma Exam 8

17. A mass m is resting at equilibrium suspended from a vertical spring of natural length L and springconstant k inside a box as shown:

The box begins accelerating upward with acceleration a. How much closer does the equilibriumposition of the mass move to the bottom of the box?

(a) (a/g)L

(b) (g/a)L

(c) m(g + a)/k

(d) m(g a)/k(e) ma/k

18. A uniform circular ring of radius R is fixed in place. A particle is placed on the axis of the ringat a distance much greater than R and allowed to fall towards the ring under the influence of therings gravity. The particle achieves a maximum speed v. The ring is replaced with one of the same(linear) mass density but radius 2R, and the experiment is repeated. What is the new maximumspeed of the particle?

(a) 12v

(b) 12v

(c) v

(d)2v

(e) 2v

19. A car has an engine which delivers a constant power. It accelerates from rest at time t = 0, and att = t0 its acceleration is a0. What is its acceleration at t = 2t0? Ignore energy loss due to friction.

(a) 12a0

(b) 12a0

(c) a0

(d)2a0

(e) 2a0


2008 F = ma Exam 9

20. The Youngs modulus, E, of a material measures how stiff it is; the larger the value of E, themore stiff the material. Consider a solid, rectangular steel beam which is anchored horizontally tothe wall at one end and allowed to deflect under its own weight. The beam has length L, verticalthickness h, width w, mass density , and Youngs modulus E; the acceleration due to gravity isg. What is the distance through which the other end moves? (Hint: you are expected to solve thisproblem by eliminating implausible answers. All of the choices are dimensionally correct.)

(a) h exp(gLE

)

(b) 2gh2

E

(c)2Lh

(d) 32gL4

Eh2

(e)3ELgh

21. Consider a particle at rest which may decay into two (daughter) particles or into three (daughter)particles. Which of the following is true in the two-body case but false in the three-body case?(There are no external forces.)

(a) The velocity vectors of the daughter particles must lie in a single plane.

(b) Given the total kinetic energy of the system and the mass of each daughter particle, itis possible to determine the speed of each daughter particle.

(c) Given the speed(s) of all but one daughter particle, it is possible to determine the speedof the remaining particle.

(d) The total momentum of the daughter particles is zero.

(e) None of the above.

22. A bullet of mass m1 strikes a pendulum of mass m2 suspended from a pivot by a string of lengthL with a horizontal velocity v0. The collision is perfectly inelastic and the bullet sticks to thebob. Find the minimum velocity v0 such that the bob (with the bullet inside) completes a circularvertical loop.

(a) 2Lg

(b)5Lg

(c) (m1 +m2)2Lg/m1

(d) (m1 m2)Lg/m2

(e) (m1 +m2)5Lg/m1


2008 F = ma Exam 10

23. Consider two uniform spherical planets of equal density but unequal radius. Which of the followingquantities is the same for both planets?

(a) The escape velocity from the planets surface.

(b) The acceleration due to gravity at the planets surface.

(c) The orbital period of a satellite in a circular orbit just above the planets surface.

(d) The orbital period of a satellite in a circular orbit at a given distance from the planetscenter.

(e) None of the above.

24. A ball is launched upward from the ground at an initial vertical speed of v0 and begins bouncingvertically. Every time it rebounds, it loses a proportion of the magnitude of its velocity due to theinelastic nature of the collision, such that if the speed just before hitting the ground on a bounce isv, then the speed just after the bounce is rv, where r < 1 is a constant. Calculate the total lengthof time that the ball remains bouncing, assuming that any time associated with the actual contactof the ball with the ground is negligible.

(a) 2v0g

11r

(b) v0g

r1r

(c) 2v0g

1rr

(d) 2v0g

11r2

(e) 2v0g

11+(1r)2

25. Two satellites are launched at a distance R from a planet of negligible radius. Both satellites arelaunched in the tangential direction. The first satellite launches correctly at a speed v0 and entersa circular orbit. The second satellite, however, is launched at a speed 12v0. What is the minimumdistance between the second satellite and the planet over the course of its orbit?

(a) 12R

(b) 12R

(c) 13R

(d) 14R

(e) 17R


2008 F = ma Exam 11

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ANSWERS TO 2008 Fnet = ma CONTEST

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EA25


Quarter Final Contest

2008

2008 Quarter-final Exam 1


2008 Quarter-Final Exam

4 QUESTIONS - 60 MINUTES

INSTRUCTIONS


Show all your work. Partial credit will be given.

Start each question on a new sheet of paper. Put your name in the upper right-hand corner ofeach page, along with the question number and the page number/total pages for this problem.For example,

Doe, Jamie

Prob. 1 - P. 1/3

A hand-held calculator may be used. Its memory must be cleared of data and programs. Youmay use only the basic functions found on a simple scientific calculator. Calculators may notbe shared.

Cell phones may not be used during the exam or while the exam papers are present. Youmay not use any tables, books, or collections of formulas.

Each of the four questions is worth 25 points. The questions are not necessarily of the samedifficulty. Good luck!

In order to maintain exam security, do not communicate any information aboutthe questions (or their answers or solutions) on this contest until after March 10,2008.



1. A charged particle with charge q and mass m is given an initial kinetic energy K0 at themiddle of a uniformly charged spherical region of total charge Q and radius R. q and Q haveopposite signs. The spherically charged region is not free to move. Throughout this problemconsider electrostatic forces only.

R

(a) Find the value ofK0 such that the particle will just reach the boundary of the sphericallycharged region.

(b) How much time does it take for the particle to reach the boundary of the region if itstarts with the kinetic energy K0 found in part (a)?

2. A uniform pool ball of radius r and mass m begins at rest on a pool table. The ball is given ahorizontal impulse J of fixed magnitude at a distance r above its center, where 1 1.The coefficient of kinetic friction between the ball and the pool table is . You may assumethe ball and the table are perfectly rigid. Ignore effects due to deformation. (The moment ofinertia about the center of mass of a solid sphere of mass m and radius r is Icm =

2

5mr2.)

r

rJ

(a) Find an expression for the final speed of the ball as a function of J , m, and .

(b) For what value of does the ball immediately begin to roll without slipping, regardlessof the value of ?



3. A block of mass m slides on a circular track of radius r whose wall and floor both havecoefficient of kinetic friction with the block. The size of the block is small compared to theradius of the track. The floor lies in a horizontal plane and the wall is vertical. The block isin constant contact with both the wall and the floor. The block has initial speed v0.

(a) Let the block have kinetic energy E after traveling through an angle . Derive anexpression for dE

din terms of g, r, , m and E.

(b) Suppose the block circles the track exactly once before coming to a halt. Determine v0in terms of g, r, and .

4. Two beads, each of mass m, are free to slide on a rigid, vertical hoop of mass mh. The beadsare threaded on the hoop so that they cannot fall off of the hoop. They are released withnegligible velocity at the top of the hoop and slide down to the bottom in opposite directions.The hoop remains vertical at all times. What is the maximum value of the ratio m/mh suchthat the hoop always remains in contact with the ground? Neglect friction.

vertical circular hoop

Ground

beads released with negligible velocity, slide on hoop wire


2008 Quarter-final Exam - Solutions 1

2008 Quarter-Final Exam Solutions

1. A charged particle with charge q and mass m starts with an initial kinetic energy K at themiddle of a uniformly charged spherical region of total charge Q and radius R. q and Q haveopposite signs. The spherically charged region is not free to move.

(a) Find the value of K0 such that the particle will just reach the boundary of the sphericallycharged region.

(b) How much time does it take for the particle to reach the boundary of the region if itstarts with the kinetic energy K0 found in part (a)?

Solution:

Assume that q is negative and Q is positive.

(a) Apply Gausss Law to a spherical shell of radius r where r < R. Then,

E4pir2 =4pir3

0

where the charge density = 3Q4piR3

Solving for the Electric Field at a distance r from the center, we find

E =r

30

We now find the potential difference between the center of the sphere and the outer boundaryof the charged cloud.

V = R0Edr

Substituting in our expression for E, we now find

V = R2

60

Using conservation of energy,

Klost = Ugained

K0 = qV = qQ8piR0

(b) Apply Newtons Second Law to the object of charge q when it is located a distance raway from the center of the sphere.



Fnet = ma

(q)E = md2r

dt2

d2r

dt2= qr

30m

We recognize that this is the differential equation for simple harmonic motion

d2r

dt2= 2r

where =

q30m

.

Since the charge has the minimum kinetic energy needed to reach the surface, the trip fromthe center of the sphere to the outer boundary is one-fourth of a cycle of SHM.

Therefore,

t =T

4=

pi

2=pi

2

30mq

.

Substituting, in for rho, we find

t =pi

2

4pi0mR3

qQ.

2. A uniform pool ball of radius r begins at rest on a pool table. The ball is given a horizontalimpulse J of fixed magnitude at a distance r above its center, where 1 1. Thecoefficient of kinetic friction between the ball and the pool table is . You may assume theball and the table are perfectly rigid. Ignore effects due to deformation. (The moment ofinertia about the center of mass of a solid sphere of mass m and radius r is Icm = 25mr

2.)

(a) Find an expression for the final speed of the ball as a function of J , m, and .

(b) For what value of does the ball immediately begin to roll without slipping, regardlessof the value of .

.

Solution 1:

(a) Consider an axis perpendicular to the initial impulse and coplanar with the table. (Through-out this solution we consider only torques and angular momenta with respect to this axis.)After the initial impulse, the torque about this axis is always zero, so angular momentum isconserved. The initial impulse occurs a perpendicular distance ( + 1) r from the axis, so theangular momentum is

L = ( + 1) rJ

After the ball has skidded along the table for a certain distance, it will begin to roll withoutslipping. At that point, = vfr .



Meanwhile, its moment of inertia about this axis is (by the parallel-axis theorem) I = Icm +mr2 = 75mr

2, so that its final angular velocity is given by

L = I

( + 1) rJ =75mr2

( + 1) J =75mvf

.

Therefore,

vf =5J7m

(1 + )

.

(b) The ball acquires linear momentum J as a result of the horizontal impulse, so its initialvelocity v is given by

mv = J

We want the initial angular velocity and initial velocity to satisfy the no-slip condition v = r;thus

( + 1) J =75J

=25

.

Solution 2:.

Consider torques and angular momenta about the center of mass. If the horizontal impulse islarge compared with the horizontal impulse from friction during the time that the cue stickis in contact with the ball, then angular impulse = change in angular momentum becomes:

Jr =25mr20

.

Linear impulse = change in linear momentum yields:

J = mv0

.

(b) We want the initial angular velocity and initial velocity to satisfy the no-slip conditionv0 = 0r; thus

J =25mv0 =

25J

. =

25

.



(a) In the case that the ball does not immediately begin to roll without slipping, friction willexert an angular impulse, ftr, about the center of mass of the ball as the ball skids along thesurface:

ftr =25mr2(f 0)

.

Friction will also exert a linear impulse ft that will cause a change in linear momentum:

ft = m(vf v0)

.

Combining the last two equations:

m(vf v0) = 25mr(f 0)

After the ball has skidded along the table for a certain distance, it will begin to roll withoutslipping. At that point, = vfr .

vf + v0 = 25vf 25r0

Now, using the relationships between the impulse J and the initial angular and linear mo-mentum,

vf + Jm

=25vf J

m

.

J

m(1 + ) =

75vf

.

vf =5J7m

(1 + )

.

3. A block of mass m slides on a circular track of radius r whose wall and floor both havecoefficient of kinetic friction with the block. The floor lies in a horizontal plane and thewall is vertical. The block is in constant contact with both the wall and the floor. The blockhas initial speed v0.

(a) Let the block have kinetic energy E after traveling through an angle . Derive anexpression for dEd in terms of g, r, , m and E.

(b) Suppose the block circles the track exactly once before coming to a halt. Determine v0in terms of g, r, and .



Solution Let v be the speed of the block after it has traveled through an angle . When weapply Newtons Second Law to the block as it is traveling around the circular path of radiusr at a speed v, we find that the force of the wall on the block, Fw, is

Fw =mv2

r

and the force of the floor on the block, FN , is

FN = mg

Therefore, the total force of friction on the block is

FFric = m(v2

r+ g)

.

Let dE denote the loss of kinetic energy as a result of the work W that friction does on theblock when it has traveled through an angle d. The block will travel a distance ds = rd asit travels through an angle d.

Then,dE = W = FFricds = FFricrd

.

Now, substituting in the expressions for FFric, we find

dE = m(v2 + gr)d

.dE

d= (mv2 +mgr) = (2E +mgr)

where we have simply used the fact that the kinetic energy E is E = mv2

2 .

(b) Now, we separate variables and solve the differential equation.dE

2E +mgr=d

.12

ln |2E +mgr| = + C.

Exponentiating both sides, yields

2E +mgr = Ce2

Let E0 denote the initial kinetic energy. Then, using initial conditions, we find that C =2E0 +mgr.

Now, using the fact that E = 0 when = 2pi, we obtain

mgr = (2E0 +mgr)e4pi



.

Solving for E0,

E0 =12mgr(e4pi 1)

.

Then, using E0 =mv202 , and we find that

v0 =gr(e4pi 1)

.

4. Two beads, each of mass m, are free to slide on a rigid, vertical hoop of mass mh. The beadsare threaded on the hoop so that they cannot fall off of the hoop. They are released withnegligible velocity at the top of the hoop and slide down to the bottom in opposite directions.What is the maximum value of the ratio m/mh such that the hoop always remains in contactwith the ground? Neglect friction.

Solution 1: Draw a free-body diagram for each bead; let FN be the (inward) normal forceexerted by the hoop on the bead. Let be the angular position of the bead, measured fromthe top of the hoop, and let the hoop have radius r. We see that

FN +mg cos = mv2

r

FN = mv2

rmg cos

The (downward) vertical component FNy is given by

FNy = FN cos

From Newtons third law, the two beads together exert an upwards vertical force on the hoopgiven by

Fu = 2FNy

Fu = 2m cos (v2

r g cos

)noting that the beads clearly reach the same position at the same time.

Meanwhile, when each bead is at a position it has moved through a vertical distancer (1 cos ). Thus from energy conservation,

12mv2 = mgr (1 cos )

v2

r= 2g (1 cos )

Inserting this into the previous result,

Fu = 2m cos (2g (1 cos ) g cos )

Fu = 2mg(2 cos 3 cos2 )



If the beads ever exert an upward force on the hoop greater than mhg, the hoop will leavethe ground; i.e., the condition for the hoop to remain in contact with the ground is that forall ,

Fu mhg

We can replace the left side by its maximum value. Letting s = cos ,

Fu = 2mg(2s 3s2)

d

dsFu = 2mg (2 6s)

The derivative is zero at s = 13 , where

Fu(max) =23mg

So our condition is23mg mhgm

mh 3

2

Solution 2: As before, we apply energy conservation to find the speed of the beads:

v =

2gr (1 cos )

The vertical (downward) component of the beads velocity is thus

vy = v sin

vy = sin

2gr (1 cos )While the hoop is in contact with the ground, the beads are the only part of the system inmotion, so the momentum of the system is simply the beads momentum:

py = 2mvy

py = 2m sin

2gr (1 cos )

The net (downward) force on the system is

Fnet =dpydt

=dpyd

d

dt=dpyd

v

r

Fnet = 2m

2grd

dt

(sin

1 cos ) 1r

2gr (1 cos )

Fnet = 4mg(

cos

1 cos + sin 12

1 cos sin )

1 cos

Fnet = 4mg(

cos cos2 + 12

sin2 )

Fnet = 2mg(2 cos 3 cos2 + 1)



This downward force is provided by gravity and the normal force upward on the hoop:

Fnet = 2mg +mhg FNIf the hoop is to remain in contact with the ground, the normal force can never be negative:

FN 0

2mg +mhg Fnet 02mg +mhg 2mg

(2 cos 3 cos2 + 1) 0

2mg(2 cos 3 cos2 ) mhg

and we proceed as above.



Semi Final Contest

2008

2008 Semifinal Exam 1


Semifinal Exam

DO NOT DISTRIBUTE THIS PAGE

Important Instructions for the Exam Supervisor

This examination consists of three parts.

Part A has four questions and is allowed 90 minutes.

Part B has two questions and is allowed 90 minutes.

Part C has one question and is allowed 20 minutes. The answer for Part C will not be used for teamselection, but will be used for special recognition from the Optical Society of America.

The first page that follows is a cover sheet. Examinees may keep the cover sheet for all three parts ofthe exam.

The three parts are then identified by the center header on each page. Examinees are only allowed todo one part at a time, and may not work on other parts, even if they have time remaining.

Allow 90 minutes to complete Part A. Do not let students look at Part B or Part C. Collect the answersto Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutesbreak between parts A and B.

Allow 90 minutes to complete Part B. Do not let students look at Part C or go back to Part A. Collectthe answers to part B before allowing the examinee to begin Part C. Examinees are allowed a 10 to 15minutes break between Parts B and C.

Allow 20 minutes to complete Part C. This part is optional; scores on Part C will not be used to selectthe US Team. Examinees may not go back to Part A or B.

Ideally the test supervisor will divide the question paper into 4 parts: the cover sheet (page 2), PartA (pages 3-7), Part B (pages 8-10), and Part C (page 11). Examinees should be provided the partsindividually, although they may keep the cover sheet.

The supervisormust collect all examination questions, including the cover sheet, at the end of the exam,as well as any scratch paper used by the examinees. Examinees may not take the exam questions. Theexamination questions may be returned to the students after March 31, 2008.

Examinees are allowed calculators, but they may not use symbolic math, programming, or graphicfeatures of these calculators. Calculators may not be shared and their memory must be cleared of dataand programs. Cell phones, PDAs or cameras may not be used during the exam or while the exampapers are present. Examinees may not use any tables, books, or collections of formulas.

Please provide the examinees with graph paper for Part A.


2008 Semifinal Exam Cover Sheet 2


Semifinal Exam

INSTRUCTIONS


Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25points. Do not look at Parts B or C during this time.

After you have completed Part A you may take a break.

Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points.Do not look at Parts A or C during this time.

Show all your work. Partial credit will be given. Do not write on the back of any page. Do not writeanything that you wish graded on the question sheets.

Start each question on a new sheet of paper. Put your school ID number, your name, the questionnumber and the page number/total pages for this problem, in the upper right hand corner of eachpage. For example,

School ID #

Doe, Jamie

A1 - 1/3

A hand-held calculator may be used. Its memory must be cleared of data and programs. You may useonly the basic functions found on a simple scientific calculator. Calculators may not be shared. Cellphones, PDAs or cameras may not be used during the exam or while the exam papers are present.You may not use any tables, books, or collections of formulas.

Questions with the s

usapho problems (2007-2014)

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