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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

Physics 8.01 Fall Term 2008

Practice Problems for Exam 2 Solutions Part I Concept Questions: Circle your answer. 1) A spring-loaded toy dart gun is used to shoot a dart straight up in the air, and the dart reaches a maximum height hm . The same dart is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the dart go this time, neglecting friction and assuming an ideal spring?

a) hm b) hm / 2 c) hm / 4 d) 2hm e) 4hm f) escapes to infinity

Answer: (c). The energy of the spring-dart-earth system is constant so for the first case if x is the

distance compressed then

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kx2 = mghm . For the second case, 12

k(x / 2)2 = mgh . So

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12

kx2⎛⎝⎜

⎞⎠⎟= mgh or

14

(mghm ) = mgh . Thus h =14

hm .

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2) Two wheels, suspended with frictionless bearings in the hubs, start from rest, with forces 1Fr

and 2Fr

applied as shown. The larger wheel has twice the radius and mass as the smaller wheel. Assume the mass of the hubs and spokes are very small compared to the mass of the rims, so that the moment of inertia about the axle is I = mR2 .

In order to impart identical angular accelerations, the magnitude of the force 2F

r applied to the

larger wheel must be

a) one fourth b) one half c) equal to d) twice e) four times

the magnitude of the force 1F

r applied to the smaller wheel.

Answer: (e). (Larger wheel has times greater moment of inertia, so it requires 4 times greater torque. As radius is 2 times greater, torque requires is 4 times greater force.)

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3) To make a projectile escape from the Earth, we must give it an initial speed larger than the magnitude of the escape velocity. This escape velocity, at the surface of the Earth, depends on

a) the radius of the Earth. b) the mass of the Earth. c) the mass of the projectile. d) a) and b). e) a) and c). f) b) and c). g) all of the above

Answer: (d). The magnitude of the escape velocity satisfies 12

mvesc2 =

Gmme

Re2 , so

vesc =

2Gme

Re2 .

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4) A dumbbell consists of two identical objects of mass m attached at each end of a rod of length l . A force F

r in the x -direction is applied to the dumbbell for a time interval Δ t . It is first

applied at the center of the dumbbell, as shown in (a), and the same force is then applied directly to one of the two objects, as shown in (b). Assume that the time Δt is so short that the dumbbell does dot move significantly during Δt.

The dumbbell in case (b) has

a) larger center of mass speed and larger kinetic energy

b) equal center of mass speed and larger kinetic energy

c) smaller center of mass speed and larger kinetic energy

d) larger center of mass speed and equal kinetic energy

e) equal center of mass speed and equal kinetic energy

f) smaller center of mass speed and equal kinetic energy

g) larger center of mass speed and smaller kinetic energy

h) equal center of mass speed and smaller kinetic energy

i) smaller center of mass speed and smaller kinetic energy

than the dumbbell in case (a) Answer: (b). (Just after the impulse FΔt , case (a) has both masses moving to the right at speed v = FΔt / (2m) while case (b) has the top mass with speed v = FΔt / m and the bottom mass at rest. Thus the speed of the CM is the same in both cases, but case (b) has two times the kinetic energy.)

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5) The potential energy function U (x) for a particle with total mechanical energy E is shown below.

The x-component of the velocity of the particle as a function of time is given by

vx (t) = Dcos(ωt) − Dsin(ωt) . The particle reaches the position 5 when

a) ωt = 0

b) ωt = π / 4

c) ωt = π / 2

d) ωt = 3π / 4

e) ωt = π

f) ωt = 5π / 4

g) ωt = 3π / 2

h) ωt = 7π / 4 Answer: (b). (Position 5 has maximum displacement and zero velocity. Cases (b) and (f) both have v = 0 , but (f) corresponds to position 1.)

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6) Which, if any, of the following dynamical equations give rise to simple harmonic motion? For any that do, give the angular frequencyω . All of the parameters represented by Greek letters are positive and non-zero.

a)

dxdt

= αx

b)

dxdt

= −αx

c) d 2xdt2 = βx

d)

d 2xdt2 = −βx

e)

dxdt

= γ x2

f)

dxdt

= −γ x2

g)

d 2xdt2 = δx2

h) d 2xdt2 = −δx2

Answer: d). ω = β

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7)) Consider two carts, of masses m and 2m , at rest on an air track. If you first push one cart for a time interval Δt and then the other for the same length of time, exerting equal force on each, the kinetic energy of the light cart is

a) one fourth

b) one half

c) equal to

d) twice

e) four times the kinetic energy of the heavy car. Answer: (d). The impulse is the same, so the heavier cart has half the speed and half the kinetic energy.

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Part II Analytic Problems: Problem 1: Particle 1 of mass M collides with particle 2 of mass 2M . Before the collision particle 1 is moving along the x-axis with a speed v1,0 and particle 2 is at rest. After the collision,

particle 2 is moving in a direction 30o below the x-axis with a speed v1,0 / 3 . (Note: sin 30o = 1 / 2 .) Particle 1 is moving upward at an angle θ to the x-axis and has a speed v1, f .

a. What is the angle θ that particle 1 makes with the x-axis after the collision?

b. What is the magnitude of the velocity v1, f of particle 1 after the collision?

c. Is the collision elastic or inelastic? Justify your answer.

Solutions: We are not given that the collision is elastic, but in the absence of external forces that would change the momentum, the vector momentum is the same before and after the collision. Using the coordinate directions as given in the figure, the x - and y -components of momentum before and after the collision are:

( )

( )( )

,0 1,0

, 1, 1,0

1, 1,0

,0

, 1, 1,0

1, 1,0

cos 2 / 3 cos30

cos

0

sin 2 / 3 sin30

sin / 3

x

x f f

f

y

y f f

f

p M v

p M v M v

M v M v

p

p M v M v

M v M v

θ

θ

θ

θ

=

= + °

= +

=

= − °

= −

(1.1)

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where sin 30 1/ 2° = , cos30 3 / 2° = have been used. Setting initial and final x - components of momentum equal and canceling the common factor of M ,

1,0 1, 1,0

1,

cos

0 cos .f

f

v v v

v

θ

θ

= +

= (1.2)

Setting initial and final y - components of momentum equal and canceling the common factor of M ,

1,01,

1,01,

0 sin3

sin .3

f

f

vv

vv

θ

θ

= −

= (1.3)

All parts of the problem involve simultaneous solution of the second equations in (1.2) and (1.3). The method presented here follows the ordering of the parts of the problem. a) Squaring both equations and adding, using 2 2cos sin 1θ θ+ = , yields

21,02

1,

1,01,

3

.3

f

f

vv

vv

=

= (1.4)

b) The second equation in (1.2) gives cos 0θ = , 90θ = ° immediately. It should be noted that the angle θ , the result of part b), can be found immediately by noting, as in (1.2), that the incident particle has no x - component of momentum after the collision, and hence must be moving perpendicular to the original direction of motion. Then, using sin 1θ = in (1.3) gives 1, fv as in (1.4) quite readily. c) The initial kinetic energy is 2

1,0(1/ 2)iK Mv= and the final kinetic energy is

( )

( )

2 21, 0,

2 21,0 1,0

21,0

1 1 22 21 1 22 23 31 ;2

f f fK Mv M v

v vM M

Mv

= +

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

=

(1.5)

the collision is elastic. The fact that the particles have the same final speed is mere coincidence.

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Problem 2: Momentum and Impulse A superball of m1 , starting at rest, is dropped from a height h0 above the ground and bounces back up to a height of hf . The collision with the ground occurs over a time

interval Δtc .

a) What is the momentum of the ball immediately before the collision? b) What is the momentum of the ball immediately after the collision?

c) What is the average force of the ground on the ball?

d) What impulse is imparted to the ball?

e) What is the change in the kinetic energy during the collision?

Solutions: For all parts, take the upward direction to be the positive j direction, and use the ground to be the zero point of gravitational potential energy. a) The initial mechanical energy is 0 1 0E m gh= ; all potential, no kinetic energy. The energy when the ball first hits the ground is 2

1 1 1 / 2E m v= , all kinetic, no potential energy. Equating these energies (with the assumption that there are no nonconservative forces) and solving for the speed 1v gives 1 02v gh= , for a momentum ( )1 1 0

ˆ2m gh= −p jr . (1.6)

b) In order to find the momentum, we need to know the speed. The speed is found, again, through conservation of mechanical energy. Denote the upward speed after the collision as 2v , with corresponding mechanical energy 2

2 1 2 / 2E m v= . This energy must be the same as the potential energy at the top of the bounce, 3 1 fE m g h= . Solving for the

rebound speed 2v gives 2 2 fv g h= , for a momentum

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( )2 1ˆ2 fm gh=p jr . (1.7)

c) While the ball is in contact with the ground, the average net force is the change in momentum divided by the time interval,

02 1net, ave 1

c c

ˆ2 fh hm g

t t t+Δ −

= = =Δ Δ Δp p pF jr r rr

. (1.8)

However, there are two forces on the ball, the contact force that the ground exerts on the ball and the gravitational force. The net force is the sum of these forces, and so

net ball, ground grav ball, ground 1

ˆm g= + = −F F F F jr r r r

. (1.9)

The gravitational force must be presumed to be constant, and hence equal to its average value. Combining the terms (watch the subscripts),

0ground, ball, ave net, ave 1 1

c

ˆ ˆ2 fh hm g m g g

t

⎛ ⎞+⎜ ⎟= + = +⎜ ⎟Δ⎝ ⎠

F F j jr r

. (1.10)

d) The impulse is the change in momentum, ( )2 1 1 0

ˆ2 fm g h hΔ = − = +p p p jr r r . (1.11)

e) The change in kinetic energy during the collision is the difference between the energies found from the heights before and after the collision, ( )0 2 1 0 1 1 0f fK E E m gh m gh m g h hΔ = − = − = − . (1.12)

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Problem 3: Center of Mass Two small particles of mass m1 and mass m2 attract each other with a force that varies at the inverse cube of their separation. At time t0 , m1 has velocity

rv1,0 directed

towards m2 , which is at rest a distance d away. At time t1 , the particles collide. How far does m1 travel in time interval t1 − t0 ?

Solution: Let’s consider the particle 2 (and everything in it) and the particle 1 as the system. The system is isolated in that there are no other external forces acting on it. Therefore the center of mass of the system does not accelerate, hence the velocity of the center of mass is constant. Choose an origin at the location of the particle 1 at t = t0 as shown in the figure below.

Let

rr1(t) = x1(t) i denote the position vector of the particle 1 at time t . Let rr2 (t) = x2 (t) i

denote the position vector of the particle 2 at time t . At time t = t0 , the particle 1 has position vector 1 0( )t t= =r 0

rr and the particle 2 has position vector rr2 (t = t0 ) = d i . So at t = t0 the center of mass is located at

1 1 2 2 20

1 2 1 2

ˆ( )cmm m m dt t

m m m m+

= = =+ +

r rR ir rr

.

The velocity of the center of mass is the derivative with respect to time

1 1 2 2

1 2cm

m mm m+

=+

v vVr rr

.

The center of mass velocity remains constant because there are no external forces acting on the system. At time t = t0 , the particle 2 is at rest 2 0( )t t= =v 0

rr and the particle 1 has

velocity 1 0 1,0ˆ( )t t v= =v ir , the center of mass velocity is

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1 1,0

1 2

ˆcm

m vm m

=+

V ir

.

The position of the center of mass as a function of time is therefore

2 1 1,0 00 0

1 2

( ) ˆ( ) ( ) ( )cm cm cm

m d m v t tt t t t t

m m+ −⎛ ⎞

= = + − = ⎜ ⎟+⎝ ⎠R R V ir r r

.

The collision takes place at the location of the center of mass at the time t = t1

2 1 1,0 1 01

1 2

( ) ˆ( )cm

m d m v t tt

m m+ −⎛ ⎞

= ⎜ ⎟+⎝ ⎠R ir

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Problem 4: Rocket Problem A rocket ascends from rest in a uniform gravitational field by ejecting exhaust with constant speed u relative to the rocket. Assume that the rate at which mass is expelled is given by dm / dt = γ m , where m is the instantaneous mass of the rocket and γ is a constant. The rocket is retarded by air resistance with a force F = bmv proportional to the instantaneous momentum of the rocket where b is a constant. What constant speed does the rocket approach (terminal speed) after a very long time has elapsed?

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Problem 5 A massless spring with spring constant k is attached at one end of a block of mass M that is resting on a frictionless horizontal table. The other end of the spring is fixed to a wall. A bullet of mass mb is fired into the block from the left with a speed v0 and comes to rest in the block. (Assume that this happens instantaneously). How fast is the block moving immediately after the bullet comes to rest?

The resulting motion of the block and bullet is simple harmonic motion.

a) Find the amplitude of the resulting simple harmonic motion.

b) How long does it take the block to first return to the position x = 0 ?

c) What fraction of the original kinetic energy of the bullet is stored in the harmonic oscillator?

d) Now suppose that instead of sliding on a frictionless table during the resulting

motion, the block is acted on by the spring and a weak friction force of constant magnitude f . Suppose that when the block first returned to the position x = 0 , the speed of the block was found to be one half the speed immediately after the collision that you found in part (a). How far did the block travel? (Your answer may include the symbol va , the speed of the block immediately after the bullet comes to rest, whether or not you have answered part (a) correctly.)

Solution: The collision is completely inelastic so

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mbv0 = (mb + M )va (1.13) Thus the speed immediately after the collision is

va =mb

(mb + M )v0 (1.14)

The energy of the spring-object system is constant since there are no external work done on the system (no fricition), therefore

12

(mb + M )va2 =

12

kxmax2 (1.15)

where the maximum displacement

xmax =

(mb + M )k

va =1

k(mb + M )mbv0 (1.16)

is the amplitude of the simple harmonic motion. Alternatively, the position of the system is given by the solution to the simple harmonic equation: x(t) = Acosωt + Bsinωt (1.17) where the angular frequency is given by

ω =k

mb + M (1.18)

The x-component of the velocity is given by vx (t) = −ωAsinωt +ωBcosωt (1.19)

At t = 0 , x(t = 0) = A = 0 , and vx (t = 0) = ωB =mb

(mb + M )v0 , so

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B =1ω

mb

(mb + M )v0 =

mb + Mk

mb

(mb + M )v0 =

1k(mb + M )

mbv0 (1.20)

The system reaches maximum amplitude when ωt = π / 2 . Thus from eq. (1.17)

x(t = π / 2ω ) = Bsin(π / 2) = B =

1k(mb + M )

mbv0 . (1.21)

(b) It takes the block half a period to return to the position x = 0 . The period is

T =2πω

= 2πmb + M

k (1.22)

So the block returns to x = 0 at time

t1 =T2=πω

= πmb + M

k (1.23)

c) The energy in the spring-bullet-block system is given by

Ea =12

(mb + M )va2 =

12

mb2

(mb + M )v0

2 =mb

mb + M⎛

⎝⎜⎞

⎠⎟12

mbv02 =

mb

mb + M⎛

⎝⎜⎞

⎠⎟Kb (1.24)

where the kinetic energy of the bullet before the collision is

Kb =12

mbv02 (1.25)

So the fraction of the original kinetic energy of the bullet that is stored in the harmonic oscillator is the ratio mb / (mb + M ) . d)

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The work done by the friction force is given by W = − fd (1.26) where d is the distance traveled by the system. This work is equal to the change in kinetic energy of the spring-bullet-block system, W = E1 − Ea (1.27) The energy when the block returns to x = 0 is

E1 =

12

(mb + M )va

2⎛

⎝⎜⎞

⎠⎟

2

=14

12

(mb + M )va2⎛

⎝⎜⎞⎠⎟=

14

Ea (1.28)

So Eq. (1.27) becomes

− fd =

14

Ea − Ea = −34

Ea = −34

12

(mb + M )va2⎛

⎝⎜⎞⎠⎟= −

38

mb2

(mb + M )v0

2 (1.29)

We can now solve for the distance traveled by the system

d =38

mb2

(mb + M ) fv0

2 (1.30)

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Problem 6 Cart B of mass m is initially moving with speed vB,0 to the right, as shown below. It collides and sticks to a second identical cart A of mass m that is initially at rest.

a) What is the speed of the two carts immediately after the collision? Express your answers in terms of m and vB,0 as needed.

The two conjoined carts then collide elastically at x = 0 with an ideal spring of spring constant k . (You may assume that the spring has negligibly small mass.) While the spring is compressing, the spring and carts undergo simple harmonic oscillation.

b) How far does the spring compress when the spring and carts first come to a stop?

Express your answers in terms of m , k , and vB,0 as needed.

c) How long does it take the right end of cart A to first return to the position x = 0 ? Express your answers in terms of m , k , and vB,0 as needed.

d) Set t = 0 to be the time immediately after the collision. Write down an expression

for the position of the right end of cart A as a function of time for the interval immediately after the collision until the right end of cart A first returns to the position x = 0 ? Express your answers in terms of m , k , and vB,0 as needed.

Solution: The system consisting of the two carts has no external forces acting on it so momentum is constant. Therefore

mBvB,0 = (mA + mB )va = 2mBva , (1.31)

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so

va =12

vB,0 (1.32)

b) There is no external work done on the carts-spring system during compression so the energy is constant, hence

12

(2m)va2 =

12

kxmax2 (1.33)

Thus

xmax =2mk

va =2mk

vB,0

2. (1.34)

c) It takes the block half a period to return to the position x = 0 . The angular frequency is

ω =k

2m (1.35)

The period is

T =2πω

= 2π2mk

(1.36)

So the block returns to x = 0 at time

t1 =T2=πω

= π2mk

(1.37)

d) The equation for the position is generally x(t) = C cos(ωt) + Dsin(ωt) (1.38) At t = 0 , x(t = 0) = 0 = C . (1.39) The x-component of the velocity is

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vx (t) =

dx(t)dt

= −ωC sin(ωt) +ωDcos(ωt) (1.40)

At t = 0 ,

vx (t = 0) = va =

12

vB,0 = ωD . (1.41)

Therefore substituting Eq. (1.35) into Eq. (1.41) and solving for D yields

D =

12ω

vB,0 =12

2mk

vB,0 . (1.42)

So substituting Eq. (1.39) and Eq. (1.42) into Eq. (1.38) yields

x(t) =12

2mk

vB,0 sink

2mt

⎛

⎝⎜

⎞

⎠⎟ (1.43)

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Problem 7 A projectile of mass m is fired vertically from the earth’s surface with an initial speed that is equal to the escape velocity. The radius of the earth is Re , the mass of the earth is Me , and the universal gravitational constant is G . Express your answers to the questions below in terms of Me , Re , m , and G as needed.

a) What is the initial speed of the projectile when it is launched from the surface of the earth? When the projectile is a distance 2Re from the center of the earth, it collides with a satellite of mass m that is orbiting the earth in a circular orbit. After the collision the two objects stick together. Assume that the collision is instantaneous.

b) What is the speed of the projectile, just before the collision, when it is a distance 2Re from the center of the earth? c) What is the speed of the satellite, just before the collision, when it is in a circular orbit of radius 2Re ? d) What is the speed of projectile and satellite immediately after the collision

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The escape speed of the projectile occurs when the energy of the projectile-earth system is zero, So

12

mpvp,esc2 −

Gmpme

Re

= 0

So the escape speed is

vp,esc =2Gme

Re

when the satellite reaches a height r = 2Re , the energy is

12

mpvp2 −

Gmpme

2Re

= 0 .

So the speed is

vp =Gme

Re

c) Newton’s Second Law for the satellite becomes

Gmsme

(2Re )2 =msvs

2

2Re

So the velocity of the satellite is

vs =Gme

2Re

d) Momentum is constant during the collision so Choose horizontal and vertical directions and θ to be the angle with respect to the horizontal that the combined objects emerge after the collision. If the masses are equal

mp = ms = m . Then the momentum equations become

vertical : mvp = 2mv f sinθ

horizontal : mvs = 2mv f cosθ

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Square and add these two equations yields

m2 (vs

2 + vp2 ) = 4m2v f

2 So

v f =

12

(vs2 + vp

2 )1/ 2 .

Substituting for the two speeds yields

v f =

12

Gme

2Re

⎛

⎝⎜⎞

⎠⎟+

Gme

Re

⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟

1/ 2

=3Gme

8Re

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Problem 8: An extreme skier is accelerated from rest by a spring-action cannon, skis once around the inside of a vertically oriented circular loop, then comes to a stop on a carpeted up-facing slope. Assume the cannon has a spring constant k and a cocked displacement x0 , the loop has a radius R , and the slope makes an angle θ to the horizontal. The only surface with friction is the carpet, represented by a friction constantμ . Gravity acts downward, with acceleration g , as shown. Express all your answers in terms of k , x0 , θ , the mass of the skier m , and g .

a) What is the velocity of the skier just before she enters the loop?

b) What is the maximum radius the loop can have and still allow the skier to remain in contact with its surface at the top of the loop?

c) What is the linear distance d the skier travels on the carpet before coming to rest?

d) What restriction must be placed on the parameters θ and μ to make sure the

skier is brought to a stop on the carpet and doesn’t slide back down?

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