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PHYSICS 101 MIDTERMOctober 27, 2005 2 hours

Please Circle your section1 9 am Galbiati 2 10 am Wang3 11 am Hasan 4 12:30 am Hasan5 12:30 pm Olsen

Problem Score1 /162 /163 /164 /185 /166 /18

Total /100

Instructions: When you are told to begin, check that this examination bookletcontains all the numbered pages from 2 through 17. The exam contains 6 problems.Read each problem carefully. You must show your work. The grade you get dependson your solution even when you write down the correct answer. BOX your finalanswer. Do not panic or be discouraged if you cannot do every problem; there areboth easy and hard parts in this exam. If a part of a problem depends ona previous answer you have not obtained, assume it and proceed. Keepmoving and finish as much as you can!

Possibly useful constants and equations are on the last page, which you maywant to tear off and keep handy

Rewrite and sign the pledge: I pledge my honor that I have not violated the Honor Codeduring this examination.

Signature

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Problem 1: Grab Bag

(a) [4 pts] A bowling ball and golf ball have the same initial momentum. The same constantforce F is applied to slow down both balls.Which ball stops faster? Box your answer and detail your reasoning in writing on theside.

1. The golf ball stops over a shorter time.

2. The bowling ball stops over a shorter time.

3. They stop over the same time.

4. They don’t stop.

The correct answer is answer 3 .

F∆t = ∆p

and therefore∆t = ∆p/F

Since F and ∆p are the same for the two balls, also ∆t is the same and they stopover the same time.

(b) [4 pts] A bowling ball and golf ball have the same initial momentum. The same constantforce F is applied to slow down both balls.Which ball stops in the shortest distance? Box your answer and detail your reasoningin writing on the side.

1. The golf ball stops over a shorter distance.

2. The bowling ball stops over a shorter distance.

3. They stop over the same distance.

4. They don’t stop.

The correct answer is answer 2 .

W = −Fs = KEf −KEi = −p2i /(2m)

Therefore:s = p2

i /(2mF )

and the distance is shorter for the ball with the larger mass.

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(c) [4 pts] Massless strings are guided through four massless and frictionless pulleys asshown. The masses of the pulleys are negligible.What is the relation between m and M so that the system stays in equilibrium?Explain your answer in detail. Box your answer.

The tension in the red rope is Mg.The tension in the green rope is Mg/2.The tension in the yellow rope is Mg/4.The tension in the blue rope is Mg/8 = mg.Therefore:

m = M/8

Note: the color code of the ropes is not essential for solving the problem, is justan easy way to identify the ropes in the solution. If the copy of the solutions youare handed is black and white, please refer to the file posted on blackboard tosee the color code.

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(d) [4 pts] A tennis ball of mass m is held just above a basketball of mass M . You canassume M >> m throughout the problem. With their centers vertically aligned, they arereleased from rest at the same time, to fall through a distance h = 6m. Note: the sizes ofboth the tennis ball and the basketball are negligible with respect to h.To what height does the tennis ball rebound?Explain your answer in detail. Box your answer.

M

m

h

When the basketball has just hit the floor, it has velocity gt directed upward.The tennis ball is still traveling downward and has a velocity gt. At this moment,they collide.The basketball is much more massive, so the center of mass of the system is,effectively, the basketball. With respect to the center of mass and just prior tothe collision, the tennis ball is falling with velocity 2gt directed downward. Justafter the collision, the velocity of the tennis ball with respect to the center ofmass flips of sign: it is 2gt directed upward. The velocity of the tennis ball in thelab frame is obtained by summing the velocity of the center of mass (unchangedin the collision), i.e. 3gt directed upward.The tennis ball, after colliding with the basketball, has un upward velocity threetimes higher than the downward velocity just before the collisions. So the tennis

ball raises to an height 32 = 9 times higher than the distance it fell from, i.e.54m.

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Problem 2: Golfing

Tiger Woods hits a golf ball with an initial velocity v0 at an angle of 45◦ with respect to theground (neglect air resistance in this problem).

(a) [5 pts] If the ball lands 400 m away, what is v0?Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

Let’s call the range of the ball D = 400 m.

t =D

v0 cos θ

∆y = 0 = v0 sin θt− 1

2gt2 ⇒ t =

2v0 sin θ

g

D

v0 cos θ=

2v0 sin θ

g

v0 =[

gD2 sin θ cos θ

]1/2=

[(9.8 m/s2)(400 m)

2 sin 45◦ cos 45◦

]1/2

= 62.6m/s

(b) [3 pts] Suppose Tiger decides to spend some of his earnings for a trip to the first golftournament on the moon. What is the acceleration of gravity on the moon? (Recall thatMmoon = 7.35× 1022 kg and Rmoon = 1.74× 106 m.)Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

F = GmMmoon

R2moon

= mgmoon

gmoon = GMmoon

R2moon

= (6.67× 10−11 m3kg−1s−2)7.35× 1022 kg

(1.74× 106 m)2= 1.6 m/s2

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(c) [4 pts] If he hits a tee shot with the same initial velocity ~v0 as part (a), how far wouldit travel on the moon?Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

Rearranging the solution to part (a) of this problem:

D =2v2

0 sin θ cos θ

g

Only g is different, so:Dmoon

Dearth

=gearth

gmoon

This implies:

Dmoon = Dearthgearth

gmoon= (400 m)

9.8 m/s2

1.6 m/s2 = 2450m

(d) [4 pts] What would be the maximum height of the ball above the lunar surface?Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

Time it takes to reach max altitude is:

t =1

2

D

v0 cos θ= 27.7 s

Therefore:∆y = v0 sin θt− 1

2gmoont

2 = 612 m

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Problem 3: Airplane Toy

A toy wooden airplane (M = 0.2 kg) ishung to the ceiling by a string and movesin a circular trajectory at constant speed.The distance h with the horizontal is0.25m. The length L of the string is 0.5m.

θ

h

R

L

(a) [6 pts] What is the period for the circular motion of the toy airplane?Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

We call F the tension in the rope and T = 2πR/v the period of the motion. Theconditions for equilibrium are:

Mg = F sin θ

Mv2/R = F cos θ

Taking the ratio of the two equations we get:

tan θ =gR

v2=

h

R⇒ R

v=

√h

g

Therefore:

T = 2πRv

= 2π√

hg

= 2π

√0.25 m

9.8 m/s2= 1.0 s

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(b) [4 pts] What is the velocity of the airplane? If you did not answer to part (a), makean educated guess for the period of the motion.Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

Let’s calculate first the radius of the trajectory:

R =√

L2 − h2 =√

0.52 − 0.252 m = 0.43 m

Therefore the velocity is:

v = 2πRT

= 2π0.43 m

1 s= 2.70m/s

The airplane is hit from behind in the tail by a bullet of mass 10 g. The speed of the bullet,relative to the ground, is 20m/s. The bullet sticks into the airplane.

(c) [6 pts] What is the momentum of the airplane right after the collision? Attention: thebullet stays into the airplane!Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

M is the mass of the toy airplane, v = 3.14m/s its velocity before the collision,m is the mass of the bullet and u its velocity before the collision. We will call wthe final velocity of the two bodies after the inelastic collision.

Mv + mu = (M + m)w

Therefore:

w = Mv+muM+m

=(2.70× 0.2 + 20× 0.01) kg m/s

0.21 kg= 3.52m/s

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Problem 4: Roller Coaster

A roller coaster of mass m = 50 kg travels on the track shown below. At point C the coasterenters a frictionless loop of radius R1=20m. After exiting the loop, the coaster travels thedistance L = 40m from C to D with friction, and then onto a straight frictionless ramp thatleads to a circular arc of radius R2=10m.

x

=60o

A

B c

h

D

R1

R 2

E

F

L

!

(a) [6 pts] Assume no friction for the A-B-C segment of the track, from what height h1 mustthe coaster start to stay barely on the track at point E? (Use this value of h1 in part (b).)Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

When the coaster barely stays on track at E, the normal force N is zero. Thenthe centripetal force is:

mv2

E

R1

= mg => v2E = gR1

Applying the principle of energy conservation:

mg(h1 − 2R1) =1

2mv2

E =1

2mgR1

Which can be simplified into:

h1 − 2R1 =R1

2

Therefore:h1 = 5

2R1 = 50m

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(b) [6 pts] Determine the kinetic friction coefficient µk of segment C–D if the roller coasterstays barely on the track at point F. If you did not complete part (a), assume that h1 = 40m.Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

When the coaster barely stays on track at F, the normal force N is zero. Thenthe centripetal force is:

mv2

F

R2

= mg => v2F = gR2

The work done by friction is:

WNC = Ef − Ei

Therefore:

−µkmgL = mgR2 +1

2mv2

F −mgh1 = mgR2 +1

2mgR2 −mgh1 =

µk = −3R2+2h1

2L=−3× 10 m + 2× 50 m

80 m= 0.88

(c) [4 pts] If the coefficient of kinetic friction is 0.1 from A to B, from what height h2 mustthe coaster start to stay barely on the track at point E.Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

The work done by friction is:

WNC = Ef − Ei

Therefore:

−mgµk cos θh2

sin θ= 2mgR1 +

1

2mv2

E −mgh2 = 2mgR1 +1

2mgR1 −mgh2

Taking the first and last member of the above expression, and eliminating thecommon factors, we obtain:

h2 (1− µk cot θ) =5

2R1

h2 = 5R1

2(1−µk cot θ)=

5(20 m)

2 (1− 0.1 cot 60◦)= 53m

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(d) [2 pts] What must be the new µk of segment C–D if the the coaster stays barely on thetrack at point F?Provide a detailed answer. Box your answer.

Since the coaster has the same mechanical energy at C as in part b, µk stays the same

(µk=0.88) for the coaster to stays barely on the track at point F.

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Problem 5: Platforms on the ice

A man of mass M = 60 kg stands in the center of a platform one of mass m = 40 kg thatrests on ice. Another identical platform floats at a distance L = 1.5m away. The contact ofthe two platforms with the ice is frictionless. The man jumps from the first to the secondplatform, and lands in the center of the second platform.

x

1 2

0

(a) [6 pts] Locate the center of mass of the two-platform and man system before the manjumps. Use the center of mass of platform one as origin.Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

xcm = x1,i(m+M)+x2,im

2m+M=

0× (40 kg + 60 kg) + 1.5 m× 40 kg

2× 40 kg + 60 kg= 0.43m

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(b) [4 pts] What is the velocity vcm of the center of mass of the two-platform and mansystem when the man is midway between the two platforms? Detail your work andprovide both an analytical AND a numerical solutions. Box your answers.

The answer is zero . The two-platform and man system is an isolated system.Since there is no net external force acting on the system, the velocity of the centerof mass is conserved. The initial value of vcm is zero. Therefore vcm stays zero atany time.

(c) [6 pts] How far apart are the two platforms when the man lands?Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

We will use the fact that the center of mass does not move. Even in the finalinstant, xcm = 0.43m.Let’s call with x1,f and x2,f the final positions of the platforms, at the momentthe man lands on the second platform. The position of the second platform hasnot changed yet and therefore x2,f = x2,i = 1.5m. Our unknown is x1,f . Thedistance between the platforms is x2,f − x1,f .

xcm =x1,fm + x2,f (m + M)

2m + M

x1,f =xcm(2m+M)−x2,f (m+M)

m

x1,f =0.43 m× (2× 40 kg + 60 kg)− 1.5 m(40 kg + 60 kg)

40 kg= −2.25 m

x2,f − x1,f = (1.5 + 2.25) m = 3.75m

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Problem 6: Well and Bucket

A farmer raises a water-filled bucket (total mass m =5 kg) from the bottom of a well by turning a hand crack,as the figure illustrates. The bucket is attached to amassless rope that is wound around a solid cylinder withmass M of 10 kg and a diameter d of 20 cm. The masslesshand crank has a turning radius r of 40 cm.

d = 0.2m

h = 5m

r = 0.4m

(a) [2 pts] How many turns of the crank does it take to raise the bucket a height h = 5 m?

Provide a detailed answer. Box your answer.

One turn raises the bucket by:

2π(d/2) = 0.62 m

So it takes:N = 5 m/0.62 m ≈ 8 turns

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(b) [4 pts] What is the magnitude of the torque on the cylinder due to the weight of thebucket?Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

τ = Fd/2 = mgd/2 =(5 kg)(9.8 m/s2)(0.2 m)

2= 4.9N m

(c) [4 pts] How much force does the farmer have to apply to the end of the crank to raisethe bucket at constant speed? (Assume the force is applied perpendicular to the lever armof the crank.)Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

∑τ = 0⇒ τW = τhand = Fr

F = τW

r=

4.9 Nm

0.4 m= 12.3N

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(d) [4 pts] While the bucket is suspended at rest from a height of 5 m, the farmer loses hisgrip on the crank. As the bucket falls, the cylinder rotates and the rope unwinds withoutslipping. What is the angular acceleration of the cylinder while the bucket is falling? (Hint:recall that the moment of inertia of a solid cylinder is I = 1

2MR2, where M is the mass and

R is the radius.)Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

Sum the forces in the y direction (down is positive):

mg − T = ma

Using a = d2α, we have T = m

(g − d

2α). Sum the torques around the axis of the

cylinder:

τ = Td

2= Iα

Substitute for T :

m

(g − d

2α

)d

2=

1

2M

(d

2

)2

α

α = mg

( 12M+m) d

2

=(5 kg)(9.8 m/s2)(

1210 kg + 5 kg

)(0.1 m)2

= 49 rad/s2

(e) [4 pts] What is the ratio KErot/KEbucket of the rotational kinetic energy of the cylinderand the linear kinetic energy of the bucket while the bucket is falling?Detail your work and provide both an analytical AND a numerical solutions.Box your answers.

KErot

KEbucket

=12Iω2

12mv2

=Iω2

m(rω)2=

I

mr2=

12Mr2

mr2= M

2m=

20 kg

2(10 kg)= 1

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POSSIBLY USEFUL CONSTANTS AND EQUATIONSYou may want to tear this out to keep at your side

L = Iω I = Σmir2i x = x0 + v0t + at2/2

PE = mgh KE = 12Iω2 KE = 1

2mv2

ω = ω0 + αt ω2 = ω20 + 2α∆θ ∆θ = ω0t + 1

2αt2

v = v0 + at F∆t = ∆p F = −GMm/r2

F = µN s = Rθ τ = F` sin θΣτ = Iα v = Rω p = mvac = v2/r W = Fs cos θ v2 = v2

0 + 2a∆xWnc = ∆KE + ∆PE a = Rα I = 1

2mr2 [disk]

I = 25mr2 [sphere]

REarth = 6400 km MEarth = 6.0× 1024 kg G = 6.67× 10−11 Nm2/kg2