Your Name: PHYSICS 101 MIDTERM. Please Circle your section 1 9 am Galbiati 2 10 am Wang 3 11 am Hasan 4 12:30 am Hasan 5 12:30 pm Olsen

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1 1 Your Name: PHYSICS 101 MIDTERM October 27, hours Please Circle your section 1 9 am Galbiati 2 10 am Wang 3 11 am Hasan 4 12:30 am Hasan 5 12:30 pm Olsen Problem Score 1 /16 2 /16 3 /16 4 /18 5 /16 6 /18 Total /100 Instructions: When you are told to begin, check that this examination booklet contains 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 depends on your solution even when you write down the correct answer. BOX your final answer. Do not panic or be discouraged if you cannot do every problem; there are both easy and hard parts in this exam. If a part of a problem depends on a previous answer you have not obtained, assume it and proceed. Keep moving and finish as much as you can! Possibly useful constants and equations are on the last page, which you may want to tear off and keep handy Rewrite and sign the pledge: I pledge my honor that I have not violated the Honor Code during this examination. Signature

2 2 Problem 1: Grab Bag (a) [4 pts] A bowling ball and golf ball have the same initial momentum. The same constant force F is applied to slow down both balls. Which ball stops faster? Box your answer and detail your reasoning in writing on the side. 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. and therefore F t p t p/f Since F and p are the same for the two balls, also t is the same and they stop over the same time. (b) [4 pts] A bowling ball and golf ball have the same initial momentum. The same constant force F is applied to slow down both balls. Which ball stops in the shortest distance? Box your answer and detail your reasoning in 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 F s KE f KE i p 2 i /(2m) Therefore: s p 2 i /(2mF ) and the distance is shorter for the ball with the larger mass.

3 3 (c) [4 pts] Massless strings are guided through four massless and frictionless pulleys as shown. 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 just an easy way to identify the ropes in the solution. If the copy of the solutions you are handed is black and white, please refer to the file posted on blackboard to see the color code.

4 4 (d) [4 pts] A tennis ball of mass m is held just above a basketball of mass M. You can assume M >> m throughout the problem. With their centers vertically aligned, they are released from rest at the same time, to fall through a distance h 6 m. Note: the sizes of both 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 to the collision, the tennis ball is falling with velocity 2gt directed downward. Just after the collision, the velocity of the tennis ball with respect to the center of mass flips of sign: it is 2gt directed upward. The velocity of the tennis ball in the lab frame is obtained by summing the velocity of the center of mass (unchanged in the collision), i.e. 3gt directed upward. The tennis ball, after colliding with the basketball, has un upward velocity three times higher than the downward velocity just before the collisions. So the tennis ball raises to an height times higher than the distance it fell from, i.e. 54 m.

5 5 Problem 2: Golfing Tiger Woods hits a golf ball with an initial velocity v 0 at an angle of 45 with respect to the ground (neglect air resistance in this problem). (a) [5 pts] If the ball lands 400 m away, what is v 0? Let s call the range of the ball D 400 m. v 0 [ t D v 0 cos θ y 0 v 0 sin θt 1 2 gt2 t 2v 0 sin θ g gd 2 sin θ cos θ D v 0 cos θ 2v 0 sin θ g [ ] 1/2 (9.8 m/s 2 ] 1/2 )(400 m) 62.6 m/s 2 sin 45 cos 45 (b) [3 pts] Suppose Tiger decides to spend some of his earnings for a trip to the first golf tournament on the moon. What is the acceleration of gravity on the moon? (Recall that M moon kg and R moon m.) g moon G Mmoon R 2 moon F G mm moon R 2 moon mg moon ( m 3 kg 1 s 2 ) kg 1.6 ( m/s2 m) 2

6 6 (c) [4 pts] If he hits a tee shot with the same initial velocity v 0 as part (a), how far would it travel on the moon? Rearranging the solution to part (a) of this problem: D 2v2 0 sin θ cos θ g Only g is different, so: This implies: D moon D earth g earth g moon D moon D earth g earth g moon 9.8 m/s2 (400 m) 1.6 m/s m (d) [4 pts] What would be the maximum height of the ball above the lunar surface? Time it takes to reach max altitude is: t 1 D 2 v 0 cos θ 27.7 s Therefore: y v 0 sin θt 1 2 g moont m

7 7 Problem 3: Airplane Toy A toy wooden airplane (M 0.2 kg) is hung to the ceiling by a string and moves in a circular trajectory at constant speed. The distance h with the horizontal is 0.25 m. The length L of the string is 0.5 m. L R θ h (a) [6 pts] What is the period for the circular motion of the toy airplane? We call F the tension in the rope and T 2πR/v the period of the motion. The conditions for equilibrium are: Mg F sin θ Mv 2 /R F cos θ Taking the ratio of the two equations we get: tan θ gr v h 2 R R h v g Therefore: T 2πR v 2π h g 0.25 m 2π 9.8 m/s 1.0 s 2

8 8 (b) [4 pts] What is the velocity of the airplane? If you did not answer to part (a), make an educated guess for the period of the motion. Let s calculate first the radius of the trajectory: Therefore the velocity is: R L 2 h m 0.43 m v 2πR T 2π 0.43 m 1 s 2.70 m/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 20 m/s. The bullet sticks into the airplane. (c) [6 pts] What is the momentum of the airplane right after the collision? Attention: the bullet stays into the airplane! M is the mass of the toy airplane, v 3.14 m/s its velocity before the collision, m is the mass of the bullet and u its velocity before the collision. We will call w the final velocity of the two bodies after the inelastic collision. Therefore: Mv + mu (M + m)w w Mv+mu M+m ( ) kg m/s 0.21 kg 3.52 m/s

9 9 Problem 4: Roller Coaster A roller coaster of mass m 50 kg travels on the track shown below. At point C the coaster enters a frictionless loop of radius R 1 20 m. After exiting the loop, the coaster travels the distance L 40 m from C to D with friction, and then onto a straight frictionless ramp that leads to a circular arc of radius R 2 10 m. A h E F θ60 o B R 1 c L D R 2 x (a) [6 pts] Assume no friction for the A-B-C segment of the track, from what height h 1 must the coaster start to stay barely on the track at point E? (Use this value of h 1 in part (b).) When the coaster barely stays on track at E, the normal force N is zero. Then the centripetal force is: m v2 E R 1 mg > v 2 E gr 1 Applying the principle of energy conservation: Which can be simplified into: mg(h 1 2R 1 ) 1 2 mv2 E 1 2 mgr 1 Therefore: h 1 2R 1 R 1 2 h R 1 50 m

10 10 (b) [6 pts] Determine the kinetic friction coefficient µ k of segment C D if the roller coaster stays barely on the track at point F. If you did not complete part (a), assume that h 1 40 m. When the coaster barely stays on track at F, the normal force N is zero. Then the centripetal force is: The work done by friction is: m v2 F R 2 mg > v 2 F gr 2 Therefore: W NC E f E i µ k mgl mgr mv2 F mgh 1 mgr mgr 2 mgh 1 µ k 3R 2+2h 1 2L 3 10 m m 80 m 0.88 (c) [4 pts] If the coefficient of kinetic friction is 0.1 from A to B, from what height h 2 must the coaster start to stay barely on the track at point E. The work done by friction is: Therefore: W NC E f E i mgµ k cos θ h 2 sin θ 2mgR mv2 E mgh 2 2mgR mgr 1 mgh 2 Taking the first and last member of the above expression, and eliminating the common factors, we obtain: h 2 5R 1 2(1 µ k cot θ) h 2 (1 µ k cot θ) 5 2 R 1 5(20 m) 2 (1 0.1 cot 60 ) 53 m

11 11 (d) [2 pts] What must be the new µ k of segment C D if the the coaster stays barely on the track at point F? Provide a detailed answer. Box your answer. Since the coaster has the same mechanical energy at C as in part b, µ k (µ k 0.88) for the coaster to stays barely on the track at point F. stays the same

12 12 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 that rests on ice. Another identical platform floats at a distance L 1.5 m away. The contact of the two platforms with the ice is frictionless. The man jumps from the first to the second platform, and lands in the center of the second platform x (a) [6 pts] Locate the center of mass of the two-platform and man system before the man jumps. Use the center of mass of platform one as origin. x cm x 1,i(m+M)+x 2,i m 2m+M 0 (40 kg + 60 kg) m 40 kg 2 40 kg + 60 kg 0.43 m

13 13 (b) [4 pts] What is the velocity v cm of the center of mass of the two-platform and man system when the man is midway between the two platforms? Detail your work and provide both an analytical AND a numerical solutions. 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 center of mass is conserved. The initial value of v cm is zero. Therefore v cm stays zero at any time. (c) [6 pts] How far apart are the two platforms when the man lands? We will use the fact that the center of mass does not move. Even in the final instant, x cm 0.43 m. Let s call with x 1,f and x 2,f the final positions of the platforms, at the moment the man lands on the second platform. The position of the second platform has not changed yet and therefore x 2,f x 2,i 1.5 m. Our unknown is x 1,f. The distance between the platforms is x 2,f x 1,f. x cm x 1,fm + x 2,f (m + M) 2m + M x 1,f x 1,f xcm(2m+m) x 2,f (m+m) m 0.43 m (2 40 kg + 60 kg) 1.5 m(40 kg + 60 kg) 40 kg x 2,f x 1,f ( ) m 3.75 m 2.25 m

14 14 Problem 6: Well and Bucket d 0.2m r 0.4m 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 a massless rope that is wound around a solid cylinder with mass M of 10 kg and a diameter d of 20 cm. The massless hand crank has a turning radius r of 40 cm. h 5m (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

15 15 (b) [4 pts] What is the magnitude of the torque on the cylinder due to the weight of the bucket? τ F d/2 mgd/2 (5 kg)(9.8 m/s2 )(0.2 m) N m (c) [4 pts] How much force does the farmer have to apply to the end of the crank to raise the bucket at constant speed? (Assume the force is applied perpendicular to the lever arm of the crank.) τ 0 τw τ hand F r F τ W r 4.9 Nm 0.4 m 12.3 N

16 16 (d) [4 pts] While the bucket is suspended at rest from a height of 5 m, the farmer loses his grip on the crank. As the bucket falls, the cylinder rotates and the rope unwinds without slipping. 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 2 MR2, where M is the mass and R is the radius.) Sum the forces in the y direction (down is positive): mg T ma Using a d 2 α, we have T m ( g d 2 α). Sum the torques around the axis of the cylinder: τ T d 2 Iα Substitute for T : m ( g d 2 α ) d M ( d 2 ) 2 α α mg ( 1 2 M+m) d 2 (5 kg)(9.8 m/s 2 ) ( 1 10 kg rad/s2 2 kg) (0.1 m) 2 (e) [4 pts] What is the ratio KE rot /KE bucket of the rotational kinetic energy of the cylinder and the linear kinetic energy of the bucket while the bucket is falling? KE rot KE bucket 1 2 Iω2 m(rω) 2 Iω2 1 mv2 2 I 1 mr 2 Mr2 M 2 mr 2 2m 20 kg 2(10 kg) 1

17 17 POSSIBLY USEFUL CONSTANTS AND EQUATIONS You may want to tear this out to keep at your side L Iω I Σm i ri 2 x x 0 + v 0 t + at 2 /2 PE mgh KE 1 2 Iω2 KE 1 2 mv2 ω ω 0 + αt ω 2 ω α θ θ ω 0 t αt2 v v 0 + at F t p F GMm/r 2 F µn s Rθ τ F l sin θ Στ Iα v Rω p mv a c v 2 /r W F s cos θ v 2 v a x W nc KE + PE a Rα I 1 2 mr2 [disk] I 2 5 mr2 [sphere] R Earth 6400 km M Earth kg G Nm 2 /kg 2