1 Your Name: PHYSICS 101 MIDTERM October 26, 2006 2 hours Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon Problem Score 1 /13 2 /20 3 /20 4 /15 5 /18 6 /14 Total /100 Instructions: When you are told to begin, check that this examination booklet contains all the numbered pages from 2 through 15. 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 Problem 1: Grab Bag (a) [3 pts] A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater centerof-mass speed? Box your answer. 1. (a) 2. (b) 3. no difference F (a) F (b) The correct answer is answer 3. Because the force acts for the same time interval in both cases, the change in momentum P must be the same in both cases, and thus the center-of-mass velocity must be the same in both cases. (b) [3 pts] An air track cart initially at rest is put in motion when a compressed spring is released and pushes the cart. Earth and the cart constitute an isolated system. The change in the cart s kinetic energy is different in the frame of reference of Earth and in a frame moving at constant speed relative to Earth because in the moving frame: 1. conservation of energy does not apply 2. the amount of energy released by the spring is different 3. the change in the kinetic energy of Earth is different Box your answer. The correct answer is answer 3. Because the system is isolated, conservation of momentum holds in any inertial frame. The amount of energy released by the spring is a measure for the change in its physical state, which must be independent of the reference frame. Answer 3 is the only option that satisfies conservation of energy.
3 (c) [3 pts] A cannonball is dropped off a high tower. Its initial velocity is null. Simultaneously, an identical cannoball is launched directly upward into the air. The velocity of the center of mass of the two cannonballs: 1. is always zero 2. is always directed downward 3. is always directed upward 4. is initially directed upward, then directed downward 5. is initially directed downward, then directed upward Box your answer. The correct answer is answer 4. The velocity of the center of mass is initially directed upward. It gets to zero only when the velocity of the falling cannonball is equal and opposite to the one of the cannonball launched upwards. From that point on, the velocity of the center of mass is directed downward.
4 (d) [4 pts] A baseball of mass m is held just above a small basketball of mass 4m. With their centers vertically aligned, they are released from rest at the same time, to fall for a time t through a distance h = 1 2 gt2. Note: the sizes of both the baseball and the basketball are negligible with respect to h. What is the velocity of the baseball just after it bounces off the small basketball? Quote your answer in units of gt. Assume the collision is elastic! m 4m h=(1/2)gt 2 When the basketball has just hit the floor, it has velocity gt directed upward. The baseball is still traveling downward and has a velocity gt. This is the instant which is immediately preceding the collision. The center of mass velocity is: 4m gt m gt = 3 5m 5 gt In the center of mass frame of reference, the velocity of the basketball is: And the velocity of the baseball is: 2 5 gt 8 5 gt Immeditely after the collision, the two velocities, relative to the center of mass frame of reference, are opposite to the ones just before the collision. In particular, the velocity of the baseball is: 8 5 gt Transferring back to the laboratory frame, the velocity of the baseball becomes: 8 5 gt + 3 5 gt = 11 5 gt
5 Problem 2: Rolling Cylinder A solid and uniform cylinder of mass M=1 kg and radius R=0.1 m is rolling without slipping on the horizontal track shown in the figure. Its total kinetic energy (translational + rotational kinetic energy) while on the horizontal track is 300 J. The track continues in a ramp oriented at an angle θ=30 above the horizontal axis. The height of the ramp is h=10 m. You can neglect friction and air resistance throughout the problem. y x θ h (a) [5 pts] What is the velocity of the cylinder while on the horizontal track? KE i = 1 2 Mv2 + 1 2 Iω2 = 1 2 Mv2 + 1 2 ( 1 2 MR2 ) ( v R ) 2 = 3 4 Mv2 v i = 4 KEi 4 300 J 3 M = 3 1 Kg = 20 m/s (b) [5 pts] What is the total kinetic energy (translational + rotational) at the point where the cylinder leaves the ramp? KE i + PE i = KE f + PE f We fix the origin of the vertical axis on the horizontal track, therefore PE i =0. KE f = KE i PE f = KE f Mgh = 300 J 1 kg 9.8 m/s 2 10 m = 202 J
6 (c) [5 pts] What is the linear velocity of the cylinder at the point where the cylinder leaves the ramp? Give both x and y components! As outlined in the solution to part a, we first calculate the speed: v f = 4 KEf 4 202 J 3 M = 3 1 Kg = 16.4 m s We then calculate the x and y components: v f,x = v f cos 30 = 14.2 m/s v f,y = v f sin 30 = 8.2 m/s (d) [5 pts] What is the maximum height reached by the cylinder during its trajectory after it leaves the ramp? Note: y=0 on the horizontal part of the track! The maximum height above the top of the ramp is: d = v2 f,y 2g = (8.2 m/s)2 2 9.8 m/s 2 = 3.4 m Therefore the maximum height above the track is: d + h = 3.4 m + 10.0 m = 13.4 m
7 Problem 3. A cart of mass M = 1000 kg is to be pulled up a 50-m-long slope in a mine, rising a total height of 30 m in the process. To accomplish this task the miners rig a mass m to the cart using a (massless) cable of length 50 m, which passes over a (massless) pulley, as shown in the figure. (a) [5 pts] How much work is done by the cable on the cart during the lift, assuming that the velocity of the cart is negligible at all times and that friction may be ignored? The work done by the cable equals the change in energy of the cart during the lift. The cart has no kinetic energy by assumption, while its change in potential energy is MgH, where H = 30 m is the change in height of the cart. Hence, the work done is: W = Mgh = 1000 9.8 30 = 2.94 10 5 J. Alternatively, the work done on the cart is given by W = T s, where T is the tension in the cable, and s = 50 m. Because the cart acquires no kinetic energy, the tension T is essentially the same as if the cart were at rest. Then, T balances the component of the gravitational force along the slope, T = Mg sin θ = Mg(3/5). Again, we have: W = T s = Mg(3/5)50 = 1000 9.8 30 = 2.94 10 5 J. (b) [5 pts] How much energy becomes available due to lowering of mass m (which is attached to the other end of the cable)? Express you answer in terms of the as-yet-unknown mass m. Mass m is lowered by height h = 50 m, so the gravitational potential energy liberated by this action is: P.E. = mgh = m 9.9 50 = 490m J.
8 (c) [5 pts] What is the minimum mass m that could lift the cart up the slope? If no energy is lost to friction or kinetic energy, then the potential energy of lowering mass m must equal the potential energy needed to raise mass M. That is, MgH = mgh, or: m = MH/h = 1000 30/50 = 600 kg. Alternatively, we need the weight of mass m to balance the tension T. Then, mg = T = (3/5)Mg, recalling the result of part a, and so: m = (3/5)M = (3/5)1000 = 600 kg. (d) [5 pts] Suppose instead that the miners choose to use m = M. What would be the velocity v of the cart when it reached the top of the slope, assuming that it started from rest at the bottom? Conservation of energy tells us that K.E. = P.E. That is, 2Mv 2 /2 = mgh MgH = Mg(h H), so that v 2 = g(h H), and v = 9.8 (50 30) = 14 m/s. Alternatively, v 2 = 2as, where s = 50 m and a is the acceleration of the cart up the slope. The equation of motion of the cart is Ma = T Mg sin θ = T (3/5)Mg. The tension T is related to the equation of motion of the falling mass m, whose acceleration is also a, but which is directed downwards. That is, ma = mg T. Adding the two equations of motion, we find (M + m)a = [m (3/5)M]g. For the case that m = M = 1000 kg, this becomes 2Ma = [1 (3/5)]Mg = (2/5)Mg, so that a = g/5. Finally, v = 2as = 2 (9.8)/5) 50) = 9.8 20 = 14 m/s.
9 Problem 4. A washing machine has a drum of mass M = 9 kg whose radius is a = 0.3 m. The load may be approximated by a mass m = 1 kg that is concentrated at a single point on the outer edge of the drum during the spin cycle, when the angular velocity of the drum is forced to be ω = 20 rad/s by a motor. In general, such an unbalanced load would cause the drum to vibrate badly during the spin cycle. This problem is mitigated by attaching the (vertical) shaft of the drum to a fixed point via a spring of constant k = 1, 000 N/m. That is, if r is the distance between the shaft and the fixed point, the spring exerts a force F = kr on the shaft, pulling it towards the fixed point. After a few turns of the drum, the system moves into a configuration in which the shaft, the fixed point and the load m all lie along a straight line, which line rotates with constant angular velocity ω about the fixed point. Thus, the shaft moves in a circle of radius r, and the load moves in a circle of radius a r. (a) [5 pts] What is the distance d of the center of mass of the drum plus load from the shaft of the drum? The center of mass of the drum itself is at the shaft, so the center of mass of the drum plus load is described by (M + m)d = ma. Hence, d = m a = 1 0.3 = 0.03 m, M+m 9+1 where d is the distance from the shaft to the CM of the system.
10 (b) [5 pts] What is the distance r between the shaft and the fixed point, when the system is rotating with angular velocity ω as shown in the figure? Note that the centripetal force experienced by mass m produces a reaction force on M; a second force (F = kr) acts on M. Masses m and M are in uniform circular motion with angular velocity ω and distances a r and r from the fixed point. Hence, the force on mass m, which is due to the drum, must be F m = mω 2 (a r). This force is directed from m to the fixed point. The drum experiences two forces, both pulling the shaft towards the fixed point. First, there is the reaction force F m. Second, there is the spring force F s = kr. Together, these forces provide the centripetal force on the drum: Mω 2 r = F drum = mω 2 (a r) + kr. r = mω 2 a (M+m)ω 2 k = 1 202 0.3 (9+1)20 2 1000 = 0.04 m. (c) [5 pts] Where is the center of mass of the drum plus load in the limit of very large angular velocity ω? Referring to the solution of part b, if ω is very large, then (M + m)ω 2 k (M + m)ω 2, and the position of the shaft becomes r am/(m + m) = d. That is, the center of mass of the system is at the fixed point in the lab, and the load is perfectly balanced.
11 Problem 5. Two masses M 1 and M 2 are connected by a system of pulleys as shown in the figure. Mass M 2 is 16 kg. The pulley on the left (near M 1 ) is fixed in position and has mass M p = 1 kg and radius R = 0.10 m. The other pulley (near M 2 ) can move vertically and is massless. The ropes are massless. The pulleys are frictionless. fixed pulley M 1 M 2 (a) [4 pts] What should mass M 1 be so that the system is in equilibrium? In equilibrim, there is no torque on the fixed pulley (as well as the moving pulley which is massless). Therefore, the tension on the string attached to M 1 is the same at all points. Let s call this tension T. In equilibrium, the accelerations are zero, hence: T M 1 g = 0, 2T M 2 g = 0 From this, we obtain: M 1 = (1/2)M 2 = 8 kg.
12 (b) [4 pts] Now assume that M 1 = 10 kg. What is the relation between the accelerations a 1 and a 2 of masses M 1 and M 2? Draw vectors a 1 and a 2 on the figure and give the algebraic relation between a 1 and a 2. Let y 1 and y 2 be the upward vertical displacements of M 1 and M 2, respectively. Since M 1 and M 2 are connected by a string of fixed length, we always have y 1 = 2y 2. Therefore, we obtain: a 1 = 2 a 2. The direction of a 1 is downward vertical because M 1 > (1/2)M 2. (c) [10 pts] Always under the assumption that M 1 = 10 kg, find the accelerations of the masses M 1 and M 2, and the tension of each segment of the rope. Ignoring the inertia of the pulley attached to M 2, T 2 = T 3, but T 1 T 2. M 1 a 1 = T 1 M 1 g M 2 a 2 = 2T 2 M 2 g a 1 = 2a 2 (T 1 T 2 )R = Iα = I a 1 R = 1 2 M pr 2 a 1 R We have 4 unknowns (a 1, a 2, T 1 and T 2 ) with 4 equations. Solving the equations, we obtain: ( ) ( ) 4M1 + 2M 2 a 1 = g = 1.352 m/s 2 2M 1 M 2, a 2 = g = 0.676 m/s 2 4M 1 + M 2 + 2M p 4M 1 + M 2 + 2M p ( ) ( ) 3M1 M 2 + 2M 1 M p 3M1 M 2 + M 2 M p T 1 = g = 84.48 N, T 2 = g = 83.81 N 4M 1 + M 2 + 2M p 4M 1 + M 2 + 2M p
13 Problem 6. A pendulum bob of mass M = 2 kg is suspended by a tight string of length l = 1.0 m one end of which is fixed to the ceiling. A metal ball with mass m = 0.1 kg travelling horizontally with v 0 = 105 m/s collides with the bob. (Throughout this problem, ignore any air resistance and friction. Also ignore the effect of finite sizes of the ball and the bob.) (a) [4 pts] What is the speed of the center of mass (CM) of the ball-bob system before collision? The velocity of the center of mass of a system is v CM = (m i v i ) / m i. Before collision, the motion of this system is by the bullet alone and 1-dimensional. v CM = mv 0 M + m = 5.0 m/s
14 (b) [6 pts] Assuming that the collision is completely inelastic, find the velocity of the bob and the ball immediately after the collision. mv 0 = (M + m)v V 0.1 105 = = 5.0 m/s 2.0 + 0.1 Note that V is the same as v CM. This is no surprise because after collision the system moves in one body. (c) [4 pts] Does the bob acquire enough energy to reach the ceiling? Justify your conclusion quantitatively. To reach the ceiling, the potential energy must increase by (M + m)gl = 2.1 9.8 1.0 = 20.6 (J) at the expense of kinetic energy. The kinetic energy immediately after collision is 1 2 (M + m)v 2 = 26.3 > 20.6. Therefore, YES, it has acquired enough energy to reach the ceiling.
15 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 = ω0 2 + 2α θ θ = ω 0 t + 1 2 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 = v0 2 + 2a x W nc = KE + PE a = Rα I = 1 2 mr2 [disk] I = 2 5 mr2 [sphere] R Earth = 6400 km M Earth = 6.0 10 24 kg G = 6.67 10 11 Nm 2 /kg 2