PHYS 211 – MT3 Fall 2012 Sample 2 1. Danielle and Susan stand on a frictionless cart on one side of the room while Austin stands on a second frictionless cart on the opposite side of the room. Both carts are initially at rest, and a long rope is held between them. The girls pull as hard as they can on the rope, while Austin merely holds on to his end. A short while later the carts collide. Which of the following must be true? A. Austin's cart would have moved a greater distance if he'd been the one pulling on the rope. B. Danielle & Susan's cart would have moved a greater distance if Austin had been the one pulling on the rope. C. The point of collision will be approximately halfway between the initial separation. D. The point of collision depends upon who pulls the rope. E. The point of collision will be approximately the location of center of mass of the system. Treating the two as a system, their center of mass can’t change as there is no external force. 2. A woman lifts a barbell 2.0 m in 5.0 s. If she lifts it the same distance in 10 s, the work done by her is: A. B. C. D. E. one-fourth as great one-half as great the same two times as great four times as great Power changes with time, not work 3. A skydiver of mass 80 kg falls through a distance of 1000 m at a constant terminal velocity of 100 m/s. What is the net work done on the skydiver? (Use g = 10 m/s2) A. B. C. D. E. F. G. -8 MJ -800 kJ -80 kJ 80 kJ 800 kJ 8 MJ None of the above No change in velocity  none in kinetic energy  no net work Page 1 of 8 PHYS 211 – MT3 Fall 2012 Sample 2 4. The graph above shows the force on an arrow due to a bow string as a function of distance as the arrow is released. What is the kinetic energy of the arrow just after it has left the bow? You may ignore friction and assume that the bow is fired horizontally. A. B. C. D. E. 60 J 80 J 100 J 120 J 160 J W   Fdx  400N  40cm  1m  160J 100cm 5. Two boxes, A and B, slide along a frictionless floor with equal magnitude momenta (pA = pB). If the mass of A one fourth the mass of B (mA = ¼mB), how do their kinetic energies compare? A. B. C. D. E. KA = KB KA = 2 KB KA = 4 KB KA = 8 K B KA = 16 KB K A p A2 2mA p A2 mB m  2  2  1 B 4 K B pB 2mB pB mA 4 mB Page 2 of 8 PHYS 211 – MT3 Fall 2012 Sample 2 6. Suppose rain falls straight down into an open bucket sliding along a frictionless, horizontal surface. The rain slowly fills up the moving bucket, and after it is full, the bucket is still moving. The empty bucket has an initial momentum po and an initial kinetic energy Ko. The full bucket (including the water it contains) has a final momentum pf and a final kinetic energy Kf. Which of the following is true? A. B. C. D. E. F. po = pf and Ko < Kf po < pf and Ko = Kf po < pf and Ko < Kf po = pf and Ko = Kf po = pf and Ko > Kf None of the above No external force  no change in momentum (it will slow down as it has to accelerate the rain) KE  p2 will shrink, as m increases but p doesn't change 2m 7. The following 5 “matching” questions are all based on the following set-up and are each worth two points (so these 5 questions together are worth 2 of the other multiple choice questions, 10 points) Below are 5 types of problems that you have been asked to solve. Note that the problems aren’t well enough defined to actually get a solution. However, you should be able to determine which of the below approaches you would use if you were given the necessary information. For each problem pick one: A. B. C. D. E. Only conservation of momentum Only conservation of mechanical energy Work-Energy Theorem Conservation of both momentum and energy None of the above can be used a) Playing billiards, a cue ball hits another ball off center (elastically), and they both go off at given angles. How fast were they moving? D (elastic) b) A bullet is fired at a wood block, hanging from a string. It embeds in the block. To what angle does the block swing upward. D (momentum 1st, then energy) c) A bullet is fired at a wood block, hanging from a string. It passes through the block, and the block swings upward a known amount. How fast does the bullet exit the block? D (momentum 1st, then energy) d) You push a block with a given constant force across a rough table. How fast is it going when it gets to the other side of the table? C (work is change in KE) e) A ball slides along horizontal surface and collides with a second identical ball. Given the final speed of the first ball, what is the final speed of the second ball? A (have to use only momentum , aren’t told it is elastic) Page 3 of 8 PHYS 211 – MT3 Fall 2012 Sample 2 8. A potential energy curve is shown below. If a 4 kg object has a speed of 2 m/s at x = 12 m, what is the largest x value it can reach? A. B. C. D. E. F. 8m 38 m 40 m 42 m 44 m 46 m At x = 12 m it has a potential energy of 28 J. It has a kinetic energy K  12 mv 2  1 2  4kg   2 ms  2  8J  E mech  36J This corresponds to the red horizontal line on the plot, which intersects the potential energy curve at x = 44 m Page 4 of 8 PHYS 211 – MT3 Fall 2012 Sample 2 9. The figure below shows the force exerted on a particle as a function of time. What is the impulse? A. B. C. D. E. 20 Ns 100 Ns 200 Ns 300 Ns 400 Ns J   F dt  1 2  20N  35s-5s   300Ns 10. About how long will it take a 2000 W motor to lift a 300 kg piano from the ground to a window 20 m above the ground? A. B. C. D. E. 10 s m W W mg y  300kg  10 s2   20m  P  t     30s 30 s t P P 2000W   100 s 133 s None of the above Page 5 of 8 PHYS 211 – MT3 Fall 2012 Sample 2 11. Each of the following true-false questions is worth 2 points (so these 5 questions together are worth 2 of the multiple choice questions, 10 points) A. T or F: If the dot product of two nonzero vectors is zero, the vectors must be perpendicular to each other. B. T or F: If two nonzero vectors point in the same direction, their dot product must be zero. It will be their length squared C. T or F: The value of the dot product of two vectors depends on the particular coordinate system being used. ABcos no matter the coordinates D. T or F: A  B  B  A Dot products are based on multiplication, so commutative E. T or F: If a force always acts perpendicular to an object's direction of motion, that force cannot change the object's kinetic energy. The work is zero so it can’t change the kinetic energy 12. A 2,000. kg car sits on frictionless, horizontal road surface when it is struck from behind by a 4,000. kg truck. If the two vehicles' bumpers lock so they stick together and travel at 2.0 m/s immediately after the collision, how fast was the truck traveling when it hit the car? Conservation of momentum in this inelastic collision: A. 2.0 m/s B. 6 m/s p0  mt v0  p f   mc  mt  v f  v0  C. 3.0 m/s D. 6.0 m/s E. None of the above  mc  mt  v mt f   6000kg  2 m  3 m s  4000kg  s 13. A ball of mass m is thrown with speed vo at a block of mass M, which sits on a frictionless surface. In which situation is the average force on the block from the ball the greatest? A. The ball sticks to the block, the collision lasts a time 4T. B. The ball sticks to the block, the collision lasts a time T. C. The ball bounces back off the block with about the same speed it originally had, the collision lasts a time 4T. D. The ball bounces back off the block with about the same speed it originally had, the collision lasts a time T. Bouncing delivers more impulse (more p). Shorter time means larger average force (J=Ft) E. The ball goes through the block (e.g., like a bullet going through a block of styrofoam), the collision lasts a time T. Page 6 of 8 PHYS 211 – MT3 Fall 2012 Sample 2 14. A shell, at rest, explodes into two fragments, one fragment 25 times heavier than the other. If any gas from the explosion has negligible mass, then A. the momentum change of the lighter fragment is 25 times as great as the momentum change of the heavier fragment. B. the momentum change of the heavier fragment is 25 times as great as the momentum change of the lighter fragment. C. the momentum change of the lighter fragment is exactly the same as the momentum change of the heavier fragment. (NO, p25m  pm ) D. the kinetic energy change of the heavier fragment is 25 times as great as the kinetic energy change of the lighter fragment. E. the kinetic energy change of the lighter fragment is 25 times as great as the kinetic energy change of the heavier fragment. p 2 2m , same p, small m  big K F. More than one of the above 15. A man sits in the back of a small boat in still water. He then moves forward to the front of the boat and sits there. Afterwards the boat: A. B. C. D. E. is forward of its original position and moving forward. is forward of its original position and moving backward. is forward of its original position and not moving. is behind its original position and moving backward. is behind its original position and not moving. No external F, xcm stays same 16. A 1/2 kg block is sliding down a frictionless incline which makes an angle of  = 30.0° with the horizontal. It then slides on a frictionless horizontal surface until it slides into a relaxed horizontal spring with a spring constant k = 250. N/m. The block becomes attached to the spring and continues moving, compressing the spring 1/2 m before momentarily stopping. What is the speed of the block just before it hits the spring? (Use g = 10 m/s2) Kinetic energy before hit transforms to spring potential energy after hit A. 5 m/s B. 10 m/s 1 2 mv 2  12 k  s   v  2 k 250 N s  1 m 12 m  m 2 kg 1 m 2 s 5  100  5 5 ms C. 5 5 m/s D. 10 5 m/s E. 115 m/s Page 7 of 8 PHYS 211 – MT3 Fall 2012 Sample 2 17. Two carts on a frictionless air track, each of mass 0.5 kg, are held together at rest with a spring compressed between them. The spring has a spring constant k = 2500 N/m and is compressed 0.1 m from its relaxed length. The carts are released simultaneously. Find the speed of one cart with respect to the air track. Same magnitude momentum and same mass  same velocity, same K A. 5 m/s 2 K f  2 12 mv 2  12 k  s  B. 5 2 m/s C. 10 5 m/s v D. 10 10 m/s E. None of the above 2 k 2500 mN s  0.1m  50 1s  0.1m  5 ms 2m 2  12 kg 18. Sarah Jane Smith rides a frictionless rollercoaster with an initial hill of height H (on top of which the car is barely moving). At the bottom, where the track has a radius of curvature R, the car has a speed V and her apparent weight is 25% heavier than her actual weight FBD Acc aC If the hill were four times as tall (i.e., 4H), the magnitude of the normal force acting on her at the bottom would be, in terms of the force of gravity acting on her Fg: A. B. C. D. E. 1.5Fg 2Fg 2.5Fg 4Fg 5Fg “Apparent Weight” = Normal Force! Draw FBD, Acc diagrams (above) to see that N  FG  macentripetal  R1 mv 2  2 1 R 2 mv 2  R2 K If we start out 4 times higher, we will have 4 times the KE at the bottom and hence require 4 times the net force. So instead of N  FG  0.25FG we will have N  FG  FG  N  2FG Page 8 of 8