Final EXAM. Physics 203. Prof. Daniel A. Martens Yaverbaum. John Jay College of Criminal Justice, the CUNY. Sections 1, 2. Wednesday, 5/21/14

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1 Final EXAM Physics 203 Prof. Daniel A. Martens Yaverbaum John Jay College of Criminal Justice, the CUNY Sections 1, 2 Wednesday, 5/21/14 (Revised, Clarified, 5/28/14) Name: Section # : SCORE:

2 DIRECTIONS 1) Many of the questions in this exam are written in a manner and sequence designed to show you a physics thought process: 2) It s not cheating to find yourself feeling that you re just about to understand something out for the first time right in the middle of a test. 3) It s not cheating to find an answer by reading between the lines of a question as long as that question is on your copy of the exam. 4) It is cheating, however, to cheat. JUST DO NOT DO IT. 5) The questions are NOT intended to be tricks: When in doubt, trust your instincts. 6) ANSWER everything even if you make up an answer for some Part (A) so that you have something to use in a Part (B).. 7) When in doubt, write down things that are true and hope that they are relevant rather than the other way around. MAY THE NET FORCE BE WITH YOU: 2

3 g 10 m s 2 3

4 I. A Ballistic Pendulum (25 pts). *** FEEL FREE TO RIP OUT THIS PAGE! ** This experiment begins at the instant a bullet, mass m, is shot with a purely horizontal velocity. It heads straight for a block, 4m, that is precisely 4 times heavier than the bullet. This block hangs vertically from a long, light string, length L. The block is stable; it rests in a place called equilibrium. The bullet hits the block. They experience a perfectly inelastic collision so that the bullet permanently embeds itself in the block. Together, the bullet/block combination swings up in an arc just like a pendulum. Somewhere above its starting point, this bullet/block-bob runs out of speed and begins heading back down. Its journey is almost immediately interrupted by a little ramp ratcheted so as only to allow upward motion. It is caught. At the instant the experiment ends, the bullet/block-bob is sitting fixed at a final and measurable angle. This final angle is measured from the vertical; 0 refers to the equilibrium position. A Bigger Angle A Greater Height above equilibrium. Why worry about both? In short: Heights are easier to use (in calculations), but more difficult to measure. In short: Angles are easier to measure, but more difficult to use. The mass of the bullet = m = 2 kg. The mass of the block = 4m = 8 kg. The length of the string = L =.500 m (500 cm). The maximum angle = θ (max) = 15. The maximum height = h =??? 4

5 a. Draw a PURE free-body diagram of ONLY THE BLOCK before the bullet collides with it (3 pts). NOTE: This diagram should depict the forces acting on the block during an early instant while the block hangs vertically at its equilibrium position. b. Choose the ONE description regarding WORK that seems most likely to be FALSE (3 pts): (a) Planet Earth did some work on a block essentially means Planet Earth pulled a block through some amount of space. (b) If Planet Earth did some negative work on a block, then the block transferred some of its kinetic energy to Planet Earth. (c) Assume that a block is definitely moving (undergoing displacement) from one location to another; then any force found in the free-body diagram of that block is definitely doing work on the block. (d) Assume that a block is definitely moving (undergoing displacement) from one height to another; then Planet Earth is necessarily doing work on that block. Just CHOOSE ONE and enter the LETTER of choice below the question: *** ANSWER: 5

6 c. Choose the ONE description regarding the force of GRAVITY that seems most likely to be FALSE (3 pts). NOTE: In this exam, no meaningful distinction is intended between the terms Planet Earth and Gravity; in this context, they are being used interchangeably. (a) As the b/b-bob, mass 5m, swings up through a height h, the b/b-bob gains 5mgh Joules of kinetic energy. (b) As the bullet/block-bob, mass 5m, swings up through a height h, the work done on the bob by the force of gravity is -5mgh. (c) When the b/b-b eventually swings down through the same height h, the work done on the bob by the force of gravity is +5mgh. (d) If a block has been displaced from equilibrium by some amount, then the potential energy stored by the block is precisely equal to the amount of work that a conservative force must do so as to return the block to equilibrium. *** ANSWER: d. At the instant after the bullet hits the block, in which direction does the bullet/block combination begin to move: That is, what is the direction of the bullet/block-bob s initial velocity (2 pts)? (a) LEFT (b) RIGHT (c) UP (d) DOWN (e) Diagonal:UP-LEFT (g) Diagonal: UP-RIGHT (f) Diagonal: Diagonal: DOWN-LEFT (h) Diagonal: DOWN-RIGHT *** ANSWER: 6

7 e. At the instant considered in (d), above, how is the bullet/block-bob s direction of motion related to the direction in which the (string) tension pulls (3 pts)? (a) The tension force acting on the bullet/block-bob is directed parallel to the b/b-bob s motion; the two vectors form a 0 angle, (b) The tension force is directed opposite ( anti-parallel ) to the motion; the two vectors form a 180 angle, (c) The tension force is perpendicular to the bob s motion; the two vectors form a 90 angle, (d) The force and the motion form a 45 angle with each other. *** ANSWER: f. Now consider some arbitrary (i.e.: not special) instant during the bullet/block-bob s pendulum swing. Draw a PURE free-body diagram at this arbitrary point during the swing (4 pts). Note again: You are depicting a moment at which the bullet/block-bob is higher than equilibrium but lower than its maximum. 7

8 g. Considering your responses above (particularly those to (e) and (f)), choose the ONE statement about the string tension that is necessarily TRUE (3 pts): (a) The string tension is a conservative force that does work on the bullet/block-bob, (b) The string tension is a conservative force that does not do work on the bullet/blockbob, (c) The string tension is a non-conservative force that does work on the bullet/block-bob, (d) The string tension is a non-conservative force that does not do work on the bullet/blockbob. *** ANSWER: h. Considering the above, choose the ONE statement about the swing of this pendulum that is necessarily TRUE (3 pts): (a) Linear momentum is conserved, but total mechanical energy is not conserved. (b) Total mechanical energy is conserved, but linear momentum is not conserved. (c) Both linear momentum and total mechanical energy are conserved, but only one of the two quantities can be computed for all points in the swing. (d) Neither linear momentum nor total mechanical energy is conserved, but both of the quantities can be measured at every point in the swing. *** ANSWER: 8

9 i. Let v(bottom) stand for the initial speed of the bullet/block-bob (the speed at the very first instant of pendulum swing) relative to the lab. Let h stand for the maximum height to which the bullet/block-bob swings. --Note: we have not measured h; therefore, we do not yet know h. We do know θ (max), the maximum angle to which the b/b-b swings. Starting with one of the principles that you considered in (f), above, show how the initial speed of the bullet/block-bob can always be determined IF we know the maximum height to which the bullet/block-bob swings. That is: Show how to: SHOW ALL WORK: find v (bottom) as a function of h. In this case: Your expression may also involve m and/or g (Some of the above constants might not end up in your final answer, but no other constants or variables are permitted.) What is v (bottom) as a function of h (and/or m, g) (5 pts)? 9

10 j. Now use the diagram below and SHOW ALL WORK in order to: find h as a function of θ (max) (5 pts). Your answer will be expressed in terms of m, g, L and θ (max). Again, all constants are permitted but not all necessary. 10

11 k. Substitute your result from (j), above, into your result from (i), above, so as to express: v (bottom) as a function of θ (max) (5 pts). l. Imagine you have some pendulum swinging back and forth. Before letting it swing, you measure the length of the pendulum string. You get some number. After it s been swinging for some time, you use a photo-gate to measure the instantaneous speed of the pendulum each time it passes through the equilibrium position at the bottom. You get some number. After a while, you re confident of your measurements so you take the pendulum string and QUADRUPLE its LENGTH, while holding all other values constant. According to your result in (k), above, (a) The instantaneous speed at the bottom should now be two times larger than before, (b) The instantaneous speed at the bottom should now be four times larger than before, (c) The instantaneous speed at the bottom should now be eight times smaller than before, (e) The instantaneous speed at the bottom should now be precisely the same as before, *** ANSWER: (2 pts). 11

12 m. Note the specific numerical measurements provided in the table at the beginning of this problem. We now seek a specific numerical result. Apply your general functions to these particular measurements: In meters per second, rounded to three significant digits, relative to the lab, WHAT IS v (bottom)? That is, compute the instantaneous speed at which this bullet/block-bob BEGINS to swing as a pendulum (5 pts.). In short, Plug In! *** NOW BEGIN CONSIDERING THE COLLISION: n. In approximately one complete sentence of English, what does it mean to state something like A bullet exerted an impulse on a block (3 pts)? 12

13 o. Choose the ONE description regarding the COLLISION between bullet and block which seems most likely to be FALSE (3 pts): (a) The impulse exerted by the bullet on the block is of the same magnitude but opposite direction as the impulse exerted by the block on the bullet. (b) The linear momentum gained by the block is precisely the same amount as the linear momentum lost by the bullet. (c) Although the bullet and the block each conserve linear momentum individually, the total amount of linear momentum for the pair decreases the instant after they start sharing one velocity. (d) When analyzing a collision between two objects, comparing two distinct points in time tends to be a more successful problem-solving strategy than comparing two distinct points in space. *** ANSWER: p. Compute another NUMERICAL answer: In kg - meters per second, rounded to three significant digits, relative to the lab, Find the total linear momentum (i.e. mass x velocity) for THE PAIR of objects AFTER they collide (3 pts). q. Again, numerically: Find the total linear momentum for the pair of objects BEFORE they collide (5 pts). Hint: Is Linear Momentum Conserved during a collision?! If so, what does that mean? 13

14 r. RELATIVE TO THE LAB (in the reference frame of the lab, as measured by a researcher standing still in the lab etc.): Find vcm(after): the velocity of the pair s center of mass AFTER they collide (3 pts). HINT: Think of vcm as a weighted average of velocities. OR: Think about the fact that everything is moving at one and only one velocity AFTER the collision... DON T FORGET: Velocity means Speed AND Direction! s. RELATIVE TO THE LAB: Find vcm(before): the velocity of the pair s center of mass BEFORE they collide (2 pts). HINT: Total linear momentum is (total) mass x velocity. Velocity of the center of mass is the one velocity that best represents the whole like an average. What happens to total linear momentum from the instant before to the instant after a collision? t. RELATIVE TO THE CENTER OF MASS (in the center of mass reference frame): What is the velocity (speed and direction) of the lab (5 pts)? 14

15 u. In meters per second, rounded to three significant digits, relative to the lab: *** WHAT IS v (bullet)? *** That is, compute a NUMERICAL value for the instantaneous speed at which the bullet is SHOT right at the beginning of the experiment. That s the point of this whole madcap adventure, no (5 pts.)? v. Look back at everything you did and express v (bullet) as as a GENERAL FUNCTION OF: m, g, L and θ (max)! SHOW ALL WORK (7 pts)! 15

16 w. Now imagine this exact same experiment no numbers changed at all with just ONE difference: After approaching the block with whatever speed you found in (v), above, The bullet collided PERFECTLY ELASTICALLY with the BLOCK: (Sure: The block (4m) swung up by itself --but you have no idea what maximum angle it attained. And, honestly, I don t think you care.) Given all same numbers but a PERFECTLY ELASTIC COLLISION, In meters per second, rounded to three significant digits, relative to the lab: Find the final velocities For BOTH the individual BULLET AND the individual BLOCK (8 pts)!(?) Just do anything and everything you can to solve this! YES: We all believe in partial credit. 16

17 x. Now imagine a bullet/block combination of 10kg is traveling West through the lab at 5 m/s. A horizontal spring is relaxed and waiting at equilibrium. The spring is described by a K (spring constant) of 200 Newtons/Meter. The bullet/block combo smashes into the spring and keeps moving until it compresses the spring a maximum distance (and everything stops for an instant). NUMERICALLY: What is this maximum distance of spring compression (5 pts)? y. The following four statements compare a moment of maximum compression for this mass/spring system to a moment of maximum height for the pendulum system (discussed in part [i]). Choose the ONE comparative statement that is most probably TRUE (3 pts): (e) The mass/spring system, at its moment of maximum spring compression, has the same amount of total mechanical energy as the pendulum (discussed in part (i)), at its moment of maximum height; reason: in both cases, the system is sitting still. (f) The mass/spring system, at its moment of maximum spring compression, has a greater amount of total mechanical energy than the pendulum (discussed in part (i)), at its moment of maximum height; reason: compare the numbers. (g) The mass/spring system, at its moment of maximum spring compression, has a smaller amount of total mechanical energy than the pendulum (discussed in part (i)), at its moment of maximum height; reason: the gravitational force exerted by a planet is stronger than the elastic force exerted by a bullet-sized spring. (h) The amount of energy in a spring cannot be compared to the amount of energy in a pendulum; it s like comparing a velocity to an acceleration. *** ANSWER: Z. Zee you Zome Zummer! (Take + 5 pts for good and certain measure). 17