Free Response Question#1 : Some of these components of this question were marked wrong or marked half-wrong on a test. I need the correct answers and explanations for test corrections.

( Green color ink - Teacher's grading and comments)

me_ Hadley Fralin Period 4 4 AP PHYSICS-1 (2021-2022) Date 2- 9- 22 Time Limit: 40 Minutes TEST 4: Energy- FREE RESPONSE Calculator Formula & Sheet NOT Allowed On my honor, I have neither given nor received any aid on this exam. fladay fualin 1. time 20 minutes) This question is a long free-response question. Show your work for each part of the question. (15 points, suggested 62-5 PET 11250 P= work izl KEV 125 OS K= 2 muz FN D TYPE = KE i PE = KE 10 m PE=Ugh ( Inputted ) PET KET1 Iramp F =M. a -x--1 PEN Work = FX Ax leads to PE smis ( imputted ) F= = H. N A box is pressed against a horizontal spring, compressing the spring from its relaxed length. The box is then KE released, and the spring launches the box horizontally along a track that ends in a ramp, as shown above. The box has enough speed to leave the ramp, and the box reaches a maximum vertical height above the floor. Assume there s negligible friction between the box and the track and air resistance is negligible. k is the spring stiffness constant KE = /2 (5 ) (5 ) 2 1 / 12 ) (5 ) 2 PE = 5 ( 10) (10) PE= 2(10)(10) X is the distance the spring is compressed s ( 100) = 200 M is the mass of the box 12 ( 5) ( 25 ) 1/2 50 12125 0 is the angle of the incline PE = SOO -62.5 = 25 h is the maximum height reached by the box M is increasing so the spring provides asmaller acceleration, by each box so the box does not aqil que as fast The scenario is repeated using a different box with a greater mass. The spring is compressed the same distance X. (a) Indicate how h in this second scenario compares to h in the original scenario, and explain why without mathematically deriving a relation for h. PET with more mass - since the blockhas a greater mass than the prior block, KET with more mass the height the block is going to gain,due to will be shorter than the the height reached the prior block. This is because there is a greater (b) Students derive an equation for h in the original scenario, h = (sin?) : gravitational force on the 2Mg , which may or may not be correct. Is block ? 0/ 2 this equation consistent with your claim in part (a)? Explain why or why not. P= work K= = 1/muz ( sin ?0 ) Kx 2 PE = wgh ZMg Scantron M-882-E-5:41 work = Fxly Two students are discussing the work done on the system consisting of only the box between the instant just before the spring begins accelerating the box and the instant the box reaches the end of the ramp. The students make the following statements. KEXPE Student 1: At first, the mechanical energy of the system increases because the spring does positive work on the system. Then, the mechanical energy of the system increases again as the box travels up the ramp because the box's potential energy increases.W = FX Ax GOOD Assume the spring is ideal, the plunger has negligible mass, the block loses contact with the plunger immediately after it reaches the end of the table, and there is non-negligible friction between the tabletops and the blocks. a) Without manipulating equations, explain why the block launched from table 2 could have a greater speed when it leaves the table than the block launched from table 1 does. $/3 - Table 2 will be launched at a greater speed because the displacement is shorter than takel I, causing the plunger to immediately, hit the blocks with a shorter amount of time. ( leading to more power) (b) Does the block launched from table 2 spend more, less, or the same amount of time in the air than the block launched from table 1 does? Explain your reasoning. 13 - They stay the savce time in the air since they are both acted on by gravity ! But table 2 is shorter. (c) Consider the system consisting of only the block. Determine the change, if any, in the total mechanical energy of the system from the instant the block leaves the table to the instant immediately before it reaches the ground. of from the spring to the ramp, the ME is relatively the same, but as the block forthurs the ground the ME decreases since the HE decreases. No height can't say/use potential to multiply The students correctly derive an equation for the speed v of the block when it is at the edge of the tabletop, in terms of s (the distance between the launcher and the edge of the tabletop): FF = M - N (2Ls - $2) - 2ugs. (d) Does this equation for v support your argument from part (a) that the block launched from table 2 could have a larger speed when it leaves the table than the block on table I does? Briefly explain why or why not. (a) yes, it is possible since the xbetween the plunger and the blocks is shorter than table!. FF = M - N As"s' increases v decreases. (e) The students now repeat the scenario with identical setups except that the coefficient of kinetic friction between the blocks and the table is larger. Does this change in the coefficient of friction make it more likely or less likely that the block launched from table 1 has a greater speed when it leaves the table than the block on table 2 does? Briefly explain your reasoning. - Ramp I has a slower speed since as the Plunger glides across the table, the friction acts on it causing it to slow down. Since there is less distance to act against the plunger on table to its speed in greater. x Less likely