A. Incorrect! Remember that momentum depends on both mass and velocity. B. Incorrect! Remember that momentum depends on both mass and velocity.

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1 AP Physics - Problem Drill 08: Momentum and Collisions No. 1 of A car and motor bike are travelling down the road? Which of these is a correct statement? (A) The car will have a higher momentum. (B) The bike has a higher momentum. (C) They both have the same momentum. (D) There is not sufficient information to say whose momentum is greatest. (E) Since they are not accelerating there is no momentum. Remember that momentum depends on both mass and velocity. Remember that momentum depends on both mass and velocity. Remember that momentum depends on both mass and velocity. D. Correct! Since we know nothing about their velocities, we cannot know about their relative momentums. Remember that momentum depends on velocity not acceleration. Momentum is given my P=mv, so although the motorbike has a lower mass it could still have the same or higher momentum than the car, if it s velocity is greater than the cars. The correct answer is (D).

2 No. 2 of A 0.2 kg ball is traveling towards you with a momentum of 5.0 kg m/s, what is the magnitude of velocity of the ball? (A) 0 m/s (B) 2.6 m/s (C) 9.8 m/s (D) 25 m/s (E) 25 m s The formula for momentum is P=mv, you need to rearrange the formula to find velocity. Momentum depends on mass not weight. The formula for momentum is P=mv, you need to rearrange the formula to find velocity. D. Correct! Rearrange the formula for momentum to give v=p/m. Be careful when reading the answers to look for correct units. Known: Mass, m = 0.2 kg Momentum, P = 5 kg m/s Unknown: Velocity, v=? m/s Define: P = mv v P m Output: v 5 kg m/s 25 m/s 0.2 kg Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (D).

3 No. 3 of A baseball player steps up to the plate and hits a.150 kg fast ball traveling at 36.0 m/s, the impact caused the ball to leave his bat with a velocity of 45.0 m/s in the opposite direction. If the impact lasted for.002sec, what force did he exert on the baseball? (A) 6080 N (B) 3250 N (C) 1130 N (D) 675 N (E) 3380 N A. Correct! Use f t=m v. Be sure to use the correct value for the change in velocity. The two velocities given in the problem are in opposite directions. Consider it the difference between -36m/s and +45 m/s. That is a change in velocity of 81m/s. Use f t=m v. Be sure to use the correct value for the change in velocity. Use f t=m v. Be sure to use the correct value for the change in velocity. The two velocities given are in opposite directions. Don t make both of them positive since that isn t represented in the situation or problem. One of the velocities given must be negative. Use f t=m v Be sure to use the correct value for the change in velocity. Known: Mass, m = kg Time, t =.002 s Initial velocity, v i = m/s (for sign convention us final direction as positive) Final velocity, v f = 45.0 m/s Unknown: F =? N Define: Use the Impulse equation. F t m v mδv m(v v ) f i Rearrange: F = Δt Δt Output: F m( v - v ) kg (45.0 m/s -(-36.0 m/s) kg m/s Δt s.002 s 6080 kg m fig) Remember 1 kg m = 1 N F=6080 N f i (3 sig Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (A).

4 No. 4 of A 1200 kg car is moving at 30 km/h hits a tree and comes to a halt in 2s. What is the impulse on the car? (A) kg m/s (B) kg m/s (C) kg m/s (D) kg m/s (E) 6000 kg m/s You must convert to m/s. B. Correct! Use J = m v, negative the impulse in the direction opposite to that of the car. Use J = m v Use J = m v Use J = m v, and check the sign of your answer, the final velocity is 0 m/s. Known: Initial velocity = 30 km/h Final velocity = 0 m/s Time = 2s Unknown: Impulse, J =? J/s Define: First covert 30 km/hr to m/s Output: the car. Impulse J = Ft = m(v f -v i ) km 1000 m 1 hr m/s hr km 60 x 60 s J = 1200 kg (0 10 m/s) = kg m/s = kg m/s. Negative that the impulse is in the opposite direction to the motion of Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (B).

5 No. 5 of A bullet of mass m and velocity v collides with a wooden block that rests on a horizontal surface. A similar bullet collides with a rubber block that has the same mass as the wooden block. In which case does the block move more? (A) The wooden block moves more. (B) The rubber block moves more. (C) They move equal amounts. (D) It cannot be determined with the information given. (E) The blocks will not move. Since the bullet would embed in the wood block; that would be a smaller change in momentum than if it had bounced off the rubber block. B. Correct! Since the rubber block will cause the bullet to bounce back, its momentum will be changed more. Thus, the momentum of the block will change more. This means it will slide a greater distance. Since the rubber block would cause the bullet to change its motion even more than simply stopping, there would be a greater impulse on the bullet and block. Although exact numbers aren t given for the masses and velocities, we can predict the results based on a relative change in momentum. There must be conservation of momentum. In comparing the two situations, the important thing to note is the change in momentum. For the wooden block, the bullet will embed and come to a stop. This results in a change in momentum for the bullet, and a corresponding change in momentum for the block. This causes the block to slide. However, the rubber block will cause its bullet to bounce back instead of just coming to a halt. This will result in a greater change in momentum, thus the block feels a greater impulse and slides farther. The correct answer is (B).

6 No. 6 of Three students are arguing one says that, student 1 says Energy and momentum are conserved in an inelastic collision Student 2 says Momentum is conserved, but kinetic energy is lost in an inelastic collision Student 3 says Momentum is conserved and kinetic is conserved in an elastic collision Who is correct? (A) Student 1 only. (B) Student 2 only. (C) Student 2 & 3. (D) Student 3. (E) All students. Student 1 is correct energy is conserved but they are not the only one. Student 1 is correct energy is conserved but they are not the only one. Student 1 & 2 are both correct. Student 3 is correct but they are not the only one. E. Correct They are all correct! Both energy and momentum are conserved in collisions, for inelastic collisions the kinetic energy will be transformed into some other type, such as sound or heat. For an elastic collision the kinetic energy is conserved. They are all correct! Both energy and momentum are conserved in collisions, but for inelastic collisions the kinetic energy will be transformed into some other type, such as sound or heat. For an elastic collision the kinetic energy is conserved. The correct answer is (E).

7 No. 7 of If you were in a car that collided head on with a wall which car would you rather be in and why? (You can assume that all the cars described would be travelling at the same velocity and have the same mass). (A) A car with a crumple zone, because the crumple zone decreases the impulse. (B) A car with a crumple zone and air bags, because they both lower the impulse. (C) A car with airbags because they increases the impulse. (D) A car with a crumple zone and air bags, they both increase the time taken to change the cars momentum and so reduce the average stopping force required to stop the car. (E) A car with a crumple zone and airbag because the crumple zone increases the impulse. Consider the formula for impulse, J = Ft = m(v f -v i ). Look at the assumption you were told to make, and consider whether the impulse changes for a car with a crumple zone. Consider the formula for impulse, J = Ft = m(v f -v i ). Look at the assumption you were told to make, and consider whether the impulse changes for a car with a crumple zone and airbag. Consider the formula for impulse, J = Ft = m(v f -v i ). Look at the assumption you were told to make, and consider whether the impulse changes for a car with airbags. D. Correct! Consider the formula for impulse, J = Ft = m(v f -v i ). A car of the same mass, travelling with the same velocity experiences the same momentum change, (impulse) whether or not it has an airbag and/or crash zone. The effect of the airbags and crash zone is to change the time taken for the collision; this reduces the force that is needed to change the momentum and reduces risk of injury. Consider the formula for impulse, J = Ft = m(v f -v i ). Look at the assumption you were told to make, and consider whether the impulse changes for a car with a crumple zone and airbag. Consider the formula for impulse, J = Ft = m(v f -v i ). A car of the same mass, travelling with the same velocity experiences the same momentum change, (impulse) whether or not it has an airbag and/or crash zone. The effect of the airbags and crash zone is to change the time taken for the collision; this reduces the force that is needed to change the momentum and reduces risk of injury. The correct answer is (D).

8 No. 8 of Sally has a mass of 50 kg. She is rolling along on roller skates at 5 m/s. Her friend Debbie has a mass of 70 kg (also on roller skates) and is moving towards her at 1 m/s. They meet, and hang onto each other. What is the magnitude and direction of their velocity after they collide? (A) 2.66 m/s (B) 180 m/s (C) 3 m/s (D) 1.5 m/s (E) None of the above When you use the velocity of Debbie, 1m/s that must be the opposite sign of Sally since they are moving in opposite directions. That would be the momentum of the combination of skaters, but that isn t their velocity. Use P=mv and conservation of momentum. That would be the average of the two initial speeds, but since they have differing masses, you must do more than just average the two together. Use P=mv and conservation of momentum. D. Correct! Use P=mv and conservation of momentum. This is like a hit and stick collision. m 1 v 1 +m 2 v 2 =(m 1 +m 2 )v 3 you are finding the last velocity, which is the speed of the two skaters as they hold onto each other. Be sure to make one of the two initial velocities negative since they are moving in opposite directions. You could pick either direction as positive, but your final answer would have the sign of Sally s initial movement. Use P=mv and conservation of momentum. Known: negative) Sally s mass, m 1 = 50 kg Sally s velocity, v 1 = 5 m/s Debbie s mass m 2 = 70 kg Debbie s velocity, v 2 = -1 m/s (assume that the Sally s velocity is positive so Debbie s velocity is This is a hit and stick collision. Unknown: Velocity after v 3 =? m/s Define: For a hit and stick collision m 1 v 1 +m 2 v 2 =(m 1 +m 2 )v 3 Output: 50 kg(5 m/s) + (70 kg)(-1 m/s) = (50 kg + 70 kg) v kg m/s - 70 kg m/s = (120 kg )v 3 180kgm/s=(120 kg )v 3 v 3 =1.5 m/s The positive sign of your answer indicates the skater combination is moving in the initial direction of Sally. Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (D).

9 No. 9 of A 300 kg motorboat is turned off as it approaches a dock and it coasts in toward the dock at.50 m/s. Joe, mass is 62.0 kg, jumps off the front of the boat with a speed of 5 m/s. What is the velocity of the boat after Joe jumps? (A) -0.43m/s (B) -0.53m/s (C) 1.5m/s (D) 0 m/s, stationary (E) 5 m/s A. Correct! Set the momentum of Joe and the boat before the jump equal to their momentum after the jump. Solve for the single variable you don t know. P of boat before +P of Joe before = P of boat after+ P of Joe after. Substitute all the information you have. Find the velocity of the boat after the collision. Since all objects are moving the in the same direction initially, you may assign all the velocities given as positive. Initially, the boat and Joe are moving at.5m/s. Be sure to account for both of their momentums before the jump. Set the momentum of Joe and the boat before the jump equal to their momentum after the jump. Solve for the single variable you don t know. Set the momentum of Joe and the boat before the jump equal to their momentum after the jump. Solve for the single variable you don t know. This is similar to an explosion collision. The momentum of the system before is still equal to the momentum of the system after the jump, m J v J + m b v Bb = m J v Ja + m b v Ba the mass of the boat and Joe has not changed so if Joes velocity changes what happens to the boats velocity. Known: Joe s mass, m J = 62 kg Joe s velocity before, v J b = m/s Joe s velocity before, v J a = 5.0 m/s Boat mass, m B = 300 kg Boat velocity before, V Bb = m/s Unknown: Velocity of boat after Joe jumps, V Ba =? m/s Define: This is similar to an explosion collision. However, there is movement before the explosion. The momentum before is still equal to the momentum after the jump. P before = P after m J v J + m b v Bb = m J v Ja + m b v Ba Output : (62 kg).5 m/s + (300 kg).5m/s = 62 kg (5.0 m/s)+ 300 kg (v Ba ) 31kg m/s kg m/s = 310 kg m/s kg (v Ba ) 181 kg m/s = 310 kg m/s kg (v Ba ) -129 kg m/s = 300 kg (v Ba ) v Ba =-0.43m/s The negative sign means the boat moves in the opposite direction compared to its initial movement that we made the positive direction. Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (A).

10 No. 10 of An explosive detonates and breaks into three 1 kg chunks. One piece flies north at 50 m/s. Another piece flies west at 70 m/s. What is the speed of the final chunk? (A) 120 m/s (B) 86 m/s (C) 45 m/s (D) 20 m/s (E) Insufficient information The direction must be taken into account. The particles aren t flying in the same direction so you can t just add the speeds as scalars. B. Correct! Add the momentum vectors of the two given chunks. The final piece must be equal and opposite to that resultant since the total of all three must equal 0. The momentum of the explosive was zero before it went off. The direction must be taken into account. The particles aren t flying in the same direction so you can t just add the speeds as scalars. The direction must be taken into account. The particles aren t flying in the same direction so you can t just add the speeds as scalars. Combine the two given chunks as momentum vectors. The third piece must be equal but opposite to that resultant. Known: m 1 =m 2 =m 3 =1 kg Initial momentum = 0 kg m/s v 1 = 50 m/s north v 2 = 70 m/s west R To find R align vectors head to tail Piece 1: 50kg m/s Unknown: Velocity v 3 = m/s Piece 2:70 kg m/s Piece #3 momentum Define: Draw a picture to help. Before the device explodes, it isn t moving so there is zero momentum. Even after it explodes, the momentum of all the pieces must add up to zero since momentum is conserved. Find the resultant of the first two. The momentum of the third piece must be equal and opposite to this resultant, R. Use the Pythagorean theorem, R= ((m 1 v 1 ) 2 + (m 2 v 2 ) 2 ) = m 3 v 3 v 3 = R/ m 3 Output: R= (50kgm/s) 2 + (70kgm/s) 2 = 86 kg m/s v 3 = 86 kg m/s/ 1 kg v 3 = 86 m/s Substantiate: Units are correct, sig figs are correct, Magnitude is correct. The correct answer is (B).

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