[ Questions #1#13  Questions #14#26  Questions #27#37 ] [ #27  #28  #29  #30  #31  #32  #33  #34  #35  #36  #37 ] 
27. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F. Assuming the same velocity change of v, the force experienced by a mass of (1/2)M in a time of (1/2)t is
a. 2F 
b. 4F 
c. (1/2)*F 
d. (1/4)*F 
e. none of these 
Answer and Explanation: 
Answer: E The impulsemomentum change theorem states that F*t = m*(Delta vel.). This equation can be rearranged to locate the F by itself on one side of the equation; rearranging yields The equation shows that force is directly related to the mass, directly related to the change in velocity, and inversely related to the time. So any change in mass will result in the same change in force; and any change in time will result in the inverse effect upon the force. In this case, halving the mass (from M to 1/2M) will half the force and halving the time (from t to 1/2t) will double the force. The combined effect of these two changes will make the new force the same size as the old force. This is a case of where equations can be a guide to thinking about how a change in one variable (or two variables) impacts other dependent variables. 
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28. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F. Assuming the same velocity change of v, the force experienced by a mass of (1/2)M in a time of 4t is
a. 2F 
b. 8F 
c. (1/2)*F 
d. (1/8)*F 
e. none of these 
Answer and Explanation: 
Answer: D The impulsemomentum change theorem states that F*t = m*(Delta vel.). This equation can be rearranged to locate the F by itself on one side of the equation; rearranging yields The equation shows that force is directly related to the mass, directly related to the change in velocity, and inversely related to the time. So any change in mass will result in the same change in force; and any change in time will result in the inverse effect upon the force. In this case, halving the mass (from M to 1/2M) will halve the force and quadrupling the time (from t to 4t) will quarter the force. The combined effect of these two changes will make the new force eight times smaller (i.e., oneeighth the size) than the old force. This is a case of where equations can be a guide to thinking about how a change in one variable (or two variables) impacts other dependent variables. 
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29. A 0.5kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. If the impusle occurs for a time duration of 0.01 s, then the average force (magnitude only) acting upon the ball is
a. 0.14 N 
b. 150 N 
c. 350 N 
d. 500 N 
e. none of these 
Answer and Explanation: 
Answer: C This is a relatively simple plugandchug into the equation with m=0.5 kg, t=0.01 s and Delta vel.=7 m/s. (The change in velocity is 7 m/s since the ball must first slow down from 5 m/s to 0 m/s and then be thrown back in the opposite direction at 2 m/s.) Using these numbers and solving for force yields 350 N. The magnitude of the force is 350 N and the "" sign indicates the direction of the force. 
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30. If mass and collision time are equal, then impulses are greater on objects which rebound (or bounce).
a. TRUE 
b. FALSE 
Answer and Explanation: 
Answer: A The impulse is equal to the momentum change. And when there is a rebound, the momentum change is larger since there is a larger velocity change. For instance, a ball thrown at a wall at 5 m/s may rebound at 3 m/s yielding a velocity change of 8 m/s. An egg thrown at the same wall at the same speed of 5 m/s hits and stops, thus yielding a velocity change of 5 m/s. More velocity change means more momentum change and thus more impulse. 
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31. An unfortunate bug strikes the windshield of a bus in a headon collision. Which of the following statements are true?
Answer and Explanation: 
Answer: The true statements are D and E. In any collision between two objects, the force, impulse, and momentum change are the same for each object. (This makes statements A, B, and C false.) However, the smaller mass object encounters a greater acceleration and velocity change. (This makes statements D and E true). 
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32. A 0.80kg ball strikes a wall moving at 5.0 m/s and rebounds in the opposite direction at 3.5 m/s. If the collision with the wall endures for a total time of 0.0080 s, then determine the average force acting upon the ball. PSYW
Answer and Explanation: 

Answer: 850 N

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33. A 16.0kg ball is thrown with a speed of 22.0 m/s to a 55kg clown who is at rest on ice. The clown catches the ball and glides across the ice. Determine the velocity of the clown (and ball) immediately following the catch. PSYW
Answer and Explanation: 

Answer: 4.96 m/s Momentum conservation principles must be used to solve this collision problem. Set the initial momentum of the system equal to the final momentum of the system and solve for the unknown velocity. The information can be organized inside of a momentum table as shown below. (NOTE: units have been left off quantities to avoid "messiness".)
352 = 71*v v = (352/71) = 4.96 m/s 
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34. A 16.0kg ball is thrown with a speed of 22.0 m/s to a 55kg clown on ice. At the time that the clown catches the ball, she is moving with a speed of 3.0 m/s in the same direction as the ball. The clown catches the ball and continues to glide across the ice. Determine the velocity of the clown (and ball) immediately following the catch. PSYW
Answer and Explanation: 

Answer: 7.28 m/s Momentum conservation principles must be used to solve this collision problem. Set the initial momentum of the system equal to the final momentum of the system and solve for the unknown velocity. The information can be organized inside of a momentum table as shown below. (NOTE: units have been left off quantities to avoid "messiness".)
517 = 71*v v = (517/71) = 7.28 m/s 
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35. A 0.050kg billiard ball moving at 1.2 m/s strikes a second 0.050kg billiard ball which is moving in the same direction with a speed of 0.40 m/s. If the faster ball slows down to a speed of 0.65 m/s, then what is the speed of the second ball? PSYW
Answer and Explanation: 

Answer: 0.95 m/s Momentum conservation principles must be used to solve this collision problem. Set the initial momentum of the system equal to the final momentum of the system and solve for the unknown velocity. The information can be organized inside of a momentum table as shown below. (NOTE: units have been left off quantities to avoid "messiness".)
0.0475 = 0.05*v v = (0.0475/0.05) = 0.95 m/s 
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36. A 0.050kg billiard ball moving at 1.5 m/s strikes a second 0.050kg billiard ball which is at rest on the table. If the first ball slows down to a speed of 0.10 m/s, then what is the speed of the second ball? PSYW
Answer and Explanation: 

Answer: 1.40 m/s Momentum conservation principles must be used to solve this collision problem. Set the initial momentum of the system equal to the final momentum of the system and solve for the unknown velocity. The information can be organized inside of a momentum table as shown below. (NOTE: units have been left off quantities to avoid "messiness".)
0.070 = 0.05*v v = (0.070/0.05) = 1.4 m/s 
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37. A 70kg hockey player moving at 5.6 m/s collides headon with an 80kg player who is heading in the opposite direction with a speed of 3.5 m/s. The two players entangle and move together across the ice. Determine their aftercollision speed. PSYW
Answer and Explanation: 

Answer: 0.747 m/s Momentum conservation principles must be used to solve this collision problem. Set the initial momentum of the system equal to the final momentum of the system and solve for the unknown velocity. The information can be organized inside of a momentum table as shown below. (NOTE: units have been left off quantities to avoid "messiness".)
112 = 150*v v = (112/150) = 0.747 m/s 
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Units 4 Review  Answers: [ Questions #1#13  Questions #14#26  Questions #27#37 ] 



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This page last updated on December 13, 1999.