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Unformatted text preview: Niu (qn269) HW06 Tsoi (58020) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The graph below shows the force on an object of mass M as a function of time. Time (s) Force(N) 1 2 3 4 10 10 For the time interval 0 to 4 s, the total change in the momentum of the object is 1. p = 20 kg m / s . 2. p = 40 kg m / s . 3. p = 20 kg m / s . 4. Indeterminable unless the mass M of the object is known 5. p = 0 kg m / s . correct Explanation: The Newtons second law of motion, in one dimension, is F = M a = M dv dt . From this, we obtain F dt = M dv , = ( M v ) = integraldisplay F dt, where M v is the momen tum of the object. So from the graph above, the change in the momentum is zero . 002 10.0 points Two identical balls are on a frictionless, hori zontal tabletop. Ball X initially moves at 10 meters per second, as shown in figure on the lefthand side. It then collides elastically with ball Y , which is initially at rest. After the col lision, ball X moves at 6 meters per second along a path at 53 to its original direction, as shown in on the righthand side. 10 m/s X before Y after 6 m/s X 0 m/s b b 53 Which of the following diagrams best repre sents the motion of ball Y after the collision? 1. after 6 m/s Y 37 b 2. after 8 m/s Y 37 b correct 3. 6 m/s Y after b 4. after 4 m/s Y 53 b 5. 4 m/s Y after b 6. after 6 m/s Y 53 b 7. 0 m/s Y b after Niu (qn269) HW06 Tsoi (58020) 2 8. after 8 m/s Y 53 b 9. 8 m/s Y after b 10. after 4 m/s Y 37 b Explanation: From conservation of momentum, m 1 vectorv 1 i + m 2 vectorv 2 i = m 1 vectorv 1 f + m 2 vectorv 2 f mvectorv = mvectorv 1 + mvectorv 2 vectorv = vectorv 1 + vectorv 2 . This relation leads to the vector diagram be low, where it is used that when two equal mass have an elastic collision, and one of the masses was originally at rest, they move per pendicular to each other after the collision. vectorv 2 b vectorv 1 b vectorv Now 90 = + , and with = 53 = 90  53 = 37 Then  vectorv  cos =  vectorv 2   vectorv 2  = (10 m / s) cos 37 = 8 m / s 003 (part 1 of 2) 10.0 points Given: Each railroad car has a mass of 44000 kg. Seven coupled railroad car moving with a speed of 3 . 3 m / s collide and couples with two other coupled railroad cars moving in the same direction with an initial speed of 1 . 8 m / s. What is the speed of all nine cars after the collision? Correct answer: 2 . 96667 m / s. Explanation: Let: The mass of a railroad car be m = 44000 kg, the speed of the faster 7 cars be v 1 = 3 . 3 m / s, the speed of the slower 2 cars be v 2 = 1 . 8 m / s, and the speed of all 9 cars after the collision be v f ....
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 Spring '09
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 Physics, Force, Mass

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