HW6solutions - Niu (qn269) HW06 Tsoi (58020) 1 This...

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Unformatted text preview: Niu (qn269) HW06 Tsoi (58020) 1 This print-out should have 24 questions. Multiple-choice 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 left-hand 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 right-hand 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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HW6solutions - Niu (qn269) HW06 Tsoi (58020) 1 This...

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