HW6solutions - Niu(qn269 HW06 Tsoi(58020 This print-out...

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Niu (qn269) – HW06 – Tsoi – (58020) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd 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) ±orce (N) 0 1 2 3 4 10 - 10 ±or the time interval 0 to 4 s, the total change in the momentum oF the object is 1. Δ p = 20 kg m / s . 2. Δ p = 40 kg m / s . 3. Δ p = - 20 kg m / s . 4. Indeterminable unless the mass M oF the object is known 5. Δ p = 0 kg m / s . correct Explanation: The Newton’s second law oF motion, in one dimension, is F = M a = M dv dt . ±rom this, we obtain F dt = M dv , = Δ( M v ) = i 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 fgure 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
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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 vV 1 i + m 2 2 i = m 1 1 f + m 2 2 f mvV = 1 + 2 = 1 + 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. 2 θ b 1 φ Now 90 = φ + θ , and with φ = 53 θ = 90 - 53 = 37 Then | | cos θ = | 2 | | 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 . From conservation of momentum Δ p = 0 7 mV 1 + 2 2 = 9 f V f = 7 1 + 2 2 9 m = 7 (44000 kg) (3 . 3 m / s) 9 (44000 kg) + 2 (44000 kg) (1 . 8 m / s) 9 (44000 kg) = 2 . 96667 m / s . 004 (part 2 of 2) 10.0 points How much energy is lost in the collision? Correct answer: 77000 J. Explanation: The initial kinetic energy is K i = 2 1 2 + 2 2 2 2 = 7 (44000 kg) (3 . 3 m / s) 2 2 + 2 (44000 kg) (1 . 8 m / s) 2 2 = 1 . 81962 × 10 6 J .
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HW6solutions - Niu(qn269 HW06 Tsoi(58020 This print-out...

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