Homework 21 - homework 21 BAUTISTA ALDO Due 4:00 am 1...

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homework 21 – BAUTISTA, ALDO – Due: Mar 22 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. Two identical balls are on a frictionless hor- izontal 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 53 Which of the following diagrams best repre- sents the motion of ball Y after the collision? 1. after 8 m/s Y 37 correct 2. 4 m/s Y after 3. after 6 m/s Y 53 4. after 4 m/s Y 37 5. after 6 m/s Y 37 6. 0 m/s Y after 7. 6 m/s Y after 8. after 4 m/s Y 53 9. after 8 m/s Y 53 10. 8 m/s Y after Explanation: ±rom conservation of momentum, m 1 ~v 1 i + m 2 ~v 2 i = m 1 ~v 1 f + m 2 ~v 2 f m~v = m~v 1 + m~v 2 ~v = ~v 1 + ~v 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. ~v 2 θ ~v 1 φ ~v Now 90 = φ + θ , and with φ = 53 θ = 90 - 53 = 37 Then | ~v | cos θ = | ~v 2 | | ~v 2 | = (10 m / s) cos 37 = 8 m / s
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homework 21 – BAUTISTA, ALDO – Due: Mar 22 2006, 4:00 am 2 Question 2 Part 1 of 3. 10 points. A hockey puck of mass 0.6 kg with an ini- tial speed of 2 . 76 m / s collides elastically with another equally massive puck initially at rest. After the collision, the two hockey pucks are observed to have the same speed, each recoil- ing symmetrically at equal magnitude but op- positely directed angles from the initial puck’s direction. v 0 v Φ Before After What is the angle Φ between the two recoil- ing pucks? Correct answer: 90 (tolerance ± 1 %). Explanation: x y Φ θ 1 θ 2 v 0 By momentum conservation x : m v 0 = m v cos θ 1 + m v cos θ 2 y : 0 = m v sin θ 1 - m v sin θ 2 and so, from the second equation, θ 1 = θ 2 . By energy conservation in an elastic collision: 1 2 m v 2 0 = 1 2 m v 2 + 1 2 m v 2 , so v 2 = v 2 0 2 , or v = v 0 2 . Inserting these results into the equation for the x component of momentum, we get v 0 = 2 v cos θ 1 = 2 v 0 cos θ 1 and then cos θ 1 = 1 2 , which means that θ 1 = 45 . Using that Φ + θ 1 + θ 2 = 180 , we get Φ = 90 . Question 3
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Homework 21 - homework 21 BAUTISTA ALDO Due 4:00 am 1...

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