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Homework 9 Solutions

# Homework 9 Solutions - homework 09 ALIBHAI ZAHID Due 4:00...

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homework 09 – ALIBHAI, ZAHID – Due: Mar 28 2007, 4:00 am 1 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 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 53 Which of the following diagrams best repre- sents the motion of ball Y after the collision? 1. 8 m/s Y after 2. after 4 m/s Y 53 3. 6 m/s Y after 4. after 8 m/s Y 53 5. 0 m/s Y after 6. after 6 m/s Y 37 7. after 8 m/s Y 37 correct 8. after 4 m/s Y 37 9. 4 m/s Y after 10. after 6 m/s Y 53 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 θ vectorv 1 φ vectorv Now 90 = φ + θ , and with φ = 53 θ = 90 53 = 37 Then | vectorv | cos θ = | vectorv 2 | | vectorv 2 | = (10 m / s) cos 37 = 8 m / s Question 2 part 1 of 1 10 points

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homework 09 – ALIBHAI, ZAHID – Due: Mar 28 2007, 4:00 am 2 Two objects of mass 5 g and 4 g , respec- tively, move parallel to the x -axis, as shown. The 5 g object overtakes and collides with the 4 g object. Immediately after the collision, the y -component of the velocity of the 5 g object is 2 m / s upward. v i 5 g v i 4 g What is the y -component of the velocity vectorv of the 4 g object immediately after the collision? 1. vectorv y = 6 m / s downward 2. vectorv y = 12 . 8 m / s downward 3. vectorv y = 2 . 5 m / s downward correct 4. vectorv y = 7 . 2 m / s downward 5. vectorv y = 15 m / s downward 6. vectorv y = 7 m/s upward 7. vectorv y = 10 . 6667 m / s downward 8. vectorv y = 5 m/s upward Explanation: Let : m 1 = 5 g , m 2 = 4 g , and v 1 y = 2 m / s . Momentum vectorp is conserved for collision m 1 vectorv 1 = m 2 vectorv 2 . In particular, the y -component of the mo- mentum must be the same before or after the collision. Since the two objects move horizon- tally along the x -axis, the y -component of the momentum before the collision is zero. Thus, to make the y -component of the momentum after the collision zero, the y - component of the velocity is 0 = m 1 v 1 y + m 2 v 2 y vectorv 2 y = m 1 m 2 v 1 y = (5 g) (4 g) (2 m / s) = 2 . 5 m / s , and the direction is downward. Question 3 part 1 of 1 10 points Two ice skaters approach each other at right angles. Skater A has a mass of 71 . 5 kg and travels in the + x direction at 1 . 96 m / s. Skater B has a mass of 55 . 1 kg and is moving in the + y direction at 1 . 09 m / s. They collide and cling together.
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Homework 9 Solutions - homework 09 ALIBHAI ZAHID Due 4:00...

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