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Unformatted text preview: 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 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. 8 m/s Y after b 2. after 4 m/s Y 53 b 3. 6 m/s Y after b 4. after 8 m/s Y 53 b 5. 0 m/s Y b after 6. after 6 m/s Y 37 b 7. after 8 m/s Y 37 b correct 8. after 4 m/s Y 37 b 9. 4 m/s Y after b 10. after 6 m/s Y 53 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 Question 2 part 1 of 1 10 points 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 xaxis, as shown. The 5 g object overtakes and collides with the 4 g object. Immediately after the collision, the ycomponent of the velocity of the 5 g object is 2 m / s upward. v i 5 g v i 4 g What is the ycomponent 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 ycomponent of the mo mentum must be the same before or after the collision. Since the two objects move horizon tally along the xaxis, the ycomponent of the momentum before the collision is zero. Thus, to make the ycomponent 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....
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 Spring '08
 Turner
 Friction, Work

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