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Unformatted text preview: Ph1a Solutions 1 Chefung Chan (email@example.com), Fall 2007 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 1.1 QP1 (5 points) 1.1.a (0.5 points) h = 1 2 gt 2 , so t = radicalBig 2 h g . For h = 2 m, we get t = 0 . 64 sec. 1.1.b (1 point) This situation is symmetric that in part a). The ball will bounce back to the original height h = 2 m since the collision is elastic, and so the time it takes to go up is the same t = 0 . 64 sec. You can double check this by calculating the velocity right before impact in part a, reversing its direction, and finding the time to stop on the way back up. 1.1.c (1.5 points) To find the time the ball collides with the floor, we express the height of the elevator floor, z fl ( t ), and the height of the ball, z b ( t ), and solve for when they are equal. We measure z as the height above the level of the 2nd floor. z fl ( t ) = v e t where v e = 0 . 5m / s For the motion of the ball, it is just falling under gravity. It is important to note that the ball initially is moving upwards with velocity v e since both it and the person holding it are travelling upwards at the same speed as the elevator. z b ( t ) = z + v e t 1 2 gt 2 where z = h = 2m v e t = h + v e t 1 2 gt 2 Solving this gives the same solution as parts a & b above, t = 0 . 64 sec. Another way to see this is to change reference frames to the one moving upwards at constant velocity with the elevator. Since this is an inertial (ie non-accelerated) frame, the laws of physics are unchanged and the situation is equivalent to a & b. 1.1.d (1 point) The elevator is initially moving upwards with the same v e = 0 . 5 m/s as in part c. Now, as the ball is dropped, the elevator accelerates with a = kt 2 where k = 4m / s 4 . As in part c, the height of the ball measured above the 4th floor is the same: z b ( t ) = h + v e t 1 2 gt 2 We must now solve for z fl ( t ) from the given acceleration....
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- Fall '07