As part of a carnival game, a 0.553-kg ball is thrown at a stack of 23.3-cm tall
ID: 1348458 • Letter: A
Question
As part of a carnival game, a 0.553-kg ball is thrown at a stack of 23.3-cm tall, 0.423-kg objects and hits with a perfectly horizontal velocity of 11.1 m/s. Suppose the ball strikes the topmost object as shown to the right. Immediately after the collision, the ball has a horizontal velocity of 4.85 m/s in the same direction, the topmost object now has an angular velocity of 4.03 rad/s about its center of mass and all the objects below are undisturbed. If the object's center of mass is located 16.3 cm below the point where the ball hits, what is the moment of inertia of the object about its center of mass? What is the center of mass velocity of the tall object immediately after it is struck?Explanation / Answer
ANgular momentum before hitting = MvR = 0.553 * 11.1* 0.233
L1 = 1.43 Kgm/s
angular momentum of objects = 0
angular momentum of ball after hitting
L2 = 0.553 * 0.233 * 4.85 = 0.625 Kgm/s
total angular momentum of objects = I W
L = 0.625 + I * 4.03 = 1.43
therefore moment of inertia (I) = 0.199 k g m^2 ---<<<<Answer to aprt A
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let velocity of center of mass = v
as V = r W
V = 4.03 * (0.233 - 0.163)
V = 0.282 m/s--------------<<<<<<<<<<<<<<<<Answer to part B
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