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As part of a carnival game, a 0.453-kg ball is thrown at a stack of 21.8-cm...

As part of a carnival game, a 0.453-kg ball is thrown at a stack of 21.8-cm tall, 0.443-kg objects and hits with a perfectly horizontal velocity of 10.3 m/s. Suppose the ball strikes the very top of the topmost object as shown to the right. Immediately after the collision, the ball has a horizontal velocity of 4.35 m/s in the same direction, the topmost object now has an angular velocity of 1.63 rad/s and all the objects below are undisturbed. If the object\'s center of mass is located 15.3 cm below the point where the ball hits, what is the moment of inertia of the object?What is the center of mass velocity of the tall object immediately after it is struck?
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Answer #1
conserving momentum,
m1v1 = m1v1' + m2v2 '
0.453 X10.3 = 0.453 X 4.35 + 0.443 X v2'
v2'= 6.08 in the same direction.
mvr = IW
0.443 X 6.08 X 0.153= I X 1.63

I= 0.2529
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Answer #2

2ad part) Conserve the liner momentum, F7Zz Substitute 0.453 kg for m, 0.443 kg for m, 10.3 m/s for and 4.35 m/s for (0.453)

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