As part of a carnival game, a 0.478-kg ball is ... Question As part of a carniva

Question

As part of a carnival game, a 0.478-kg ball is ... Question As part of a carnival game, a 0.478-kg ball is thrown at a stack of 15.3-cm tall, 0.343-kg objects and hits with a perfectly horizontal velocity of 13.1 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 4.03 m/s in the same direction, the topmost object now has an angular velocity of 4.85 rad/s about its center of mass and all the objects below are undisturbed. If the object's center of mass is located 10.7 cm below the point where the ball hits, (1) what is the moment of inertia of the object about its center of mass?

And also (2) What is the center of mass velocity of the tall object immediately after it is struck?

Explanation / Answer

It is conservation of momentum
Linear momentum of ball = m * v
Angular momentum of object = I *

Initial linear momentum of the ball = 0.478 * 13.1
Initial momentum of the object = 0
Final momentum of the ball = 0.478 * 4.03
Final momentum of the object = I * 4.85
Total final momentum = total initial momentum
0.478 * 4.03 + I * 4.85 = 0.478 * 13.1
1.926 + I * 4.85= 6.2618
I * 4.85 = 6.2618 – 1.926 =
I * 4.85 = 4.3358
I = 4.3358 / 4.85 = 0.894kg m2

0.478kg * 13.1m/s = 0.478kg * 4.03m/s + 0.343kg * v
6.262 = 1.926+ 0.343*v

v= (6.262-1.926)/0.343 =12.64m/s

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