As part of a carnival game, a 0.528-kg ball is thrown at a stack of 23.3-cm tall

Question

As part of a carnival game, a 0.528-kg ball is thrown at a stack of 23.3-cm tall, 0.403-kg objects and hits with a perfectly horizontal velocity of 10.6 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 4.35 m/s in the same direction, the topmost object now has an angular velocity of 4.83 rad/s about its center of mass and all the objects below are undisturbed.

1) 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?

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

Explanation / Answer

mass of ball , m = 0.528 Kg

mass of object , M = 0.403 Kg

let the moment of inertia is I

1)

Using conseration of angular momentum

angular momentum of system before collision = angular momentum of system after collision

0.163 * 0.528 * 10.6 = I * 4.83 + 0.163 * 0.528 * 4.35

solving for I

I = 0.111 Kg.m^2

the moment of inertia of ball is 0.111 Kg.m^3

2)

let the velocity of centre of mass is v

Using conservation of momentum ,

0.528 * (10.6 - 4.35) = 0.403 * v

v = 8.19 m/s

the centre of mass velocity is 8.19 m/s

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