As part of a carnival game, a 0.603-kg ball is thrown at a stack of 24.3-cm tall

Question

As part of a carnival game, a 0.603-kg ball is thrown at a stack of 24.3-cm tall, 0.423-kg objects and hits with a perfectly horizontal velocity of 12.6 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 4.85 m/s in the same direction, the topmost object now has an angular velocity of 2.03 rad/s about its center of mass and all the objects below are undisturbed. If the object's center of mass is located 17.0 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

Given that

The ball of mass (m1) =0.603kg

The another object of mass (m2) =0.423kg

The initial velocity of the ball (ui) =12.6m/s

The final velocity of the ball after collison with stack is (v1) =4.85m/s

The angular velocity of the top most object (w) =2.03rad/s

First from the conservation of momentum

m1u1+m2u2 =m1v1+m2v2

m1u1+0 =m1v1+m2v2

m1u1-m1v1/m2 =v2

m1(u1-v1)/m2 =v2

v2 =0.603(12.6-4.85)/0.423 =11.047m/s

We know that angualr momentum is given by

mvr =L =IW

(0.423kg)(11.047)(0.17m) =I(2.03rad/s)

Then the moment of inertia is I =0.3913kg.m2

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