As part of a carnival game, a 0.628-kg ball is thrown at a stack of 19.3-cm tall
ID: 1505149 • Letter: A
Question
As part of a carnival game, a 0.628-kg ball is thrown at a stack of 19.3-cm tall, 0.403-kg objects and hits with a perfectly horizontal velocity of 11.6 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 3.85 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 13.5 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
Appliying conservation of momentum
Total final momentum = total initial momentum
Initial linear momentum of the ball(m*v1= 0.628 * 11.6 ) +Initial momentum of the object (I*w1= 0)
= Final momentum of the ball(m*v2 = 0.628 * 3.85) +Final momentum of the object I*w2= I * 4.03)
0.628 * 3.85 + I * 4.03 = 0.628 * 11.6
2.4178 + I * 4.03 = 7.2848
I * 4.03 = 4.867
I = 1.2077 kg/m2
Final peed of the object = v,
Hence, m'*v = I*w =1.2077*4.03
Speed v = 4.8670/ 0.403
=12.077 m/s.
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