As part of a carnival game, a 0.553-kg ball Is thrown at a stack of 15.3-cm tall

Question

As part of a carnival game, a 0.553-kg ball Is thrown at a stack of 15.3-cm tall, 0.343-kg objects and hits with a perfectly horizontal velocity of 13.3 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 10.7 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

m = mass of ball = 0.553 kg

Vi = initial horizontal velocity of ball = 13.3 m/s

Vf = final horizontal velocity of ball after collision = 3.85 m/s

r = distance of point of collision of ball above the center of mass = 10.7 cm = 0.107 m

I = moment of inertia of object

w = angular speed of object = 4.03 rad/s

using conservation of angular momentum

m Vi r = m Vf r + IW

0.553 (13.3) (0.107) = 0.553 (3.85) (0.107) + I (4.03)

I = 0.13875 kgm2

using conservation of momentum

m Vi = m Vf + M V

0.553 (13.3) = 0.553 (3.85) + (0.343) V

V = 15.24 m/s

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