Question

As part of a carnival game, a 0.528-kg ball is thrown at a stack of 18.8-cm tall, 0.383-kg objects and hits with a perfectly horizontal velocity of 12.8 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 3.60 m/s in the same direction, the topmost object now has an angular velocity of 1.63 rad/s about its center of mass and all the objects below are undisturbed. If the object\'s center of mass is located 13.2 cm below the point where the ball hits, what is the moment of inertia of the object about its center of mass? Answer in kg.m.

What is the center of mass velocity of the tall object immediately after it is struck? Answer in m/s

Answer 1

m = mass of ball = 0.528 kg

V_i = initial velocity of ball = 12.8 m/s

V_f = final velocity of ball = 3.60 m/s

r = distance from center of mass = 0.132 m

M = mass of top object = 0.383 kg

I = moment of inertia of Top object

w = angular velocity = 1.63 rad/s

using conservation of angular momentum about centre of mass

mVi r = m V_f r + Iw

(0.528) (12.8) (0.132) = (0.528) (3.60) (0.132) + I (1.63)

I = 0.39 kgm²

V_cm = center of mass velocity = rw = 0.132 x 1.63 = 0.22 m/s