A bullet of mass mB = 0.01 kg is moving with a speed of 126 m/s when it collides

Question

A bullet of mass mB = 0.01 kg is moving with a speed of 126 m/s when it collides with a rod of mass mR = 9 kg and length L = 5 m (shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating. Find the angular velocity, ?, of the bullet?rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass.

Explanation / Answer

conservation of angular momentum about centre of rod
mBv(L/4) = I

I =mLL2/12 + mB(L/4)2 = 9*5*5/12 + 0.01(5/4)2 =18.765

0.01*126*(5/4) = 18.765*

=0.0839 rad/s

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