A bullet of mass m B = 0.0201 kg is moving with a speed of 103 m/s when it colli

Question

A bullet of mass mB = 0.0201 kg is moving with a speed of 103 m/s when it collides with a rod of massmR = 7.99 kg and length L = 1.08 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.

a) 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.

b) How much kinetic energy is lost in the collision?
(Hint: Please enter a positive number for your answer! If K < 0, then the amount of kinetic energylost is a positive number.)

Explanation / Answer

conserving the angular momentum about the hinge
mvl/4 = I+m(l/4)2

.0201*103*1.08/4=7.99*1.03*1.03*/12+.0201**(1.08/4)2

.559=.706+.0015

=.79 rad/sec

K.E= 1/2 * I *2

=.5*(7.99*1.03*1.03/12 + .0201*(1.08/4)2)*.792

=.225 joule

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