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A bullet of mass mB = 0.0199 kg is moving with a speed of 104 m/s when it collid

ID: 2153727 • Letter: A

Question


A bullet of mass mB = 0.0199 kg is moving with a speed of 104 m/s when it collides with a rod of mass mR = 7.79 kg and length L = 1.01 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, ? (omega), 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 energy lost is a positive number.)

A bullet of mass mB = 0.0199 kg is moving with a speed of 104 m/s when it collides with a rod of mass mR = 7.79 kg and length L = 1.01 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, ? (omega), 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

Explanation / Answer

Moment Of Inertia Of Rod =I1 = mL2/12= 7.79*1.012/12 = 0.6622 kgm2

Moment of Inertia of Bullet Just before Collision =I2 = m*(L/4)2 = 0.0199*1.012/16 = 1.269*10-3 kgm2

Moment of Inertia Of Bullet & rod = I = 0.6635 kgm2

Initial Angular Momentum = I2*V/(L/4) = 1.269*10-3*104*4/1.01 = 0.5227

Final Angular Momentum = 0.6635*

(a) By Conservation of momentum

0.6635 = 0.5227

=> = 0.7878 rad/s

(b) Initial K = (1/2)*0.0199*1042 = 107.6192 J

Final K = (1/2)I2 = 0.206 J

So K = -107.41 J

So Loss In K.E. = 107.41 J