A 22-g bullet moving at 1,200 m/s is fired through a one-kg block of wood emerging at a speed of 100 m/s.What is the change in the kinetic energy of the bullet-block system as a result of the collision assuming the block is free to move?

Question

Quizplus · Accepted Answer

The initial kinetic energy of the bullet is (1/2)(0.022 kg)(1200 m/s)^2 = 396 J. 
The final kinetic energy of the bullet is (1/2)(0.022 kg)(100 m/s)^2 = 11 J. 
The change in kinetic energy of the bullet is therefore -385 J. 
The block is free to move, so its final velocity can be found using conservation of momentum: 
(0.022 kg)(1200 m/s) = (1 kg + 0.022 kg)(v) 
v = 22.08 m/s 
The initial kinetic energy of the block is zero, and its final kinetic energy is (1/2)(1 kg)(22.08 m/s)^2 = 256.9 J. 
The change in kinetic energy of the block is therefore 256.9 J. 
The total change in kinetic energy of the system is the sum of the changes in kinetic energy of the bullet and the block: 
-385 J + 256.9 J = -128.1 J, which is approximately -15.4 kJ.

A 22-G Bullet Moving at 1,200 M/s Is Fired Through