A 24-g bullet moving at 900 m/s is fired through a one-kg block of wood emerging at a speed of 200 m/s.What is the kinetic energy of the block that results from the collision if the block had not been moving prior to the collision and was free to move?

Question

Quizplus · Accepted Answer

First, we can use conservation of momentum to find the velocity of the block after the collision:
p = mv
Total momentum before collision = (24 g)(900 m/s) = 21600 g*m/s
Total momentum after collision = (1024 g)(v block) where v block is the velocity of the block after the collision
21600 g*m/s = (1024 g)(v block)
v block = 21.09375 m/s

Now we can use the kinetic energy equation to find the kinetic energy of the block:
KE = 1/2 * m * v^2
KE = 1/2 * (1000 g) * (21.09375 m/s)^2
KE = 14,062.5 J or 0.14 kJ

A 24-G Bullet Moving at 900 M/s Is Fired Through