A 26.0-g bullet moving at 1 200 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?
A)6.6E+2 J
B)4.9E+2 J
C)3.4E+2 J
D)7.4E+2 J
E)1.3E+3 J

Question

Quizplus · Accepted Answer

Conservation of momentum tells us that the momentum of the bullet before the collision is equal to the momentum of the block and bullet after the collision. 
p_before = p_after 
m_bullet * v_bullet_before = (m_bullet + m_block) * v_after

Solving for v_after, we get: 
v_after = (m_bullet * v_bullet_before) / (m_bullet + m_block) 
v_after = (26.0 g * 1200 m/s) / (26.0 g + 1000 g) 
v_after = 22.46 m/s

The kinetic energy of the block after the collision can be found using the formula: 
KE = (1/2) * m_block * v_after^2 
KE = (1/2) * 1000 g * (22.46 m/s)^2 
KE = 3.37 x 10^2 J

Therefore, the answer is choice C.

A 260-G Bullet Moving at 1 200 M/s Is Fired Through