A 3.72 g bullet moving at 290 m/s enters and stops in an initially stationary 3.30 kg wooden block on a horizontal frictionless surface.

What fraction of the bullet's kinetic energy was lost in this perfectly inelastic collision?
f=?
How much work was done in stopping the bullet?
W=?

Question

A 3.72 g bullet moving at 290 m/s enters and stops in an initially stationary 3.30 kg wooden block on a horizontal frictionless surface.

What fraction of the bullet's kinetic energy was lost in this perfectly inelastic collision? 
f=?
How much work was done in stopping the bullet? 
W=?

mani_jha · Answer

Use the law of conservation of momentum. Do you know it?

jamesj · Answer

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anonymous · Answer

Adding to what Mani Jha said, this is the Law of Conservation of Momentum:

$$m_1u_1 +m_2u_2 = m_1v_1 + m_2v_2$$
where u = initial velocity.