Please see the attached file.

A 0.150 kg projectile is fired with a velocity of 715 m/s at a 2.00 kg wooden block that rests on a frictionless table. The velocity of the block, immediately after the projectile passes through it, is 40.0 m/s. What is the velocity with which the projectile exits from the block?
Total momentum before impact = m1v1 + m2*0 = m1v1 = 0.15*715 = 107.25 kg-m/s
Total momentum after impact = m2v2 + m1v1' = 2*40 + 0.15*v1' = 80 + 0.15v1'
By Law of conservation of momentum, 107.25 = 80 + 0.15v1'
Therefore, v1' =