# Conservation of mometum

Let m1 (mass of bullet) = 20g = 0.02kg

m2 (mass of wooden block) = 980g = 0.98 kg

u1 = 200 m/s (Initial velocity of the bullet)

u2 = 0 m/s (Initial velocity of the block of wood)

After the collision, the block as well as the bullet move witha common welocity. Let this velocity be v.

By Law of Conservation of Momentum

m1u1 + m2u2 = m1v + m2v

0.02*200 + 0*0.98 = v(m1+m2)

4 = v(0.98 + 0.02)

4 = v

Therefore the velocity with which both the wooden block and the bullet move is 4 m/s.