Question

Suppose you are firing a bullet of mass 10g which is moving with a velocity of 400m/s and penetrates into a freely suspended wooden block of mass 900 g. What is the velocity acquired by the block?

Answer 1

Law of conservation of momentum
ie. momentum before collision = momentum after collision

10 (400)+ 900 (0)= 10 (0)+ 900 (v)
4000= 900v
v =40/9
So v= 4.4 m/s

Answer 2

as momentum is conserved in every collision, momentum before the collision is equal to the momentum after the collision.
momentum of bullet before collision:0.01 kg*400m/s=4kg.m/s
momentum of wood before collision:0kg.m/s
momentum of bullet after collision: 0kg.m/s(as the bullet penetrates into the block it's velocity is zero hence momentum is also zero)
momentum of wood after collision=mass of wood*velocity acquired by it.= 0.9kg*v
as momentum is conserved
4kg.m/s+0=0+ 0.9kg*v
v= 4kg.m/s/0.9kg=> 4.44 m/s