a physics question?

Question

Two 500 g blocks of wood are 2.0 m apart on a frictionless table. A 10 g bullet is fired at 400m/s toward the blocks. It passes all the way through the first block, then embeds itself in the second block. The speed of the first block immediately afterward is 6.0m/s.What is the speed of the second block after the bullet stops?



I know i use conservation of momentum but I just can't figure it out...maybe i might be plugging in numbers wrong, but could i please get some advice or help on this one?

Answer 1

Momentum
p=mv

Bullet:
(10g)(400m/s) = 4,000 g m/s

First block
(500g)(6m/s) = 3,000 g m/s

The difference is 1,000 g m/s

v = p/m
v = (1,000 g m/s) / (500g + 10g)
v ≈ 1.96 m/s

If the bullet had not embedded, adding its mass to the second block, the answer would be 2.0 m/s.

Answer 2

You have the momentum of the first block and the momentum of the second block whose mass includes the bullet. Both these momentums equal the original momentum of the bullet alone.