Abullet with mass 4.00 g and speed 650.0 m/s is fired at a block of wood with mass 0.0950 kg. the block rests on a "frictionless" surface and is thin enough so that the bullet passes completely through it. immediately after the bullet exits the block, the speed of the block is 23.0 m/s. (a) what is the speed of the bullet after it exits the block? ( 104 m/s)

26.4 m/s

Explanation:

4 g = 0.004 kg

By the law of momentum conservation, the momentum before and after the impact must be the same

So the speed of the block after the impact is 26.4 m/s

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