Question

A 0.14 kg bullet traveling at speed hits a 7.9 kg block of wood and stays in the wood. The block with the bullet imbedded in it moves forward with a velocity of 1284.0. What was the velocity (speed) of the bullet immediately before it hit the block?

Answer 1

You have not mentioned the unit of the velocity.

I am assuming this cm/s.

So, the final velocity of the bullet and wooden block is 1284 cm/s = 1.284 m/s.

suppose initial velocity of the bullet is v m/s.

so apply conservation of momentum -

m*v = (m +M)*1.284

=> 0.14*v = (0.14 + 7.9)*1.284

=> v = (8.04 x 1.284) / 0.14 = 73.74 m/s

This is your answer. I have considered the measuring unit of the velocity as cm/s. You can also take as m/s or as per your text book.