A bullet of mass 0.062 kg traveling horizontally at a speed of 150 m/s embeds itself in a block of mass 90 kg that is sitting at rest on a nearly frictionless surface

Question

What is the speed of the block after the bullet embeds itself in the block?

Answer 1

conserve momentum:

.062*150 + 90*0 = (.062+90)v
v = 0.103 m/s

Answer 2

Teri 9211