Question

A 50g bullet with a velocity of 600m/s hits a 5kg block of wood which is at rest. the bullet becomes embedded in the block. what is the speed of the block after it is hit by a bullet?

Answer 1

conservation of momentum:
\[m_bv_b=(m_b+m_B)v\]

Answer 2

so v after is:
\[v=\frac{ m_bv_b }{ m_b+m_B }\]
50g=0,05kg
so
\[v=\frac{ 0,05*600 }{ 0,05+5 }=5,94 m/s\]