Question

A 3.0g bullet moving at 2.0km/s strikes an 8.0kg wooden block at rest on a frictionless table. The bullet lodges in the wooden block. How fast does the block move across the table after being struck?

Answer 1

The velocity is `=0.75ms^-1`

Explanation:

There is conservation of momentum

`(m_bxxv_b)+Mxx0=(m_b+M)xxv`

The mass of the bullet `m_b=0.003kg`

The velocity of the block `v_b=2000ms^-1`

The mass of the block is `M=8.0 kg`

The velocity of the block and bullet is

`v=(m_bxxv_b)/(m_b+M)=(0.003*2000)/(8.003)`

`=6/8.003=0.75ms^_1`