Chapter 7, Problem 12P

Solution

The velocity of block after the bullet emerges.

  $v=0.69$ m/sec

Given:

Mass of bullet is 23 gm.

Initial speed is 230m/sec

Mass of wood block is 2 kg.

At 170 m/sec bullet emerges.

Formula used:

According to law of conservation of momentum,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, m₁and m₂are the mass of bullet and block.

u₁and u₂are initial velocities of bullet and block.

v₁and v₂are the final velocities of bullet and block.

Calculation:

Applying the law conservation of momentum

Momentum of the system before collision = Momentum of the system after collision

Mass of bullet × initial velocity of bullet + mass of block × initial velocity of the block = mass of bullet × final velocity of the bullet + mass of the block × final velocity of the block

  $\begin{array}{l}{m}_B\times v_{iB}+m_{Bl}\times v_{iBl}=m_B\times v_{oB}+m_{Bl}\times v_{oBl}\\ \frac{23}{1000}\times 230+0=\frac{23}{1000}\times 170+2.0\times v\\ 5.29=3.91+2v\\ v=0.69\text{m}/\text{s}\end{array}$