Chapter 9.2, Problem 19PP

To determine

The speed of the bullet before it struck the lumber.

Answer to Problem 19PP

  $1.237\times {10}^3\text{m/s}$

Explanation of Solution

Given:

The mass of the bullet, $m_b=35.0\text{g}=0.035\text{g}$

The mass of the stationary piece of lumber, $m_l=5.0\text{kg}$

The bullet embeds itself in the wood and then both the lumber and bullet fly off at the speed, $v_f=8.6\text{}\text{m/s}$

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

And, law of conservation of linear momentum: Momentum is conserved in a closed and isolated system and is given by,

  $p_i=p_f$

Calculation:

Consider the bullet and wood as a system, then it will be a closed and isolated system and law of conservation of momentum can be applied to it.

Now, since the lumber was stationary initially, thus initial velocity of lumber is,

  $v_{li}=0\text{m/s}$

Let, the speed of the bullet before it struck the lumber be $v_{bi}$ , then using the law of conservation of linear momentum,

  $\begin{array}{c}{p}_i=p_f\\ m_l{v}_{li}+m_b{v}_{bi}=\left(m_l+m_b\right)v_f\end{array}$

Using given values in above,

  $\begin{array}{c}\left(5\right)\left(0\right)+\left(0.035\right)v_{bi}=\left(5+0.035\right)8.6\\ v_{bi}=1.237\times {10}^3\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the bullet before it struck the lumber is $1.237\times {10}^3\text{m/s}$ .