Chapter 7, Problem 33P

(a)

To determine

To derive: A formula for the fraction of kinetic energy lost, $\raisebox{1ex}{$\Delta KE$}\!\left/ \!\raisebox{-1ex}{$KE$}\right.$ , for the ballistic pendulum collision.

Answer to Problem 33P

The formula for the fraction of kinetic energy lost is $\frac{\Delta KE}{KE}=-\frac{M}{m+M}$

Explanation of Solution

Introduction:

According to the conservation of principle of momentum, the total momentum before and after collision remains constant, that is

  $\begin{array}{l}{p}_i=p_f\\ {\displaystyle \sum m_i{u}_i}={\displaystyle \sum m_f{u}_f}\end{array}$

This gives

  $v\text{'}=\frac{m}{m+M}v$

The formula to calculate kinetic energy is $KE=\frac{1}{2}m{v}^2$

Initial kinetic energy is calculated as $K{E}_i=\frac{1}{2}m{v}^2$ and final kinetic energy is $K{E}_f=\frac{1}{2}(m+M)v{\text{'}}^2$

Thus, to calculate the fraction of difference in final and initial kinetic energy & initial kinetic energy is

  $\begin{array}{l}\frac{\Delta KE}{KE}=\frac{K{E}_f-K{E}_i}{K{E}_i}=\frac{\frac{1}{2}(m+M)v{\text{'}}^2-\frac{1}{2}m{v}^2}{\frac{1}{2}m{v}^2}\\ \frac{\Delta KE}{KE}=\frac{(m+M){\left(\frac{mv}{m+M}\right)}^2-m{v}^2}{m{v}^2}\\ \frac{\Delta KE}{KE}=\frac{\frac{m^2{v}^2}{m+M}-m{v}^2}{m{v}^2}=\frac{m}{m+M}-1\\ \frac{\Delta KE}{KE}=-\frac{M}{m+M}\end{array}$

Conclusion:

Therefore, the fraction of lost kinetic energy is $\frac{\Delta KE}{KE}=-\frac{M}{m+M}$

(b)

To determine

The fraction of kinetic energy lost.

Answer to Problem 33P

The $96\%$ fraction of lost kinetic energy in the collision.

Explanation of Solution

Given:

Masses are provided as $m=14g$ , and $M=380g$ .

Formula used:

Fraction of lost kinetic energy is $\frac{\Delta KE}{KE}=-\frac{M}{m+M}$

Calculation:

The lost kinetic energy is

  $\begin{array}{l}\frac{\Delta KE}{KE}=-\frac{M}{m+M}\\ \frac{\Delta KE}{KE}=-\frac{380}{14+380}=-0.96=96\%\end{array}$

Negative sign included since the energy is lost

Conclusion:

Thus, $96\%$ of kinetic energy is lost in the collision.