Question 13.159: To apply shock loading to an artillery shell, a 20-kg pendul...

First impact: A impacts B. \quad m_{A}=20 \mathrm{~kg}, m_{B}=?

Impulse-momentum: \quad\Sigma m \mathbf{v}+\Sigma \mathbf{I m} \mathbf{p}_{1 \rightarrow 2}=\Sigma m \mathbf{v}_{2}

Components directed left: 

\begin{aligned}& m_{A} \nu_{0}=m_{A} \nu_{A}^{\prime}+m_{B} \nu_{B}^{\prime} \\& 20 \nu_{0}=20 \nu_{A}^{\prime}+m_{B} \nu_{B}^{\prime}\quad\text{(1)}\end{aligned}

Coefficient of restitution:

\begin{aligned}\nu_{B}^{\prime}-\nu_{A}^{\prime} & =e\left(\nu_{A}-\nu_{B}\right) \\\nu_{B}^{\prime}-\nu_{A}^{\prime} & =e \nu_{0} \\\nu_{A}^{\prime} & =\nu_{B}^{\prime}-e \nu_{0}\quad\text{(2)}\end{aligned}

Substituting Eq. (2) into Eq. (1) yields

\begin{aligned}20 \nu_{0} & =20\left(\nu_{B}^{\prime}-e \nu_{0}\right)+m_{B} \nu_{B}^{\prime} \\20 \nu_{0}(1+e) & =\left(+m_{B}\right) \nu_{B}^{\prime} \\\nu_{B}^{\prime} & =\frac{20 \nu_{0}(1+e)}{20+m_{B}}\quad\text{(3)}\end{aligned}

Second impact: B impacts C. \quad m_{B}=?, m_{C}=1 \mathrm{~kg}

Impulse-momentum: \quad\Sigma m \mathbf{v}_{2}+\Sigma \mathbf{I m p}_{2 \rightarrow 3}=\Sigma m \mathbf{v}_{3}

Components directed left:

\begin{aligned}& m_{B} \nu_{B}^{\prime}=m_{B} \nu_{B}^{\prime \prime}+m_{C} \nu_{C}^{\prime \prime} \\& m_{B} \nu_{B}^{\prime}=m_{B} \nu_{B}^{\prime \prime}+\nu_{C}^{\prime \prime}\quad\text{(4)}\end{aligned}

Coefficient of restitution:

\begin{aligned}& \nu_{C}^{\prime \prime}-\nu_{B}^{\prime \prime}=e\left(\nu_{B}^{\prime}-\nu_{C}^{\prime}\right) \\& \nu_{C}^{\prime \prime}-\nu_{B}^{\prime \prime}=e \nu_{B}^{\prime} \\& \nu_{B}^{\prime \prime}-\nu_{C}^{\prime \prime}=e \nu_{C}^{\prime}\quad\text{(5)}\end{aligned}

Substituting Eq. (4) into Eq. (5) yields

\begin{aligned}m_{B} \nu_{B}^{\prime} & =m_{B}\left(\nu_{C}^{\prime \prime}-e \nu_{B}^{\prime}\right)+m_{C} \nu_{C}^{\prime \prime} \\m_{B} \nu_{B}^{\prime}(1+e) & =\left(1+m_{B}\right) \nu_{C}^{\prime \prime} \\\nu_{C}^{\prime \prime} & =\frac{m_{B} \nu_{B}^{\prime}(1+e)}{1+m_{B}}\quad\text{(6)}\end{aligned}

Substituting Eq. (3) for \nu_{B}^{\prime} in Eq. (6) yields

\nu_{C}^{\prime \prime}=\frac{20 m_{B} \nu_{0}(1+e)^{2}}{\left(20+m_{B}\right)\left(1+m_{B}\right)}

The impulse applied to the shell C is

m_{C} \nu_{C}^{\prime \prime}=\frac{(1)(20) m_{B} \nu_{0}(1+e)^{2}}{\left(20+m_{B}\right)\left(1+m_{B}\right)}

To maximize this impulse choose m_{B} such that

Z=\frac{m_{B}}{\left(20+m_{B}\right)\left(1+m_{B}\right)}

is maximum. Set d Z / d m_{B} equal to zero.