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A $5 \mathrm{~kg}$ rifle fires a $6 \mathrm{~g}$ bullet with $S$ peed of $500 \mathrm{~m} / \mathrm{s}$. Find the ratio of the distance the rifle move backward while the bullet is in the barrel to the distance the bullet moves forward?
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Verified Answer
The correct answer is:
$1 / 1000$
By conservation of momentum, since no external impulses force acts on the system
$\begin{aligned} & \mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}} \\ & \text { Or } \\ & 0=\mathrm{m}_{\mathrm{r}} \mathrm{v}_{\mathrm{r}}+\mathrm{m}_{\mathrm{b}} \mathrm{v}_{\mathrm{b}} \cdots(1) \\ & \mathrm{v}_{\mathrm{r}}=\frac{-\mathrm{m}_{\mathrm{b}}}{\mathrm{m}_{\mathrm{f}}} \mathrm{v}_{\mathrm{b}}\end{aligned}$
From equation (1)
$\begin{aligned} & 0=m_r \frac{\Delta x_r}{\Delta t}+m_b \frac{\Delta x_b}{\Delta t} \\ & 0=m_r \Delta x_r+m_b \Delta x_b\end{aligned}$
Which shows
$\left|\frac{\Delta x_{\mathrm{f}}}{\Delta x_b}\right|=\frac{m_b}{m_t}=\frac{6}{6000}=\frac{1}{1000}$
$\begin{aligned} & \mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}} \\ & \text { Or } \\ & 0=\mathrm{m}_{\mathrm{r}} \mathrm{v}_{\mathrm{r}}+\mathrm{m}_{\mathrm{b}} \mathrm{v}_{\mathrm{b}} \cdots(1) \\ & \mathrm{v}_{\mathrm{r}}=\frac{-\mathrm{m}_{\mathrm{b}}}{\mathrm{m}_{\mathrm{f}}} \mathrm{v}_{\mathrm{b}}\end{aligned}$
From equation (1)
$\begin{aligned} & 0=m_r \frac{\Delta x_r}{\Delta t}+m_b \frac{\Delta x_b}{\Delta t} \\ & 0=m_r \Delta x_r+m_b \Delta x_b\end{aligned}$
Which shows
$\left|\frac{\Delta x_{\mathrm{f}}}{\Delta x_b}\right|=\frac{m_b}{m_t}=\frac{6}{6000}=\frac{1}{1000}$
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