A moving body with mass ' $\mathrm{m}_1$ ' strikes a stationary mass ' $m_2$ '. What should be the ratio…
A moving body with mass ' $\mathrm{m}_1$ ' strikes a stationary mass ' $m_2$ '. What should be the ratio $\frac{m_1}{m_2}$ so as to decrease the velocity of first by (1.5) times the velocity after the collision?
$1: 25$
$1: 5$
$5: 1$.
$25: 1$
Solution
Let initial velocity of mass $m_1$ be $v_1$ and final velocity of mass $m_2$ be $v_2$
According to the given condition,
Final velocity of mass $\mathrm{m}_1$ is $\frac{\mathrm{v}_1}{1.5}=\frac{2}{3} \mathrm{v}_1$
Coefficient of restitution,
$\mathrm{e}=\frac{\text { Velocity after collision }}{\text { Velocity before collision }}$
$\begin{array}{ll}
1 & 1=\frac{\left(v_2-\frac{2 v_1}{3}\right)}{v_1} \ldots(e=1, \text { for elastic collision }) \\
\therefore \quad & v_2=\frac{5 v_1}{3}...(i)
\end{array}$ By following conservation of momentum
$\begin{aligned}
& \mathrm{m}_1 \mathrm{v}_1=\frac{\mathrm{m}_1 \mathrm{v}_1}{1.5}+\mathrm{m}_2 \mathrm{v}_2 \\
& \mathrm{~m}_1 \mathrm{v}_1=\mathrm{m}_1 \frac{2}{3} \mathrm{v}_1+\mathrm{m}_2 \frac{5}{3} \mathrm{v}_1 \\
& \frac{1}{3} \mathrm{~m}_1 \mathrm{v}_1=\mathrm{m}_2 \frac{5}{3} \mathrm{v}_1 \Rightarrow \frac{\mathrm{~m}_1}{\mathrm{~m}_2}=\frac{5}{1}
\end{aligned}$