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?

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
\(\begin{aligned}
& \mathrm{m}_1 \mathrm{v}_1=\frac{\mathrm{m}_1 \mathrm{v}_1}{1.5}+\mathrm{m}_2 \mathrm{v}_2 \\
& \mathrm{~m}_1 \mathrm{v}_1=\mathrm{m}_1 \frac{2}{3} \mathrm{v}_1+\mathrm{m}_2 \frac{5}{3} \mathrm{v}_1 \\
& \frac{1}{3} \mathrm{~m}_1 \mathrm{v}_1=\mathrm{m}_2 \frac{5}{3} \mathrm{v}_1 \Rightarrow \frac{\mathrm{~m}_1}{\mathrm{~m}_2}=\frac{5}{1}
\end{aligned}\)