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A body of mass $m_1$ collides elastically with another body of mass $m_2$ at rest. If the velocity of $m_1$ after collision becomes $2 / 3$ times its initial velocity, the ratio of their masses is
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$5: 1$
Let $v$ be the initial velocity of body of mass $m_1$ before collision.
Since collision is elastic both momentum and kinetic energy are conserved.
$\therefore m_1 v=m_1 \frac{2}{3} v+m_2 v_2$ or $m_1\left(v-\frac{2}{3} v\right)=m_2 v_2$
Also, $\frac{1}{2} m_1 v^2=\frac{1}{2} m_1\left(\frac{2}{3}\right)^2 v^2+\frac{1}{2} m_2 v_2^2$ or $m_1\left(v^2-\frac{4}{9} v^2\right)=m_2 v_2^2$
or $\frac{m_1}{m_2}=\frac{9 v_2^2}{5 v^2}$
Equating (i) and (ii), we get
$\begin{aligned} & \frac{9 v_2^2}{5 v^2}=\frac{3 v_2}{v} \text { or } \frac{v_2}{v}=\frac{5}{3} \\ & \therefore \frac{m_1}{m_2}=\frac{3 \times 5}{3}=\frac{5}{1}\end{aligned}$
Since collision is elastic both momentum and kinetic energy are conserved.
$\therefore m_1 v=m_1 \frac{2}{3} v+m_2 v_2$ or $m_1\left(v-\frac{2}{3} v\right)=m_2 v_2$
Also, $\frac{1}{2} m_1 v^2=\frac{1}{2} m_1\left(\frac{2}{3}\right)^2 v^2+\frac{1}{2} m_2 v_2^2$ or $m_1\left(v^2-\frac{4}{9} v^2\right)=m_2 v_2^2$
or $\frac{m_1}{m_2}=\frac{9 v_2^2}{5 v^2}$
Equating (i) and (ii), we get
$\begin{aligned} & \frac{9 v_2^2}{5 v^2}=\frac{3 v_2}{v} \text { or } \frac{v_2}{v}=\frac{5}{3} \\ & \therefore \frac{m_1}{m_2}=\frac{3 \times 5}{3}=\frac{5}{1}\end{aligned}$
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