Question

A body of mass $m_1$ collides head on elastically with a stationary body of mass $m_2$. If velocities of $m_1$ before and after the collision are $v$ and $–v/3$ respectively then the value of $m_1/m_2$ is

• A. $2$
• B. $0.5$
• C. $4$
• D. $1$

Answer 1

The correct option is B $0.5$

Given:
Initial velocity of $m_1,=v$
Final velocity of $m_1,=-v/3$
Initial velocity of $m_2,=0$

From conservation of momentum

$P_i=P_f$
$m_{1}v+m_2.0=m_1\left(\frac{-v}{3}\right)+m_2{v}_2$
$\frac{4}{3}{m}_{1}v=m_2{v}_2$
$\frac{m_1}{m_2}=\frac{3v_2}{4v}...\left(1\right)$

Form energy conservation

$\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2\times (0{)}^2=\frac{1}{2}{m}_1{\left(\frac{v}{3}\right)}^2+\frac{1}{2}{m}_2{v}_2^2$
$\frac{1}{2}{m}_1{v}^2-\frac{1}{18}{m}_1{v}^2=\frac{1}{2}{m}_2{v}_2^2$
$\frac{8m_1{v}^2}{18}=\frac{1}{2}{m}_2{v}_2^2$
$\frac{8}{9}{m}_1{v}^2=m_2{v}_2^2$
$\frac{8}{9}\frac{m_1}{m_2}={\left(\frac{v_2}{v}\right)}^2$
$\frac{v_2}{v}=\sqrt{\frac{8m_1}{9m_2}}$

Now in equation $\left(1\right)$

$\frac{m_1}{m_2}=\frac{3}{4}\times \sqrt{\frac{8m_1}{9m_2}}$
$\frac{m_1}{m_2}=\frac{1}{2}=0.5$

Hence, option $\left(C\right)$ is correct.

OR

For elastic collision,

Velocity of $m_1$ after collision is $v_1=\frac{m_1-m_2}{m_1+m_2}{u}_1+\frac{2m_2}{m_1+m_2}{u}_2$

As $u_1=v,u_2=0\&v_1=-v/3$

$-\frac{1}{3}=\frac{m_1-m_2}{m_1+m_2}$

$-m_1-m_2=3m_1-3m_2$

$\frac{m_1}{m_2}=\frac{1}{2}=0.5$

Hence, option $\left(C\right)$ is correct.