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. 1
• B. 2
• C. 0.5
• D. 4

Answer 1

Answer: (C)

From conservation of momentum
$P_i=P_f$
$m_{1}v=-m_1\frac{v}{3}+m_2{v}_2$
$\frac{4}{3}{m}_{1}v=m_2{v}_2$
$\frac{m_1}{m_2}=\frac{3v_2}{4v}\dots$(1)
Form energy conservation
$\frac{1}{2}{m}_1{v}^2=\frac{1}{2}{m}_1\frac{v^2}{9}+\frac{1}{2}{m}_2{v}_2^2$
$\frac{1}{2}{m}_1{v}^2-\frac{1}{18}{m}_1{v}^2=\frac{1}{2}{m}_2{v}_2^2$
$\frac{9m_1{v}^2-m_1{v}^2}{18}=\frac{1}{2}{m}_2{v}_2^2$
$\frac{48m_1{v}^2}{18}=\frac{1}{2}{m}_2{v}_2^2$
$\frac{8}{9}{m}_1{v}^2=m_2{v}_2^2$
$\frac{8}{9}\frac{m_1}{m_2}={\left(\frac{v_2}{v}\right)}^2$
$\frac{v_2}{v}=\sqrt{\frac{8m_1}{9m_2}}$
Now in equation (1)
$\frac{m_1}{m_2}=\frac{3}{4}\times \sqrt{\frac{8m_1}{9m_2}}$
${\left(\frac{m_1}{m_2}\right)}^2=\frac{9}{16}\times \frac{8}{9}\left(\frac{m_1}{m_2}\right)$
$\frac{m_1}{m_2}=\frac{1}{2}=0.5$