A particle of mass $m_1$ moving along the $X$-axis collides with a stationary particle of mass $m_2$ and deviates by an angle $30^{\circ}$ to the $X$-axis as shown in the figure. If the percentage change in kinetic energy of the combined system of these two particles reduces by $50 \%$, then the ratio of the masses $\frac{m_2}{m_1}$ is

• Correct answer
  A

  8
• B

  6
• C

  $\frac{8}{7}$
• D

  $\frac{1}{6}$

Solution

Momentum is conserved in both $x$ and $y$ directions separately $$ \begin{array}{ll} \Rightarrow & m_1 u=m_2 v_2 \cos 30^{\circ} \\ \text { and } & m_1 v_1=m_2 v_2 \sin 30^{\circ} \end{array} $$ Also, loss of K E $=50 \%$ $$ \Rightarrow \quad 50=\frac{\frac{1}{2} m_1 u^2-\frac{1}{2} m_1 v_1^2-\frac{1}{2} m_2 v_2^2}{\frac{1}{2} m_1 u^2} $$ Solving Eqs. (i), (ii) and (iii), we get $$ \frac{m_2}{m_1}=8 $$

Hence, the correct answer is option: 'A'.