A body mass of $1 \mathrm{~kg}$, initially at rest explodes and breaks into three parts. The masses of the parts are in the ratio $1: 1: 3$. The two pieces of equal mass fly off perpendicular to each other with a speed of $30 \mathrm{~m} / \mathrm{s}$ each. The velocity of the heavier part in $\mathrm{m} / \mathrm{s}$ is

• Correct answer
  A

  $10 \sqrt{2}$
• B

  6
• C

  3
• D

  $6 \sqrt{2}$

Solution

Initial momentum before explosion is zero. Final momentum after explosion.
$$ \sqrt{2} \frac{m}{4} \times 30-\frac{3 m}{4} v^{\prime} $$ From conservation of momentum Initial momentum $=$ Final momentum $$ \begin{aligned} \Rightarrow \quad \sqrt{2} \frac{m}{4} \times 30-\frac{3 m}{4} v^{\prime} & =0 \\ \Rightarrow \quad v^{\prime} & =10 \sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned} $$

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