A mass $3 m$, initially at rest at the origin explodes into three fragments of equal mass. Two of the fragments have speed $v$ each and move perpendicular to each other. The third fragment will move with a speed

• A

  $\frac{v}{\sqrt{2}}$
• B

  $\frac{v}{2}$
• C

  $v$
• Correct answer
  D

  $v \sqrt{2}$

Solution

From law of conservation of momentum
$\begin{aligned} & m v-m v^{\prime} \cos \theta=0...(i) \\ & m v-m v^{\prime} \sin \theta=0...(ii) \end{aligned}$
From Eqs. (i) and (ii), we get $\begin{aligned} & \theta=45^{\circ} \\ & v^{\prime}=\frac{v}{\cos \theta}=v \sqrt{2} \end{aligned}$

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