Kaboom!

Let the velocity of $m_{1}$ be $v_{1}$ and the velocity of $m_{2}$ be $v_{2}$ . So what we have here is, $\frac{1}{2} m_{1}v_{1}^2 = 2 \times \frac{1}{2} m_{2}v_{2}^2$ $\Rightarrow m_{1}v_{1}^2=2m_{2}v_{2}^2$ Also from since no external forces are acting on our system other than the internal force which broke the object (explosion) therefore we can conserve the momentum of the object. Since object is initially at rest and we are assuming that final masses went in opposite directions then we shall have, $0=m_{1 }v_{1 }-m_{2 }v_{2 }$ $\Rightarrow m_{1 }v_{1 }=m_{2 }v_{2 }$ Solve both equations to get $m_{2}=2m_{1}$