A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms^$-$1 and 40 ms^$-$1 respectively. The velocity of the third piece will be :

Given problem a body of mass $M$ explodes into three pieces of mass ratio $1: 1: 2$ <br/><br/>$\therefore$ Mass of fragments will be $x, x, 2 x$ <br/><br/>Hence, $M=x+x+2 x=4 x \mathrm{~kg}$ <br/><br/>As in the process of explosion no external forces are involved and explosion occurs due to internal forces. Thus, momentum of the system will be conserved. <br/><br/>$p_{\text {initial }}=p_{\text {final }}$ <br/><br/>By law of conservation of momentum, <br/><br/>$$ M \times 0=\frac{M}{4} \times 30 \hat{i}+\frac{M}{4} \times 40 \hat{j}+\frac{2 M}{4} \vec{v} $$ <br/><br/>Where $\vec{v}$ is the velocity of the third fragment. <br/><br/>$\frac{M \vec{v}}{2} =-\frac{M}{4}(30 \hat{i}+40 \hat{j})$ <br/><br/>$\Rightarrow$ $\vec{v} =-15 \hat{i}-20 \hat{j}$ <br/><br/>Thus, magnitude of $\vec{v}=|\vec{v}|=\sqrt{v_x^2+v_y^2}$ $=\sqrt{(-15)^2+(-20)^2}$ <br/><br/>$|\vec{v}|=\sqrt{625}=25 \mathrm{~m} / \mathrm{s}$