Let the masses of the three fragments be $2M$, $2M$, and $M$ respectively. Initial momentum of the shell is zero. According to the law of conservation of linear momentum, the final momentum of the system must also be zero . $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$ The two fragments of equal mass $2M$ fly off in mutually perpendicular directions with speed $v$. Let them move along the x and y axes. $\vec{p}_1 = 2Mv \hat{i}$ $\vec{p}_2 = 2Mv \hat{j}$ The resultant momentum of these two fragments is: $|\vec{p}_{12}| = \sqrt{(2Mv)^2 + (2Mv)^2} = \sqrt{4M^2v^2 + 4M^2v^2} = \sqrt{8M^2v^2} = 2\sqrt{2}Mv$ For the total momentum to be zero, the third fragment must have a momentum equal and opposite to this resultant momentum. $|\vec{p}_3| = |\vec{p}_{12}|$ $M v' = 2\sqrt{2}Mv$ $v' = 2\sqrt{2}v$ Therefore, the speed of the third, lighter fragment is $2\sqrt{2}v$.