A bomb of mass $5m$ initially at rest explodes and breaks into three pieces of masses in the ratio $1:1:3$. The two pieces of equal mass fly off perpendicular to each other with a speed of $v_0$. The energy released in explosion $\frac{x}{y}m{v}_0^2$. Find $x+y$.

Draw rough figure for the given situation.

Find speed of the third part.
Given, mass of the bomb, $=5m$

Initially at rest, so initial momentum will be zero.

Total mass is split in a ratio $1:1:3$,

Hence masses of three pieces are
$m_1=m,m_2=m$ and $m_3=3m$

Let the velocity of heavier piece is 𝑣.Momentum of the heavier part (let $P$) will be equal and opposite to the direction of resultant of the momentum of the two equal masses.

As given the two equal masses fly off perpendicular to each other with speed $v_0$,

So, $P=\sqrt{2}p$

$3mv=\sqrt{2}m{v}_0$

$v=\frac{\sqrt{2}{v}_0}{2}$

Find kinetic energy in explosion of the bomb.
Initial $K.E.=0$
Final $K.E.$

$=\frac{1}{2}m{v}_0^2+\frac{1}{2}m{v}_0^2+\frac{1}{2}\times 3m{\left(\frac{\sqrt{2}{v}_0}{3}\right)}^2$

$=m{v}_0^2+\frac{1}{2}m{v}_0^2$

$=\frac{4}{3}m{v}_0^2$

$K.E.$ released in explosion

$=\frac{4}{3}m{v}_0^2$

So,
$(x+y)=3+4=7$