An object of mass $5 \mathrm{kg}$ is projected with a velocity $20 \mathrm{m} {\mathrm{s}}^{-1}$ at an angle $60°$, to the horizontal. At the highest point of its path, the projectile explodes and breaks up into two fragments of masses $1 \mathrm{kg}$ and $4 \mathrm{kg}.$ The fragments separate horizontally after the explosion, which releases internal energy such that the $KE$ of the system af the highest point is doubled. The separation between the two fragments when they reach the ground is

Given, $m=5 \mathrm{kg},v=20 \mathrm{m} {\mathrm{s}}^{-1},\theta =60°$

Vertical component of velocity, $v_y=v \sin 60°$

$=20\times \frac{\sqrt{3}}{2}=10\sqrt{3} \mathrm{m} {\mathrm{s}}^{-1}$

Time taken to reach the highest point $=$ Time taken to reach the ground from highest point.

$t=\frac{v \sin \theta }{g}=\frac{v_y}{g}=\frac{10\sqrt{3}}{9.8}=1.77 \mathrm{s}$

At the highest point, $m$ splits up into two parts of masses $m_1=1 \mathrm{kg}$ and $m_2=4 \mathrm{kg}$

If their velocities are $v_1$ and $v_2$ respectively, then applying the principle of conservation of linear momentum, we get

$m_1{v}_1+m_2{v}_2=mv \cos \theta$

$v_1+4v_2=5\times 20\times \frac{1}{2} \left[\because \theta =60°\right]$

$v_1+4v_2=5\times 10=50 ....\left(\mathrm{i}\right)$

Initial $KE=\frac{1}{2} m {\left(v \cos \theta \right)}^2$

$=\frac{1}{2}\times 5\times {\left(10\right)}^2=250 \mathrm{J}$

Final $KE=2$ (initial $KE$)$=2\times 250=500 \mathrm{J}$

$\therefore \frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=500$

or $\frac{1}{2}\times 1\times v_1^2+\frac{1}{2}\times 4\times v_2^2=500$

or $v_1^2+4v_2^2=1000 ....\left(\mathrm{ii}\right)$

Solving Eq. (i) and (ii), we get

$v_1=30 \mathrm{m} {\mathrm{s}}^{-1}, v_2=5 \mathrm{m} {\mathrm{s}}^{-1}$

Hence, the separation between the two fragments

$=\left(v_1-v_2\right)\times t=\left(30-5\right)\times 1.77 \mathrm{m}=44.25 \mathrm{m}$