Question

A shell of mass $m$ is at rest initially. It explodes into three fragments having masses in the ratio $2:2:1$. The fragments having equal masses fly off along mutually perpendicular direction with speed $v$. What will be the speed of the third (lighter) fragment?

Answer 1

Total mass of the shell is $m$.

The ratio of the masses of the fragments is $2:2:1$

$\Rightarrow$$2x+2x+x=m$

$\Rightarrow x=0.2m$

Let the velocity of the lighter fragment be $v^{\prime }$

$p^{\prime }=\sqrt{p_1^2+p_2^2}$

$\Rightarrow$$0.2m{v}^{\prime }=\sqrt{(0.4mv{)}^2+(0.4mv{)}^2}$

$\Rightarrow 0.2m{v}^{\prime }=\sqrt{2}\times 0.4mv$

$\Rightarrow v^{\prime }=2\sqrt{2}v$