Question

A particle of mass $4m$ initially at rest explodes into three fragments. If two fragments each of mass $m$ move mutually perpendicular to each other with speed $v$, then the total energy released in the process of explosion will be

• A. $\frac{3}{2}m{v}^2$
• B. $\frac{2}{3}m{v}^2$
• C. $4m{v}^2$
• D. $v^2$

Answer 1

The correct option is A $\frac{3}{2}m{v}^2$

Let the speed of third fragment is $v^{\prime }$
$\therefore 2m{v}^{\prime }\cos \theta =mv...\left(i\right)$
and $2m{v}^{\prime }\sin \theta =mv...\left(ii\right)$
$\Rightarrow \sin \theta =\cos \theta =\frac{1}{\sqrt{2}}$
so, from Eq.(i)
$2m{v}^{\prime }\left(\frac{1}{\sqrt{2}}\right)=mv$
$v^{\prime }=\frac{v}{\sqrt{2}}$
$\therefore$ Total $KE$
$=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}\left(2m\right){\left(\frac{v}{\sqrt{2}}\right)}^2=\frac{3}{2}m{v}^2$