Question

A particle of mass $4m$ which is at rest explodes into masses $m,m$ & $2m$. Two of the fragments of masses $m$ and $2m$ are found to move with equal speeds $v$ each in opposite directions. The total mechanical energy released in the process of explosion is

• A. $m{v}^2$
• B. $2m{v}^2$
• C. $\frac{1}{2}m{v}^2$
• D. $4m{v}^2$

Answer 1

The correct option is B $2m{v}^2$

Assume right direction$=$positive

using momentum conservation

$p_i=p_f$

$o=2m(-v)+m{v}_1+mv$

$v_1=v$

K.E final$=\frac{1}{2}2m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2$

$=\frac{1}{2}m{v}^2\left(4\right)$

$=2m{v}^2$.