Question

A body of mass $4m$ at rest explodes into three fragments. Two of the fragments each of mass $m$ move with speed $v$ in mutually perpendicular directions. Total energy released in the process is

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

Answer 1

Answer: (B)

Here initial momentum $\vec{p}=0$. Since no external force exists, hence momentum must remain conserved
i.e, ${\vec{p}}_1+{\vec{p}}_2+{\vec{p}}_3=0$
As two fragments of mass $m$ each are moving with speed $v$ each at right angles, so
$\left|{\vec{p}}_1+{\vec{p}}_2\right|=m\sqrt{v^2+v^2}=\sqrt{2}mv$
$\therefore \left|{\vec{p}}_3\right|=\left|{\vec{p}}_1+{\vec{p}}_2\right|=\sqrt{2}mv$
The mass of third fragment is $2m$.
$\therefore$ Kinetic energies of three fragments are
$K_1=\frac{p_1^2}{2m}=\frac{1}{2}m{v}^2,K_2=\frac{p_2^2}{2m}=\frac{1}{2}m{v}^2$
and $K_3=\frac{p_3^2}{2(2m)}=\frac{1}{2}m{v}^2$
Total energy released during explosion
$=K_1+K_2+K_3=\frac{3}{2}m{v}^2$