Question

A particle of mass $4m$ which is at rest explodes into three fragments. Two of the fragments each of mass $m$ are found to move with a speed $v$ each in mutually perpendicular directions. The energy released in the process of explosion is:

• A. $\frac{3}{2}m{v}^2$
• B. $3m{v}^2$
• C. $2m{v}^2$
• D. $\frac{1}{2}m{v}^2$

Answer 1

Answer: (A)

$\frac{3}{2}m{v}^2$

Here momentum of third fragment is
$p^3=\sqrt{p_1^2+p_2^2}$
or $p^3=\sqrt{(mv{)}^2+(mv{)}^{\overline{2}}}$
$=\sqrt{2}mv$
Final K.E. of the system
$=\frac{p_1^2}{2m}+\frac{p_2^2}{2m}+\frac{p_3^2}{2\left(2m\right)}$
$=\frac{3}{2}m{v}^2+\frac{3}{2}m{v}^2+\frac{3}{2}mv$
$=\frac{3}{2}m{v}^2$
Since initial K.E. $=$ 0 therefore energy released $=\frac{3}{2}m{v}^2$