Question

A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each with same KE, $E_0$. The energy of explosion will be:

• A. $6E_0$
• B. $\frac{4E_0}{3}$
• C. $4E_0$
• D. $8E_0$

Answer 1

The correct option is A $6E_0$

Let the three mutually perpendicular directions be along x, y and z axis respectively,
$\overrightarrow{p_1}=m{v}_0\hat{i}$
$\overrightarrow{p_2}=m{v}_0\hat{j}where,\frac{1}{2}m{v}_0^2=E_0$
$\overrightarrow{p_3}=m{v}_0\hat{k}$
and $\overrightarrow{p_4}=m\overrightarrow{v}$
By linear momentum conservation,
$0=\overrightarrow{p_1}+\overrightarrow{p_2}+\overrightarrow{p_3}+\overrightarrow{P_4}$
or $\overrightarrow{v}=-v_0(\hat{i}+\hat{j}+\hat{k})$
or $v=v_0\sqrt{1^2+1^2+1^2}=v_0\sqrt{3}$
energy $=3\left(\frac{1}{2}m{v}_0^2\right)$
total energy $=3\left(\frac{1}{2}m{v}_0^2\right)+\frac{1}{2}m{v}^2$
$=3E_0+3E_0$
$=6E_0$.