Question

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

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

Answer 1

Answer: (A)

$6E_0$

Let the three fragments move along x, y, z direction with velocities $V_a\hat{i},V_b\hat{j},V_c\hat{k}$

hence,$m_a{V}_a\hat{i}+m_b{V}_b\hat{j}+m_c{V}_c\hat{j}+m_d\overrightarrow{V_d}=0$

As $m_a=m_b=m_c=m_d$

so,$\overrightarrow{V_d}=-V_0(V_a\hat{i}+V_b\hat{j}+V_c\hat{k})$

Again as K.E are same and mass are same

so,$V_a=V_b=V_c=V_0$

Hence,$\overrightarrow{V_d}=-V_0(\hat{i}+\hat{j}+\hat{k})$

The energy of explosion$=K.E_f-K.E_i$

$=\frac{1}{2}m{V_d}^2+\left[\frac{1}{2}m{V_0}^2{X}^3\right]$

$=\frac{1}{2}m{V_0}^2\left(3\right)+\frac{1}{2}m{V_0}^2\left(3\right)$

$=3m{V_0}^2$

$\hat{P}\left(initial\right)=\hat{P}\left(final\right)=6E_0$