Question

A stationary body explodes into three identical fragments such that two of them fly off mutually perpendicular to each other, each with same kinetic energy, $E_o$. The energy of the explosion is

• A. $\sqrt{3}{E}_o$
• B. $3E_o$
• C. $4E_o$
• D. $(2+\sqrt{2})E_o$

Answer 1

The correct option is C $4E_o$

Two fragments have same kinetic energy and mass, then their speed must be same.
Let speed of both fragments is $v$
Both are perpendicular, then consider velocity of first fragment is $x-$ axis direction while velocity of second is $y-$ axis direction.
Therefore, $\overrightarrow{v_1}=v\hat{i},\overrightarrow{v_2}=v\hat{j}$
Third fragment will fly in a direction opposite to the direction of resultant of other two particles. So, the total momentum of the system after explosion also remains zero.
$\therefore$ Applying momentum conservation for all the three fragments,
$m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}+m_3\overrightarrow{v_3}=0$
$\overrightarrow{v_1}+\overrightarrow{v_2}+\overrightarrow{v_3}=0$
$\Rightarrow v\hat{i}+v\hat{j}+\overrightarrow{v_3}=0$
$\overrightarrow{v_3}=-(v\hat{i}+v\hat{j})\Rightarrow \left|v_3\right|=\sqrt{2}v$

Now, kinetic energy of the first and second fragment $=E_0=\frac{1}{2}m{v}^2$

$\therefore$ Total K.E $=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}m(\sqrt{2}v{)}^2$
$=E_0+E_0+2E_0$
$\Rightarrow$ Total KE $=4E_0$