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# A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other with the same Kinetic energy. Find the energy of the explosion.A) $7{E_o}$ B) $6{E_o}$ C) $8{E_o}$ D) $9{E_o}$

Last updated date: 04th Mar 2024
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Hint: We can use conservation of momentum to find the momentum of the fourth fragment and then use conservation of energy to find the total energy of explosion.

Let the mass of each fragment after the explosion be $m$ . Since the mass is the same for all fragments , the kinetic energy gets equally divided between the four fragments.
Thus, ${E_1} = {E_2} = {E_3} = E$
Momentum of each fragment $\left( p \right)$ = $\sqrt {2mE}$
The fourth fragment after explosion will fly off in a direction opposite to the direction of resultant of the other three fragments. This we know by the conservation of momentum, i.e., the total momentum of the system is zero before and after the explosion.
Momentum of fourth fragment is thus ${p_4}$
${p_4} = \sqrt {p_1^2 + p_2^2 + p_3^2}$
${p_4} = \sqrt {2mE + 2mE + 2mE}$
${p_4} = \sqrt {6mE}$
Kinetic energy of the fourth fragment can then be found in the following way:
$p = mv$ and $E = \dfrac{1}{2}m{v^2}$
Solving these two by eliminating $v$ from both the equations we get
${E_4} = \dfrac{{p_4^2}}{{2m}}$
${E_4} = \dfrac{{6mE}}{{2m}}$
${E_4} = 3E$
Therefore energy of explosion is the sum of energy of the four fragments = ${E_1} + {E_2} + {E_3} + {E_4}$
$= E + E + E + 3E$
$= 6E$
Thus the answer is option $B$ .

Note: The conservation of momentum can be applied here since net external force is zero. Similar is the reason for conserving energy. We also should take care of the calculations since tricky questions like this have huge chances of calculation error.