Question

A body of mass (4m) is lying in the X-Y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (u). The total kinetic energy generated due to explosion is
A. $m{v}^{2}$
B. $\dfrac {3}{2} m{v}^{2}$
C. $2m{v}^{2}$
D. $4m{v}^{2}$

Answer 1

Hint: To solve this problem, use the law of conservation of momentum. Take the magnitude of momentum of heavier piece and equate it with the magnitude of sum of the momentum of each piece of mass m. Rearrange it to get the velocity of the third piece in terms of velocity of another two pieces. Then, use the formula for Kinetic energy. Add up the kinetic energy of all the three pieces. Substitute the velocity of the third piece in terms of velocity of the other two pieces. Solve it and get the total kinetic energy generated due to the explosion.

Complete step by step answer:
According to the question, the two pieces have mass m each. So, their total mass is 2m. The third piece will have the remaining mass of 2m. The third part moves back because the total momentum of the system after the explosion must remain zero.
Let the velocity of the third piece be ${v}^{‘}$.
According to law of conservation of momentum,
$\sqrt {{2m{v}^{‘}}^{2}}= \sqrt {{mv}^{2}+ {mv}^{2}}$
$\Rightarrow 2m{v}^{‘}= \sqrt{2}mv$
$\Rightarrow {v}^{‘}= \dfrac {\sqrt{2}v}{2}$
$\Rightarrow {v}^{‘}= \dfrac {v}{\sqrt{2}}$ …(1)
Kinetic energy generated is given by,
$K.E.= \dfrac {1}{2}m{v}^{2}+\dfrac {1}{2}m{v}^{2}+\dfrac {1}{2}2m{{v}^{‘}}^{2}$
$\Rightarrow K.E.= \dfrac {1}{2} (m{v}^{2}+ m{v}^{2}+ 2m {{v}^{‘}}^{2})$
Substituting equation. (1) in above equation we get,
$K.E.= \dfrac {1}{2} (m{v}^{2}+ m{v}^{2}+ 2m \dfrac {{v}^{2}}{2})$
$\Rightarrow K.E.= \dfrac {1}{2}(m{v}^{2}+ m{v}^{2}+ m{v}^{2})$
$\Rightarrow K.E.= \dfrac {1}{2}\times 3m{v}^{2}$
$\Rightarrow K.E.= \dfrac {3}{2} m{v}^{2}$
Hence, the total kinetic energy generated due to explosion is $\dfrac {3}{2} m{v}^{2}$.

So, the correct answer is “Option B”.

Note: To solve these types of questions, students must have a clear understanding of the laws of conservation. Law of conservation of momentum is an important consequence of Newton's law of motion. When the force acting on a system is zero, the total linear momentum of the system is either conserved or remains constant. Momentum can neither be created nor be destroyed.