Question

A particle of mass 4m initially at rest explodes into three fragments. If two fragments each of mass m move perpendicular to each other with speed v, then the total energy released in the process of explosion will be
A. $\dfrac{3}{2}m{v^2}$
B. $\dfrac{2}{3}m{v^2}$
C. $4m{v^2}$
D. ${v^2}$

Answer 1

The initial momentum of the particle is zero. Use the law of conservation of momentum after the explosion and calculate the velocity of the third fragment. The total energy released in the explosion is the sum of kinetic energy of each fragment after the explosion.

Formula used:
The linear momentum of the particle of mass m and velocity v is,
$P = mv$
The kinetic energy of the body of mass m moving with velocity v is,
$K.E. = \dfrac{1}{2}m{v^2}$

Complete step by step answer:
We assume the first fragment moves in the direction of the x-axis and the second fragment moves in the direction of the y-axis as they go off to right angles to each other. We know that, x-axis is represented by the unit vector $\hat i$ and the y-axis is represented by the unit vector $\hat j$.
We have the initial momentum of the particle is zero. According to the conservation of momentum, we can write,
$0 = {\vec P_1} + {\vec P_2} + {\vec P_3}$
$\Rightarrow \left| {{{\vec P}_3}} \right| = \left| {{{\vec P}_1}} \right| + \left| {{{\vec P}_2}} \right|$
Here, ${P_1}$ is the momentum of the first fragment, ${P_2}$ is the momentum of the second fragment and ${P_3}$ is the momentum of the third fragment.
In the above equation, we have taken the magnitude of the momentum because we don’t know the direction of the third fragment after the explosion.
We know that the momentum of the body is the product of its mass and velocity. Therefore, we can write the above equation,
${m_3}{\vec v_3} = {m_1}{v_1}\hat i + {m_2}{v_2}\hat j$
We have given that the velocity of the first two fragments moved right angles to each other is v. Also, the mass of the third fragment is 2m while the mass of each of the first two fragments is m.
Therefore, we can write the above equation as follows,
$2m{\vec v_3} = mv\hat i + mv\hat j$
$\Rightarrow 2{\vec v_3} = \sqrt {{v^2} + {v^2}}$
$\Rightarrow 2{\vec v_3} = \sqrt {2{v^2}}$
$\Rightarrow {\vec v_3} = \dfrac{v}{{\sqrt 2 }}$
Therefore, the velocity of the third fragment is $\dfrac{v}{{\sqrt 2 }}$.
Now, we can calculate the total energy released in the explosion as follows,
$T.E = {\left( {K.E} \right)_1} + {\left( {K.E} \right)_2} + {\left( {K.E} \right)_3}$
$\Rightarrow T.E = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}m{v^2} + \dfrac{1}{2}\left( {2m} \right){\left( {\dfrac{v}{{\sqrt 2 }}} \right)^2}$
$\Rightarrow T.E = \dfrac{1}{2}m{v^2}\left( {1 + 1 + 1} \right)$
$\Rightarrow T.E = \dfrac{3}{2}m{v^2}$
Therefore, the total energy released in the explosion is $\dfrac{3}{2}m{v^2}$.

So, the correct answer is “Option A”.

Note:
If the direction of bodies after the collision is not linear, students should never ignore the direction of the bodies. While solving these types of questions, one should assign a proper direction for each particle. In the above solution, we have ignored the direction of the third fragment and taken its magnitude.