Question

A body of mass $(4m)$ is lying in $x-y$ plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass $(m)$ move perpendicular to each other with equal speeds $(v)$. The total kinetic energy generated due to explosion is

• A. $mv^2$
• B. $\frac{3}{2}mv^2$
• C. $2mv^2$
• D. $4mv^2$

Answer 1

The Correct Option is B

 To solve this problem, we need to apply the principles of conservation of momentum and energy. The mass of the object before the explosion is \(4m\). After the explosion, it divides into three pieces, with the third piece having a mass of \(2m\) since two pieces of mass \(m\) each are given.

Step 1: Conservation of Momentum

Initially, the body is at rest, so its total momentum is zero. After the explosion, the momentum of the system must still be zero. Let's assume that the two pieces move along the x-axis and the y-axis. If their speeds are \(v\) along these axes, the third piece must have a certain velocity to keep the momentum conserved.

Let the third piece's velocity be \(\mathbf{v_3} = (v_{3x}, v_{3y})\).

• For the x-component of momentum:
• For the y-component of momentum:

Step 2: Calculation of Velocity of the Third Piece

The resultant velocity of the third piece is:

\(v_3 = \sqrt{v_{3x}^2 + v_{3y}^2} = \sqrt{\left(\frac{v}{2}\right)^2 + \left(\frac{v}{2}\right)^2} = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{v^2}{2}} = \frac{v}{\sqrt{2}}\)

Step 3: Calculation of Kinetic Energy

The kinetic energy of each piece after the explosion is calculated as follows:

• Kinetic energy of the two \(m\) mass pieces: \(KE_1 = KE_2 = \frac{1}{2}mv^2\)
• Kinetic energy of the third \(2m\) mass piece:

Thus, the total kinetic energy after the explosion is:

\(KE_{\text{total}} = KE_1 + KE_2 + KE_3 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{mv^2}{2} = \frac{3}{2}mv^2\)

Therefore, the correct answer is \(\frac{3}{2}mv^2\).