Question

A particle of mass 5 m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with speed v each. the energy released during the process is :

• A. $\frac{3}{5}mv^{2}$
• B. $\frac{5}{3}mv^{2}$
• C. $\frac{3}{2}mv^{2}$
• D. $\frac{4}{3}mv^{2}$

Answer 1

The Correct Option is D

To solve this problem, we need to understand the concept of conservation of momentum and kinetic energy. The particle initially at rest suddenly breaks into three fragments. Let's analyze the steps involved:

1. Initially, the particle is at rest, so the total momentum is zero.
2. It breaks into three fragments: two fragments of mass m each moving along mutually perpendicular directions with speed v each.
3. For the system to conserve momentum, the third fragment must have a momentum equal and opposite to the vector sum of the first two fragments.

Since the first two fragments are moving perpendicularly, their resultant momentum is given by:

Resultant momentum magnitude: \sqrt{(mv)^2 + (mv)^2} = mv\sqrt{2}

Thus, the third fragment of mass 3m has momentum -mv\sqrt{2}.

The speed of the third fragment (mass 3m) can be calculated as follows:

Momentum: 3mv_{\text{third}} = mv\sqrt{2}

Speed of the third fragment: v_{\text{third}} = \frac{v\sqrt{2}}{3}

Let's calculate the kinetic energy released:

1. Kinetic energy of first fragment: \frac{1}{2}mv^2

2. Kinetic energy of second fragment: \frac{1}{2}mv^2

3. Kinetic energy of third fragment:

\(\frac{1}{2}(3m)\left(\frac{v\sqrt{2}}{3}\right)^2 = \frac{1}{2}(3m)\left(\frac{2v^2}{9}\right) = \frac{m}{3}v^2\)

Total kinetic energy released:

\(\frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{m}{3}v^2 = mv^2 + \frac{m}{3}v^2 = \frac{3}{3}mv^2 + \frac{1}{3}mv^2 = \frac{4}{3}mv^2\)

Thus, the energy released during the process is \frac{4}{3}mv^{2}.