Question

A stationary body explodes into two parts of masses \( M_1 \) and \( M_2 \). They move in opposite directions with velocities \( v_1 \) and \( v_2 \). The ratio of their kinetic energies is

• A. \( \frac{M_2}{M_1} \)
• B. \( \frac{M_2}{M_1} \times \frac{1}{2} \)
• C. \( \left( \frac{M_1}{M_2} \right)^2 \)
• D. \( \left( \frac{M_2}{M_1} \right)^2 \)

Hint:

In an explosion, the total momentum is conserved. Use this principle to find the relationship between the velocities of the two parts, and then use it to calculate the ratio of their kinetic energies.

Answer 1

The Correct Option is A

Step 1: Using conservation of momentum.
In an explosion, the total momentum before and after the explosion is conserved. Since the body was initially stationary, the total momentum after the explosion must be zero. Thus, the momentum of the two parts must cancel each other out, giving the relationship: \[ M_1 v_1 = M_2 v_2 \] This can be written as: \[ \frac{v_1}{v_2} = \frac{M_2}{M_1} \]
Step 2: Relating kinetic energies.
The kinetic energy of the two parts is given by: \[ K_1 = \frac{1}{2} M_1 v_1^2 \quad \text{and} \quad K_2 = \frac{1}{2} M_2 v_2^2 \] We can express \( v_1 \) and \( v_2 \) in terms of their ratio. From the conservation of momentum, we have: \[ K_1 = \frac{1}{2} M_1 \left( \frac{M_2}{M_1} v_2 \right)^2 = \frac{M_2^2}{2 M_1} v_2^2 \] Similarly, for \( K_2 \): \[ K_2 = \frac{1}{2} M_2 v_2^2 \]
Step 3: Finding the ratio.
The ratio of the kinetic energies is: \[ \frac{K_1}{K_2} = \frac{\frac{M_2^2}{2 M_1} v_2^2}{\frac{1}{2} M_2 v_2^2} = \frac{M_2}{M_1} \] Thus, the correct answer is (A) \( \frac{M_2}{M_1} \).