Question

If a body having mass m1 (2kg), moving with 5 m/s approaches another mass,m2 (3kg) with speed of 1 m/s in same direction, relative speed of approach is 4 m/s. Relative speed of separation after collision will be:

Options

A
4 m/s
B
2 m/s
C
6 m/s
D
depends on masses

Explanation

In the given problem, we have two masses, m1 = 2kg and m2 = 3kg, approaching each other with speeds of 5 m/s and 1 m/s, respectively. The relative speed of approach is calculated as the difference in their speeds, which is 5 m/s - 1 m/s = 4 m/s. In an elastic collision, the principle states that the relative speed of separation after the collision is equal to the relative speed of approach. Hence, the relative speed of separation will also be 4 m/s, making Option A the correct answer. The other options are incorrect due to the principles of momentum and energy conservation in elastic collisions, which do not allow for arbitrary differences in speeds post-collision.

This is the correct answer because, in an elastic collision, the relative speed of separation is equal to the relative speed of approach.

This option is incorrect because it does not follow the principle of conservation of momentum and kinetic energy in elastic collisions.

This option is incorrect as it suggests a higher speed than the initial relative speed of approach, which contradicts the laws governing elastic collisions.

This option is misleading; while masses affect the final velocities, the relative speed of separation remains equal to the relative speed of approach in elastic collisions.