If a body having mass m1 (2 kg) moving with 5 m/s approaches another mass m2 (3 kg) 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:
Correct answer: A. 4 m/s
- A. 4 m/s
- B. 2 m/s
- C. 6 m/s
- D. Depends on masses
Explanation
For a perfectly elastic collision the relative speed of separation equals the relative speed of approach, whatever the masses, which follows from conserving both momentum and kinetic energy. So the 4 m per second closing speed becomes a 4 m per second separating speed. In an inelastic collision the separation speed would be smaller, and in a perfectly inelastic one it would be zero.
This question appeared on the UHS MDCAT 2025 paper, which you can sit online with every answer explained.
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About Collisions
Collisions involve the conservation of linear momentum when bodies interact for a short time. The topic covers impulse, elastic and inelastic collisions, kinetic energy changes, and the coefficient of restitution, with calculations usually based on one dimensional motion. Momentum remains conserved in isolated collisions, but kinetic energy is conserved only in elastic collisions.
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