A body of mass 2 kg moving with a speed of 4 m/s. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial peed. The speed of the two body centre of mass is x/10 m/s. Then the value of x is ___________.

Solution

Physics sir · 3 steps

Find the unknown mass first; after that the centre of mass speed is just total momentum over total mass.

1. Elastic collision. In a head-on elastic collision, e = 1, so the speed of separation equals the speed of approach. The first body slows to 4/4 = 1 m s⁻¹, so v₂ - 1 = 4 - 0, which gives v₂ = 5 m s⁻¹.
2. Momentum gives mass. Momentum is conserved: 2 × 4 = 2 × 1 + m₂ × 5. So 5m₂ = 6 and m₂ = 1.2 kg.

   v_(cm) = (m₁ u₁ + m₂ u₂)/(m₁ + m₂) = (2 × 4 + 1.2 × 0)/(2 + 1.2) = 8/3.2 = 2.5 m s⁻¹

3. So. x/10 = 2.5, so x = 25. No external force acts, so this speed stays the same before, during and after the collision.

The common slip: 2.5. Reporting v_(cm) itself gives 2.5, not 25. The question asks for x, where v_(cm) = x/10.

The idea to keep. For a 1D elastic collision, pair momentum conservation with e = 1 (separation speed equals approach speed). It avoids squaring anything, and v_(cm) never changes in a collision.

Answer from the NTA final answer key. Solution by PrepWiser.