Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are m, 2m and 2m respectively. A moves towards B with a speed of 9 m/s and makes an elastic collision with it. Thereafter B makes a completely inelastic collision with C. All motions occur along same straight line. The final speed of C is :

Solution

Physics sir · 3 steps

Do it in two collisions: an elastic one hands B its speed, then B and C stick and share it.

1. A hits B elastically. B is at rest, so after an elastic collision B moves off with v_B = 2m_A/(m_A + m_B) u. Here m_A = m, m_B = 2m and u = 9 m s⁻¹.

   v_B = 2m/(m + 2m) × 9 = 6 m s⁻¹

2. B sticks to C. In a completely inelastic collision B and C move together, so only momentum is conserved: 2m × 6 = (2m + 2m) v. This gives v = 12m/4m = 3 m s⁻¹.
3. So. C finally moves at 3 m s⁻¹, which is option (d).

The idea to keep. In a chain of collisions, solve one collision at a time. Elastic means momentum plus kinetic energy, so use the standard result. Perfectly inelastic means the bodies share one common velocity.

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