Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Body 'P' having mass M moving with speed 'u' has head-on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, body 'Q' will have a maximum speed equal to '2u' after collision.
Reason R : During elastic collision, the momentum and kinetic energy are both conserved.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Physics sir · 4 steps

A heavy body barely slows down, and Q must leave it at speed u relative to it, so Q ends up at 2u.

1. Use both laws. Write momentum conservation and kinetic energy conservation for the head-on collision and solve them together. This gives the final speed of the body that was at rest.

   v_Q = 2M/(M+m) u

2. Take m much smaller. When m ≪ M, M + m ≈ M, so v_Q ≈ 2M/Mu = 2u. This is the largest speed Q can get, because the factor 2M/(M+m) grows towards 2 as m shrinks. Assertion A is correct.
3. Check the physics. In an elastic collision the speed of separation equals the speed of approach, which is u. The heavy P keeps almost all of its speed u, so Q moves off at u + u = 2u. Q's kinetic energy 1/2m(2u)² = 2mu² is still tiny next to 1/2Mu², so energy is not violated.
4. So. Reason R is correct, and the 2u result comes straight from using momentum and kinetic energy conservation together. So R is the correct explanation of A, which is option (c).

The idea to keep. In a 1D elastic collision with the target at rest, v₂ = 2m₁/(m₁+m₂)u. A very heavy body hitting a very light one sends it off at almost 2u, never more.

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