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Relative Speed and Momentum Conservation in Elastic Collision

STATEMENT-1
In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.
because
STATEMENT-2
In an elastic collision, the linear momentum of the system is conserved.

Official Notice: Marks Awarded to All

This question was dropped / full marks were awarded to all candidates in the official answer key by the exam conducting body due to an ambiguity or error in the question or options.

Options

A

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C

Statement-1 is True, Statement-2 is False

D

Statement-1 is False, Statement-2 is True

Step-by-Step Solution

To determine the correct option, let us analyze each statement:

1. Analysis of Statement-1:

  • Case I: One-Dimensional (Head-On) Elastic Collision
    In a head-on elastic collision between two bodies of masses m1m_1 and m2m_2 with initial velocities u1u_1 and u2u_2, and final velocities v1v_1 and v2v_2:

    • By conservation of linear momentum:
      m1u1+m2u2=m1v1+m2v2  ⟹  m1(u1−v1)=m2(v2−u2)m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \implies m_1(u_1 - v_1) = m_2(v_2 - u_2)
    • By conservation of kinetic energy:
      12m1u12+12m2u22=12m1v12+12m2v22  ⟹  m1(u12−v12)=m2(v22−u22)\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \implies m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)
      Dividing the kinetic energy equation by the momentum equation gives:
      u1+v1=v2+u2  ⟹  ∣v2−v1∣=∣u1−u2∣u_1 + v_1 = v_2 + u_2 \implies |v_2 - v_1| = |u_1 - u_2|
      Thus, the relative speed after the collision equals the relative speed before the collision (e=1e = 1).
      Under this interpretation, Statement-1 is True.
  • Case II: General (Oblique / Multi-Dimensional) Elastic Collision
    In an oblique collision, the coefficient of restitution e=1e = 1 applies only to the velocity components along the line of impact (common normal), while the tangential velocity components remain unchanged. Consequently, the total relative speed after the collision is generally not equal to the total relative speed before the collision, even though the total kinetic energy of the system is conserved:
    ∣v⃗1−v⃗2∣≠∣u⃗1−u⃗2∣(in general)|\vec{v}_1 - \vec{v}_2| \neq |\vec{u}_1 - \vec{u}_2| \quad \text{(in general)}
    Under this general interpretation, Statement-1 is False.


2. Analysis of Statement-2:

In any collision (elastic or inelastic), in the absence of external net forces on the two-body system, the total linear momentum of the system is always conserved:
P⃗initial=P⃗final\vec{P}_{\text{initial}} = \vec{P}_{\text{final}}
Thus, Statement-2 is True.


3. Evaluation of Explanation (Relation between Statements):

Even if we consider Statement-1 to be true (head-on collision):
Linear momentum is conserved in all types of collisions (elastic, inelastic, and perfectly inelastic). The equality of relative speeds arises fundamentally from the conservation of kinetic energy in an elastic collision, not from momentum conservation alone.
Therefore, Statement-2 is NOT a sufficient or correct explanation for Statement-1.


Conclusion:

  • If the collision is considered to be one-dimensional (head-on), Statement-1 is True, Statement-2 is True, and Statement-2 is NOT the correct explanation for Statement-1, leading to (B).
  • If the collision is considered in a general multi-dimensional sense (oblique collision), Statement-1 is False and Statement-2 is True, leading to (D).

Hence, both (B) and (D) are valid answers.