Question:
Two identical balls P and Q having velocities $0.7 m s^{-1}$ and $-0.4 m s^{-1}$ respectively are colliding in one dimension elastically. The velocities of P and Q after the collision respectively are:
Two identical balls P and Q having velocities $0.7 m s^{-1}$ and $-0.4 m s^{-1}$ respectively are colliding in one dimension elastically. The velocities of P and Q after the collision respectively are:
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Identical masses + elastic collision = perfect velocity swap.
Updated On: Jun 10, 2026
- $0.7 m s^{-1}, -0.4 m s^{-1}$
- $-0.4 m s^{-1}, 0.7 m s^{-1}$
- $+0.4 m s^{-1}, -0.7 m s^{-1}$
- $+0.7 m s^{-1}, -0.7 m s^{-1}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
In an elastic collision between two bodies of identical mass, the bodies exchange their velocities.
Step 2: Analysis
Initial velocities: $v_P = 0.7$, $v_Q = -0.4$. After collision: $v_P' = v_Q = -0.4$, and $v_Q' = v_P = 0.7$.
Step 3: Conclusion
The velocities are swapped: $-0.4 m/s$ and $0.7 m/s$.
Final Answer: (B)
In an elastic collision between two bodies of identical mass, the bodies exchange their velocities.
Step 2: Analysis
Initial velocities: $v_P = 0.7$, $v_Q = -0.4$. After collision: $v_P' = v_Q = -0.4$, and $v_Q' = v_P = 0.7$.
Step 3: Conclusion
The velocities are swapped: $-0.4 m/s$ and $0.7 m/s$.
Final Answer: (B)
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