Question:
Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides with A. The maximum compression in the spring is
Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides with A. The maximum compression in the spring is
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Assume the collision of C with A is elastic: equal masses exchange velocities, so A moves with v. Then conserve momentum and energy for the A-B-spring system at maximum compression.
Updated On: Nov 4, 2026
- \(\sqrt{\frac{mv}{K}}\)
- \(\sqrt{\frac{m}{2K}}\)
- \(\sqrt{\frac{mv}{2K}}\)
- \(v\sqrt{\frac{m}{2K}}\)
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The Correct Option is D
Solution and Explanation
C and A have equal masses and the collision is taken as elastic and instantaneous, so velocities are exchanged: C stops and A moves with speed v.
Maximum compression occurs when A and B move with the same velocity V. Momentum: \(mv = 2mV\), so \(V = \frac{v}{2}\).
Energy: \(\frac{1}{2}mv^2 = \frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 + \frac{1}{2}Kx^2\).
\(\frac{1}{2}Kx^2 = \frac{1}{2}mv^2 - \frac{1}{4}mv^2 = \frac{1}{4}mv^2\), so \(x^2 = \frac{mv^2}{2K}\).
\(x = v\sqrt{\frac{m}{2K}}\)
Answer: (4) \(v\sqrt{\frac{m}{2K}}\)
Maximum compression occurs when A and B move with the same velocity V. Momentum: \(mv = 2mV\), so \(V = \frac{v}{2}\).
Energy: \(\frac{1}{2}mv^2 = \frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 + \frac{1}{2}Kx^2\).
\(\frac{1}{2}Kx^2 = \frac{1}{2}mv^2 - \frac{1}{4}mv^2 = \frac{1}{4}mv^2\), so \(x^2 = \frac{mv^2}{2K}\).
\(x = v\sqrt{\frac{m}{2K}}\)
Answer: (4) \(v\sqrt{\frac{m}{2K}}\)
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