Question:
Two blocks each of mass \(m\) are connected to a spring of spring constant \(K\). If both are given velocity \(V\) in opposite directions as shown in the figure, then the maximum elongation of the spring is
center
Two blocks each of mass \(m\) are connected to a spring of spring constant \(K\). If both are given velocity \(V\) in opposite directions as shown in the figure, then the maximum elongation of the spring is
center
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At maximum compression or elongation of a spring, relative velocity becomes zero and all kinetic energy converts into spring potential energy.
Updated On: Jun 17, 2026
- \( \sqrt{\dfrac{mV^2}{K}} \)
- \( \sqrt{\dfrac{2mV^2}{K}} \)
- \( \sqrt{\dfrac{mV^2}{2K}} \)
- \( \sqrt{\dfrac{mV^2}{4K}} \)
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The Correct Option is B
Solution and Explanation
Concept:
When two identical masses move in opposite directions, the spring stores all the kinetic energy at maximum elongation.
At maximum extension:
\[
\text{Initial K.E.}=\text{Spring Potential Energy}
\]
Step 1: Calculate total initial kinetic energy. Each block has kinetic energy: \[ \frac12 mV^2 \] For two blocks: \[ 2\times \frac12 mV^2 \] \[ =mV^2 \]
Step 2: Use energy conservation. At maximum elongation \(x\): \[ \frac12 Kx^2=mV^2 \] Multiply both sides by 2: \[ Kx^2=2mV^2 \] \[ x^2=\frac{2mV^2}{K} \] \[ x=\sqrt{\frac{2mV^2}{K}} \] Hence, \[ \boxed{\sqrt{\frac{2mV^2}{K}}} \]
Step 1: Calculate total initial kinetic energy. Each block has kinetic energy: \[ \frac12 mV^2 \] For two blocks: \[ 2\times \frac12 mV^2 \] \[ =mV^2 \]
Step 2: Use energy conservation. At maximum elongation \(x\): \[ \frac12 Kx^2=mV^2 \] Multiply both sides by 2: \[ Kx^2=2mV^2 \] \[ x^2=\frac{2mV^2}{K} \] \[ x=\sqrt{\frac{2mV^2}{K}} \] Hence, \[ \boxed{\sqrt{\frac{2mV^2}{K}}} \]
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