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Acceleration

Two blocks ar...

Question

Two blocks are connected by a spring of spring constant k. The mass $m_{2}$ is given a sharp impulse so that it acquires a velocity $v_{0}$ towards right. The maximum elongation in the spring will be.

v_{0} \sqrt{\frac{2 m_{1} m_{2}}{(m_{1} + m_{2}) k}}

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v_{0} \sqrt{\frac{m_{1} m_{2}}{(m_{1} - m_{2}) k}}

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v_{0} \sqrt{\frac{m_{1} m_{2}}{(m_{1} + m_{2}) k}}

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v_{0} \sqrt{\frac{3 m_{1} m_{2}}{(m_{1} + m_{2}) k}}

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Solution

The correct option is C $v_{0} \sqrt{\frac{m_{1} m_{2}}{(m_{1} + m_{2}) k}}$
$v_{C M} = \frac{m_{1} x_{0} + m_{2} x v_{0}}{m_{1} + m_{2}}$
$= \frac{m_{2}}{m_{1} + m_{2}} v_{0}$
$K E_{(i)} = \frac{1}{2} m_{2} v_{0}$
$K E_{f} = \frac{1}{2} (m_{1} + m_{2}) v_{C M}^{2}$
$= \frac{1}{2} (m_{1} + m_{2}) {(\frac{m_{2}}{m_{1} + m_{2}} v_{0})}^{2}$
$= \frac{1}{2} \frac{m_{2}^{2}}{m_{1} + m_{2}} v_{0}^{2}$ .
From work energy theorem
$W_{s p r i n g} = Δ K E = k_{f} - k_{i}$
$\Rightarrow 0 - \frac{1}{2} k x^{2} = \frac{1}{2} \frac{m_{2}^{2}}{m_{1} + m_{2}} v_{0}^{2} - \frac{1}{2} m_{2} v_{0}^{2}$
$= \frac{1}{2} m_{2} v_{0}^{2} [\frac{m_{2}}{m_{1} + m_{2}} - 1]$
$\Rightarrow - k x^{2} = m_{2} v_{0}^{2} \cdot \frac{(m_{2} - m_{1} - m_{2})}{m_{1} + m_{2}}$
$\Rightarrow x^{2} = \frac{m_{1} m_{2}}{(m_{1} + m_{2}) k} v_{0}^{2}$
$∴ x = \sqrt{\frac{m_{1} m_{2}}{(m_{1} + m_{2}} k} \cdot v_{0}$ .

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Figure

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Two blocks are connected by a spring of spring constant k. The mass m2 is given a sharp impulse so that it acquires a velocity v0 towards right. The maximum elongation in the spring will be.

Two blocks are connected by a spring of spring constant k. The mass $m_{2}$ is given a sharp impulse so that it acquires a velocity $v_{0}$ towards right. The maximum elongation in the spring will be.