A spring of spring constant k connects two blocks of masses $m_{1}$ and $m_{2}$ as shown in the figure. The block of mass $m_{2}$ is given a sharp impulse so that it acquires a velocity $V_{o}$ towards right. Find the maximum elongation of the spring.

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Solution

x => maximum elongation of spring

Change of Kinetic Energy = Potential stored in spring

$\frac{1}{2} m_{2} {v_{o}}^{2} - \frac{1}{2} (m ₁ + m ₂) \times \frac{m ₂ v ₀^{2}}{(m ₁ + m ₂)} = \frac{K}{x^{2}}$

$m_{2} {v_{0}}^{2} [1 - \frac{m_{2}}{(m ₁ + m ₂)}] = K x^{2}$

$x = \sqrt{\frac{m_{1} m_{2}}{(m_{1} + m_{2})^{2}}} \times v_{o}$

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Figure

Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of constant k (given figure).The block of mass $m_{2}$ is given a sharp impulse so that it acquires a velocity $v_{0}$ towards right.Find (a) the velocity of the centre of mass,(b) the maximum elongation that the spring will suffer.