Two block masses $m_{1}$ and $m_{2}$ are connected with the help of a spring of spring constant $k$ initially the spring in its natural length as shown. A sharp impulse is given to mass $m_{2}$ so that it acquires a velocity $v_{0}$ towards right. If the system is kept on smooth floor then find :
(a) the velocity of the centre of mass
(b) the maximum elongation that the spring will suffer?

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Solution

a) velocity of centre of mas $= \frac{(m_{2} v_{0} + m_{1} \times 0)}{m_{1} + m_{2}} = \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

b) The spring will attain maximum elongation when both velocity of two blocks will attain the velocity of centre of mass.

$x \Rightarrow$ maximum elongation of spring change of k.E $-$ p.E stored in spring

$\frac{1}{2} m_{2} v_{0}^{2} - \frac{1}{2} (m_{1} + m_{2}) (m_{2} v_{0} / (m_{1} + m_{2}))^{2} = \frac{1}{2} k x^{2}$

$m_{2} v_{0}^{2} [1 - (m_{2} / (m_{1} + m_{2})] = k x^{2}$

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

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