Two block of mass $m_{1}$ and $m_{2}$ are connected with the help of a 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

Mass of first block $= m_{1}$
Mass of second block $= m_{2}$
Velocity given to mass $m_{2} = v_{0}$
(a) So the velocity of centre of mass
$= \frac{m_{2} v_{0} + m_{1} \times 0}{m_{1} + m_{2}}$
$= \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

(b) Let $x$ be the maximum elongation in the spring.
So, change in kinetic energy = potential energy stored in spring
$\frac{1}{2} m_{2} {v_{0}}^{2} - \frac{1}{2} (m_{1} + m_{2}) {(\frac{m_{2} v_{0}}{m_{1} + m_{2}})}^{2} = \frac{1}{2} k x^{2}$
$x = \sqrt{\frac{1}{k} (m_{2} {v_{0^{2}}}^{2} - \frac{m_{2} v_{0}}{m_{1} + m_{2}})}$

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Figure

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