A block of mass $m$ moving with a velocity $v_{0}$ collides with a stationary block of mass $M$ at the back of which a spring of spring constant $k$ is attached, as shown in the figure. Select the correct alternatives(s)

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When the spring is in the state of maximum compression the kinetic energy in the centre of mass frame is zero.

The initial kinetic energy of the system in the centre of mass frame is $\frac{1}{4} (\frac{m}{m + M}) {v^{2}}_{0}$

The velocity of centre of mass is $(\frac{m}{m + M}) v_{0}$

The maximum compression in the spring is $v_{0} \sqrt{\frac{m}{(m + M)} \frac{1}{k}}$

answer is A, C, D.

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Detailed Solution

Clearly, the velocity of centre of mass $= (\frac{m}{m + M}) v_{0}$

When the compression in the spring is maximum ,

kinetic energy of the system $K_{c m}$

$= \frac{1}{2} (m + M) v_{c m}^{2} = \frac{1}{2} (m + M) {[(\frac{m}{m + M}) v_{0}]}^{2} = \frac{1}{2} \frac{m^{2} v_{0}^{2}}{m + M}$

The maximum compression $(x_{m})$ in spring is given by

$\frac{1}{2} k x_{m}^{2} = E - K_{c m} o r \frac{1}{2} k x_{m}^{2} = \frac{1}{2} m v_{0}^{2} - \frac{1}{2} \frac{m v_{0}^{2}}{m + M} o r k x = m v_{0}^{2} (1 - \frac{m}{m - M}) = m v_{0}^{2} (\frac{M}{m - M}) = (\frac{m M}{m - M}) v_{0}^{2} \Rightarrow x_{m} = v_{0} \sqrt{\frac{m}{(m + M)} \frac{1}{k}}$