# A spring of mass m lies on a smooth table. One end of it is sclamped to the vertical wall and the other end is attached to a mass m and is placed on a horizontal frictionless surface. If the Block of mass M is given a velocity V towards right, find the KE of the system

I can assume that the mass of the spring and block is same.

Since the velocity of the spring is zero and the velocity of the block(m) is V.

Velocity of the Center of the mass of the system is:

${V}_{cm}=\frac{m\times v+m\times 0}{m+m}=\frac{v}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}K.Eofthesystem=\frac{1}{2}\times Mas{s}_{system}\times {{V}^{2}}_{cm}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 2m\times (\frac{v}{2}{)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{m{v}^{2}}{4}Ans.$

Regards

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