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There is a long narrow and smooth groove in a horizontal table. Two identical blocks A and B each of mass m are placed inside the groove at some separation. An ideal spring is fixed to A as shown. Block A is given a velocity u to the right and it interacts with B through the spring of force constant K.
During their course of interaction what is the minimum kinetic energy of the system?

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$\frac{1}{8} m u^{2}$

$\frac{1}{16} m u^{2}$

$\frac{1}{9} m u^{2}$

$\frac{1}{4} m u^{2}$

answer is D.

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Detailed Solution

It is just an elastic collision between two blocks of equal mass.
Kinetic energy of the system is minimum when compression produced in the spring is maximum and at that instant the blocks will move with a common velocity V .

Conservinng momentum ,2mV=mu $\Rightarrow V = \frac{u}{2}$

So minimum kinetic energy = $\frac{1}{2} . 2 m . {(\frac{u}{2})}^{2} = \frac{` 1}{4} m u^{2}$

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