Consider the system shown below, with two equal masses `m` and a spring with spring constant `K`. The coefficient of friction between the left mass and horizontal table is `mu=1//4`, and the pulley is frictionless. The spring connecting both the blocks is massless and inelastic. The system is held with the spring at its unstretched length and then released.

The extension in spring when the masses come to momentary rest for the first time is

Options

A. `(2mg)/(3K)`
B. `(mg)/(2K)`
C. `(mg)/(K)`
D. `(1mg)/(3K)`

Answer

C

Explanation

AT the instant is cut, let the extension in spring be `x_(0)`. The maximum compression `x` will occur for spring when left block comes to rest first time after the string is cut.
`:. ` From work energy Theorem `DeltaW =0`
`=(1)/(2)kx_(0)^(2)-(1)/(2)kx^(2)-mu mg(x+x_(0))=0`
`x_(0)=(3mg)/(2k)` and `mu=(1)/(4)`
solving we get `x=(mg)/(k)`