A block of mass m is kept on a smooth surface. It is also in contact with a thin plank of mass M=2m, one end of which is clamped to a wall as shown in the figure. The length of the plank is L. The block is given an initial velocity `v_(0)`, towards right and initially the block is at a distance `x_(0)` from hinge. Coefficient of friction between the plank and block is `mu`.
Assume that througout the motion block remains, below the rod, then at what position from the fixed point it will stop?
(GIven `beta=(v_(0)^(2))/(2mugL))`

Options

A. `x_(0)e^(2beta)`
B. `x_(0)e^(beta)`
C. `x_(0)e^(-2beta)`
D. `x_(0)e^(-beta)`

Answer

B

Explanation

`a=-mu(gL)/(2xm)=v(dv)/(dx)`