A body of mass m, having momentum p, is moving on a rough horizontal surface. If it is stopped in a distance x, the coefficient of friction between the body and the surface is given by

Options

A. `mu=(p^(2))/(2gm^(2)x)`
B. `mu=(p^(2))/(2mgx)`
C. `mu=(p)/(2mgx)`
D. `mu=(p)/(2gm^(2)x)`

Answer

A

Explanation

Force of friction = `mu`mg.
Therefore, retardation a = `mu`mg/m= `mu`g.
Also 2ax= `v^(2)or2am^(2)x=m^(2)v^(2)`. But p =mv
Therefore,
But `2am^(2)x=p^(2)`
But a = `mu`g. Therefore, 2`mugm^(2)x=p^(2)`
or `mu=(p^(2))/(2gm^(2)x)`