Question

A vehicle of mass m is moving on a rough horizontal road with momentum p. If the coefficient of friction between the tyres and the road be u, then the stopping distance is given by..??

Answer 1

Initial velocity `u=p/m`
Final velocity `v=0`, (as the vehicle must stop)
Force of friction `=mumg`
(where `g` is acceleration due to gravity.)
Acceleration due to friction `= -(mumg)/m=-mug`
(`-`ve sign shows that it is retardation.)

Using the kinematic expression

`v^2-u^2=2as`

and inserting various values we get stopping distance `s`

`(0)^2-p^2/m^2=2(-mug)s`
`=>s=p^2/(2m^2mug)`