A vehicle of mass 'M' is moving with momentum 'P' on a rough horizontal road. The coefficient of friction between the tyres and the horizontal road is '\(\mu\)'. The stopping distance is (g \(=\) acceleration due to gravity)

Initial velocity \(\mathrm{u}=\frac{\mathrm{p}}{\mathrm{m}}\)
Final velocity \(\mathrm{v}=0\) (as the vehicle must stop)
Force of friction \(=\mu \mathrm{mg}\)
(where \(\mathrm{g}\) is acceleration due to gravity)
Acceleration due to friction \(=-\frac{\mu \mathrm{mg}}{\mathrm{m}}=-\mu \mathrm{g}\)
(-ve sign shows that it is retardation )
Using the kinematic expression
\(\mathrm{v}^{2}=\mathrm{u}^{2}=2 \mathrm{as}\)
and inserting various values we get stopping distance \(s\)
\(\begin{array}{l}(0)^{2}-\frac{\mathrm{p}^{2}}{\mathrm{~m}^{2}}=2(-\mu \mathrm{g}) \mathrm{s} \\\Rightarrow \mathrm{s}=\frac{\mathrm{p}^{2}}{2 \mu^{2} \mu \mathrm{g}}\end{array}\)