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 $\mu$, then the stopping distance is

• A. $\frac{p}{2\mu mg}$
• B. $\frac{p^2}{2\mu mg}$
• C. $\frac{p}{2\mu m^{2}g}$
• D. $\frac{p^2}{2\mu m^{2}g}$

Answer 1

The correct option is D $\frac{p^2}{2\mu m^{2}g}$

Initial velocity $u=\frac{p}{m}$

Final velocity $v=0$ (as the vehicle must stop)

Force of friction $=\mu mg$

(where $g$ is acceleration due to gravity)

Acceleration due to friction $=-\frac{\mu mg}{m}=-\mu g$

$(-ve$ sign shows that it is retardation $)$

Using the kinematic expression

$v^2=u^2=2as$

and inserting various values we get stopping distance $s$

$(0{)}^2-\frac{p^2}{m^2}=2(-\mu g)s$

$\Rightarrow s=\frac{p^2}{2{\mu }^2\mu g}$