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A vehicle of mass m is moving on a rough horizontal road with momentum P. If the co-efficient of friction between the tyres and the road be $μ$ , then the stopping distance is,

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$\frac{p^{2}}{2 μ m g}$

$\frac{p^{2}}{2 μ m^{2} g}$

$\frac{p}{2 μ m^{2} g}$

$\frac{p}{2 μ m g}$

answer is C.

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Detailed Solution

The body will stop when the work done by the frictional force will be equal to the kinetic energy of the body

$W_{s} = f . s; K . E = \frac{P^{2}}{2 m}$

$\Rightarrow s = \frac{P^{2}}{2 {μm}^{2} g}$

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