First slide

Kinetic friction

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 rod be $μ$ , then stopping distance is

Moderate

A

$\frac{p}{2 μmg}$
B

$\frac{p}{2 {μm}^{2} g}$
C

$\frac{p^{2}}{2 μmg}$
D

$\frac{p^{2}}{2 {μm}^{2} g}$

Solution

Let x be the stopping distance. Then using the
$equation v^{2} = u^{2} + 2 ax, we have v^{2} = 0 + 2 (F / m) x (where F / m = a, retardation) ∴ x = \frac{\frac{1}{2} {mv}^{2}}{F} = \frac{(p^{2} / 2 m)}{μmg} = \frac{p^{2}}{2 {μm}^{2} g}$

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