Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$\frac{p^2}{\mu mg}$

$\frac{p^2}{2\mu mg}$

$\frac{p^2}{\mu m^2 g}$

$\frac{p^2}{2\mu m^2 g}$

Step-by-Step Solution

Kinetic Energy and Momentum: The kinetic energy ( $K$ ) of a body can be expressed in terms of its momentum ( $p$ ) and mass ( $m$ ) as: $K = \frac{p^2}{2m}$ [Source 57, 78].
Work Done by Friction: When the vehicle stops, the work done by the frictional force ( $f$ ) opposes the motion and dissipates the kinetic energy. The frictional force on a horizontal road is $f = \mu N = \mu mg$ . The work done over stopping distance $s$ is $W = f \cdot s = \mu mg s$ [Source 64, 80].
Work-Energy Theorem: According to the theorem, the work done by the retarding force equals the change in kinetic energy (magnitude wise): $\mu mg s = K = \frac{p^2}{2m}$
Solving for Stopping Distance ( $s$ ): $s = \frac{p^2}{2m(\mu mg)} = \frac{p^2}{2 \mu m^2 g}$

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