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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)
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Verified Answer
The correct answer is:
$\frac{\mathrm{P}^{2}}{2 \mu \mathrm{gM}^{2}}$
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 $s$
$$
\begin{array}{l}
(0)^{2}-\frac{\mathrm{p}^{2}}{\mathrm{~m}^{2}}=2(-\mu \mathrm{g}) \mathrm{s} \\
\Rightarrow \mathrm{s}=\frac{\mathrm{p}^{2}}{2 \mu^{2} \mu \mathrm{g}}
\end{array}
$$
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 $s$
$$
\begin{array}{l}
(0)^{2}-\frac{\mathrm{p}^{2}}{\mathrm{~m}^{2}}=2(-\mu \mathrm{g}) \mathrm{s} \\
\Rightarrow \mathrm{s}=\frac{\mathrm{p}^{2}}{2 \mu^{2} \mu \mathrm{g}}
\end{array}
$$
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