Question

A car of mass m is moving with momentum $p$. If $ \mu $ be the coefficient of friction between the tyres and the road, what will be stopping distance due to friction alone?

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

Answer 1

Answer: (C)

If the car stops after covering distance $s$, then
$F \times s=\frac{p^{2}}{2 m}$
where $F=$ friction
Now $F=\mu m g$
So, $\mu m g \times s=\frac{p^{2}}{2 m}$
$\Rightarrow s=\frac{p^{2}}{2 \mu m^{2} g}$