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

Question

A car of mass m is moving with momentum p. If μ be the coefficient of friction between the tyres and the road, what will be stopping distance due to friction alone? (A)  (p2/2μ g)  (B)  (p2/2mμ g)  (C)  (p2/2m2μ g)  (D)  (p2/2mg)

Tardigrade Admin · Accepted Answer

If the car stops after covering distance s, then F × s=(p2/2 m) where F= friction Now F=μ m g So, μ m g × s=(p2/2 m) ⇒ s=(p2/2 μ m2 g)

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

$\frac{p ^{2}}{2 μg}$

$\frac{p ^{2}}{2 m μg}$

$\frac{p ^{2}}{2 m ^{2} μg}$

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

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

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

2μgp2​

2mμgp2​

2m2μgp2​

2mgp2​