Question

50.
A vehicle of mass M is moving on a rough horizontal road with a momentum P. If the coefficient of
friction between the tyres and the road is x then the stopping distance is


1. p/ 2xMg
2. p2/2xMg
3. p2/2xM2g
4.p/2xM2g​

Answer 1

Answer:

Given:

Mass of car = M , moving with momentum P ,

Coefficient of Friction between tyres and road = x

To find:

Stopping distance

Concept:

The road - Tyre interface produces friction . It's a type of opposing , dissipational force against the direction the Velocity vector.

Calculation:

Frictional Force :

$\sf{friction = \mu \times N \: }$

$\sf{ = > friction = x \times (mg) }$

$\sf{ = > a_{f} \: = xg }$

Now , we know that :

$\sf{momentum = mass \times velocity}$

$\sf{ = > P = mu}$

$\sf{ = > u = \dfrac{P}{M}}$

Now applying , equation of motion :

$\sf{ {v}^{2} = {u}^{2} + 2as}$

$\sf{= > {0}^{2} = { \bigg( \dfrac{P}{M} \bigg)}^{2} - 2(xg)s}$

$\sf{= > s = \dfrac{ {P}^{2} }{2 {M}^{2} xg}}$

So final answer :

$\boxed{ \sf{ \red{ \huge{ \bold{s = \dfrac{ {P}^{2} }{2 {M}^{2} xg}}}}}}$