Question

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
A. $\dfrac{P}{{2\mu mg}}$
B. $\dfrac{{{P^2}}}{{2\mu mg}}$
C. $\dfrac{P}{{2\mu {m^2}g}}$
D. $\dfrac{{{P^2}}}{{2\mu {m^2}g}}$

Answer 1

We will use frictional force expression when the vehicle moves, giving us the relationship between the frictional force and its normal. We will also use the equilibrium concept to establish the relationship for the retardation of the given vehicle.

Complete step by step answer:
Given:
The coefficient of friction between the tyres and the road is $\mu$.
The mass of the vehicle is m.

Assume:
The initial velocity of the vehicle is u.
The final velocity of the vehicle is v.

We are required to calculate the value of stopping potential.

Let us write the expression for the vehicle's initial momentum when it is moving with u velocity.
$P = mu$
On rearranging the above expression to get the value of velocity u, we get:
$u = \dfrac{P}{m}$

We know that the vehicle's final velocity is equal to zero because it is coming to the rest position.
$v = 0$

Let us write the expression for the normal friction force between tyres of a vehicle and road.
$N = mg$
Here g is the acceleration due to gravity

We know that the ratio of friction force and its normal gives us the value of the coefficient of friction so we can write:
$\dfrac{F}{N} = \mu$
On substituting $mg$ for N in the above expression, we get:

\dfrac{F}{{mg}} = \mu \\\ F = \mu mg \end{array}$$……(1) We can also write the expression for force from the concept of Newton's second law of motion. $$F' = ma$$……(2) Here a is the acceleration due to frictional force. Using the equilibrium concept, we can equate the negative value of equation (1) with equation (2) to find the acceleration to frictional force. $$\begin{array}{c} ma = - \mu mg\\\ a = - \mu g \end{array}$$ Let us write the expression for the third equation of motion for the given vehicle. $${v^2} = {u^2} + 2as$$ Here s is the distance covered by the vehicle. On substituting $$\dfrac{P}{m}$$ for u, 0 for v and $$ - \mu g$$ for a in the above expression, we get: $$\begin{array}{l} {0^2} = {\left( {\dfrac{P}{m}} \right)^2} + 2\left( { - \mu g} \right)s\\\ s = \dfrac{{{P^2}}}{{2\mu {m^2}g}} \end{array}$$ Therefore, the expression for stopping distance is _$$\dfrac{{{P^2}}}{{2\mu {m^2}g}}$$_ **So, the correct answer is “Option D”.** **Note:** The negative sign in the expression for acceleration due to frictional force indicates that the acceleration is retarding in nature; hence it is termed as retardation. We also know that the vehicle is coming to rest, so we can say that the body is in retarding motion from intuition.