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 μ, then the stopping distance is

The correct answer is P²/2μm��g.

We are given a vehicle of mass m moving on a rough horizontal road with momentum P, and the coefficient of friction between the tyres and the road is μ. We need to find the stopping distance.

Step 1: Find the initial velocity from momentum.

We know that linear momentum is defined as:

P=mv

Solving for the initial velocity v:

v=\frac{P}{m}

Step 2: Find the deceleration due to friction.

The only horizontal force acting on the vehicle is the kinetic friction force, which opposes motion. The magnitude of this friction force is:

f_k=\mu mg

Using Newton's second law (F = ma), the deceleration (negative acceleration) is:

a=\frac{f_k}{m}=\frac{\mu }{m}=\mu g

Step 3: Apply the kinematic equation to find stopping distance.

We use the third equation of motion:

v^2=u^2-2as

When the vehicle comes to a stop, the final velocity = 0. Here, u is the initial speed and s is the stopping distance:

0=u^2-2\left(\mu g\right)s

s=\frac{u^2}{2\mu g}

Step 4: Substitute the expression for velocity.

Replace u with \frac{P}{m}:

s=\frac{{\left(\frac{P}{m}\right)}^2}{2\mu g}=\frac{\frac{P^2}{m^2}}{2\mu g}=\frac{P^2}{2\mu m^{2}g}

Result:

The stopping distance is:

s=\frac{P^2}{2\mu m^{2}g}

Notice that the stopping distance is proportional to the square of the momentum and inversely proportional to the square of the mass, the coefficient of friction, and gravitational acceleration.