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 `mu` is then the stopping distance is .

Options

A. `(P)/(2muMg)`
B. `(P^(2))/(2muMg)`
C. `(P^(2))/(2muM^(2)g)`
D. `(P^(2))/(2muM^(2)g)`

Explanation

To find the stopping distance of a vehicle of mass \( M \) moving on a rough horizontal road with a momentum \( P \) and a coefficient of friction \( \mu \), we can follow these steps:

### Step 1: Relate momentum to velocity
The momentum \( P \) of the vehicle is given by the formula:
\[ P = M \cdot v \]
where \( v \) is the velocity of the vehicle. We can express the velocity in terms of momentum:
\[ v = \frac{P}{M} \]

### Step 2: Determine the force of friction
The frictional force \( F_f \) acting on the vehicle, which is responsible for stopping it, can be calculated using the coefficient of friction \( \mu \) and the normal force. On a horizontal road, the normal force is equal to the weight of the vehicle, which is \( M \cdot g \) (where \( g \) is the acceleration due to gravity). Thus, the frictional force is:
\[ F_f = \mu \cdot M \cdot g \]

### Step 3: Apply Newton's second law
According to Newton's second law, the net force acting on the vehicle is equal to the mass of the vehicle multiplied by its acceleration \( a \):
\[ F_f = M \cdot a \]
Substituting the expression for frictional force, we have:
\[ \mu \cdot M \cdot g = M \cdot a \]
Dividing both sides by \( M \) (assuming \( M \neq 0 \)):
\[ a = \mu \cdot g \]

### Step 4: Use kinematic equations to find stopping distance
The vehicle comes to a stop from an initial velocity \( v \) with a constant deceleration \( a \). We can use the kinematic equation:
\[ v^2 = u^2 + 2a s \]
where:
- \( v = 0 \) (final velocity, since the vehicle stops),
- \( u = \frac{P}{M} \) (initial velocity),
- \( a = -\mu g \) (deceleration, negative because it is opposite to the direction of motion),
- \( s \) is the stopping distance.

Substituting the values into the equation:
\[ 0 = \left(\frac{P}{M}\right)^2 + 2(-\mu g)s \]
This simplifies to:
\[ 0 = \frac{P^2}{M^2} - 2\mu gs \]
Rearranging gives:
\[ 2\mu gs = \frac{P^2}{M^2} \]
Solving for \( s \):
\[ s = \frac{P^2}{2\mu g M^2} \]

### Final Answer
The stopping distance \( s \) is given by:
\[ s = \frac{P^2}{2\mu g M^2} \]

---