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 br mu, then the stopping distance is:

Options

A. `(P)/(2mumg)`
B. `(P^(2))/(2mumg)`
C. `(P)/(2mum^(2)g)`
D. `(P^(2))/(2mum^(2)g)`

Explanation

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

### Step 1: Understand the relationship between momentum and velocity
The momentum \( P \) of the vehicle is given by the formula:
\[ P = m \cdot u \]
where \( u \) is the initial velocity of the vehicle. From this, we can express the initial velocity \( u \) as:
\[ u = \frac{P}{m} \]

### Step 2: Determine the frictional force acting on the vehicle
The frictional force \( F_f \) that will stop the vehicle is given by:
\[ F_f = \mu \cdot N \]
where \( N \) is the normal force. On a horizontal surface, the normal force \( N \) is equal to the weight of the vehicle, which is \( mg \). Therefore, we have:
\[ F_f = \mu \cdot mg \]

### Step 3: Calculate the acceleration due to friction
The acceleration \( a \) (which will be negative since it is deceleration) can be found using Newton's second law:
\[ F = m \cdot a \]
Thus, we can set the frictional force equal to the mass times the acceleration:
\[ \mu mg = m \cdot a \]
Dividing both sides by \( m \) gives:
\[ a = -\mu g \]
(Note: The negative sign indicates that the acceleration is in the opposite direction of the motion.)

### Step 4: Use the equations of motion to find the stopping distance
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity (0, since the vehicle stops),
- \( u \) is the initial velocity (\( \frac{P}{m} \)),
- \( a \) is the acceleration (\(-\mu g\)),
- \( s \) is the stopping distance.

Substituting the known values into the equation:
\[ 0 = \left(\frac{P}{m}\right)^2 + 2(-\mu g)s \]
Rearranging gives:
\[ \left(\frac{P}{m}\right)^2 = 2\mu gs \]

### Step 5: Solve for the stopping distance \( s \)
Now we can solve for \( s \):
\[ s = \frac{P^2}{2\mu mg} \]

### Final Answer
Thus, the stopping distance \( s \) is:
\[ s = \frac{P^2}{2\mu mg} \]