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/(2mu mg)`
B. `(p^2)/(2mu mg)`
C. `p/(2mu m^2 g)`
D. `p^2/(2mu m^2 g)`

Answer

To find the stopping distance of a vehicle moving on a rough horizontal road with a given 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:
\[ P = M \cdot u \]
where \( M \) is the mass of the vehicle and \( u \) is its initial velocity. From this, we can express the initial velocity \( u \):
\[ u = \frac{P}{M} \]
This is our **Equation 1**.

### Step 2: Determine the acceleration due to friction
The frictional force \( F \) that stops the vehicle is given by:
\[ F = \mu \cdot N \]
where \( N \) is the normal force. On a horizontal road, the normal force \( N \) is equal to the weight of the vehicle, \( N = M \cdot g \). Therefore, the frictional force becomes:
\[ F = \mu \cdot M \cdot g \]

According to Newton's second law, the net force acting on the vehicle is equal to the mass times its acceleration \( a \):
\[ F = M \cdot a \]
Setting the frictional force equal to the mass times acceleration gives:
\[ \mu \cdot M \cdot g = M \cdot (-a) \]
Here, the negative sign indicates that the acceleration is in the opposite direction of the motion (deceleration). We can simplify this to find the acceleration:
\[ a = -\mu \cdot g \]

### Step 3: Use the equation of motion to find stopping distance
We can use the equation of motion:
\[ v^2 - u^2 = 2as \]
where \( v \) is the final velocity (which is 0 when the vehicle stops), \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the stopping distance. Substituting the known values:
\[ 0 - u^2 = 2(-\mu g)s \]
This simplifies to:
\[ -u^2 = -2\mu g s \]
or:
\[ u^2 = 2\mu g s \]

### Step 4: Substitute \( u \) from Equation 1
Now, substituting \( u = \frac{P}{M} \) into the equation:
\[ \left(\frac{P}{M}\right)^2 = 2\mu g s \]
This can be rearranged to solve for \( s \):
\[ s = \frac{P^2}{2\mu g M^2} \]

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