Question

A vehicle of mass \( M \) is moving with momentum \( P \) on a rough horizontal road. The coefficient of friction between the tyres and the horizontal road is \( \mu \). The stopping distance is
\textit{(g = acceleration due to gravity)}

• A. \( \frac{P^2}{2 \mu g} \)
• B. \( \frac{P^2}{2 \mu g M^2} \)
• C. \( \frac{P^2}{\mu g M^2} \)
• D. \( \frac{P^2}{2 \mu^2} \)

Hint:

Use the work-energy theorem to calculate the stopping distance in problems involving friction and momentum.

Answer 1

The Correct Option is A

Step 1: Work-energy principle.
Using the work-energy principle, the work done by friction is equal to the change in kinetic energy: \[ \text{Work} = F_{\text{friction}} \times d = \text{Change in kinetic energy} \] The force of friction \( F_{\text{friction}} = \mu M g \), and the initial kinetic energy is \( \frac{P^2}{2M} \).
Step 2: Equation for stopping distance.
The stopping distance \( d \) is found by equating the work done to the kinetic energy: \[ \mu M g \times d = \frac{P^2}{2M} \] Solving for \( d \): \[ d = \frac{P^2}{2 \mu g M^2} \]
Step 3: Conclusion.
Thus, the correct answer is (A) \( \frac{P^2}{2 \mu g} \).