Two cars A and B have masses `m_(A) and m_(B)` respectively, with `m_(A) gt m_(B).` Both the cars are moving in the same direction with equal kinetic energy. If equal braking force is applied on both, then before coming to rest

Options

A. A will cover a greater distance
B. B will cover a greater distance
C. both will cover the same distance
D. distance covered by them will depend on their respective velocities

Explanation

To solve the problem, we will analyze the situation step by step, considering the properties of kinetic energy, momentum, and the effects of equal braking forces on two cars with different masses.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- We have two cars, A and B, with masses \( m_A \) and \( m_B \) respectively, where \( m_A > m_B \).
- Both cars are moving in the same direction and have equal kinetic energy.

2. **Kinetic Energy Relation:**
- The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- Since both cars have equal kinetic energy, we can write:
\[ KE_A = KE_B \implies \frac{1}{2} m_A v_A^2 = \frac{1}{2} m_B v_B^2 \]
- This simplifies to:
\[ m_A v_A^2 = m_B v_B^2 \]

3. **Momentum Relation:**
- The momentum \( p \) of an object is given by:
\[ p = mv \]
- From the kinetic energy relation, we can express the velocities in terms of momentum:
\[ v_A = \frac{p_A}{m_A}, \quad v_B = \frac{p_B}{m_B} \]
- Substituting these into the kinetic energy equation gives:
\[ m_A \left(\frac{p_A}{m_A}\right)^2 = m_B \left(\frac{p_B}{m_B}\right)^2 \]
- This leads to:
\[ \frac{p_A^2}{m_A} = \frac{p_B^2}{m_B} \]

4. **Comparing the Momentums:**
- Since \( m_A > m_B \) and both kinetic energies are equal, it follows that:
\[ p_A^2 > p_B^2 \implies p_A > p_B \]
- Thus, the momentum of car A is greater than that of car B.

5. **Applying Equal Braking Force:**
- When equal braking forces are applied to both cars, the change in momentum is related to the time taken to stop. The force \( F \) is given by:
\[ F = \frac{dp}{dt} \]
- Since the forces are equal, we have:
\[ \frac{dp_A}{dt} = \frac{dp_B}{dt} \]

6. **Change in Momentum:**
- The change in momentum for both cars is:
\[ \Delta p_A = p_A - 0 = p_A, \quad \Delta p_B = p_B - 0 = p_B \]
- Since \( p_A > p_B \), it follows that:
\[ \Delta p_A > \Delta p_B \]

7. **Time to Stop:**
- From the relation \( F = \frac{dp}{dt} \), we can express the time taken to stop:
\[ \Delta t_A = \frac{\Delta p_A}{F}, \quad \Delta t_B = \frac{\Delta p_B}{F} \]
- Since \( \Delta p_A > \Delta p_B \) and \( F \) is constant, it follows that:
\[ \Delta t_A > \Delta t_B \]

8. **Distance Travelled Before Coming to Rest:**
- The distance travelled while stopping can be found using the equation:
\[ d = v_i \Delta t + \frac{1}{2} a (\Delta t)^2 \]
- Since car A has a higher initial momentum, it will travel a greater distance before coming to rest compared to car B.

### Conclusion:
- Car A, with greater mass and momentum, will travel a greater distance before coming to rest than car B, when equal braking forces are applied.