Two blocks A and B of same material and of masses m and 2m moving with same kinetic energy on a surface enter in a region of identical rough surface. Choose the correct statement from the following options.

Options

A. A will move more distance before stopping
B. B will move more distance before stopping
C. Both will move same distance before stopping
D. Both will never stop

Answer

To solve the problem, we need to analyze the motion of two blocks A and B, which have different masses but the same kinetic energy. Let's denote the mass of block A as \( m \) and the mass of block B as \( 2m \).

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
Both blocks have the same kinetic energy. The kinetic energy \( KE \) is given by the formula:
\[ KE = \frac{1}{2}mv^2 \]
For block A:
\[ KE_A = \frac{1}{2} m V_1^2 \]
For block B:
\[ KE_B = \frac{1}{2} (2m) V_2^2 \]
Since \( KE_A = KE_B \), we have:
\[ \frac{1}{2} m V_1^2 = \frac{1}{2} (2m) V_2^2 \]

2. **Canceling Common Terms**:
We can cancel \( \frac{1}{2} \) and \( m \) from both sides:
\[ V_1^2 = 2 V_2^2 \]

3. **Finding the Relationship Between Speeds**:
Rearranging gives:
\[ V_1 = \sqrt{2} V_2 \]

4. **Deceleration Due to Friction**:
The deceleration \( a \) due to friction for both blocks can be expressed as:
\[ a = \mu g \]
where \( \mu \) is the coefficient of friction and \( g \) is the acceleration due to gravity.

5. **Calculating the Stopping Distance**:
The stopping distance \( S \) for each block can be calculated using the formula:
\[ S = \frac{V^2}{2a} \]
For block A:
\[ S_1 = \frac{V_1^2}{2\mu g} \]
For block B:
\[ S_2 = \frac{V_2^2}{2\mu g} \]

6. **Substituting the Speeds**:
Substitute \( V_1 = \sqrt{2} V_2 \) into the equation for \( S_1 \):
\[ S_1 = \frac{(\sqrt{2} V_2)^2}{2\mu g} = \frac{2 V_2^2}{2\mu g} = \frac{V_2^2}{\mu g} \]
Now, substituting for \( S_2 \):
\[ S_2 = \frac{V_2^2}{2\mu g} \]

7. **Finding the Ratio of Distances**:
Now, we can find the ratio of the distances:
\[ \frac{S_1}{S_2} = \frac{\frac{V_2^2}{\mu g}}{\frac{V_2^2}{2\mu g}} = \frac{2}{1} = 2 \]
This means:
\[ S_1 = 2 S_2 \]

8. **Conclusion**:
Since \( S_1 \) is the distance covered by block A and \( S_2 \) is the distance covered by block B, we conclude that:
\[ \text{Block A will move more distance before stopping compared to Block B.} \]

### Final Answer:
**Block A will move more distance before stopping.**