Two blocks A and B having different kinetic energies `K_(A)` and `K_(B)(gt K_(A))` are released on rough horizontal ground. Coefficient of friction for both of them is same. Then

Options

A. momentum of B is greater than momentum of A
B. more work has to be done by friction to stop B
C. B will travel more distance before stopping
D. from the given data we can not compare the distance travelled by them before stopping

Answer

To solve the problem step by step, we need to analyze the information given and apply the concepts of kinetic energy, friction, and motion.

### Step 1: Understand the given information
We have two blocks, A and B, with kinetic energies \( K_A \) and \( K_B \) respectively, where \( K_B > K_A \). Both blocks are placed on a rough horizontal surface with the same coefficient of friction.

### Step 2: Analyze the kinetic energy
The kinetic energy of an object is given by the formula:
\[ K = \frac{1}{2} m v^2 \]
where \( m \) is the mass and \( v \) is the velocity of the object. Since \( K_B > K_A \), we can infer that block B has either a greater mass or a greater velocity (or both) compared to block A.

### Step 3: Relate kinetic energy to momentum
The momentum \( P \) of an object is given by:
\[ P = m v \]
From the kinetic energy expressions, we can express momentum in terms of kinetic energy:
\[ K_A = \frac{P_A^2}{2 m_A} \quad \text{and} \quad K_B = \frac{P_B^2}{2 m_B} \]
This implies:
\[ P_A = \sqrt{2 K_A m_A} \quad \text{and} \quad P_B = \sqrt{2 K_B m_B} \]

### Step 4: Compare the momenta
Since we do not have a direct relation between the masses \( m_A \) and \( m_B \), we cannot definitively conclude that \( P_B > P_A \) just because \( K_B > K_A \). Therefore, we cannot say that the momentum of block B is greater than that of block A.

### Step 5: Work done by friction
The work done by friction to stop an object is given by:
\[ W = f \cdot d \]
where \( f \) is the frictional force and \( d \) is the distance traveled before stopping. The frictional force \( f \) can be expressed as:
\[ f = \mu m g \]
where \( \mu \) is the coefficient of friction and \( g \) is the acceleration due to gravity. Since both blocks have the same coefficient of friction, the work done by friction to stop block B will indeed be greater than that for block A, as block B has greater kinetic energy.

### Step 6: Distance traveled before stopping
The distance \( d \) traveled before stopping can be calculated using the work-energy principle. The work done by friction is equal to the change in kinetic energy:
\[ W = K - 0 = K \]
Thus, for block B, the distance it travels before stopping can be expressed as:
\[ K_B = \mu m_B g d_B \]
And for block A:
\[ K_A = \mu m_A g d_A \]
Since \( K_B > K_A \), it follows that the work done by friction on block B is greater, indicating that it will travel a longer distance before stopping, but we cannot definitively compare the distances without knowing the masses.

### Conclusion
- We cannot determine the relationship between the momenta of blocks A and B due to the lack of information about their masses.
- More work is done by friction to stop block B since it has greater kinetic energy.
- We cannot definitively conclude which block travels further before stopping without knowing the masses.