A bullet of mass m hits a block of mass M. The transfer of energy is maximum when

Options

A. `m gt gt M `
B. `M gt gt m`
C. `M =2m`
D. `M=m`

Answer

To solve the problem of when the transfer of energy is maximum when a bullet of mass \( m \) hits a block of mass \( M \), we can analyze the situation using the principles of conservation of momentum and kinetic energy.

### Step-by-Step Solution:

1. **Understand the Scenario**:
- A bullet of mass \( m \) is moving with an initial velocity \( v \) and strikes a block of mass \( M \). We need to determine the condition under which the transfer of kinetic energy from the bullet to the block is maximized.

2. **Conservation of Momentum**:
- According to the law of conservation of momentum, the total momentum before the collision must equal the total momentum after the collision.
- The equation can be expressed as:
\[ mv = mv_1 + MV_f \]
where \( v_1 \) is the velocity of the bullet after the collision, and \( V_f \) is the velocity of the block after the collision.

3. **Kinetic Energy Before Collision**:
- The initial kinetic energy of the bullet is given by:
\[ KE_{initial} = \frac{1}{2} mv^2 \]

4. **Kinetic Energy After Collision**:
- The kinetic energy after the collision can be expressed as:
\[ KE_{final} = \frac{1}{2} mv_1^2 + \frac{1}{2} MV_f^2 \]

5. **Condition for Maximum Energy Transfer**:
- The maximum transfer of energy occurs when the bullet comes to rest after the collision, meaning \( v_1 = 0 \). Under this condition, all the kinetic energy of the bullet is transferred to the block.
- Therefore, the kinetic energy transferred to the block is:
\[ KE_{transferred} = \frac{1}{2} MV_f^2 \]

6. **Setting Up the Equation**:
- From the conservation of momentum, if the bullet comes to rest, we have:
\[ mv = MV_f \]
- Rearranging gives:
\[ V_f = \frac{mv}{M} \]

7. **Substituting into Kinetic Energy**:
- Substitute \( V_f \) into the kinetic energy equation:
\[ KE_{transferred} = \frac{1}{2} M \left(\frac{mv}{M}\right)^2 = \frac{1}{2} \frac{m^2 v^2}{M} \]

8. **Maximizing Energy Transfer**:
- To maximize the transferred energy, we need to consider the ratio of the masses. The maximum energy transfer occurs when the mass of the block \( M \) is equal to the mass of the bullet \( m \):
\[ M = m \]

### Conclusion:
The transfer of energy is maximum when the mass of the block is equal to the mass of the bullet, i.e., \( M = m \).