Object A strikes the stationary object B with a certain give speed u head-on in an elastic collision. The mass of A is fixed, you may only choose the mass of B appropriately for following cases. Then after the collision:

Options

A. For B to have the greatest speed, choose `m_(B) = m_(A)`
B. For B to have the greatest momentum, choose `m_(B) lt lt m_(A)`
C. For B to have the greatest speed, choose `m_(B) lt lt M_(A)`
D. For the maximum fraction of kinetic energy transfer, choose `m_(B) = m_(A)`

Explanation

To solve the problem of an elastic collision between two objects, A and B, we will analyze the situation step by step.

### Given:
- Object A strikes stationary object B with speed \( u \).
- Mass of A is \( m_A \) (fixed).
- Mass of B is \( m_B \) (to be chosen).

### Step 1: Understand the Collision
In an elastic collision, both momentum and kinetic energy are conserved. We need to derive the equations for the final velocities of both objects after the collision.

### Step 2: Write the Conservation of Momentum Equation
The conservation of momentum before and after the collision can be expressed as:
\[ m_A u + m_B \cdot 0 = m_A v_1 + m_B v_2 \]
Where:
- \( v_1 \) is the final velocity of A.
- \( v_2 \) is the final velocity of B.

### Step 3: Write the Conservation of Kinetic Energy Equation
The conservation of kinetic energy can be expressed as:
\[ \frac{1}{2} m_A u^2 + \frac{1}{2} m_B \cdot 0 = \frac{1}{2} m_A v_1^2 + \frac{1}{2} m_B v_2^2 \]

### Step 4: Simplify the Equations
From the momentum equation, we can simplify:
\[ m_A u = m_A v_1 + m_B v_2 \quad \text{(1)} \]
From the kinetic energy equation, we can simplify:
\[ m_A u^2 = m_A v_1^2 + m_B v_2^2 \quad \text{(2)} \]

### Step 5: Use the Formulas for Final Velocities
For a head-on elastic collision, the final velocities can be derived using the formulas:
\[ v_1 = \frac{m_A - m_B}{m_A + m_B} u \]
\[ v_2 = \frac{2 m_A}{m_A + m_B} u \]

### Step 6: Analyze Conditions for Maximum Speed and Kinetic Energy Transfer
1. **For B to have the greatest speed**:
- We want \( m_B \) to be much less than \( m_A \). This maximizes \( v_2 \):
\[ v_2 \approx 2u \quad \text{(if \( m_B \ll m_A \))} \]

2. **For maximum kinetic energy transfer**:
- Set \( m_B = m_A \). In this case, A will come to rest, and B will move with speed \( u \):
\[ v_1 = 0 \quad \text{and} \quad v_2 = u \]

3. **For B to have the greatest momentum**:
- Again, \( m_B \) should be less than \( m_A \). Since momentum \( p = m v \), if \( v_2 \) is maximized, \( p_B \) is also maximized.

### Conclusion
- For B to have the greatest speed, choose \( m_B \ll m_A \).
- For maximum kinetic energy transfer, choose \( m_B = m_A \).
- For B to have the greatest momentum, choose \( m_B \ll m_A \).

### Final Answer
- The correct choices for the mass of B depend on the specific conditions outlined above.