Consider an one dimensional elastic collision between a given incoming body A and body B, initially at rest. The mass of B in comparison to the mass of A in order that B should move with greatest kinetic energy is

Options

A. `m_(B) gt gt m_(A)`
B. `m_(a) lt lt m_(A)`
C. `m_(B) = m_(A)`
D. can't say anything

Explanation

To solve the problem of determining the mass of body B in comparison to the mass of body A such that body B moves with the greatest kinetic energy after a one-dimensional elastic collision, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We have two bodies, A and B. Body A has mass \( m_A \) and is moving with an initial velocity \( u \). Body B has mass \( m_B \) and is initially at rest (velocity = 0). We need to find the ratio of the masses \( \frac{m_B}{m_A} \) such that body B moves with the greatest kinetic energy after the collision.

2. **Conservation of Momentum**:
In an elastic collision, both momentum and kinetic energy are conserved. The conservation of momentum 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 body A and \( v_2 \) is the final velocity of body B.

3. **Conservation of Kinetic Energy**:
The conservation of kinetic energy can be expressed as:
\[ \frac{1}{2} m_A u^2 + 0 = \frac{1}{2} m_A v_1^2 + \frac{1}{2} m_B v_2^2 \]

4. **Expressing Final Velocities**:
For elastic collisions, we can derive the final velocities using the following formulas:
\[ v_1 = \frac{m_A - m_B}{m_A + m_B} u \]
\[ v_2 = \frac{2 m_A}{m_A + m_B} u \]

5. **Kinetic Energy of Body B**:
The kinetic energy of body B after the collision is given by:
\[ KE_B = \frac{1}{2} m_B v_2^2 \]
Substituting \( v_2 \):
\[ KE_B = \frac{1}{2} m_B \left( \frac{2 m_A}{m_A + m_B} u \right)^2 \]
Simplifying this, we get:
\[ KE_B = \frac{2 m_B m_A^2 u^2}{(m_A + m_B)^2} \]

6. **Maximizing Kinetic Energy**:
To maximize \( KE_B \), we can differentiate it with respect to \( m_B \) and set the derivative to zero. However, we can also use a physical insight: for maximum kinetic energy transfer, body A should come to rest after the collision. This occurs when:
\[ v_1 = 0 \implies \frac{m_A - m_B}{m_A + m_B} u = 0 \]
This implies \( m_A = m_B \).

7. **Conclusion**:
Therefore, for body B to move with the greatest kinetic energy after the collision, the mass of B must be equal to the mass of A:
\[ \frac{m_B}{m_A} = 1 \]