A particle of mass `m_1` moving with velocity `v` in a positive direction collides elastically with a mass `m_(2)` moving in opposite direction also at velocity v. If `m_(2)gt gtm_(1)`, then

Options

A. the velocity of `m_(1)` immediately after collision is nearly `3v`
B. the change in momentum of `m_(1)` is nearly `4m_(1)v`
C. the change in kinetic energy of `m_(1)` is nearly `4mv_(2)`
D. all of the above

Explanation

To solve the problem step by step, we will analyze the elastic collision between two particles with masses \( m_1 \) and \( m_2 \) and their respective velocities before and after the collision.

### Step 1: Understand the Initial Conditions
- We have two particles:
- Particle 1 (mass \( m_1 \)) is moving with velocity \( v \) in the positive direction.
- Particle 2 (mass \( m_2 \)) is moving with velocity \( -v \) (opposite direction).
- Given that \( m_2 > m_1 \).

### Step 2: Apply Conservation of Momentum
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. The momentum before the collision can be expressed as:
\[ p_{\text{initial}} = m_1 v + m_2 (-v) = m_1 v - m_2 v = (m_1 - m_2)v \]

Let \( v_1' \) and \( v_2' \) be the velocities of \( m_1 \) and \( m_2 \) after the collision, respectively. The momentum after the collision is:
\[ p_{\text{final}} = m_1 v_1' + m_2 v_2' \]

Setting the initial momentum equal to the final momentum:
\[ (m_1 - m_2)v = m_1 v_1' + m_2 v_2' \]

### Step 3: Apply Conservation of Kinetic Energy
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. The kinetic energy before the collision is:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 = \frac{1}{2} (m_1 + m_2) v^2 \]

The kinetic energy after the collision is:
\[ KE_{\text{final}} = \frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2')^2 \]

Setting the initial kinetic energy equal to the final kinetic energy:
\[ \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2') \]

### Step 4: Solve for Final Velocities
We can use the equations derived from conservation of momentum and kinetic energy to solve for \( v_1' \) and \( v_2' \).

Using the formulas for elastic collisions, we can derive:
\[ v_1' = \frac{(m_1 - m_2)v}{m_1 + m_2} + \frac{2m_2 v}{m_1 + m_2} \]
\[ v_2' = \frac{(m_2 - m_1)(-v)}{m_1 + m_2} + \frac{2m_1 v}{m_1 + m_2} \]

### Step 5: Calculate Change in Momentum and Kinetic Energy
1. **Change in Momentum of \( m_1 \)**:
\[ \Delta p_1 = m_1 v_1' - m_1 v = m_1 (v_1' - v) \]

2. **Change in Kinetic Energy**:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = \left(\frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2')^2\right) - \frac{1}{2} (m_1 + m_2) v^2 \]

### Final Result
After calculating \( v_1' \) and \( v_2' \), we can find the change in momentum and kinetic energy.