A body of mass `2 kg` is projected upward from the surface of the ground at `t = 0` with a velocity of `20 m//s`. One second later a body B, also of mass `2 kg`, is dropped from a height of `20m`. If they collide elastically, then velocities just after collision are :-

Options

A. `v_(A) = 5 m//s` downward, `v_(B) = 5 m//s` upward
B. `v_(A) = 10 m//s` downard, `v_(B) = 10 m//s` upward
C. `v_(A) = 10 m//s` upward, `v_(B) = 10 m//s` downward
D. both move downward with velocity `5 m//s`

Explanation

To solve the problem, we need to analyze the motion of both bodies A and B, find their velocities just before the collision, and then apply the principles of elastic collision to find their velocities just after the collision.

### Step-by-Step Solution:

1. **Identify the Motion of Body A:**
- Body A is projected upwards with an initial velocity \( u_A = 20 \, \text{m/s} \).
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \) acts downwards.
- The height \( h \) covered by body A after time \( t \) seconds can be expressed using the equation of motion:
\[ h = u_A t - \frac{1}{2} g t^2 \]
- Therefore, the height of body A after \( t \) seconds is:
\[ h = 20t - 5t^2 \]

2. **Identify the Motion of Body B:**
- Body B is dropped from a height of \( 20 \, \text{m} \) with an initial velocity \( u_B = 0 \).
- The distance fallen by body B after time \( t \) seconds is:
\[ d = u_B t + \frac{1}{2} g t^2 \]
- Therefore, the height of body B after \( t \) seconds is:
\[ 20 - d = 20 - 5t^2 \]

3. **Set Up the Collision Condition:**
- For the bodies to collide, the heights must be equal:
\[ 20t - 5t^2 = 20 - 5t^2 \]
- Simplifying this gives:
\[ 20t = 20 \implies t = 1 \, \text{s} \]

4. **Calculate Velocities Just Before Collision:**
- For body A:
\[ V_A = u_A - g t = 20 - 10 \times 1 = 10 \, \text{m/s} \, \text{(upwards)} \]
- For body B:
\[ V_B = u_B + g t = 0 + 10 \times 1 = 10 \, \text{m/s} \, \text{(downwards)} \]

5. **Apply the Elastic Collision Formula:**
- Since both bodies have equal mass \( m = 2 \, \text{kg} \), the velocities after an elastic collision can be determined using the formula:
\[ V_{A, \text{final}} = \frac{(m_A - m_B)V_A + 2m_B V_B}{m_A + m_B} \]
\[ V_{B, \text{final}} = \frac{(m_B - m_A)V_B + 2m_A V_A}{m_A + m_B} \]
- Plugging in the values:
\[ V_{A, \text{final}} = \frac{(2 - 2)(10) + 2(2)(10)}{2 + 2} = \frac{40}{4} = 10 \, \text{m/s} \, \text{(downwards)} \]
\[ V_{B, \text{final}} = \frac{(2 - 2)(10) + 2(2)(10)}{2 + 2} = \frac{40}{4} = 10 \, \text{m/s} \, \text{(upwards)} \]

### Final Velocities:
- Velocity of body A after collision: \( 10 \, \text{m/s} \, \text{(downwards)} \)
- Velocity of body B after collision: \( 10 \, \text{m/s} \, \text{(upwards)} \)