Consider the elastic collision f two bodies A and B of equal mass. Initially B is at rest and A moves with velocity `v`. After the collision -

Options

A. the body A trace, its path back with the same speed
B. the body A comes to rest and B moves aways in the direction of A' s approach with the velocity u
C. both the bodies stick together and are at rest
D. B moves along with velocity u/2 and A retraces its path with velocity u/2

Answer

To solve the problem of an elastic collision between two bodies A and B of equal mass, where initially B is at rest and A moves with velocity \( v \), we can follow these steps:

### Step 1: Define the Initial Conditions
- Let the mass of both bodies be \( m \).
- Initial velocity of body A, \( u_1 = v \).
- Initial velocity of body B, \( u_2 = 0 \) (since it is at rest).

### Step 2: Understand Elastic Collision
In an elastic collision, both momentum and kinetic energy are conserved. The coefficient of restitution \( e \) for a perfectly elastic collision is equal to 1.

### Step 3: Apply Conservation of Momentum
The conservation of momentum before and after the collision can be expressed as:
\[ m u_1 + m u_2 = m v_A + m v_B \]
Substituting the known values:
\[ m v + m \cdot 0 = m v_A + m v_B \]
This simplifies to:
\[ v = v_A + v_B \quad \text{(1)} \]

### Step 4: Apply Conservation of Kinetic Energy
The conservation of kinetic energy can be expressed as:
\[ \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 \]
Substituting the known values:
\[ \frac{1}{2} m v^2 + \frac{1}{2} m \cdot 0 = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 \]
This simplifies to:
\[ v^2 = v_A^2 + v_B^2 \quad \text{(2)} \]

### Step 5: Solve the Equations
From equation (1), we can express \( v_B \) in terms of \( v_A \):
\[ v_B = v - v_A \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ v^2 = v_A^2 + (v - v_A)^2 \]
Expanding the right side:
\[ v^2 = v_A^2 + (v^2 - 2vv_A + v_A^2) \]
Combining like terms:
\[ v^2 = 2v_A^2 - 2vv_A + v^2 \]
Subtracting \( v^2 \) from both sides:
\[ 0 = 2v_A^2 - 2vv_A \]
Factoring out \( 2v_A \):
\[ 0 = 2v_A(v_A - v) \]
This gives us two solutions:
1. \( v_A = 0 \) (body A comes to rest)
2. \( v_A = v \) (not possible since body B was initially at rest)

Thus, we conclude:
\[ v_A = 0 \quad \text{and} \quad v_B = v \]

### Final Result
After the collision:
- Body A comes to rest (\( v_A = 0 \)).
- Body B moves with velocity \( v \) in the same direction as A was initially moving.