A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground choose the correct option

Options

A. the direction of motion of the ball changes every 10 seconds
B. speed of ball changes every 10 seconds
C. average speed of ball over any 20 seconds interval is fixed
D. the acceleration of the ball is the same as from the train.

Explanation

To solve the problem, we need to analyze the motion of the ball as seen from the ground, considering both the motion of the ball within the train and the motion of the train itself.

### Step-by-Step Solution:

1. **Understanding the Motion of the Ball**:
- The ball is bouncing elastically between two walls of a railway compartment that is 10 meters long.
- The speed of the ball relative to the train is 1 m/s.

2. **Motion of the Train**:
- The train is moving at a constant velocity of 10 m/s in a direction parallel to the motion of the ball.

3. **Velocity of the Ball with Respect to the Ground**:
- When the ball is moving towards the front wall of the train, its speed relative to the ground is:
\[ v_{\text{ball, ground}} = v_{\text{ball, train}} + v_{\text{train, ground}} = 1 \, \text{m/s} + 10 \, \text{m/s} = 11 \, \text{m/s} \]
- When the ball bounces back towards the rear wall, its speed relative to the ground is:
\[ v_{\text{ball, ground}} = -1 \, \text{m/s} + 10 \, \text{m/s} = 9 \, \text{m/s} \]

4. **Direction of Motion**:
- The direction of the ball does not change; it is always moving forward and backward between the walls. However, the speed changes from 11 m/s to 9 m/s.

5. **Average Speed Calculation**:
- Over a time interval of 20 seconds, the ball will have traveled back and forth multiple times.
- The average speed of the ball over any 20-second interval can be calculated based on the distance traveled by the train:
- The train moves 200 meters in 20 seconds (10 m/s × 20 s).
- The average speed of the ball with respect to the ground is:
\[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{200 \, \text{m}}{20 \, \text{s}} = 10 \, \text{m/s} \]

6. **Acceleration of the Ball**:
- Since both the ball and the train are moving at constant velocities, the acceleration of the ball with respect to both the ground and the train is zero.

### Conclusion:
- The correct options based on the analysis are:
- **Option B**: The speed of the ball changes every 10 seconds.
- **Option C**: The average speed of the ball over any 20 seconds interval is fixed.
- **Option D**: The acceleration of the ball is the same as from the train (which is zero).