Question

Two parallel walls are 4.0 m apart. A ball is projected from the foot of one of the walls such that in absence of the opposite wall its range would have been 14.0 m. The ball collides elastically with the walls several times and finally lands on the ground. Assuming height of the walls sufficient enough to restrict the ball in between them. Find the total number of collisions of the ball with the walls and the point where the ball hits the ground.

Answer 1

Total number of collisions = 3, Landing point = 2.0 m from the starting wall

Answer 2

1. Understanding the problem:

The ball is projected from the foot of one wall. The walls are parallel and 4.0 m apart. In the absence of walls, the ball's range would be 14.0 m. Collisions with walls are elastic. We need to find the total number of collisions and the final landing position.

2. Key Principle - Independence of Motion and Elastic Collisions:

• In projectile motion, horizontal and vertical motions are independent.
• The total time of flight ($T$) depends only on the vertical component of initial velocity ($u_y$) and gravity ($g$). Since the walls are vertical, collisions with them do not affect the vertical motion or the total time of flight. So, the total time of flight remains the same as if there were no walls.
• For elastic collisions with vertical walls, the magnitude of the horizontal velocity component ($u_x$) remains constant, only its direction reverses.

3. Total Horizontal Distance (Effective Range):

Since the time of flight and the magnitude of horizontal velocity are unaffected by elastic collisions with vertical walls, the total horizontal distance covered by the ball in its flight, irrespective of collisions, is equal to its range in the absence of walls. Given, Range ($R_0$) = 14.0 m.

4. Analyzing Horizontal Motion and Collisions:

Let the starting wall be Wall A (at $x=0$) and the opposite wall be Wall B (at $x=4.0$ m). The ball is projected from Wall A. The ball's horizontal motion can be visualized by "unfolding" the path, as if the walls were mirrors. The ball effectively travels a straight horizontal path of length $R_0 = 14.0$ m.

Let's track the ball's horizontal movement and collisions:

• Initial state: Ball at $x=0$ (Wall A), moving towards Wall B.
• First segment: The ball travels 4.0 m to the right to reach Wall B.
  • Horizontal distance covered = 4.0 m.
  • 1st collision occurs at Wall B ($x=4.0$ m). The ball's horizontal velocity reverses direction (now moving left).
• Second segment: The ball travels 4.0 m to the left to reach Wall A.
  • Horizontal distance covered = 4.0 m + 4.0 m = 8.0 m.
  • 2nd collision occurs at Wall A ($x=0$ m). The ball's horizontal velocity reverses direction (now moving right).
• Third segment: The ball travels 4.0 m to the right to reach Wall B.
  • Horizontal distance covered = 8.0 m + 4.0 m = 12.0 m.
  • 3rd collision occurs at Wall B ($x=4.0$ m). The ball's horizontal velocity reverses direction (now moving left).

5. Determining the Landing Point and Total Collisions:

• After 3 collisions, the ball has effectively covered a horizontal distance of 12.0 m.
• The total effective range is 14.0 m.
• Remaining horizontal distance to be covered = $14.0 \text{ m} - 12.0 \text{ m} = 2.0 \text{ m}$.
• After the 3rd collision, the ball is at Wall B ($x=4.0$ m) and is moving towards Wall A (left).
• It travels an additional 2.0 m to the left before landing on the ground.
• Therefore, the landing point is at $x = 4.0 \text{ m} - 2.0 \text{ m} = 2.0 \text{ m}$ from Wall A.

Conclusion:

• Total number of collisions with the walls = 3.
• The point where the ball hits the ground is 2.0 m from the wall from which it was projected.