Question

Elastic Collision with a Wall: Figure shows a boy throwing a ball toward wall B at speed 20 m/s at $\theta = 15^\circ$. If all collisions of ball with walls are considered elastic, find where the ball will finally land on ground.

Range of ball $R = \frac{u^2\sin2\theta}{g} = \frac{(20)^2\sin(30^\circ)}{10} = 20 m$.

Answer 1

0m

Answer 2

The problem describes a projectile motion scenario where a ball is thrown with an initial speed $u = 20 \, \text{m/s}$ at an angle $\theta = 15^\circ$ with the horizontal. The acceleration due to gravity is $g = 10 \, \text{m/s}^2$. The ball is thrown towards wall B. There are two walls: wall A at a distance of 4m to the left of the boy, and wall B at a distance of 6m to the right of the boy. All collisions with the walls are elastic.

First, let's calculate the horizontal range of the ball if there were no walls. The formula for the range is: $R = \frac{u^2 \sin(2\theta)}{g}$ Given $u = 20 \, \text{m/s}$ and $\theta = 15^\circ$: $R = \frac{(20)^2 \sin(2 \times 15^\circ)}{10} = \frac{400 \sin(30^\circ)}{10}$ Since $\sin(30^\circ) = 0.5$: $R = \frac{400 \times 0.5}{10} = \frac{200}{10} = 20 \, \text{m}$ So, the total horizontal distance the ball would travel in its flight time is 20m.

Now, let's analyze the path of the ball considering the walls. We can set the initial position of the boy (where the ball is thrown from) as $x=0$. Based on the figure:

• Wall A is at $x_A = -4 \, \text{m}$.
• Wall B is at $x_B = 6 \, \text{m}$. The ball is thrown towards wall B, meaning it initially moves in the positive $x$ direction.

For elastic collisions with vertical walls, the horizontal component of the ball's velocity reverses direction, while its magnitude remains constant. The vertical motion is unaffected by the walls. This means that the total horizontal distance covered by the ball during its flight (sum of magnitudes of horizontal displacements) will be equal to the range $R$.

Let's track the ball's horizontal position and the effective horizontal distance covered:

1. Initial launch to Wall B: The ball starts at $x=0$ and moves towards wall B at $x=6 \, \text{m}$. It travels a horizontal distance of $d_1 = 6 \, \text{m}$. Upon hitting wall B, the horizontal velocity reverses, and the ball starts moving towards the left (negative $x$ direction). Effective horizontal distance covered so far: $D_1 = 6 \, \text{m}$. Remaining effective horizontal distance to cover: $R - D_1 = 20 - 6 = 14 \, \text{m}$.

2. From Wall B to Wall A: The ball is now at $x=6 \, \text{m}$ and moves left. It will encounter wall A at $x=-4 \, \text{m}$. The distance traveled in this segment is $d_2 = |x_A - x_B| = |-4 - 6| = 10 \, \text{m}$. Upon hitting wall A, the horizontal velocity reverses again, and the ball starts moving towards the right (positive $x$ direction). Effective horizontal distance covered so far: $D_2 = D_1 + d_2 = 6 + 10 = 16 \, \text{m}$. Remaining effective horizontal distance to cover: $R - D_2 = 20 - 16 = 4 \, \text{m}$.

3. From Wall A to Landing: The ball is now at $x=-4 \, \text{m}$ and moves right. It needs to cover an additional effective horizontal distance of $4 \, \text{m}$ to complete its total range. The ball travels $d_3 = 4 \, \text{m}$ to the right. The final landing position will be $x_f = x_A + d_3 = -4 + 4 = 0 \, \text{m}$. Effective horizontal distance covered in total: $D_3 = D_2 + d_3 = 16 + 4 = 20 \, \text{m}$. This matches the total range $R$.

Therefore, the ball will finally land on the ground at the starting point, which is $0 \, \text{m}$ from the boy's initial position.