A ball starting from point $A$, after bouncing off a wall 5 m away from $A$, reaches point $B$, which is 2 m away from the wall. Here it also hits the wall, and bounces back to point $A$. The distance $AB$ is 9 m. The initial speed of the ball is $v_{0}=5 \mathrm{~m} \mathrm{sec}^{-1}$, and its constant acceleration is $a=-0.4 \mathrm{~m} \mathrm{sec}^{-2}$.

a) What is the speed of the ball when it returns to $A$?

b) What initial speed would the ball need to be launched with under the same conditions to stop exactly at $A$ upon its return?
Considering perfectly elastic collisions, the collision does not affect the ball's speed, only changes the direction of motion according to the known theorem: the angle of incidence is equal to the angle of reflection.

Denoting the first reflection point by $T$,

$$
A T+T B=A T+T B^{\prime}=A B^{\prime}
$$

where $B^{\prime}$ is the mirror image of point $B$ with respect to the first wall (see the 1st figure, which also shows the labeling).

![](https://cdn.mathpix.com/cropped/2024_05_02_0656574db34753854ec4g-1.jpg?height=461&width=656&top_left_y=469&top_left_x=732)

From the right triangle $A C B^{\prime}$, by the Pythagorean theorem,

$$
A B^{\prime 2}=B^{\prime} C^{2}+A C^{2}=B^{\prime} C^{2}+7^{2}
$$

But $B^{\prime} C=B D$, and from the right triangle $A D B$,

$$
B D^{2}=A B^{2}-A D^{2}=9^{2}-3^{2}
$$

and thus

$$
A B^{\prime 2}=9^{2}-3^{2}+7^{2}=121
$$

from which

$$
A B^{\prime}=A T+T B=11 \mathrm{~m}
$$

Therefore, the total distance

$$
s=A T+T B+B A=11+9=20 \mathrm{~m}
$$

a) First, let's calculate the time $t$ required to cover the total distance. Using the known formula $s=v_{0} t+\frac{a}{2} t^{2}$,

$$
5 t-0.2 t^{2}=20
$$

or

$$
0.2 t^{2}-5 t+20=0
$$

from which

$$
t_{1}=5
$$

and

$$
\left[t_{2}=20 .\right]
$$

The $t_{2}=20 \mathrm{sec}$ does not fit the problem, because according to the formula $v_{t}=v_{0}+a t$, $v_{20}=v_{0}+20 a=5-20 \cdot 0.4<0$, meaning that after $20 \mathrm{sec}$ the speed is already negative.

Therefore, the ball arrives back at $A$ after 5 seconds, and at that time its speed is

$$
v_{5}=5-5 \cdot 0.4=5-2=3 \mathrm{~m} \mathrm{sec}^{-1}
$$

b) Let the desired initial speed be denoted by $v_{0}^{\prime}$, and the time elapsed until stopping at $A$ by $t^{\prime}$, then

$$
v_{0}^{\prime}+a t^{\prime}=0
$$

from which

$$
v_{0}^{\prime}=-a t=0.4 t^{\prime}
$$

The time $t^{\prime}$ is calculated using the formula

$$
s=v_{0}^{\prime} t^{\prime}+\frac{a}{2} t^{\prime 2}
$$

Substituting the value of $v_{0}^{\prime}$ from (1) into this,

$$
0.4 t^{\prime 2}-0.2 t^{\prime 2}=20
$$

from which

$$
t^{\prime}=\sqrt{\frac{20}{0.2}}=10 \mathrm{sec}
$$

and thus from (1)

$$
v_{0}^{\prime}=0.4 \cdot 10=4 \mathrm{~m} \mathrm{sec}^{-1}
$$

Note: Many solvers incorrectly reasoned that if the ball returns with a speed of $3 \mathrm{~m} \mathrm{sec}^{-1}$ when the initial speed is $5 \mathrm{~m} \mathrm{sec}^{-1}$, then with an initial speed of $5-3=2 \mathrm{~m} \mathrm{sec}^{-1}$, it will return with a speed of $3-3=0 \mathrm{~m} \mathrm{sec}^{-1}$.