Initially 340 Given two points $A$ and $B$ on a straight line $l$, the distance between them is $10000 \mathrm{~cm}$. At points $A$ and $B$, there are two movable barriers, designated as Barrier 1 and Barrier 2, respectively. Assume there is a ping-pong ball between $A$ and $B$, moving along the straight line $l$ at a uniform speed. Initially, Barrier 1 and the ping-pong ball start from point $A$ and move along the line $l$ towards point $B$ at uniform speeds of $4 \mathrm{~cm} / \mathrm{s}$ and $5 \mathrm{~cm} / \mathrm{s}$, respectively, and assume the speed of Barrier 1 remains constant. When the ping-pong ball first hits Barrier 2, it reverses direction, and its speed decreases by $20 \%$, while Barrier 2 quickly moves $10 \mathrm{~cm}$ along the direction of $A B$, so the distance from Barrier 2 to point $A$ is now $10010 \mathrm{~cm}$. When the ping-pong ball first contacts Barrier 1 (referred to as the first contact between the ping-pong ball and Barrier 1) and reverses direction, its speed increases by $25 \%$. When the ping-pong ball hits Barrier 2 for the second time, it continues to move in the same manner, and Barrier 2 quickly moves $9 \mathrm{~cm}$ along the direction of $A B$. Each subsequent time the ping-pong ball hits Barrier 2, it continues to move in the same manner, and Barrier 2 moves $1 \mathrm{~cm}$ less than the previous time, until the number of times the ping-pong ball hits Barrier 2 exceeds 10, at which point Barrier 2 no longer moves. When the distance between Barrier 1 and Barrier 2 first equals $1 \mathrm{~cm}$, how many times has the ping-pong ball contacted Barrier 1?
Let's assume that when the ping-pong ball contacts the 2nd board for the $n$-th time, the position of the 2nd board after it moves quickly is $B_{n}$; when the ping-pong ball contacts the 1st board for the $n$-th time, its position is $A_{n}$, and we set
$$
\begin{array}{l}
B_{0}=B, A_{n} B_{n-1}=x_{n}, A_{n} B_{n}=y_{n} . \\
\text { Then } \frac{10000}{5}+\frac{x_{1}}{5(1-20 \%)}=\frac{10000-x_{1}}{4} \\
\Rightarrow x_{1}=\frac{10000(5-4)(1-20 \%)}{4+5(1-20 \%)}=1000 .
\end{array}
$$

Thus, $y_{1}=1010$.
Similarly, $x_{2}=101, y_{2}=110$;
$$
\begin{array}{l}
x_{3}=11, y_{3}=19 ; \\
x_{4}=1.9, y_{4}=8.9 .
\end{array}
$$

When the ping-pong ball hits the 2nd board for the 5th time, the distance between the 1st board and the 2nd board is
$8.9 \times\left(1-\frac{4}{5}\right)=1.78>1$.
Thus, $x_{5}=0.89, y_{5}=6.89$.
When the ping-pong ball hits the 2nd board for the 6th time, the distance between the 1st board and the 2nd board is
6. $89 \times\left(1-\frac{4}{5}\right)=1.378>1$.
Thus, $x_{6}=0.689, y_{6}=5.689$.
When the ping-pong ball hits the 2nd board for the 7th time, the distance between the 1st board and the 2nd board is
$5.689 \times\left(1-\frac{4}{5}\right)=1.1378>1$.
Thus, $x_{1}=0.5689, y_{7}=4.5689$.
When the ping-pong ball hits the 2nd board for the 8th time, the distance between the 1st board and the 2nd board is
4. $5689 \times\left(1-\frac{4}{5}\right)=0.91378<1$.
Therefore, when the distance between the 1st board and the 2nd board is first equal to $1 \mathrm{~cm}$, the 1st board has contacted the ping-pong ball 7 times.