Two vertical blocks of height h = 0.5 m leave a narrow…

Two vertical blocks of height $h = 0.5\ \mathrm{m}$ leave a narrow vertical gap of width $d = 3\ \mathrm{cm}$. A small steel ball of radius $R = 0.3\ \mathrm{cm}$ rolls toward the gap, perpendicular to it, with speed $v_0 = 1\ \mathrm{m/s}$, falls in, and bounces elastically between the walls until it reaches the bottom. How many times does it hit the walls?

Explanation of Method

A ball bouncing down a narrow gap.

Step 1: What an elastic hit does

At each elastic hit only the horizontal velocity reverses; its size stays $v_0$, and the vertical motion is not affected. So vertically the ball simply falls freely, while horizontally it goes back and forth at 1 m/s.

Step 2: Fall time

\[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{1}{9.81}} \approx 0.319\ \mathrm{s}. \]

Step 3: Horizontal path and wall-to-wall distance

In that time the ball travels $v_0 t \approx 0.319\ \mathrm{m}$ horizontally (back and forth). Its centre can move only $d - 2R = 3 - 0.6 = 2.4\ \mathrm{cm}$ between the walls, and each such crossing ends with one hit:

\[ n = \frac{v_0 t}{d - 2R} = \frac{0.319}{0.024} \approx 13.3. \]

\[n = 13 \ \text{hits}\]

$v_0 = 1\ \mathrm{m/s}$ $d = 3\ \mathrm{cm}$ $h = 0.5\ \mathrm{m}$ $13 \text{ hits}$

Not to scale (the gap is drawn much wider). Each elastic hit only reverses the horizontal motion, so the ball zig-zags down while falling freely; the zig-zags get longer as it speeds up.

Result

\[\boxed{\,n = 13 \ \text{hits}\,}\]

Remark

Note. This assumes the ball's centre enters the gap right at the first wall. The zig-zags are longer near the bottom only because the ball falls faster there; the time between hits stays the same, $0.024/1 = 0.024\ \mathrm{s}$.