# Analogy between reflected rays and elastic collisions of an object

I came across a problem which is as follows:

In free space, a particle is projected from a point P on-axis of a fixed rigid cone AOB, at an angle α = 37° with the axis (see figure). The distance of point P from the apex O is x = 10 cm and the apex angle of the cone is β = 20°. All the collisions of the ball with the cone are perfectly elastic. Find the number of collisions of the ball with the cone and the distance of the closest approach of the ball from the apex.

MY ATTEMPT : First approach

It seems like a mechanics problem and I tried it by calculating the angle at which the ball collides with the cone. After the fifth collision, the ball starts moving away from the apex of the cone. But calculating the distances with nonstandard angles like 10 degrees etc. doesn't seem possible to me.

SECOND APPROACH :

But it seems to me that such a problem can be solved using geometrical optics probably using the concept of image formation by mirrors inclined at an angle but I don't really know how? I know elastic collision trace the path of reflected rays if we were to consider the sides of the cones as mirrors but to what extent can the analogy between light and the elastic collisions of the ball be drawn? A ray of light would form 17 images with the mirrors but that doesn't match with the answer given for the number of collisions of the ball.

Any help will be greatly appreciated!

• There should not be any difference between the reflection of a light ray and a particle. If you can work out that it is 17 reflections for the former then it should also be 17 for the latter. Commented Jun 26, 2021 at 15:18
• @Thomas Okay thanks ! The answer given is 7 that may be a typing error. But then how to find the distance of closest approach? I wasn’t able to think of an analogy between the ball and light to find it.
– Nil
Commented Jun 26, 2021 at 16:49

The distance can be calculated using the sine rule, for example if the ball first collides with the cone at $$Q$$ $$\frac{10}{\sin 133}= \frac{OQ}{\sin 37}$$

Then continue finding the distance from the apex each time the ball hits the cone. The shortest distance is shortly after the 3rd bounce, when the direction of travel is at right angles to the direction from the particle to $$O$$.

An optical way to check would be to set up an experiment with a laser and mirrors. Use a powder e.g. chalk dust, or smoke to show the path of the laser and measure the distance of closest approach.

There is probably a graphical package that could also be used to model the situation and check the calculation.

• thanks, an image will be posted shortly. first is blue path, then black, then red and then part way along the second blue is the closest distance to $O$ Commented Jun 26, 2021 at 19:31
• Thanks! I calculated all the angles and using the sine rule as you suggested the answer surprisingly comes out to be very simple as all the complicated angle terms cancelled out .
– Nil
Commented Jun 27, 2021 at 8:55
• Out of curiosity, what answer did you get?, there is an answer from this end too, hope it's right... Commented Jun 27, 2021 at 9:00
• I got 10 (sin 37 ) that is 6 cm . What answer did you get ?
– Nil
Commented Jun 27, 2021 at 9:42
• Yes, my answer was 6.018cm (same as your 10 (sin 37 )), got from distance OT = 6.0264 times cos(3), (where T is where the red and green lines cross). Interesting how things cancelled to leave the 10 (sin 37 ), might have a look how that happened. Measurements on the diagram confirm the answers. All the best. Commented Jun 27, 2021 at 9:46

There's a very elegant way to approach this, which also explains how the answer turns out to be something simple like $$10\sin 37^{\circ} \mathrm{cm}$$.

Suppose that the ball instead of being reflected continued on a straight line path (shown in green) and instead the cone itself was rotated by an angle $$\beta$$ about its apex.
The green path is a reflection of the actual path of the ball, about the wall of the cone. With a little bit of geometry it can be verified that this new situation is equivalent to the cone being fixed and the ball colliding.

Now note that at any given time the distance of the ball from the apex is the same as the distance of its "imaginary" counterpart traveling on the green path.
So rather than finding the minimal distance of the ball on it's original we may find the minimal distance on the green line.

We get the minimal distance by dropping a perpendicular as shown above. In the right angled triangle $$OMP$$, we know $$OP = 10$$ cm and $$\angle OPM = 37^{\circ}$$ so $$PM = \boxed{10\sin{37^{\circ}} \mathrm{cm}}$$