# Elastic collision between a circle and a point [duplicate]

In a 2D environment, I have a circle with velocity v, a stationary point (infinite mass), and I am trying to calculate the velocity of the circle after a perfectly elastic collision with the point.

This is what I've came up with:

$$p$$ is the position of the point

$$c$$ is the position of the circle

On a collision, the new velocity is $$normalize(p - c) * velocity.magnitude$$. Would this approach be right? If not, how do I correctly determine the velocities after a collision?

• Is it a straight ahead head-on collision or at an angle. And what the mass and final velocity of the point mass. – TechDroid Mar 4 '19 at 13:01
• @TechDroid It can be at an angle or head-on, the velocities are random, and the point is fixed. – Sarah Mar 4 '19 at 13:03
• There are many questions on this site of a similar nature, which have already been answered. I've indicated one in my comment above: you just need to look at the accepted answer, and insert your conditions that the mass of one of the particles is infinite, and its velocity is zero. You should see that the impulse, not the final velocity, should be in the direction of the centre of the moving particle. Otherwise, even a grazing collision would produce a 90-degree change in direction of velocity. – user197851 Mar 4 '19 at 13:05
• Note that the answer is the same as if the circle had the infinite mass and the point was incident. This is a standard scattering problem. – Jon Custer Mar 4 '19 at 19:42
• I see that you also posted this question at math.stackexchange.com/questions/3134854/… and in fact you have cross posted all your Physics SE questions (so far) at Maths SE. Please be aware that this is not generally encouraged: it makes more work for people who look at your questions, especially if they don't know that they might have been answered elsewhere. I would recommend picking one site, which you judge to be best suited to the question. If you don't get a response, you can always delete the question and post it elsewhere. – user197851 Mar 5 '19 at 11:59

I like this question because a variant of it was on a recent problem set in my quantum class! So I’ll use a modified form of my solution to explain to you how this works. See my drawing for reference. We refer to the initial angle of the circle as $$\phi$$, and the final angle as $$\phi’$$. Of course, $$\phi’$$ is some function of $$\phi$$: $$\phi’(\phi)$$, which is what we hope to find.

Note that the incoming and outgoing velocity vectors both form an angle $$\theta$$ with the normal force exerted on the ball by the pin.

Now, from $$\Delta AOB$$, we can find $$\pi - \phi’ + \phi + \pi - 2\pi = \pi$$ $$\implies \phi’ = \pi + \phi -2\theta$$

You can convince yourself that $$\tan\theta = \frac{b}{\sqrt{R^2-b^2}}$$ which we use to see $$\phi’ = \pi + \phi -2\tan^{-1}\left(\frac{b}{\sqrt{R^2-b^2}}\right)$$

For the record, the geometric quantity $$b$$ which I show in the diagram is referred to as the impact parameter of the collision.