More collision, but this time it's a physics question

[HarryW]

Well I've figured out my collision detection problem between two balls(from my previous thread in the maths forum) and now I have to figure out the physics side of the collision.

I know that the force exerted on each ball will be a scalar product of the vector between the centres of the balls. So I know the direction of the force. However, I am uncertain how to calculate the velocities of the balls after the collision, conserving energy and momentum. I have the following equations, derived from the energy and momentum before and after the collision:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

These equations can be solved, giving two sets of solutions for v₁ and v₂, but they don't take into account the angle that the balls collide at. I need to combine the knowledge of the direction in which the force acts with these equations somehow.

I'm sure someone knows how to do this, I think I must be having some kind of mental block or something. I'll keep at it anyway in the meantime, but I appreciate any kind of help.

[Parke]

Your problem is that you have written the momentum equation as a scalar when in fact it is a vector equation.
m1u1x + m2u2x = m1v1x + m2v2x
m1u1y + m2u2y = m1v1y + m2v2y

(I cannot seem to get my subs to work I must know less than I thought.)

Hope this helps,
Parke

[HarryW]

I know that, I have seperated the momentum equation into two vector components already.

I would really like to be able to do this without trig, but I'm not sure if I know enough vector geometry to do so. Currently I'm thinking of using the dot product of two of the vectors involved to get the angle at which the balls collide, and getting two new vector components for velocity, Vn and Vt, n being the normal axis and t being the tangential axis. I'm sure there's a way to do it without trig (well pretty sure) but I don't know how

[Parke]

If you are doing two dimensional collisions, you are going to be using trig in the sense of sin and cos. For me, the easiest way is to pick one of the original directions as the x-axis. The y-axis is then normal to that. This eliminates one component along the y-axis.

The equations for this are in Physics texts. Unfortunately I am moving and all mine are packed. What are you looking for? You know that the original momentum before the collision equals the momentum after the collision. So another trick is to call the original P vector the x-axis which means the sum of the y components is zero.

Unfortunately, I am not certain if I am helping.

Parke

[HarryW]

Heh

Well I know what you're saying. The line of impact (is that what it's called? I know there's a technical term for it) is not in the same direction as the velocities of either sphere though; if you consider a collision where both balls are moving at different angles you'll see what I mean. So in order to eliminate any change in velocity in one axis, I need to make the two axes the normal and tangential vectors to the point of collision.

Are you really certain trig is necessary? I mean the way I am doing it at the moment involves trig, but I have a feeling there is a way to do this with vector maths only. Ah well.