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Physics of simple collisions

I have 2 photos of the balls, one before the collision and one after the collision. They do a elastic collision. I want to know how is the reflected value calculated. I know it uses dot product and such stuff. But it all is very vague to me. Can anybody show me a working calculation and explain me based on this figures? I have been trying to understand this for past 2 days but I can't get how it all works.

This is the image before collision:

enter image description here

This is the image after collision:

enter image description here

Thanks in advance :)

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marked as duplicate by Mark Eichenlaub, David Z Jun 8 '11 at 17:57


What "reflected value" do you want to know specifically? I imagine you mean a physical value of the system after the interaction. –  Alan Rominger Jun 7 '11 at 2:54
@Zassounotsukushi: Reflected value means I want to know how will the objects after collision move. Please let me know if I am still not clear. –  TCM Jun 7 '11 at 3:01
""I know it uses dot product and such stuff."" This is wrong. All You need is vector addition. –  Georg Jun 7 '11 at 9:46
@Georg: "All You need is vector addition". How? If you know the answer can you please post it along with some same data and calculations? –  TCM Jun 7 '11 at 15:06

1 Answer 1

up vote 2 down vote accepted

Anthony, even after your parlay with Georg and Mark, I am still not confident I know exactly what you want, but I'll give it a stab...

First, the the velocities of the balls, both before and after collision, are broken into two components: velocities tangential to the point of contact and velocities normal to the point of contact. What happens in each of these orthogonal axes does not affect what happens to the other. This is the first really important concept to grasp: For momentum, what happens in tangential stays in tangential; and what happens in normal stays in normal.

Speaking of tangential: Notice that the tangential velocity of each ball is exactly the same, both before and after collision. This would be expected if we assume (and we do) no friction. Another way to look at it: If the tangential velocity of each ball was the ONLY component of velocity, and they barely gave each other a frictionless kiss in passing, wouldn't we expect the velocity of each ball to remain unchanged? Yes. OK, then, we have established that the tangential velocity (and momentum, by the way) of each ball remains unchanged throughout the collision.

Now we go on to the normal component of velocities: We know the velocities prior to collision, and we know the masses throughout the collision. We also know that kinetic energy and momentum are conserved (since this is an elastic collision). Furthermore, we also know that the tangential components of velocities have remained unchanged. We are left with a boat load of equations and only two unknowns: Normal velocities of the masses (m1 and m2) after the collision (v1an and V2an, respectively).

Velocities before collision: v1b = SQRT((v1bn^2)+(v1bt^2)) and v2b = SQRT((v2bn^2)+(v2bt^2)).

Velocities after collision: v1a = SQRT((v1an^2)+(v1at^2)) and v2a = SQRT((v2an^2)+(v2at^2)).

Since momentum is conserved --> (m1)(v1bn) + (m2)(v2bn) = (m1)(v1an) + (m2)(v2an) ... and the normal velocities must remain normal, it is a short step of reasoning to deduce, for m1=m2, that |v1an| = |v1bn|, but are in the opposite direction, and likewise for |v2an| = |v2bn|.

For the more general case, where m1 does not equal m2 and |v1| does not equal |v2|, the equations are:

(v1an) = (v1bn)(((m1-m2)+((2m2)(v2bn))/(m1+m2))

(v2an) = (v2bn)(((m2-m1)+((2m1)(v1bn))/(m1+m2))

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