In a 1D elastic collision, it is well-known that the relative velocities of the two objects (before and after the collision) are reversed. What is the extension of this result to 2D or higher? Is there an analogous result for 2D or 3D collisions (the two objects are not travelling along the same line)?


1 Answer 1


In a 1D elastic collision, it is well-known that the relative velocities of the two objects (before and after the collision) are reversed.

When you say reversed do you mean that each object keeps their own velocity just with a change of sign?

That would not always be the case for a 1D situation. If a resting block $v_{1,before}=0$ is hit by a moving block $v_{2,before}>0$ of equal mass, then the resting block will start to move and the other one will stop, $v_{1,after}>0$ and $v_{2,after}=0$. If the masses are not equal, then even if they move towards each other they may after the collision be moving in the same direction at different speeds. See the GIF's in this this link.

What is the extension of this result to 2D or higher

For a 2D or 3D case you can split the velocity into components. If two objects hit each other with an angle, the perpendicular velocity components will behave as in the 1D case, while the other component will remain unaltered.

enter image description here

See this link for a good demonstration.


enter image description here

Each ball might have its own $x$ and $y$ components (see the left drawing on the illustration).

We place the coordinate system so that they each have one component pointing towards each other while the other components are parallel. This will always to be possible to do.

The straight-on component - the perpendicular component - causes the collision. If that component was the only one present (middle drawing) the balls would hit head-on as in the 1D case. This component will be altered at a collision because of momentum conservation as in the 1D case.

If the other component - the parallel component - was the only one present (right drawing), then there would never be a collision. This component will not be altered at any collision caused by any other component.

It is always possible to position the coordinate system in a manner so that one component only acts in the perpendicular direction to the other ball. And so, this component alone will cause collision and this component alone will experience changes due to conservation of momentum.

Therefore, any collision in 1D, 2D or 3D will happen as was it 1D. I hope this helps.

Another addition

To give an example consider this situation of two balls. Equal mass to keep it simple, but might have different velocities:

enter image description here

The momentum of each ball is constant in the y-direction (perpendicular to the "axis" of collision shown). Only the momentum-components that are perpendicular to the surfaces (along the "axis" of collision) are changes, as only they interact with each other. For an elastic collision momentum will be conserved and:


For the billard table example you bring up in the comments, the same situation is the case, but one ball is initially at rest:

enter image description here

The y-components are still constant and the x-components are not. (Masses are assumed equal.) If the ball was hit head-on it would move straight away backwards. But as in this case, by letting the balls collide with an angle they will roll off in directions other than the original one.

  • $\begingroup$ Steeven, thanks for taking the time to provide this Answer. You stated "while the other component will remain unaltered". I don't think that's true in general. I can think of scenarios where both components change. $\endgroup$
    – JHN
    Feb 8, 2015 at 2:58
  • $\begingroup$ @user12707 Can you provide an example? $\endgroup$
    – Steeven
    Feb 8, 2015 at 8:37
  • $\begingroup$ Steeven, one example is the oblique collision of two billiards balls - one stationary, one moving, then the two will have a component of momentum in the perpendicular direction, unless it's exactly head on. $\endgroup$
    – JHN
    Feb 9, 2015 at 6:59
  • $\begingroup$ @user12707 You say there is a "perpendicular direction". This is one dimension. No matter how the balls move you can always consider them as having one component in this direction (you can define your coordinate system to fit to this). So in this example as in any other, the collision in itself will always be along one axis (1 dimension). $\endgroup$
    – Steeven
    Feb 9, 2015 at 19:38
  • 1
    $\begingroup$ Steeven, yes, you have convinced me :) Beer and billiards are on me if we do meet someday! $\endgroup$
    – JHN
    Feb 15, 2015 at 23:32

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