So in a 1d elastic collision, you can derive \begin{align} v_{1,i} + v_{1,f} = v_{2,i} +v_{2,f} \end{align} from conservation of momentum and energy. Can this equation be used in 2d elastic collisions by splitting the velocities into components or will it not work because energy is not conserved in each direction?

  • $\begingroup$ Possible duplicate physics.stackexchange.com/questions/352793/… $\endgroup$
    – Anu3082
    Dec 28, 2021 at 6:57
  • $\begingroup$ are you sure about the first statement? I don't believe is true in general case, unless the mass is the same for all particles $\endgroup$
    – Numenorean
    Dec 28, 2021 at 7:42

1 Answer 1


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In $2D$-collisions, it is useful to work in a cartesian frame one of whose axis is along the line joining the two centers of the pucks.

The components of velocities of the masses $m_1$ and $m_2$ before and after collision are ("perpendicular to the tangent, "along the tangent") (primes denote velocities after collision):

  1. Before Collision \begin{equation} \begin{split} m_1:& \left(v_{1\perp}, v_{1\parallel}\right) \\ m_2:& \left(v_{2\perp}, v_{2\parallel}\right) \end{split} \end{equation}

  2. After Collision \begin{equation} \begin{split} m_1:& \left(v_{1\perp}', v_{1\parallel}'\right) \\ m_2:& \left(v_{2\perp}', v_{2\parallel}'\right) \end{split} \end{equation}

Now it is important that we assume that there are no forces along the tangent (no friction essentially) so that:

\begin{equation} \begin{split} v_{1\parallel} &= v_{1\parallel}' \\ v_{2\parallel} &= v_{2\parallel}' \end{split} \end{equation}

Thus conservation of energy equation essentially becomes one-dimensional: just erase out the parts containing the tangential velocities by using the above equality. The two-dimensional collision problem becomes one-dimensional collision in the direction perpendicular to the tangent.

Along the normal: \begin{equation} \begin{split} \frac{1}{2} m_1 v_{1 \perp}^2 + \frac{1}{2} m_1 v_{2 \perp}^2 &= \frac{1}{2} m_1 v_{1 \perp}'^2 + \frac{1}{2} m_1 v_{2 \perp}'^2 \\ m_1 v_{1 \perp} + m_2 v_{2 \perp} &= m_1 v_{1 \perp}' + m_2 v_{2 \perp}' \end{split} \end{equation}

Now if you will solve this similar to a one-dimensional elastic collision problem, you will obtain, \begin{equation} v_{1 \perp} + v_{1 \perp}' = v_{2 \perp} + v_{2 \perp}' \end{equation}

just what you asked for.

So two-dimensional elastic collisions is essentially a one-dimensional problem along the centers joining the two pucks. Just do not use your ordinary $x$ and $y$ coordinates, and you are all set.

Also, this is a very good video on elastic collisions: https://youtu.be/K_nSJtrO890


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