I know that the equations for conservation of momentum and energy

$m_1v_{i1}+m_2v_{i2} = m_1v_{f1}+m_2v_{f2},\;\frac{1}{2}\epsilon(m_1v_{i1}^2+m_2v_{i2}^2) = \frac{1}{2}(m_1v_{f1}^2+m_2v_{f2}^2)$,

where $\epsilon$ is the efficiency of the collision, and the masses and initial velocities are known, is easily soluble in 1D, as there are 2 equations and 2 variables.

However, in n-dimensions, there are $n+1$ equations--since the energy equation is scalar-valued rather than vector-valued--and $2n$ variables.

Therefore, are there multiple solutions to, for instance, the 3D collision problem?

If there are, is there a systematic way to solve the problem?

  • $\begingroup$ Do you know how to derive the impulse-momentum theorem? It might help to take a look how you could extrpolate to the third dimension. $\endgroup$ Commented Dec 23, 2014 at 21:18
  • $\begingroup$ Impulse-momentum, do you mean $F\mathrm{d}t=\mathrm{d}\rho$? $\endgroup$
    – k_g
    Commented Dec 23, 2014 at 21:20
  • $\begingroup$ Generically, two straight lines in dimension > 2 don't intersect, which may explain why the problem is ill-posed. If you are willing to assume (given that the collision occurs) that the resultant momentum $p_f$ lies in the same plane as the initial momentum $p_i$, then it reduces to the 2-d problem (up to a rigid transformation). $\endgroup$
    – amomin
    Commented Dec 23, 2014 at 21:59
  • $\begingroup$ Oh, I see. I thought that by only talking about velocities, I had already dealt with that. So, just transform the velocity functions onto XY [or whatever], then calculate in XY, then transform back? It seems as if that problem still has 4 variables and 3 equations, however... $\endgroup$
    – k_g
    Commented Dec 23, 2014 at 22:03
  • $\begingroup$ OK, so I figured out how to simplify the problem to 2D, just define a new basis using the unit vectors $\hat{v_1},\hat{v_2}$. But then the problem is still that there are 2 momentum and 1 energy equation for 4 variables... $\endgroup$
    – k_g
    Commented Dec 23, 2014 at 23:09

1 Answer 1


Let's say that you have two objects colliding, one with momentum $p_1$ and the other momentum $p_2$. You are treating the collision as instantaneous, and so you are applying an impulse $J_i$ (instantaneous change of momentum, i.e. Dirac delta of force) to each body.

Conservation of momentum dictates that $J_1 = J, J_2 = -J$ for some $J$; this is the instantaneous analogue of Newton's third law. If you know the direction of $J$, conservation of energy gives you its magnitude.

But, as you say, the direction is not uniquely determined, and you need additional assumptions to solve for $J$. This assumption is that the impulse acts only in the contact normal direction $J = \lambda \hat{n}$ for some unknown scalar (determined by conservation of energy) $\lambda$. There are at least two ways to intuit why this should be the case:

  1. The least change to the momenta of the two objects that stops them from interpenetrating is to apply an impulse in the contact normal direction. The impulse should not apply any unnecessary extra change to the tangential components of the momenta.
  2. You can think of instantaneous impact as a limiting case of soft contact, where you replace the impulse with a force acting over finite time. The contact potential is proportional volume of overlap of the two bodies, and the force due to this potential is its gradient, which for small amounts of penetration acts in approximately the contact normal direction. In the limit of this force becoming infinitely stiff, you get an impulse acting in the contact normal direction.

So with the extra assumption that the impulse acts in the contact normal direction on each body you can solve for a unique impulse and hence post-collision momentum for each body. Note that this is the case only when two bodies collide: in high school you might have seen Newton's cradle, where a line of stationary balls is hit by one moving ball (https://www.youtube.com/watch?v=0LnbyjOyEQ8), and heard the claim that the physically correct solution is the one where the striking ball comes to rest, the line of balls stays still, except for the last ball which starts moving. You might find it interesting that there is no first-principle justification for this claim (and in fact if you look at the video you see it's clearly not the case, as the second ball separates from the first and third after impact): the post-impact velocities of the balls are under-determined. You must model the balls as soft bodies, and make assumptions about how the materials deform etc. in order to be able to uniquely solve for the motion of the balls. In fact if you build a Newton's cradle out of rubber balls, or out of steel balls, you will see a dramatic difference in the behavior of the cradle. This is not the case for just two balls colliding, where conservation laws (and the assumption that the contact force acts in the normal direction) uniquely determine the motion.

  • $\begingroup$ I see. <accept> for being the only answer. +1 for actually acknowledging the underconstrainedness of the situation. $\endgroup$
    – k_g
    Commented Jan 10, 2015 at 5:46

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