(For the purpose of this question, "calculating a collision" means: given the velocities and masses of two objects in a collision, figuring out the new velocities of both objects after the collision).

I know how to calculate a totally elastic collision, and how to calculate a totally inelastic collision.

But I don't know how to calculate a collision which is part elastic and part inelastic. I don't know where to start.

Guidance will be appreciated.

(Go easy on the math please).

  • $\begingroup$ Many texts seem to assume that an elastic collision conserves energy, and in an in-elastic collision, the objects stick together. I like to call the other alternative “non-elastic”. This one conserves momentum only, and normally requires some information from after the collision (or a coefficient of restitution). $\endgroup$
    – R.W. Bird
    Commented Sep 8, 2019 at 20:13

4 Answers 4


There is not such thing as a "partially elastic" collision. Classical collisions between particles can be separated into two categories: elastic and inelastic. Elastic collisions are defined as collisions in which no energy leaves the system (i.e. $E_i = E_f$). All other collisions are inelastic, as some energy is lost ($E_i > E_f$). A perfectly inelastic collision is a type of inelastic collision where all the kinetic energy of the system is lost ($E_f = 0$).

Edit: I should mention that these definitions apply to a CM (center-of-mass) frame of reference. For a non-CM frame, a perfectly inelastic collision becomes one where the maximal amount of kinetic energy is lost. Thanks to David Z. for mentioning this.

  • $\begingroup$ Note: I assume all speeds are slow enough such that it is safe to neglect relativistic effects. Also, if you give me a problem I'd be happy to walk you through it. $\endgroup$
    – Ultima
    Commented Jul 16, 2014 at 23:49
  • $\begingroup$ I see. What factors affect the amount of energy that is lost? I.e. what makes a collision perfectly inelastic? $\endgroup$
    – Aviv Cohn
    Commented Jul 16, 2014 at 23:51
  • 1
    $\begingroup$ I think it'd be worth mentioning in the answer that a perfectly inelastic collision only has $E_f = 0$ in the center-of-mass frame of the colliding objects. That makes it equivalent to the other definition I've heard for "perfectly inelastic" or "totally inelastic," which is one where the objects stick together after colliding. (Unless you really did mean $E_f = 0$ in whatever frame you're observing the collision from, but that would be an odd definition I think.) $\endgroup$
    – David Z
    Commented Jul 16, 2014 at 23:57
  • 3
    $\begingroup$ By the way, @Nathan, since you mentioned a problem walkthrough in your first comment: if you haven't already, you might want to take a look at our homework policy. $\endgroup$
    – David Z
    Commented Jul 17, 2014 at 0:01
  • $\begingroup$ Thanks for the tip, @DavidZ. I'm new to this marvelous site, so the information is helpful. $\endgroup$
    – Ultima
    Commented Jul 17, 2014 at 0:06

Perfectly elastic and perfectly inelastic collisions are just limiting cases on a scale of how much kinetic energy is retained. As noted in @Nathan's answer, if you work in the center-of-mass frame, a perfectly inelastic collision results in 0% of the kinetic energy retained, while perfectly elastic collisions have 100% of kinetic energy retained. So, you just need a way of scaling between those two limiting points.

In all cases, momentum will be conserved. If we choose to work in the CM frame, then it's always zero, and $ v_1m_1 = -v_2m_2 $ before the collision and $ v_1'm_1' = -v_2'm_2' $ after the collision.

Within those constraints, the actual values of $ v_1, v_1', v_2 $ and $ v_2' $ depend on the kinetic energy. KE before and after the collision is

$ KE = (m_1v_1^2 + m_2v_2^2)/2 $

$ KE' = (m_1'v_1'^2 + m_2'v_2'^2)/2 $

Then you choose the degree of inelasticity. If $ \alpha $ is the fraction of kinetic energy that's retained (so $ \alpha = 0 $ corresponds to perfect inelasticity), then

$ KE' = \alpha KE $

Once you've chosen a value for $ \alpha $, you can calculate $ KE' $ from that and the initial velocities. Given known values of $ KE' $ and 0 for kinetic energy and momentum, respectively, you're left with two equations (for net kinetic energy and net momentum) and two unknowns ($ v_1` $ and $ v_2' $), which you can solve algebraically. Then you just need to add and subtract velocities to transform in and out of the CM frame, and you're done.

  • $\begingroup$ Thank you. I tried the calculation before and it seemed hopeless, using the zero momentum frame of reference made all the difference. $\endgroup$
    – Rimilel
    Commented May 22, 2019 at 3:27

To parameterize the degree of inelasticity you use the "coefficient of restitution" which is 1 for elastic processes and 0 for completely inelastic processes.

For a two body collision this is described by $$ \text{coef. of restitution} = c_R = \frac{\text{final relative speed}}{\text{initial relative speed}} = \frac{v_2 - v_1}{u_1 - u_2} \,. \tag{*} $$ More generally it is the square root of the fractional bulk kinetic energy.

This also tells you how to compute the final state of such events (assuming the coefficient of restitution is known). You use (*) as an additional constraint on your system just as you would have used equality of kinetic energy for a elastic collision.


From a mathematical standpoint, any collision in which no mass is lost is described by two equations:

Conservation of energy: $ m_1v_1^2 + m_2v_2^2 = m_1v_1'^2 + m_2v_2'^2 + E $

Conservation of momentum: $ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' $

You know the masses and initial velocities, so this reduces to a system of two equations in three unknowns ($ v_1', v_2', E $). In order to solve it, you need to know one of those parameters: the energy lost through inelasticity, or one of the final velocities.

  • $\begingroup$ $2E$? $\frac{m v^2}{2}$? $\endgroup$
    – user121330
    Commented Jun 7, 2017 at 20:19

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