I am teaching a PHY class at the high school level. I wanted to check that this is a valid way to get the relative velocities of objects after an elastic collision to wit:

-- Usually we are told (for example on a Regent's exam) the velocity of $m_1$ and $m_2$ before the collision, and given one of the two velocities afterwards. This means that you can use conservation of momentum and solve for only one variable. (This is using stuff like the relation $m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$ ) That's great, it's a simple problem. BUT...

What if I wanted to know the velocities after a collision without knowing one of the velocities? That is, knowing only the velocities before collision could I figure out the velocities afterwards knowing only $m_1$, $v_1$, $v_2$ and $m_2$?

From the Newton's Cradle problem, it says that KE and p both have to be conserved. So my thought was that, for the general case, you assume that is true and treat them as two simultaneous equations:

$$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$$ $$\frac{1}{2}m_1v^2_1 + \frac{1}{2}m_2v^2_2=\frac{1}{2}m_1v'^2_1 + \frac{1}{2}m_2v'^2_2 $$

One thing that also struck me was that you can simplify this a bit because the $\frac{1}{2}$ cancels out

$$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$$ $$m_1v^2_1 + m_2v^2_2=m_1v'^2_1 + m_2v'^2_2 $$

And that leaves us with two equations that should be soluble. The other funny thing I noticed was that when I plotted these on a coordinate plane I got two solutions, with the shape being a circle intersecting a line.

In any case, as an example, consider a 2kg mass moving at 3 m/s and a 5 kg mass moving in the opposite direction at 2 m/s. I want to know the velocities after they collide.

For momentum:

$$(2kg)(3m/s)+(5kg)(-2m/s) = 2v'_1 + 5v'_2$$

$-4 kg \cdot m/s = 2v'_1 + 5v'_2 \tag{1}$

For KE: $$m_1v^2_1 + m_2v^2_2=m_1v'^2_1 + m_2v'^2_2 $$

$$(2kg)(3 m/s)^2 + (5kg)(-2 m/s)^2 = 2v'^2_1 + 5v'^2_2$$ $$(2kg)(9 m^2/s^2)+(5kg)(4 m^2/s^2) = 2v'^2_1 + 5v'^2_2$$ $$(18 J)+(20 J) = 2v'^2_1 + 5v'^2_2$$ $$38 J = 2v'^2_1 + 5v'^2_2$$

And I can treat these as I would $2x +5y = -4 $ and $2x^2 + 5y^2 = 38$ and just substitute to get one of my velocities.

Given that $v'_1 = \frac{-5}{2}v'_2-2 $ if I substitute in I get

$$38 J = 2(\frac{-5}{2}v'_2-2)^2 + 5v'^2_2$$ $$38 J = 2(\frac{25}{4}v'^2_2+10v'_2+4) + 5v'^2_2$$ $$38 J = (\frac{25}{2}v'^2_2+20v'_2+8) + 5v'^2_2$$ $$38 J = (\frac{35}{2}v'^2_2+20v'_2+8)$$

which is a perfectly legitimate quadratic I can solve. I can even simplify it further: $$\frac{35}{2}v'^2_2 + 20 v'_2 -30 = 0$$ $$\frac{1}{2}(35v'^2_2 + 40 v'_2 -60) = 0$$

and there are roots at $v_2 = 0.86 m/s$ and $v_2 = -2 m/s$ and then it's just a matter of seeing which root makes physical sense. I can check that with the momentum equations as well.

The whole point of this long post is to see if this is a legitimate way of doing the problem. If I have made some horrible error somewhere please do tell me.

And yes, I understand the definition of elastic collision, and all that; I realize this would be a perfectly idealized case. I am thinking in terms of physics for high school students who are NOT doing calculus. The point I am getting at is whether this is a good approximation method, or if I am missing some fundamental point.

  • $\begingroup$ "The whole point of this long post is to see if this is a legitimate way of doing the problem. If I have made some horrible error somewhere please do tell me." Check-my-work questions are off topic here. $\endgroup$ Dec 24, 2020 at 19:11
  • 2
    $\begingroup$ Wouldn't that apply to just about any question where you show work at all? $\endgroup$
    – Jesse
    Dec 24, 2020 at 19:19
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    $\begingroup$ @BioPhysiscist OP is asking for a better approach to momentum problem. "The whole point of this long post is to see if this is a legitimate way of doing the problem. If I have made some horrible error somewhere please do tell me." refers to the actual approach of the problem. $\endgroup$
    – user256872
    Dec 24, 2020 at 19:45
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    $\begingroup$ I updated my answer.... Note that both solutions to the quadratic are physically sensible, but the trivial solution involving no changes in velocities (as if there was no interaction) isn't very helpful since you knew that already. $\endgroup$
    – robphy
    Dec 24, 2020 at 20:15
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    $\begingroup$ I added a desmos visualization $\endgroup$
    – robphy
    Dec 24, 2020 at 20:32

2 Answers 2


Solving for velocities in 1D an elastic collision for two objects $A$ and $B$ boils down to solving the following system of equations,

$$ \left\{ m_Av_A + m_Bv_B = m_Av_A' + m_Bv_B'\quad \text{cons. of momentum} \atop \dfrac1 2m_Av_A^2 + \dfrac1 2m_Bv_B^2 = \dfrac1 2m_Av_A'^2 + \dfrac1 2m_Bv_B'^2\quad \text{cons. of KE} \right.$$ The above system yields the following solution when $v_A$, $v_B$ and the ratio of masses are known:

$$v_A' = \dfrac{m_A - m_B}{m_A+m_B} v_A + \dfrac{2 m_B}{m_A+m_B}v_B$$ $$v_{B}' = \dfrac{2m_A}{m_A+m_B} v_{A} - \dfrac{m_A-m_B}{m_A+m_B}v_{B}$$

These above equations are obtained using the relative velocity shortcut -- for an elastic collision, the following is true: $$v_B' + v_B = v_A ' + v_A\quad \mathrm{or}\quad v_B' -v_A' = -\left(v_B-v_A\right)$$ Then, said shortcut is substituted into the momentum equation to solve for final velocities.

More can be found here https://www.khanacademy.org/science/physics/linear-momentum/elastic-and-inelastic-collisions/v/deriving-the-shortcut-to-solve-elastic-collision-problems.

  • $\begingroup$ Thanks, this is quite helpful; I was playing around with the math a bit and thinking of how this gets applied to Newton's cradle problems. And trying to work out for myself a few whys (one of which I got from a student)... $\endgroup$
    – Jesse
    Dec 24, 2020 at 19:47
  • $\begingroup$ For Newton's cradle, you can obtain the theoretical result easily using the equations I provided. As for a more conceptual explanation, I'd explore: en.wikipedia.org/wiki/Newton%27s_cradle, especially the animations on the right side of the screen. $\endgroup$
    – user256872
    Dec 25, 2020 at 0:16
  • $\begingroup$ I've seen the Wiki, thanks. The point is that few of these really get into the math, and why (from that perspective) it works as it does. There's a hyperphysics page that touches on it, and that's one reason I got the idea for this particular exercise. $\endgroup$
    – Jesse
    Dec 25, 2020 at 1:24

Suppose you are given a general collision problem in 1-dimension with two known masses and their initial velocities.

Conservation of total momentum is an equation with two unknowns, and thus can't alone determine the final velocities.

You need additional information... like a final velocity...
or a relation between the final velocities.

For a totally inelastic collision, the equality of the final velocities provides such a relation.

For an elastic collision, conservation of total kinetic energy can provide a different relation.

To avoid the quadratic equation, it's best to write each conservation law in gain-loss form: $$\Delta K_1 =-\Delta K_2$$ $$\Delta p_1 = -\Delta p_2$$

Note that the kinetic-energy equation in gain-loss form is a difference of squares on each side.

Furthermore, you'll notice that you can simplify $\Delta K_1/\Delta p_1$ and obtain an expression involving the initial and final velocities of $m_1$.

I'll leave the rest to you.

If you solve the system using the method in the OP and you get two solutions, one of those solutions should be that each velocity is unchanged, as if there was no collision. (Both are valid solutions... but this trivial one isn't very helpful since we already knew this.)

No calculus was needed here.
However, some mathematical maturity (solving a system of linear equations, maybe dividing equations or their equivalent) is necessary.


Here is a Desmos visualization of this system of equations for an elastic collision


I'll leave it to you to get the non-trivial solution.

  • $\begingroup$ Thanks for this as well, it makes it a lot simpler to explain to non-calculus students as well! $\endgroup$
    – Jesse
    Dec 24, 2020 at 19:51
  • $\begingroup$ And yes I noticed that one solution involved not changing the v of $m_2$ :-) $\endgroup$
    – Jesse
    Dec 24, 2020 at 20:22

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