I want to predict how a collision will go with physics. Say I have a perfectly elastic 1-dimensional system with a frictionless ground and two objects of masses say $m_1$ and $m_2$ moving with initial velocities $u_1$ and $u_2$ respectively. With only this knowledge, (no knowledge of the final velocities of either object) the law of conservation of momentum or the law of conversation of energy doesn't seem to help.

Say I were to take to objects with some mass and some initial velocity and collide then and measure their final velocities, I would for sure get the same results if I repeat the experiments, again and again, my point being - I seem to have everything fixed, what am I missing?

Let: $m_1$ = 1 kg; $m_2$ = 2 kg; $u_1$ = 5 m/s; $u_2$ = 0 m/s.

To find final velocities say $v_1$ and $v_2$: $$ p_\text{before} = p_\text{after} $$ Therefore: $$ \begin{gather} m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \\ 5 \text{ kgm/s} = m_1 v_1 + m_2 v_2 \end{gather} $$ I seem to have no idea what to do next. I have a similar fate with the law of conversation of Energy: $$ 12.5 \text{ J} = \frac 1 2 m_1 v_1^2 + \frac 1 2 m_2 v_2^2 $$ I could try to simplify but all my attempts end in vain. What am I missing?


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    $\begingroup$ Isolate one of the velocities in the momentum conservation law and substitute in the conservation of energy. After one is determined the other should be easy. $\endgroup$ – ErickShock Feb 14 at 16:49
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    $\begingroup$ You have two equations (momentum and energy conservation) and two unknown values (v1 and v2). Why do you think you can't determine these two unknown values then? $\endgroup$ – Aaron Stevens Feb 14 at 16:51
  • $\begingroup$ Wow i seem to have solved this $\endgroup$ – Anonymous Feb 14 at 16:58
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    $\begingroup$ @Anonymous, also have a look at the center of mass reference frame. That reference frame makes your elastic collision problem almost trivial. $\endgroup$ – David White Feb 14 at 17:18

As mentioned in the comments, you can in fact solve these equations to determine the final velocities after collision. Let's try it using your example. We'll start by solving the momentum equation for one of the velocities, say $v_1$. \begin{equation} v_1 = \frac{5-m_2 v_2}{m_1} \end{equation} Now plug this into $v_1$ in your energy expression. \begin{equation} 12.5 = \frac{1}{2}m_1 (\frac{5-m_2 v_2}{m_1})^2+ \frac{1}{2}m_2 v_2^2 \end{equation} This is a quadratic equation in $v_2$, which can be solved using, say, the quadratic formula. Let's use the same masses for before and after the collision. (assuming the blocks stay intact!) Then you will find that this equation has two solutions. \begin{equation} v_2 = 0, \,\, v_2 = \frac{10}{3} = 3.33333 \,\text{m/s} \end{equation} Plugging back in to $v_1$ gives $v_1=5$ in the first case, $v_1 = -\frac{5}{3}$ in the second. These are both valid possibilities. So which one is right?

Well, it turns out you can't glean this from considering momentum or energy conservation alone. You must also know something about the type of interaction itself. (Can you guess what type of interaction the $v_2 = 0$ case might correspond to?) This isn't much of a barrier here, but in more than one dimension, you would find not just two possible trajectories but, in fact, and infinite number of them! Nevertheless, it is interesting that you can learn so much just from energy and momentum without knowing how the masses "talk" to each other.

  • $\begingroup$ I'm guessing v2 = 3.33? I say this because if the first case were true, object 1 would have to phase through object 2. Hold on I tried the same thing with other values and got a quadratic equation in v2 but with two positive answers. How do I know which one is right? $\endgroup$ – Anonymous Feb 14 at 19:12
  • $\begingroup$ My values were: m1 = 2kg; u1 = 5m/s; m2 = 5kg; u2=2m/s. I got v2 = 2 or v2=25/7. Either seems probable. $\endgroup$ – Anonymous Feb 14 at 19:24
  • $\begingroup$ Right, so which one corresponds to no interaction? What would you expect to see in that case, and does it match one of the two solutions you got? $\endgroup$ – Jacob Feb 14 at 19:26
  • $\begingroup$ I don't think I understand what you mean by 'no interaction' $\endgroup$ – Anonymous Feb 14 at 19:27

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