I know that there are three different collisions elastic, inelastic and perfectly inelastic. I wanted to know how the three velocities would rank respective to the type of collision. For example, if a ball were to collide with marble, which type of collision would maximize and/or minimize the velocity of the marble.

From my understanding, the velocities from greatest to least should go like- Elastic, Inelastic, Perfectly Inelastic.

  • $\begingroup$ This is a rather vague. Which velocity are you talking about, exactly? $\endgroup$
    – QuIcKmAtHs
    Commented Nov 19, 2018 at 8:52
  • $\begingroup$ It's a very very very vague question. Please try to reframe the question . $\endgroup$ Commented Nov 19, 2018 at 14:07

1 Answer 1


To make your example more concrete, let's assume we are looking at a 1D collision where an object of mass $m_1$ moving with velocity $v$ collides with an object of mass $m_2$ initially at rest (I will leave the analysis to you for the case where the second object is initially moving). Afterwards, the velocities of the objects are, respectively, $v_1$ and $v_2$.

Since momentum is conserved in any of our collisions, the following will always hold: $$m_1v=m_1v_1+m_2v_2$$

Let us consider each type of collision:

Elastic Collision

In elastic collisions, kinetic energy is also conserved, so we have $$\frac12m_1v^2=\frac12m_1v_1^2+\frac12m_2v_2^2$$

Through doing some fun algebra, we can show that our two conservation conditions leads to the result that the relative velocity between the two objects switches directions. In other words: $$v=v_2-v_1$$

Putting this back into our momentum conservation equation, we end up getting: $$v_2=\frac{2m_1v}{m_1+m_2}$$

Perfectly Inelastic Collision

In this collision, the objects stick together and move with the same velocity after the collision, so $v_1=v_2$ and our momentum conservation equation can be rewritten as $$m_1v=(m_1+m_2)v_2$$ or, $$v_2=\frac{m_1v}{m_1+m_2}$$

Inelastic Collision

For this final collision, there isn't a way to know what will happen exactly. It depends on how much energy leaves the system. We can, however, determine bounds on $v_2$.

Our momentum conservation still holds, but now we have a loss of energy: $$\frac12m_1v^2>\frac12m_1v_1^2+\frac12m_2v_2^2$$

We can follow the same algebra as in the elastic case wile being careful with the inequality to arrive at: $$v>v_2-v_1$$ which leads to $$v_2<\frac{2m_1v}{m_1+m_2}$$

We cannot get a lower bound larger than $0$ on $v_2$ in this case. All we can say is the the kinetic energy that is lost will be less than what is lost in the perfectly inelastic collision.

This (as well as everything discussed above) can be seen in an example plot. If we plot the change in kinetic energy as a function of $v_2$ assuming that momentum is conserved we get this: change in K vs. v2

(in this plot I used $m_1 = 3\ \rm{kg}$, $m_2=1\ \rm{kg}$, $v=1\ \rm{m/s}$, and by momentum conservation, $v_1=\frac{m_1v-m_2v_2}{m_1}$)

As you can see, we get the most kinetic energy loss for the perfectly inelastic case when $v_2=\frac{m_1v}{m_1+m_2}$ ($v_2=0.75\ \rm{m/s}$ in this case). However, $v_2$ can actually be less than this value and momentum still be conserved with energy loss.

Therefore, all you can say is that the velocity of the second object after an elastic collision is larger than it is for an inelastic collision.


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