Take this example problem:

A 2 kg ball moving east at 3 m/s that enters in collision with another 1 kg ball moving west at 2 m/s. After the collision, the 2 kg ball has an eastward speed of 10 m/s. What is the final speed of the second ball?

The answer given is 6.7 m/s [E]. I can find this with

$v_2' = (\frac{2m_1}{m_1+m_2})v_1 $

where $v_1$ is the speed from the frame of reference of $m_2$.

But, when I use the equation

$\Delta p=0\\m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$

and solve for $v_2'$, the answers are not the same. How do I know when to use one equation or the other?

  • 1
    $\begingroup$ Every collision satisfies the momentum conservation. Thus $\Delta p = 0$ is always valid. Please check how you derived your first formula. $\endgroup$
    – Semoi
    Nov 27, 2019 at 21:23
  • 1
    $\begingroup$ How did the heavier ball get so much faster after collision? Is it a typing error $\endgroup$ Nov 27, 2019 at 21:34
  • $\begingroup$ For the example problem to be correct, the two balls would need to pass through each other, with some internal structure on one ball giving a kick to the other ball... $\endgroup$
    – DJohnM
    Nov 28, 2019 at 2:36
  • $\begingroup$ Velocity is a vector, so it has a direction. This means that final velocity can be positive or negative. Since mass is always positive, your top equation doesn't allow for the case where the final velocity can have a different sign than $v_1$. $\endgroup$ Nov 28, 2019 at 3:48

1 Answer 1


TL;DR: The first equation was incorrect and the example collision is problematic. You should generally use conservation of momentum.

The first equation should be:

$$v'_2 = \frac{2m_1}{m_1+m_2}v_1+\frac{m_2-m_1}{m_1+m_2}v_2$$


However, this equation is only true for perfectly elastic collisions (i.e. total kinetic energy is conserved). Your example problem isn't an example of a perfectly elastic collision though:

  • Initial kinetic energy of the system = $0.5\cdot2\cdot3^2+0.5\cdot1\cdot(-2)^2=11J$.

  • Final kinetic energy of the 2kg ball is $0.5\cdot2\cdot100^2=100J$, which is a lower bound for the final kinetic energy. $100J>>11J$!

This is only possible if the 1kg ball has an internal structure that pushed the 2kg ball when they collided (or else conservation of energy is violated).

If this were a perfectly elastic collision, the final velocity of the 2kg ball would be $-\frac{1}{3}ms^{-1}$ (taking eastwards as the positive direction), as given by the formula:

$$v'_1 = \frac{m_2-m_1}{m_1+m_2}v_1+\frac{2m_1}{m_1+m_2}v_2$$

In most situations, we don't know if the collision is elastic or inelastic. So, use conservation of momentum.


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