# Which collision equation is appropriate to use?

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?

• Every collision satisfies the momentum conservation. Thus $\Delta p = 0$ is always valid. Please check how you derived your first formula. Nov 27, 2019 at 21:23
• How did the heavier ball get so much faster after collision? Is it a typing error Nov 27, 2019 at 21:34
• 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... Nov 28, 2019 at 2:36
• 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 $v_1$. Nov 28, 2019 at 3:48

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:

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

Source

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.