A question about $y$-axis momentum in an elastic collision involving billiard balls of equal mass

Question

My physics textbook asks me to prove that when a billiard ball moving along the x-axis hits another billiard ball of equal mass in an elastic, non-head on, collision, the two balls will move away from each other at a 90 degree angle (see attached picture).

I understand their explanation except for one detail: the analysis - as one would expect - starts with the formulae for momentum for both the $x$- and the $y$-ingredients of the balls' motion. For the one dealing with the y-ingredient, they give: $v_1\sin\Theta_1-v_2\sin\Theta_2=0$.

This is of course based on the fact that we defined the x-axis according to the initial movement of the moving ball, which means that there was no y-component of momentum, so the y-component needs to total 0 after the collision as well.

My problem is the following: as a formula I learnt that $m\mathbf{V}_1 + m\mathbf{V}_2 = m\mathbf{v}_1 + m\mathbf{v}_2$, not $m\mathbf{V}_1 + m\mathbf{V}_2 = m\mathbf{v}_1 - m\mathbf{v}_2$.

I assume that the minus in the answer is there because we know that the velocity of the second ball along the y-axis is in the opposite direction of the other ball.

But that leads to another oddity, namely that if I rearrange the formula, I get: $v_1\sin\Theta_1=v_2\sin\Theta_2$ rather than $v_1\sin\Theta_1=-v_2\sin\Theta_2$.

So the velocities end up being identical, despite the fact that they are going in opposite directions…

Could somebody clear up the confusion?

Thank you!

The answer:

Answer 1

Both equations are "correct".

In this problem you are using the fact that the initial momentum in the $\hat y$ direction is zero and so the final momentum in the $\hat y$ direction must also be zero.

$m \,\vec 0 = m \,\vec v_{\rm 1y} + m \,\vec v_{\rm 2y}$

$\vec v_{\rm 1y}$ is no problem because $\vec v_{\rm 1y}= v_{\rm 1y} \, \hat y$ where $v_{\rm 1y}$ is either the component in the $\hat y$ direction or the magnitude of the vector $\vec v_{\rm 1y}$ and both representations will produce a positive value for $v_{\rm 1y}$.

Now what about $\vec v_{\rm 2y}$?

You could say that $\vec v_{\rm 2y} = v_{\rm 2y} \hat y$ where $v_{\rm 2y}$ is the component of $\vec v_{\rm 2y}$ in the $\hat y$ direction and when the calculations are done it will be found that $v_{\rm 2y}$ is a negative quantity, ie $\vec v_{\rm 2y}$ is in the $(-\hat y)$ direction.
Using this notation $$\vec 0 = \vec v_{\rm 1y} + \vec v_{\rm 2y} \Rightarrow 0 \,\hat y= v_{\rm 1y}\,\hat y + v_{\rm 2y} \, \hat y \Rightarrow v_{\rm 1y}= - v_{\rm 2y}$$

On the other hand you could say that $\vec v_{\rm 2y} = v_{\rm 2y} (-\hat y)$ where $v_{\rm 2y}$ is the component of $\vec v_{\rm 2y}$ in the $(-\hat y)$ direction and when the calculations are done it will be found that $v_{\rm 2y}$ is a positive quantity.
Using this notation $$\vec 0 = \vec v_{\rm 1y} + \vec v_{\rm 2y} \Rightarrow 0 \,\hat y= v_{\rm 1y}\,\hat y + v_{\rm 2y} \, (-\hat y) \Rightarrow v_{\rm 1y}= v_{\rm 2y}$$

An equivalent statement in this case is to say that $\vec v_{\rm 2y} = (-v_{\rm 2y}) \hat y$ where $v_{\rm 2y}$ is the magnitude of $\vec v_{\rm 2y}$ and when the calculations are done it will be found that $v_{\rm 2y}$ is a positive quantity which must be so as magnitudes are always positive.
Using this notation $$\vec 0 = \vec v_{\rm 1y} + \vec v_{\rm 2y} \Rightarrow 0\, \hat y= v_{\rm 1y}\,\hat y + (-v_{\rm 2y}) \, \hat y \Rightarrow v_{\rm 1y}= v_{\rm 2y}$$

In the first method you are making no judgment about the direction of the velocity $\vec v_{\rm 2y}$ and when you have done the calculation because you find that component $v_{\rm 2y}$ is negative you then know that it is in the $(-\hat y)$ direction.

In the second and third methods you are making a judgment about the direction of the velocity $\vec v_{\rm 2y}$ as being in the $(-\hat y)$ direction and when you have done the calculation you expect the numerical value of $v_{\rm 2y}$ to be positive.

PS To prove the angle is a right angle for an elastic collision (kinetic energy conserved)

$$\frac 12 m v^2 = \frac 12 m v_1^2 + \frac 12 m v_2^2 \Rightarrow v^2 = v_1^2 + v_2^2$$

ie the velocity triangle $\vec v = \vec v_1 + \vec v_2$ is right-angled - Pythagoras.