Finding speed of two billiard balls in elastic collision

[I_Try_Math]

Homework Statement

Two identical billiard balls collide. The first one is initially traveling at $(2.2m/s)\hat{i}−(0.4m/s)\hat{j}$ and the second one at $−(1.4m/s)\hat{i}+(2.4m/s)\hat{j}$. Suppose they collide when the center of ball 1 is at the origin and the center of ball 2 is at the point (2R,0) where R is the radius of the balls. What is the final velocity of each ball?

Relevant Equations

$p_i = p_f$
$KE_{i} = KE_{f}$

$\vec{v_{1i}} = \langle 2.2, -0.4 \rangle$
$\vec{v_{2i}} = \langle -1.4, 2.4 \rangle$

Let $\theta$ and $\phi$ be the angles made by $\vec{v_{1i}}$ and $\vec{v_{2i}}$ with the x axis, respectively.
Let $\theta'$ and $\phi'$ be the angles made by $\vec{v_{1f}}$ and $\vec{v_{2f}}$ with the x axis, respectively.

$p_{ix} = p_{fx}$
$mv_{1ix} + mv_{2ix} = mv_{1fx} + mv_{2fx}$
$v_{1ix} + v_{2ix} = v_{1fx} + v_{2fx}$
$0.8 = v_{1f}\cos(\theta') + v_{2f}\cos(\phi')$


$p_{iy} = p_{fy}$
$v_{1iy} + v_{2iy} = v_{1fy} + v_{2fy}$
$2 = v_{1f}\sin(\theta') + v_{2f}\sin(\phi')$


$KE_{i} = KE_{f}$
$\frac{1}{2}mv_{1i}^2 + \frac{1}{2}mv_{2i}^2 = \frac{1}{2}mv_{1f}^2 + \frac{1}{2}mv_{2f}^2$
$v_{1i}^2 + v_{2i}^2 = v_{1f}^2 + v_{2f}^2$
$12.75 = v_{1f}^2 + v_{2f}^2$

If I understand correctly I should be able to solve it using the the three blue equations. The problem is everytime I try the algebra gets kind of unwieldy and complicated, making me think my work is wrong or there is a different approach to the problem which is conceptually simpler.

Am I on the right track with this problem solving strategy? If so is there is a trick to make the algebra easier? Perhaps a trig identity?

 

[kuruman]

Make a good drawing of the two balls at the moment of the collision. Note that the momentum of each ball changes only in the direction along the line joining their centers.

 

[hutchphd]

The easiest way is to go to the Center of mass frame. For equal masses this is equivalent to defining vector coordinates $\mathbf v_1 +\mathbf v_2$ and $\mathbf v_1 - \mathbf v_2$.

 

[I_Try_Math]

Make a good drawing of the two balls at the moment of the collision. Note that the momentum of each ball changes only in the direction along the line joining their centers.

Okay that sounds like an interesting observation. I'll give the drawing a go and see what I can come up with.

 

[haruspex]

I should be able to solve it using the the three blue equations.

No, because you have four unknowns.
What you are not using is that the impact, being presumed frictionless, can only transfer momentum normally to the contact plane, i.e. along the line joining the centres at that instant.

 

[I_Try_Math]

No, because you have four unknowns.
What you are not using is that the impact, being presumed frictionless, can only transfer momentum normally to the contact plane, i.e. along the line joining the centres at that instant.

So this would mean the y components of the final velocities would be equal to the y components of the initial velocities, correct?

Even if that's right I can't figure out how to use it to solve the equations.

 

[haruspex]

So this would mean the y components of the final velocities would be equal to the y components of the initial velocities, correct?

Yes.

Even if that's right I can't figure out how to use it to solve the equations.

It reduces it to a one dimensional problem. Not only is the X momentum conserved, the X component of the KE is.
Don't introduce angles, just work with the momentum and KE in the X direction.

A useful trick is to use Newton's Experimental Law:
$v_{1f}-v_{2f}=R(v_{2i}-v_{1i})$ where R is the coefficient of restitution.
Note the sign reversal between before and after.
This is important mostly in inelastic collisions, but is useful also in one dimensional elastic ones, where R=1. It can be deduced from momentum and KE conservation but avoids the quadratics.

 

[I_Try_Math]

Yes.

It reduces it to a one dimensional problem. Not only is the X momentum conserved, the X component of the KE is.
Don't introduce angles, just work with the momentum and KE in the X direction.

A useful trick is to use Newton's Experimental Law:
$v_{1f}-v_{2f}=R(v_{2i}-v_{1i})$ where R is the coefficient of restitution.
Note the sign reversal between before and after.
This is important mostly in inelastic collisions, but is useful also in one dimensional elastic ones, where R=1. It can be deduced from momentum and KE conservation but avoids the quadratics.

I was finally able to solve it with your hints after hours of trying. Interestingly I don't think I've ever needed to break up kinetic energy in x and y components but it came in very handy for this problem. Also I'll definitely check out Newton's Experimental Law. Anyway thanks a bunch for your help.

 

[kuruman]

If so is there is a trick to make the algebra easier?

Now that you solved the problem, yes there is a trick that exploits the symmetry afforded by the equality of the masses. If you have solved the general 1D perfectly elastic collision problem before, you might have noticed that when the colliding masses are equal, they simply exchange velocities.

In this particular case, the center to center distance is conveniently along the x-axis. That's also the direction along which momentum is exchanged. So you simply write down the final velocities with the x-components swapped and the y-components the same as before the collision. Clearly, momentum and energy are conserved. No algebra involved.

 

[I_Try_Math]

Now that you solved the problem, yes there is a trick that exploits the symmetry afforded by the equality of the masses. If you have solved the general 1D perfectly elastic collision problem before, you might have noticed that when the colliding masses are equal, they simply exchange velocities.

In this particular case, the center to center distance is conveniently along the x-axis. That's also the direction along which momentum is exchanged. So you simply write down the final velocities with the x-components swapped and the y-components the same as before the collision. Clearly, momentum and energy are conserved. No algebra involved.

That's a very clever way to break it down. Certainly seems obvious in hindsight.