What Is the Final Direction and Speed of Each Billiard Ball After Collision?

[physicsss]

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. One is moving upward along the y-axis at 3 m/s, and the other is moving to the right along the x-axis with speed 4 m/s. After the collision (assumed elastic), the second ball is moving along the positive y axis.



(a) What is the final direction of the first ball?

(b) What are their two speeds?

OK, I use two different for each x and y component:
x-direction: m*4=m*v1'*cos(theta)
y-direction: m*3=m*v2'+m*v1'*sin(theta)

Now I can cancel the masses, but still, I have two equations but 3 unknowns...

 

[Doc Al]

Now I can cancel the masses, but still, I have two equations but 3 unknowns...

There is a third equation you can use: Realize that the collision is elastic.

 

[physicsss]

No matter what I do, I can't get the answer.(yes, I used conservation of energy as my third equation) Can I assume that ball a will move along the positive x-axis after collision?

 

[Doc Al]

No matter what I do, I can't get the answer.(yes, I used conservation of energy as my third equation)

So you should have 3 equations with 3 unknowns.

Can I assume that ball a will move along the positive x-axis after collision?

Of course not!

Here's a trick to simply the two momentum equations and eliminate theta: Rewrite your x-direction equation to isolate v1cos(theta) and your y-direction equation to isolate v1sin(theta). Square both sides of each equation and add them. Use a simple trig identity. Now combine with energy conservation to solve for V1 and V2. (Then go back and solve for theta.)

 

[Skomatth]

In an elastic collision the the relative speed before the collision is equal to the opposite of the negative relative speed after the collision v1-v2=-(v1-v2)
I'm not sure how this applies in 2 dimensions though, could someone explain?