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Billiard ball physics homework

  • Thread starter physicsss
  • Start date
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

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physicsss said:
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.
 
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

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physicsss said:
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! :smile:

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.)
 
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?
 

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