# Momentum in a game of pool

by VitaX
Tags: game, momentum, pool
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 P: 184 1. The problem statement, all variables and given/known data In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at 3.10 m/s along a line making an angle of 27.0° with its original direction of motion, and the second ball has a speed of 2.30 m/s. Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. 2. Relevant equations Conservation of Momentum 3. The attempt at a solution Not sure how to solve this at all. Only thing I am sure of is my drawing. What steps do I use to go about finding the angle?
 Sci Advisor HW Helper Thanks P: 26,148 Hi VitaX! Call the angle θ, do x and y components of momentum (separately), and solve.
P: 184
 Quote by tiny-tim Hi VitaX! Call the angle θ, do x and y components of momentum (separately), and solve.
My teacher told me this through an email:

m v1i cos 0 + m(0) = m (3.1) cos 27 + m (2.3) cos (theta 2) ... (1)

m v1i sin 0 + m(0) = m (3.1) sin 27 - m (2.3) sin (theta 2) ... (2)

I see that here mass is irrelevant so it all cancels out. Why in the 2nd equation is it minus 2.3m*sin(theta) instead of plus? Is it just to make sure the angle I get comes out to be positive instead of negative?

Sci Advisor
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P: 26,148
Momentum in a game of pool

Hi VitaX!

(what happened to that θ i gave you? )
 Quote by VitaX Why in the 2nd equation is it minus 2.3m*sin(theta) instead of plus? Is it just to make sure the angle I get comes out to be positive instead of negative?
ah, I expect you drew the diagram logically, and measured both angles anti-clockwise from the right, as usual …

that makes one angle positive, and the other (θ2) negative.

But usuallly we draw the diagram realistically, with the the balls (or whatever) going the way we expect, ie one ball goes to the left, so the other must go to the right, so we draw it that way …

that of course makes θ2 positive (which, as you suggested, is the purpose)!

It really makes no difference … if you substitute -θ2 for θ2 in the equation, you get the same actual results.
(btw, what did you not see before the teacher emailed you? would you be able to do a similar problem in future, or is there still something worrying you about it?)
P: 184
 Quote by tiny-tim Hi VitaX! (what happened to that θ i gave you? ) ah, I expect you drew the diagram logically, and measured both angles anti-clockwise from the right, as usual … that makes one angle positive, and the other (θ2) negative. But usuallly we draw the diagram realistically, with the the balls (or whatever) going the way we expect, ie one ball goes to the left, so the other must go to the right, so we draw it that way … that of course makes θ2 positive (which, as you suggested, is the purpose)! It really makes no difference … if you substitute -θ2 for θ2 in the equation, you get the same actual results. (btw, what did you not see before the teacher emailed you? would you be able to do a similar problem in future, or is there still something worrying you about it?)
I knew it was conservation of momentum. But I would not have known that you have to write the components for x and y utilizing the angles given. That was where I was confused. I suspect that the angle is 0 at the initial because that's the path the ball is taking up until it hits the ball at rest.
Sci Advisor
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Thanks
P: 26,148
 Quote by VitaX I knew it was conservation of momentum. But I would not have known that you have to write the components for x and y utilizing the angles given. That was where I was confused. I suspect that the angle is 0 at the initial because that's the path the ball is taking up until it hits the ball at rest.
Yes: general rule …

give unknown things (like angle or velocity) letters such as θ or v, and just plough ahead, writing all the equations.

Often it's messy, and you have two or more equations that you have to solve simultaneously, and you can't see where you're going until you get there!

But you can always help by choosing your original direction so as to make the equations simpler … in this case by choosing (and it is a choice, it's not compulsory!) 0 angle as the initial direction of the cue ball.

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