Billard Ball Momentum Problem

  • Thread starter Bones
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  • #1
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Homework Statement


Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is moving upward along the y axis at 1.5 m/s, and ball B is moving to the right along the x axis with speed 4.0 m/s. After the collision (assumed elastic), the second ball is moving along the positive y axis (see figure).

http://www.webassign.net/gianpse4/9-43alt.gif

(a) What is the final direction of ball A?
______° (counterclockwise from the +x axis is positive)

(b) What are their two speeds?
v'A =____ m/s
v'B =____ m/s

Homework Equations





The Attempt at a Solution


a) 90+45=135°

b) I am not sure where to begin....I know that KE is conserved, so momentum before is equal to momentum after.
 
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Answers and Replies

  • #2
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Any takers??
 
  • #3
hage567
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Start with conservation of momentum. You will have to consider this in both the x and y directions.
 
  • #4
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Y direction: mA(1.5m/s)=mB2(VB2)
X direction: mB1(4.0m/s)=?

I am not sure how to do that when there are 2 different directions.
 
  • #5
hage567
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Have you drawn a diagram of the collision? You know the "before" and "after" of ball B. You only know the "before" of ball A. So it is going to move off at some angle relative to its original direction of motion. You need to resolve this vector into x and y components, so you can apply conservation of momentum in each direction.
 
  • #6
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I understand but I am not sure how to do that.
 
  • #8
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1.5m/ssin45 y component=1.061
1.5m/scos45 x component=1.061
|A|=sqrt 1.061^2+1.061^2
|A|=1.5
but that doesn't seem right...
 
  • #9
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Wouldn't the y component be 1.5m/s and
Asin45=1.5
A=2.12
 
  • #10
hage567
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Wouldn't the y component be 1.5m/s and
Asin45=1.5
A=2.12
The 1.5 m/s is used for the initial momentum of ball A. You are trying to find a way to express its final momentum.

I don't understand where you are getting the 45 degrees. You do not know the angle that ball A is moving after the collision. This is an unknown, and you should call it θ.

So, draw a vector at an angle theta from the y axis. This vector will represent the final momentum of ball A. You can make this into a right angled triangle by drawing a line "across" to the y-axis. So the y component of this vector can be found using trig,
so [tex]cos\theta = \frac{P_{ay}^'}{P_a^'}[/tex]
arranging this gives [tex]P_{ay}' = P_a^{'}cos\theta [/tex], where [tex]P_{ay}^'[/tex] is the y-component of A's final momentum, and [tex]P_a'[/tex] is A's total final momentum.

Now do the same thing for the other side of the triangle (the x-component).
 
  • #11
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Pax=Pa(sinθ)
 
  • #12
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I know the before and after are set equal to each other because the collision is elastic and I know that the the velocity of ball B before is 4.0m/s and is is moving to the right along the x axis and the velocity of ball A before is 1.5m/s and is moving up along the y axis. I know ball B moves up the y axis after collision. I know ball A is going to move at an angle and I need to break it up into it's x and y components, but I am not sure how to set up the actual equations for the before and after of the x and y directions for the problem. Can someone please help me with this.
 
  • #13
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Ok, I think I got the setup right.
X direction: mB(4.0m/s)=mA'vA'Cosθ
Y direction: mA(1.5m/s)=mB'vB'+mA'vA'Sinθ
Do I need to take into account the change in kinetic energy? Also, can someone help me solve this, it is due today.
 
  • #14
hage567
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Those equations look reasonable to me. Conservation of energy becomes useful as it gives you another equation that allows you relate unknowns. You will need to use it because you have three unknowns (vA', vB' and θ), and currently only two equations.
 
  • #15
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Ok.
1/2mB(4.0m/s)^2+1/2mA(1.5m/s)^2=1/2mB'vB'^2+1/2mA'vB'^2

Is that correct?
 
  • #16
hage567
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Ok.
1/2mB(4.0m/s)^2+1/2mA(1.5m/s)^2=1/2mB'vB'^2+1/2mA'vB'^2

Is that correct?

Yes, that's correct. Now you should be able to get your final answers.
 

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