# Kinetic Energy Poll Ball Question

1. Dec 11, 2008

### lilbunnyf

1. The problem statement, all variables and given/known data
Two identical billiard balls are at rest on a frictionless, level surface, touching each other at one common point. A third, identical ball, the cue ball, is approaching along the common tangent with a constant speed of 20m/s. Assuming a completely elastic collision, with no spin on any balls, and making(reasonable) assumptions about symetry, calculate the velocity of the cue ball after the collision. The graph is as shown below

(the masses of the 3 balls are the same, but the value is unknown)
o------->8
20m/s

2. Relevant equations
I think two equations could be used in this question, one is
Pt = (20m/s)M = Pt= (v1 m/s)M + (v2 m/s)M + (v3 m/s)M
the other one is The Conservation of Energy
Ek= Ek

(M(20M/S)^2)/2 = (M(v1 m/s)^2)/2 + (M(v2 m/s)^2)/2 + (M(v3 m/s)^2)/2

3. The attempt at a solution
The only thing i can do is assume that ball1 and ball2 have the same speed after the collision
so that i got 2 equations

no.1 20M= 2(V1)M + (Vc)M, where V1 is the velocity of ball1 and ball2,Vc is the value we r looking for

no.2 (20^2)M/2 = 2M(V1^2)/2 + (Vc^2)M/2

and by substituting 20M-2V1=Vc in no.1 into no.2, I got V1 = 13.33, which will result in Vc= - 6.67m/s

but the answer in the book says Vc=-4m/s ... Did i do something wrong with my equations? Helps/Suggestions are really appreciated, Thanks in advanced!!

2. Dec 12, 2008

### rl.bhat

[QUOTE=lilbunnyf;1998392]1.
2. Relevant equations
I think two equations could be used in this question, one is
Pt = (20m/s)M = Pt= (v1 m/s)M + (v2 m/s)M + (v3 m/s)M
the other one is The Conservation of Energy
Ek= Ek

(M(20M/S)^2)/2 = (M(v1 m/s)^2)/2 + (M(v2 m/s)^2)/2 + (M(v3 m/s)^2)/2

QUOTE]

V1, V2 and V3 are not in the same direction. Take the camponents of V2 and V3 in the direction and perpendicular to the direction of cue ball.

3. Dec 12, 2008

### lilbunnyf

V2 and V3 are in opposite direction since they are perpendicular to V1? Btw why would they be perpendicular?

4. Dec 12, 2008

### cepheid

Staff Emeritus
No he's not saying they're perpendicular, he's saying that v2 and v3 are vectors with some general directions (after the collisions) that you don't know, and therefore you're going to have to break up these vectors into *components* parallel to, and perpendicular to, the cue ball's initial direction.

5. Nov 14, 2011

### mbig

But even after breaking them up into components, you don't know the angles. How do you solve this question?

I get vc = vb1cosθ1 + vb2cosθ2 + vc' and 0 = vb1sinθ1 + vb2sinθ2 where vb1 and vb2 are the billiard balls and vc is cue ball

6. Nov 14, 2011

### Redbelly98

Staff Emeritus
Welcome to Physics Forums.

Are you currently working on this problem yourself? This question was originally posted about 3 years ago.

At any rate, you don't have to use vb1, θ1, vb2, θ2 to express the velocities. It may be easier instead to use vb1x, vb1y, vb2x, vb2y.

7. Nov 14, 2011

### mbig

Yes, I am working on this problem...but I did break them into components just labeled differently.

x-component : vc = vb1cosθ1 + vb2cosθ2 + vc OR vc = vb1x + vb2x + vc'
y-component : 0 = vb1sinθ1 + vb2sinθ2 OR 0 = vb1y + vby2

But I still don't know vb1x and vb2x.

8. Nov 14, 2011

### Staff: Mentor

In the instant that all three balls are touching, their centers form an equilateral triangle. In what direction will the impulses delivered to each stationary ball by the cue ball be delivered?

9. Nov 14, 2011

### PeterO

Why is it that common tangent the cue ball is approaching along? I envisaged the cue ball approaching roughly parallel to the two stationary balls, off-set by half a radius???

10. Nov 14, 2011

### Staff: Mentor

I see it thus:

#### Attached Files:

• ###### Fig1.jpg
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11. Nov 14, 2011

### mbig

Thanks for all your input, but how do I setup the equations here?

12. Nov 14, 2011

### Staff: Mentor

Conservation of momentum in the horizontal direction, conservation of energy overall. You should be able to deduce the angle at which the balls depart the collision.