If you assume KE is convserved, I think you will find a solution is impossible, assuming one could hit both balls simultaneously.
The conservation of momentum and energy will give you the following variables:
v', v2, v3, theta1, theta2 (v' is the velocity of the first ball after collision, v2, v3 are those balls final velocitys, and theta is the respective angles.
So five unknowns, three equations (momentum in two directions, energy).
Finally, if you consider KE of spin of the balls, it is a unsolvable mess. Not all problems have solutions.
See http://forums.xkcd.com/viewtopic.php?f=17&t=21898
To solve something like this, you have to assume many things: v2=v3, theta1=-theta2, and you really have little basis of those assumptions, but that reduces the unknowns to three: v', v2;v3, theta
So if you then assume v' is zero, you now have two unknowns.
But how can a physicist assume all those things are true?
considering 3 identical balls. ball 1 with initial speed V collides elastically with balls 2 & 3 whose centres are on a line perpendicular to the initial velocity of ball 1 and ball 2 & 3 are in contact with each other. ball 1 is aimed directly at the contact point and all motion is frictionless. After collision what are velocities of each ball?
2 answers
I am going to make a lot of symmetry assumptions and assume these are hockey puck with no rotational energy
initial:
V = vx i
call ball mass = 1 kg
momentum = vx i
energy = .5 vx^2
Final
ball 1
V1 = V1x i assuming symmetry so V1y = 0
momentum = V1x i
energy = .5 Vix^2
ball 2
V2 = V2x i + V2y j
momentum = V2x i + V2y j
energy = .5 (V2x^2 +V2y^2)
ball 3
assume by symmetry V3x = V2x
and V3y = -V2y
momentum = V2x i - V2y j
energy = .5 (V2x^2 + V2y^2) same as 2
initial momentum = final momentum
along i (x direction)
Vx = V = V1x + 2 V2x
along j (y direction) 0=0
we already used this to say V3y = - V2y
initial energy = final energy
.5 V^2 = .5 [ V1x^2 + 2 V2x^2 + 2V2y^2]
============
so
============
V1x = V - 2 V2x
V^2 = V^2-4VV2x+4V2x^2 +2V2x^2 +2V2y^2
4VV2x=4V2x^2 +2V2x^2 +2V2y^2
Now I am going to assume that the only force between these ball when they hit is between ball one and each of the other two at the contact point and toward the center.
If you draw a line between the three centers you see that an equilateral triangle is formed and ball two and ball 3 go off at 30 degrees to the initial path of ball 1
So
V2y/V2x = tan 30 = 1/sqrt3
V2y = V2x/sqrt 3
then back
4VV2x=4V2x^2 +2V2x^2 +(2/3)V2x^2
4V = 4 V2x +2 V2x + (2/3)V2x = (20/3)V2x
so finally
V2x = (12/20) V = (3/5)V
V2y = (sqrt 3)V/5
V1x = V- 2 V2x = V - 6/5 V = (-1/5)V
initial:
V = vx i
call ball mass = 1 kg
momentum = vx i
energy = .5 vx^2
Final
ball 1
V1 = V1x i assuming symmetry so V1y = 0
momentum = V1x i
energy = .5 Vix^2
ball 2
V2 = V2x i + V2y j
momentum = V2x i + V2y j
energy = .5 (V2x^2 +V2y^2)
ball 3
assume by symmetry V3x = V2x
and V3y = -V2y
momentum = V2x i - V2y j
energy = .5 (V2x^2 + V2y^2) same as 2
initial momentum = final momentum
along i (x direction)
Vx = V = V1x + 2 V2x
along j (y direction) 0=0
we already used this to say V3y = - V2y
initial energy = final energy
.5 V^2 = .5 [ V1x^2 + 2 V2x^2 + 2V2y^2]
============
so
============
V1x = V - 2 V2x
V^2 = V^2-4VV2x+4V2x^2 +2V2x^2 +2V2y^2
4VV2x=4V2x^2 +2V2x^2 +2V2y^2
Now I am going to assume that the only force between these ball when they hit is between ball one and each of the other two at the contact point and toward the center.
If you draw a line between the three centers you see that an equilateral triangle is formed and ball two and ball 3 go off at 30 degrees to the initial path of ball 1
So
V2y/V2x = tan 30 = 1/sqrt3
V2y = V2x/sqrt 3
then back
4VV2x=4V2x^2 +2V2x^2 +(2/3)V2x^2
4V = 4 V2x +2 V2x + (2/3)V2x = (20/3)V2x
so finally
V2x = (12/20) V = (3/5)V
V2y = (sqrt 3)V/5
V1x = V- 2 V2x = V - 6/5 V = (-1/5)V
2 answers