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?

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?

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

To determine the velocities of the three balls after the collision, we can apply the law of conservation of momentum and the law of conservation of kinetic energy. Here's how we can approach it step by step:

1. Assign variables:
Let V be the initial speed of ball 1 and v1, v2, and v3 be the velocities of balls 1, 2, and 3 after the collision, respectively.

2. Law of Conservation of Momentum:
In an elastic collision, the total momentum before and after the collision remains conserved. The initial momentum is given by:
Initial momentum = mass of ball 1 * velocity of ball 1
(Since all three balls are identical, their masses are the same.)
Therefore, the initial momentum is m * V, where m is the mass of any ball.

After the collision, the final momentum is given by:
Final momentum = mass of ball 1 * velocity of ball 1 + mass of ball 2 * velocity of ball 2 + mass of ball 3 * velocity of ball 3

Since the balls are in contact and moving in a line perpendicular to the initial velocity of ball 1, the momentum of ball 2 and ball 3 together is zero, as they move together.

Therefore, the final momentum simplifies to:
Final momentum = mass of ball 1 * velocity of ball 1 + 0 + 0
= mass of ball 1 * velocity of ball 1
= m * v1 (since v2 = v3 = 0)

Setting the initial and final momenta equal, we get:
m * V = m * v1
Canceling out the mass, we obtain:
V = v1 ... Equation 1

3. Law of Conservation of Kinetic Energy:
In an elastic collision, the total kinetic energy before and after the collision remains conserved.
The initial kinetic energy is given by:
Initial kinetic energy = (1/2) * mass of ball 1 * (velocity of ball 1)^2
= (1/2) * m * V^2

After the collision, the final kinetic energy is given by:
Final kinetic energy = (1/2) * mass of ball 1 * (velocity of ball 1)^2 + (1/2) * mass of ball 2 * (velocity of ball 2)^2 + (1/2) * mass of ball 3 * (velocity of ball 3)^2

Since ball 2 and ball 3 move together, their velocities have the same magnitude but in opposite directions.

Therefore, the final kinetic energy simplifies to:
Final kinetic energy = (1/2) * mass of ball 1 * (velocity of ball 1)^2 + (1/2) * mass of balls 2 & 3 * (velocity of ball 2)^2

Substituting the values, we have:
Final kinetic energy = (1/2) * m * (v1)^2 + (1/2) * 2m * (v2)^2 (since mass of ball 2 = mass of ball 3 = m)

Simplifying further, we get:
Final kinetic energy = (1/2) * m * (v1)^2 + m * (v2)^2

Since the balls are frictionless, the kinetic energy remains conserved.
Therefore, the initial and final kinetic energies are equal, giving us:
(1/2) * m * V^2 = (1/2) * m * (v1)^2 + m * (v2)^2

Cancelling out the mass and rearranging, we have:
V^2 = (v1)^2 + 2 * (v2)^2 ... Equation 2

4. Solving the equations:
We have Equation 1: V = v1
And Equation 2: V^2 = (v1)^2 + 2 * (v2)^2

Substituting V = v1 in Equation 2, we get:
(v1)^2 = (v1)^2 + 2 * (v2)^2

Simplifying, we find:
0 = 2 * (v2)^2

Since (v2)^2 cannot be negative, the only solution is v2 = 0.

Using v2 = 0 in Equation 1, we have:
V = v1 => v1 = V

Therefore, the velocities of the three balls after the collision are:
v1 = V
v2 = 0
v3 = 0