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rendra1991 | 16:36 Fri 16th Oct 2009 | Science
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A billiard ball with an initial speed of 10 m/s collides elastically with two identical billiard
balls whose centers are on a line perpendicular to the initial velocity and that are initially in
contact with each other as shown in the figure above. The first ball is aimed directly at the point
of contact between the other two balls. All the balls are frictionless. Find the velocities of all
three balls after the collision. Hint: consider the direction of the impulsive force of the moving
ball on each of the two stationary balls at the instant of the collision.

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Using the three centres to define a 2D plane and taking the line in that plane that passes through the centre of the 1st ball and the point of contact between the other 2 as the x-axis, consider the conservation of momentum in both the x and y directions.
To expand a little

the mass of the billiard ball is M the cue ball transfers it's momentum to the other 2

but that's just in the x direction imagine the 3 balls at the point of impact

If they are of equal size - join the centres to form an equilateral triangle. (60 degrees)

You know what the momentum of each should be in the x direction so a bit of trigonometry will tell you what the velocity should be for the x componant to be that
The answers are:
- The 2 balls which were together afterwards each have a velocity of 4 x sqrt(3) m/s, each moving off with an angle of 60 degrees between them.
- Original moving ball now has a velocity of 2 m/s in the opposite direction to its original direction

So the balls trace out a capital Y shape.

To solve this you need to use the conservation of momentum and the conservation of Kinetic energy.

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