- #151

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So, angular momentum about the point (0,0) before the collision is: M/2*v1*L/sqrt(3) = M*L*v1/2sqrt(3)

After the collision the angula momentum is: 3M*v2*L/sqrt(3) = M*L*v2*sqrt(3) where v2 is velocity of the incoming ball in the center of mass frame.

>>Conservation of angular momentum means these must be equal so: v2 = v1/6.

The linear momentum before the collision is M*v1/2.

After the collision, the linear momentum is 3M*vcm, where vcm is the velocity of the center of mass.

>>Conservation of momentum means that these must be equal so: vcm = v1/6

So, immediately after the collision, the two stuck together balls at the top are moving with velocity: v2 + vcm = v1/3 to the right.

The x and y components of the velocity of the bottom left ball are:

vy = v2*sqrt(3)/2 = v1*sqrt(3)/12 and vx = vcm - v2/2 = v1/12

and for the bottom right ball are:

vy = -v1*sqrt(3)/12 and vx = v1/12

Subsequently, all four balls continue to rotate at a speed of v1/6 around the center of mass. This uniform circular motion (in the cm frame) has a constant velocity component v1/6 to the right (+x direction) added to it at all times.

How's that?