The final radius (distance between particle centers) is not zero. It is twice the radius.
Subtract the final potential energy of the complete system from the initial P.E.
The difference will be the final kinetic energy, which will be evenly divided between the three particles (because of symmetry).
If a is the distance between particle centers in an equilateral triangle configuration, the PE of the system is
-Gm^2/a -2Gm^2/a = -3Gm^2/a
P.E change = 3Gm^2[1/.136 - 1/9.9]
1/3 of that is the KE at collision, per particle.
Suppose that each of the three masses in the figure below has a mass of 5.05 kg, a radius of 0.0679 m, and are separated by a distance of 9.9 m. If the balls are released from rest, what speed will they have when they collide at the center of the triangle? Ignore gravitational effects from any other objects.
The objects in this question are arranged at the vertexes of an equilateral triangle.
So far the only ways I have found to solve this problem involve calculus but we are expressly forbidden from using and calculus.
I have tried using conservation of energy (.5mv^2 - G(m)(m)/r) but always end up with the final radius as zero, which seems wrong because then I am dividing by zero.
Any help wold be greatly appreciated!
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