Three identical solid spheres each of mass “m” and radius “R” are released from positions as shown in the figure (assume no external gravitation). What would be the speed of any of three spheres just before they collide.

Since no external force is involved, the mechanical energy of the system at the time of release should be equal to mechanical energy just before the collision. In other words, the mechanical energy of the system is conserved. The initial potential energy of system is given by, U_i = −3Gm²/a

Let “v” be the speed of any sphere before collision. The configuration just before the collision is shown in the figure. We can see that linear distance between any two centres of two identical spheres is “2R”. Hence, potential energy of the configuration before collision is,

U_F = −3Gm²/2R

Applying conservation of mechanical energy, K_i + U_i = K_F + U_F

0 − 3Gm²/a= 1/2mv² − 3Gm²/2R

Therefore, v = √ {Gm (1/R− 2/a)}