Answer:
[tex]v_2 = \sqrt{\frac{GM}{3R}}[/tex]
[tex]v_1 = 2\sqrt{\frac{GM}{3R}}[/tex]
Explanation:
As we know by energy conservation that change in gravitational potential energy of the system = change in kinetic energy of the two ball
So here we can say
[tex]-\frac{GM(2M)}{12R} + 0 = -\frac{GM(2M)}{4R} + \frac{1}{2}Mv_1^2 + \frac{1}{2}(2M)v_2^2[/tex]
Also since there is no external force on the system of two masses so here total momentum of the two balls will remains conserved
[tex]0 = Mv_1 + 2Mv_2[/tex]
[tex]v_1 = -2v_2[/tex]
now we have
[tex]\frac{GM^2}{2R} - \frac{GM^2}{6R} = \frac{1}{2}M(-2v_2)^2 + \frac{1}{2}(2M)v_2^2[/tex]
[tex]\frac{GM^2}{3R} = Mv_2^2[/tex]
[tex]v_2 = \sqrt{\frac{GM}{3R}}[/tex]
[tex]v_1 = 2\sqrt{\frac{GM}{3R}}[/tex]
