Three solid spheres each of mass 1 kg and ra | vedclass

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Question

(A) Let $m = 1 \ kg$ be the mass of each sphere and $r = 2 \ m$ be the radius. The initial distance between the centers of any two spheres is $d_i = 1

Vedclass · Accepted Answer

$\sqrt{\frac{3 G}{10}}$

Vedclass · Answer

(A) Let $m = 1 \ kg$ be the mass of each sphere and $r = 2 \ m$ be the radius. The initial distance between the centers of any two spheres is $d_i = 10 \ m$.At the time of collision,the spheres touch each other. Since they are identical,the distance between the centers of any two spheres at collision is $d_f = 2r = 2 	imes 2 = 4 \ m$.By the law of conservation of energy,the total energy at the initial position equals the total energy at the final position: $U_i + K_i = U_f + K_f$.Initially,$K_i = 0$. The potential energy of the system of three spheres is $U = -3 	imes \frac{G m^2}{d}$.So,$-3 \frac{G m^2}{d_i} = -3 \frac{G m^2}{d_f} + 3 	imes (\frac{1}{2} m v^2)$,where $v$ is the speed of each sphere.Dividing by $3m/2$,we get $v^2 = 2 G m (\frac{1}{d_f} - \frac{1}{d_i})$.Substituting the values: $v^2 = 2 	imes G 	imes 1 	imes (\frac{1}{4} - \frac{1}{10}) = 2 G (\frac{5-2}{20}) = 2 G (\frac{3}{20}) = \frac{3 G}{10}$.Therefore,$v = \sqrt{\frac{3 G}{10}}$.

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