Two uniform solid spheres of equal radii R, but mass M and 4M have a centre to centre separation 6R, the two spheres are held fixed on a horizontal floor. A projectile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.

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$v = {(\frac{3 R}{5 GM})}^{1 / 2}$

$v = {(\frac{8 R}{5 GM})}^{1 / 2}$

$v = {(\frac{5 GM}{3 R})}^{1 / 2}$

$v = {(\frac{3 GM}{5 R})}^{1 / 2}$

answer is B.

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Detailed Solution

The neutral point N in at a distance x from mass M, given by
$x = \frac{d}{\sqrt{\frac{m_{2}}{m_{1}}} + 1} = \frac{6 R}{\sqrt{\frac{4 M}{M}} + 1} = 2 R$
The mechanical energy at the surface of sphere of mass M is
$E = \frac{1}{2} {mv}^{2} - \frac{GMm}{R} - \frac{4 GMm}{5 R}$
At neutral point N, the speed approaches zero. The mechanical energy at N purely potential.
$E_{N} = - \frac{GMm}{2 R} - \frac{4 GMm}{4 R} = - \frac{GMm}{2 R} - \frac{GMm}{R}$
From the principle of conservation of mechanical energy
$\begin{array}{l} \frac{1}{2} {mv}^{2} - \frac{GMm}{R} - \frac{4 GMm}{5 R} = - \frac{GMm}{2 R} - \frac{GMm}{R} \\ \frac{v^{2}}{2} = \frac{GM}{R} (\frac{4}{5} - \frac{1}{2}) \\ \Rightarrow v = {[\frac{2 GM}{R} \times \frac{3}{10}]}^{\frac{1}{2}} = {(\frac{3 GM}{5 R})}^{\frac{1}{2}} \end{array}$