A hollow non conducting sphere $A$ and a solid non conducting sphere $B$ of equal radius ' $R$ ' and masses $m$ and $2 m$ are kept a large distance apart on a rough horizontal surface. Charge on the two spheres $A$ and $B$ are $Q$ and $- 2 Q$ respectively. Charges are distributed uniformly and remain constant and uniform as the spheres come closer. Friction is sufficient to support pure rolling and the kinetic energy of the two spheres just before collision is $K_{A}$ and $K_{B}$ . Find $100 \times \frac{K_{A}}{K_{B}} .$

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answer is 168.

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Detailed Solution

Let $v_{1}$ and $v_{2}$ be the speeds of sphere A and B just before collision, then from conservation of energy,

$0 = \frac{1}{2} m v_{1}^{2} (1 + \frac{2}{3}) + \frac{1}{2} (2 m) v_{2}^{2} (1 + \frac{2}{5}) + \frac{K . Q (- 2 Q)}{2 R} 0 = \frac{5}{6} m v_{1}^{2} + \frac{7}{2} m v_{2}^{2} - \frac{K . Q^{2}}{R} \frac{5}{6} m v_{1}^{2} + \frac{7}{2} m v_{2}^{2} = \frac{K . Q^{2}}{R} ............... (i)$

From conservation of angular momentum about any point ‘O’

$0 = m v_{1} R + \frac{2}{3} m R^{2} (\frac{v_{1}}{R}) - 2 m v_{2} R - \frac{2}{5} (2 m) r^{2} (\frac{v_{2}}{R}) v_{1} = \frac{42}{25} v_{2} .............. (i i)$