Question

A solid sphere and sperical shell of same mass m and same radius R are given charges +Q and -Q. Both spheres are kept on rough surface where friction is sufficient to avoid sliding. Initial seperation between spheres is 10 R. If m = 126 kg, $Q^2=23\times {10}^{-9}{C}^2$ and R = 1m.

(A)	Speed of solid sphere just before collision is	1) $\frac{3}{5}$ m/s
(B)	Speed of spherical shell just before collision is	2) 0 m/$s^2$
(C)	Speed of center of mass just before collision is	3) $\frac{5}{7}$ m/s
(D)	Acceleration of center of mass	4) Towards left
		5) Towards right

• A. A $\to$ 3, B $\to$ 1, 4, C $\to$ 5, D $\to$ 5
• B. A $\to$ 1, B $\to$ 3, 4, C $\to$ 4, D $\to$ 5
• C. A $\to$ 1, B $\to$ 1, 4, C $\to$ 5, D $\to$ 4
• D. A $\to$ 3, B $\to$ 3, 4, C $\to$ 4, D $\to$ 4

Answer 1

The correct option is A A $\to$ 3, B $\to$ 1, 4, C $\to$ 5, D $\to$ 5


By conservation of energy.
$\frac{-k{Q}^2}{10R}=\frac{-k{Q}^2}{2R}+\frac{1}{2}{I}_A{\omega }_A^2+\frac{1}{2}{I}_A{\omega }_B^2$ ---(1)
$I_A=\frac{7}{5}m{R}^{2}and{I}_B=\frac{5}{3}m{R}^2$
are MOI about IAOR.

By conservation of angular momentum about any point on surface $\Rightarrow I_A{\omega }_A=I_B{\omega }_B\Rightarrow \frac{7}{5}m{R}^2{\omega }_A=\frac{5}{3}m{R}^2{\omega }_B$ ---(2)

Solving (1) and (2) we get ${\omega }_A=\frac{25}{21}{\omega }_B$ and and${\omega }_B^2=\frac{21\times 6}{25\times 23}\frac{9\times {10}^9{Q}^2}{m{R}^3}\Rightarrow {\omega }_B=\frac{3}{5}rad/s$
and $v_B=R{\omega }_B=\frac{3}{5}m/salso{\omega }_A=\frac{5}{7}rad/s$
and $v_A=\frac{5}{7}m/s$

Since $f_B>f_A$ acceleration of COM is towards right.