Question

Three identical small ball each of mass m=1 kg and $Q={10}^{-4}C$ are connected by insulating string of length L=1 m and are placed on frictionless horizontal surface as shown in figure. Middle ball B is given velocity V=10 m/s, then minimum seperation between ball A and C can be

Answer 1

By COM $\Rightarrow mV=3m{V}^{\prime }\Rightarrow V^{\prime }=\frac{V}{3}$

By COE $\Rightarrow \frac{1}{2}m{V}^2+$

$\frac{K{Q}^2}{L}\times 2+\frac{K{Q}^2}{2L}$

$=\frac{1}{2}m{V}^{\prime 2}\times 3+\frac{K{Q}^2}{L}\times 2+\frac{K{Q}^2}{x}$

$\frac{1}{2}m{V}^2-\frac{1}{2}m{V}^{\prime 2}\times 3+\frac{K{Q}^2}{2L}=\frac{K{Q}^2}{x}\Rightarrow \frac{K{Q}^2}{x}=\frac{K{Q}^2}{2L}+\frac{m{V}^2}{3}$

$x=\frac{6LK{Q}^2}{3K{Q}^2+2Lm{V}^2}=\frac{54}{47}$