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        Two smooth spherical non conducting shells A and B each of radius R having uniformly distributed charge Q and -Q on their surfaces are released on a smooth non conducting surface when the distance between their centres is 10R. The mass of A is m and that of B is 2m. What is the speed of A just before it collides with B? (Neglect gravitational interaction)Please help as soon as possible as jee mains is very near. Thankyou
3 years ago

Pushkar Prateek
23 Points
							Here we will use two concepts, considering both spheres as a single system.
1. Loss in electrostatic P.E of the system = Gain in K.E. of the system.
2. Since no net external force acts on the system so linear momentum of the system remains conserved.
From 1.
½ m Va2 + ½  2m Vb2   =  KQ2/2R  – KQ2/10R     (Separation between the spheres just before collision will be 2R)
From 2.
m Va = 2m Vb
Solving above two equations we get
Va = $\sqrt{8KQ^2/15R}$

3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions