Two identical particles of mass m carry a charge Q each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed v. The closest distance of approach be :

Answer

`Q^(2)/(mpi in_(0)V^(2))`

Explanation

Conservation of momentum (COM)
`0+mV=(m+m)V_(f), V_(f)=V/2`
COME: `1/2 mV^(2)=1/2 m(V/2)^(2)+1/2 m (V/2)^(2)+(Kq^(2))/d`
`d=Q^(2)/(pim in_(0) v^(2))`