Given information:
- Particle A has charge q and mass m.
- Particle B has charge 4q and mass 4m.
- Particle A moves towards particle B with speed V.
- Particle B is initially at rest.
- Particle A approaches B up to the closest distance.

To find:
- The speed of particle B at the moment when particle A is closest to it.

Solution:

1. Conservation of energy:
- The initial kinetic energy of particle A is 1/2*m*V^2.
- At the closest distance, all the kinetic energy of particle A is converted into potential energy due to the interaction between A and B.
- The potential energy between two charges q and 4q at a distance r is given by k*q*4q/r, where k is Coulomb's constant.
- The closest distance between A and B is when they both stop moving towards each other and start moving away due to repulsion.
- At that moment, the potential energy between A and B is converted back into kinetic energy of both particles.
- Therefore, the final kinetic energy of particle A and B is 1/2*m*V^2.

2. Conservation of momentum:
- The initial momentum of the system is (m+4m)*V = 5m*V.
- At the closest distance, the direction of momentum of particle A is reversed due to repulsion.
- The momentum of particle B is in the same direction as before, but its magnitude has changed.
- Let the final speed of particle B be v. Then, the final momentum of the system is m*(-V) + 4m*v = -m*V + 4m*v.

3. Equating kinetic energies and momentum:
- From step 1, we have 1/2*m*V^2 = 1/2*m*v^2. Cancelling 1/2*m from both sides, we get V^2 = v^2.
- From step 2, we have 5m*V = -m*V + 4m*v. Simplifying, we get V = 4v/5.

4. Final answer:
- Therefore, the speed of particle B at the moment when particle A is closest to it is v = V/5 = (4/5)*V.
- Substituting the given value of V, we get v = V/5 = (4/5)*sqrt(k*q/5m).
- The answer is in terms of given variables and physical constants.