A battery of six cells each of e.m.f . 2V and internal resistance `0.5Omega` is being charged by D.C. mains of e.m.f. 220V by using an external resistance of `10Omega`. What is the potential difference across the battery?

Explanation

(a) There is no net external force in the verticle direction and in the horizontal direction, only external force friction is non-impulsive, therefore momentum of the bullet-block system during their interaction remains conserved.
(b) Let us denote velocities of the bullet before it hits the block and immediately after it oierces through the block by `v_(b0)` and `v_(b)`, velocity of the block immediately after the bullet pierces through it is `v_B` and masses of the bullet and the block by m and M respectively. These are shown in the adjacent figure.Applying principle of momentum for horizontal component, we have `mv_(b0) = mv_(b) + Mv_(B) rarr v_(B) = (m(v_(b0
))-V_(b))/M`
Substituting the given values, we have `v_(B) = 10m//s`

(c) To calculate distance traveled by the block before it stops, work kinetic energy theorem has to be applied.
DUring sliding of the box on the ground only the force of kinetic friction does work.
`W_(1rarr2) = K_2-K_1rarr " " -muMgx=0-1/2Mv_B^(2)`
`x = (v_(B)^(2))/(2mug)`
Substituting given values, we have `x=20m`