Four identical cells each of emf 2V, are joined in parallel providing supply of current to external circuit consisting of two `15 Omega` resistors joined in parallel. The terminal voltage of the cells ae read by an ideal voltmeter is 1.6V. Calculate the internal resistance of each cell.

Explanation

When the bullet hits the block, in a negligible time interval, it becomes embedded in the block system starts moving horizontally. During this process, net force acting on the bullet-lock system in vertical direction is zero and no force acts in the horizontal direction. Therefore, momentum if the bullet-block system remains conserved.

Let us denote momentum of the bullet-block system immediately after the bullet becomes embedded in the block by p and apply principle of conservation of momentum to the system for horizontal component of momentum.
`p = mv_(0)`
Using equation `K= p^(2)/2m`, we can find kinetic energy `K_(1)` of the bullet-block system immediately after the bullet becomes embedded in the block.
`K_(1) = (mv_(0))^(2)/(2(M+m))`
During swing, only gravity does not work on the bullet-block system. Applying work-kinetic energy theorm during swing of the bullet-block system, we have

`W_(1 rarr 2) = K_2-K_(1) rarr " " -(M+m)g h = 0 - (mv_(0))^(2)/(2(M+m))`
Rearringing terms, we have `v_(0) = ((M+m))/(m)sqrt(2gh)`