Four identical cells each of e.m.f. 2V and joined in parallel providing supply of current to external circuit consisting of two `15Omega` resistors joined in parallel. The terminal voltage of the equivalent cell as read by an ideal voltmeter is 1.6 V 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)`