The `n` rows each containinig `m` cells in series are joined parallel. Maximum current is taken from this combination across jan external resistance of `3 Omega` resistance. If the total number of cells used are 24 and internal resistance of each cell is `0.5 Omega` then

Explanation

Denoting the first block by A and the second block by B velocities immediately defore and after the impact are shown in the figure.

Applying principle of conservation of momentum, we have
`m_(B)v_(B) + m_(A)v_(A) = m_(A)u_(A) + m_(B)u_(b) rarr 3v_(B) + 5VA = 5 xx 2 + 3 xx (-4)`
`3v_(B) + 5v_(A) = -2` ...(i)
Applying equation of coefficient of restitution, we have `v_(B) - v_(A)= e(u_(A)-u_(B)) rarr v_(b) - v_(A) = e{2-(-4)}`
`v_(B)- v_(A) = 6e` ...(ii)
(a) For perfectly elastic impact `e=1`. using this value in equation (ii), we have
`v_(B) - v_(A) =6` ...(ii)
Now from equation (i) and(iia), we obtain
`v_(A) = -2.5 m//s` and `v_(b) = 3.5m//s`
(b) For value `e= 0.6`, equation 2 is modified as
`v_(B)-v_(A)= 3.6` ....(iib)
Now from equation (i) and (iib), we obtain
`v_(A) = - 1.6m//s` and `v_(B) = 2.0m//s`
Block A reverse back with speed 1.6 m/s and B also move in opposite direction to its original direction with speed `2.0 m//s`.