Two batteries of e.m.f. `E_(1)` and `E_(2)` and internal resistance `r_(1)` and `r_(2)` are connected in parallel. Determine their equivelent e.m.f.

Explanation

(a) Let us consider the ball as the body A and the wall as the body B. Since the wall has infinte,y large inertia (mass) as compared to the ball, the state of motion of the wall, emains unaltered during the impact i.e. the wall remain stationary.
Now we show velocities of the ball and its t and n-components immediately before and after the impact. For the purpose we have assumed velocity of the ball after the impact v.

Component along t-axis Components of momentum along t-axis of the ball is conserved. Hence, t-component of velocities of each of the bodies remains unchanged. `v_t=u_t=usintheta` ...(i)
Component along -axis Concept of coefficient of restitution e is applicable only for the n-component velocities.
`v_(Bn)-v_(An)=e(u_(An)-u_(Bn)) rarr -v_(n) = eU_(n)` ,
From equations (i) and (ii), the t and n-components of velocity of the ball after the impact are
`v_(t) = u sin theta` and `v_(n) = eu sin theta`
(b) If the impact is perfectly elastic, we have `v_(t) = u sin theta, v_(n) = usintheta` and `theta = theta` The ball will rebound with the same speed making the same angle with the vertical at which it has collided. In other words, a perfectly elastic collsion of a ball with a wall follows the same laws as light follows in reflection at a plane mirror.