A heavy block A is made to move uniformly along  a smooth floor with velocity V = 0.01 m/s towards  left. A ball of mass m = 50 g is projected towards 
the block with a velocity of u = 100 m/s. The ball keeps bouncing back and forth between the block A and fixed wall B. Each of the collisions is elastic. After the ball has made 1000 collisions with the block and wall each, the distance between the block and the wall was found to be L = 1.2 m. Calculate the average force being experienced by the block due to collision at this instant. All collision are instantaneous.
 

Block move uniformly at speed V = 0.01 m/s. After each collision with A the speed of the ball will increase by 2V [Collision with fixed wall B does not cause  change in speed]  After n = 1000 collisions, speed of the balls is 
${\mathrm{u}}_{\mathrm{n}}=\mathrm{u}+2\mathrm{nV}=100+2\times 1000\times 0.01=120\mathrm{m}/\mathrm{s}$  
Notice that speed of the ball changes marginally during each collision but the change becomes significant after a  large number of collisions. Change in momentum of the ball during n^th collision with the body is
$\mathrm{\Delta P}=\mathrm{m}[\mathrm{u}+2\mathrm{nV}]+\mathrm{m}[\mathrm{u}+2(\mathrm{n}-1\left)\mathrm{V}\right]=\mathrm{m}[2\mathrm{u}+4\mathrm{nV}-2\mathrm{V}]$
For n as large as 1000, we can neglect 2V
$$\begin{gathered} \therefore \mathrm{\Delta P}=\mathrm{m}[2\mathrm{u}+4\mathrm{nV}] \\ =\frac{50}{1000}\times [2\times 100+4\times 1000\times 0.01] \end{gathered}$$
= 12 kg m/s 
At this instant frequency of collision of the ball with body A is $\mathrm{f}=\frac{120\mathrm{m}/\mathrm{s}}{1.2\mathrm{m}}=100{\mathrm{s}}^{-1}$
$\therefore \text{ Force }=\mathrm{\Delta Pf}$
$\simeq 1200N$