A steel ball of mass `m=50g` falls from the height `h=1.0m` on the horizontal surface of a massive slab. Find the cumulative momentum that the ball imparts to the slab after numerous bounces, if every impact decreases the velocity of the ball `eta=1.25` times.

Explanation

Velocity of the ball, with which it hits the slab, `v=sqrt(2gh)`
After first impact, `v^'=ev` (upward) but according to the problem `v^'=(v)/(eta)`, so `e=1/eta` (1)
and momentum, imparted to the slab,
`=-mv-(-mv^')=mv(1+e)`
Similarly, velocity of the ball after second impact,
`v^('') =ev^'=e^2v`
And momentum imparted `=m(v^'+v^(''))=m(1+e)ev`
Again, momentum imparted during third impact,
`=m(1+e)e^2V`, and so on,
Hence, net momentum, imparted `=mv(1+e)+mve(1+e)+mve^2(1+e)+....`
`=mv(1+e)(1+e+e^2+...)`
`=mv((1+e))/((1+e))`, (from summation of G.P.)
`=sqrt(2gh)((1+1/eta))/((1-1/eta))=msqrt(2gh)//(eta+1)//(eta-1)` (Using Eq. 1)
`=0.2kg m//s`. (On substitution)