A bouncing ball of mass 200 g falls from the height of 5m on a horizontal ground. After every impact with the ground, the velocity of the ball decreases by `(1)/(2)` times. The total momentum the ball imparts on to the ground after 3 impacts is (Let `g=10m//s^(2)`)

Options

A. `(14)/(4)kgm//s`
B. `(20)/(6)kgm//s`
C. `(26)/(12)kgm//s`
D. `(21)/(4)kgm//s`

Answer

D

Explanation

Given mass of ball, m =200 g =0.2 kg and height of the ball, h =5m
`:.` Velocity of ball at the time of first impact,
`v_(1)=sqrt(2gh)=sqrt(2xx10xx5)=10m//s`
The recoil velocity of the ball after first impact,
`v'_(1)=(v_(1))/(2)=(10)/(2)=5m//s`
`:.` Momentum imparted by the ball in first impact,
`p_(1)=mv_(1)-m(v'_(1))`
`=mv_(1)+mv'_(1)`
`=0.2xx10+0.2xx5`
`p_(1)=3kgm//s`
Similarly, velocity of the ball at the time of second impact and its recoil velocity will be 5 m/s and `(5)/(2)` m/s Hence, momentum imparted by ball in second impact,
`p_(2)=mxx5-m(-(5)/(2))`
`=0.2xx5+0.2xx(5)/(2)=1.5=(3)/(2)kgm//s`
Again, velcoity of ball at the time of thired impact and its recoil velocity will be `(5)/(2)` m/s and `(5)/(4)` m/s.
Hence, momentum imparted by ball in third impact,
`p_(3)=m((5)/(2))-m(-(5)/4)`
`p_(3)=0.2xx(5)/(2)+0.2xx(5)/(4)=(3)/(4)kg.m//s`
`:.` Total momentum imparted by the ball to the ground,
`p=p_(1)+p_(2)+p_(3)`
`=3+(3)/(2)+(3)/(4)=(21)/(4)kgm//s`