A man of mass m climbs a rope of length L suspended below a balloon of mass M. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed `v_("rel")`. (relative to rope) in what direction and with what speed (relative to gound) with the balloon move ?

Explanation

The change in momentum `Delta p` of the ball during `Delta = 0.01` sec is known as the impulse `(F Delta t)` of the force of impact
`rArr F Delta t = Delta p`
`|Delta vec(p)|=|m vec(v)_(2)-m vec(v)_(1)|= m (v_(1)+v_(2))`
`because m vec(v)_(1) & m vec(v)_(2)` are antiparallel
`rArr F delta t = m (v_(1)+v_(2)) " "` ...(1)

Since, the ball falls through a height h, and its velocity just before striking the surface is `sqrt(2gh)`. As the ball losses 75percent of its total mechanical energy that is knetic energy.
`rArr 1//2 mv_(2)^(2)=1//2 m v_(1)^(2)(1-75//100)`
`rArr v_(2)=(v_(1))/(2)=(sqrt(2gh))/(2)=sqrt((gh)/(2))`
Substitution of the values of `v_(1)` and `v_(2)` in (1) yields
`F Delta t = 50xx10^(-3)xx3 sqrt((gh)/(2))=50xx10^(-3)xx 3 sqrt(((9.8)(10))/(2))=1.05` N - sec.