A body of mass m=1kg falls from a height h=20m from the ground level .
(a) What is the magnitude of total change in momentum of the body before it strikes the ground?
(b) What is the corresponding average force experienced by it ?`(g=10ms^(-1))`

Explanation

(a) Since the body falls from rest (u = 0) through a distance h before striking the ground, the speed v of the body is given by kinematical equation.
`v^(2)=u^(2)+2` as, Putting a = g and s = h
We obtain `v=sqrt(2gh)`
`rArr ` The magnitude of total change in momentum of the body
`Deltp =|mv-0|=mv, " where "v=sqrt(2gh)`
`rArr Delta p=msqrt(2gh)=(1) sqrt((2xx10xx20)) `kg-m/sec
`rArr Deltap=20` kg-m/sec.
( b) The average force experienced by the body `=F_(av)=(Deltap)/(Deltat)`, where `Delta t=` time of motion of the body = t(say). We know `Deltap=20kg` m/sec. Therefore we will have to find t using the given data. From kinematics,
`s=ut+(1)/(2) at^(2) rArr h =(1)/(2) g t^(2)" "(u=0)`
`rArr t=sqrt((2h)/(g))`
`therefore F_(av)=(Deltap)/(Delta t)=(Deltap)/(t)`
Putting the general values of `DeltaP` & t, we obtain
`F_(av)= (m sqrt(2gh))/(sqrt((2h)/(g)))=mg`
`rArr F_(av)=mg`
where mg is the weight (W) of the body and g is always directed vertically downward. Therefore the body experiences a constant vertically downward force of magnitude mg.