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Question

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=10ms2)

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Solution

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:v2=u2+2as;
Putting a=g and s=h, we obtain v=2gh
Thus, the magnitude of the total change in momentum of the body just before it strikes the ground=ΔP=|mv0|=mv
Where v=2ghΔP=m2gh
or ΔP=(1)(2×10×20)kgms1
=20kgms1......(i)
We define average force experienced by the body,
Fav=ΔPΔt;Δt= time of motion of the body = t(say).
We know ΔP=20kgms1 from (i). Now, let us find t using the facts given in problem. From kinematics, we know
s=ut+12at2 (here u=0,s=h, and a=g)
h=12gt2 or t=2hg
Fav=ΔPΔt=ΔPt
Putting the general values of ΔP and t, we get
Fav=m2gh2h/g=mg
Fav=mg
Where mg is the weight of the body and g is directed vertically downward. Therefore, the body experiences a constant vertically downward force of magnitude mg.

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