When a rain falls from the sky, what happens to its momentum as it hits the ground? Is your answer also valid for Newton‘s famous apple?

To answer this question, the momentum (p) equation will be
invaluable:

p = mv

With the
rain drop, m is the mass of the rain drop, and v is its terminal velocity (zero
acceleration) through the air.

When the rain drop hits the
ground the momentum changes.  This can be conceptualized by taking the derivative of the
momentum equation with respect to time.  Remember that "d" simply means "change
in":

dp/dt = m (dv/dt)

Since
dv/dt is the change in velocity over time which is acceleration we can replace it with
"a" for acceleration:

dp/dt =
ma

And as we know, F =
ma

Because the drop decelerates when it hits the ground, F
= ma tells us an impact force is created.  This impact force causes the raindrop to
break apart and splatter in all different directions on the
ground.

However, the conservation of momentum rule
(momentum of the whole drop is equal to the sum of the momentum of the splattered drops)
is not valid in this case because there are energy loses due to the impact
force.

This effect seen with the rain drop is likely what
you would see with an apple.  Depending on the apple's coefficient of elasticity (e),
the ratio of bounce to splatter will differ (an e=1.0 would mean a perfect bounce, where
as an e=0.5 would mean some splatter, and thus, momentum loss due to the impact
force).