How is linear momentum conserved in case of a freely falling body? When an object is experiencing free fall, it has a constant acceleration and hence an increasing velocity (neglecting friction). Thus its momentum is increasing. But according to law of conservation of momentum, shouldn't there be a corresponding decrease in momentum somewhere else ?
Where is it ?
 A: Sciencisco's is the best, but I thought I would add one thought: the external potential $V = mgy$ does not exhibit translational symmetry in the $y$ direction. Noether's theorem says that each symmetry gives a conservation law. Furthermore, if you don't have a symmetry, then you don't have the associated conservation law. Translational symmetry gives us conservation of momentum. Because this potential is not translationally invariant in the $y$ direction, momentum is not conserved in that direction.
A: Linear Momentum is conserved only in systems with net external force equal to zero. For a body falling on Earth, it experiences Earth's gravitational force so its linear Momentum increases.
But if you include Earth in your system then definitely, momentum is conserved, as an equal amount of momentum of Earth is increased in upward direction. But individually for both it's not conserved, there is an external force of gravity on each.
A: Linear momentum of a system remains conserved unless an external force acts on it. Since during free fall, a gravitational force acts on the body, it's momentum will not remain conserved. However, if we change the reference in such a manner that the gravitational force becomes an internal force of the system, i.e. regard both the body and Earth together as a system, and consider this system to be isolated in the universe, with no other body present near the system, we can now apply the law of conservation of linear momentum as there are no external forces acting on the system now.
A: Momentum is not conserved for you alone, because an external force acts on you. But if you consider both the earth and you. Then since $F_{ext}= 0 $, net momentum change is definitely zero. 
Let's say you are starting from rest. Now let force by earth on you be $F$. So $a=\frac{F}{m_{you}}$ and your velocity after time $t$ is $v= \frac{Ft}{m_{you}}$ Now your momentum is $m_{you}v= Ft$ .
Similarly $F$ on earth by you is $-F$. {Negative as the direction is opposite}.So $A_{earth}=-\frac{F}{M_{earth}}$ and earth's velocity after time $t$ is $V= -\frac{Ft}{M_{earth}}$ Now earth's  momentum is -$M_{earth}V= -Ft$. Thus net $\Delta P = Ft-Ft=0$.
A: Neither energy nor momentum is individually conserved during free-fall.
During free-fall the ratio of energy to momentum changes - full stop.
Kinetic Energy is Momentum in disguise.
The idea of GPE being converted to KE is born out of ignorance and double accounting (somehow thinking KE is more than just a different representation of momentum).
When a mass is in free-fall
$dE/ds = d\rho/dt$
There is nothing more needed, apart from some scaling.
Did Einstein's famous mass-energy equation teach you guys nothing about the relationships between energy, momentum, time, distance and light speed?
