Conserving momentum along $y$-axis We know that momentum of a system is conserved if no external force acts on it and since gravity acts along the $y$ axis,momentum of a system excluding the earth cannot be conserved in the $y$ axis. But i have a doubt on this.
What difference does it make to include or exclude the earth?If we include earth in our system,gravity is an internal force. Then we are just supposed to take into account,the momentum of earth before or after,right?The earth is so huge that neither its mass changes nor velocity. So won't the initial and final momentum of the earth always remain conserved? Doesn't that mean we can always conserve momentum in the $y$ axis as well?
Mathematically suppose we have a vertically projectiled body and a falling body having a collision in $y$ axis. Taking earth into the system,
$m_1u_1+m_2u_2+M_{\mathrm{earth}}u_{\mathrm{earth}}=m_1v_1+m_2v_2+M_{\mathrm{earth}}v_{\mathrm{earth}}$
Here the terms $M_{\mathrm{earth}}u_{\mathrm{earth}}$ and $M_{\mathrm{earth}}v_{\mathrm{earth}}$ get cancelled since $u_{\mathrm{earth}}=v_{\mathrm{earth}}$. So it doesn't seem to make any difference conserving momentum when gravity is an external force. Hence,can't we say that momentum is always conserved when gravity is an external force?
I may be wrong. Please enlighten me.
 A: 
Can't we say that momentum is always conserved when gravity is an external force?

No. An object at rest, like a stone after I drop it, will be accelerated by gravity meaning $\dot P \neq 0$, meaning momentum is not conserved, if we take the earth out of the equation.
What you got right though is that the change in earth's momentum is very small, but if we decided to neglect it, that means that even if we take Earth into the system momentum is not conserved not the other way around.

The earth is so huge that neither its mass changes nor velocity.

This is not true, it's a good approximation but, saying that is like saying Newton's third law does not apply to earth, which is the same as saying gravity is not an interaction force, it's a spooky force from outside, an external force. Of course then we don't have conservation of momentum, as you well know "momentum of a system excluding the earth cannot be conserved".
To finish with the calculation that shows that the total sum of momentum does not change:
Let's say you let an apple ($m$) fall and look at the forces between apple and earth ($M$), down being the negative direction,
$$F_{{e\to a}}=-\frac{G m M}{ r^2}=-F_ {{a\to e}}$$
Acceleration of the apple: $$a_a=F_{{e\to a}}/m$$ acceleration of earth $$a_e=F_{{a\to e}}/M$$
The mass of both don't change so we have a change in momentum for both that is just $\dot P = m a$:
$$\dot P_a = m a_a=m F_{{e\to a}}/m = -\frac{G m M}{ r^2} $$
And the earth:
$$\dot P_e = M a_e=M F_{{a\to e}}/M= \frac{G m M}{ r^2} $$
We see that the change in momentum is zero. $$\dot P = \dot P_e + \dot P_a =0 $$
I.e. momentum is conserved.  If you now took $\dot P_e =0$, you of course have a nonzero change in momentum i.e. no momentum conservation.
