Conservation laws (testing my understanding) 
Suppose we have the following system, the ball lies on a frictionless horizontal plane. At $t=0$ the ball is on top, at $t>0$, the ball starts to drop due to gravitational force. Now,
First of all,  If we consider the ball a point mass, the only work done is by the gravitational force? If I don't ignore the air resistance, then the friction is also doing work there, is that correct?
Then, According to Goldstein's book, if the total external force is zero, the total linear momentum is conserved, since there is an external force (gravitational force) the linear momentum is not conserved? If there is no air resistance, the total kinetic energy is also conserved? My professor said we should consider the momentum in $x$ and $y$ directions separately, i dont see how that is relevant...
 A: Let's start with physical principles holding in classical mechanics, and then look at your problem.
Physical principles and equations

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*second principle of dynamics (dynamical equation for translation of the center of mass): time derivative of the momentum of a system is equal to the resultant of the external force,
$\dfrac{d\mathbf{Q}}{dt} = \mathbf{R}^e$,
where $\mathbf{Q}$ is the momentum of the system and $\mathbf{R}^e$ is the resultant of the external force acting on the system.
This is a vector equation, and thus it is equivalent to 3 scalar equations, in 3D physical space. Thus, you may need to consider all the space direction.


*theorem of the kinetic energy: the time derivative of the kinetic energy of a system is equal to the total power $P^{tot}$ of forces and moments acting on the system, both from internal and external forces
$\dfrac{d K}{dt} = P^{tot} = P^{int} + P^{ext}$.
Exercise

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*If you neglect air drag, gravitational force is the only external force that does work on the system composed by the two masses; the reaction of the ground does no work, since the displacement of the body with mass $M$ in vertical direction is zero.


*If you consider air drag, it produces negative work on the system since drag is in the opposite direction w.r.t. the motion, $P^{drag} < 0 $, and thus it makes the mechanical energy of the system decrease.


*Goldstein is right, as you can see above in the second principle of dynamics: if $\mathbf{R}^e = \mathbf{0}$ then $\dfrac{d \mathbf{Q}}{dt} = \mathbf{0}$ and thus $\mathbf{Q}(t) = \overline{\mathbf{Q}} = \text{const.}$


*even if you neglect air drag, the total kinetic energy is not conserved, since gravity introduces some power $P^g$ in the system
$\dfrac{d K}{dt} = P^{g}$
A: $M$=The ramp
$m$=mass moving on the ramp
When $M$ is free to slide on the ground with no friction:
Momentum:
When you consider $m$ and $M$ together as one system, forces like frictional forces, reaction forces etc between $m$ and $M$ are simply internal forces. (assuming no air resistance, as that would count as an external force). External force(gravity) acts only in the minus y direction, so along the y direction linear momentum is not conserved. But along the x direction there is no external force so linear momentum is conserved along the x direction. Think of this intuitively, as $m$ starts moving down the ramp from rest, it's picking up speeds in both x and y directions. But, as the ramp is also free to move, it moves in minus x direction due to the normal reaction of $m$ on the ramp. So momentum in the x direction cancels out as $m$ and $M$ possess equal and opposite momentum in the x direction, so total momentum in x direction stays zero. If however the system as a whole were moving with speed $v\hat{i}$ before $m$ is even allowed to move, then the momentum in the x direction would stay constant at $(M+m)v\hat{i}$ as $m$ moves down the ramp.
Energy:
If there are no frictional forces anywhere, then $mg\Delta y$=KE of $m$ + KE of the ramp(Conservation of Mechanical Energy). If friction was present anywhere, then we will have to consider the work done by the friction as well, since [potential energy+Kinetic energy=Mechanical energy] would not be conserved anymore. In such cases you need to write the general Work-Energy theorem.
When $M$ is fixed on the ground:
Momentum:
when the ramp is fixed on the ground, It won't move at all when $m$ slides on it. Now when you take $m$ and $M$ as the system, there is external influence in y direction(gravity=$-mg\hat{y}$), But unlike previous case there is external influence in the x direction as well, namely the reaction forces from the ground(or from the nuts and bolts used to hold $M$ in place). Clearly, linear momentum will not be conserved at all: As ramp is fixed so it would possess no momentum as $m$ moves its course, but $m$ possesses momentum along the direction that's tangent to the ramp throughout it's journey down the ramp. At any instant in the journey, Momentum of system=Momentum of $m$ can be resolved in x and y directions, both of these components change at every point on the ramp.
Energy:
If no frictional forces are present between $m$ and $M$: $mg\Delta y$=KE of $m$ + KE of the ramp=KE of $m$ + zero. (If you're confused about these energy equations, it's always better to write the general work energy theorem, which would simply reduce to conservation of mechanical energy when only conservative forces are present).
