Q:

Feynman Lectures Vol1 : Question on relativistic mass derivation

http://www.feynmanlectures.caltech.edu/I_16.html, 
In chapter 16, under section 16–4 Relativistic mass, Feynman makes the below statement.

Now the question is, what is the vertical velocity $u\tan \alpha$? If we knew that, we could get the correct expression for the momentum, using the law of conservation of momentum in the vertical direction. Clearly, the horizontal component of the momentum is conserved: it is the same before and after the collision for both particles, and is zero for particle 1.

I am not able to understand why he says that horizontal component of the momentum is conserved. Why we cannot say the same for the vertical component of momentum and instead go on to derive it? Is it because the Lorentz transformation of velocity applies only to the vertical component of the velocity of the obliquely moving particle? I feel like I did not understand something more fundamental than what I have asked in the question. Any explanation would be much appreciated.

A:

You see I struggled a bit there. But then, I saw, Feynman and Einstein are both right because they believed physics is true in all inertial reference frames blah blah blah...
Here's the trick. 
We already know there is conservation of momentum because we need that assumption ( it is also mentioned in the book) in all reference frames. It is conserved in both horizontal and vertical components. 
Now, when hopped inside the car, we see particle 1 moving directly upwards and directly downwards with the same velocity. There is no, sideways velocity seen. So, we need only concern with the vertical portion of the momentum conservation. There is no distrubance given to particle 2 by 1 in the horizontal component. So the particle 2 is not influenced horizontally. 
But it is possible to look for changes in the vertical direction of particle 2 as there is a vertical input from particle 1.
So we need to work out the conservation equation as it needs to be conserved (because it is conserved in all inertial reference frames) in the vertical direction of the system and ended up with the increasing mass formula! 
$Update1:$ Also the collision has to be elastic in both the reference frames.
$Update2:$ Elastic collision in 2D basics. This all boils down to Newton's 2nd law, i.e., momentum can only be changed by an applied force, keeping this in mind.
We look at the collision with this perspective. 
In 2D collision, as they touch, the force is imparted only in the direction of the common normal but not in the tangential direction. Therefore, Newton's second law tells us that momentum will change in the direction of the common normal but not in the tangential direction. So velocity doesn't change in the tangential direction, in this case the horizontal component of particle 2. Therefore momentum is conserved in the horizontal direction both for particle 1 and 2,(particle 1 doesn't even have a horizontal momentum component, and particle 2 doesn't change its momentum horizontally). 
Hence, we need only concern about the common normal direction where force is being imparted.

