How Feynman proves momentum is conserved in this example?

[PLAGUE]

TL;DR Summary

Two masses moving with same but opposite velocity will stop dead if they collide.

Here is what Feynman says, "Suppose we have two equal masses, one moving with velocity v and the other standing still, and they collide and stick; what is going to happen? There is a mass 2m altogether when we are finished, drifting with an unknown velocity. What velocity? That is the problem. To find the answer, we make the assumption that if we ride along in a car, physics will look the same as if we are standing still. We start with the knowledge that two equal masses, moving in opposite directions with equal speeds v, will stop dead when they collide. Now suppose that while this happens, we are riding by in an automobile, at a velocity -v. Then what does it look like? Since we are riding along with one of the two masses which are coming together, that one appears to us to have zero velocity. The other mass, however, going the other way with velocity v, will appear to be coming toward us at a velocity 2v. Finally, the combined masses after collision will seem to be passing by with velocity v. We therefore conclude that an object with velocity 2v, hitting an equal one at rest, will end up with velocity v, or what is mathematically exactly the same, an object with velocity v hitting and sticking to one at rest will produce an object moving with velocity v/2. "

Feynman said, "two equal masses, moving in opposite directions with equal speeds, will stop dead when they collide." If so, then the car we are riding, moving with velocity -v, and the other mass, moving with velocity v, collide, they must stop dead as their masses are same. Then why do they keep moving with half of the velocity?

https://www.feynmanlectures.caltech.edu/I_10.html

 

[kuruman]

TL;DR Summary: Two masses moving with same but opposite velocity will stop dead if they collide.

If so, then the car we are riding, moving with velocity -v, and the other mass, moving with velocity v

That's not what Feynman said about what we see when we ride in the car moving with velocity -v.

Feynman said according to your post ( I highlighted in red the discrepancy)

Then what does it look like? Since we are riding along with one of the two masses which are coming together, that one appears to us to have zero velocity. The other mass, however, going the other way with velocity v, will appear to be coming toward us at a velocity 2v.

 

[Dale]

He is doing a transformation of the coordinates. Let $v_1\big\rvert_{t<0}$ be the velocity of object 1 before the collision, $v_2\big\rvert_{t>0}$ be the velocity of object 2 after the collision, etc. Then, in the reference frame of the center of mass we have $$V_1\Big\rvert_{t<0}=V$$$$V_2\Big\rvert_{t<0}=-V$$$$V_1\Big\rvert_{t>0}=0$$$$V_2\Big\rvert_{t>0}=0$$ where the capital letters indicate quantities in the center of mass frame.

Now, if we have another frame where the center of mass is moving at $u$, then a simple Galilean transform gives us the velocities $$v_1\Big\rvert_{t<0}=u+V_1\Big\rvert_{t<0}=u+V$$$$v_2\Big\rvert_{t<0}=u+V_2\Big\rvert_{t<0}=u-V$$$$v_1\Big\rvert_{t>0}=u+V_1\Big\rvert_{t>0}=u$$$$v_2\Big\rvert_{t>0}=u+V_2\Big\rvert_{t>0}=u$$

So, we simply apply this formula to the problem. We have before the collision that $$v_1\Big\rvert_{t<0}=u+V_1\Big\rvert_{t<0}=u+V=v$$$$v_2\Big\rvert_{t<0}=u+V_2\Big\rvert_{t<0}=u-V=0$$So we have two equations in $u$ and $V$ which we solve to get $V=v/2$ and $u=v/2$. Then we simply plug those into the after the collision equations to obtain$$v_1\Big\rvert_{t>0}=u+V_1\Big\rvert_{t>0}=u=v/2$$$$v_2\Big\rvert_{t>0}=u+V_2\Big\rvert_{t>0}=u=v/2$$

This is what he is doing in great detail. He just glossed over the math, but this is the concept he is trying to convey.