Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 63

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Mischek:
Well, skateboards can't go the same way. I see them under my window every day, for fuck's sake. There you go again with the spherical horses in a vacuum.
There's a perpetual motion - we push the trolley to infinity and it just keeps going on and on.)
 
Avals:
There is a perpetual motion - we push the cart to infinity, and it keeps going on and on)))
I wanted to write the same thing to Matemat, but I was afraid that the moderator might banish him
 
TheXpert:

There is no point in going any further.

but to talk ((

Sir, I will defend myself.

I agree with you about surface friction, with skis on ice in a vacuum, but my skateboard has Italian rolling bearings (and Bochev plugs).

So that your similar skateboard will go farther.

 
Avals:
(It's a perpetual motion machine - we push the cart to infinity, and it just keeps going and going))
Where do you get that from?
 

Another TV challenge (not with braingames, quite challenging and interesting).

Mr and Mrs megabrains play coin toss. Mr. megabrain has a fair coin, Mrs. has 0.4 probability of tails (for eagle: 1 - 0.4 = 0.6), and she knows it. Mega-brains flip their coins the same number of times, and the one with the most tails at the end of the game wins. Mrs. Megamind realizes that her chances of winning are less than her husband's and she can decide how many times in the game the coin is flipped before the winner is determined.

Question: what number of flips must Mrs. Megamogs set in order to have the maximum chance of winning? Is this number different from 1?

 
TheXpert:

we will add (and clear) by dm of snow

then for the non-cleared trolley

_______________

momentum MV

after adding snow (M + dm) V1 ; V1 = MV/(M + dm)

after next addition of snow, speed (M + 2dm)V2; V2 = (M + dm)V1/(M + 2dm) = MV/(M + 2dm)

It is clear here. Mass increases, momentum is constant - hence velocity decreases.

_______________

for cleared

_______________

MV momentum

after addition of snow (M + dm) V1' ; V1' = MV/(M + dm)

after dumping, momentum M*V1' = M^2*V/(M + dm)

after the next addition of snow, velocity (M + dm)V2'; V2' = (M)V1'/(M + dm) = M^2*V/(M + dm)^2

V2 - V2' = V(M/(M + 2dm) - M^2/(M + dm)^2) = MV*((M + dm)^2 - M*(M + 2*dm))/((M + 2dm)*(M + dm)^2)

(M + dm)^2 - M*(M + 2*dm) = dm^2 > 0 --> V2 - V2' > 0

And so for every iteration you can prove that not brushing is more efficient.

Now here's where you've messed up something. It's like there's no snow for the one being cleared. Adding dm reduces the speed a bit, but the megamosk resets it (perpendicular to the movement!) - and returns the original speed. Nothing has changed, for the law of conservation of momentum.

And another thing: for both carts, the snow has no effect on the momentum - because its own momentum is perpendicular to the cart's motion!!!

Mischek:

Or let's simplify it this way.

You are riding on a train platform, the platform passes the station. At the station there is a suitcase of 1 000 kg .

You drive past it and grab it by the handle.

Now that tonne is coming with you. It was standing and now it's moving. It shifted and picked up the speed of the railway platform, taking away some of its energy.

Now spinning back, not a suitcase, but a snowflake, not from the station, but from the sky.

The momentum of the platform has not changed, although the speed has changed - it has fallen.

P.S. Addendum: the friction force is bluntly proportional to the weight of the cart (coefficient - mu).

 
Mathemat:

There's something you've messed up here. For the one being cleared, there is no snow, as it were.

Where does it go? It just gets dropped. Same increase in mass.

Only if it falls at zero velocity, then it is dropped (perpendicularly) to the velocity of the trolley if you count along the axis of the trolley.

Adding dm reduces the speed a bit, but the megamotor discards it (perpendicular to the motion!) - and returns the original speed.

No! Perpendicular to the motion! There is no return.

And another thing: for both carts, the snow has no effect on the momentum - because its own momentum is perpendicular to the cart's motion!!!

That's exactly what we share the momentum when dropping it.

 
Mathemat:


There's something you've messed up here. It's not as if there is any snow for the cleared one. Adding dm lowers the speed a little, but the megamosk resets it (perpendicular to the movement!) - and returns the original speed

You can't drop something that isn't there and you can't even talk about jet power, it's perpendicular.
 

You both don't get it.

There is a cart, momentum mv. There is no friction yet.

The snow has fallen. Strictly vertical. This means that nothing happened on the horizontal axis. The momentum remains the same. (m+dm)v' = mv, where v' = v*m/(m+dm).

Now one MM has dropped the snow strictly orthogonal to the motion. Nothing has changed horizontally. The momentum is the same. mv'' = (m+dm)v'. Hence v'' = v'*(m+dm)/m = v*m/(m+dm) * (m+dm)/m = v.

TheXpert: Нет! Перпендикулярно движению! Никакого возвращения.

Well yes, orthogonal. But the velocity is back to what it was before the snow fell. The momentum hasn't changed anywhere.

 
Mathemat:

You both misunderstand.

There is a cart, momentum mv. There is no friction yet.

The snow has fallen. Strictly vertical. That means,

You misunderstand the meaning of "upright"

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