Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 114
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OK, M1 > M2 for carts. dm -- mass of snow over dt. mu -- coefficient of friction. V0 is the initial velocity.
Consider time dt
V1dt = (V0 - mu*g*dt)*M1/(M1 + dm)V2dt = (V0 - mu*g*dt)*M2/(M2 + dm)
dv = V1 - V2 = (V0 - mu*g*dt)*(M1/(M1 + dm) - M2/(M2 + dm)) =
(M1/(M1 + dm) - M2/(M2 + dm)) = (M1*M2 + M1*dm - M1*M2 - M2*dm)/((M1 + dm)*(M2 + dm)) = dm*(M1 - M2)/((M1 + dm)*(M2 + dm)) > 0
From this we can say that for the same initial velocity, the cart with the smaller mass will always brake more. Therefore it will travel less.
You see -- friction is removed from the comparison altogether. Only the change in velocity due to impact is involved.
OK, M1 > M2 for carts. dm -- mass of snow over dt. mu -- coefficient of friction. V0 is the initial velocity.
Consider the time dt
V1dt = (V0 - mu*g*dt)*M1/(M1 + dm)V2dt = (V0 - mu*g*dt)*M2/(M2 + dm)
This is only for the initial time moment, and for it M1=M2 - unlike your assumption. And if for arbitrary?
And where is the ejection of snow by a working megamotor?
This is only for an initial moment in time, and for it M1=M2 - unlike your assumption. And if for arbitrary?
And where is the ejection of snow?
This is the solution to this problem --
There are two carts. One with mass M the other with mass m < M.
Both start driving at the same speed, snow falls on them. Which one goes the furthest?Strictly speaking, you still have to prove that with the same masses, the lower speed will remain with the cart with the lower speed. But I think it's obvious.
Anyway, I've run out of steam, and I don't understand what you don't understand. I won't bother with this problem any more.
It's not a solution, Andrei. You've only shown the first moment in time.
But the original problem is very easily reduced to this one.
I've been trying for a few days now, and I can't figure it out.
Moreover, without friction it will go infinitely further, because the momentum of the cart with the sloth will not change at all, i.e. the speed changes according to the law 1/(ax+b), and the integral of it (path) is infinite.
I didn't write it right-
без того трения , которое ты пытаешься учесть
In this problem you don't need to count and account for friction.
This is not a solution, Andrei. You have only shown the first moment of time.
Still as a solution. With a caveat.
Strictly speaking, you still have to prove that with the same masses, the lower speed will remain with the cart with the lower speed. But I think this is obvious.
After that you can strictly by induction prove the ratio of velocities (more-lower) for any moment in time.
I'm done, Alexey, I'll take it from here.
OK, for those who don't like physics, I remind you of the balloon problem. I got exactly 2 weighing always.
The proof that one is not enough is elementary and fits in a couple of lines. The most difficult thing is to find an algorithm for exactly two weighings.
P.S. Finally found an oh-so-beautiful solution to the cart problem!
Friction is essential, it cannot be thrown out under any circumstances. But the equation of motion of the cart with the worker is reduced to the equation for the lazy one, i.e. it is possible to make him not throw the snow.
With balls - or with carts?
We make an equation for the sloth based on dp/dt = m(t)dv/dt + vdm/dt = -mu m(t) g. I.e. we reveal the momentum explicitly.
Compose the equation for the worker, considering both forces acting on the cart.
We note their almost complete similarity.
And make it complete by multiplying the equation for the worker by the integrating factor equal to 1 at zero.
It turns out that the new equation for the worker can be interpreted as follows: the former worker now does not dump snow, but also lies on the cart and does nothing. But the snow increases the mass of the cart according to a different law - not linear, but exponential. Further the proof is obvious, given that the integrating factor is an exponent equal to 1 at zero andgreater than a linear function.
Next (2)(if you know the answer - don't write!!! ):
Despicable invaders have taken over a village of megabrains, lined them up one after another in a column so that each successive one sees all the previous ones. They put a black or white hood on each megabrain so that no megabrain can see his own hood. Starting with the very last one (the one who sees everyone but himself), each megabrain is asked the colour of his cap in turn. If he is wrong, he is killed. But just in case, the megabrains have agreed in advance how to minimize the number of people killed. What did the mega-brains agree on?
Note: each responder can only say "black" or "white". No intonation, whistling, squatting or anything else will carry any information. In short, just one bit. Can't be silent either - they'll be killed.