Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 112

 
MetaDriver:


// Pisa smokes.

Pisa will fall with envy
 
Mischek:
Pisa will fall with envy.

No, that's not it. I'm going to recalculate.

Maybe there'll be some corrections and the line will add up.

 
MetaDriver:

No, that's not it. I'm going to recalculate.

Maybe there'll be a correction and the numbers will add up.

That's right! Pisa's taking vitamins and he's on the mend.

The series is quite convergent, all the way down to one. 1/2+1/4+1/8+1/16 +(1/2^n)

That's all there is to it.

As it was initially expected, the maximum shift is one brick at any tower height.

Ramine.

------------------------------------------------

Once again, it was all wrong. The final series: 1/2+1/4+1/6+1/8+1/10+...1/(2*n)

It diverges at infinity, i.e. the maximum shift is infinite.

Pisa is in intensive care after an overdose, chances are doubtful.

 
MetaDriver:


It's diverging at infinity, i.e. the maximum deflection is infinite.

Pisa in intensive care after an overdose, the chances are doubtful.

This terrible news has caused inexplicable shifts in the foundations of a number of famous towers, resulting in deviations from the vertical.

By 6am Moscow was able to make some measurements, but applications from towers continue to come in from almost all over the world

As we can see from the report, even London's vaunted Big Ben failed. But Pisa is still of particular concern. If the deflection dynamics are maintained, it will collapse by lunchtime.

 

Here it is, the real top - quite thematic (not counting Humour, i.e. looking only at the first two topics). We will soon catch up with the first topic, but we don't stand a chance with the second one:

P.S. The answer to the trolley problem did not count.

 
Mathemat:

P.S. The answer to the trolley problem did not count.

:) post your solution and we'll dissect it :)) You post my solution.
 
TheXpert: :) post your solution we'll dissect :)) You post my solution.

I don't know your complete solution, it wasn't here. The friction force has to be taken into account anyway.

Here's my version (slightly adjusted, as I was talking about recoil in the beginning):

Let's assume that the snow falls with constant speed, so the mass of the cart with MM, if the snow is not dumped from it, would grow by the law

m(t) = m_0 + alpha * t.

The general equation of motion is the same for both carts (on the left is the derivative of the cart's momentum):

dP/dt = - F.

However, each cart is subject to different braking forces.

The "lazy" cart is only affected by the increasing frictional force, equal to

F_fr = mu *g * (m_0 + alpha * t).

The working man's cart is subject to a similar frictional force -

F_frr = mu * m_0 * g,

If during the time dt the mass alpha * dt of snow fell on the cart going with speed v, then by transferring the same mass of snow and during the same time sideways (so that the process is continuous), MM gives impulse dp = alpha * v * dt to the snow along the cart motion.

Since, according to the condition of the problem, the friction is very small, and "the carts gradually but slowly slow down from friction", it is suspected that the main events unfold closer to the finale than at the beginning. Consider the laws by which each of the braking forces acts on the motion of the cart.

1. The variable frictional force on the cart of the sloth during the time from the beginning of its motion to the moment t will take away from it the momentum equal to

mu * m_0 * g * t + alpha * mu * g * t^2/2.

Thus this function of time is increasing and concave, i.e. it grows "with acceleration".

2. A constant frictional force on the working cart for time t will take away momentum

mu * m_0 * g * t.

3. MM when throwing snow will take away from the cart an impulse equal to

alpha * S(t) (see the blue expression above).

Here S(t) is the distance travelled by the cart. Since the cart slows down, this function is increasing and convex with time, and at longer times it grows slower than both of the functions considered.

Thus, of the three functions considered, the function from item 1 is "asymptotically" the fastest (when time is long enough). So, momentum will be taken away the fastest by the lazy one, and it will stop earlier.

Longer the cart with the working one will go.

Dissect it. I don't know what to do anymore. The only thing left to do is solve the diphires. And a moderator just drones on and on: "the reasoning is (at least) wrong".

In short, I see a fundamental error. I am comparing times, and I should be comparing distances.

 
Mathemat:
Prepare it.
I'll try it tonight. It didn't work out. In (1) (2) everything seems to be correct, in (3) I have to figure it out.
 

By the way, the balloon problem has been dropped. It was either 2 or 3 weighings - and then it's unclear. I mean, like 3.

I have an unambiguous solution. Shall we solve it?

 
Mathemat:

In short, I see a fundamental error. I'm comparing times, I'm comparing distances.

You're just drawing the wrong conclusion. You cannot draw conclusions "asymptotically", because you do not even know the type of function, and there you get a diffura, because speed is a function of time, and you have to take an integral with it.

In short. I'll say it again -- friction force can be ignored at all, since it gives a constant inverse acceleration to a cart, irrespective of its mass. Further, see my very first post. The difference depends only on momentum transfer.