Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 66
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I looked a couple of pages ago at the formulas that Mishek and Andrei were putting under the skateboard. It turns out that friction can be ignored, it doesn't care about mass.
Sliding friction doesn't matter (the formulas above are quite true: these forces are proportional to mass and, therefore, at any value of mass they will give the body the same acceleration).
Rolling friction is a slightly different story as it is a physically different process. Its essence is that the wheel constantly has to crush under itself the lying inelastic surface ahead (a layer of snow), which is equivalent to a slow as if rolling down a very gentle hill. The layer of snow in front of both carts is the same, and it is easy enough to check that neither cart has a range advantage: the height of the equivalent slide that a cart can roll up is determined from the law of conservation of energy mgH = mv^2/2, where mass, as we see, decreases regardless of even whether it is constant or variable.
My answer is that both carts will travel exactly the same distance.
The moderator claims that the formula looks the same as for sliding friction.
So everything about friction figured out?
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What about the acceleration of falling snow? Any specific objections to my logic (and Andreev's)?
What's up with dispersing falling snow? Any specific objections to my logic (and Andreev's)?
MD: . ..snow falling on both carts brakes both carts because it falls vertically, i.e. it has a velocity vector directed against motion (in the carts' coordinate systems). That is, the carts have to accelerate the "stationary" snow to its speed.
I don't want to move to carts coordinate systems, which are non-inertial in principle - simply because snow, by increasing the mass of the carts, slows them down. What the hell do I need all this trouble for?
Since the non-cleared cart is heavier and the quantity of snow (by default) falls on them is approximately equal, then the heavy cart will be braked less.
So it brakes less, so what? But there's more friction on it. And it's heavier, so it goes slower. Still not convincing, Volodya.
So you both insist that the snow not only brakes the cart indirectly (through increased friction force), but also directly? I don't see that. In the ground system, the snow falls strictly vertically and does not impart any momentum to the carts in the direction of travel. What you call "the need to accelerate the stationary snow" is purely the law of conservation of momentum, from which the braking of the cart is directly derived if the snow is not smeared on the ground.
In short, there's no way I can absorb the problem to the point where I can understand it without formulas.
P.S. That's what has come to my mind - and this factor has not been taken into account by none of us yet. Snow creates constant "dynamic pressure from above" - simply because it has mass and speed. This pressure increases the 'weight' of the cart - even if the snow is then removed.
It's easier to understand it if you consider that it's not snow, but elastic balls that constantly bombard the cart from above. The pressure on the support (asphalt) is higher than just the weight of the cart. The cart will decelerate faster because there will be more support reaction, i.e. the friction will also be higher.
P.P.S. A couple more simple ones:
(4) There are two electric kettles with thermostats. They work like this: when the temperature drops to 70°, the heating coil turns on and a light bulb comes on; when the temperature reaches 90°, both the heating device and the light bulb turn off. On one of the kettles the light bulb comes on, on the other it does not. Which one is most likely to have hotter water and why?
(4) A tributary forms a sharp angle as it flows into the river. On land, inside the corner, stands Megamozg's shack. Every day, Megamozg leaves it, goes to the tributary, meets the sunrise, then goes to the river, meets the sunset and returns to the shack. How does Megamozg need to plot a route so that the distance he walks each day is minimal? Count the banks of the river and a tributary as straight lines.
Now, you've got something wrong here. It's like there is no snow for the snow being cleared. Adding dm slightly reduces the speed, but the megamotor drops it (perpendicular to the motion!) - and returns the original speed. Nothing has changed, for the law of conservation of momentum.
Doesn't the ejected snow have momentum?
When the snow hits the cart, the total momentum is conserved, but the snow also gets momentum. When snow is thrown, its momentum is also thrown. It's like when billiard balls collide - the momentum adds up, and after the collision it splits. A stationary ball hit by another one is snow)), and a brother-in-law is a cart. Clearly, the momentum of the brother-in-law after the collision will be less than before (the snow was dumped). And the speed will decrease accordingly. The variant when the snow was not dumped is when the balls stick together during the collision. It's clear that the speed after the collision will be lower because the mass will increase, but the total momentum will remain the same. And the whole question is whether the momentum is lower in the first case, or the mass is higher and the friction is higher in the second one.
And what kind of friction to take into account? Rolling or sliding. If rolling, it depends on properties of the wheel. If you are talking about an abstract wheel that is completely solid like a surface, this is one thing; if it deforms under the mass of a body like a real one, this is another. But a cart rides on snow and there is no perfectly solid surface. A heavy cart can get stuck in snow))) Anyway, how does the friction force change with increasing mass. Apparently it's linear, otherwise you can't solve the problem.
The moderator claims that outwardly the formula looks the same as for sliding friction.
What about acceleration of falling snow? Any objections to my (and Andreev's) logic?
So the snow falls equally on both carts, where does the difference in motion come from? And at dumping, the cart is not affected by any impact on the axis of motion, hence nothing affects the equations in this projection: the mass was reduced and is reduced as well...
I think the task is quite accessible for a good understanding, I don't see a problem)
Avals:
Doesn't the ejected snow have momentum?
possesses, but in the projection on the acceleration of the launch its velocity is equal to that of the cart, hence the latter is unaffected by resetting. If you want, write down the law of conservation, it's all obvious.
has, but in the projection on the launch ramp its speed is equal to that of the bogie, so the latter does not change when it is dropped. If you want, write down the law of conservation, it's obvious.
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Most likely the water is hotter in the kettle with the light bulb on.
Heating and cooling occur non-linearly, the heating curve does not coincide with the cooling curve and is higher.
So the snow falls equally on both carts, where does the difference in motion come from? And at dumping the cart does not experience any influences along the axis of motion, hence nothing affects the equations in this projection: the mass was reduced and is reduced as well...
By dumping the snow, MM reduces the energy of the system. Does that make sense?
Therefore, the snow does not fall in the same way.
velocity is not reduced by dumping, but momentum is reduced. Speed is reduced when the snow hits the cart in both cases. But does the distance travelled to a stop depend only on speed and not on mass?
Yes, it only depends on speed, a well-known fact.
And the second "batch" of snow will reduce the speed of the trolley being cleared more significantly than the one not being cleared, because the trolley weights will be different.
TheXpert:
By dumping the snow, MM reduces the energy of the system. Does it make sense?