Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 61
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Justify it, Andrew. I have the same answer, but the moderator doesn't accept it.
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 velocity (M + 2dm)V2 ; V2 = (M + dm)V1/(M + 2dm) = MV/(M + 2dm)
_______________
for cleared
_______________
momentum MV
after the addition of snow (M + dm) V1' ; V1' = MV/(M + dm)
after resetting 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 each iteration you can prove that not brushing is more efficient.
Write down the equations of motion. I'm talking specifically about momentum, not cart speeds.
Or simplify it this way.
You're riding on a train platform, the platform passes the station. There is a 1,000 kg suitcase at the station.
You pass it and grab its 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 spin it back around, not a suitcase, but a snowflake, not from the station, but from the sky.
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 velocity (M + 2dm)V2 ; V2 = (M + dm)V1/(M + 2dm) = MV/(M + 2dm)
_______________
for cleared
_______________
momentum MV
after the 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
Why isn't it taken into account that the friction force is greater on the heavier cart?
It is wrong to solve the problem in terms of momentum. Energy is not added, but escapes through a single mechanism - friction. And increasing mass increases friction. And, accordingly, it takes more energy to travel the same distance.
You can simplify it without any harm to the task like this. We divide the path into two sections.
At the beginning of the first section both carts received the same momentum and drove to the end of the first section, accumulating snow in themselves and not removing it.
At the end of the first part (at the beginning of the second part), the snow is removed from the second cart in one stroke perpendicular to the movement. The snow stopped falling from the sky. Who will get the furthest.
The energy of the second cart has decreased by the mass of the snow thrown off; it will pass through less.
// there is a nuance, the friction is equal in the same conditions (masses)
At the end of the first section (beginning of the second section) the second cart dropped snow in one fell swoop perpendicular to the traffic. The snow stopped falling from the sky. Who will get the furthest.
The second cart's energy is reduced by the mass of snow it has thrown, so it will pass through less.
No, both will travel the same :)
It is wrong to solve the problem in terms of momentum. Energy is not added, it escapes through a single mechanism - friction. And increasing mass increases friction. And, accordingly, it takes more energy to travel the same distance.
without regard to friction?