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Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 67

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alsu: The formulas show that the distance travelled will be exactly the same. I insist on that.
Then the formulas are in the studio.

I write a rigorous solution.

Let there be some cart with variable mass m(t) (both the first and the second fit this definition). Let us write down Newton's second law:

m(t)*x'(t) = F(t),

where F is the net result of all forces acting on the trolley. Only the friction force Ftr(t) = - k*N = - k*m(t)*g, where k is the friction coefficient (combined, taking into account both sliding and rolling), N is the support reaction force, which by Newton's 3rd law is numerically equal to the cart weight, g is the free fall acceleration. Minus corresponds to the direction of the force against motion. So,

m(t)*x'(t) = -k*m(t)*g

As we see, the mass decreases, whence

x''(t) = -k*g = const,

since the acceleration of free fall is constant and the coefficient of friction depends only (!) on the material of the wheel and the surface.

So the cart moves at an equal acceleration regardless of how its mass changes. Hence, the distance travelled is exactly the same.

alsu:

So the cart is moving at an equal rate regardless of how its mass changes.

Bravissimo, cap.

Where are the impulses when snow falls?

I said right off the bat that you can lay off friction and only compare the effect of snow falling on velocity.

TheXpert:

Bravissimo, cap.

Where are the impulses in snowfall scheduled?

they are accounted for in the mass variable
alsu:
They're accounted for in the mass variable.

where are they accounted for in the speed variable? if the snow didn't take some of the momentum, it would be right. It doesn't make sense.

The path is independent of speed?

TheXpert:

and where are they accounted for in the speed variable? if the snow didn't take some of the momentum, it would be right. But it's a mess.

The path doesn't depend on speed?

I think

Yes, lied in the second law. The correct way would be:

p'(t) = F(t)

(m(t)*v(t))' = -k*m(t)*g

m(t)*v'(t) + m'(t)*v(t) = -k*m(t)*g

v'(t) + m'(t)/m(t)*v(t) = -k*g

v'(t) = a(t) = -k*g - v(t)*[ln(m(t)]'

That is, the deceleration (negative acceleration) of the system has two components - 1) a constant plus 2) a variable additive proportional to the current velocity and the derivative of the logarithm of mass. Obviously, to answer the problem one must analyze the second summand.

I remove my previous answer from the discussion, it is obviously wrong))

alsu:

I write a rigorous solution.

Suppose there is some cart with variable mass m(t) (both the first and the second fit this definition). Let us write down Newton's second law of motion:

m(t)*x''(t) = F(t),

Or maybe actually dP/dt = - F_frict?

On the left is the derivative of momentum. But in the case of a lazy megamotor (no snow dumping) the mass is increasing.

In short, the equation comes out about the same as for reactive motion (although there is none).

P.S. One more point. A megamotusk dumping snow orthogonal to the motion creates a support pressure component perpendicular to the motion (it pushes the cart orthogonal to the motion). Does this not affect the support reaction?

P.P.S. You are already corrected.

In the evening I'll try to derive an explicit expression for the speed of the cart on time, the whole complexity here is unknown m(t), but I'll try to at least get a qualitative solution and, consequently, the final answer to the problem.

Mathemat:

Won't this affect the support reaction?

It does :) one side will exert more pressure and the other less. And the total friction force should increase.

But it's very fleeting. Probably can be neglected.

The farther into the woods...

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