Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 29
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But it will jump not by half, but by a quarter.
My explanation is very simple: the spring is stretched uniformly. Its upper end is moving with speed v and its lower end is stationary. So the velocity of the centre of mass is v/2. But it will jump not by half, but by a quarter, for the maximum height is proportional to the square of the initial velocity.
There! Now I agree. Yes, by 0.25, right. But this is only true for the spring. You have to consider exactly the motion of the centre of mass, not halve the mass of the spring by two. That's the error in my calculations.
BUT! By the terms of the problem the ball is elastic to a pulp and so hard and incompressible as a paratrooper's point during the jump that it has no velocity during the bounce of the brick as it has never been compressed at all and therefore does not spring from the table surface.
BUT! By the terms of the problem, the ball is elastic to a pulp and so hard and incompressible as a paratrooper's point during the jump that it has no speed during the bounce of the brick, as it never compresses at all, and therefore never springs from the table surface.
As I understand it, "absolute elasticity" and "incompressibility" are two big differences, even two and a half.
The condition stipulates elasticity and omits incompressibility. You also made up something about hardness. There was no such letter.
As I understand it, "absolute elasticity" and "incompressibility" are two big differences, even two and a half.
The condition stipulates elasticity and omits incompressibility. You also made that up about hardness. There was no such letter.
Oh, right, a perfectly elastic body is compressible and reverts to its original shape after the load is removed without residual deformation. Pardon me.
My remark:
ZS. Don't try to calculate the force acting on a perfectly elastic body at the moment of impact. It tends towards infinity. That's why they never use the perfectly elastic body model in strength calculations for impact.
The condition does not refer to a perfectly elastic body, but to a perfectly solid body.
So the ball and brick problem is solved and the height of the ball bounce is exactly 0.25m. Amen.
The blot problem, I take it, does not interest anyone. Is the solution interesting or not? Or will you try? It's really very simple (although it's 5 points).
On a plane with a rectangular grid with step n, ink is poured in the form of lots of blots of different size and shape. The total area of the ink spots is less than n². Prove that it is possible to shift the grid in such a way that no node of the grid is flooded with ink.
The blot problem, I take it, does not interest anyone. Is the solution interesting or not? Or would you like to try? It really is very simple.
That's just awful, it smells a lot like integrals. But the solution is interesting, of course.
Not a single integral, I swear. Sixth or seventh grade high school level.
MD, try it. The most useful thing is just the kind of problems you've never taken on before.
For blots to be guaranteed to cross at least one node of the grid, the area of the blots must be greater than n*n (the area of the cell), and that area is less than n*n. So the mesh can always be placed in such a way that no node falls on a blot.
Of course, it's a bullshit solution, but that's how it is. :)
Yeah, that's bullshit. More creativity is needed here, Andrei (I myself have problems with creativity).
Here's a blot for you. The area is clearly smaller than the area of the square:
You wrote only the conclusion, but how it turned out - did not write.
There is a simpler solution: it is a tetrahedron inscribed into a cube with edge 1 and one vertex at the origin (a cube in the positive octant of the three-dimensional Cartesian coordinate system).
The coordinates of the vertices are (1,0,0), (0,1,0), (0,0,1) and (1,1,1).
It is clear that the solution is shiftable, scalable and rotatable. i.e. there are many integer solutions.
// But my centre of gravity is at zero.