Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 28
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No, it has nothing to do with it. The vertices are in integers, but the lengths of the sides are not necessarily integers.
// Самое простое - проверять именно на моей последней "полушар-полупружина" модели. Практически нет шансов запутаться, и никаких интегралов не светит.
Nope, there's a nasty point there: the bottom half has non-zero energy, as it's compressible too. You've oversimplified it. A spring would be fairer.
I'll try and work it out with a clean spring.
Nah, there's a nasty bit there: the bottom half has non-zero energy, as it is also compressed. You've oversimplified it. A spring would be fairer.
I'm going to try to figure out a pure spring.
It's up to you, but I think you'll still get the same thing. Go for it.
And anyway, my condition says that halves of the ball are incompressible and inelastic, and between them there is a weightless perfectly elastic spring. Take it as a base - you can't think of a better model.
No, it has nothing to do with her. The vertices are in integers, but the side lengths aren't necessarily integers.
Yeah, I've already found it. Oddly enough, there are at least a dafetag of tetrahedrophs. Example, the vertices are (0,x,2x),(0,x,-2x),(2x,-x,0),(-2x,-x,0) , where x is any integer
It's easy to see that all the points are equidistant.
I'm getting sick of the subject. Maybe I just didn't get enough sleep (I had to stay at work for 24 hours - urgent order, I only slept for two hours...)
Really, my brain is on edge.
System states:
1. A brick at a height of 1m has potential energy:
EPk=Mk*G*Hk; Vk=0.
2. The brick has reached the surface of the spring, all the potential energy of rest has transferred to kinetic energy of motion:
EKk=(Mk*Vk^2)/2
at this point the velocity is maximal:
Vk=(Vk0^2+2*a*s)^0.5=(0+2*9.81*1)^0.5=4.429 m/c
EKk=(Mk*4.429^2)/2=Mk*9.81
3. Kinetic energy of brick passes into potential energy of compression of spring, spring is compressed by distance depending on stiffness of spring, velocity of brick is equal to 0.
4. At the point when all the energy of the brick has passed into the compressive energy of the spring and the velocity of the brick is equal to 0, the unclamping of the spring begins. The maximum velocity of the end of the spring touching the brick and of the brick itself is 4.429 m/s at the point of initial contact. It is from this point and exactly at this velocity that the brick must start moving upwards to reach a height of 1m.
5. The velocity of the end of the spring is exactly 4.429 m/c. The moving half of the mass of the spring drags the other half behind it. Since half, the height of the jump is equal to half the distance of 1m, i.e. 0.5m.
Now, imagine instead of a spring a perfectly elastic body, our ball from the problem. It is incompressible by definition, so the brick changes direction instantly at the point of contact, the velocity vector changes sign but does not change magnitude, and the velocity is exactly the same as its initial velocity, for the body to reach a height of 1m. But the ball will not jump because it has not been compressed, as it is absolutely elastic.
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 on impact.
Where did I go wrong, why do I have a figure of 0.5 instead of 0.25?
Where did I go wrong, why do I have a figure of 0.5 instead of 0.25?
Because the jump height is supposedly proportional to the square of the velocity.
// It says so in Newton's precepts. I don't believe it either, but to avoid expulsion from the collective, I prefer to pretend to agree... Forgive me, Truth....
You manipulate the halves of the ball, you have to manipulate the spring.
Yeah, I've already found it, oddly enough, there's at least a dafega of these points. Example, the vertices are (0,x,2x),(0,x,-2x),(2x,-x,0),(-2x,-x,0) , where x is any integer
Nope, it doesn't work like a stone flower. The 1-2 edges are 4x = sqrt(0 + 0 + (4x)^2 ),
and 2-3 is sqrt( (2x)^2 + (2x)^2 + (2x)^2 )= 4x.
Because the height of the jump is supposedly proportional to the square of the velocity.
// It says so in Newton's precepts. I don't believe it either, but to avoid expulsion from the collective, I prefer to pretend to agree... Forgive me,truth.....
Geez, so mgh = mv^2/2. What are you unhappy about?
And it's not in his precepts, it's just a consequence of them.
joo: The velocity of the end of the spring is exactly 4.429 m/s. Moving half the mass of the spring drags the other half with it. Since half, the height of the jump is equal to half the distance of 1m, i.e. 0.5m.
You're strong, Andryukha. But the conclusion (in blue) is too bold.
You manipulate the halves of the ball, you manipulate the spring.
Nope, not like that. Edge 1-2 is 4x = srqt(0 + 0 + (4x)^2 ),
and 2-3 is srqt( (2x)^2 + (2x)^2 + (2x)^2 )= 4x.
Because the height of the jump is supposedly proportional to the square of the velocity.
// I don't believe it either, but to avoid expulsion from the collective, I prefer to pretend to agree...
This is true. But not for all bodies in the system in question. A spring has the same speed when the brick bounces, but only one end (half the mass), and the other half of the mass has to be pulled behind it as well. Otherwise, the spring would fly the same distance as the brick, but only half the distance.
yeesh, phiségés. :)
Geez, so mgh = mv^2/2. What are you unhappy about?
And it's not in his commandments, it's just a consequence of them.
This is true. But not for all bodies in the system in question. A spring has the same speed when the brick bounces, but only one end (half the mass), and the other half of the mass has to be pulled behind it as well. Otherwise, the spring would fly the same distance as the brick, but only half the distance.