Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 26

 
Mathemat:

What's so wild about it? It's just a simple conspiracy, a Newtonian one. That's not the gimmick, it's that you have to look at the spring as a whole body.

Where did you put the balloon? ?? We were all staring at it, and suddenly it's gone, and now you're putting a spring in its place. It wasn't bouncing at all.


And what's the inarticulate question at the end?

This one:
And explain now why, if this picture is put vertically, a ball having twice the speed will jump to a height four times the height.

And anyway, I'm already dead. Please don't disturb my good memory.

Until the resurrection.

 
MetaDriver: What did you do with the balloon already? ?? We were all staring at it, and suddenly it's gone, and now you're putting a spring in its place. It wasn't bouncing at all.
And now it's jumping. That's the centerpiece of the switch.
 

(5 points; who knows the answer - do not write!!!!)

Is it possible to arrange a regular tetrahedron in Cartesian coordinate system so that all its vertices lie at points with integer coordinates?

 
Mathemat:
And now it is bouncing. This is the centre substitution.

OK, thimble man, let's get more specific with the details. The jump height of the mysterious ball spring, according to the Newtonian myth, = 25 cm - (almost / 4), right?

Now honestly take the ball back out and let's figure out its mass already, at least in relation to the brick.

Its velocity, as the public disassembly found out, is half the velocity of the brick at the moment of jumping.

Suppose the mass of the brick = x, the mass of the ball = x

The impact energy of the fall (E) is = X * Vdo^2

The same energy is then distributed between the ball and the brick. i.e. X * V before^2 = X*V after^2 + x * (V after/2)^2

We know that the brick has jumped 100 cm - almost

I'm confused further. How do I make the right equation (proportion)?

 
MetaDriver:

OK, thimble man, let's get more specific with the details. The jump height of the mysterious ball spring, according to the Newtonian myth, = 25 cm - (almost / 4), right?

Now honestly take the ball back out and let's find out its mass, at least in relation to the brick.

Don't give a fuck about mass. You really want to find out how much of the brick's energy is taken up by the ball-spring?

I doubt there's enough data. A spring can weigh 10 grams but be stiff enough not to break. It will syrano jump 25 centimetres. It will continue to oscillate during the jump. But that's how much of the energy of the original brick these vibrations have - I don't know.

Have you looked at my solution? There's nowhere to go without the spring stiffness, if you want to know anything else. We can only say with certainty that the shortfall in the height of the brick is the total energy of the spring.

 
Mathemat:

Don't give a fuck about mass. You really want to find out how much of the energy of the brick is taken up by the spring ball?

I doubt there is enough data. A spring can weigh 10 grams but be stiff enough not to break. It will syrano jump 25cm. It will continue to oscillate during the jump. But that's how much of the energy of the original brick these vibrations have - I don't know.

Unconvincing. I didn't really account for the energy trapped in the vibrations. But it doesn't look hopeless. If the height of the jump can be calculated, then the energy can be figured out as well.

// And leave the spring alone already, I remember exactly that there was a ball in the beginning!

 
MetaDriver:

Unconvincing. The energy trapped in the oscillation I did not take into account. But still the case does not seem to be hopeless. If the height of the jump can be calculated, then this energy can be dealt with.

// And leave the spring alone already, I remember exactly that there was a ball in the beginning!

That's the beauty of the problem: we don't know how the energies will be distributed, but we know how the velocities will be distributed.

In fact, the ball spring will make a complex motion - translational and oscillatory. How the energies of these movements will be distributed - I don't know yet.

And don't go back to the balloon, you'll be tortured to calculate its internal vibrations.

The total energy of the ball (spring) is equal to the lack of energy of the brick to reach exactly one meter. That's where you dance from. But it is the sum of two quantities - energy of translational motion, which depends on mass, and energy of natural oscillations, which depends on stiffness and amplitude.

Well, that's about it:

M_brick * g*H = m_ball*g*H / 4 + k*x^2 / 2 + M_brick * g*(H-delta)

On the left is the total energy of the brick in the beginning, on the right is the energy distribution after the bounce. Cool result.

M_brick * g*delta = m_ball*g*H / 4 + k*x^2 / 2

Hence the total vibrational energy of the spring is equal to:

k*x^2 / 2 = M_brick * g*delta - m_ball*g*H / 4

 
Mathemat:

Did you look at my solution? There's nowhere to go without spring stiffness if you want to know anything else. We can only say with certainty that the shortfall in the height of the brick is the total energy of the spring.

OK, then.

E of system = X * V before^2 = X*V after^2 + x * (V after/2)^2 + E of ball spring oscillation

Vdo = (2gH)^(1/2) = (2 * 9.8 * 1) ^(1/2)// constructed from what I found on the internet

V after = (2 * 9.8 * (1 - almost)) ^(1/2)

substitute

Ecis = X * (2 * 9.8 * 1) ^(1/2) = X*((2 * 9.8 * (1 - almost)) ^(1/2))^2 + x * (((2 * 9.8 * (1-most)^(1/2)) /2)^2 + Ecosys

Is that it?

One more thing. an important consideration. it seems to me that the "vibrational" energy of the ball spring is strictly equal to its kinetic energy at the time of the jump. this is a speculative consideration, but no holes can be seen. Proceeding from the observation that at the moment the brick bounces off the ball, the top of the ball is moving at the same speed as the brick, but the ball ends up moving at half the speed. That is, the other half of the velocity is stolen by the processes all of whose energy is residual oscillation.

If this consideration is correct, then the formula is glib.

Clever people, please correct it.

 
MetaDriver:

Also. an important consideration. it seems to me that the "vibrational" energy of the ball spring is strictly equal to its kinetic energy at the moment of jumping. this is a speculative consideration, but no holes can be seen. Proceeding from the observation that at the moment the brick bounces off the ball, the top of the ball is moving at the same speed as the brick, but the ball ends up moving at half the speed. That is, the other half of the velocity is stolen by the processes all of whose energy is residual oscillation.

If this consideration is correct, then the formula is glib.

I'm not sure, but I think there's a theorem somewhere in theorising about the distribution of vibrational energy between vibrational and translational. But I do not remember it.

It is usually used when calculating the heat capacity of gases.

 
Mathemat:

That's the beauty of the problem: we don't know how the energies will be distributed, but we know how the velocities will be distributed.

In fact, the ball spring will make a complex movement - translational and oscillatory. How the energies of these movements will be distributed - I don't know yet.

And don't go back to the ball, you'll be tortured to calculate its internal vibrations.

The total energy of the ball (spring) is equal to the lack of energy of the brick to reach exactly one meter. That's where you dance from. But it is the sum of two quantities - energy of translational motion, which depends on mass, and energy of natural oscillations, which depends on stiffness and amplitude.

It goes something like this:

M_brick * g*H = m_ball*g*H / 4 + k*x^2 / 2 + M_brick * g*(H-delta)

On the left is the total energy of the brick at the beginning, on the right is the energy distribution after the rebound. That's pretty cool.

Yeah, your formula's about the same. Now, think about how the energy of vibration has to depend on stiffness and amplitude. I don't believe it. Think again. It doesn't look like anything. We know the ball is perfectly elastic. That's enough. How exactly the waves walk in it, unlike the spring, is absolutely one-dimentional invariant - it does not influence on the amount of energy conserved in the vibrations.