This was a high school competition problem in my country. Can?

This was a high school competition problem in my country. Can IQfy solve it?

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  1. 2 months ago
    Anonymous

    question does not specify the nature of the force which would cause any change in position, so the question is unanswerable without assumptions. should we assume classical universal gravitation, general relativity, constant acceleration from gravity near the Earth's surface or some other source, ...? in the limit as the downward force approaches zero, the velocity approaches zero as well, by simple conservation of energy. or we can simply assume no force is applied (none is mentioned) and thus it's a trick question and the ball never hits. now what's my prize?

    • 2 months ago
      Anonymous

      That's a lot of cope for not being able to solve it. Obviously the force is constant acceleration from gravity near the Earth's surface, even if it's not explicitly stated.

      • 2 months ago
        Anonymous

        >obvious
        source?

        • 2 months ago
          Anonymous

          >implied
          source?

          Keep coping because you can't solve it lmao

      • 2 months ago
        Anonymous

        But earths atmosphere has air resistance. I think this is more likely on the moon

        • 2 months ago
          Anonymous

          Even space has air resistance.

          2sqrt(11gr/15) probably doing something wrong?

      • 2 months ago
        Anonymous

        peer reviewed fully funded journal study citation for that, chuddy?

    • 2 months ago
      Anonymous

      Op here, constant downwards gravitational accelation g is implied in such problems.

      • 2 months ago
        Anonymous

        >implied
        source?

        • 2 months ago
          Anonymous

          OP already stated it was a high school competition. Tried some common sense, lately?

    • 2 months ago
      Anonymous

      Crazy cope. Goes without saying the ball is affected by gravity and with the starting distance not being 0 and there being no friction it will hit the ground.

      https://i.imgur.com/rZxmVnx.png

      This was a high school competition problem in my country. Can IQfy solve it?

      So uh, he ball falls 2r - eps. It will horizontally displace both cubes by r-eps/2. You do see how the eps causes alot of unnecessary trouble ?
      At first almost all force is used to accelerate the cubes. At the time the ball has fallen r the force to accelerate the cubes has reached 0, the cubes will continue their outward trajectories and the ball will fall as normal.
      This would mean one needs to come up with a term for the speed the ball is starting it's free fall with after dropping 1r.

    • 2 months ago
      Anonymous

      You're the type to write walls of text on an exam question that asks for an explanation of an answer where two sentences would suffice. Please, worthless b***h, let's not be stupid here.

    • 2 months ago
      Anonymous

      moronic pseud

    • 2 months ago
      Anonymous

      >there is no friction
      >sitting on midpoint of 2 planes
      Oof

  2. 2 months ago
    Anonymous

    Apply assumption that if all objects are perfectly rigid, and the implication of a collision with the phrasing of "hits the ground," the ground must be an impermeable entity, and by virtue of this, any object in contact with the ground moving towards it will have 0 velocity in the component orthogonal to the ground's surface, as the ball cannot tunnel into the ground, or deform at the moment of impact

    • 2 months ago
      Anonymous

      Why does everyone in this thread feel the need to act smart. This is a standardly phrased mechanics question. Clearly we are talking about the instant right before it hits the ground, i.e. epsilon time before.

      Also you are wrong, for a perfectly rigid body, it transitions from velocity V right before hitting to velocity -V right after hitting. There is never an instant when the velocity is 0, as that would violate convervation of energy. If anything it is undefined at the moment when it contacts the ground.

      • 2 months ago
        Anonymous

        People do this a lot when faced with any kind of riddle or brainteaser, not just science/mathematics problems. They'd rather "outsmart" the question by nitpicking some small detail or loophole, thereby absolving themselves of having to make a good-faith attempt at the problem and risk the worst fate imaginable (getting the wrong answer).

      • 2 months ago
        Anonymous

        >OP posts a high school math problem
        >surprised that underage smartasses flood the thread

      • 2 months ago
        Anonymous

        >Why does everyone in this thread feel the need to act smart.
        It's because barely anyone on this board actually is smart. You can see this pretty much any time a problem is posted, 80% of the thread will be homosexuals like this.
        One look at the catalog should've told you about the average intelligence of this board.

  3. 2 months ago
    Anonymous

    Sum of energies constant.
    Simple pythag relating ball height and block displacement.
    Eliminate variable.
    Solve differential equation.
    Check when contact is lost.
    Simple after that.

    • 2 months ago
      Anonymous

      Okay, do it?

    • 2 months ago
      Anonymous

      I'm getting sqrt(4gr) unless there is something tricky happening.
      Basically except you do not need to solve the differential equation.

    • 2 months ago
      Anonymous

      no diff eq

      https://i.imgur.com/frMhiNW.jpeg

      [math]sqrt((1 +pi/2)mgr)[/math]

      units are bad

      I'm getting sqrt(4gr) unless there is something tricky happening.
      Basically except you do not need to solve the differential equation.

      this is the same if the block isnt there

      I integrated the force along the path the ball takes which I then set be to kinetic energy to obtain this value. KE = mgr + mgr*(integral{0, pi/2} of cos(t)-cos^5(t)dt).

      integral over 0 -> pi/2? Also, idk where ur getting mgcos^2 from. Im picturing the diagram in my head and I dont see what's the point of using a vector of that length.

      question is inconsistent with itself. on the one hand, you're lead to believe that the ball will push the squares aside, thus causing them to have a velocity. but if you write down the generalized lagrangian, you'll see the squares have no horizontal component of potential energy, and thus the euler lagrange equations imply their velocities are constant. the initial condition is their velocity is zero, so the squares don't move and the ball never hits the ground. the only way out is to assume quasistatic motion of the squares such that the ball is moving at infinitesimal speeds at all times and that it hits the ground with speed zero, which is patently absurd. you need to include friction for this problem to work..

      Suppose that the left block is removed. The ball is in unstable eq., so suppose it falls to the left. Ur saying the block aint gonna move to the right to counter the fall to the left?

      https://i.imgur.com/BwdjZNU.png

      Insufficient information regarding to change in conditions from the OP question so here I am to give you three velocity when it hits the ground.

      If you want to be more technical then the first v = 0 isn't quite answering the "hits the ground" question because it will always be there. Frick this poorly worded stupid bait ass question.

      u need to learn about equilibriums and stability

      • 2 months ago
        Anonymous

        Here's a diagram (the symmetry is implied). i may probably be double counting some cos thetas, not sure.

        • 2 months ago
          Anonymous

          Ok, so from that diagram, starting at Fn*cos, what length is Fn? So now, you have an Fn and an Fn*cos^2 that are in the same direction. What are each supposed represent? Do u see the problem here?

        • 2 months ago
          Anonymous

          you're multiplying by cos twice to get [math]cos^2 [/math] for no reason, I think you can directly decompose [math]F_N = F_x + F_g = F*cos(theta) + mg [/math]. You're not creating a triangle whose hypothenuse is Fn, you're projecting Fn over cos and sine and solving back using Fg

      • 2 months ago
        Anonymous

        >u need to learn about equilibriums and stability
        And you, a massive moron on the left of the dunning-kruger curve, need to get a bachelor's in physics and a brain for basic reading comprehension to understand the OP is a shitty ass incomplete bait question with no conditions specified for epsilon.

        Gosh really IQfy is full of moronic STEM dropouts

        • 2 months ago
          Anonymous

          ?
          u solve in terms of epsilon tardy

          • 2 months ago
            Anonymous
          • 2 months ago
            Anonymous

            Now give me my pip

          • 2 months ago
            Anonymous

            I know how to whip up a storm - simple in some eyes - judging by the size of my wiener - my attempts at sex - and theirs in comparison. You distinguish it this way. This is falsehood.

          • 2 months ago
            Anonymous

            I suffered quite immensely for 15 years. Prior to that I suffered characteristically for 10. You claim the profit I earned from this position must be lawed in response, even a simple reversal. Come on...

          • 2 months ago
            Anonymous

            They would have never let go, if the tables were reversed it would be the max of immorality. That's your only real argument, that immorality doesn't match morality.

          • 2 months ago
            Anonymous

            Hell doesn't scorn me, neither does heaven. Life does, life does not know all yet. So we should suffer for the unchanging?

          • 2 months ago
            Anonymous

            Your forgiven - (joke) - I'll take your hell - etc.

            End.

  4. 2 months ago
    Anonymous

    the ball will free fall r distance. becouse after r distance the 2 blocks are shoved away for free fall.
    so its root(2r/g)
    i ignore start velocity becouse its a highschool question.

    • 2 months ago
      Anonymous

      wait i used wrong formel. thats the fall time.

      • 2 months ago
        Anonymous

        It looks like to find the fall time, you would have to solve a nonlinear differential equation of the form y'' = ay^2 + by. I cant see any other way outside of a numerical approx to evaluate this, at least according to my google search

    • 2 months ago
      Anonymous

      How do you know it won't lose contact before then?

      • 2 months ago
        Anonymous

        i dont. maybe moving masses without friction without changing high dont even need energy. then would it free fall from start.

        • 2 months ago
          Anonymous

          It will keep adding kinetic energy to the blocks while it is in contact so it won't be free fall from the start.

          • 2 months ago
            Anonymous

            maybe that the trick the ball transfers his fall energy 100% in the first r length fall.
            So that you have the remaining r free fall without start energy.

          • 2 months ago
            Anonymous

            maybe that the trick the ball transfers his fall energy 100% in the first r length fall.
            So that you have the remaining r free fall without start energy.

            maybe the kinetic energy from the ball equals the kinetic energy from the 2 quebes.

  5. 2 months ago
    Anonymous

    maybe its a trick question. isnt the max fall speed in vaccum very soon gained ?

    so the answer is 9.8 m/s

  6. 2 months ago
    Anonymous

    I genuinely don't understand the question. Nothing is moving, so the ball doesn't fall? Did you forget to specify that someone pushes the ball or something?

    • 2 months ago
      Anonymous

      That is true if epsilon is 0. If epsilon is not 0 the ball exerts a slight horizontal force on the boxes based on the angle of the point of contact, meaning the boxes will move away from eachother and increase this angle at the point of contact, making the ball fall.

    • 2 months ago
      Anonymous

      Gravity.

      • 2 months ago
        Barkon

        IQ low

    • 2 months ago
      Anonymous

      Theres a space epsilon between the squares so they would slide out from under the ball

      • 2 months ago
        Anonymous

        https://i.imgur.com/Abjenyt.png

        [...]
        did i do it right?

        The right box would rotate counter-clockwise while it moves right and the left box would rotate clockwise while it moves left as you have a non-zero torque.
        With epsilon near zero the torque for each box is something like mgr/2. You can't ignore that.

  7. 2 months ago
    Anonymous

    [math]sqrt((1 +pi/2)mgr)[/math]

    • 2 months ago
      Barkon

      No.

  8. 2 months ago
    Anonymous

    2sqrt(11gr/15) probably doing something wrong?

    • 2 months ago
      Anonymous

      I integrated the force along the path the ball takes which I then set be to kinetic energy to obtain this value. KE = mgr + mgr*(integral{0, pi/2} of cos(t)-cos^5(t)dt).

  9. 2 months ago
    Anonymous

    >This was a high school competition problem in my country
    what, are you 12 or something? Imagine caring about high school comp problems kek

  10. 2 months ago
    Anonymous

    question is inconsistent with itself. on the one hand, you're lead to believe that the ball will push the squares aside, thus causing them to have a velocity. but if you write down the generalized lagrangian, you'll see the squares have no horizontal component of potential energy, and thus the euler lagrange equations imply their velocities are constant. the initial condition is their velocity is zero, so the squares don't move and the ball never hits the ground. the only way out is to assume quasistatic motion of the squares such that the ball is moving at infinitesimal speeds at all times and that it hits the ground with speed zero, which is patently absurd. you need to include friction for this problem to work..

    • 2 months ago
      Anonymous

      Sure they will move because the ball pushes them apart it's just obvious. The problem is to ask what is the limit of the end velocity of the ball when the gap size approaches zero.

    • 2 months ago
      Anonymous

      Amateur.
      I reject the premise because it implies a sphere has equal mass to a cube. This is absurd because nobody could reproduce a cube to the tolerance of pi. Had they included some error metric, then I would be more inclined. As it stands, I don't waste time solving problems in universes that don't exist.

  11. 2 months ago
    Anonymous

    uuh... what's the inverse of the Jacobi amplitude function?

    • 2 months ago
      Anonymous

      >what's the inverse of the Jacobi amplitude function?
      The noitcnuf edutilpma ibocaJ

  12. 2 months ago
    Anonymous

    (2+sqrt(2))*sqrt(gr)

    • 2 months ago
      Anonymous

      shouldn't any value larger than 2sqrt(gr) not be possible as it implies the system gains energy?

      • 2 months ago
        Anonymous

        Did you account for dark energy?

      • 2 months ago
        Anonymous

        oops you're right of course, that was moron algebra.
        How about sqrt(8/3*gr)?

        • 2 months ago
          Anonymous

          your result is the same as this one

          I integrated the force along the path the ball takes which I then set be to kinetic energy to obtain this value. KE = mgr + mgr*(integral{0, pi/2} of cos(t)-cos^5(t)dt).

          but with cos^3 instead of cos^5 i think. I may have been applying the forces wrong.

          • 2 months ago
            Anonymous

            *almost the same

  13. 2 months ago
    Anonymous

    0. The ball will never hit the ground because they are all at rest and no force is acting on them. There's no gravity in this problem and everyone who assumed otherwise just got filtered.

    • 2 months ago
      Anonymous

      There is mass. This implies gravitational potential between all three. Sorry pseud, you failed to see that the system is in a vacuum and we are looking for the final orientation of the composite elements.
      The tendencies are the following:
      1. The sphere drives the cubes apart, to centralize the mass
      2. The cubes rotate against the sphere and stagger / lock into eachother
      3. The edges / corners never perfectly balanced against one another so the system remains dynamic, the ball consistently pushes into the cubes.
      4. This cyclic interaction builds rotation energy into the system which eventually balances against this.
      5. The chirality ultimately determines the final configuration, but the ball is going to lean into a cube going in the same direction of the rotating mass.
      6. Notice that in a zero friction state, the rotation is going to be very slow, otherwise displacement will kick the system into orbital mechanics and elastic collision. This change in behavior is controlled by proportion of m to r.

      • 2 months ago
        Anonymous

        >This implies gravitational potential between all three.
        okay. apply einstein's law of gravitation. there is no earth in the problem.

      • 2 months ago
        Anonymous

        >There is mass. This implies gravitational potential between all three.
        Mass and weight are completely separate things and mass does not imply gravity. Sorry pseud, you failed to see that the system is in a vacuum.

  14. 2 months ago
    Anonymous

    learn calculus they said

    then you end up like this. frick maths

  15. 2 months ago
    Anonymous

    It's just going to fall to the ground. Nothingburger.

  16. 2 months ago
    Anonymous

    All of those morons in the thread feverishly doing OP's homework. Kek, they'll never learn.

  17. 2 months ago
    Anonymous

    Insufficient information regarding to change in conditions from the OP question so here I am to give you three velocity when it hits the ground.

    If you want to be more technical then the first v = 0 isn't quite answering the "hits the ground" question because it will always be there. Frick this poorly worded stupid bait ass question.

    • 2 months ago
      Anonymous

      Schizo. Take the limit of the answer as epsilon goes to 0. There are standard conventions for succinctly communicating such problems to non-morons.

    • 2 months ago
      Anonymous

      i came to the same conclusion here

      question is inconsistent with itself. on the one hand, you're lead to believe that the ball will push the squares aside, thus causing them to have a velocity. but if you write down the generalized lagrangian, you'll see the squares have no horizontal component of potential energy, and thus the euler lagrange equations imply their velocities are constant. the initial condition is their velocity is zero, so the squares don't move and the ball never hits the ground. the only way out is to assume quasistatic motion of the squares such that the ball is moving at infinitesimal speeds at all times and that it hits the ground with speed zero, which is patently absurd. you need to include friction for this problem to work..

      [...]

      >It is completely obvious the ball will push them aside. What are you on?
      meds. the point of my post was that despite this "obviousness", the information given in the problem indicates the cubes have a constant zero velocity. don't believe me? write the lagrangian and solve the euler lagrange equations.

      • 2 months ago
        Anonymous

        Clearly impossible. If everything is stationary then the ball must be getting pushed up by the boxes with force G/2 from each. But since epsilon is positive, there must be a positive component of the force in the x direction, so the boxes accelerate.

        • 2 months ago
          Anonymous

          here let's try this instead. in reality, the sphere will transfer some of its stored gravitational potential energy to the kinetic energy of the cubes. this is done via friction. mathematically, friction couples the cubes to the sphere and gives a way for the cubes to acquire kinetic energy. now remove friction from the problem. what physical parameter couples the sphere to the cubes? nothing. so there's no way the cubes can acquire kinetic energy, which is why the euler-lagrange equations indicate the cubes have a constant zero-velocity. ironically enough by removing friction from the problem, you've forced the system into static equilibrium lol

          • 2 months ago
            Anonymous

            >this is done via friction.
            lol, moron alert. The force acts on the edges that touch the sphere's curvature in one point at a time. It has as little to do with friction as billiard balls transferring momentum to each other. The lack of friction ensures the edges move along the sphere till they stop touching it, and that the boxes move apart.

  18. 2 months ago
    Anonymous

    this gets really ugly when you try to account for the fact the ball makes the boxes rotate

    • 2 months ago
      Anonymous

      It doesn't though? If you work out a solution, you will see that at no point do they stop touching the ground with their whole bases.

  19. 2 months ago
    Anonymous

    all the potential energy of the ball is converted into kinetic energy of the cubes. now you can calculate how much would a force transfering the same energy to the ball slow it down

  20. 2 months ago
    Anonymous

    If I'm not mistaken, this mostly reduces to a pendulum starting out inverted. If [math]theta(t)[/math] is the angular distance between the bottom of the ball and the box corners (with [math]theta(-infty) = 0[/math]), we get
    [math]
    dot{theta}^2 = frac{2g}{r}(1 - cos theta).
    [/math]
    This holds as long as the velocity of a box (which is [math]v = partial_t(r sin theta)[/math]) is increasing: for [math]dot{v} > 0[/math], the ball is pushing the boxes, [math]dot{v} < 0[/math] would correspond to the ball pulling the boxes closer together. The separation point [math]theta_1[/math] is thus at [math]dot{v} = 0[/math], which you can solve for
    [math]
    cos theta_1 = frac{2}{3}.
    [/math]
    Then you calculate the energy of the ball at separation time (after which it is in free fall), and eventually get a final velocity of
    [math]
    frac{10}{3sqrt{3}} sqrt{gr}.
    [/math]

    Or something like that, I haven't really double-checked anything.

    • 2 months ago
      Anonymous

      Let h be the height of the ball center above the top of boxes.
      Let x be the distance of the boxes from the center line.
      From energy:
      2g(r-h) = x'^2 + h'^2
      -gh' = x'x'' + h'h''
      From geometric constraint:
      r^2 = x^2 + h^2.
      xx' + hh' = 0
      xx'' + hh'' + x'^2 + h'^2 = 0

      Using intuition from we want to find the state where x''=0.
      Let's impose this on the equations above.
      -gh' = h'h''
      hh'' + x'^2 + h'^2 = 0
      This gives x'^2 + h'^2 = gh
      Plugging this into the first equation gives:
      2(r-h)=h
      h=2r/3
      using this you can get h' = -sqrt(10gr/27)
      Do free fall from there to get v^2 = 100gr/27.

      The only thing I'm not sure of is if the ball can re-contact the boxes or not.

      • 2 months ago
        Anonymous

        I guess you can just check whether the didtance is ever < r after the free fall starts. If not, I guess the collisions become discrete events anyway, so you just have to enumerate them and compute the trajectories of stuff bouncing off.

      • 2 months ago
        Anonymous

        >recontact
        ooooh did not even think about this. Ok i did the calc. and no, it does not rehit, but it fricking narrowly misses by a little less than r/8.

        oops you're right of course, that was moron algebra.
        How about sqrt(8/3*gr)?

        I noticed that this is is 3 times the final speed of the blocks

        https://i.imgur.com/YwNBW2t.jpeg

        >u need to learn about equilibriums and stability
        And you, a massive moron on the left of the dunning-kruger curve, need to get a bachelor's in physics and a brain for basic reading comprehension to understand the OP is a shitty ass incomplete bait question with no conditions specified for epsilon.

        Gosh really IQfy is full of moronic STEM dropouts

        >massive moron on the left of the dk curve
        >need to get a bach in phys.
        this is really ironic, lol. Dude, at least 3 people itt already got the answer, looks like u got a skill issue

      • 2 months ago
        Anonymous

        Solving in terms of an arbitrary initial [math]epsilon[/math], I get [math]v = sqrt{2g[r+sqrt{4r^2-{epsilon^2}}(frac{23}{54}+frac{1}{54}frac{epsilon^2}{r^2}})][/math]. As [math]epsilonto0[/math], [math]vtosqrt{frac{100}{27}gr}approxsqrt{3.703gr}<sqrt{4gr}[/math], which agrees with

        [math] sqrt{4gr - gr cdot tfrac{2}{3} + gr cdot tfrac{10}{27} } [/math]

        The ball is slowed down during the first r/3 units of distance falling, then it falls at normal acceleration.
        If there is no blocks, then at most the speed would've been [math] sqrt{4gr} [/math], which yall prob already know.
        The first term in the sqrt is the normal speed, the second term is what was lost since it's not accelerating normally during the first r/3 units, and the last term is the gain of the weird acceleration during the first r/3 units.

        https://i.imgur.com/8k7qEL2.jpeg

        If I'm not mistaken, this mostly reduces to a pendulum starting out inverted. If [math]theta(t)[/math] is the angular distance between the bottom of the ball and the box corners (with [math]theta(-infty) = 0[/math]), we get
        [math]
        dot{theta}^2 = frac{2g}{r}(1 - cos theta).
        [/math]
        This holds as long as the velocity of a box (which is [math]v = partial_t(r sin theta)[/math]) is increasing: for [math]dot{v} > 0[/math], the ball is pushing the boxes, [math]dot{v} < 0[/math] would correspond to the ball pulling the boxes closer together. The separation point [math]theta_1[/math] is thus at [math]dot{v} = 0[/math], which you can solve for
        [math]
        cos theta_1 = frac{2}{3}.
        [/math]
        Then you calculate the energy of the ball at separation time (after which it is in free fall), and eventually get a final velocity of
        [math]
        frac{10}{3sqrt{3}} sqrt{gr}.
        [/math]

        Or something like that, I haven't really double-checked anything.

        . Also, as [math]epsilonto2r[/math], [math]vtosqrt{2gr}[/math], which is the velocity expected if the sphere starts free falling with its center at the same height as the top of the cubes. The units also work, so hopefully it's correct.

      • 2 months ago
        Anonymous

        Solving in terms of an arbitrary initial [math]epsilon[/math], I get [math]v = sqrt{2g[r+sqrt{4r^2-{epsilon^2}}(frac{23}{54}+frac{1}{54}frac{epsilon^2}{r^2}})][/math]. As [math]epsilonto0[/math], [math]vtosqrt{frac{100}{27}gr}approxsqrt{3.703gr}<sqrt{4gr}[/math], which agrees with [...] [...]. Also, as [math]epsilonto2r[/math], [math]vtosqrt{2gr}[/math], which is the velocity expected if the sphere starts free falling with its center at the same height as the top of the cubes. The units also work, so hopefully it's correct.

        Yeah these are correct.

        Given there have been 94(95 now) posts and not one agreed upon answer, plus OPs homosexual 'You should be able to solve this' addition means there is no real way to solve this since there is a variable or something else missing. Gj op at being a homosexual, as per usual.

        You are moronic if you think there isn't enough info. The full initial configuration is there. How could you think it's not enough?

      • 2 months ago
        Anonymous

        >Let h be the height of the ball center above the top of boxes.
        >h=2r/3
        i find this sus. i've drawn the image to scale, so that you all can visualize the claim being made for when the sphere detaches from the cubes.

        • 2 months ago
          Anonymous

          >i find this sus
          Then you aren't good at finding.

        • 2 months ago
          Anonymous

          The boxes accelerate much faster than the ball, because they have to move further sideways than the ball has to move down.
          In your image, the ball has only travelled by r/3, but the boxes have gone way more to the side than that. So the boxes must be moving faster.

          • 2 months ago
            Anonymous

            this can't be right. the constraint was supposed to be valid for h > 2r/3, meaning the ball keeps contact with the cubes until precisely that height. therefore the picture i drew must be valid (they're in contact). and that's why i think the proposed solution is wrong, and mine is right (and consistent).

            I gave this problem a closer look and triple checked my results. Nobody in the thread has my answer yet. We use Lagrangian mechanics. The relevant Lagrangian is
            [math]mathcal{L} = frac{1}{2}mleft(dot{x}^2 + dot{y}^2right) + mgy[/math]
            with a holonomic constraint of
            [math]f(x,y,lambda) = y - r - sqrt{r^2 - x^2} = 0[/math]
            here x is the horizontal distance traveled by one block, and y is the height of the sphere with respect to the cubes' centers of masses. The Euler-Lagrange equations read
            [math]begin{align}
            m&ddot{x} - xlambda = 0 \
            m&ddot{y} - mg - (y-r)lambda = 0
            end{align}
            [/math]
            Some ridiculous simplifications occur, giving
            [math]begin{align}
            ddot{x} &= xg/r \
            ddot{y} &= yg/r
            end{align}
            [/math]
            the implication is that the cubes acquire the same acceleration as the sphere, which should make sense given the symmetry of the masses. One could solve these directly, or invoke symmetry. Nevertheless you find the speed of each block and the sphere are all the same:
            [math] v_{text{block}} = v_{text{cube}} = sqrt{rg} [/math]
            note this is when the sphere loses contact with the cubes, which is at [math]y=r[/math]
            The rest free fall, or energy conservation yielding a final speed of the sphere before impact as
            [math]v=sqrt{3gr}[/math]
            note that energy is conserved at every stage here.

          • 2 months ago
            Anonymous

            >and mine is right
            What is the definition of the lagrangian?
            What did you write here

            I gave this problem a closer look and triple checked my results. Nobody in the thread has my answer yet. We use Lagrangian mechanics. The relevant Lagrangian is
            [math]mathcal{L} = frac{1}{2}mleft(dot{x}^2 + dot{y}^2right) + mgy[/math]
            with a holonomic constraint of
            [math]f(x,y,lambda) = y - r - sqrt{r^2 - x^2} = 0[/math]
            here x is the horizontal distance traveled by one block, and y is the height of the sphere with respect to the cubes' centers of masses. The Euler-Lagrange equations read
            [math]begin{align}
            m&ddot{x} - xlambda = 0 \
            m&ddot{y} - mg - (y-r)lambda = 0
            end{align}
            [/math]
            Some ridiculous simplifications occur, giving
            [math]begin{align}
            ddot{x} &= xg/r \
            ddot{y} &= yg/r
            end{align}
            [/math]
            the implication is that the cubes acquire the same acceleration as the sphere, which should make sense given the symmetry of the masses. One could solve these directly, or invoke symmetry. Nevertheless you find the speed of each block and the sphere are all the same:
            [math] v_{text{block}} = v_{text{cube}} = sqrt{rg} [/math]
            note this is when the sphere loses contact with the cubes, which is at [math]y=r[/math]
            The rest free fall, or energy conservation yielding a final speed of the sphere before impact as
            [math]v=sqrt{3gr}[/math]
            note that energy is conserved at every stage here.

            ?
            Cope and seethe

          • 2 months ago
            Anonymous

            change the potential energy to a negative sign then, i still get the same EOM keeping y'' the same sign as y.

            How is y'' positive? Isn't gravity pulling down?

            who said y'' is positive? it has the same sign as y (as it should).

  21. 2 months ago
    Anonymous

    [...]

    2sqrt(11gr/15) comes from integrating the forces acting on the ball up until it hits the ground, the integral is split into two parts one when it has contact with the two balls and one when it doesnt, the (r-length) integral when it doesnt have contact is just mgr. I then integrate across the r-length segment where it does have contact which I express in terms of the angle from the vertical.

    so the integral for that segment is int{0 to pi/2} of rcos(theta)f(theta) where f gives the force acting on the ball. Im not sure if my f is correct so feel free to criticize this reasoning, but heres what I did: the downwards force on the (both)blocks from the ball at angle theta should mgcos^2(theta) which is what the normal force resists so then we find the blocks are pushing mgcos^2(theta)cos^2(theta) = mgcos^4(theta) upwards on the ball in return. and then I just add the force of gravity mg acting on the block so for some given angle theta the force acting on the ball (might) be mg - mgcos^4(theta) so our integral is rcos(theta)(mg - mgcos^4(theta)). I then took the integral and set it to be the kinetic energy of the ball and just rearranged in terms of v.

    someone else in the thread got a similar result but with my cos^5 term changed to cos^3 so Im not sure.

    • 2 months ago
      Anonymous

      *two blocks

    • 2 months ago
      Anonymous

      Why mgcos^2(theta)?

  22. 2 months ago
    Anonymous

    0 m/s.

    I however cannot tell you its speed immediately before it hits the ground.

  23. 2 months ago
    Anonymous

    [...]

    The fall time tends to infinity -- the limit does not exist.

    • 2 months ago
      Anonymous

      Fall time yes, but velocity at impact does not.

  24. 2 months ago
    Anonymous

    [math] sqrt{4gr - gr cdot tfrac{2}{3} + gr cdot tfrac{10}{27} } [/math]

    The ball is slowed down during the first r/3 units of distance falling, then it falls at normal acceleration.
    If there is no blocks, then at most the speed would've been [math] sqrt{4gr} [/math], which yall prob already know.
    The first term in the sqrt is the normal speed, the second term is what was lost since it's not accelerating normally during the first r/3 units, and the last term is the gain of the weird acceleration during the first r/3 units.

    • 2 months ago
      Anonymous

      man, this reminds me of the stuff i had to do for college phys. Don't miss it, but sorta fun

  25. 2 months ago
    Anonymous

    I would just to assume that the ball moves slowly until its center reaches 2r, then it will be free falling for r.
    Converting its potential energy mgr into kinetic energy gives me ½mv2.
    Then solving for v gives me sqrt(2gr).

  26. 2 months ago
    Anonymous

    The boxes absolutely do start to rotate while getting pushed since the force is not applied to the center of mass of the boxes.

    • 2 months ago
      Anonymous

      That would be true if there was no ground on which they're sitting.

    • 2 months ago
      Anonymous

      That would be true if there was no ground on which they're sitting.

      To clarify. Obviously it is possible to apply some sort of force to the boxes that would make them spin even though there is a ground, but you can compute what happens and it doesn't happen here. IT is pretty intuitive, so you can just assume it and then check once you have the answer for the trajectories.

  27. 2 months ago
    Anonymous

    Given there have been 94(95 now) posts and not one agreed upon answer, plus OPs homosexual 'You should be able to solve this' addition means there is no real way to solve this since there is a variable or something else missing. Gj op at being a homosexual, as per usual.

  28. 2 months ago
    Anonymous

    >gets answer with forces
    >gets contradiction with lagrangian mechanics
    >lagrangian mechanics equivalent to force analysis
    see the problem?? write the lagrangian and you'll see the problem cannot be solved.

    • 2 months ago
      Anonymous

      Post your lagrangian, you ridiculous pseud. I bet it's wrong.

      • 2 months ago
        Anonymous

        it isn't a hard lagrangian to write. x represents the blocks' kinetic energy while y represents the sphere's kinetic energy.
        [math]mathcal{L} = frac{1}{2}mleft(dot{x}^2 + dot{y}^2right) - mgy [/math]

        • 2 months ago
          Anonymous

          yeah, thanks for this lagrangian for a free-falling sphere and blocks randomly moving in some place, now try the one for the actual problem. Maybe look up holonomic constraints.

        • 2 months ago
          Anonymous

          this guy

          yeah, thanks for this lagrangian for a free-falling sphere and blocks randomly moving in some place, now try the one for the actual problem. Maybe look up holonomic constraints.

          is right btw. This is what makes it a harder HS problem

          • 2 months ago
            Anonymous

            yeah, thanks for this lagrangian for a free-falling sphere and blocks randomly moving in some place, now try the one for the actual problem. Maybe look up holonomic constraints.

            yeah i forgot about that.
            [math]f(x,y,lambda)=y - r sqrt{r^2-x^2}[/math]
            which yields an ODE for the EOM of the cubes as
            [math]ddot{x} = xgsqrt{1-left(frac{x}{r}right)^2}[/math]
            find the velocity of the cubes when the sphere loses contact. apply conservation of energy to find velocity of sphere when it loses contact. then apply free fall.

          • 2 months ago
            Anonymous

            ur units are off, but close enough

          • 2 months ago
            Anonymous

            forgot a negative sign is all.
            [math]f(x,y,lambda)=y - r -sqrt{r^2-x^2} [/math]
            i mean it changes a lot. i probably fricked up a constant somewhere since i found energy isn't conserved while the ball is in contact with the cubes.

          • 2 months ago
            Anonymous

            >energy isn't conserved in this frctionless problem
            wrong

          • 2 months ago
            Anonymous

            esl? since my answer violated conservation of energy, i made an error. i'll look at it tomorrow with a fresher mind.

  29. 2 months ago
    Anonymous

    It's something between 2 and 300 m/s

  30. 2 months ago
    Anonymous

    Why is the ball going to hit the ground? Do the cubes vanish? What idiot write this question?

    • 2 months ago
      Anonymous

      Ignore the other answers in the thread
      Do the same problem but with an upside down equilateral triangle with a side length of r.

      Why would it hit the ground?

  31. 2 months ago
    Anonymous

    Why would it hit the ground?

  32. 2 months ago
    Anonymous

    i think it's
    [math]
    sqrt{ 4 g r}
    [/math]
    tho i may have messed up some vectors, or forgotten my vector calculus

  33. 2 months ago
    Anonymous

    When it hits the ground, it's speed will be reduced to 0, followed either by a bounce or a lack thereof depending on whether or not the collision is elastic or inelastic.

  34. 2 months ago
    Anonymous

    Well I wrote down the energy conservation equation. That's gotta count for something

  35. 2 months ago
    Anonymous

    The ball never hits the ground

  36. 2 months ago
    Anonymous

    I gave this problem a closer look and triple checked my results. Nobody in the thread has my answer yet. We use Lagrangian mechanics. The relevant Lagrangian is
    [math]mathcal{L} = frac{1}{2}mleft(dot{x}^2 + dot{y}^2right) + mgy[/math]
    with a holonomic constraint of
    [math]f(x,y,lambda) = y - r - sqrt{r^2 - x^2} = 0[/math]
    here x is the horizontal distance traveled by one block, and y is the height of the sphere with respect to the cubes' centers of masses. The Euler-Lagrange equations read
    [math]begin{align}
    m&ddot{x} - xlambda = 0 \
    m&ddot{y} - mg - (y-r)lambda = 0
    end{align}
    [/math]
    Some ridiculous simplifications occur, giving
    [math]begin{align}
    ddot{x} &= xg/r \
    ddot{y} &= yg/r
    end{align}
    [/math]
    the implication is that the cubes acquire the same acceleration as the sphere, which should make sense given the symmetry of the masses. One could solve these directly, or invoke symmetry. Nevertheless you find the speed of each block and the sphere are all the same:
    [math] v_{text{block}} = v_{text{cube}} = sqrt{rg} [/math]
    note this is when the sphere loses contact with the cubes, which is at [math]y=r[/math]
    The rest free fall, or energy conservation yielding a final speed of the sphere before impact as
    [math]v=sqrt{3gr}[/math]
    note that energy is conserved at every stage here.

    • 2 months ago
      Anonymous

      by v_block, of course i mean v_sphere.

    • 2 months ago
      Anonymous

      >y-r-y=0

      • 2 months ago
        Anonymous

        [math]yneqsqrt{r^2-x^2)[/math]
        to be more precise, the holonomic constraint is simply recasting
        [math]y = r + sqrt{r^2-x^2}[/math]
        notice that when x = 0, y = 2r as expected and when x = r, y = r again as expected.

    • 2 months ago
      Anonymous

      >note this is when the sphere loses contact with the cubes, which is at y=r
      This has been deboonked.

      • 2 months ago
        Anonymous

        only true if the cubes acquire speed at a faster rate than the sphere. which i don't think is true, or at least isn't reflected from the equations of motion.

        • 2 months ago
          Anonymous

          If they were still touching at y=r, the ball would have to be much faster than the box.
          The sides of the ball would be pretty much vertical at the connection point.

          • 2 months ago
            Anonymous

            idk if you're esl or not, because i cannot parse your sentence at all. the equations of motion don't reflect the velocities being disproportionately gained. furthermore i fail to see how energy could be conserved in your scenario.

          • 2 months ago
            Anonymous

            I am too lazy to check exactly what you wrote. But it is obvious that they lose contact way before y=r.

            Imagine epsilon before this moment. At that point the ball is basically vertical at the contact point, so it is moving down dows not increase its width at that height. However, the boxes are moving outwards, so moving further away from the ball. So if the ball is touching the boxes at some time eps before y=r, it can't also be touching them then.

            Basically what you can see is that at some point, even though the ball is getting faster, its rate of "widening" at the contact height start getting slower, while the boxes only start moving apart faster and faster. Therefore, at that moment they must not be in contact.

            Others' answers have determined the point when they lose contact. I'd be interested to see what your mistake is once you find it.

          • 2 months ago
            Anonymous

            the other results don't say what the speed of the blocks are, and i suspect they violate conservation of energy

          • 2 months ago
            Anonymous

            I computed stuff from scratch following the described approaches and got the same thing. Can you find the error there?

          • 2 months ago
            Anonymous

            >other results don't say what the speed of the blocks are
            Because it isn't needed.
            Use

            Let h be the height of the ball center above the top of boxes.
            Let x be the distance of the boxes from the center line.
            From energy:
            2g(r-h) = x'^2 + h'^2
            -gh' = x'x'' + h'h''
            From geometric constraint:
            r^2 = x^2 + h^2.
            xx' + hh' = 0
            xx'' + hh'' + x'^2 + h'^2 = 0

            Using intuition from we want to find the state where x''=0.
            Let's impose this on the equations above.
            -gh' = h'h''
            hh'' + x'^2 + h'^2 = 0
            This gives x'^2 + h'^2 = gh
            Plugging this into the first equation gives:
            2(r-h)=h
            h=2r/3
            using this you can get h' = -sqrt(10gr/27)
            Do free fall from there to get v^2 = 100gr/27.

            The only thing I'm not sure of is if the ball can re-contact the boxes or not.

            >h=2r/3
            >h' = -sqrt(10gr/27)
            and plug them in
            >x'^2 + h'^2 = gh
            x'^2 = 8gr/27

            When all of the potential (2mgr) is converted to kinetic, you get:
            (m/2)*(100gr/27 + 8gr/27) = 2mgr

          • 2 months ago
            Anonymous

            >dont say the speed of the blocks
            see

            >recontact
            ooooh did not even think about this. Ok i did the calc. and no, it does not rehit, but it fricking narrowly misses by a little less than r/8.

            [...]
            I noticed that this is is 3 times the final speed of the blocks

            [...]
            >massive moron on the left of the dk curve
            >need to get a bach in phys.
            this is really ironic, lol. Dude, at least 3 people itt already got the answer, looks like u got a skill issue

        • 2 months ago
          Anonymous

          It could happen. Like a rally car can bounce off from the road, and lagragian mechs goes unpredictable

        • 2 months ago
          Anonymous

          What if the cubes were much lighter than the ball?

          • 2 months ago
            Anonymous

            the two blocks combined have the same mass as the sphere so that's where symmetry comes from

    • 2 months ago
      Anonymous

      How is y'' positive? Isn't gravity pulling down?

      • 2 months ago
        Anonymous

        He triple checked it, stop questioning him, you esl or something?

  37. 2 months ago
    Anonymous

    Trick question. The ball is sitting on two cubes, not travelling toward the ground. The ball doesn't hit the ground. The ball can't have any velocity when it hits the ground because it's not going to hit the ground anyway. Why would it?

  38. 2 months ago
    Anonymous

    moron here. So obviously (r-h)^2 + x^2 = r^2. But why can't we differentiate this equation? If you do, you get something wrong that says when h = r, dx/dt = 0

    • 2 months ago
      Anonymous

      the block has already left the ball by h=r, so the eq aint valid there

      • 2 months ago
        Anonymous

        ok but what about h = r - q where q is very small

        • 2 months ago
          Anonymous

          read the thread to find the final speed for both objects, and when it detaches, and how to find it

          • 2 months ago
            Anonymous

            There is nothing in the thread that answers my question.

          • 2 months ago
            Anonymous

            You have no question

          • 2 months ago
            Anonymous

            You just don't understand my question. You're an idiot.

          • 2 months ago
            Anonymous

            If you are so smart, then explain what you meant

          • 2 months ago
            Anonymous

            I’m the guy who responded to his question. I already understand it. Stop responding to him, his comments are dumb and aren’t worth typing something up.

    • 2 months ago
      Anonymous

      ok but what about h = r - q where q is very small

      You can differentiate the equation, it just becomes invalid when the ball loses contact. This happens when the contact force is 0, which implies x’’ = 0. When you differentiate the equation, apply conservation of energy, and solve for x’, you get a function for x’ that increases from 0, reaches a maximum, and then decreases back to 0 at x = r. The maximum of x’ is where x’’=0 and the ball loses contact with the cubes. So the equation is invalid after the maximum, where x’’ < 0 and x’ is decreasing. The force on the cubes is only ever outwards. They accelerate until they lose contact with the ball, after which they continue on with constant velocity.

      • 2 months ago
        Anonymous

        Thanks, I geddit now

  39. 2 months ago
    Anonymous

    The ball is inert

  40. 2 months ago
    Anonymous

    gravity is splitting between the squares applying a force on the corners creating a rotational spin, but the square wont spin because it's on the ground so only the perpendicular forces will act
    this is just my guess and I don't know how to actually compute an exact analytical solution.

    • 2 months ago
      Anonymous

      Don't try to solve the DE. You just need to find what everything is when x'' is 0.

      Let h be the height of the ball center above the top of boxes.
      Let x be the distance of the boxes from the center line.
      From energy:
      2g(r-h) = x'^2 + h'^2
      -gh' = x'x'' + h'h''
      From geometric constraint:
      r^2 = x^2 + h^2.
      xx' + hh' = 0
      xx'' + hh'' + x'^2 + h'^2 = 0

      Using intuition from we want to find the state where x''=0.
      Let's impose this on the equations above.
      -gh' = h'h''
      hh'' + x'^2 + h'^2 = 0
      This gives x'^2 + h'^2 = gh
      Plugging this into the first equation gives:
      2(r-h)=h
      h=2r/3
      using this you can get h' = -sqrt(10gr/27)
      Do free fall from there to get v^2 = 100gr/27.

      The only thing I'm not sure of is if the ball can re-contact the boxes or not.

      • 2 months ago
        Anonymous

        I don't understand it's doesn't start that way though

        >applying a force on the corners creating a rotational spin
        That would only happen with friction.

        what does friction have to do with inertia?

    • 2 months ago
      Anonymous

      >applying a force on the corners creating a rotational spin
      That would only happen with friction.

      • 2 months ago
        Anonymous

        Torque is applied since the force between the ball and the boxes is not vertical. In fact it could be enough to tip the boxes outward if the ball is heavy enough (perhaps not with the masses specified in OP).

  41. 2 months ago
    Anonymous

    0 m/s

  42. 2 months ago
    Anonymous

    Wow thread is finally done with schizo posters.

  43. 2 months ago
    Anonymous

    by energy conservation,
    mg(3r)=0.5 * m * v^2
    v= sqrt(6*g*r)

  44. 2 months ago
    Anonymous

    Why do soientists think they can know everything and why do they try to brainwash everyone's children into their delusional cults?

  45. 2 months ago
    Anonymous

    Trick question. The ball will never hit the ground because none of the objects are under the influence of any forces.
    >source: gravity, electromagnetism, etc, etc, weren't mentioned in the question

  46. 2 months ago
    Anonymous

    g(4r^2)

    • 2 months ago
      Anonymous

      https://i.imgur.com/rZxmVnx.png

      This was a high school competition problem in my country. Can IQfy solve it?

      did i do it right?

    • 2 months ago
      Anonymous

      the inititial height is 2r because it stops when the bottom of the ball hits the ground

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