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A very very difficult Pendulum Problem from BR, for the masters here.

Discussion in 'MCAT Study Question Q&A' started by hellocubed, May 6, 2012.

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  1. hellocubed

    hellocubed

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    Basically, a block the mass 1 is traveling with velocity 1 towards the stationary spring system with mass 2. How much will the spring system compress?

    I originally thought the answer was C. (highlight)

    1/2mv^2=1/2Kx^2
    (mv^2)/K=x^2
    Then equation C

    But NO!

    The answer is D (highlight)

    For the initial KE, they set the formula to accompany the second mass.

    m1v1=(m1+m2)vf
    vf=m1v1/(M1+m2)

    1/2(m1+m2)v^2
    1/2(m1+m2)(m1v1/(M1+m2))^2
    1/2(m1v1)^2

    1/2(m1v1)^2=1/2Kx^2
    Then derive equation D.


    Why was is it not my first choice?
    I have no idea why they decided to incorporate m2 into the KE equation. The mass 1 is the only mass with initial velocity, and 1/2m1v^2 IS THE COMPLETE ENERGY being applied in the system. It does not need mass 2.

    I have no idea why they decided to solve it the way they did.


  2. EnginrTheFuture

    EnginrTheFuture

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    I'm sorry I wasn't able to read your entire explanation because I'm in a bit of a hurry but this is the ideal problem type for limiting cases. Deriving any of these answers is a bad idea unless you recognize very obvious formula patterns in all of them and you believe it can be done quickly.

    Here you clearly have inelastic collision formula and some sort of oscillatory formula.

    This is exactly what I did in my head... exactly :laugh::

    1) Can I derive a hybrid formula with conservation of energy, inelastic collision and oscillation formulas in less than 1 minute? Probably not, let me see if they made any stupid answers available for my taking.

    2) Answer A... if my spring constant (rigidity of the spring) goes significantly up should x increase or decrease? Definitely decrease. If my blocks are extremely energetic but my k states that it is essentially un-squishable, x is going to become tiny.

    3) Answer B.... nice, they are stupid for giving me two freebies. Same logic as above.

    At this point I am definitely looking at not deriving anything since I narrowed to two and simply have to find fault with one of the two left

    4) Answer C... So it looks like in answer C and D they are assuming an inelastic collision since they both include m_1 + m_2. No need to think about if elastic vs inelastic is what distinguishes them. C seems wrong right off the bat because it does not have the m_1/ (m_1 + m_2) that is typically seen in inelastic collisions, they omit the single m_1 term. Non-the-less what else? If I increase m_2 infinitely should the energy m_1 have before inelastically colliding move such a huge mass further than a lighter m_2? No... it will effect it less and less as I increase its mass. It must be in the denominator. C is out

    5) D matches all the tests above so it is not eliminated and I didn't mess up, nice :thumbup:
  3. chiddler

    chiddler

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    When the masses collide, they bunch together so the effective mass of the block attached to the spring is (m1+m2). With what you wrote

    1/2mv^2=1/2Kx^2
    (mv^2)/K=x^2

    you are only accounting for one mass.

    To account for the other, you have to find the velocity right after the two collide. That velocity will tell you how much compression there is in the spring. Thus the use of momentum to find that velocity. And then they did what you did above but with this new velocity.
  4. Morsetlis

    Morsetlis SGU MS-4

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    Energy is related to velocity, but velocity changes to conserve momentum. Momentum incorporates m2 when the two masses collide. The finite amount of energy you started out with now must move 2 objects instead of one, so m2 has to be in the equation somehow (because v is changed, and to figure out the new v, you need m1+m2).
  5. hellocubed

    hellocubed

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    I am confused because
    1/2m1v1^2 FULLY incorporates ALL of the energy added into the system.
    And mass 2 is accounted for in 1/2Kx^2
    ....................................................................................oh.................



    haha, I just realized that Mass2 is no where in the PE of spring equation.
    Alright that makes sense.


    So if it was a mgh problem, I could just write it:
    1/2m1v1^2=(m1+m2)gh

    Because the PE equation incorporates the 2 masses?
  6. Morsetlis

    Morsetlis SGU MS-4

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    The only mgh problem with momentum I can think of would be 1 cube sliding down a slope colliding with another cube in a perfectly inelastic collision.

    In that case, mgh = 0.5mv^2 = 0.5(m1+m2)(m1v1 / (m1+m2))

    Of course you would still have to find v1, which involves mgh and an angle.
  7. milski

    milski 1K member

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    If you do it in reverse, it becomes a modified bullet velocity determination.

    Shoot a bullet with mass m, velocity v at a heavy cube with mass M on a ramp. Measure the max vertical displacement of the cube -> KE of the cube -> initial velocity of the cube -> total momentum in the system -> initial velocity of the bullet.
  8. SaCkO

    SaCkO

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    yes the problem is you didn't acount for momentum.. Velocity that will be used in equation KE=Elastic potential energy is not going to be V1 directly, it is V2 which is the velocity of both masses after the collision since they are together.
    So You need to derive V2 interms of V1 which would be = m1v1/m1+m2
    then plug in the original equation and you will get the answer
    I like how Engineer used his theoretical approach, i went for the mathematical approach..
  9. hellocubed

    hellocubed

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    I am sure the problem was that I didn't account for the second mass... not momentum.

    The only thing that is important in this problem is Energy.

    1.) The first 1/2mv^2 account for the COMPLETE energy input. But...
    2.) The spring PE 1/2Kx^2 does not account for the second weight
    3.) that is why you have to account for the second weight in the first equation.


    You should not need momentum specifically. Momentum does not change the input of the total energy. You have to account for all of the masses however.

    So if the PE equation did account for mass (like (m1+m2)gh) you would not need to include momentum at all.
  10. SaCkO

    SaCkO

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    There is no gravitational potential energy here.. it isa change between elastic potential energy and kinetic energy

    i dont know why your are using the equation for GPE
  11. johnwandering

    johnwandering

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    no no no no no

    He is saying that if there was GPE, you would not need to use momentum equation.


    Because there is no GPE, and because 1/2Kx^2 doesn't include m2, you have to include m2 somewhere.
    You are not using momentum in 1/2(m1+m2)(m1v1/(M1+m2))^2 BECAUSE you "need momentum."

    You are ONLY using it because m2 is not elsewhere in the equation, so you are using momentum as a "shortcut" to edge in m2 into the equation somewhere.
  12. EnginrTheFuture

    EnginrTheFuture

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    When it comes down to it though... on the MCAT, you shouldn't be worried about anything but eliminating the wrong answers. Take m_2 and k to infinite separately, follow your intuitive nose and move on to the rocking the next PS problem :D

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