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Spring Propels Two Blocks Up Two Ramps

  1. Nov 17, 2011 #1
    The problem statement, all variables and given/known data

    A spring of negligible mass is compressed between two masses on a frictionless table with sloping ramps at each end. The masses are released simultaneously. The masses have the same volume, but the density of M1 is greater than that of M2.

    G-Greater than, L-Less than, E-Equal to.

    A) The duration of the force exerted by the spring on M2 is ... the time the force acts on M1.
    B) The momentum of M2 is ... the momentum of M1 once they both lose contact with the spring.
    C) The final height up the ramp reached by M1 is ... the height reached by M2.
    D) The force exerted by the spring on M2 is ... the force it exerts on M1.
    E) The speed of M2 is ... the speed of M1 once they both lose contact with the spring.
    F) The kinetic energy of M2 is ... the kinetic energy of M1 once they both lose contact with the spring.

    The picture is attached below.

    The attempt at a solution

    This is how I attempted to answer this question. There are flaws in my logic, and I don't know where or what they are.

    A) L - M1 has the same volume but is less dense, therefore it is lighter. It would take less force for the spring to return to equilibrium send end it out at a higher velocity.
    B) E - Momentum is mv, right? So despite their difference in weight, shouldn't that cancel each other out? M2 would have a larger weight and less velocity, M1 would have a small weight and a heigher velocity?
    C) G - M1 would be moving faster, therefor it would go further up the ramp.
    D) G - M2 is larger, therefore it would need more force exerted against it to move.
    E) L - I think I've beaten my thought on this into the ground. M1 is lighter, so shouldn't it be moving faster?
    F) L - Kinetic energy is (1/2)mv^2. M1 may be lighter, but it would be moving at a higher velocity.
     

    Attached Files:

  2. jcsd
  3. Nov 17, 2011 #2
    Imagine you are in position to do a push up. We will call equilibrium the point at which your arms are completely stretched so that your chest is at its highest point from the ground. Now you let yourself down until your chest almost touches the ground and then come back up. Because of Newton's 3rd law we know that the force of you pushing on the ground is exactly equal in magnitude and opposite in direction to the force of the earth pushing on you. During your push up, was the duration of the force you exerted on the earth any shorter than the duration of the force exerted by the earth on you? Note that the earth is many orders of magnitude more massive than you.

    Also note that M1 is more massive than M2, i think you have it backwards.

    Yes, this fact comes from the conservation of momentum.

    Consider the total energy of each mass to relate the height each mass travels to their velocities.

    Going back to the push up problem... which force is greater, that exerted by you on the earth? Or the force of the earth pushing you up?

    I think you have the two mixed up. The density of M1 is greater than the density of M2, while they both have the same volume. Therefore, M1 is more massive than M2. So then which has the larger speed?

    Use conservation laws to solve these types of problems. Conservation of Momentum and Conservation of Energy. Have you learned these?
     
  4. Nov 17, 2011 #3
    You are right, I read the question incorrectly. Which is a little ridiculous considering I retyped it all out.

    Taking that into account, this is what I came up with, however this is also incorrect.
    A) E
    B) E
    C) L - Potential energy is mgh, and would I be correct in assuming PE for both should be equal? So, because M1 is denser, therefore heavier, it wouldn't go as far up the ramp.
    D) E
    E) G - M1 is heavier and should therefore be moving at a lower velocity.
    F) Okay, so, for momentum to me conserved: m1v1i + m2v2i = m1v1f + m2v2f, and for kinetic energy to be conserved, would (1/2)m1v^2i + (1/2)m2v^2i = (1/2)m1v^2i + (1/2)m2v^2i. I don't really know where this leaves me about the answer, however.

    My previous attempts for this question were the first series of letters I posted, and EELEGE. Both were incorrect.
     
  5. Nov 17, 2011 #4

    ehild

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    Homework Helper
    Gold Member

    Think:
    KE=1/2 mv^2=1/2 mv * v = p^2/(2m ).

    The momenta are equal, so the KE of the object with lower mass is higher.

    ehild
     
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