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Homework Help: 2 boxes attached to sprint. concept questions.

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data

    A spring of negligible mass is compressed between two masses (M1 is less massive than M2) on a frictionless table with sloping ramps at each end. The masses are released simultaneously. Select the appropriate symbol for each statement: G (Greater than), L (Less than), or E (Equal to).

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

    2. Relevant equations

    Well, there are no numbers so I don't really need to use equations but:


    3. The attempt at a solution

    This is what I put, but at least one of these are wrong.

    A) Greater because the same force is being applied to a smaller mass.
    B) Less because the smaller mass would have a greater velocity
    C) duration of m1 is greater because it would travel farther.
    D) the KE is equal because they are pushed at the same force and even though one is going faster, it's a lesser mass, so it balances out.
    E) the force exerted is equal because they are being pushed by the same force.
    F) the momentum is the same because the force is the same. The lower mass balances with the higher velocity.

    Any help would be appreciated, thanks!
  2. jcsd
  3. Nov 23, 2008 #2


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    How do you reconcile C) and E)?
  4. Nov 23, 2008 #3

    For (D), we have (1) 1/2.m1.v1^2 = 1/2.m2.v2^2 ie kinetic energies equal

    For (F) we have momentum is equal - which must be the case because momentum is conserved - which is m1v1 = m2v2

    From (F) we can derive that if m1 is smaller than m2, v1 is bigger than v2. If we take (F) to be true, then it follows that KE of smaller mass is greater than that of the larger mass. So I think (D) is less than

    (C) is tricky, but I don't think it's right. Because there's less resistance from the smaller mass, the half of the spring pushing that mass would fully extend faster. The half of the spring pushing against the larger mass would extend slower, so I'm inclined to say Less for that.

    My 2 cents - I'm not positive.
  5. Nov 23, 2008 #4
    Well, I wasn't too sure on C because I don't fully understand when the force stops acting on it I guess. My reasoning was because M1 would be in motion longer than M2.

    For E, I've been assuming that since it's getting pushed by the same spring that the force exerted on M1 and M2 are equal.
  6. Nov 23, 2008 #5


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    So in E) the force is the same on both.

    But in C) it acts for a longer time on one and not the other?

    Doesn't that seem like a contradiction?
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