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Homework Help: Linear momentum (2 blocks and spring question)

  1. Nov 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Here is the problem:

    "Two blocks of masses 1m and 3m are placed on a frictionless surface. A light spring is attached to the massive block, and the blocks are pushed together with the spring between them. A string holding the 2 blocks together is cutt, sending the 3m black to the right with speed of 2 m/s.

    Find the systems original elastic potential Energy, taking m = .350kg ?"

    I cant seem to find out what forumla to use to get the original elastic potential energy... ?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 9, 2011 #2
    What is the original linear momentum of the WHOLE system?
  4. Nov 9, 2011 #3
    Wouldnt it be 0 ?
  5. Nov 9, 2011 #4

    Hence what is the FINAL linear momentum of the WHOLE system?
  6. Nov 9, 2011 #5
    FINAL linear momentum of the WHOLE system? - I am not sure :(
  7. Nov 9, 2011 #6
    What is the NET force on the WHOLE system?
  8. Nov 9, 2011 #7
    Net force on the whole system? You mean as in F = MA ?

    How would I find the net force if I dont know the acceleration of both objects?
  9. Nov 9, 2011 #8
    The spring causes block 1 to push on block 2 and block 2 to push on block 1. Hence by Newton's ......law the NET force on the WHOLE system is .....
  10. Nov 9, 2011 #9
    Newtons 3rd law correct?

    So, the sum of all forces acting on the system will be the Net force.

    m1 + m2 = 3.350kg - Not sure what to do with that to find net force of system or original elastic potential energy.
  11. Nov 9, 2011 #10
    We are not talking about the total mass but we are talking about the net force on the whole system. By Newton's 3rd law (as you said) the net force is zero because the two blocks push on each other with forces equal in magnitude but opposite in direction.

    Since the net force is zero, there is no change in momentum for the WHOLE system. Hence the final momentum is equal to the initial i.e. zero.
  12. Nov 9, 2011 #11
    Ohh ok, I see.

    But as the question states, how do I figure out the systems original Elastic Potential Energy ?

    So, the net force is 0, and the linear momentum is 0, because they are at rest. But the systems original Elastic Potential Energy ?
  13. Nov 9, 2011 #12
    Thinking in Physics is best done... step by step.

    Net force on whole system is zero.
    Therefore total momentum must not change.
    hHence final momentum is equal to the initial momentum i.e. zero.
    Hence the momentum of 1m must be equal and opposite to that of the 3m so that the final momentum will be zero.
    This will give you the KE of both masses.
    Add the KEs and you get the original elastic PE.
  14. Nov 9, 2011 #13
    Ok... One thing im not getting. So, the momentum will be zero, because initially when the system is at rest the momentum is zero.

    However to find the KE of both blocks, would I not go?

    1/2mv^2 for each of the blocks?

    But, I only know the velocity of one?

    Here is what I did:

    1/2(.350)(2.0)^2 + 1/2(3)(2.0)^2 - BUt I got 6.7, and the correct answer is 8.40
  15. Nov 9, 2011 #14
    What is the momentum of the 3m after spring is released?
  16. Nov 9, 2011 #15
    momentum would be mass*A so it would be 6m/s correct?
  17. Nov 9, 2011 #16
    momentum of the 3m = mass x velocity
    = 3x0.350x2
    = ...kgm/s
  18. Nov 9, 2011 #17
    3x0.350x2 - why x2 at the end? Sorry if thats a dumb question.

    so 3 as in the mass of mass2 x 0.350 as in the mass of mass1 x 2 as in the velocity?

    why are you multiplying 3 and .350 and not adding them up, and multiplying that by the velocity, which is 2?
  19. Nov 9, 2011 #18
    momentum = mass x velocity

    mass of 3m = 3 x 0.350kg
    velocity of 3m = 2m/s

    so momentum = 3 x 0.350 x 2
  20. Nov 9, 2011 #19
    oh ok, so the reason 3m has a mass of 3 + .350 is because they are connected together correct?. Because 1m has a mass of .350 and 3m has a mass of 3

    Im not getting this. The reason we are figuring out the momentum of 3m is why?

    This gives us 2.1 N

  21. Nov 9, 2011 #20
    I think that 3m means that its mass is (3 times 0.350kg) and not (3 + 0.350)kg.
    (3 + 0.350)kg would have been written as (3 + m) and not 3m.

    Now since we found the final momentum of 3m, which is 2.1kgm/s (as you said), then the final momentum of m will be equal in magnitude to 2.1kgm/s but opposite in direction so that the final total momentum will be zero.(remember that momentum is a vector)

    From this we can find the velocity of mass m.
  22. Nov 9, 2011 #21
    So the momentum of m would be -2.1 J ?

    How can I find the velocity with that?
  23. Nov 9, 2011 #22
    momentum = mass x velocity
    - 2.1 = 0.350 x v
    v = ?
  24. Nov 9, 2011 #23
    -6 correct?
  25. Nov 9, 2011 #24
    C O R R E C T!

    now you can find the ke of both masses and add them up to give the original elastic PE of spring since energy is being assumed to be conserved.
  26. Nov 9, 2011 #25
    Ok, so the KE of m block is: 1/2(.350)(-6)^2 = 6.3

    KE of 3m block is: 1/2(3.350)(2)^2 = 6.7

    This cannot be correct?
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