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Falling masses

  1. Apr 9, 2005 #1

    WY

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    I have been mulling over this question for ages and I just don't know where to start!
    Question: Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance s, its speed is v. If the total mass of the two blocks is M, what is the mass of the more massive block?
    Take free fall acceleration to be g

    I started off using conservation of potential energy and kinetic energy to try and relate the masses to get that and choosing the height from where the mass fell to be the relative 0-
    for the larger mass: mgs = 1/2mv^2
    but then I realised that I couldn't relate that to the other mass
    so then i decided to use the equations of motion to find how long the mass took to fall the distance of s getting
    sqrt(2*s*g) = t

    Now I am stumped as to what to do - or I could be completely off track - can someone help me out please?
     
  2. jcsd
  3. Apr 9, 2005 #2
    Couldn't you just apply the Newton's laws, or is it a requirement to use conservation of energy?
    You know [tex]v_0[/tex] (0), v, and s, so you can determine the acceleration a. Using Newton's laws and forces try to express a in terms of [tex]m_1[/tex], [tex]m_2[/tex], and [tex]g[/tex]. Also you could use [tex]m_2 = km_1[/tex], so that [tex]M = (k+1)m_1[/tex].
     
  4. Apr 9, 2005 #3

    SpaceTiger

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    Staff Emeritus
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    Remember that when you're conserving energy, you want to do it for the whole system, not just that one block. What can you say about the other block's potential and kinetic energy?
     
  5. Apr 9, 2005 #4

    Doc Al

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    Staff: Mentor

    Another hint: Since the two masses are connected by a rope:
    if one descends a distance s, what must happen to the other?
    if one moves with speed v, what must be the speed of the other?​
     
  6. Apr 9, 2005 #5

    WY

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    O I feel like such an idiot... I missed the most obvious things!!!
    Thanks for the hints guys!
     
  7. Apr 9, 2005 #6
    I've been trying to use the hints you gave out but i'm just not getting anywhere. I don't know how to go about dealing with the conservation of energy in the SYSTEM. And the 2nd part of the question is to find the mass of the lighter block and I'm pretty sure I can't do that without first understanding how to find the mass of the larger block.
     
  8. Apr 10, 2005 #7

    Doc Al

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    The energy of the system is just the sum of the energies of each mass. Since the two masses start from rest, take the initial energy of the system to be zero. As one mass falls, the other rises, and both speed up. Conservation of energy tells us that [itex]\Delta \mbox{KE} = - \Delta \mbox{PE}[/itex] for the system.

    Try it.
     
  9. Apr 10, 2005 #8
    I tried it using that formula but I'm pretty sure I stuffed it up somewhere.
    So I got: 1/2*m_1*v^2 + 1/2*m_2*v^2 = -m_1*g*h - m_2*g*h
    Now I took the height of m_1 (the larger mass) to now be -s (since it moves down) and the height of m_2 (the smaller mass) to be s (since it moves up). I'm not sure if this was the right thing to do though.

    Anyway I simplified all of that and substituted M-m_1 for m_2 and ended up with the answer: m_1 = (v^2M+2Mgs)/4gs have i stuffed up somewhere or is this actually right?
     
  10. Apr 10, 2005 #9

    Doc Al

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    Looks good to me.
     
  11. Apr 10, 2005 #10
    Excellent. Thanks for clearing up my confusion :)
     
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