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Homework Help: Coservation of energy and impulse

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    A stream of elastic glass beads, each with a mass of 0.46 g, comes out of a horizontal tube at a rate of 108 per second. The beads fall a distance of 0.54 m to a balance pan and bounce back to their original height. How much mass must be placed in the other pan of the balance to keep the pointer at zero?

    2. Relevant equations

    3. The attempt at a solution
    What I attempted was solving for the velocity of the ball as it hits the pan, then I multiplied by the mass, but Im not sure of how to account the rate.
  2. jcsd
  3. May 5, 2010 #2
    Rate is necessary as force is momentum change per unit time. You need not think in terms of energy rather think of force acting on the pan as momenta lost by the beads PER UNIT TIME.
  4. May 5, 2010 #3
    The force on the pan will be 2*n*d/dt(m*v)
    n being the number of beads hitting the pan per unit time.
    v being the VERTICAL velocity of a bead as it hits the pan.
  5. May 5, 2010 #4
    n is the rate here so don't you think d/dt(mv) is misleading?
  6. May 5, 2010 #5
    Yes it is. It should be 2*n*m*v?
  7. May 5, 2010 #6
  8. May 5, 2010 #7
    Would I just use the conservation of energy to solve for the velocity then? Or momentum?
  9. May 5, 2010 #8
    Use conservation of energy to find the velocity. Then use the velocity in the momentum calculation.
  10. May 5, 2010 #9
    so I would use the following setup mgh=1/2mv^2 correct?
    I calculated my velocity with that setup
    ended up with 3.254....
    plugged that into the formula suggested which was 2(n)(m)(v):
    2(108)(0.46)(3.254)=323g, but that is giving me the incorrect answer?
    Where am I going wrong in my calculations?
    Last edited: May 5, 2010
  11. May 5, 2010 #10
    0.46g is in g; should be in kg.
  12. May 5, 2010 #11
    Ok, I switched out 0.46g into 0.00046kg, did the same calculation ended up with .010227kg, and my answer wants it in grams, so I multiplied it by 1000, for 10.227g, and its still incorrect. Im really confused now.
  13. May 5, 2010 #12
    Re: conservation of energy and impulse

    Ok, first of all, you should get a value of 0.323, but that is in Newtons.
    Find the mass that is needed to balance that force.
  14. May 5, 2010 #13
    Ok that makes sense, appreciate everyone's help.
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