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Conservation of Energy Help!

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Assume there is no air friction. A ball of mass m = 3.39 kg is dropped from rest at height h = 1.51 m. It hits the ground, losing energy ΔE = 1/11 of its total energy in the collision. How high will the ball bounce upward before it comes momentarily to rest?

    2. Relevant equations

    U = mgh
    KE = ½mv²

    3. The attempt at a solution

    mgh = ½mv_f² [At the start before the first bounce]
    v_f = √(2gh) ≈ 5.44 m/s

    ½mv_f² = 10/11 * mgh_n where h_n is the new height
    h_n = 11/20 * v_f²/g ≈ 1.66 m, which is wrong.
     
  2. jcsd
  3. Oct 12, 2012 #2
    well so the moment it hits the ground it has some velocity which you determined

    right after it hits the ground it loses 1/11 of that energy

    what you have here: ½mv_f² = 10/11 * mgh_n where h_n is the new height

    is setting the energy right before impact equal to 10/11ths of some other amount of energy

    so if you move the 10/11 to the other side you get

    [itex]\frac{11}{10}\frac{1}{2}mv_{f}^{2} = mgh_{n}[/itex]

    what does that look like to you
     
  4. Oct 12, 2012 #3
    It's what I already did. Still wrong answer.
     
  5. Oct 12, 2012 #4
    I know, but think about what that equation means.

    [itex]\frac{11}{10}\frac{1}{2}mv_{f}^{2} = mgh_{n}[/itex] has kinetic energy on the left and potential energy on the right

    the new potential energy is what we want and it's on the right

    but on the left we have the kinetic energy right before impact *multiplied by 11/10*

    if you multiply something by 11/10 does it get bigger or smaller?
     
  6. Oct 12, 2012 #5
    First of all, you don't need all the intermediate steps. Why do you need to calculate the speed at impact?

    Second, the energy after collision is less than that before.
    You wrote your equation as it is larger!
    You have KE_after=(1-1/11) KE_before =10/11 KE_before.
    And then, the final potential energy (mgh) is equal to which one, the KE_before or KE_after?
     
  7. Oct 12, 2012 #6
    Bigger.
     
  8. Oct 12, 2012 #7
    SO this means you need...

    mgh = 11/10 * mgh_1! Oops!
     
  9. Oct 12, 2012 #8
    You do not need to even worry about the velocity that the ball has when it hits the ground. The only information that you need to consider is that E_initial=(11/10)E_final. So think of what the energy is due to when the ball is "momentarily at rest" and you should have your answer.
     
  10. Oct 12, 2012 #9
    so basically you want to take 10/11ths of the energy right before impact and use conservation of energy to determine how high that kinetic energy will take the ball
     
  11. Oct 12, 2012 #10
    Thanks! You are right this time.
     
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