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Need help with the impulse problem.

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A rubber ball of mass m = 0.225 kg is dropped from a height ho = 5.92 m, and it rebounds to a height hf = 4.58 m. Assume there is no air friction.

    Find J, the magnitude of the impulse exerted by the ground on the ball.

    2. Relevant equations

    I = Ft
    Energy conservations formula: mgh or ½mv²?

    3. The attempt at a solution

    Ugh... Don't think I have the right type of approach. -__-

    I first found that the energy lost is 2.96 J, which is right. I tried to use this form:

    mgh = ½mv² + ΔE

    But seems like there is no other way around.
     
  2. jcsd
  3. Oct 18, 2012 #2

    lewando

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    Gold Member

    Impulse is also defined as the change of momentum of a body.
     
  4. Oct 18, 2012 #3
    Unclear of what you are trying to say here. Yes, I do get that impulse is same as the change of momentum, but how can I approach this problem? Is it just by using the impulse and momentum form or solve that problem using energy laws and then, the momentum rules? Still confused. Not helpful.
     
  5. Oct 18, 2012 #4
    Hi,
    On the other hand I=Δv*m so you need to calculate the speeds of the ball after dropping and after bouncing using the equation mgh=½mv^2. When the ball is dropped its potential energy changes into kinetic energy and after the bounce vice versa.
    I hope this helps.
     
  6. Oct 18, 2012 #5
    Still stumped. Still not sure how to approach this problem. You might want to give the reasons for such explanation. -__-
     
  7. Oct 18, 2012 #6
    Impulse is change in momentum. You can figure out the velocities from immediately before and after the ball bounces using suvat equations; from these you know the momentum before and after the ball bounces.
    Difference is the impulse.
     
  8. Oct 18, 2012 #7
    I suppose this was your homework but...
    I=Δv*m=m(v_b-v_a), where v_b=velocity before the bounce
    v_a=velocity after the bounce
    h_before=5.92m
    m=0.225kg h_after=4.58m

    Let's solve for v
    mgh=½mv^2 |:m |*2
    2gh=v^2
    v=√(2gh)
    v_b=√(2gh_b)
    ... and v_a=√(2gh_a)

    I=m*((√(2gh_b))-(√(2gh_a)))...
     
  9. Oct 18, 2012 #8
    Still not working. Not sure how you prove this.
     
  10. Oct 18, 2012 #9
    Why not working?
     
  11. Oct 18, 2012 #10
    Even though your calculation is right, it seems that I obtain the wrong value. Yes, I did 0.292 J*s, but it's incorrect. Don't get quite why. Maybe, that is because you are determining the impulse of the ball.
     
  12. Oct 18, 2012 #11

    lewando

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    Gold Member

    Show your work.
     
  13. Oct 18, 2012 #12
    I believe that is...

    mgh_0 = ½mv0²
    v0 = √(2gh_0)

    ½mvf² = mgh_f
    vf = √(2gh_f)

    That is by some kind of energy conservation.

    Then, I guess that...

    P = m√(2gh_f) - m√(2gh_0)

    So there is negative impulse.
     
  14. Oct 18, 2012 #13
    for right before v = -10.772, for right after v = +9.47....

    change in momentum is (9.47)(.225) - (-10.772)(.225) - = 4.55 = impulse
     
  15. Oct 18, 2012 #14
    bdh, you gave away the answer. Anywho, thanks.
     
  16. Oct 18, 2012 #15
    I'm sorry you were on the right track and i wasn't patient enough but just try to remember momentum is a vector so the velocities will have a direction attached to them...
     
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