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Elastic Collision between projectile and a pendulum, what's the initial speed?

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A 29.0 g ball is fired horizontally with initial speed v0 toward a 100 g ball that is hanging motionless from a 1.10 m-long string. The balls undergo a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle θmax = 50.0°
    see image for question

    2. Relevant equations
    Ki + Ui = Kf + Uf
    vp = √[2*g*L*(1 - cosθ)]
    mb*v0 = mp*vp + mb*vb



    3. The attempt at a solution
    I suspect you have to use the mass ratio and it's relation to the ratio of momentum or kinetic energy, but I have been unsuccessful in getting the correct answer. The tried answers are availible along with the question in the image.
     

    Attached Files:

  2. jcsd
  3. Dec 1, 2009 #2

    Delphi51

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    Homework Helper

    I got a different answer. But I can't just tell you - that would be spoiling your learning. If you show your calculation, I can suggest improvements.
     
  4. Dec 1, 2009 #3
    I think the problem with my calculations was that I used a mass ratio which I think only applies to inelastic but here's what I did
    0.5*mp*vp² = mp*g*L*(1 - cosθ)

    vp = √[2*g*L*(1 - cosθ)]

    0.5*mb*v0² = 0.5*mb*vb² + 0.5*mp*vp²
    mb*v0² = mb*vb² + mp*vp²
    mb*vb² = mb*v0² - mp*vp²
    vb = √[v0² - (mp/mb)*vp²]

    v0 = √[v0² - (mp/mb)*vp²] + (mb/mp)*vp


    mb = 27 g
    mp = 100 g
    vp = √[2*g*L*(1 - cosθ)]; L = 1.1 m, θ = 50º

    vp = 2.775 m/s

    mp/mb = 3.70

    v0 = √[v0² - (3.7)*7.701] + (3.7)*2.775
    v0 = √[v0² - 28.5] + 10.27
    v0 - 10.27 = √[v0² - 28.5]
    v0² - 20.54*v0 + 105.4 = v0² - 28.5
    - 20.54*v0 + 105.4 = - 28.5

    v0 = 6.52 m/s

    I'm pretty sure the vp(final velocity of the pendulum is right)
     
  5. Dec 1, 2009 #4

    ideasrule

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    Homework Helper

    This came from the conservation of momentum equation, but you made a mistake here. If you write out the full momentum equation, you'll see what it should be.
     
  6. Dec 1, 2009 #5
    I got it!
    vp=[2mb/(mb+mp)]*vb
    0.5*mvbf^2+mghf=0.5*mvbi^2+mghi
    gL(1-costheta)=).5(Vo^2)(3364/16641)
    Vo=6.17m/s

    although when I actually did it there were a few more intermeadiate steps but I didn't feel it necessary to type out the algebra since square roots and fractions are hard to type.
     
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