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Homework Help: Trebuchet help!

  1. Dec 10, 2009 #1
    Trebuchet help: Calculating velocity

    1. The problem statement, all variables and given/known data
    I'm new to this site, so bear with me!
    OK, so I'm doing a trebuchet for my physics project. I have done three trials with counterweights of different masses. I have all of the distances that a golf ball of mass=0.045kg has traveled. I also have the masses of the counterweights.
    Just as a starting point for the first trial:
    mass of the counterweight=10.937
    h=0.631m

    The average distance for this trial was 34.33m.



    2. Relevant equations
    PEg=mgh
    KE=(1/2)mv2


    3. The attempt at a solution
    I've done the calculations; however, my answers were nowhere near what the velocity should be. For one trial, the mass of the counterweight was 10.937kg. I did the calculations and the velocity came out to be 3.53m/s which seemed a little slow.

    Obviously, I'm doing something wrong. I wonder if it is because the mass of the golf ball needs to be taken into account somehow??? Anyways, thanks in advance!

    I've used the example from https://www.physicsforums.com/showthread.php?t=250335&highlight=trebuchet and I still can't get it. I don't even understand how the answer was calculated in that example.
     
    Last edited: Dec 10, 2009
  2. jcsd
  3. Dec 10, 2009 #2
    Well, I think I might have it! I used that example in the post I linked to in the previous post, and I just changed the values to fit my case. I did the other example to get the right answer that was provided. So, I'm assuming that my answer is correct

    Here are the givens:
    m1=10.937
    m2=.045
    h1=.076
    h2=.838

    I used:
    m1gh1-m2gh2=(1/2)m1v12+(1/2)m2v22

    And I used the proportion:
    v1 .076
    -- = --
    v2 .838

    And just solved for v2.

    My final answer was 10.742 m/s which sounds correct, but I'd like to see if I made any mistakes.
     
  4. Dec 10, 2009 #3
    That proportion I used, v1/v2= 0.076/0.838, what exactly is that?
     
  5. Dec 11, 2009 #4
    OK, I think that solved for velocity correctly. Using the calculated velocity and the average distance traveled of the golf balls in a trial, I calculated a time of about 3.3 seconds. When I launched golf balls and manually timed it, the results were close enough.
     
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