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Bob throwing a rock

  1. Sep 17, 2008 #1
    SOVLED

    1. The problem statement, all variables and given/known data

    Bob, who has a mass of 75 kg, can throw a 600 g rock with a speed of 35 m/s. The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.6 m.

    (a) What constant force must Bob exert on the rock to throw it with this speed?
    229.7 N

    (b) If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?
    ___ m/s


    2. Relevant equations

    1. F=ma
    2. Vf=Vo+at
    3. X=Vot+.5a(t^2)
    4. Vf^2=(Vo)^2+2ax

    3. The attempt at a solution

    Well, I got the first part of the problem correct by using the fourth equation and solving for acceleration which equals 382.8. I then applied that to the first equation, because the mass (.6) was given, to find the force which equals 229.7. But, I can't seem to figure out how to start the second equation. I thought that the force would be the same for Bob because of Newton's third law and so I found the acceleration using equation 1, and then applied that to the fourth equation again to find the final velocity. That number kept coming out around 3.13 but that answer isn't correct. Any idea what I'm doing incorrectly and what I should be doing? Thank you in advance!
     
    Last edited: Sep 17, 2008
  2. jcsd
  3. Sep 17, 2008 #2
    The second part of the question is simpler than the first one. Just apply conservation of momentum.
    0.6*35=75*v
    v=0.28 m/sec
    right??
     
  4. Sep 17, 2008 #3
    Whew, it's taking me a while to rap my head around the conservation of momentum. Thank you very much. I don't think we've gone over this in class yet, so you've been a lifesaver.

    Thank you so much!
     
  5. Sep 17, 2008 #4
    It can also b don using Newton's law but conservation of momentum is the easy way round. You can learn it here:
    http://en.wikipedia.org/wiki/Momentum
     
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