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Force and recoil problem.

  1. Oct 13, 2013 #1

    vac

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    1. The problem statement, all variables and given/known data
    Bob, who has a mass of 75 kg, can throw a 500 g rock with a speed of 30 m/s. The distance through which his hand moves as he accelerates the rock from rest until he releases it is 1.0 m.

    2. Relevant equations
    a. What constant force must Bob exert on the rock to throw it with this speed?
    b. If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?

    3. The attempt at a solution

    Known
    vi = 0
    vf = 30 m/s
    delta x = 1 m

    [itex]v^2 = vi^2 + 2ax = a = 450 m/s^2[/itex]
    [itex]F_{net} = m_{total} * a = (75 kg + 0.500 kg)*450 m/s^2 = 33975 N[/itex]

    recoil have no idea.
     
  2. jcsd
  3. Oct 13, 2013 #2
    You'll want to check that again!!!...its not correct. Acceleration is right.

    For recoil conserve momentum.
     
  4. Oct 13, 2013 #3

    vac

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    I know it is not correct, can you please show me the mistake I have done?
     
  5. Oct 13, 2013 #4
    you have taken mass, incorrectly. What is being accelerated????
     
  6. Oct 13, 2013 #5

    vac

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    For the rock is this 0.500kg(450m/s^2) = 225 N
    But the other force I can't get correctly.
     
  7. Oct 13, 2013 #6
    You don't have to!!!...I hope you have studied momentum conservation. because the force applied is internal to the system (Bob+rock), their momentum is conserved before and after the releasing the rock. I hope this makes it clear!
     
  8. Oct 13, 2013 #7

    vac

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    I hate to disappoint you but I actually have to.
    The textbook answer is: 2.3x10^2 N = 230 N
    I just don't know how to get that.

    Now I am guessing that they round it up just to through me in a loop.
     
  9. Oct 13, 2013 #8
    for part (a), you got the right answer....its just rounded off ##225≈230##, because you only have two significant figures.

    part (b), you need to calculate velocity not force, velocity can be found out easily be applying conservation of momentum

    if its something else, I really don't know what you are talking about!!!:rolleyes:

    Edit: I guess you meant to find some other force which was applying 5 Newtons. You'll have learn about this, in almost every other scientific/science calculation there is rounding off. Read about significant figures and scientific notation to learn more.
     
  10. Oct 13, 2013 #9

    vac

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    Yes I have, I actually took three semester chemistry and significant figures date back to algebra. How do you apply this conservation of momentum? I know that I studied it before but I just don't remember how?
     
  11. Oct 14, 2013 #10
    The formula is,

    ##m_{1}.v_{1i}+m_{2}.v_{2i}=m_{1}.v_{1f}+m_{2}.v_{2f}##

    You just have to fill in the data, since Bob is on frictionless ice, no horizontal external force would act on system (Bob+rock). ask yourself, what was the velocity of rock and Bob before the rock was thrown??
     
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