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Three equations of motion homework

  1. Sep 18, 2015 #1
    1. The problem statement, all variables and given/known data
    ck_55fc3aad67048.png

    2. Relevant equations
    ## 2as = v^2-u^2 ##
    ## v = 0 .... \text{at maximum height}##
    ## \therefore s = \frac{-(u)^2}{2a} \\ a = -g \\ \therefore s = \frac{-(u)^2}{-2g} = \frac{(u)^2}{2g} ##
    Please explain using only the Three equations of motion and not anything related to Energy or other formulae.
    3. The attempt at a solution
    Is my answer correct:-
    ck_55fc3ab91ae5e.png
     
  2. jcsd
  3. Sep 18, 2015 #2

    Doc Al

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    You are using the kinematic equations for uniformly accelerated motion. Do you think they would apply?
     
  4. Sep 18, 2015 #3
    Yeah. Because they are uniformly accelerated to ## g ##
     
  5. Sep 18, 2015 #4

    Doc Al

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    The acceleration is ## g ## when the balls are close to the earth's surface. But those speeds are pretty high.
     
  6. Sep 18, 2015 #5
    Ohh then please tell me how to find the Heights.
     
  7. Sep 18, 2015 #6

    Doc Al

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    Use conservation of energy.
     
  8. Sep 18, 2015 #7
    We haven't learnt that. Can you please tell me the formula.
    Some other sites states the following:-
    ##KE + PE \text{ at the surface} = PE \text{ at the maximum height}##
    Can you please explain in simpler words because I am very new to this.
     
  9. Sep 18, 2015 #8

    Doc Al

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    That's exactly the equation you need. Of course you need to learn how to express the gravitational PE.

    If you haven't studied this yet, where did you get the problem?
     
  10. Sep 18, 2015 #9
    We havent learnt it completely yet. But this question was in our chapter called Gravitation.
     
  11. Sep 18, 2015 #10

    Doc Al

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    Good. So it should cover how you calculate the gravitational PE at some distance from a planet.
     
  12. Sep 18, 2015 #11
    No we learnt only these equations:-
    ##F = \frac{GMm}{R^2}\\ g = \frac{GM}{R^2}##
    Do you mean this:-
    ## PE = mgh##
    ## PE = \frac{GMmh}{R^2}##
     
  13. Sep 18, 2015 #12

    Doc Al

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  14. Sep 18, 2015 #13
  15. Sep 18, 2015 #14
  16. Sep 18, 2015 #15

    Doc Al

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    How did you determine that?
     
  17. Sep 18, 2015 #16
    Using
    ## KE + PE \text{ at surface} = PE \text{ at maximum height} ##
    ## \frac{1}{2} m *(2\sqrt{\frac{gR}{3}})^2 - \frac{GMm}{R}= \frac{-GMm}{R+h}\\
    h = 2 R\\ \text{For the other ball:-} \\ h' = \frac{R}{2}\\ \frac{h}{h'} = \fbox{4:1}\\ ##
     
  18. Sep 19, 2015 #17

    haruspex

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    Looks right to me. Well done.
     
  19. Sep 19, 2015 #18

    Doc Al

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    Good work! (Just wanted to check. :smile:)
     
  20. Sep 19, 2015 #19
    Have a doubt. How exactly do we know when to use Gravitational Potential Energy. Please explain in the simplest words. Thanking you in advance.
     
  21. Sep 19, 2015 #20

    Doc Al

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    That's difficult to answer. Just stating the obvious: When gravity is involved and a massive object changes position, you may be able to make use of gravitational PE. But it depends on the problem and what you need to find.
     
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