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Work and energy

  1. May 29, 2016 #1
    1. The problem statement, all variables and given/known data
    two masses of 1g and 4g are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is :
    a: 4:1
    b root2: 1
    c; 1:2
    d; 1:16

    2. Relevant equations
    KE=1/2MV^2
    work=FD

    3. The attempt at a solution
    no idea what to do.. give me a hint please
     
    Last edited: May 29, 2016
  2. jcsd
  3. May 29, 2016 #2

    ProfuselyQuarky

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    There is yet another formula that you have to use along with the one you've posted. Which do you think it is?
     
  4. May 29, 2016 #3
    pe=ke?
    pe=mgh?
     
  5. May 29, 2016 #4
    w= fd?
     
  6. May 29, 2016 #5

    ProfuselyQuarky

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    None of those, I'm afraid.
    The problem has nothing to do with work.
     
  7. May 29, 2016 #6

    ProfuselyQuarky

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    You're looking for the ratio of the magnitudes of the masses linear momentum.
     
  8. May 29, 2016 #7
    yes .... but what equation or formula i should use....?
     
  9. May 29, 2016 #8

    ProfuselyQuarky

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    What is the typical equation for momentum?
     
  10. May 29, 2016 #9
    momentum=mass*velocity
     
  11. May 29, 2016 #10

    ProfuselyQuarky

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    Yep! That’s the one. Now, we are already given the mass of both objects, so we have to “equal” the energies of them, using the equation in you OP. How do you think you can do that?
     
  12. May 29, 2016 #11
    but we dont know velocity?

    p=mv
    p=1*v

    what now? :--
     
  13. May 29, 2016 #12

    ProfuselyQuarky

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    We've got:

    ##E_1=E_2##
    ##\frac {1}{2}(1)v^2=\frac {1}{2}(4)v^2##

    What does canceling the 1/2 give you?
     
  14. May 29, 2016 #13
    for E1 it is 1/2 or 0.5
    and for E2 it is 2 ?

    your way of make me understand is awsome!!
     
  15. May 29, 2016 #14

    ProfuselyQuarky

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    Well, we’re not dividing both sides by two, or else, we’d have to divide ##v^{2}_{1}## and ##v^{2}_{2}## by two, as well. We don't want to do that. We’re just canceling out the ½ because it’s on both sides.

    That said, what are we left with?
     
  16. May 29, 2016 #15
    1/2=2

    that means it is close to option c 1:2 ? and it is the right answer in the book.
    thank you..
    we are left with 1 and 2 !!!!! great am i right because 1/2 is canceled
     
  17. May 29, 2016 #16

    ProfuselyQuarky

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    Indeed, the answer is 1:2

    Without ½ we get that ##(1)v^{2}_{1}=4v^{2}_{2}##. Square both sides and we derive ##v_1=2v_2##. Remember how ##p=mv## give us the momentum? We can now plug in the values.

    ##p_1=(1)(v_1)##
    ##p_2=(2)(v_1)##

    We’re looking for ratios, thus:

    ##\frac{p_1}{p_2}=\frac{(1)(v_1)}{(2)(v_1)}##

    Canceling out the ##v_1## gives us 1:2
     
  18. May 29, 2016 #17
    ohh! great job... thanks
     
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