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The momentum and the kinetic energy

  1. Jun 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Multiple choice question:
    The momentum of a body has increased by 25%, then its kenitic energy will roughly increase by.....
    1. 25% 2. 5% 3. 38% 4. 56% 5. 65%

    2. Relevant equations
    Pl = mv (Pl is the momentum, m is mass and v is velocity)
    K.E = 0.5 mv^2

    3. The attempt at a solution
    I don't know how to start, so I will appreciate some hints.
     
    Last edited: Jun 9, 2017
  2. jcsd
  3. Jun 9, 2017 #2

    gneill

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    Start by completing the template... what are the relevant equations that pertain to the problem?
     
  4. Jun 9, 2017 #3

    BvU

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    Do you have expressions for momentum and kinetic energy ?
    List them under "2. relevant equations" !
     
  5. Jun 9, 2017 #4
    I added the equations, then what?
     
  6. Jun 9, 2017 #5

    BvU

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    Take a body; let its mass be ##m##. If it has a speed ##v##, the momentum will be ##mv## and the kinetic energy will be ##{1\over 2} mv^2##, right ?
    Now increase its momentum to ##p'## with ##p'= 1.25\, p##. It's still the same body, so the ##m## remains the same. What changes is ##v## -- let the new value be ##v'##. Now work out ## {1\over 2} mv'^{\, 2} / {1\over 2} mv^ 2 \ ##.
    That's all.
     
  7. Jun 9, 2017 #6
    OK, I got 0.64. That means the percentage will be 64%, and the correct answer is 65%. Right?
     
  8. Jun 9, 2017 #7

    BvU

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    No. Show your steps, not just he answer.

    And: it would be strange if the kinetic energy went from 100 % to 64 % (how do you get 0.64 ?)
     
  9. Jun 9, 2017 #8

    cnh1995

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    No. What is the new velocity if the original velocity was v?
     
  10. Jun 9, 2017 #9
    Ok, here is my steps, where is the mistake?
    ontIM.jpg
     
  11. Jun 9, 2017 #10

    cnh1995

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    Initially, the momentum was p and the velocity was v. Now the momentum is 1.25p, with mass unchanged. So what is the new velocity?
     
  12. Jun 9, 2017 #11
    The new velocity is
    v = 1.25p/ m
    isn't it?
     
  13. Jun 9, 2017 #12

    cnh1995

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    What is the new velocity in terms of v?
    Or use actual numbers.
    Initial momentum=1, m=1, v=1. Now the momentum is 1.25 with m still equal to 1. What is the new velocity now?
     
  14. Jun 9, 2017 #13
    the new velocity (v2) is (1.25v1) where v1 is the original velocity, right?
    Or if you want it just in numbers it will be 1.25 since m is still equal to 1, right?
     
  15. Jun 9, 2017 #14

    BvU

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    The original question asks for KE2 / KE1 x 100 % - 100 % (the increase in KE in percent)

    You had the right answer (1.56 * 100 % - 100 % ) but handled it the wrong way around !

    Case solved, onto the next exercise !
     
  16. Jun 9, 2017 #15

    cnh1995

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    Yes. Even if m is not 1, still v2=1.25v1.

    Now compute kinetic energy E2. You know E1=1.
     
  17. Jun 9, 2017 #16
    I don't understand this line, why did you interpret the question as this?
     
  18. Jun 9, 2017 #17

    cnh1995

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  19. Jun 9, 2017 #18
    @cnh1995
    E2 = 0.5 m (1.25v)^2 = 0.5 m (1.5625 v^2) = 1.5625 E1
    Right?
     
  20. Jun 9, 2017 #19

    cnh1995

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    Right.
     
  21. Jun 9, 2017 #20
    Ok, how could I get the percentage and choose the correct answer?
     
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