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Momentum and impulse of jump problem

  1. Dec 15, 2016 #1
    1. The problem statement, all variables and given/known data
    After winning a prize on a game show, a 72 kg contestant jumps for joy.
    A.
    If the jump results in an upward speed of 2.1 m/s, what is the impulse experienced by the contestant?
    b. 0.0
    c. 75.6 kg.m/s
    d. 151.2 kg.m/s
    e. 226.8 kg.m/s
    B.
    Before the jump, the floor exerts an upward force of magnitude mg on the
    contestant. What additional average upward force does the floor exert if the contestant pushes down on the floor
    for 0.36 s during the jump?
    a. 0.0
    b. 210.0 N
    c. 420.0 N
    d. 630.0 N

    2. Relevant equations
    P=mv
    delta p= f*delta t

    3. The attempt at a solution
    For part A: Since the contestant is on the ground initially, he has zero momentum, and after he jump his final speed is 2.1m/s. P=mv, I calculated 151.2 kg*/s as his change in momentum. However, the answer for this part should be 75.6kg*m/s...
    For part B I used the impulse divide by time to get the average force, and the value is 420N. The answer is 420 N but I thought that I should use mg substruct 420N to get the "additional force"

    Any suggestions please?
     
  2. jcsd
  3. Dec 15, 2016 #2

    PeroK

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    You're correct on part a). You can easily calculate a) from the answer to b).

    What do you want to do with the answer of ##420N## for b)?
     
  4. Dec 15, 2016 #3
    I think the answer for part A is 151.2kg*m/s and the 420N is the additional force as well as the net force.
     
  5. Dec 15, 2016 #4
    Since the question mentioned that the floor exerts an upward force of mg, and it is asking what ADDITIONAL upward force does the floor exert. By using the impulse expression, the average force can be found and the value is 420 N. I thought mg+ additional force =420N, therefore, I need to use mg-420N to find the additional force?
     
  6. Dec 15, 2016 #5
    420N=upward force exerted by floor - mg ; If the floor exerts mg as upward force then there is no net force to push constestant.
     
  7. Dec 15, 2016 #6

    PeroK

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    What direction is positive in this calculation?
     
  8. Dec 15, 2016 #7
     
  9. Dec 15, 2016 #8
    I think upward is positive
     
  10. Dec 15, 2016 #9

    PeroK

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    So, the additional force on the man is downwards, which causes him to accelerate upwards?

    ##mg - 420N \approx -350N##
     
  11. Dec 15, 2016 #10
    I am afraid I am losing you now...
    I thought the additional force is 285.6 N( mg- additional force=average force)
     
  12. Dec 15, 2016 #11

    PeroK

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    Let's suppose the additional force is ##285N##. You have three forces on the man: 1) this additional force 2) The floor pushing up with ##mg## and 3) Gravity pushing down with ##-mg##.

    The net force on the man is, therefore, ##285N + mg - mg = 285N##

    So, the acceleration and impulse must be calculated using this force.

    But, you've already calculated that you need a force of ##420N##. Not ##285N##.
     
  13. Dec 15, 2016 #12
    I see! Many thanks for your excellent explanation!
     
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