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Impulse of person jumping on a force platform

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  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Starting from rest, a 65 kg athlete jumps down onto the force platform from a height of .6 m. While she is in contact with the platform during time interval 0s-.8s, the she exerts on it is described by the function F=9200t-11500t^2 where F is in N and t is in s.
    a) What impulse did the athlete receive from the platform?
    b) With what speed did she reach the platform?
    c) With what speed did she leave it?
    d) To what height did she jump upon leaving the platform?

    2. Relevant equations
    I=∫ΣFdt
    kinematic equations
    I=ΔP
    ΔP=MVf-MVi
    3. The attempt at a solution
    For part a) i just integrated and got I=983.
    For part b) I used kinematic equations and got v=3.43 m/s.
    Part c and d I am confused how to solve them. For part c) I did I=ΔP, 983=65Vf-65(3.43)= 18.5 m/s which seems like it would be way too high, but I used that for part d) where I did .5mv^2=mgh and I ended up with 17.5m which I am assuming is way wrong.
     
    Last edited: Nov 13, 2014
  2. jcsd
  3. Nov 13, 2014 #2

    Simon Bridge

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    Please show your reasoning:
    For (a) you used ##I = \int_T F(t)\;dt## ... this is correct. I don't see any working though.
    For (b) you did not say how you used the information supplied to get that figure.
    For (c) and (d) you should be using the information provided.
    So (c) impulse = change in momentum, you know the momentum before and the change in momentum you just calculated...
    ... this is what you seem to have done.
    For (d) you did not say how you were calculating that height.

    So the possibilities are: you have the wrong specific impulse, you have the wrong initial speed, or your calculation for the final height was wrong, or everything is correct and the problem just has unrealistic numbers.
     
  4. Nov 13, 2014 #3
    Sorry I will be more specific. For part a) I did ∫(0,.8)9200t-11500t^2 which ended up as 4600t^2-(11500/3)t^3 |(0,.8) = 983.
    For b) I used the initial height which is .6, the acceleration which is g, and Vi is zero. Vf^2=2g(.6), Vf=3.43.
    Part c I showed, and part d I plugged in my numbers into .5mv^2=mgh, (.5)(65)(18.5^2)=65gh. h ended up being 17.5m
     
  5. Nov 13, 2014 #4

    Simon Bridge

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    in part (b) you have assumed that the person jumping was falling freely from zero speed - but they jumped - so they may have put some energy into it, so the initial speed may not have been zero.

    I'd think about it like this:
    At t=0 the person has some momentum mu ... and at t=0.8s the person has some momentum mv.
    Since v points in the opposite direction as u, the momentum must have been zero at some time.
    ... at which time is the person's momentum going to be zero?

    Mind you I havn't had my coffee yet :)
     
  6. Nov 13, 2014 #5
    I may have missed the very first part of the question when writing it on here heh. The question starts with "Starting from rest..." I'll edit that into the post.
    And wouldn't it be when mv=mu
     
    Last edited: Nov 13, 2014
  7. Nov 13, 2014 #6

    Simon Bridge

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    OK - if mv=mu, then how does the final height compare with the initial height?
     
  8. Nov 13, 2014 #7
    I'm not really sure how get height by thinking of mv=mu, but if i think of it with potential and kinetic energy, then the initial height would be equal to the final when the velocities are the same, unless I did that wrong.
     
  9. Nov 13, 2014 #8

    Simon Bridge

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    That's correct - since ##K=p^2/2m##.

    So if v=u, writing in terms of v, then what is the equation for the change in momentum?
     
  10. Nov 13, 2014 #9
     
  11. Nov 13, 2014 #10

    haruspex

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    Maybe I'm missing something, but I don't see where you are going with this. The OP method looked OK to me, just went wrong in the details.
    Simon defined u and v as the landing and take-off velocities respectively, then asked about some intermediate time. v and u are not varying with time, so there's no time at which those are equal.
    Note that I wrote "velocities". Velocity and momentum are vectors. Are you taking up as positive or down as positive? Think again about this equation:
     
  12. Nov 13, 2014 #11

    Simon Bridge

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    ... and yet the person is acted on by an unbalanced force which ultimately changes her direction: momentum is a vector. Lets say v=u=v, then u=-v.

    @haruspex: feel free to point out the detail of the mistake :)
     
  13. Nov 13, 2014 #12
    Ah that is a silly mistake. So if I take the y axis to be positive upwards, it would be 983=65Vf-65(-3.43), Vf=11.7 m/s. I'm still not sure how reasonable that is, but I suppose I am not an athlete. But then for calculating the final height with .5mv^2=mgh I get a height of 6.98 meteres, which still seems unreasonable. Unless I still didn't do part c right.
     
  14. Nov 13, 2014 #13

    haruspex

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    There is just one more thing wrong. Was gravity turned off for those 0.8 seconds?
     
  15. Nov 13, 2014 #14
    The only type of problems I have done with conservation of momentum before this one have been under very small time intervals where we basically just ignored gravity, so I actually had no idea I had to take into account. But would it be 983=65Vf-65(-3.43)+(65*9.8*.8)? Which equals 3.85 m/s and would also give a reasonable height.
     
  16. Nov 13, 2014 #15

    haruspex

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    Yes. Think of it like this: when you stand still the floor exerts a force mg upwards on you. For you to gain any upwards momentum it must exert a greater force; Or like this: while the platform is exerting the stated upward force, gravity is exerting a downward force.
     
  17. Nov 13, 2014 #16
    OK that makes sense, thank you for the help! Also thank you Simon, though haruspex's explanation made more sense for me personally.
     
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