Impulse of person jumping on a force platform

In summary: If v=u, wouldn't the change in momentum equal zero since mass is the same?No, the change in momentum would be equal to the initial momentum minus the change in momentum.
  • #1
timnswede
101
0

Homework Statement


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?

Homework Equations


I=∫ΣFdt
kinematic equations
I=ΔP
ΔP=MVf-MVi

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:
Physics news on Phys.org
  • #2
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.
 
  • #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
 
  • #4
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 :)
 
  • #5
Simon Bridge said:
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 :)
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:
  • #6
OK - if mv=mu, then how does the final height compare with the initial height?
 
  • #7
Simon Bridge said:
OK - if mv=mu, then how does the final height compare with the initial height?
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.
 
  • #8
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.
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?
 
  • #9
Simon Bridge said:
That's correct - since ##K=p^2/2m##.

If v=u, wouldn't the change in momentum equal zero since mass is the same?
 
  • #10
Simon Bridge said:
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?
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.
timnswede said:
And wouldn't it be when mv=mu
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:
timnswede said:
983=65Vf-65(3.43)
 
  • #11
If v=u, wouldn't the change in momentum equal zero since mass is the same?
... and yet the person is acted on by an unbalanced force which ultimately changes her direction: momentum is a vector. Let's say v=u=v, then u=-v.

@haruspex: feel free to point out the detail of the mistake :)
 
  • #12
haruspex said:
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:
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.
 
  • #13
timnswede said:
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.
There is just one more thing wrong. Was gravity turned off for those 0.8 seconds?
 
  • #14
haruspex said:
There is just one more thing wrong. Was gravity turned off for those 0.8 seconds?
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.
 
  • #15
timnswede said:
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.
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.
 
  • #16
haruspex said:
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.
OK that makes sense, thank you for the help! Also thank you Simon, though haruspex's explanation made more sense for me personally.
 

Related to Impulse of person jumping on a force platform

1. How does the impulse of a person jumping on a force platform affect their jump height?

The impulse of a person jumping on a force platform is directly related to their jump height. The greater the impulse, the higher the jump will be. This is because impulse measures the change in momentum, and a greater change in momentum results in a greater vertical displacement.

2. How is impulse calculated on a force platform?

Impulse is calculated by multiplying the force exerted on the platform by the time it is exerted for. This can be visualized as the area under the force-time curve on a graph. The unit for impulse is Newton-seconds (N*s).

3. Can the impulse on a force platform be used to measure the power of a jump?

Yes, the impulse on a force platform can be used to calculate the power of a jump. Power is defined as the rate of doing work, and since work is equal to force times distance, power can be calculated by dividing the impulse by the time of the jump.

4. Is the impulse of a person jumping on a force platform affected by their body weight?

No, the impulse of a person jumping on a force platform is not affected by their body weight. The force platform measures the force exerted on it, and not the weight of the person. However, the person's weight can affect their jump height and therefore the impulse.

5. How does the impulse of a jump compare for different types of jumps (e.g. vertical jump vs. long jump)?

The impulse of a jump can vary for different types of jumps, depending on the force exerted and the time it is exerted for. In general, a vertical jump will have a greater impulse than a long jump, as the vertical displacement is greater and therefore requires a greater change in momentum. However, this can also vary based on the technique and strength of the individual performing the jump.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
652
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
3K
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
10K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top