Calculating Impulse on Bungee Jumper

[shadowice]

Homework Statement

A 68.0 kg bungee jumper, tied to a 31.0 m long cord, leaps off a bridge whose deck is 61.0 m above the water. He falls to 7.0 m above the water before the bungee cord pulls him back up. What is the magnitude of the impulse exerted on the bungee jumper while the cord stretches?

M= 68kg
Y(Length of cord) = 31m
H= 61m
y3(at v3=0) = 7m
V3(cord fully extended)= 0

Homework Equations

I = dP = Pfinal - Pint = m(v3-v2)
y=yint+vint*t+1/2Ay*t^1/2
t=(2y)^1/2/g

The Attempt at a Solution

v2 = g((2y)^1/2/g)
v2 = 9.8(2(31)/9.8)^1/2=7.87m/s

then i used that in conservation of energy
mgh- 1/2mV^2
m's cancel out
gh- 1/2V^2
9.8*61 - 1/2*7.87^2= 566.83 kg*m/s
but i get Impulse is the change in momentum. You know the jumper's momentum at the end of the jump. Free fall should tell you the momentum before the bungee cord starts to pull.

 

[Delphi51]

This is quite complicated! The jumper accelerates at 9.81 until he reaches the end of the cord, then accelerates less and less until acceleration reverses and stops him.
You can't use mgh = 1/2mv^2 because the PE is converted into both KE and spring energy.
Maybe work at the bottom, where all of the PE is converted into spring energy, with no KE. Yes, that should allow you to calculate the k for the spring and then you can work with forces. Can you do integrals?

 

[shadowice]

havent learned how to do intergrals yet.

 

[Delphi51]

Okay, no problem - an average will do.
Have you found k yet?

 

[shadowice]

K Final + P final = K int + P int
K fin = 0
P fin = 68*9.8*54 = 35985.6
K int = 1/2*68*7.87^2= 2105.85
P int = 68*9.8*31=20658.4

not sure about those h values the 54 is the initial 61- 7
and the 31 is the length of the cord.
Im using initial as once the cord is fully extended before it stretches.

but when i did those clacs, i got -13221.35 am i getting any closer

 

[Delphi51]

I fear I have mislead you. I couldn't finish the problem to find F*delta t - I could not find the time to stop the jumper without calculus because it is non-constant acceleration.

Back to m*delta v, which I think you were onto before I messed you up!
This doesn't look promising to me, either. The impulse is not only m*delta v but also the F*delta t working against gravity while it is stopping. I'm stumped here, too - can't find the time without calculus.

I hope someone will help us out, here!

 

[shadowice]

we already got time

t=(2y)^1/2/g

 

[Delphi51]

That would be for a free fall, height y.
This is not a free fall because the cord is exerting an upward force.
That force varies with the stretch of the cord, so it is all very complicated.
I can't even do the calculus - it is a nasty differential equation.

How about a numerical solution on a spreadsheet?
EDIT: this worked out easily. The time to stop is 1.31 seconds.
I can easily coach you on making the spreadsheet if you are interested.

 

[Delphi51]

Perhaps I am being too picky.
If you just use your mgh = 1/2*m*v^2 and pretend there is no cord, you could find the speed at the bottom of a 54 m free fall. If you do m*delta v with that, you ought to be pretty close.

 

[LowlyPion]

I don't think you are going to need to find k or t even.

The problem is worded such that they want the impulse provided on the jumper during the stretching regime of the downward jump.

That regime starts at 31m below the bridge when the relaxed length gets taut and the stretching continues until 7m. Determine the V at 31m from the v = √(2gh) That times the mass is the initial momentum.

The momentum at the bottom is apparently 0. So the way I see it ...

I = Δp = 0 - (m*-v) = mv

I think then this is the Impulse on the jumper.

 

[Delphi51]

That would certainly be true if gravity is ignored during the stretching.
Perhaps that is what the question is looking for. According to my numeric calc, the impulse needed to hold back mg for the 1.3 seconds is 867 - which is quite significant.

 

[LowlyPion]

I think the jumper experiences the net of gravity and the force of the bungee. It's the net force that affects the momentum. Taken over the interval of the fall then, momentum continues to increase even for a time before the bungee force exceeds mg and then it actually begins to reverse speed to 0. So answering the question as stated suggests that the interval of the integral should be from the moment it starts to stretch, until the moment it starts to contract. The momentum change then is the initial momentum, since it goes to 0 at the bottom, before reversing as it contracts.

That's my thinking anyway.

 

[shadowice]

Thank you both of you, lowlypion was right it wanted velocity from 31m instead of the 68 i had been using. I plugged that in and it accepted it

 

[LowlyPion]

Unfortunately it's a problem that lends itself to over-thinking I think.

Glad it worked out in the end.

 

[Delphi51]

The other way of looking at it is informative.
F = mg - kx = ma so a = g - k/m*x
I put this on a spreadsheet with columns for time, stretch, velocity, acceleration and force.
Calculate stretch from the previous row's velocity, acceleration from the stretch and so on.
With time increasing by .01 second at a time, it should be pretty accurate.
It takes 1.31 seconds to stop and the sum of the F*dt's is 2745 N.s
With the m*dv method using a fall of 31 m, you get 1671.
With a fall of 54 m, you get 2213.
It is rather annoying me that this last 2213 is not closer to the 2745.

Kind of interesting to see it go . . .