# Impulse in a Bungee Jump?

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1. Feb 12, 2016

### ZCV99

1. The problem statement, all variables and given/known data

An small doll is attached to an elastic cord which is suspended from a support as shown. An accelerometer is attached to the doll. The doll is allowed to fall and produces the graph shown. When the recording system is started the doll is stationary at the upper support and released approximately 5 sec later.

https://bb9.waukesha.k12.wi.us/bbcswebdav/pid-692076-dt-content-rid-2436426_1/xid-2436426_1 [Broken] {Copy of image added by moderator}

-The graph can also be found here: https://docs.google.com/document/d/1k0hjDdK4kqCp05ZFZt-aNn_T9xFhHaF-teH7M08jH7Q/edit?usp=sharing

A) At what time on the graph does the elastic cord begin to pull upward on the doll?

B) At what time on the graph is the doll at its lowest point?

D) Calculate the approximate impulse given to the doll during the first falling portion of the jump.

E) How would each of the following be different if a larger mass doll were attached to the same elastic cord?

• maximum speed of the doll before the cord acted
• total impulse given to the doll during the falling portion of the fall
• width of the pulse on the graph for the first oscillation of the doll.

2. Relevant equations
I=mvf-mvi

3. The attempt at a solution

A. The cord pulls upwards on the doll beginning at 8 seconds, because the acceleration shifts from negative to positive

B. The doll is at its lowest point at 10 seconds because this is when the positive acceleration it experiences from the bungee cord reaches its peak.

D. I am not entirely sure how to find the impulse without knowing the mass of the doll, but I would assume that I could find it by finding its change in velocity, and multiply each of these by the constant m. However, I don't know how to find the initial velocity of the doll before the jump, since it appears that its negative acceleration was not a constant 9.8m/s^2 between 5-8 sec. Does this meant that the cord was working on the doll before 8 seconds and my answer to A is wrong as well? I am assuming that the final velocity can be found by estimating the area under the curve from about 8 to 11 seconds, which would be

vf=area under max height (26m/s^2)

0.5*26*3s=39m/s

But how do I find the initial velocity? Is my approach and my thought process entirely wrong?

E.

-The maximum speed of a larger mass for the doll would not be any different from a smaller mass because acceleration due to gravity is constant regardless of the mass of an object

-The total impulse would be higher because impulse is equal to mvf-mvi

-I'm not entirely sure what this is asking, but I would guess that it is referring to the time between each positive acceleration due to the bungee jump and the negative acceleration due to gravity during freefall, which would be unaffected by the mass of the object in question.

Ultimately, I am only fairly confident in my answer for B. I am especially thrown off by the fact that the acceleration of the doll does not shift immediately from negative to positive and vice versa. Does this mean that the cord begins pulling on the doll when the acceleration begins to increase even when it is still negative, or when the acceleration shifts from negative to positive?

Last edited by a moderator: May 7, 2017
2. Feb 12, 2016

### BvU

Good.
Now about a: what would the graph look like if there was no chord ?
You pick the point where a = 0 , meaning $\Sigma F = 0$. Why ?

3. Feb 12, 2016

### ZCV99

Oh, so the tension of the cord is always pulling on the doll, and acceleration is based on the Net Force acting on the doll. That would mean that the cord starts to pull upward at 5 seconds when the doll is first dropped, right?

Since the force of gravity acting on the doll increases when mass increases (F=ma), does that mean that the upward tension force of the cord would require more time to overcome this larger force, so the distance between each oscillation would be longer, and that the doll would reach a smaller maximum velocity?

Can I find the impulse by estimating the average force acting on the doll (halfway between the initial minimum acceleration and the maximum acceleration times the mass) and multiplying this by the change in time?

4. Feb 12, 2016

### BvU

No, only when the chord it taut. At least if the weight of the chord can be ignored. So initially the doll undergoes free fall over the length of the chord. Then the chord starts to stretch and the more it stretches, the harder it pulls.

5. Feb 12, 2016

### ZCV99

During freefall the acceleration of the doll would be a constant -9.8m/s^2, right? Yet the acceleration does not stay constant, which is why I thought that the cord was taut from the beginning. Does it actually become taut when the net force and the acceleration of the object is zero?

6. Feb 12, 2016

### ZCV99

I'm sorry but I am still extremely confused. It appears that the acceleration starts increasing from its initial -9.8m/s^2 almost as soon as the doll is dropped, which means that the cord was taut from this moment.

Can I find impulse by finding the acceleration halfway between the minimum and maximum accelerations of the doll and multiply this by time and the unknown mass, m, of the doll?

Are my answers in post #3 correct? Would the doll reach a smaller maximum velocity if its mass was increased?

7. Feb 12, 2016

### haruspex

The graph is a bit strange at the start. Yes, you would expect to see it go from 0 straight to -9.8 and stay there briefly before starting to return to 0. Instead it goes significantly more negative, as though either the doll was thrown down or started so far above the anchor point of the cord that the cord pulled it down. But in either of those scenarios there should be a period when the cord is slack. So I would dismiss the values below -9.8 as noise. This leads me to agree with your conclusion that the cord starts taut, but not stretched. So yes, the answer to A is 5 seconds.

8. Feb 12, 2016

### ZCV99

Would the impulse then be 54m N*s, since the maximum velocity is 39m/s, outlined above, and the minimum velocity is the area between the x axis and the curve is about 0.5*3seconds*-10m/s^2=-15m/s, so 39m/s*m--15m/s*m=54m N*s?

Also, what about question E? I know that the magnitude of the impulse would increase as mass increases using the above formula, but how do I answer the other two parts?

9. Feb 12, 2016

### haruspex

I assume you mean 54Ns.
Don't confuse tension in the cord with net force.
For max speed before cord acts, didn't you deduce that the cord acts immediately?
For period, do you have a standard equation for that?