Which Has Greater Impulse: Sand or Water?

[TheHamburgler1]

Jumping onto sand vs. jumping into water results in which of the following as each substance brings you to a halt?
a) The sand exerts a greater impulse than the water
b) The water exerts a greater impulse than the sand
c) The sand and water exert the same impulse, but the sand exerts a greater average force
d) the sand and water exert the same impulse, but the water exerts a greater average force
e) Each substance exert the same impulse and same average force

I thought the answer was d but that was ultimately a guess

 

[berkeman]

What is the mathematical definition of impulse? Does sand or water stop you more quickly?

 

[TheHamburgler1]

J=F change T= change P= m change V

sand stops you more quickly

 

[TheHamburgler1]

This therefore means that the correct answer would be B correct?

 

[berkeman]

J=F change T= change P= m change V

sand stops you more quickly

Correct. So what is the correct answer to the original question? BTW, here is the LaTex version of what you typed. You can use the "QUOTE" button to see the LaTex source. There is a LaTex tutorial in the Math & Science Tutorials section

https://www.physicsforums.com/forumdisplay.php?f=151

$$J = F \Delta T= \Delta P = m \Delta V$$

 

[berkeman]

This therefore means that the correct answer would be B correct?

Why do you say that?

 

[TheHamburgler1]

The correct answer is B) Water exerts a greater impulse than the sand because time is larger for the water

Correct?

 

[berkeman]

The correct answer is B) Water exerts a greater impulse than the sand because time is larger for the water

Correct?

Is that statement consistent with your previous (correct) statement?

sand stops you more quickly

 

[TheHamburgler1]

I believe it is or am i missing something big

 

[berkeman]

Hmmm, I think I see a source (of at least my) confusion. I was thinking in terms of peak impulse and peak force, not the total impulse required to stop you in either sand or water. You are correct that the lower force exerted by water is exerted longer. So the question is whether sand or water or neither exerts the larger total implulse to stop the person...

$$I = \int F dt$$

Now I have to re-think my answers...

 

[berkeman]

I believe it is or am i missing something big

You may indeed be right, but I think it takes one more step rather than just saying the the water takes longer to stop the person. If the F * t is the same for both (longer time, proportionally lower force), then the impulses would be equal. How can we figure out what the proportionality is?

 

[TheHamburgler1]

You may indeed be right, but I think it takes one more step rather than just saying the the water takes longer to stop the person. If the F * t is the same for both (longer time, proportionally lower force), then the impulses would be equal. How can we figure out what the proportionality is?

Based upon this information wouldn't you think the answer would be C)The sand and water exert the same impulse, but the sand exerts a greater average force

This seems to work with what you stated

 

[TheHamburgler1]

and obviously the greater force is due to the lower time which together equal the waters impulse which has longer time and lower force

 

[berkeman]

I think it depends on what model you choose for the deceleration that you experience in sand versus in water. If the deceleration is assumed to be constant for each, then that gives one answer (quiz question -- which answer? Not C, BTW). If the deceleration is not constant (like water resistance is velocity-dependent), that gives a different answer (quiz question -- which one?). Do you have any more information on which model we should be using for the deceleration due to sand versus water?

 

[TheHamburgler1]

No this is the only given information. Could you explain to me why C is incorrect please

 

[TheHamburgler1]

Can anyone help me with what the real answer is? Because now I am kinda lost

 

[berkeman]

Well, another input would be helpful if we can get it, but in the mean time, how about anwering my Quiz Question #1. I'm thinking that if they didn't give you any assumptions about the character of the decelerations in sand and water, that they probably want you to assume a constant deceleration for both. That makes the answer straightforward. If the two decelerations are each constant, what does that do to the impulse integrals?

 

[TheHamburgler1]

When I look at this problem I still see the answer being that the sand and water exert the same impulses but the sand exerts greater average force. This is because larger force but less time and therefore smaller force but more time for water. What is the error in my logic?

 

[moose]

Alright, impulse is change in momentum, correct? You assume that the initial momentum is the same.

The change in momentum is also equal to force * time.

I don't see where you are confused.

EDIT: TheHamburgler1, I think you are correct. Forget what I said above

 

[berkeman]

When I look at this problem I still see the answer being that the sand and water exert the same impulses but the sand exerts greater average force. This is because larger force but less time and therefore smaller force but more time for water. What is the error in my logic?

Okay, upon further review, I think your are correct with (C) for the case where the declerations are constant (but different), and for the reason you state. Sorry that I confused you, but you figured it out correctly and stuck to your guns. That's good (and my bad).

I'm still thinking about the case where the sand has a constant deceleration and the water is more accurately modeled as velocity-dependent (so it decreases with time). But it sounds like the simpler case is probably what you are being asked to solve.

I promise to do better on your next thread...

 

[TheHamburgler1]

Hey its alright, thanks a lot for the help I really appreciate it

 

[OlderDan]

Based upon this information wouldn't you think the answer would be C)The sand and water exert the same impulse, but the sand exerts a greater average force

This seems to work with what you stated

This is correct. The impulse equals the change in your momentum, so it is the same for both sand and water. The average force means the time average. The sand stops you in far less time than water, so the average force is considerably greater for the sand than for water.

Contemplate the effect of doing a 10 meter platform dive into a sandbox.