# Application: using energy and momentum to calculate estimated weight

1. Jul 15, 2011

### Hobbit300

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

An arctic hunter goes hunting polar bears and shoots one. He finds it lying dead on a large patch of almost frictionless ice. He has no scales or weighing device of any kind (and of course, there's nothing to hang it on) but he does have a long rope. How can he get a pretty good estimate of the weight of the bear? Give a detailed answer, perhaps with a sample calculation. Reasonable assumptions may be made. Answer does not have to be a specific number, but may be a re-arranged formula for the unknown with all other variables being easily substituted for by the hunter.

2. Relevant equations
While force equations may be used (F=ma), this question is meant to be answered using equations involving momentum and velocity.

total initial momentum= total final momentum
therefore, m1v1i + m2v2i = m1v1f + m2v2f, where m1= mass of hunter, v1i=initial velocity of hunter, m2= mass of polar bear (unknown), v2i=initial velocity of polar bear, and v(1 or 2)f = final velocity of hunter or polar bear respectively.

Ek=1/2mv^2
Ep=mgh

Et = (Ekf-Eki) + (Epf-Epi)

Other possible equations:
Ep=-GMm/r

3. The attempt at a solution

Honestly, I have absolutely no idea where to start. I have attempted to use the momentum formula above, making the assumptions that the hunter knows his own mass, as well as that he can judge his own velocity and that of the polar bears (if he runs or drags the polar bear).
If i set vi for both the polar bear and hunter to 0, my masses would also be zero, so instead, I attempted the problem by making v1i=2m/s and v2i=2m/s also because if the hunter tied one end of the rope to the polar bear and ran with it, pulling with even tension, the polar bears speed would also equal the hunters.
when substituting a random number for the hunter's mass, and 5m/s for the v1 and v2 f I kept ending up with m2=-8 ( negative the hunter's mass).

Then, calculating weight, W=mg, =(-8)(-9.8) i get 78.4kg - a reasonable weight for a polar bear - but how can m1=m2??? the masses of a polar bear and hunter i'm assuming will be very different - so they cannot have the same mass or the same weight.

I have no idea where to go from here and I've been working on this application question for a couple days now - any ideas or changes in scenario? (a.k.a other things the hunter may do with the rope?)

I thought of a different scenario - one where the hunter could tie the ends of the rope to himself and the polar bear, and then going and standing right next to it. If he starting running till the point where he reached the end of the rope and the rope was taught - would the hunter's kinetic energy at that point = polar bear's potential energy? Even so, how does h from (Ep=mgh) factor in, if the polar bear is being held at no height?

Please help! This is due Monday and I'm nowhere near being able to understand how to solve this! I did try asking my teacher but she refused to give any hints or answer any clarification questions, saying to just read the question over.

2. Jul 15, 2011

### tiny-tim

Welcome to PF!

Hi Hobbit300! Welcome to PF!
Why all this running?

What happens if he just sits down, and pulls the rope?

3. Jul 15, 2011

### Hobbit300

If the hunter was sitting down then his initial and final velocity would both be zero - this leaves me with m2v2i=m2v2f. if the polar bear starts with inital 0m/s or any other velocity i still end up with m2v2i = m2v2f. if v2i=0, 0=m2v2f, and whatever v2f equals, m2=0.
if v2i=2m/s for example, m2=0/something which doesn't work either.

Am i using the wrong equation or is there something i'm missing that i could use in the Ek and Ep equations?

4. Jul 15, 2011

### tiny-tim

but he's on ice

5. Jul 15, 2011

### Hobbit300

...so he would slide forward while pulling the bear closer! Thank you!!!! I think i've got it now, but would you check if i'm on the right track? so i know that v1i and v2i =0, but v1f and v2f are not. I'm thinking their velocities would not be equal because their masses are different.... so i end up with 0=m1v1f + m2v2f. If the hunter is able to judge his own velocity as well as the polar bears... so for example he knows his v1f=4, his m1=2, and v2f=-3 because the bear is being pulled in the opposite direction.. then i am able to solve for m2 which would equal in this case -8/3. and then i could calculate weight but multiplying by -9.8..which would give me the polar bear's weight. ?

6. Jul 15, 2011

### tiny-tim

Hi Hobbit300!
That's right!

(but why use 9.8?)

But there might be a simpler way, without using velocity …

what will happen when they meet?

7. Jul 15, 2011

### Hobbit300

I was using 9.8 because m2 is only the mass of the polar bear. For weight, which equals mg i would have to multiply by 9.8, or so i thought?

If they meet they would collide.. but i don't understand how i could take velocity out of the equation since their final velocities would not be equal nor would they be zero if the hunter rebounds a little, which i'm assuming he would because the polar bear weighs a lot more than him. So i'm not sure how to approach this without including velocity :S

8. Jul 16, 2011

### tiny-tim

Hi Hobbit300!

(just got up :zzz: …)
But aren't you comparing the bear's weight with the hunter's weight?
But what if he keeps hold of the rope (or of the bear)?

9. Jul 16, 2011

### the big D

but if the hunter pulls on the rope and moves why will his initial velocity be zero?