A momentum question? Not too sure

  • #1

Homework Statement



A 30 kg girl standing on slippery ice catches a 0.5 kg ball thrown with a speed of 16 m/s.
What then happens to the girl?

Homework Equations



Momentum = mass x velocity

or

p1 + p2 = (3 m/s) m <----(p is a symbol for momentum)

or

mv1 + mv2 = (3 m/s) m

or

v1 + v2 = 3 m/s

(ONLY 1 needs to be applied...not sure which one)


The Attempt at a Solution



I need help applying one of the formulas provided above to the question provided. I've tried a couple of these formulas already myself and the answers I got at the end I couldn't relate to the question thus I suspect none of these formulas might be useful in finding the answer, but these are the formulas I've been provided with to answer this question.

Also, aside from the formula dilemma, I found that slippery ice has no friction, so do be mindful of that if you'll be attempting this equation with me.

I guess the first question I need answered is which formula to use to answer this question if I am to use one of the provided formulas

Thanks in advance, whoever you are.
 

Answers and Replies

  • #2
Dick
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Homework Statement



A 30 kg girl standing on slippery ice catches a 0.5 kg ball thrown with a speed of 16 m/s.
What then happens to the girl?

Homework Equations



Momentum = mass x velocity

or

p1 + p2 = (3 m/s) m <----(p is a symbol for momentum)

or

mv1 + mv2 = (3 m/s) m

or

v1 + v2 = 3 m/s

(ONLY 1 needs to be applied...not sure which one)


The Attempt at a Solution



I need help applying one of the formulas provided above to the question provided. I've tried a couple of these formulas already myself and the answers I got at the end I couldn't relate to the question thus I suspect none of these formulas might be useful in finding the answer, but these are the formulas I've been provided with to answer this question.

Also, aside from the formula dilemma, I found that slippery ice has no friction, so do be mindful of that if you'll be attempting this equation with me.

I guess the first question I need answered is which formula to use to answer this question if I am to use one of the provided formulas

Thanks in advance, whoever you are.
Yes, it's a conservation of momentum question. Just think about it. The initial momentum of ball+girl should equal the final momentum of ball+girl. If you call the final velocity of the girl v, can you write an equation expressing that? And where did 3m/s come from???
 
  • #3
acceleration = v1 + v2?
Even if I were to use this equation, how will I answer "what happens to the girl"? Is the equation supposed to help me figure out whether or not she falls and perhaps how far she falls? What do you think?
 
  • #4
Dick
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acceleration = v1 + v2?
Even if I were to use this equation, how will I answer "what happens to the girl"? Is the equation supposed to help me figure out whether or not she falls and perhaps how far she falls? What do you think?
No, conservation of momentum. Do you know what that is? The answer they are fishing for is that after catching the ball the girl will find herself sliding over the frictionless ice at some velocity. Doesn't that make sense? What's her velocity after she catches the ball?
 
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  • #5
please give me a moment to consult with my friend's notes (since I missed this class regarding momentum)
 
  • #6
Dick
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please give me a moment to consult with my friend's notes (since I missed this class regarding momentum)
After you do just say it in words. No formulas yet.
 
  • #7
"The law of conservation of momentum: the total momentum of any system remains unchanged, regardless of interactions among the system's parts, so long as no part of the system is acted upon by forces external to that system"

I get the general idea now, although it would've been a lot clearer had it been explained to me by my instructor
 
  • #8
Dick
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"The law of conservation of momentum: the total momentum of any system remains unchanged, regardless of interactions among the system's parts, so long as no part of the system is acted upon by forces external to that system"

I get the general idea now, although it would've been a lot clearer had it been explained to me by my instructor
Maybe the instructor did explain, you missed the class. Now can you translate that into a solution??
 
  • #9
oh I know the instructor explained it, that is why I was specific with my words when I said "had it been explained to ME by my instructor". As for a solution, I first need to figure out which formula I use, which will then lead me to a solution. There are a few formulas above that I listed, I think the one I need to use is this: mv1 + mv2 = (3 m/s) m. The only problem is, I don't know what to do where the (3 m/s) comes from
 
  • #10
Dick
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oh I know the instructor explained it, that is why I was specific with my words when I said "had it been explained to ME by my instructor". As for a solution, I first need to figure out which formula I use, which will then lead me to a solution. There are a few formulas above that I listed, I think the one I need to use is this: mv1 + mv2 = (3 m/s) m. The only problem is, I don't know what to do where the (3 m/s) comes from
I don't know where the 3m/s is coming from either. I would suggest you just think about what 'conservation of momentum' means and work from there. That might be a formula from some completely different problem. What's the sum of the momenta of the girl and the ball before the catch? What's an expression for the sum after, assuming she winds up moving at velocity v? Set them equal.
 
  • #11
SteamKing
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oh I know the instructor explained it, that is why I was specific with my words when I said "had it been explained to ME by my instructor". As for a solution, I first need to figure out which formula I use, which will then lead me to a solution. There are a few formulas above that I listed, I think the one I need to use is this: mv1 + mv2 = (3 m/s) m. The only problem is, I don't know what to do where the (3 m/s) comes from
Well, think, where did this formula come from? Was it in your friend's notes? Could it be from some other problem? Why 3 m/s? What if you disregarded it, and just applied the concept of momentum to the problem as is?
 
  • #12
Yeah, this 3 m/s has gotta be from another problem. And to be honest, I don't understand momentum, nor can I see the vividly see the relation between the principle of momentum that I listed above and the question that need be answered. Except for maybe that ice removes the external force of friction
 
  • #13
Dick
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Yeah, this 3 m/s has gotta be from another problem. And to be honest, I don't understand momentum, nor can I see the vividly see the relation between the principle of momentum that I listed above and the question that need be answered. Except for maybe that ice removes the external force of friction
Take it one step at a time. What's the momentum of the girl before the catch? What's the momentum of the ball? Those should make for an easy starting point.
 
  • #14
The momentum of the girl and the ball before the catch was 0 (unmoving). And the law of momentum (as listed above) states that the "total momentum" remains unchanged. How can that be though? How can the momentum stay at 0 even though the ball was thrown and caught by a girl who then slid on the ice by a little. That's momentum changing is it not?
 
  • #15
Dick
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The momentum of the girl and the ball before the catch was 0 (unmoving). And the law of momentum (as listed above) states that the "total momentum" remains unchanged. How can that be though? How can the momentum stay at 0 even though the ball was thrown and caught by a girl who then slid on the ice by a little. That's momentum changing is it not?
Before the catch the girl isn't moving. So she has momentum 0. Good so far but the ball IS moving. What's its momentum? You aren't starting from a state of 0 momentum and you aren't going to wind up in a state of 0 momentum. Momentum will be conserved. And if there is no friction she won't slide 'a little'. She will keep right on sliding. There's nothing to stop her.
 
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  • #16
would the ball's momentum equivalent to its speed then (16 m/s)? Or would I need some sort of a formula to calculate the momentum? Sorry if this is becoming painfully obvious and I am unable to see it, I am the epitome of what my username suggests
 
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  • #17
Dick
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would the ball's momentum equivalent to its speed then (16 m/s)? Or would I need some sort of a formula to calculate the momentum? Sorry if this is becoming painfully obvious and I am unable to see it, I am the epitome of what my username suggests
Yes, you did miss the lecture. There is a formula. Momentum is equal to mass times velocity. It's not an abstract concept. And it has a direction, which isn't terribly important in this problem.
 
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  • #18
hmm, easy then.

Momentum = 0.5 x 16

Momentum = 8 m/s or 8 N?

therefore, upon interaction with the ball, the girl will begin to slide 8 m/s or 8 N?

what's the measurement unit then? my guess is m/s, right?
 
  • #19
Dick
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hmm, easy then.

Momentum = 0.5 x 16

Momentum = 8 m/s or 8 N?

therefore, upon interaction with the ball, the girl will begin to slide 8 m/s or 8 N?

what's the measurement unit then? my guess is m/s, right?
As momentum is mass*velocity the units are kg*m/s. Not m/s and not N. You need to think about this some more. And the girl isn't sliding by herself. She's holding the ball. Assuming she winds up moving at velocity v, what's the final momentum after the catch?
 
  • #20
SteamKing
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Momentum = mass * velocity

What units would you expect momentum to have?

You're not thinking these matters through carefully.
 
  • #22
Dick
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hmm, 8 kg.m/s?
Right. That's the total momentum before the collision. Now can you express total momentum after the collision in terms of v, the girls final velocity? Set the two equal and solve for v.
 
  • #23
well, I did hand in what I had for this question, in hopes of part marks if anything. But I did attempt to solve for V and figured that since momentum is conserved, then it'd be 8 kg.m/s regardless? I'm probably off but thanks for all your help, Dick.
 
  • #24
Dick
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well, I did hand in what I had for this question, in hopes of part marks if anything. But I did attempt to solve for V and figured that since momentum is conserved, then it'd be 8 kg.m/s regardless? I'm probably off but thanks for all your help, Dick.
The final momentum is the mass of the girl and ball times the velocity they are travelling. So it's (30kg+0.5kg)*v. Equate that to the initial momentum 8kg*m/s and solve for v. It's actually pretty easy to finish up.
 
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