Momentum: instantaneous or average?

[paulb203]

Homework Statement

m; 90kg
v; 3m/s
p; ?

Relevant Equations

av.S=D/T
p=mv

I get the impression that momentum, in most physics questions (at least at GCSE level), is instantaneous. Is that correct?

I tried to apply the basics to a walk to the local shops that I take regularly

Distance 0.9 miles

Time taken 0.25 hours

Avg S=D/T

Avg S=0.9/0.25

Avg S=3.6mph

Avg v=3.6mph East

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p=mv

p=90kg(3.6mph)

p=324kg m/s

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But this is based on my average velocity. Therefore it’s my average momentum, yeah?

To get my instantaneous momentum I’d need my instantaneous velocity, yeah?

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I haven’t yet come across a GCSE question about average momentum. Why is that? Are physicists/engineers generally more interested in instantaneous momentum?

And what does instantaneous momentum imply, generally? My first thought is greater momentum = greater force. If I’m walking to the shop at 3.6mph, or slightly more, with a mass of 90kg, and I bump into someone of similar mass, the force won’t be so great as to knock them over or anything like that.

But if I’m sprinting at 10m/s (around 20mph) then I’ll hit the person with a greater force and probably knock them over, yeah?

Likewise for vehicles.

*

Last question.

Going back to my average speed of 3.6mph.

If I start at zero mph, and my average is 3.6mph, my fastest speed must be greater than 3.6mph, yeah? Otherwise (if my fastest speed was 3.6mph) the zero mph, and the other earlier speeds, would bring the average speed down, yeah?

P.S. I hope this belongs in the homework forum, even though it's not actual homework set by a tutor. I think the last time I put this kind of question in the main forum it was moved here, even thought it wasn't technically homework.

 

[kuruman]

I get the impression that momentum, in most physics questions (at least at GCSE level), is instantaneous. Is that correct?

Probably. I am not familiar with physics questions at the GCSE level.

But if I’m sprinting at 10m/s (around 20mph) then I’ll hit the person with a greater force and probably knock them over, yeah?

Nah! You are putting together speed momentum and force to form a misleading conclusion. Force is the ratio of the change in momentum (mass times instantaneous velocity) $\Delta p$ over a time interval $\Delta t$.

If you are a gnat traveling at 10 m/s, you will be squashed, the person you hit will still be standing and will scrape you off the spot where you hit him. If you are a battleship traveling at 10 m/s, nothing will happen to you, the person you hit will be squashed and it will be necessary to scrape him off the pier on which he is standing, yeah?

If I start at zero mph, and my average is 3.6mph, my fastest speed must be greater than 3.6mph, yeah?

Yeah!

 

[Steve4Physics]

I get the impression that momentum, in most physics questions (at least at GCSE level), is instantaneous. Is that correct?

It's risky to generalise, but at GCSE level, most questions won't distinguish between average and instantaneous momentum. Providing there is no acceleration (no speed or direction change), the average momentum (during some time-interval) is the same as the instantaneous momentum at any instant during the time-interval.

I suppose (at GCSE level) you could be asked qualitatively about accelerating systems such as circular motion, accelerating rockets. There are then important differences between average and instantaneous values.

Distance 0.9 miles

Time taken 0.25 hours

Avg S=D/T

Avg S=0.9/0.25

Avg S=3.6mph

OK.

Avg v=3.6mph East

OK. But the question should have given the displacement as 0.9miles east.

For example if the shop is 0.5miles east of your starting point the required displacement is 0.5miles east. But the road might be bendy (total distance 0.9 miles). In this case the average speed is 3.6mph but the average velocity is 2 mph east.

p=mv

p=90kg(3.6mph)

p=324kg m/s

Aagh! No! If you want momentum in units of kgm/s (kilogram.metres/second), you must convert 3.6 miles per hour to metres per second. (It's unfortunate that miles and metres have the same symbol.)

But this is based on my average velocity. Therefore it’s my average momentum, yeah?

If you do the calculation correctly, it would be the magnitude of your average momentum. Since momentum is a vector, you need to specify its direction.

To get my instantaneous momentum I’d need my instantaneous velocity, yeah?

Yes. Your speed (and direction on a bendy road) might not be constant. You would use your speed and direction at a particular moment in time to find you instantaneous momentum at that moment.

I haven’t yet come across a GCSE question about average momentum. Why is that? Are physicists/engineers generally more interested in instantaneous momentum?

Can you give an example of a GCSE question about instantaneous momentum? I'm not clear what you mean.

Instantaneous momentum may be important if something is accelerating. For example, the momentum of a car changes from moment to moment. In such situations, knowing the instantaneous momentum at specific times may be useful. But I'd be surprised if that came up at GCSE-level. Or if you have an impact, you need the momentum at the moment of impact.

And what does instantaneous momentum imply, generally?

It doesn't imply anything. If it is changing its useful for certain calculations. Or you may need the momentum at a particular moment in time, e.g. for an impact you would want the instantaneous momentum at the time of impact.

My first thought is greater momentum = greater force.

No. The faster the momentum changes, the greater the force. Read about Newton's second law of motion if interested.

If I’m walking to the shop at 3.6mph, or slightly more, with a mass of 90kg, and I bump into someone of similar mass, the force won’t be so great as to knock them over or anything like that

But if I’m sprinting at 10m/s (around 20mph) then I’ll hit the person with a greater force and probably knock them over, yeah?

Likewise for vehicles.

Momentum is a useful tool for impact problem. In this case you would want the momentum at the moment of impact. Your movement over the previous few minutes wouldn't be relevant. Whether or not you knock the person over depends on other factors too.

If I start at zero mph, and my average is 3.6mph, my fastest speed must be greater than 3.6mph, yeah? Otherwise (if my fastest speed was 3.6mph) the zero mph, and the other earlier speeds, would bring the average speed down, yeah?

Yes. That's maths!

P.S. I hope this belongs in the homework forum, even though it's not actual homework set by a tutor. I think the last time I put this kind of question in the main forum it was moved here, even thought it wasn't technically homework.

I think it's fine. It's good to think and ask about these things.

 

[erobz]

My thoughts are that this question-just scratching the surface-are a chapter in my Physics textbook. I'm glad I didn't get through a response! I was struggling...

 

[haruspex]

greater momentum = greater force

If the duration of the force is the same, yes. Equally, the same force applied for a longer time imparts more momentum.

the force won’t be so great as to knock them over or anything like that

If you bump into your mirror image the momenta cancel at all heights, from toe to crown. Neither will fall over, regardless of the mass, speed and elasticity. So, more generally, you have to consider relative heights and mass distributions.

Likewise for vehicles.

For consideration of damage, to vehicles and passengers, the critical value is peak force. Cars have crumple zones, fore and aft, designed to spread the forces at each end as uniformly as possible over a relatively long time. The impulse to be transmitted, $\int F.dt$, is a given, so to minimise $\max F$ you want $\Delta t$ long and $F$ constant.

 

[Merlin3189]

Coming back to the question of instantaneous momentum. I'd have thought, that's the only sort of momentum there is. The mathematically minded can calculate a statistic called the average momentum (presumably mean, though maybe median or mode), but does it have any significance?
As others have said, we are interested in changes in momentum because they require a force (or impulse). Momentum might help calculate that force or its duration, or for a given force, to calculate the change in velocity.
What could the statistic of average momentum usefully tell you? You talk about bumping into someone when you have a certain average momentum. Did that happen when you just happened to be instantaneously at the average speed, or was it at a moment when you were going faster, slower or even paused waiting to cross the road? (In which last case, it was his fault!) The average is of no use to you: you have to know the momentum at that moment. (Or, as others have said, you actually want the rate of change of momentum, but that will surely be related to the instantaneous momentum at the start of the collision.)
The average momentum statistics don't tell you much about the work you do on your walk. If you could accelerate to your average speed and remain at exactly that speed until you reached your destination, you would do very little work. Unfortunately, you probably speed up and slow down many times, each time doing work to change your momentum. Some say that you also raise and lower your centre of gravity through each step cycle, gaining and losing vertical momentum, which for non-angels must average out at zero over the trip and not contribute to your average statistic.

 

[paulb203]

"OK. But the question should have given the displacement as 0.9miles east."

Ah, I remember now, thanks. Displacement puzzles me. I get the general idea, I think. Like you said, the distance could be 0.9mi, but the road could be bendy, and the A to B straight line distance (the displacement) could be 0.5mi.
But, for example, when it comes to a round trip, like, to the shops and back, where the distance might be 100m, but the displacement is zero, because you end up back where you started; where you're not actually dispaced (regards your starting point). I'm ok with this in itself, but when it leads to our velocity therefore being zero, that's when I get puzzled.
I guess it's getting used to the idea that velocity = displacement/time, not distance/time. Not being a physicist, or physics student (until now), I always saw velocity in layman's terms (a fancy word for speed) :)

For example if the shop is 0.5miles east of your starting point the required displacement is 0.5miles east. But the road might be bendy (total distance 0.9 miles). In this case the average speed is 3.6mph but the average velocity is 2 mph east.


"Aagh! No! If you want momentum in units of kgm/s (kilogram.metres/second), you must convert 3.6 miles per hour to metres per second. (It's unfortunate that miles and metres have the same symbol.)"

Ah, of course. So, 3.6mph ≈ 1.6m/s, therefore, p=90(1.6)=144 kg m/s

"If you do the calculation correctly, it would be the magnitude of your average momentum. Since momentum is a vector, you need to specify its direction."

Again; of course. I've noticed with questions online, GCSE and otherwise, they omit the direction part. Some websites stress the vector thing, then neglect in their questions/answers.

Can you give an example of a GCSE question about instantaneous momentum? I'm not clear what you mean.

They don't explicitly ask for instantaneous momentum, but they start with, for example, "A car is heading north at 30m/s..." And when I Googled, "Is momentum generally instantaneous?", a lot of the results said yes. And given that those type of questions didn't say anything else about the speed/velocity, it seemed to me that they were implicitly after the instantaneous velocity.

"No. The faster the momentum changes, the greater the force. Read about Newton's second law of motion if interested"

Ah, I see. I've now had a look at his. So crumple zones etc in cars slow the change down, therefore reducing the force.
Side question. Is the F in F=ma generally shorthand for average force?

*
Thanks a lot for such a comprehensive reply. Much appreciated.

 

[paulb203]

Coming back to the question of instantaneous momentum. I'd have thought, that's the only sort of momentum there is. The mathematically minded can calculate a statistic called the average momentum (presumably mean, though maybe median or mode), but does it have any significance?
As others have said, we are interested in changes in momentum because they require a force (or impulse). Momentum might help calculate that force or its duration, or for a given force, to calculate the change in velocity.
What could the statistic of average momentum usefully tell you? You talk about bumping into someone when you have a certain average momentum. Did that happen when you just happened to be instantaneously at the average speed, or was it at a moment when you were going faster, slower or even paused waiting to cross the road? (In which last case, it was his fault!) The average is of no use to you: you have to know the momentum at that moment. (Or, as others have said, you actually want the rate of change of momentum, but that will surely be related to the instantaneous momentum at the start of the collision.)
The average momentum statistics don't tell you much about the work you do on your walk. If you could accelerate to your average speed and remain at exactly that speed until you reached your destination, you would do very little work. Unfortunately, you probably speed up and slow down many times, each time doing work to change your momentum. Some say that you also raise and lower your centre of gravity through each step cycle, gaining and losing vertical momentum, which for non-angels must average out at zero over the trip and not contribute to your average statistic.

Is finding an obect's momentum (or more than one object's momentum) primarily with regard to potential collisions?
And 'non-angels' cracked me up, lol.

 

[Steve4Physics]

Displacement puzzles me.

Displacement is the straight line distance and direction from a start point to an end point. The route taken doesn't affect the displacement. So displacement is the 'overall' result of the movement.

But, for example, when it comes to a round trip, like, to the shops and back, where the distance might be 100m, but the displacement is zero, because you end up back where you started; where you're not actually dispaced (regards your starting point). I'm ok with this in itself, but when it leads to our velocity therefore being zero, that's when I get puzzled.

The average velocity for the whole trip would indeed be zero. Say you travel 100m east and 100m west. Your start and end positions are exactly the same as if you hadn't moved at all! The same as if your velocity had been zero for the whole time. So sayinh average velocity =0 makes sense.

I guess it's getting used to the idea that velocity = displacement/time, not distance/time. Not being a physicist, or physics student (until now), I always saw velocity in layman's terms (a fancy word for speed) :)

Yes. Same for 'acceleration. In physics this can also mean slowing down or changing direction, as well as speeding up.

Ah, of course. So, 3.6mph ≈ 1.6m/s, therefore, p=90(1.6)=144 kg m/s

Don't round too much in intermediate steps of a calculation. 3.6miles/h = 1.609m/s.
90x1.609 = 144.81 = 145kgm/s rounded.

Again; of course. I've noticed with questions online, GCSE and otherwise, they omit the direction part. Some websites stress the vector thing, then neglect in their questions/answers.

They don't explicitly ask for instantaneous momentum, but they start with, for example, "A car is heading north at 30m/s..." And when I Googled, "Is momentum generally instantaneous?", a lot of the results said yes. And given that those type of questions didn't say anything else about the speed/velocity, it seemed to me that they were implicitly after the instantaneous velocity.

They are typically giving the speed/velocity at the time of interest (e.g. just before an impact) so you can treat it as the instantaneous value at that moment.

Ah, I see. I've now had a look at his. So crumple zones etc in cars slow the change down, therefore reducing the force.

Yes.

Side question. Is the F in F=ma generally shorthand for average force?

F and a could both be instantaneous values or both be average values.

m is usually constant but doesn't have to be. E.g. consider a rocket firing its engines. The thrust, acceleration and mass will generally all be changing continuously. So (for accurate calculations) you need the instantanous values of each at every instant! (Fortunately calculus makes this calculation possible!.)

Thanks a lot for such a comprehensive reply. Much appreciated.

You're welcome!

 

[paulb203]

Thanks, Steve4Physics, for another great reply. Really helpful.