(Change of) momentum/ velocity

[alba]

What is the difference between velocity and change of velocity?. I have read that change of velocity is not acceleration, if this is true, what is it and what are its units? are they just m/s? is it just velocity?

1) If I push a bowl (m=1) at curling, and $\Delta v = 10m/s$ , haven't I accelerated it by 10m/s? and given it a change of momentum = 10 Kgm/s and Ke = 50 J?

2) Now, another more difficult aspect: I can push it by .5 or 1 m , but the change of velocity is the same, how do I reconcile it with work = 50 J in both cases?

3) the same with time, I can push it for .5 or 1 or 2 seconds, and work is still 50 J how do I reconcile it with impulse?

Why in one case we refer to work done and in another to change of momentum? Can I use every concept in every situation, or sometimes there is an obliged choice?

Thanks

 

[Nugatory]

What is the difference between velocity and change of velocity?. I have read that change of velocity is not acceleration...

Where did you read that? If you don't provide the source, we have no way of knowing what it was trying to say and therefore whether it's wrong or you misunderstood it.

 

[Dale]

What is the difference between velocity and change of velocity?

Hi alba, welcome to PF!

The change of velocity is just the final velocity minus the initial velocity. If you start at 30 m/s and end at 100 m/s then the change in velocity is 70 m/s.

. I have read that change of velocity is not acceleration, if this is true, what is it and what are its units? are they just m/s? is it just velocity?

The acceleration is a change in velocity divided by the change in time. So if I changed my velocity as above in 7 s then my acceleration is 10 m/s^2, but if I took 14 s to do the same change in velocity then my acceleration is only 5 m/s^2.

1) If I push a bowl (m=1) at curling, and $\Delta v = 10m/s$ , haven't I accelerated it by 10m/s? and given it a change of momentum = 10 Kgm/s and Ke = 50 J?

You can only calculate the acceleration if you know the time it took to push it. Your change in momentum and KE are correct assuming it started at rest.

2) Now, another more difficult aspect: I can push it by .5 or 1 m , but the change of velocity is the same, how do I reconcile it with work = 50 J in both cases?

You had to use a higher force over a shorter distance.

3) the same with time, I can push it for .5 or 1 or 2 seconds, and work is still 50 J how do I reconcile it with impulse?

You usually would try to reconcile impulse with the change in momentum, not KE.

Why in one case we refer to work done and in another to change of momentum? Can I use every concept in every situation, or sometimes there is an obliged choice?

I wouldn't say "obliged", but usually you will use change in momentum with impulse and change in KE with work.

 

[FactChecker]

I have read that change of velocity is not acceleration,

I think this distinction is to prepare you for the situation of an accelerating reference frame (rotating or accelerating linearly). These are non-Newtonian reference frames and a change of velocity measured in them is not the same as an acceleration due to a force.

 

[A.T.]

I think this distinction is to prepare you for the situation of an accelerating reference frame (rotating or accelerating linearly). These are non-Newtonian reference frames and a change of velocity measured in them is not the same as an acceleration due to a force.

Yes, it might be a "proper acceleration" vs. "coordinate acceleration" confusion.

 

[A.T.]

Yes, it might be a "proper acceleration" vs. "coordinate acceleration" confusion.

... or vs. "common language acceleration" (increase in speed, the opposite of "deceleration").

 

[alba]

... or vs. "common language acceleration" (increase in speed, the opposite of "deceleration").

I am beginning to understand: when you hop in your Ferrari you give it 'common language' and when you fall from a tree you get 'proper' acceleration?
I was about to reply to DaleSpam and tell him I understood that acceleration is only constant acceleration, was that conclusion wrong?

Can you expand a little on that in language accessible to a student?

 

[alba]

Yes, it might be a "proper acceleration" vs. "coordinate acceleration" confusion.

I couldn't edit my post: here http://en.wikipedia.org/wiki/Proper_acceleration wiki says these terms are used in SR, is it relevant with the Ferrari?
If I accelerate a car by 30 m/s in one second, change of velocity +30 divided by time 1, is it acceleration or change of velocity, ir it proper or coordinate?

 

[xAxis]

I don't think introducing "coordinate" acceleration at this point will help. Alba, read DaleSpams post No 3. Acceleration is simply difference between speeds of a body divided by the time in which that change took place.

 

[A.T.]

Can you expand a little on that in language accessible to a student?

Okay, I will try. The term "acceleration" can mean (at least) 3 different things:

1) In everyday language: "increase of speed over time", where "speed" is the magnitude of the velocity vector. This "acceleration" is a scalar value. It's opposite, a "decrease of speed over time", is called "deceleration".

2) In physics: "change of the velocity vector over time". This "coordinate acceleration" is a vector value. It depends on the chosen reference frame or coordinate chart.

3) In physics: "what an accelerometer measures". This "proper acceleration" is a vector value. It does not depend on the chosen reference frame and is the physical acceleration that you can actually "feel".

 

[alba]

Hi alba, welcome to PF!
The change of velocity is just the final velocity minus the initial velocity.
The acceleration is a change in velocity divided by the change in time.

Thanks, Dalespam, you are very kind (and you too have a nice avatar)

My Ferrari, (I don't know about yours), goes from 0 to 100 Km/h in ca. 3 seconds. That is roughly the acceleration of g.

after 3 seconds the
- change of velocity is (+30 - 0 = ) 30m/s = total increase (commonon lanuage acceleration)lan
- the average increase, unitary increase = physics acceleration is (30/3s) = 10 m/s²

But, if the acceleration lasts only 1 sec then the common language and the physics acceleration coincide.

Is this right? now suppose the car has mass 1 000 Kg

- change of momentum is the total increase of momentum = m*v = 30 000 Kg*m/s
- the average/unitary increase, the rate of change of momentum is (30000/3) = 10 000 kg*m/s

Is this right? I suppose there is no specific name for that.

Now, if I push a stone of 20Kg on a curling sheet (or a cue ball at finger billiard) I need a F of 20 N to give it an acceleration of 1 m/s, if I apply the F for 2 seconds the rock will have a net increase of v = 2m/s. P is 40 Kg*m/s, and its KE is = 40J

Supposing no friction, the rock_A will hit another rock_B with equal mass and ( in 1/10 s) give it all its v, p and Ke.,
If this is right, apart from typos and slips, can

- you show me how to integrate F on time and get J the impulse?
- explain the reasons why I should prefer J and change of momentum to Ke? or
- why should at all calculate J and the change of momentum in B when I know it is equal to P_A?

Thanks a lot

 

[FactChecker]

I have read that change of velocity is not acceleration

I just noticed one thing that I don't know if you really meant it this way or if anyone has commented on it already. Acceleration is a change of velocity per unit time. If a car goes from 0 mph to 30 mph in one second, that is fast acceleration. If it does that in 100 seconds, that is slow acceleration. So the velocity change is not the only factor in acceleration. You have to divide by the time it took for that change.

 

[DrGreg]

the rate of change of momentum is (30000/3) = 10 000 kg*m/s

I suppose there is no specific name for that.

Yes there is. Force.:)

 

[alba]

Thanks, Greg, of course m*s/t²

Can you explain what happens at curling, please?

 

[A.T.]

Can you explain what happens at curling, please?

 

[FactChecker]

2) Now, another more difficult aspect: I can push it by .5 or 1 m , but the change of velocity is the same, how do I reconcile it with work = 50 J in both cases?

You are doing the same work in less distance (force is greater but distance is shorter)

3) the same with time, I can push it for .5 or 1 or 2 seconds, and work is still 50 J how do I reconcile it with impulse?

Same work in less time. Force is greater but time is shorter.

Why in one case we refer to work done and in another to change of momentum? Can I use every concept in every situation, or sometimes there is an obliged choice?

You can apply a force against a spring. If you compress the spring for some distance and stop, then you have done work but nothing has gained momentum. In general the is more to work than just changes in momentum.

 

[alba]

Now, if I push a stone of 20Kg on a curling sheet (or a cue ball at finger billiard) I need a F of 20 N to give it an acceleration of 1 m/s, if I apply the F for 2 seconds the rock will have a net increase of v = 2m/s. P is 40 Kg*m/s, and its KE is = 40J

Supposing no friction, the rock_A will hit another rock_B with equal mass and ( in 1/10 s) give it all its v, p and Ke.,
If this is right, apart from typos and slips, can

- you show me how to integrate F on time and get J the impulse?
- explain the reasons why I should prefer J and change of momentum to Ke? or
- why should at all calculate J and the change of momentum in B when I know it is equal to P_A?
Thanks a lot

Thanks, I know the game of curling, I asked what happens at this particukar throw

 

[FactChecker]

- you show me how to integrate F on time and get J the impulse?

Impulse is related to the derivative of force (actually to the jerk = derivative of acceleration), not the integral.

 

[jbriggs444]

Impulse is the integral. It is often used when the magnitude of the force is large (and irrelevant) and its duration is brief (and also irrelevant).

 

[FactChecker]

Impulse is the integral. It is often used when the magnitude of the force is large (and irrelevant) and its duration is brief (and also irrelevant).

Yes . I stand corrected. Thanks.

 

[Dale]

Thanks, Dalespam, you are very kind (and you too have a nice avatar)

Thanks!

after 3 seconds the
- change of velocity is (+30 - 0 = ) 30m/s = total increase (commonon lanuage acceleration)lan
- the average increase, unitary increase = physics acceleration is (30/3s) = 10 m/s²

I would be very hesitant to ever use the "common language acceleration" meaning here on Physics Forums. One of the most challenging things about learning physics is understanding all of the precise technical meaning of ordinary words. Here, most people will consistently use the precise technical meaning and will assume that you are using it also. If you are not, you will cause unnecessary communication breakdowns.

But, if the acceleration lasts only 1 sec then the common language and the physics acceleration coincide.

If, for example, an acceleration of $a = 1.324 m/s^2$ lasts for 1 s then the change in velocity is $\Delta v = 1.324 m/s$. The number is the same, but the units are different. So even in that case it would be wrong to say that $a$ and $\Delta v$ coincide. The units being different means that the quantity is fundamentally different, and even the number being the same would not necessarily be true in other units.

Is this right? now suppose the car has mass 1 000 Kg

- change of momentum is the total increase of momentum = m*v = 30 000 Kg*m/s
- the average/unitary increase, the rate of change of momentum is (30000/3) = 10 000 kg*m/s

The change in momentum is $\Delta p = m \Delta v = 30000 kg\;m/s$. The average rate of change in momentum is $\Delta p/\Delta t = m \Delta v /\Delta t = 10000 kg\;m/s^2$. Note the square on the units for the rate of change.

Is this right? I suppose there is no specific name for that.

Actually, the specific name for the rate of change of momentum is the "net force".

Now, if I push a stone of 20Kg on a curling sheet (or a cue ball at finger billiard) I need a F of 20 N to give it an acceleration of 1 m/s, if I apply the F for 2 seconds the rock will have a net increase of v = 2m/s. P is 40 Kg*m/s, and its KE is = 40J

Supposing no friction, the rock_A will hit another rock_B with equal mass and ( in 1/10 s) give it all its v, p and Ke.,
If this is right, apart from typos and slips, can

- you show me how to integrate F on time and get J the impulse?

For a problem as simple as this you don't need to do an integration to determine the impulse. The impulse is simply equal to $\Delta p$. Since you have the initial and final momentum, calculating the change in momentum is just a subtraction.

- explain the reasons why I should prefer J and change of momentum to Ke?

If J is the impulse and p is the momentum then $J=\Delta p$. There is no such easy relationship between J and KE.

- why should at all calculate J and the change of momentum in B when I know it is equal to P_A

I agree. If you know the answer from conservation principles then performing a detailed calculation seems rather masochistic.

 

[alba]

For a problem as simple as this you don't need to do an integration to determine the impulse. The impulse is simply equal to $\Delta p$. Since you have the initial and final momentum, calculating the change in momentum is just a subtraction.

If J is the impulse and p is the momentum then $J=\Delta p$. There is no such easy relationship between J and KE.

I agree. If you know the answer from conservation principles then performing a detailed calculation seems rather masochistic.

Thanks, Dalespam, you have clarified almost all my doubts, I came here confused and then I was feeling slightly intoxicated.

You will greatly oblige me if you give me a detailed description of a case when you actually use/need J and the formula , do you just multiply F by the time or do you integrate?

 

[Dale]

You will greatly oblige me if you give me a detailed description of a case when you actually use/need J and the formula , do you just multiply F by the time or do you integrate?

In the scenario you gave $\Delta p$ was known, so you don't need the formula to calculate J. You would use the formula any time the force is known but $\Delta p$ is unknown.

Any time you need to calculate J it is the integral of F with respect to time. If F is constant then the integral is just F times the time, but that is just a simplification for constant force.

 

[alba]

In the scenario you gave $\Delta p$ was known, so you don't need the formula to calculate J. You would use the formula any time the force is known but $\Delta p$ is unknown.

Any time you need to calculate J it is the integral of F with respect to time. If F is constant then the integral is just F times the time, but that is just a simplification for constant force.

Thanks, I wish you could give an example (also in a link) when J is necessary, and show me exactly how it is used and how you integrate.Thanks.

 

[HallsofIvy]

I'm getting in here late so this may already have been said but "change in velocity" is NOT "acceleration". Acceleration is rate of change of velocity- it is the change in velocity divided by the change in time.