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M3 edexcel Further dynamics

Please tell me the book is in error.

This is the question 6c and their answer. I am so rubbish at mechanics that I doubt everything I do but shouldn't the answer be zero?
Reply 1
Original post by maggiehodgson
Please tell me the book is in error.

This is the question 6c and their answer. I am so rubbish at mechanics that I doubt everything I do but shouldn't the answer be zero?


Sketch the graph - you'll want the absolute area.

Look at ghostwalkers post.
(edited 8 years ago)
Reply 2
Original post by maggiehodgson
Please tell me the book is in error.

This is the question 6c and their answer. I am so rubbish at mechanics that I doubt everything I do but shouldn't the answer be zero?


This force switches sign in the interval in question
it will move you backwards and forwards for equal length of time
(sketch the graph of cos4t)

MAGNITUDE of impulse is the total area which is 1 but the workings there are not correct
Need to integrate from 0 to pi/8 then pi/8 to pi/4

the analogy I can give you is the displacement is zero but the distance is not
As Zacken said.

Edit: :holmes: Just seen TeeEm's response now.
(edited 8 years ago)
Reply 4
Original post by maggiehodgson
Please tell me the book is in error.

This is the question 6c and their answer. I am so rubbish at mechanics that I doubt everything I do but shouldn't the answer be zero?


Won't let me attach the picture in my other post:

Reply 5
Original post by ghostwalker
As Zacken said.

Edit: :holmes: Just seen TeeEm's response now.


No reply for 35 minutes, then three replies all at once. The workings of this forum. :lol:
Original post by Zacken
No reply for 35 minutes, then three replies all at once. The workings of this forum. :lol:


Well short of the record.

Having looked at the question again, I think that they've used the term "force of magnitude F" erroneously, and it should just be "force of F Newtons"

My thinking is that magnitude of impulse is then just the straight integration from start of end; then take the modulus.

So, I still go with 0 as the answer.

If we're going to mean the area between the curve and the axis as always being positive regardless of whether it's above or below the axis, then we also need to revisit part a, where F is initially going negative.
(edited 8 years ago)
Reply 7
Original post by ghostwalker
x


I'll admit to not being well versed in this area, but I think I can see where you're coming from and I do agree with you. Afterall, we say the magnitude of the displacement is 0 even though it has a non-zero distance.

I'll defer to your expertise here and delete my post.
Original post by Zacken
I'll admit to not being well versed in this area, but I think I can see where you're coming from and I do agree with you. Afterall, we say the magnitude of the displacement is 0 even though it has a non-zero distance.


Displacement is a vector quantity, as is impulse.

The only issue with the question that I see is the use of the word magnitude when refering to the force. Which would imply the use of F(t)  dt\int |F(t)|\;dt, rather than F(x)  dt\int F(x)\;dt, and I think that's just an error in the wording of the question.



I'll defer to your expertise here and delete my post.


I claim no great expertise, and your post is fine, IMO.

Edit: Posts #2 and #4 together now look rather odd - verging on an infinite loop :smile:
Edit2: It's changed again!
(edited 8 years ago)
Original post by TeeEm
This force switches sign in the interval in question
it will move you backwards and forwards for equal length of time
(sketch the graph of cos4t)

MAGNITUDE of impulse is the total area which is 1 but the workings there are not correct
Need to integrate from 0 to pi/8 then pi/8 to pi/4

the analogy I can give you is the displacement is zero but the distance is not


THANKS.

Such a lot to learn.
Original post by ghostwalker
Well short of the record.

Having looked at the question again, I think that they've used the term "force of magnitude F" erroneously, and it should just be "force of F Newtons"

My thinking is that magnitude of impulse is then just the straight integration from start of end; then take the modulus.

So, I still go with 0 as the answer.

If we're going to mean the area between the curve and the axis as always being positive regardless of whether it's above or below the axis, then we also need to revisit part a, where F is initially going negative.


That's gone over my head for the moment but I will try to understand it in due course.
Original post by Zacken
Won't let me attach the picture in my other post:



This I get but I would never have thought of it.

Many thanks.
Reply 12
Original post by ghostwalker
Displacement is a vector quantity, as is impulse.

The only issue with the question that I see is the use of the word magnitude when refering to the force. Which would imply the use of F(t)  dt\int |F(t)|\;dt, rather than F(x)  dt\int F(x)\;dt, and I think that's just an error in the wording of the question.




I claim no great expertise, and your post is fine, IMO.

Edit: Posts #2 and #4 together now look rather odd - verging on an infinite loop :smile:


Okay - so, to clear this up (for myself and the OP):

It's normally magnitude of an impulse is given by t1t2Fdt\left| \displaystyle \int_{t_1}^{t_2} F \, \mathrm{d}t \right| where FF is the force.

However, if you're only given FF as being the magnitude of the force, then your impulse is t1t2Fdt\displaystyle \int_{t_1}^{t_2} |F|\,\mathrm{d}t

Is that it?
(edited 8 years ago)
Original post by Zacken
Okay - so, to clear this up (for myself and the OP):

It's normally magnitude of an impulse is given t1t2Fdt\displaystyle \int_{t_1}^{t_2} F \, \mathrm{d}t where FF is the force.

However, if you're only given FF as being the magnitude of the force, then your impulse is t1t2Fdt\displaystyle \int_{t_1}^{t_2} |F|\,\mathrm{d}t

Is that it?


IMO.

Caveat:
The first form gives the impulse. If you want its magnitude, drop any sign after integrating.

The second form is something I've never seen nor would I expect to, but is my attempt to interpret the "magnitude of force" statement, which I think is erroneously in the question.
(edited 8 years ago)
Reply 14
Original post by ghostwalker
IMO.

Caveat:
The first form gives the impulse. If you want its magnitude, drop the sign after integrating.

The second form is something I've never seen nor would I expect to, but is my attempt to interpret the "magnitude of force" statement, which I think is erroneously in the question.


Fair enough - we can chalk this down to improper textbook wording then.

(I'm going to edit in the modulus signs into my post)

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