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# F=ma question

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1. F=ma question
I am stuck on the following question and how to work out the two unknowns. If anyone could point me in the right direction, i would be grateful!

'An experimenter wishes to calculate the mass of an object and sets up an experiment to do so. She makes measurements of the force F ( in Newtons) and acceleration a (in ms-2) and plots a graph, from which a straight line results. She measures the gradient of the line to be 4.51.

Assuming that the equation F = 3.79p.m.a is valid, where p = 2.90, calculate the value of m which she measures. [7 marks]'
2. Re: F=ma question
This question seems poorly worded to me. Is this literally how it is written? No diagram? Are we to assume that force is plotted on the x axis, since it's presumably being applied and acceleration measured? The question states they are both "measured"... strikes me as a somewhat odd experiment!

Anyway, I'm guessing the two bits of information are the relationship between a and F in para 1 (a = 4.51F presumably) and the equation at the bottom F = 10.99ma.
3. Re: F=ma question
(Original post by Pkysam)
This question seems poorly worded to me. Is this literally how it is written? No diagram? Are we to assume that force is plotted on the x axis, since it's presumably being applied and acceleration measured? The question states they are both "measured"... strikes me as a somewhat odd experiment!

Anyway, I'm guessing the two bits of information are the relationship between a and F in para 1 (a = 4.51F presumably) and the equation at the bottom F = 10.99ma.
Yes. The question is exactly copied, and is from the previous years exam, we have no diagrams, just the question. Hence you can probably see why I am a bit confused. But yes I guessed to have to assume that force is plotted on the X axis.
4. Re: F=ma question
In that case, do you now have the answer? Can I ask what the paper was?
5. Re: F=ma question
(Original post by Pkysam)
In that case, do you now have the answer? Can I ask what the paper was?
I'm struggling to find one if I am honest!
6. Re: F=ma question
(Original post by MrBlackwood)
I am stuck on the following question and how to work out the two unknowns. If anyone could point me in the right direction, i would be grateful!

'An experimenter wishes to calculate the mass of an object and sets up an experiment to do so. She makes measurements of the force F ( in Newtons) and acceleration a (in ms-2) and plots a graph, from which a straight line results. She measures the gradient of the line to be 4.51.

Assuming that the equation F = 3.79p.m.a is valid, where p = 2.90, calculate the value of m which she measures. [7 marks]'
The gradient of the line is F/a = 4.51.
p = 2.90

F = 3.79 x p x m x a
F/a = 3.79 x 2.90 x m

4.51 = 3.79 x 2.90 x m

m = 0.410 kg to 3.s.f.

Edit: The Force *must* have been on the y-axis, as if it was on the x, this would mean the gradient was a/F = a/ma = 1/m, which is not a straight line, but rather a reciprocal function. So F/a is definitely 4.51

Lovely question actually
Last edited by The Polymath; 06-05-2012 at 19:37.
7. Re: F=ma question
(Original post by Junaid96)
The gradient of the line is F/a = 4.51.
p = 2.90

F = 3.79 x p x m x a
F/a = 3.79 x 2.90 x m

4.51 = 3.79 x 2.90 x m

m = 0.410 kg to 3.s.f.

Edit: The Force *must* have been on the y-axis, as if it was on the x, this would mean the gradient was a/F = a/ma = 1/m, which is not a straight line, but rather a reciprocal function. So F/a is definitely 4.51

Lovely question actually
THANK YOU!

It was the f/a = M that I forgot! Minor brain freeze. I was like, how can you have two unknowns and be required to work out an exact answer!
8. Re: F=ma question
(Original post by MrBlackwood)
THANK YOU!

It was the f/a = M that I forgot! Minor brain freeze. I was like, how can you have two unknowns and be required to work out an exact answer!

If it was a 7 mark question, do you reckon you would have had to have stated (lovely tense there, dunno what you'd call it) the thing about it being F/a and not a/F.
9. Re: F=ma question
(Original post by Junaid96)
The gradient of the line is F/a = 4.51.
p = 2.90

F = 3.79 x p x m x a
F/a = 3.79 x 2.90 x m

4.51 = 3.79 x 2.90 x m

m = 0.410 kg to 3.s.f.

Edit: The Force *must* have been on the y-axis, as if it was on the x, this would mean the gradient was a/F = a/ma = 1/m, which is not a straight line, but rather a reciprocal function. So F/a is definitely 4.51

Lovely question actually
Hang on... if the gradient is 1/m, which is a constant (since m is a mass and not changing) then this is still a straight line and not reciprocal at all surely?
10. Re: F=ma question
(Original post by Pkysam)
Hang on... if the gradient is 1/m, which is a constant (since m is a mass and not changing) then this is still a straight line and not reciprocal at all surely?
You may be right.. Might have jumped too quickly to the reciprocal argument upon seeing the 1/m.
11. Re: F=ma question
Sorry to be the bearer of bad news! I stand by my assertion that it's a bad question!

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Last updated: May 6, 2012
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