M2 - FRI Jan 21st (Edexcel) official thread!

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dieeiervonsatan
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#41
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Does anyone remember how many marks were there going for the car and trailor question (the 2 after the deceleration part - i think it was part c and d). Was it something like 2 marks for one and 4 for the other?
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dieeiervonsatan
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#42
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(Original post by Gaz031)
(b) 1.4 i think. You considered the car and caravan together, as tension cancels.
(c) 850N IIRC. You considered only the car and the forces acting on it while acceleration stays the same.


This was not simply the work done in stopping the vehicle. It was the work done by the brakes in stopping the vehicle. The resistances also did some of the work. I can't remember my answer but you multiplied the KE loss by the proportion of the constant force that was the brakes.
For the KE loss part, did you do 0.5 * (1000+1500)*25*25? and get something like 781 kJ?
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Gaz031
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#43
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Yes thats for the KE loss.
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dieeiervonsatan
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#44
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(Original post by Gaz031)
Yes thats for the KE loss.
K.. sorry to be annoying! But did you then have to find the distance travelled before it stopped by doing v^2 = u^2 + 2as, getting 0= 25^2 - 2*1.4*s? so s = something like 223 m ish?
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Gaz031
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#45
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(Original post by dieeiervonsatan)
K.. sorry to be annoying! But did you then have to find the distance travelled before it stopped by doing v^2 = u^2 + 2as, getting 0= 25^2 - 2*1.4*s? so s = something like 223 m ish?
That would have been a valid way to find out the work done by the brakes.
I presume you're finding the distance, then mutliplying this by the braking force to find the work done?

An alternative way (that i used) was to find the work done by all forces, then multiply this by the proportion of the forces that the brakes were. I think the brakes were something like 4/7ths [brake force/total forces] or somethig of the total forces and thus the same proportion of the total work done.


Hmm. I have a full bin of paper from my now useless January exam revision work. You might say i have a full bin combination....
Actually i must apologise for that terrible D1 related pun.
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Womble548
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#46
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Think I did ok except for 6c and d. Didnt know what thrust was and didnt know how to do work done. Got the distance but didnt know what force to multiply it by.

Also, for the guy who got an e value greater than 1. I did that the first time . You probably forgot that the direction of Q changes so in the impulse equation it was m(v+u)
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dieeiervonsatan
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#47
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(Original post by Gaz031)
That would have been a valid way to find out the work done by the brakes.
I presume you're finding the distance, then mutliplying this by the braking force to find the work done?

An alternative way (that i used) was to find the work done by all forces, then multiply this by the proportion of the forces that the brakes were. I think the brakes were something like 4/7ths [brake force/total forces] or somethig of the total forces and thus the same proportion of the total work done.


Hmm. I have a full bin of paper from my now useless January exam revision work. You might say i have a full bin combination....
Actually i must apologise for that terrible D1 related pun.
Something like that, I can't remember what I did after I found the distance. I tried all sorts of things coz I really didn't know what to do, so I figured I might pick up the odd mark or so for that part of the question. I reckon thats about 2 out of 4 for that bit. So not too bad generally then.
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Womble548
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#48
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I got the speed too Multiplied it by some random force

Also, what the hell is thrust? Never covered it
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Newton
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#49
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(Original post by Womble548)
Also, what the hell is thrust? Never covered it
The car and the caravan were connected by an inextensible tow-bar. The tension from the caravan was acting towards the right, which means that it was exerting a certain force on the car itself i. e. because that force was an external force acting on the car acting in the direction of motion, according to Newton's First Law it is "accelerating" the car.

That is what they want you to understand; thrust is tension that is not acting away from the direction of motion but in the direction of motion.

So, Tension=Thrust.

Newton.
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Womble548
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#50
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Never mind, I did the aceleration in the previous answer X a random force
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dieeiervonsatan
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#51
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(Original post by Newton)
The car and the caravan were connected by an inextensible tow-bar. The tension from the caravan was acting towards the right, which means that it was exerting a certain force on the car itself i. e. because that force was an external force acting on the car acting in the direction of motion, according to Newton's First Law it is "accelerating" the car.

That is what they want you to understand; thrust is tension that is not acting away from the direction of motion but in the direction of motion.

So, Tension=Thrust.

Newton.

So what was the answer then for that?
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BloodyValentine
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#52
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i got 850N for that by doing F=ma. This cheered me up everyone seems to have got pretty much the same results as me. It was a pretty good paper overall i had lots of time to just take my time and check everything
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danmint
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#53
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#53
does anyone have a link to the jan 05 m2 paper or can they sent it through email to me?
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Gaz031
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#54
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(Original post by danmint)
does anyone have a link to the jan 05 m2 paper or can they sent it through email to me?
We wrote in the question book so i find it unlikely that anyone has the question-answer booklet.
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Gregball_87
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#55
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Remember that CofR qu, where every1 got e > 2/3, i had 9e + 4 < 15e then went to 4 < 24e, so how many mark u reckon ill drop??? think it was a 5 mark part....
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