The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

G484: Impulse (area under graph question)

Hi all,

For the Jan 2010 paper of OCR A, link - http://www.ocr.org.uk/Images/65273-question-paper-unit-g484-the-newtonian-world.pdf

on 1bi

when calculating the impulse I do so like this:

Impulse=Force*Time

milli=*10^-3

23 large squares, thrfr (23*25)=575 smaller squares.

Area of 1 mini square is 100*0.04=4units

Thrfr 575*4=2300

Thrfr 2300*(*10^-3) = 2.3 Ns

mark scheme says tolerance is 2.2Ns-2.4Ns.


My question is that as I had never seen a question ever like this, I attempted the question accordingly as shown below. Can anyone confirm if my method was right and if it merits the two marks.

Conversely this is a 2 mark question. Assuming one mark is for the realisation of milli being *10^-3 or calculating the area being one mark. It seems my method assuming its right is a lot of time taken for that particular part.

Alternatively is there another way of working it out. I assume most people would do triangle area, thrfr 0.5*2800*(1.2*10^-3) which isnt correct but warrants 1 mark from the mark scheme. The answer to said statement is 1.68.

I have spent a long time trying to figure out an another way to this question, and after going through the endeavour of sifting through past discussion one user said that he did:

1.60+2x


I have a mock soon and don't have opportunity to ask a teacher so can anyone shed light in how they would do the question, if my method was right and how to get the answer from said calculation above.

Thanks if you read all this!

Blair.
Reply 1
Avatar for EmmaBxoxo
EmmaBxoxo
OP
19
bump

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you