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AS-level physics rounding help?

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Sitting my AS physics in oct/nov, I was going through some past papers and in the mark scheme the values in the working are rounded to 2 or 3 SF and the end answer ends up being different to me. I usually leave the values as it is during working and then round to 3SF at the end (i've been doing this in school internals and I havn't been penalised at all)

What does cambridge want? Does my working have to be exact to its mark scheme??
I am referring to question 1D in W11/23 btw, I get 5.833N for T1 but the mark scheme says 5.7...

Any help is appreciated, Pls give me a hand!
 
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I think it should be exact in such cases, they do give a range for the answers as you can see they have mentioned a range for scale diagram in this question.

How did you solve?
Your answer is acceptable in case of using a scale diagram.

I am getting the exact same answer as marking scheme to three decimal places. Solve the equations and write the values to three decimal places for every working in this case. Even if you arrange the equations in sin and cos form, and solve at the end you should get the marking scheme answer.
 
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I think it should be exact in such cases, they do give a range for the answers as you can see they have mentioned a range for scale diagram in this question.

How did you solve?
Your answer is acceptable in case of using a scale diagram.

I am getting the exact same answer as marking scheme to three decimal places. Solve the equations and write the values to three decimal places for every working in this case. Even if you arrange the equations in sin and cos form, and solve at the end you should get the marking scheme answer.

Came back to this question the next morning, my working is wrong :)
Could you tell me the correct solution? I did derive T1Sin50 + T2sin40 - 7.5 = 0 but im stuck after that.

Cheers
 
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T1Sin50 + T2sin40 = 7.5
.766T1 +.643T2 = 7.5 ---> 1

T2Cos40 = T1Cos50
.766T2=.643T1
T2=.643T1/.766
T2=.839T1 ---> 2

Put T2 in equation 1.

.766T1+.643(.839T1) = 7.5
Solve for T1
 
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