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OCR AS Maths (A) Pure and Mechanics - 23/05/18 [Exam Discussion]

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How was the exam?

Leave any thoughts, feelings and answers about the exam here. How do you think you did?

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I think I might move to Australia and never look at my grade
Reply 2
Not enough time to get it all done
I know!! I was rushing all of it so I couldn’t concentrate properly and get the right answer
Reply 4
Original post by Stressedoutxxx
I think I might move to Australia and never look at my grade

That's how I feel about my Further Mechanics exam. :frown:
:backstab:
I got the exponential one correct at first and then spent 20 more minutes on it and changed the value of A

Already lost 9 marks now

OH one more now

the pulley one, I got the answer and then crossed it out.

FINE 17 MARKS
(edited 5 years ago)
Reply 6
Original post by vivienne0402
:backstab:
I got the exponential one correct at first and then spent 20 more minutes on it and changed the value of A

Already lost 9 marks now

OH one more now

the pulley one, I got the answer and then crossed it out.

FINE 17 MARKS


What were the answers?
Original post by Thanu123
What were the answers?


a=100, k=-ln 0.6 and something... I can't remember
Original post by vivienne0402
a=100, k=-ln 0.6 and something... I can't remember

a was 75, not 100.

100 = 25+ae^-kt
t= 0 therefore
100 = 25+ae^0
100 = 25+ a*1
100= 25 + a
a= 75
Original post by Name Name123
a was 75, not 100.

100 = 25+ae^-kt
t= 0 therefore
100 = 25+ae^0
100 = 25+ a*1
100= 25 + a
a= 75


I got 75 too
Original post by Name Name123
a was 75, not 100.

100 = 25+ae^-kt
t= 0 therefore
100 = 25+ae^0
100 = 25+ a*1
100= 25 + a
a= 75


OMG
then i get it right now...
ALL my friends told me it was 100
yeah!!! thank you so much
does anyone remember the equations of the tangent one?
What did you get for the value of k for the exponential?
Original post by vivienne0402
does anyone remember the equations of the tangent one?


I think it was 5y + x=k and 4y +10 = x^2 ?
Original post by Thanu123
I think it was 5y + x=k and 4y +10 = x^2 ?


How did you find the answer to that?
Original post by Stressedoutxxx
How did you find the answer to that?

Dunno if this is right but it was my best guess..
5y= k-x
y= 1/5(k-x)
y= -1/5(x-k) (in the form of y-y1 = m(x=x1)) therefore gradient = -1/5

x^2+4y= 10
4y= 10-x^2
y= 10/4 - 1/4 x^2
dy/dx = -1/2x


-1/5 = -1/2x
x= -2/5
which also = k since k is the x coordinate of the intersection
Original post by Name Name123
Dunno if this is right but it was my best guess..
5y= k-x
y= 1/5(k-x)
y= -1/5(x-k) (in the form of y-y1 = m(x=x1)) therefore gradient = -1/5

x^2+4y= 10
4y= 10-x^2
y= 10/4 - 1/4 x^2
dy/dx = -1/2x


-1/5 = -1/2x
x= -2/5
which also = k since k is the x coordinate of the intersection


Guess i lost 5 marks then😧😊
Original post by Thanu123
Guess i lost 5 marks then😧😊

To be honest, I don't think what I did was correct.
Reply 18
For the exponential question what did you get for k?

Since ky = gradient, I got 75k = 15 so k = 0.2

For the tangent question, I got k = -12.7 by using the following method:

[br]5y=kxsoy=kx5[br][br]5y = k-x so y = \dfrac{k-x}{5}[br]

sub in to other equation:

[br]x2+4(kx5)=10[br][br]x^{2} + 4(\dfrac{k-x}{5}) = 10[br]

[br]x2+45x+45k10=0[br][br]x^{2} + -\dfrac{4}{5}x + \dfrac{4}{5}k -10 = 0[br]

then since it is a tangent, there is one point of intersection so just one root hence b^2 - 4ac = 0

[br](45)24(1)(45k10)=0[br][br](-\dfrac{4}{5})^{2} - 4(1)(\dfrac{4}{5}k-10) = 0[br]

[br]1625165k+40=0[br][br]\dfrac{16}{25} - \dfrac{16}{5}k + 40 = 0[br]

[br]165k=101625[br][br]-\dfrac{16}{5}k = -\dfrac{1016}{25}[br]

[br]k=12.7[br][br]k = 12.7[br]
(edited 5 years ago)
Original post by rikkiardo
For the exponential question what did you get for k?

Since ky = gradient, I got 75k = 15 so k = 0.2

For the tangent question, I got k = -12.7 by using the following method:

[br]5y=kxsoy=kx5[br][br]5y = k-x so y = \dfrac{k-x}{5}[br]

sub in to other equation:

[br]x2+4(kx5)=10[br][br]x^{2} + 4(\dfrac{k-x}{5}) = 10[br]

[br]x2+45x+45k10=0[br][br]x^{2} + -\dfrac{4}{5}x + \dfrac{4}{5}k -10 = 0[br]

then since it is a tangent, there is one point of intersection so just one root hence b^2 - 4ac = 0

[br](45)24(1)(45k10)=0[br][br](-\dfrac{4}{5})^{2} - 4(1)(\dfrac{4}{5}k-10) = 0[br]

so k = -12.7


I think this is the right solution, cheers.

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