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

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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]subintootherequation:[br][br]x2+4(kx5)=10[br]x2+45x+45k10=0[br]thensinceitisatangent,thereisonepointofintersectionsojustoneroothenceb24ac=0[br][br](45)24(1)(45k10)=0[br]sok=12.7[br][br]5y = k-x so y = \dfrac{k-x}{5}[br][br]sub in to other equation:[br][br]x^{2} + 4(\dfrac{k-x}{5}) = 10[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][br](\dfrac{-4}{5})^{2} - 4(1)(\dfrac{4}{5}k-10) = 0[br]so k = -12.7[br]


Wasn’t the gradient of the exponential -k75e^-kt?
Reply 21
Original post by Stressedoutxxx
Wasn’t the gradient of the exponential -k75e^-kt?


There are two ways of finding the gradient of an exponential of the form y = ae^kx, one is ake^kx, the other is just ky. Since the initial gradient is 15, y must be 75 when t is 0 (I ignored the 25 for this part, I'm not sure whether I should have or not) so 75k = 15 and k = 0.2.
Original post by rikkiardo
There are two ways of finding the gradient of an exponential of the form y = ae^kx, one is ake^kx, the other is just ky. Since the initial gradient is 15, y must be 75 when t is 0 (I ignored the 25 for this part, I'm not sure whether I should have or not) so 75k = 15 and k = 0.2.

I got -75k= 15 for some reason so got -0.2 as k :frown:(
That was overall a very easy exam, however I'm not too sure on the k = -12.7 question, I got -30.25 and that gave repeated roots, so Idk
Reply 24
Original post by Stressedoutxxx
I got -75k= 15 for some reason so got -0.2 as k :frown:(


The exponential is decreasing so the power has to be negative. I think the minus was already in the equation they gave us so if k is negative, the power becomes positive because you get -(-0.2) = 0.2 but if k is 0.2 you get -(0.2) = -0.2. I completely forgot about that though so I think I was a bit lucky there.
For the exponentials question i got 0.2 but in more exact form, -ln 0.8
Reply 26
Original post by thotproduct
For the exponentials question i got 0.2 but in more exact form, -ln 0.8


What method did you use for those two questions? I'm still pretty unsure about the exponential one.
Original post by rikkiardo
What method did you use for those two questions? I'm still pretty unsure about the exponential one.


I can't even remember the equations anymore but instead of doing ky I just compared by doing T(1), little marks anyway so it won't be that deep.
Reply 28
Original post by thotproduct
I can't even remember the equations anymore but instead of doing ky I just compared by doing T(1), little marks anyway so it won't be that deep.


Ah right. The question I thought was a bit weird was the pulleys one. It asked for the distance travelled by the 3kg mass before the string would be taut again. I thought the answer was twice the distance the mass went up since it had to come back down to the same place for the string to be taut again right? So I multiplied my answer by two.
Original post by rikkiardo
Ah right. The question I thought was a bit weird was the pulleys one. It asked for the distance travelled by the 3kg mass before the string would be taut again. I thought the answer was twice the distance the mass went up since it had to come back down to the same place for the string to be taut again right? So I multiplied my answer by two.


Yeah, I had similar thoughts to you on that one, the first part was light work but I wasn't too sure what it was trying to ask in the second part? I just did something like that to get some sort of legible answer. Anyway the kinematics question last was very nice, almost as nice as the 9 mark integration Q tbh.
Reply 30
Original post by thotproduct
Yeah, I had similar thoughts to you on that one, the first part was light work but I wasn't too sure what it was trying to ask in the second part? I just did something like that to get some sort of legible answer. Anyway the kinematics question last was very nice, almost as nice as the 9 mark integration Q tbh.


Not as nice as the seven mark dimensional analysis question where everything equals 1 yesterday though haha. (assuming you do further mechanics and not one of the other modules)
(edited 5 years ago)
Original post by rikkiardo
Not as nice as the seven mark dimensional analysis question where everything equals 1 yesterday though haha. (assuming you do further mechanics and not one of the other modules)


That Further Mechanics exam was quite slick actually, I clocked it was a SUVAT equation early on and just wrote it all down, lmao in the follow up question i just wrote k = 1/2 as soon as.
What do you guys think will be the grade boundaries? Surely it won’t be 80% for an A?
Reply 33
Definitely not. The new course is meant to be tougher so probably around 75% was what my teachers were saying.
Original post by Stressedoutxxx
What do you guys think will be the grade boundaries? Surely it won’t be 80% for an A?
Also what did everyone get as the value of a for the last are under a curve question? I did all the working and ended up with a quadratic in disguise but then I kept getting a value of a which was lower than 8?
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]

[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]


How many marks could i get if I got the equation but didnt do the discriminant?
Reply 36
Yes, I got a quadratic too but rejected the one that was less than 8 (I think it was -2) and it left me with a = 27. I used the integral function on my calculator to sub it in and check and it was definitely right.

Original post by Stressedoutxxx
Also what did everyone get as the value of a for the last are under a curve question? I did all the working and ended up with a quadratic in disguise but then I kept getting a value of a which was lower than 8?
Original post by mandirapatelxo
What did you get for the integration q?


27 iirc
Original post by mandirapatelxo
how did you get that? Any chance you remember the method?


You are given a lower bound. You are given the function. You are given the area beneath the curve between the lower bound and a. You do what you would normally do with definite integrals, but doing it algebraically. You get a disguised quadratic in a^3/2 or something, you find two roots, you are told that a > 8, so reject the root does not give you a value of a above 8, the other value of a is 27, simple.
Reply 39
what other topics came up??

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