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Mathematics: Post your doubts here!
- Thread starter XPFMember
- Start date
1. I'm taking P1 and P2 in this May&June exam, and I'll take M1 and S1 at 2018 October&November session, and take P3 at 2019 May&June session. Is this possible? And if so, is my score of P2 is invalid for final Mathematics grade?
2. If I can't get satisfied in my test result, can I re-take only one component of the subject at the next exam session? (For example, if I get not good score in practical this time, can I take only practical at upcoming October?)
3. I'm taking Physics and Mathematics AS level this May&June, but not really prepared for the Chemistry, so I want to take Chemistry AS at this October&November session. And take rests of A2 components in next year. Can I take like this?
It's bit longer than I expected, but I'll be grateful if you reply me. Thx
2. If I can't get satisfied in my test result, can I re-take only one component of the subject at the next exam session? (For example, if I get not good score in practical this time, can I take only practical at upcoming October?)
3. I'm taking Physics and Mathematics AS level this May&June, but not really prepared for the Chemistry, so I want to take Chemistry AS at this October&November session. And take rests of A2 components in next year. Can I take like this?
It's bit longer than I expected, but I'll be grateful if you reply me. Thx
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i dont think you can repeat one component (only available for edexcel i guess)....and about the M1 S1 P2 issue....i dont know why you did so, i know that we should take P1 and one paper (either M1 S1 P2).
I didn't know at that time... P2 test is tmr and I can't cancel it.. My darn school didn't tell me anything about it.. Do you think I can take P1,P2 for AS and M1,S1,P3 for A2? Will I get any penalties? I hope I won't...
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i dont find going for two extra papers in A2 necessary (M1 and S1)....i'd recommend you continue as you are and take M1 next year (if you're good with physics) or S1 (if you dislike physics) but i dont find taking both necessary. And again i prefer you consult a professional in your area (am just a student as you are.)
I'll have a consult, anyway thank you
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You're saying that like you have a lawyer lol
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Can anyone post m18 p42 would be really appreciated
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Here ya go fam.
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Tysm fam
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Okay, in part one, since the car is travelling with constant speed 60 and we need to find power we apply the formula P=FxV. You have your V=60, and since the resistance is 2100(given in question as 35v, meaning 35x60) your F would be 2100 too as the car is travelling at constant speed(you can check this by apply F=ma and since there is no acceleration F-2100=0 giving F=2100). So now just multiply 2100 and 60 to get 126000W, your max power of car.
Part two: Use the formula they gave in the question, 35v to calculate your resistance(which is 1050N in this part). As they are asking you for the maximum acceleration, you would use the maximum power which you found in the previous part, 126000W to find the driving force of the car using P=FxV. 126000=Fx30, so driving force is 4200N. This is the greatest possible driving force, and now you apply F=ma to find the acceleration: 4200-1050=1200(a). And you get acceleration=2.625... TA DAAAAA
Part three: In this case you don't know the constant speed at which the car is travelling as u need to find it(duhhh). So you apply P=FxV first to get an expression for driving force(DF) of the car. Since the question says find the greatest speed of the car, use the max power of the car which we found in part one, 126000W. 126000=DFxV. DF=126000/V. Now use the expression of resistance as 35V as specified by the question and remember to take the component of weight of the car down the slope. So now the equation is DF-resistance-weight component down the slope=0(it is equals zero because the car has a constant speed so no acceleration, applying newton's 3rd law here). Just plug the expressions into the equation i mentioned and kabooom you get a quadratic which gives you v=40.
Tension P1X is 5N and tesion P2X is WN. Then you can draw a vector triangle with 7.3N. Using Pythagoras theory, you can obtain W. And use trigonometry to find angle AP1X. I got W=4.8N and angle=41.1 degree. Hope it's right.
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The tension in the string is AX is the weight of A, 5.5N, and the tension XC is W. Take the X and Y components of each force and the sum of the X components and Y components is equal to zero since the system is in equilibrium. Solve these 2 equations to get W. You can now draw a right angle triangle of AP1X containing the forces around that area and find the angleAP1X.
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Can someone help me in part (ii)
oct/2017/42
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How do we do Question 4 of the March 2018 mechanics paper?
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did anyone give mechanics 42 today how was it