A 'cap' means that for that paper, to achieve 100 UMS 75/75 raw marks will not be needed. For example getting 73-75 may get you 100 UMS. This happens when examiners feel that the paper was harder than previous years.
Just thought I’d make an unofficial mark scheme for the MM1B aqa paper. Let me know your opinions on this and whether you agree or disagree good luck everyone!
Just thought I’d make an unofficial mark scheme for the MM1B aqa paper. Let me know your opinions on this and whether you agree or disagree good luck everyone!
Terrific work on an unofficial mark scheme. Do you happen to have the paper?
Aaah I was wondering where this thread had gone - couldn't find it! I walked out of the exam almost trembling at the difficulty of it, my answers were all frantic scribbles and scattered with bits crossed out and desperate attempts to scrape by. Absolutely expected a C or D in it. Got an A! Around 85 ums I think.
First question asked you to work out R which is 520N. Second question asked you for the tension exerted on the trailer which is 1280N. Third question asked to state the tension exerted on the tractor which should be equal to the tension exerted on the trailer which is 1280N.
The full question was: 'A tractor, of mass 3500kg, is used to tow a trailer, of mass 2400kg, across a horizontal field. The trailer is connected to the tractor by a horizontal tow bar. As they move, a constant resistive force of 800 newtons acts on the trailer and a constant resistance force of R newtons acts on the tractor. A forward driving force of 2500 newtons acts on the tractor. The trailer and tractor accelerate at 0.2ms-2.
a) Find R.
b) Find the magnitude of the force that the tow bar exerts on the trailer.
c) State the magnitude that the tow bar exerts on the tractor.'
For part (a):
Using F=ma and taking the mass of both the tractor and the trailer,
Driving force - resistive forces = mass x acceleration
2500 - 800 - R = (2400 + 3500) x 0.2
Rearranging for R,
R = 2500 - 800 - (2400 + 3500) x 0.2 R = 520N
For part (b):
Using F=ma but focussing on only the trailer this time,
Forward force - resistive forces = mass x acceleration
T - 800 = 2400 x 0.2
Rearranging for T,
T = 800 + 2400 x 0.2 T = 1280N
For part (c):
The tension exerted by the tow bar on the tractor and trailer must be equal therefore the force exerted by the tow bar on the tractor is also 1280N.