Aqa as physics 2018 unofficial mark scheme

Paper 1:
1.1: Strong nuclear force
Attractive for a short range (0.5-3 fm)
Exchange particle is a pion (3 marks)

1.2: alpha decay baryon number 1 on both sides (proton to neutron)
Beta decay Baryon number 1 on both sides (neutron to proton) (3 marks)

1.3: quark structure is uds (2 marks)

1.4: weak because strangeness isnt conserved (1 mark)

1.5: Need for collaboration so can carry out experiments and make results Reproducible and repeatable (2 marks)

2.1: speed of truck is 12.5 m/s (2 marks)

2.2: velocity changes
acceleration changes constantly
Gravity no longer acts on the horizontal part unlike on the first slope. (3 marks)

2.3: speed of truck is the same as GPE at the top is the same in both. Since air resistance and friction are negligible kinetic energy at the bottom is the same in both (1 mark)

2.4:Change of momentum qs as rain falls:
Horizontal and Vertical Momentum vectors are independent of each other. The Rain falling vertically loses all it's momentum and gives it to the Truck/planet earth - so vertical momentum is conserved.

If the truck is moving initially at constant velocity with constant mass then it has a set momentum. Rain falling into it increases the total mass of the system but momentum HAS to be conserved therefore the velocity of the truck has to reduce in order to compensate for the increase in mass and keep the total horizontal momentum equal before and after the rainfall. (3 marks)

3.1: diagonal set of ticks starting from top right (1 mark)

3.2: the photon has to have energy= to e1-e2
Energy of the photon completely absorbed by the atom
Whereas an electron just transfers its kinetic energy (3 marks)

3.3: transtion C would result in photoelectrons being emitted as frequency of the photon is higher than threshold frequency. Transition A and B wouldn't cause emission (3 marks)

3.4: photoelectron speed is 1.87*10^6 m/s
Find kinetic energy of photoelectron then find speed (3 marks)

4.1: total anticlockwise moment must be = to total clockwise moment
Line of action of the weight force must be in the base/ No resultant force.
(2 marks)

4.2: point where the entire mass of an object is thought to be (1 mark)

4.3: centre of mass vertically below A because the beam has come to rest in that position.
The centre of mass is vertically below the point where a suspended object comes to rest (2 marks)

4.4: tension worked out by resolving forces forming 2 simultaneous and solving

T1- 7221.8 N
T2- 9583.6 N (4 marks)

4.5: using young modulus= stress/ strain work out strain then usr strain= change in length/ original length.

Extension is 4.07*10^-3 m (3 marks)

5.1: weight of ice cube is 0.036 N (2 mark)

5.2: vol of water dipslaced by ice cube is
(0.036÷9.81) ÷1000= 3.7×10^-6 (1 mark)

5.3: volume of iron was 3.2*10^-8 (3 marks)

6.1:you can work out the resistance in each type of lamp (A and C are the same and B and D are the same) using P=V2/R for A and C I got 6ohms and B and D i got 3ohms. then you can work out the voltage dissapated in A and C (looking at the branches separately although theyll be the same because they're basically identical) so the voltage dissapated in A and C is 6V each (you do this by looking at the ratio of resistances) that leaves 3V each for B and D, this is below the voltage they need to be at full brightness so they are slghtly dimmed but A and C are at full brightness. (4 marks)

6.2: The brightness of the other 4 bulbs remains the same however bulb E doesn't light up. The PD across bulb E is 0 because 4.5V from either end cancel out. (3 marks)

6.3: Here we need to look at the total resistance in the circuit before and after C dies. Before the total resistance is 1/(1/9 1/9) so 4.5ohms whereas after C dies its 3 1/(1/6 1/12) = 7ohms. Because the total resistance has now increased the current across the battery must have decreased. (3 marks)

7.1: there will be a central maxima composed of white light with a spectra of colour fringes on either side of the maxima all adjacent to eachother. (2 marks)

7.2: red light will have wider fringe spacing than green light. There will be alternate red and green bright fringes on the screen with dark fringes between them.
Use a calculation to show this (I put in arbitary values into the eqt) (4 marks)

7.3: w=(lambda*D)/s
Decreasing slit sparation means bigger fringe spacing so interference pattern more spread. It also means uncertainty in the length of the slit is increased as you need a measuring instrument with a higher resolution in order to accurately measure it. As a reseult uncertainty in the wavelength increases.

As you decrease D the reading decreases, and resolution stays the same, this means the uncertainty increases.
The fringe width decreases, which therefore means the reading for the fringe width decreases, resolution is the same so uncertainty increases. (6 marks)

Because no one can be bothered to redo the paper and make a markscheme now?
You have the paper, sounds like you're volenteering to do it yourself

is there a paper 2 unofficial markscheme

this is paper 2 unofficial mark scheme.
I didn't make this myself btw.

Hey, this is for the AS 2018 paper 2?? Just wanting to make sure

No but momentum must always be conserved so velocity has to decrease.

hi do you have the mark scheme for paper 2?