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AQA Physics AS 2017 Unofficial mark shceme watch

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1. If you can remember your answers please post them here.

1. Particles
a. What is particle X, udd (1 mark)
Neutron
b. What is the interaction (1 mark)
Weak interaction
c. What group of particles does W- belong in? (1 mark)
W bosons
d. What is the most stable baryon? (1 mark)
Proton
e. What are the decay products of a muon?
muon neutrino, electron, anti-electron neutrino
f. Show how charge and baryon number are conserved in this interaction, making reference to quarks.
Neutron ---> Proton + electron + anti-electron neutrino
charge before: udd = +2/3 -1/3 -1/3 = 0
charge after: udu + electron + anti-electron neutrino = +2/3 -1/3 +2/3 -1 +0 = 0

Baryon number before: udd = +1/3+1/3+1/3 = 1
Baryon number after: udu + electron + anti-electron neutrino = +1/3+1/3+1/3+0+0 = 1

2. Electricity
a. Why is the graph of a filament lamp a curve with regard to the electron motion? (4 marks)
Resistance is very low at first, therefore there is a high current. As the voltage increases, the filament heats up, the electrons and ions vibrate more, this makes it more difficult for ions to move. Therefore the resistance increases and current is low.
b. resistance of the resistor
15 Ohms
c. What is the resistance of the filament in series to the resistor when the current is 0.18?
14.5
d. What is the resistance of the filament in parallel to the resistor when the EMF is 4V?
6.25

3.

4. Forces and Motion
a. Show the power output is 100W (2 marks)
Travels 12km at 1.5m/s, therefore in 8000seconds. Using Q = It, find the current. Then use P= IV. Power was 108W, which is approximately 100W
b.
c.
d. Resistive force
72N
e. How do these factors affect the maximum range? (4 marks)
Mass: higher mass decreases the range
Speed: higher velocity increases the range

5. Young's double slit

Maximum intensity. Why is the minimum frequency between C-D (minima) not 0?
Intensity is about the number of photons arriving per second, therefore does not affect frequency.

6. Stationary waves
a. Show that the extension of the wire is less than 4mm (2 marks)
Extension was 3.125mm
b. Calculate the necessary tension in the supporting ropes so that the there is no horizontal resultant force acting on the cable.
c. Calculate the frequency of a stationary wave formed on the cable when at the third harmonic.
Use the first harmonic equation
7.4 or 7.5
d. Drawing of third harmonic on wire
4 nodes including at the fixed points, 3 antinodes
e. Explain how stationary waves are formed on the cable when it is windy, and why only certain wavelengths are possible.
f. Why does the copper wire sag when there are strong winds?
Strong winds cause a larger amplitude of the waves, therefore a larger extension. The wires suffers from plastic deformation.

7. Fluorescent tube
a. Which is an evidence for the wave nature of an electron? (1 mark)
C - Diffraction rings
b. What is the state of 0? (1 mark)
Ionisation state
c. What is the state -13.7eV? (1 mark)
Ground state
d. Why are the values negative below 0?
e. Energy of photons with the lowest wavelength of light?
1.9ev
f. Why does the fluorescent tube need a high pd supply? How does the energy level diagram show the wavelengths of light in the spectrum. Which one wavelength corresponds to 2 energy levels? (6 marks)
A high pd supply means that the free electrons in tube become fast moving free electrons, therefore collide with mercury atoms. This causes excitation and ionisation. De-excitation emits UV photons. UV absorbed by atoms on the fluorescent coating, causing excitation. De-excitation of coating atoms emit visible light photons. The energy level diagrams show the de-excitation of the electrons in in the coating atoms. The quantum jump shows the energy of the photon emitted. This can be used to find the wavelength of the photon.

Total number of marks: 70
2. Mark scheme sorry typo
3. 1.
particle x was neutron
most stable baryon = proton
muon decay products = muon neutrino, electron, anti-electron neutrino
4. Question about energy of photons used the lowest wavelength of light and E=hc/Lamba in electron volts, 3.03eV I got.
5. (Original post by Echtebilbo)
Question about energy of photons used the lowest wavelength of light and E=hc/Lamba in electron volts, 3.03eV I got.
wouldnt it be the lowest ev value jump so 0.85-0.54=0.31eV

E=hf so the lower the energy in ev the lower the frequency will be
6. (Original post by MercrutioCaius)
wouldnt it be the lowest ev value jump so 0.85-0.54=0.31eV

E=hf so the lower the energy in ev the lower the frequency will be
i think it asked for the photon with the highest frequency that corresponded to one of the lines from the visible spectrum. I think the wavelength was 656 nm which gives a frequency of 4.57 * 10^14 Hz so the photon has an energy of 3.03 * 10^-19 J = 1.90 eV
7. Can we have some clarification on the following please:
"Which one wavelength corresponds to 2 energy levels?"
I think it was maybe just asking for any drop between one energy and another that produced a line on the spectrum, but I could be wrong? What exactly did this pointer want?

1b. Show how charge and baryon number are conserved in this interaction, making reference to quarks.
5. Also asked for calculation of wavelength/frequency (can't remember which)
6a. Show that the extension is below 4mm given the tension and Young's Modulus, along with other information.
6b. Calculate the necessary tension in the supporting ropes so that the there is no horizontal resultant force acting on the cable.
6c? Calculate the frequency of a stationary wave formed on the cable when at the third harmonic.
6d. Explain how stationary waves are formed on the cable when it is windy, and why only certain wavelengths are possible.
8. (Original post by MattB00)
i think it asked for the photon with the highest frequency that corresponded to one of the lines from the visible spectrum. I think the wavelength was 656 nm which gives a frequency of 4.57 * 10^14 Hz so the photon has an energy of 3.03 * 10^-19 J = 1.90 eV
I got 1.9 eV too
9. (Original post by zxcvbnm909)
If you can remember your answers please post them here.
I've​ tidied up the thread and someone has also fixed the title
10. What was the tension in the 12m copper wire?

and what did people get for the 3rd harmonic frequency?
11. (Original post by Doonesbury)
I've​ tidied up the thread and someone has also fixed the title
Thank you!!!
12. Last years grade boundaries for paper 1
a: 52
b: 47
c: 42
d: 38
13. Were you meant to half the tension when making the horizontal resultant zero, or was it not necessary? My assumption was that the tension pulls both side equally.
14. (Original post by zxcvbnm909)
Last years grade boundaries for paper 1
a: 52
b: 47
c: 42
d: 38
As the new exam is linear, "Subject Grade Boundaries" are actually used, which differ - each paper does not get a 'real' grade boundary.

You can find the boundaries here: http://filestore.aqa.org.uk/over/sta...-JUNE-2016.PDF

It's about 65% for an A, and that document shows the actual "Subject Grade Boundary" (used for determining your grade) along with boundaries for Paper 1 and Paper 2 (last year, paper two required one less mark than paper 1 for an A).

I know that grade boundaries usually go up, but this year, I think it'll only be by one or two marks due to the difficulty?
15. (Original post by console.log)
I know that grade boundaries usually go up, but this year, I think it'll only be by one or two marks due to the difficulty?
No they don't, they can vary upwards or downwards each year.
16. (Original post by zxcvbnm909)
f. Show how charge and baryon number are conserved in this interaction, making reference to quarks.
muon ----------- > muon neutrino + electron + anti-electron neutrino
Charge -1 0 -1 0
Baryon number 0 0 0 0
Wasn't this question referring to the diagram? My answer was:
Neutron ---> Proton + electron + anti-electron neutrino
charge before: udd = +2/3 -1/3 -1/3 = 0
charge after: udu + electron + anti-electron neutrino = +2/3 -1/3 +2/3 -1 +0 = 0

Baryon number before: udd = +1/3+1/3+1/3 = 1
Baryon number after: udu + electron + anti-electron neutrino = +1/3+1/3+1/3+0+0 = 1
17. (Original post by Helium2)
Wasn't this question referring to the diagram? My answer was:
Neutron ---> Proton + electron + anti-electron neutrino
charge before: udd = +2/3 -1/3 -1/3 = 0
charge after: udu + electron + anti-electron neutrino = +2/3 -1/3 +2/3 -1 +0 = 0

Baryon number before: udd = +1/3+1/3+1/3 = 1
Baryon number after: udu + electron + anti-electron neutrino = +1/3+1/3+1/3+0+0 = 1
Oh yes! Mixed it up. I'll fix it now
18. (Original post by zxcvbnm909)
Oh yes! Mixed it up. I'll fix it now
THANK GOD! YOU GAVE ME A HEART ATTACK!

I was so scared that in this horrible paper I lost some easy marks.
19. Ahh I didn't read that you had to reference quakrs As if I lost these easy marks!!
20. 6d: Might be because copper wire has a specific length (12m), so each subsequent harmonic has a specific wavelength, hence specific frequencies.
6e: Large amplitude->Large extension
Suffers plastic deformation
7b: I just wrote 'ionisation state'.
*7d: I read somewhere that the maximum gravitational potential energy is 0J.

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