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Edexcel Physics IGCSE 1P Unofficial Mark Scheme 25th May 2016

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Tell me the questions which I'm missing out - I'll add them in. Please try your best to remember the rough question number. Thanks for all the contributions, including Advait10.

Do any of you remember which question they gave you a figure of 250000km or something like that. I'm trying to think of it but I can't remember. Tag me if you can remember it.

Do the poll!

1.a)
Constant velocity with horizontal line. (1)

1.b)
Distance travelled is area under line. (1)

1.c)
Average velocity is distance moved / time taken. (1)

2.a)
Coolant to reduce energy.Shielding to absorb radiation.Fuel rods for Uranium fuel. (2)

2.b)
Kinetic energy. (1)

2.c)
The purpose of the moderator is to slow down fast moving neutrons to become slow moving thermal neutrons. This makes it easier for neighbouring uranium nuclei to absorb the neutrons and hence go under fission. (2)

2.d)
A controlled chain reaction where only 1 out of 3 (on average) of the neutrons emitted through fission are absorbed by neighbouring uranium nuclei. This is helped by using boron control rods to absorb these remaining neutrons. This prevents an uncontrolled chain reaction. The neutron which is absorbed by a uranium nucleus becomes unstable hence splits into 2 daughter nuclei and 2-3 fast moving neutrons. This process keeps repeating. (3)

2.e)
Draw week 6 position of complete. (1)

2.f)
X between comet and Earth. (1)

2.g)
The distance from 1-3 is is roughly the same. 3-4 is increase in displacement. 4-5 distance decreases. Time is the same so change in displacement = change in speed as s = d/t.

2.h)
Orbits are closet to each other at 9. (1)

2.g)
Orbital speed = 150,000,000 * 2 * pi = 94247796.
1km365 * 24 = 8760so divide the first by the second to get 107588.7895km/hroughly equal to 108000km/h. (3)

3.a)
Pressure is P1V1 = P2V2 so: (101*1700)/12 = 14308.3kPa. (3)

3.b)
Pressure with depth = height x density x g. (1)

3.c)
113.08kPa. (3)

3.d)
113.08 + 101 = 214.08kPa. (1)

3.e)
Bubbles get larger as they rise due to the decrease in pressure with the decrease in depth as pressure = hpg, so air expands into larger volume. Also, smaller bubbles join together to form larger bubbles. (2)

4.a)
Gamma for tracing.X-rays for internal objects.Microwaves for food. (3)

4.b)
Particles carrying the energy (vibrations) travel perpendicular to the direction of the wave in motion. (1)

4.c)
You have to half the time because the time measured was for the wave travelling to the point of reflection and back as the sound wave travelled double the distance (to what you are measuring) so to get the depth from the point of emission to reflection, the time is divided by 2. (2)

4.d)
As v = frequency x wavelength, upon entering a denser medium the speed of sound increases as the wavelength increases. (2)

5.a)
Draw voltmeter parallel to LDR. (1)

6.a)
DC current is current that flows in one direction. (1)

6.b)
You can measure resistance by taking voltmeter reading (ensure it is constant) and ammeter reading and then do R = V/I (Ohms). (2)

6.c)
Graph. (4)

6.b)
The purpose for the split ring commutator and carbon brushes is so that once the coil goes past the half turn, when the coil retouches the commutator the direction of the current reverses hence magnetic field hence the effect of the force on the coil, ensuring it travels unidirectionally. (3)

6.c)
The coil continues to rotate clockwise. (1)

7.a)
Refractive index in glass is higher than air. YES
If i = 0, the ray doesn't deviate. YES
All rays that enter are totally internally reflected. NO (4)

7.b)
Draw point on graph. (1)

7.c)
It fits with correlation and gradient of graph as point is plotted above point following. (1)

7.d)
Refractive index can't be 0, only <1. (1)
Critical angle can't be 0. (1)

7.e)
Critical angle was 41.82 so rounds to 42. (3)

7.f)
Refractive index is the simplest ratio between speed of light in air: speed of light in material medium. (2)

7.g)
Draw rays and arrows. (4)

8.a)
Ruler. (1)

8.b)
Hooke's law: Measure length of spring using ruler. Add 1N mass, measure new length, subtract original length from this to get extension. Repeat this process for a range of masses. Repeat whole experiment x3 and average for increased reliability. Plot graph of extension vs load, if follows Hooke's law, the gradient line between points will be contrast as F = ke (extension is proportional to the load). (5)

9.a)
5.3 cm. (2)

9.b)
5.12cm (1)

9.c)
Differing results.
String measurements are to 0.1 cm.
Digital measurements are to 2 decimal places.
String measurements are averaged.
Digital only takes 1 measurement.
String measurement subject to parallax error due to human judgement.
Calliper more accurate.
Calliper measures diameter.
String measures circumference. (4)

10.a)
P = IV (1)

10.b)
Max power = (4000*600)/1000000 = 2.4 MJ. (2)

10.c)
Work done = Force x distance

10.d)
76000000J of energy. (3)

10.d)
35.02 seconds. (2)

10.e)
Decreasing power increases time taken for the same amount of work done. (1)

11.a)
KE = 1/2 x m x v(squared). (1)

11.b)
Velocity = 12m/s. (3)

11.c)
GPE = mxgxh. (1)

11.d)
GPE is gained as the lift moves up from platform to surface hence increase in height hence increase in GPE. (2)

11.e)
Slope up causes transfer of KE to GPE hence decrease in velocity hence deceleration hence less work required by brakes. Slope down converts GPE into KE hence increase in speed hence acceleration hence less work required by motor. Energy is conserved. (4)

12.a)
Sky has lowest temperature and snow and bear have same temperature which is warmer. (2)

12.b)
Hair acts as an insulator by trapping in a pocket of air, hence preventing heat loss via convection as convection requires for the movement of gas particles. (2)

12.c)
Snow was best at reflecting, worst at absorbing. Sky and polar bear good at absorbing, bad at reflecting. (2)

12.d)
UV is not reflected by the sky, it is absorbed. (1)

12.e)
Hence with white fur on top, it increases the surface area of the fur exposed to the air, minimising loss of thermal energy via radiation. By having the black skin underneath, it limits heat loss via radiation and any heat that is radiated is reflected back by fur hence retaining heat. (4)

12.f)
UV waves don't reach skin as they are good absorbers, bad reflectors hence do not undergo total internal reflection hence doesn't reach skin. (3)
as far as i remember .. in the last question they stated that the polar bear hair is like a tube with air in it.. so why tir dont take place..so i wrote that the medium is same for both .. for TIR to happen light must go from more dense to less dense medium... hair tubes were filled with air....

hope i m right
2. (Original post by BOBQ)
The hairs weren't dense enough to slow down the uv enough for total internal reflection to occur

(I think)
I asked my tuition teacher he said that TIR does happen on the edges of the hollow tube….there is a small gap between the edge and the hair so TIR takes place there…..however the question ask why it doesnt reach the skin so the answer is absorption…..furthermore to prove this remember they asked us about the image of the UV…..the polar bear was black suggesting it doesnt radiate UV but it is absorbed by the hair…… so its absorption…..plus its 2 marks nothing much to worry about…
3. (Original post by pro567)
I asked my tuition teacher he said that TIR does happen on the edges of the hollow tube….there is a small gap between the edge and the hair so TIR takes place there…..however the question ask why it doesnt reach the skin so the answer is absorption…..furthermore to prove this remember they asked us about the image of the UV…..the polar bear was black suggesting it doesnt radiate UV but it is absorbed by the hair…… so its absorption…..plus its 2 marks nothing much to worry about…
Well your tutor is defying the laws of physics, TIR cannot take place from a low refractive to high refractive index medium, not unless n < 0, which it is safe to say polar bear fur isn't. That was an IR camera, not UV. No creatures skin radiates substantially in UV, to do so it would have to be too hot.
4. Hey I need help please, For the question about motors, the split ring commutators and the brushes, I just put that they prevent the wire from tangling up and switch the direction of the current every half turn so the motor spins in one direction. Did I have to put exactly what the commutator and the brushes did independently or would what I wrote be correct? Thanks.
5. Hey I need help please, For the question about motors, the split ring commutators and the brushes, I just put that they prevent the wire from tangling up and switch the direction of the current every half turn so the motor spins in one direction. Did I have to put exactly what the commutator and the brushes did independently or would what I wrote be correct? Thanks.
(Original post by Bede735)
Well your tutor is defying the laws of physics, TIR cannot take place from a low refractive to high refractive index medium, not unless n < 0, which it is safe to say polar bear fur isn't. That was an IR camera, not UV. No creatures skin radiates substantially in UV, to do so it would have to be too hot.
Thanks a lot! It was stage 9.
Sorry, could you explain why?
7. (Original post by da lesh)
Hey I need help please, For the question about motors, the split ring commutators and the brushes, I just put that they prevent the wire from tangling up and switch the direction of the current every half turn so the motor spins in one direction. Did I have to put exactly what the commutator and the brushes did independently or would what I wrote be correct? Thanks.
I wouldn't dwell too much on the exam and concentrate on revision for paper 2. However, what you wrote is correct (although the order in which you wrote them would need to be the other way round if they are in the same direction as their use). I haven't seen a mark scheme so I suppose it would depend on how many marks were available for the question.
8. (Original post by amkay)
Sorry, could you explain why?
You literally just got your ruler out, and at stage 9, the planet and comet were closest to each other.
9. It really was a hard exam. Physics was one of the subjects that I could have achieved an A* in, now we're hoping for an A!
10. (Original post by VaJJGear)
It really was a hard exam. Physics was one of the subjects that I could have achieved an A* in, now we're hoping for an A!
grade boundaries will be low though
11. I think it was a challenge and the questions were a bit unusual but after looking at this, I think I did better than I thought I did. Coming out of the exam though, I thought it was very brutal.
12. It was possibly the hardest exam I've ever done!
13. Mark scheme for 1PR?
14. IF u GUYZ WANT 1P paper . i have a soft copy of it on my PC.. dont ask me how i got it.. just i have it...TSR plz tell me it isnt illegal to share... if it is then i wont share ....
15. (Original post by Hsevras)
grade boundaries will be low though
hopefully they are low
16. any predicted topics for tomorrows exam?
17. Any predictions???
18. (Original post by 1000000)
Any predictions???
physics might be in there

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19. (Original post by pk789)
physics might be in there

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banta
20. across both papers I think I got 127-134. with grade boundaries in mind do you think that will be an A

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