# OCR GCSE Physics Gateway A 9-1 UNOFFICIAL MARK SCHEME

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#2

That was easy imo, basically told u all the equations, I’m set for a 9, what abt u guys?

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#3

I ****ed it up anyway

(Original post by

That was easy imo, basically told u all the equations, I’m set for a 9, what abt u guys?

**Time2Revise**)That was easy imo, basically told u all the equations, I’m set for a 9, what abt u guys?

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#4

I’m sure you did great mate, don’t worry too much about it, we still have paper 2 to go 😬

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#5

Was really easy but I ****ed up the 5 mark one about acceleration. Got 12.3 initially (which a lot of people got supposedly) using suvat formula but I now realize I should've rearranged the formula instead of just calculating the two as separate equations which is what confused me.

But I got my final answer by using the sine rule (knew it was crazy at the time) bcuz I drew a triangle and used sine rule to find hypotenuse (1.56m) and inserted into 2as to get 10.92m/s^2, I did this because from the wording of the question I assumed the distance was vertical not slanted height.

TL;DR: I had so much time left on my hands so I got a bit carried away with the acceleration question.

Terrified that the grade boundaries will be 84 for a 9 or something crazy.

But I got my final answer by using the sine rule (knew it was crazy at the time) bcuz I drew a triangle and used sine rule to find hypotenuse (1.56m) and inserted into 2as to get 10.92m/s^2, I did this because from the wording of the question I assumed the distance was vertical not slanted height.

TL;DR: I had so much time left on my hands so I got a bit carried away with the acceleration question.

Terrified that the grade boundaries will be 84 for a 9 or something crazy.

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#6

I did the same thinking 1m was the height so used sine rule to find the length of the ramp but I got 3.96 as my final answer???

(Original post by

Was really easy but I ****ed up the 5 mark one about acceleration. Got 12.3 initially (which a lot of people got supposedly) using suvat formula but I now realize I should've rearranged the formula instead of just calculating the two as separate equations which is what confused me.

But I got my final answer by using the sine rule (knew it was crazy at the time) bcuz I drew a triangle and used sine rule to find hypotenuse (1.56m) and inserted into 2as to get 10.92m/s^2, I did this because from the wording of the question I assumed the distance was vertical not slanted height.

TL;DR: I had so much time left on my hands so I got a bit carried away with the acceleration question.

Terrified that the grade boundaries will be 84 for a 9 or something crazy.

**Jeqen**)Was really easy but I ****ed up the 5 mark one about acceleration. Got 12.3 initially (which a lot of people got supposedly) using suvat formula but I now realize I should've rearranged the formula instead of just calculating the two as separate equations which is what confused me.

But I got my final answer by using the sine rule (knew it was crazy at the time) bcuz I drew a triangle and used sine rule to find hypotenuse (1.56m) and inserted into 2as to get 10.92m/s^2, I did this because from the wording of the question I assumed the distance was vertical not slanted height.

TL;DR: I had so much time left on my hands so I got a bit carried away with the acceleration question.

Terrified that the grade boundaries will be 84 for a 9 or something crazy.

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#7

What was the answer to the final question on about the pressure due to a column of liquid?

the pressure of the liquid itself i got was 5,500 Pa. However, i thought to add in Atmospheric pressure (100,00 Pa) due to the fact that pressure from the atmosphere is also pushing down on the water/liquid.

So my final answer was 105,500 Pa. Did u get this?? or anyone else??

(Original post by

That was easy imo, basically told u all the equations, I’m set for a 9, what abt u guys?

the pressure of the liquid itself i got was 5,500 Pa. However, i thought to add in Atmospheric pressure (100,00 Pa) due to the fact that pressure from the atmosphere is also pushing down on the water/liquid.

So my final answer was 105,500 Pa. Did u get this?? or anyone else??

**Time2Revise**)

That was easy imo, basically told u all the equations, I’m set for a 9, what abt u guys?

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#8

(Original post by

What was the answer to the final question on about the pressure due to a column of liquid?

the pressure of the liquid itself i got was 5,500 Pa. However, i thought to add in Atmospheric pressure (100,00 Pa) due to the fact that pressure from the atmosphere is also pushing down on the water/liquid.

So my final answer was 105,500 Pa. Did u get this?? or anyone else??

**Rayzzz_**)What was the answer to the final question on about the pressure due to a column of liquid?

the pressure of the liquid itself i got was 5,500 Pa. However, i thought to add in Atmospheric pressure (100,00 Pa) due to the fact that pressure from the atmosphere is also pushing down on the water/liquid.

So my final answer was 105,500 Pa. Did u get this?? or anyone else??

It was 5500 because there is no need to bring in atmospheric pressure as that is not applicable. You have already used 10 (or 9.8 its negligible) so there's no need to go any further.

Either way you should get most marks (-1/2 maybe)

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#9

i added atmospheric pressure because pressure at the surface pushes down on the liquid (atmospheric pressure being applied on the liquid) which then pushes down on itself (pressure due to column of liquid) so 100,000 Pa + 5500 Pa = 105,500 Pa. So 105,500 Pa was the total pressure at the bottom of the container. Btw the question was a 4 marker so i thought that there was more to it after using the equation for column of liquid. i'm not really sure if this makes sense, but i'm wondering if anyone else thought this?

(Original post by

Why? What part of you thought it would be correct to add atmospheric pressure?

It was 5500 because there is no need to bring in atmospheric pressure as that is not applicable. You have already used 10 (or 9.8 its negligible) so there's no need to go any further.

Either way you should get most marks (-1/2 maybe)

**hmussa1**)Why? What part of you thought it would be correct to add atmospheric pressure?

It was 5500 because there is no need to bring in atmospheric pressure as that is not applicable. You have already used 10 (or 9.8 its negligible) so there's no need to go any further.

Either way you should get most marks (-1/2 maybe)

Last edited by Rayzzz_; 2 years ago

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#10

**Rayzzz_**)

What was the answer to the final question on about the pressure due to a column of liquid?

the pressure of the liquid itself i got was 5,500 Pa. However, i thought to add in Atmospheric pressure (100,00 Pa) due to the fact that pressure from the atmosphere is also pushing down on the water/liquid.

So my final answer was 105,500 Pa. Did u get this?? or anyone else??

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#11

Btw this is what I did for the question where u had to calculate acceleration of the trolley along a 40 deg raised ramp, correct me on parts if I’m wrong also I don’t remember the question exactly this is off the top of my head

EDIT:

- should be (final)^2 - (initial)^2 then rearrange

Last edited by Time2Revise; 2 years ago

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#12

I got 115920 joules for that energy transfer question

I think you had to find the charge by doing current x time, then multiplying the current by the potential difference to get the answer

I think you had to find the charge by doing current x time, then multiplying the current by the potential difference to get the answer

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#13

(Original post by

I got 115920 joules for that energy transfer question

I think you had to find the charge by doing current x time, then multiplying the current by the potential difference to get the answer

**Bignibba445**)I got 115920 joules for that energy transfer question

I think you had to find the charge by doing current x time, then multiplying the current by the potential difference to get the answer

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#14

*Forgot what the question was about, all i know wqs that they gave you the current, p.d which was 230V, and the frequency for some reason*

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#15

I got that but unfortunately... it's wrong ;(

The method we did was that we used 0.28 A from previous question (WE ROUNDED THE 0.28 SO WE USED THE ROUNDED VALUE FFS) and then used that to multiply it by 30 minutes (30 x 60 seconds). That gave us the charge flow. Then, using energy transferred = charge x pd, we used the charge value and multiplied it by the pd (230) to get 115920 joules (which is wrong!)

If we had used the rounded value, we would've got 117000J

Hopefully we got the error carried forward thingy and only lost 1 mark but we'll see..

(Original post by

I got 115920 joules for that energy transfer question

I think you had to find the charge by doing current x time, then multiplying the current by the potential difference to get the answer

The method we did was that we used 0.28 A from previous question (WE ROUNDED THE 0.28 SO WE USED THE ROUNDED VALUE FFS) and then used that to multiply it by 30 minutes (30 x 60 seconds). That gave us the charge flow. Then, using energy transferred = charge x pd, we used the charge value and multiplied it by the pd (230) to get 115920 joules (which is wrong!)

If we had used the rounded value, we would've got 117000J

Hopefully we got the error carried forward thingy and only lost 1 mark but we'll see..

**Bignibba445**)

I got 115920 joules for that energy transfer question

I think you had to find the charge by doing current x time, then multiplying the current by the potential difference to get the answer

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#16

Hopefully the grade boundaries aren't stupid high, but seeing how everyone found it ez it probably will be :c

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#17

(Original post by

I did the same thinking 1m was the height so used sine rule to find the length of the ramp but I got 3.96 as my final answer???

**tomevandawson**)I did the same thinking 1m was the height so used sine rule to find the length of the ramp but I got 3.96 as my final answer???

Last edited by Jeqen; 2 years ago

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#18

I didn’t use sine rule or Pythagoras or anything lol, just rearranged the terms so A was on its own and got it right

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