AQA physics paper 3 unofficial mark scheme - 29th June 2017

Can you someone put answers for question 2 as thats the one i got efffed by the most

Reply 9

teakhoo

Original post by revisionphysics

1.1 PIE radians
1.2 left blank
1.3 use a set square so the metal plate is perpendicular so that x is parallel and x is properly measured - reduces parallax error
1.4 GMAX=0.33 something
GMIN = 0.24 somthing
1.5 LLAMDA = GMAX+GMIN / 2=0.27
1.6 i got 43.7% as uncertainty i am certain that is wrong
1.7 y is the y intercept if you re arrange the equation
1.8 G MAX REDUCED G MIN REDUCED dont remember llamda and y

2 i got this question wrong

3.1 table
3.2table but something related to x=p-23.8cm for each value
3.3 dont remember
3.4 graph
3.5 gradient is negative and is straight line so x decreases exponentially.
3.6 logs and gradient and expotnential calcualtions

3.7
repeat measurement of p and x so spot anomalies calculate mean reduce effect of random error

use measuring instrument with higher resolution

i got lamda as reduced and y the same

Reply 10

Original post by revisionphysics

1.1 PIE radians
1.2 left blank
1.3 use a set square so the metal plate is perpendicular so that x is parallel and x is properly measured - reduces parallax error
1.4 GMAX=0.33 something
GMIN = 0.24 somthing
1.5 LLAMDA = GMAX+GMIN / 2=0.27
1.6 i got 43.7% as uncertainty i am certain that is wrong
1.7 y is the y intercept if you re arrange the equation
1.8 G MAX REDUCED G MIN REDUCED dont remember llamda and y

2 i got this question wrong

3.1 table
3.2table but something related to x=p-23.8cm for each value
3.3 dont remember
3.4 graph
3.5 gradient is negative and is straight line so x decreases exponentially.
3.6 logs and gradient and expotnential calcualtions

3.7
repeat measurement of p and x so spot anomalies calculate mean reduce effect of random error

use measuring instrument with higher resolution

Hi mate
What was the p?
1.98 cm?
And x=1.98_23.8=...?
Am I right?

Reply 11

Original post by revisionphysics

1.1 PIE radians
1.2 left blank
1.3 use a set square so the metal plate is perpendicular so that x is parallel and x is properly measured - reduces parallax error
1.4 GMAX=0.33 something
GMIN = 0.24 somthing
1.5 LLAMDA = GMAX+GMIN / 2=0.27
1.6 i got 43.7% as uncertainty i am certain that is wrong
1.7 y is the y intercept if you re arrange the equation
1.8 G MAX REDUCED G MIN REDUCED dont remember llamda and y

2 i got this question wrong

3.1 table
3.2table but something related to x=p-23.8cm for each value
3.3 dont remember
3.4 graph
3.5 gradient is negative and is straight line so x decreases exponentially.
3.6 logs and gradient and expotnential calcualtions

3.7
repeat measurement of p and x so spot anomalies calculate mean reduce effect of random error

use measuring instrument with higher resolution

I'm really worried
Could you please tell me about p?
P was 198 CM?
X was 198-23.8=???
But my graph looked strange
The gradient was negative but so gentle and flat?
Why?

(edited 6 years ago)

Reply 12

Original post by revisionphysics

Can't be bothered to do the turning points

Hi
Anybody know about last question of turning point ?
Deletion time was 7 hours ???
Not several days
So his predict was not precision
AM I right ?

Reply 13

Original post by Richard459

On the table on the last question I couldnt find x I minused 6 instead of 23 but then correctly took logs of these values to plot the graph using these logs instead (which gave the correct shape) and answered the questions. Will I still get error carried forward from using different x values???

How the graph looked like?
My graph was decreasing
Start from 5.....
But gradient was so gentle and flat
Is that looks like yours?

Reply 14

metrize

18

i got like 2% for uncertainty, i did half the range divided by my gradient

Reply 15

DrBono

Astro- Q1- Diagram of cassegrain set up Q2- Mirror B had a greater collecting power of around 33times using the ratio. 400^2/70^2 and therefore produced brighter images Q2.2-The angle in radians is proportional to 1/D and the therefore the higher the value for Diameter, the greater the diameter, the lower the minimum angular resolution and mentioned about airy discs overlapping etc. Q2.3- As the diameter of B is greater, it is able to resolve the two objects better with more detail Q3-One parsec corresponds to the distance at which the mean radius of the earth's orbit subtends an angle of one second of arc. Q3.1- The largest star is the one that's furthest out of the two (can't remember the name) as they have the same apparent magnitude and similar temp but the one furthest away will have the greatest power output and P=σAT^4 Q3.2-2900k Q3.3-Its the M class star because the absolute temperature for M class stars are 3500k< Q3.4-The one that's furthest away beginning with R is the brightest. Q4-Light curve of a type 1a supernova peaking at -19.3 at around 20 days Q4.1-A standard candle is an object whose absolute magnitude is known Q4.2-Mention about the accelerating expansion of the universe due to type 1a supernova not being as bright from billions of light years away and mention dark energy being the cause. Q5. Mention of transit method and radial velocity method and mention about how radial velocity is more accurate and the negatives involved with transit method like long orbital periods.

Reply 16

DrBono

I got Gmax to be about 0.035 and Gmin to be about 0.025 giving lamda to be about 0.028m. I then said Gmax would be unchanged as the max gradient possible was independent of the last error bar as it only went through the middle, that the min gradient had decreased, therefore so did lamda and therefore Y increased

Reply 17

Kierans1701

Original post by DrBono

I got Gmax to be about 0.035 and Gmin to be about 0.025 giving lamda to be about 0.028m. I then said Gmax would be unchanged as the max gradient possible was independent of the last error bar as it only went through the middle, that the min gradient had decreased, therefore so did lamda and therefore Y increased

I remember getting similar values

Reply 18