AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread]

Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11
(ii) Show that theta = 3.12x10 6 radians
(iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)
(iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit

2) (i) Six marker, Don't ask.
(ii) m - M = 5log25/10 = -0.0097

3)(i) T = 16111K
(ii) The star was actually a white dwarf but I messed up and put main sequence
(iii) H balmer lines because the star's atmosphere was hot enough to have hydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmer series in the absorption spectra of stars with this surface temperature.

4)(i) A black hole is an object whose escape velocity is greater than c
(ii) Schwarzschild Radius, can't remember.
(iii) 4.96x10 17 s

I cannot remember the question well enough to know I read it correctly but can you explain your working out for 1i) please?

Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit2) (i) Six marker, Don't ask.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was actually a white dwarf but I messed up and put main sequence(iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in its n = 2 state and the deep prominent absorption lines in theBalmer series in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s

I used Pythagoras and square rooted 2.75^2 and added 1 (these were the valued in Au) then multiplied by 1.5x10 11 to convert to meters

Just realised i used the parsec->metre coversion for it...damn!

1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 -6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 5-1=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s

The mass given on the Schwartzchild radius question was 1x10^10, therefore a radius of ~3x10^13 m is to be calculated.

Thanks I remember this

Were we supposed to calculate v fir the six marker?

Once again, thanks a lot for your help increating this
1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Cassegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 5-1=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius was 3x10^13 m(iii) 4.96x10 17 s

Once again, thanks a lot for your help increating this
1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Cassegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 5-1=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius was 3x10^13 m(iii) 4.96x10 17 s

I did the same thing but that's wrong. How many marks out of 3 do you guys reckon we would get for it?

1 at most. Use of the value of an AU in meters.

Is there ecf in calculations? I calculated 1(a) wrong and used it to find the angle subtended in the next part obviously.


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I think that too but how come the other two marks are for adding 1 and 2.57? That seems a little generous to me. Maybe they will also give a mark for using 2.57 instead of the other distance. Who knows.

The first mark will be for realisation of the fact that they will have to be on different sides of the sun. Although that is shown as 1+2.57 its the principle it represents. 2nd mark is for use of the meter value of AU. 3rd mark is for the correct answer.

It was 2 marks, not 3