Indicative content:
• fusion (processes in stars) produce new elements
• cloud of gas / hydrogen and dust OR nebula
• pulled together by
... [Show More] gravity
• causing increasing temperature (to start the fusion process)
• (to become a) protostar
• hydrogen nuclei fuse to form helium nuclei
• and the star becomes main sequence
• hydrogen begins to run out
• helium nuclei fuse to make heavier elements
• up to iron
• the star expands (to become a)
• red super giant
• (the star collapses rapidly) and explodes
• called a supernova
• creating elements heavier than iron
• and distributing them throughout the universe
• leaving behind a neutron star
• or a black hole.
02.5 Temperature 1 AO1 4.6.3.2
Total 13
Question 3
Question
Answers
Extra information
Mark AO / Spec. Ref.
03.1 it is harder to judge where the centre of a wider ray is
causing a larger uncertainty (in the measurements)
allow increasing random errors (in the measurements) 1
1 AO3 4.6.1.3 RPA9
03.2 line of best fit drawn and extrapolated to 80 degrees
41 (degrees)
allow 40 to 43 (degrees) 1
1 AO3 4.6.1.3 RPA9
03.3 Level 3: The design/plan would lead to the production of a valid outcome. All key steps are identified and logically sequenced.
5–6 AO1 4.6.1.3 RPA9
Level 2: The design/plan would not necessarily lead to a valid outcome. Most steps are identified, but the method is not fully logically sequenced.
3–4
Level 1: The design/plan would not lead to a valid outcome. Some relevant steps are identified, but links are not made clear.
1–2
No relevant content 0
Indicative content:
• place a glass block on a piece of paper
• draw around the glass block
• use the ray box to shine a ray of light through the glass block
• mark the ray of light entering the glass block
• mark the ray of light emerging from the glass block
• join the points to show the path of the complete ray through the block
• and draw a normal line at 90 degrees to the surface
• use a protractor to measure the angle of incidence
• use a protractor to measure the angle of refraction
• use a ray box to shine a ray of light at a range of different angles (of incidence)
• increase the angle of incidence in 10 degree intervals
• from an angle of incidence of 10 degrees to an angle of incidence of 70 degrees.
allow use of optical pins instead of a ray box
03.4 (28 + 25 + 22)
3 = 25
3 (degrees)
allow alternative method 28 – 22 = 6 (1)
= 3 (degrees) (1) 1
1 AO3 4.6.1.3
03.5 Velocity 1 AO1 4.6.2.2
Total 13
Question 4
Question
Answers
Extra information
Mark AO / Spec. Ref.
04.1 any two correct lines drawn from the top of the visitor and passing through the lens
image drawn at the correct position and with the correct orientation allow construction lines that are not dashed
mark only scores if first two marks scored.
a convex lens diagram scores 0 marks
2
1 AO2 4.6.2.5
04.2 Decreases 1 AO3 4.6.2.5
04.3 Iron 1 AO1 4.7.2.1
04.4 there is a current in the solenoid
/ circuit
creating a magnetic field
attracting the bolt allow a charge flows through the solenoid / circuit
allow the solenoid / coil is magnetised 1
1
1 AO1 4.7.2.1
04.5 1.50 cm = 0.015 m
2.88 = k × 0.015
k = 2.88 / 0.015
k = 192 (N/m)
this mark may be awarded if distance is incorrectly/not converted
this mark may be awarded if distance is incorrectly/not converted
allow a correctly calculated answer using an incorrectly/not converted distance 1
1
1
1 AO2 4.5.3
04.6 Any two from:
• increase the current (in the solenoid / circuit)
• add more turns to the solenoid
• use a spring with a lower spring constant
allow any sensible suggestion for increasing the current such as increasing the p.d. / power of the battery OR using lower resistance wire in the solenoid
do not allow increase the number of coils
allow use a weaker spring 2 AO3 4.7.2.1
Total 14
Question 5
Question
Answers
Extra information
Mark AO / Spec. Ref.
05.1 (total) momentum before = (total) momentum after allow (total) momentum stays the same 1 AO1 4.5.7.2
05.2 momentum of
player A = 585 (kg m/s)
momentum of
player B = –500.5 (kg m/s)
(–500.5 + 585) 84.5
(78 + 91) OR 169
= 0.5 (m/s)
allow 1085.5
169
this answer only 1
1
1
1 AO2 4.5.7.1
4.5.7.2
05.3 (protective pads) increase the time taken to stop (during the collision)
so the rate of change of momentum decreases
reducing the force (on the ice hockey player) allow increases impact / contact
/ collision time
do not allow slows down time
allow reduces acceleration/deceleration
allow increases the time to reduce the momentum to zero for 2 marks
allow impact for force
do not allow if linked to an incorrect explanation 1
1
1 AO1 4.5.7.3
Total 8
Question 6
Question
Answers
Extra information
Mark AO / Spec. Ref.
06.1 320 MHz = 3.2 × 108 Hz
3.0 × 108 = 3.2 × 108 × λ
3.0 × 108
λ =
3.2 × 108
wavelength = 0.9375
metres or m allow 320 000 000
this mark may be awarded if frequency is incorrectly/not converted
this mark may be awarded if frequency is incorrectly/not converted
allow correct calculation using an incorrectly/not converted frequency
allow an answer that rounds to 0.94 1
1
1
1
1 AO2 AO2
AO2
AO2
AO1 4.6.1.2
06.2 (alternating) current induced (in the electrical circuit)
with the same frequency as the radio wave allow electrons vibrate / oscillate (in the electrical circuit) 1
1 AO1 4.6.2.3
06.3 Any two from:
• (radio waves are) transverse
• (radio waves) travel at a higher speed
• (radio waves) don’t need a medium
• (radio waves are) electromagnetic
allow sound waves are longitudinal
allow a description of transverse/longitudinal waves
allow (only) radio waves travel through a vacuum
allow sound waves are mechanical 2 AO1 4.6.1.1
4.6.1.2
06.4 accelerating allow speeding up 1 AO3 4.5.6.1.4
06.5 appropriate tangent drawn
correct reading from graph for change in distance and change in time (eg 5.6 (m) and 20 (s))
gradient of tangent shown (eg 5.6/20)
0.28 (m/s)
allow correct reading from their tangent for change in distance and change in time
allow correct gradient from their tangent
this answer only
allow 0.25 to 0.30 (m/s) if the tangent is appropriate
allow 2.8 / 20 = 0.14 (m/s) for 1
mark 1
1
1
1 AO2 4.5.6.1.4
06.6 0.522 – 0.122 = 2 x 0.04 x s
0.522 − 0.122
s = 2 × 0.04
s = 3.2 (m)
0.48 = F x 3.2
0.48
F = 3.2
F = 0.15 (N)
OR
this mark may be awarded if the displacement is incorrectly calculated
this mark may be awarded if the displacement is incorrectly calculated
allow a correctly calculated F using and incorrectly calculated displacement 1
1
1
1
1
1 AO2 4.5.2
4.5.6.1.5 [Show Less]