Questions on Fields
1. (a) (i) Cp = 2 + 4 = 6 μF A1
(ii) 1/C = 1/2 + ¼ C1
Cs = 4/3 =1.33 μF A1
(b) (i) 6.0 V A1
(ii) Q = CpV C1
= 6 × 6 = 36 μC A1
(c) E = ½ CsV2 C1
= 24 × 10–6 A1
(d) (i) The capacitors discharge through the voltmeter. B1
(ii) V = V0e–t/CR
1/4 =e–t/(6×12) C1
ln 4 = t / 72 C1
t = 72 ln 4 ≈ 100 s A1
[12]
2. It is the force (of attraction) per unit mass. B1
[1]
3. (i) Suitable recognisable pattern around (not just between) the charges B1
Quality mark: symmetry, spacing, lines joined to charges B1
Consistent arrows toward B on some lines B1
(ii) Use of E = (1/4πε0)Q/r2 C1
Sum of two equal terms
E = 2 × 9 × 109 × 1.6 × 10–19 / (2.0 × 10–10)2 C1
E = 7.2 × 1010 N C–1 or V m–1 A1
(iii) The separation between the ions because this has an effect on the
breaking force. (Allow the size of ionic ‘charges’) B1
[7]
4. (i) The gravitational field strength g is not constant. B1
The student’s value would be greater than the actual value (because the
average magnitude of g is less than 9.81 m s–1). B1
(ii) KE = 1/2 mv2
v = 2πr / T C1
v = 2 × π × (6800 + 6400) × 103 / 8.5 × 103 / v = 9.76 × 103 (m s–1) C1
KE = 1/2 × 1500 × (9.76 × 103)2
KE = 7.1(4) × 1010 (J) A1
(iii) A geostationary satellite stays above the same point on the Earth and as B1
such can be used for radio communications. (the term communications to M1
be included and spelled correctly to gain the mark). A1
The satellite is not in geostationary orbit
because its period is less than 1 day / 8.6 × 104 s.
[8]
5. (a) forces FS and FG acting inwards, force FE acting outwards - all through
centre of proton;
3 forces 2/2, 2 forces 1/2, marked and labelled (2) 2
(b) FE = FS + FG;
accept FE + FS + FG = 0 allow ecf from (a) (1) 1
(c) (i) FE = Q2 / (4π ε0 r2) (1)
= (1.6 × 10–19)2 / [4π × 8.85 × 10–12 (2.8 × 10–15)2] = 29 N (1)
use of r = 1.4 × 10–15 m (–1) once only 2
(ii) FG = m2 G / r2 (1)
= (1.67 × 10–27)2 × 6.67 × 10–11 / (2.8 × 10–15)2 = 2.4 × 10–35 N (1) 2
(iii) FS = 29 N / same as FE allow ecf (1) 1
(d) FE >> FG so FG negligible / insignificant / can be ignored or AW (1) 1
(e) (i) FE = 0 (1) 1
(ii) FG = 2.4 × 10–35 N (approx.) allow ecf (1) 1
(iii) FS = 2.4 × 10–35 N (approx.) (1)
comment: FS now repulsive (not attractive) or AW
or indicated by minus sign with FS; (1) any 3 1
[12]
6. (i) r has been increased by a factor of 3 from the centre of planet. C1
g = (40/32 =) 4.4(4) (N kg–1) A1
(ii) M = gr2 / G
M = (40 × [2.0 × 107]2) / 6.67 × 10–11 C1
M = 2.4 × 1026 (kg) A1
(iii) M = ρV = 4/3 πr3 ρ M1
g = GM / r2 r3 / r2 (Hence g r) A1
[6]
7. The astronaut is accelerating / has centripetal acceleration (1)
and the space station has the same acceleration (1)
a person does not feel gravity (1)
only feels forces applied by contact with the walls of the space station (1)
no support force from the space station (as they have the same acceleration) (1) 4
MAXIMUM (4)
[4]
Tuesday, May 25, 2010
Monday, May 24, 2010
y13 Momentum questions
1. The (net) force acting on an object is (directly) proportional to the rate of
change of momentum and takes place in the direction of the force. B2
[2]
2. According to Newton’s third law: When two objects interact, the force
acting on one of objects is equal but opposite to the force acting on the
other object. B1
The time t of ‘contact’ for the objects is the same and since Δp = Ft , B1
the gain in momentum for one object is equal to the loss of momentum for
the other object. B1
[3]
3. (i) u = 32 (m s–1) v = –2/3 × 32 = –21.33 (m s–1) t = 0.50 s C1
Δp = 800(–21.33 – 32) = –4.27 × 104 kg m s–1 A1
(ii) F = Δp / Δt = / F = 4.27 × 104 / 0.50 C1
F ≈ 8.5 × 104 (N) A1
Direction: Opposite to the initial velocity / away from the wall B1
(iii) N = pV / kT = (1.0 × 105 × 3.4 × 10–2) / (1.38 × 10–23 × 293) C1
N = 8.41 × 1023 C1
mass = 8.41 × 1023 / 6.02 × 1023) × 0.014
mass = 0.0196 (kg) ≈ 0.020 kg A1
[8]
4. (a) (i) to come to rest simultaneously, total mtm. = 0 or AW (1) 1
(but initial mtm. not zero)
(ii) initial mtm. = 3 m u – 2 m u = m u (1)
when closest, mtm. = (3m + 2m) v (1) 2
so 5m v = m u (and v = u / 5)
(b) (i) initial k.e. = final k.e. + (gain of) p.e. (1) 1
(ii) k.e. = ½ m v2 (1)
total k.e. = ½ × 3 m u2 + ½ × 2 m u2 (= 2.5 m u2) (1)
= 2.5 × 1.67 × 10–27 u2 (= 4.18 × 10–27 u2) (1) 3
allow m = 1.66 × 10–27 kg for full credit
(iii) gain of p.e. = initial k.e. – final k.e.
= 4.18 × 10–27 u2 – 4.18 × 10–27 (u/5)2 (2)
1.53 × 10–13 = 4.01 × 10–27 u2 (1) algebra
u = 6.18 × 106 m s–1 (1) 4
omits - 4.18 × 10–27 (u/5)2, gets u = 6. 06 × 106 m s–1: 1/2, 1, 1 = 3/4
[11]
5. (a) (i) Mass × velocity/mv with symbols defined 1
(ii) 0 = mAvA ± mBvB or mAvA = mBvB (1)
vA/vB = ± mB/mA (1) 2
max 1 mark for final expression without line 1
(b) (i) vA = (10/5 =) 2.0 (m s–1) and vB = (10/10 =) 1.0 (m s–1) 1
(ii) t1 = 3.0/2.0 = 1.5 (s) ecf b(i) 1
(iii) x = 2.1 – 1.0 × 1.5 = 0.6 (m) 1
(iv) v = vB + (5/50)vA = 1.0 + 0.2 (= 1.2 m s–1) 1
(v) t2 = t1 + 0.6/1.2 = 2.0 (s) 1
(vi) At collision the container (and fragments) stop (1)
By conservation of momentum, total momentum is still zero/AW (1) 2
(vii) straight lines from (0,0) to (1.5,0); (1.5,0) to (2.0,0.1); (x,0.1) for all x>2 3
[13]
6. (a) (i) 3.8 ± 0.3 (N s) 1
(ii) momentum (of the ball) accept impulse 1
(iii) mv = 3.8 or v = 3.8/0.16; = 23 (m s–1) ecf a 2
(iv) use F = ma giving 24 = 0.16a; a = 150 (m s–2) 2
(b) (i) exponential 1
(ii) e.g. h1/h2 = ek = 2.1(5); giving k = 0.74 to 0.76
or substitution from a line of table; gives 0.748, 0.757 or 0.746 2
(iii) 1.5 (m) 1
(iv) ∆k.e. = mg∆h; = 0.16 × 9.8 × 0.38 (= 0.60 J) 2
[12]
change of momentum and takes place in the direction of the force. B2
[2]
2. According to Newton’s third law: When two objects interact, the force
acting on one of objects is equal but opposite to the force acting on the
other object. B1
The time t of ‘contact’ for the objects is the same and since Δp = Ft , B1
the gain in momentum for one object is equal to the loss of momentum for
the other object. B1
[3]
3. (i) u = 32 (m s–1) v = –2/3 × 32 = –21.33 (m s–1) t = 0.50 s C1
Δp = 800(–21.33 – 32) = –4.27 × 104 kg m s–1 A1
(ii) F = Δp / Δt = / F = 4.27 × 104 / 0.50 C1
F ≈ 8.5 × 104 (N) A1
Direction: Opposite to the initial velocity / away from the wall B1
(iii) N = pV / kT = (1.0 × 105 × 3.4 × 10–2) / (1.38 × 10–23 × 293) C1
N = 8.41 × 1023 C1
mass = 8.41 × 1023 / 6.02 × 1023) × 0.014
mass = 0.0196 (kg) ≈ 0.020 kg A1
[8]
4. (a) (i) to come to rest simultaneously, total mtm. = 0 or AW (1) 1
(but initial mtm. not zero)
(ii) initial mtm. = 3 m u – 2 m u = m u (1)
when closest, mtm. = (3m + 2m) v (1) 2
so 5m v = m u (and v = u / 5)
(b) (i) initial k.e. = final k.e. + (gain of) p.e. (1) 1
(ii) k.e. = ½ m v2 (1)
total k.e. = ½ × 3 m u2 + ½ × 2 m u2 (= 2.5 m u2) (1)
= 2.5 × 1.67 × 10–27 u2 (= 4.18 × 10–27 u2) (1) 3
allow m = 1.66 × 10–27 kg for full credit
(iii) gain of p.e. = initial k.e. – final k.e.
= 4.18 × 10–27 u2 – 4.18 × 10–27 (u/5)2 (2)
1.53 × 10–13 = 4.01 × 10–27 u2 (1) algebra
u = 6.18 × 106 m s–1 (1) 4
omits - 4.18 × 10–27 (u/5)2, gets u = 6. 06 × 106 m s–1: 1/2, 1, 1 = 3/4
[11]
5. (a) (i) Mass × velocity/mv with symbols defined 1
(ii) 0 = mAvA ± mBvB or mAvA = mBvB (1)
vA/vB = ± mB/mA (1) 2
max 1 mark for final expression without line 1
(b) (i) vA = (10/5 =) 2.0 (m s–1) and vB = (10/10 =) 1.0 (m s–1) 1
(ii) t1 = 3.0/2.0 = 1.5 (s) ecf b(i) 1
(iii) x = 2.1 – 1.0 × 1.5 = 0.6 (m) 1
(iv) v = vB + (5/50)vA = 1.0 + 0.2 (= 1.2 m s–1) 1
(v) t2 = t1 + 0.6/1.2 = 2.0 (s) 1
(vi) At collision the container (and fragments) stop (1)
By conservation of momentum, total momentum is still zero/AW (1) 2
(vii) straight lines from (0,0) to (1.5,0); (1.5,0) to (2.0,0.1); (x,0.1) for all x>2 3
[13]
6. (a) (i) 3.8 ± 0.3 (N s) 1
(ii) momentum (of the ball) accept impulse 1
(iii) mv = 3.8 or v = 3.8/0.16; = 23 (m s–1) ecf a 2
(iv) use F = ma giving 24 = 0.16a; a = 150 (m s–2) 2
(b) (i) exponential 1
(ii) e.g. h1/h2 = ek = 2.1(5); giving k = 0.74 to 0.76
or substitution from a line of table; gives 0.748, 0.757 or 0.746 2
(iii) 1.5 (m) 1
(iv) ∆k.e. = mg∆h; = 0.16 × 9.8 × 0.38 (= 0.60 J) 2
[12]
Wednesday, May 19, 2010
y13 Questions on magnetism and Universe
y13 Questions on magnetism and Universe
1. (a) B = F/Il with symbols explained or appropriate statement in words; (1)
explicit reference to I and B at right angles/define from F = BQv etc (1) 2
(b) (i) arrow towards centre of circle 1
(ii) field out of paper; Fleming’s L.H. rule/moving protons act as
conventional current 2
(iii) F = Bev allow BQv 1
(iv) F = mv2/r; Bev = mv2/r; (2)
B = mv/er = 1.67 × 10–27 × 1.5 × 107/(1.6 × 10–19 × 60); = 0.0026; T (3) 5
allow Wb m–2
(v) the field must be doubled; (1)
B ∞ v (as m, e and r are fixed)/an increased force is required
to maintain the same radius (1) 2
[13]
2. At least 3 field lines inside solenoid parallel to axis; (1)
Lines equally spaced over some of length of solenoid. (1)
Arrows on lines pointing left to right. (1) 3
[3]
3. (a) Positive as E-field is downwards/top plate is positive/like charges repel/AW (1) 1
(b) (i) k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2) 2
(ii) 1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1) (2) 2
(c) E = V/d; so d = V/E = 600/4 × 104 = 0.015 m (2) 2
(d) (i) semicircle to right of hole (1) ecf(a); (a) and d(i) to be consistent 1
(ii) mv2/r; = BQv; (2)
giving r = mv/BQ = 2.3 × 10–26 × 6.5 × 104/(0.17 × 1.6 × 10–19); (1)
r = 55 mm;so distance = 2r = 0.11 m (2) 5
[13]
4. Any four from:
Uniform intensity in all directions / everywhere
Structure in background intensity / ripples
Produced when matter and radiation decoupled
Originally gamma radiation
(gamma) red-shifted to microwave / originally higher energy
Evidence that universe began with big bang
Temperature corresponds to 2.7 K / 3K / that predicted by big bang model B1 × 4
Link between evidence and explanation. (1)
[5]
5. Any 5 from
red shift data for galaxies (accept stars) 1
calculate velocity from red shift 1
galaxies/ stars receding from Earth 1
distance data for galaxies/ stars 1
velocity α distance / v/r = constant / v-r graph straight line 1
universe began at a single point 1
[5]
6. (a) Any two
stars rotate around galactic centre 1
star with velocity component towards Earth 1
reference to motion/shape of galaxy 1
or other valid points eg blue shift
(b) Ho = 75/ 3 × 1019 s–1 1
t ≈ 1/ 2.5 × 10–18 1
t ≈ 4 × 1017 s 1
[5]
1. (a) B = F/Il with symbols explained or appropriate statement in words; (1)
explicit reference to I and B at right angles/define from F = BQv etc (1) 2
(b) (i) arrow towards centre of circle 1
(ii) field out of paper; Fleming’s L.H. rule/moving protons act as
conventional current 2
(iii) F = Bev allow BQv 1
(iv) F = mv2/r; Bev = mv2/r; (2)
B = mv/er = 1.67 × 10–27 × 1.5 × 107/(1.6 × 10–19 × 60); = 0.0026; T (3) 5
allow Wb m–2
(v) the field must be doubled; (1)
B ∞ v (as m, e and r are fixed)/an increased force is required
to maintain the same radius (1) 2
[13]
2. At least 3 field lines inside solenoid parallel to axis; (1)
Lines equally spaced over some of length of solenoid. (1)
Arrows on lines pointing left to right. (1) 3
[3]
3. (a) Positive as E-field is downwards/top plate is positive/like charges repel/AW (1) 1
(b) (i) k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2) 2
(ii) 1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1) (2) 2
(c) E = V/d; so d = V/E = 600/4 × 104 = 0.015 m (2) 2
(d) (i) semicircle to right of hole (1) ecf(a); (a) and d(i) to be consistent 1
(ii) mv2/r; = BQv; (2)
giving r = mv/BQ = 2.3 × 10–26 × 6.5 × 104/(0.17 × 1.6 × 10–19); (1)
r = 55 mm;so distance = 2r = 0.11 m (2) 5
[13]
4. Any four from:
Uniform intensity in all directions / everywhere
Structure in background intensity / ripples
Produced when matter and radiation decoupled
Originally gamma radiation
(gamma) red-shifted to microwave / originally higher energy
Evidence that universe began with big bang
Temperature corresponds to 2.7 K / 3K / that predicted by big bang model B1 × 4
Link between evidence and explanation. (1)
[5]
5. Any 5 from
red shift data for galaxies (accept stars) 1
calculate velocity from red shift 1
galaxies/ stars receding from Earth 1
distance data for galaxies/ stars 1
velocity α distance / v/r = constant / v-r graph straight line 1
universe began at a single point 1
[5]
6. (a) Any two
stars rotate around galactic centre 1
star with velocity component towards Earth 1
reference to motion/shape of galaxy 1
or other valid points eg blue shift
(b) Ho = 75/ 3 × 1019 s–1 1
t ≈ 1/ 2.5 × 10–18 1
t ≈ 4 × 1017 s 1
[5]
Friday, May 14, 2010
y12 Waves
Qustions on waves
1. circular arcs (penalise anything flat) B1
same constant wavelength before and after gap – judged by eye or labelled B1
this means at least 3 wavefronts need to be drawn
[2]
2. for noticeable diffraction gap size (WTTE) B1
for sound much bigger than for light (WTTE) B1
[2]
3. (i) wave sources that have a constant phase difference (WTTE) B2
{max of 1 mark for sources have same frequency/wavelength/in phase C1}
(ii) sum of displacements (= resultant displacement) (WTTE) B1
(no marks for reference to amplitude)
[3]
4. (a) (i) constant phase difference
(allow 1 mark for same phase difference or same frequency/wavelength) B2
(ii) path difference = λ/2 B1
(b) (i) evidence shown that fringe width x = 8.0 mm B1
a = λD/x = 6.4 × 10–7 × 1.5/8.0 × 10–3 = 1.2 × 10–4 m C1
(give 2 marks for using x = 4.0 mm giving a = 2.4 × 10–4 m) A1
(ii) maximum intensity when y = 0 AND minima at +4 and –4 B1
correct repeat distance, i.e. 8.0 mm with at least 2 full cycles drawn B1
[8]
5. (i) semicircular wavefronts leaving the gap B1
no change in wavelength stated OR
clearly shown (at least 3 waves needed) – judged by eye B1
(ii) LESS diffraction would occur – shown or stated B1
wavefronts mainly plane (by eye) (allow curved at edges) B1
(iii) MORE diffraction for SOUND B1
Wavelength of sound > wavelength of light (WTTE) B1
Valid comparison of wavelength of light or sound with doorway e.g.
doorway of similar size to wavelength of sound OR
wavelength of light is very small compared to door (WTTE) B1
[7]
6. (i) 1. path diff. = n (where n = 0,1,2 etc) {allow 0, OR , OR 2 etc} B1
2. path diff = (n ½) (where n = 0,1,etc) {allow = 0.5 OR,1.5, etc} B1
{do not allow answers purely about phase diff. e.g. with degrees or
used and no ref to }
(ii) recall of formula = ax/D C1
correct substitution for a, and D: e.g. = (4.86 10–7 2)/0.5 10–3 C1
x = 1.94 10–3 m (1.9 or 1.944) A1
(iii) central white fringe B1
other fringes are coloured (WTTE: e.g. allow spectrum formed) B1
[7]
7. diagram showing
laser/light source placed directly behind double slit AND screen placed
in front of slits B1
{single slit NOT required; no labelling required}
(i) D: allow any value between 30cm and 10m B1
(ii) a: allow any value between 0.1mm and 2mm B1
[3]
8. (a) (i) evidence of good practice: i.e distance for nx measured e.g.
5x = 18mm C1
x = 3.6 mm (OR 3.5 OR 3.7) A1
{x = 3.4, 3.8, 3.9, 4.0, or 4 mm, implying is directly
measured, and score 1 mark)
(ii) for O path difference = 0 B1
for A path difference = 3() B1
for B path difference = 1.5() B1
(b) recall of = ax/D OR =D/a OR B1
is smaller for blue light (than red light) hence is SMALLER (WTTE) B1
[7]
9. Maximum of 2 marks for correctly identifying the 4 errors OR stating the 2
correct notes:
i.e. errors in notes 1, 2, 3, and 6 (shown anywhere) B2
{5 or 6 or 2 or 1 notes nominated scores ZERO, 4 correct scores 2, 3 correct scores 1}
Valid corrections score 1 mark each: do not allow “NOT” corrections apart
from note 3
Note 1: In longitudinal waves vibrations are parallel to wave direction (WTTE) B1
{OR in transverse waves vibrations are perpendicular to wave
direction (WTTE}
Note 2 light (or any of the em waves) can travel through a vacuum (WTTE) B1
{allow sound/longitudinal waves cannot travel thro’ a vacuum}
Note 3: waves carry energy/disturbance (not displacement or info)
from…. (WTTE) B1
{allow “waves do not carry the medium” and “the medium carries
the waves from…….”}
Note 6: wavelength = distance from crest to crest/trough to trough/max
to max (WTTE) B1
[6]
1. circular arcs (penalise anything flat) B1
same constant wavelength before and after gap – judged by eye or labelled B1
this means at least 3 wavefronts need to be drawn
[2]
2. for noticeable diffraction gap size (WTTE) B1
for sound much bigger than for light (WTTE) B1
[2]
3. (i) wave sources that have a constant phase difference (WTTE) B2
{max of 1 mark for sources have same frequency/wavelength/in phase C1}
(ii) sum of displacements (= resultant displacement) (WTTE) B1
(no marks for reference to amplitude)
[3]
4. (a) (i) constant phase difference
(allow 1 mark for same phase difference or same frequency/wavelength) B2
(ii) path difference = λ/2 B1
(b) (i) evidence shown that fringe width x = 8.0 mm B1
a = λD/x = 6.4 × 10–7 × 1.5/8.0 × 10–3 = 1.2 × 10–4 m C1
(give 2 marks for using x = 4.0 mm giving a = 2.4 × 10–4 m) A1
(ii) maximum intensity when y = 0 AND minima at +4 and –4 B1
correct repeat distance, i.e. 8.0 mm with at least 2 full cycles drawn B1
[8]
5. (i) semicircular wavefronts leaving the gap B1
no change in wavelength stated OR
clearly shown (at least 3 waves needed) – judged by eye B1
(ii) LESS diffraction would occur – shown or stated B1
wavefronts mainly plane (by eye) (allow curved at edges) B1
(iii) MORE diffraction for SOUND B1
Wavelength of sound > wavelength of light (WTTE) B1
Valid comparison of wavelength of light or sound with doorway e.g.
doorway of similar size to wavelength of sound OR
wavelength of light is very small compared to door (WTTE) B1
[7]
6. (i) 1. path diff. = n (where n = 0,1,2 etc) {allow 0, OR , OR 2 etc} B1
2. path diff = (n ½) (where n = 0,1,etc) {allow = 0.5 OR,1.5, etc} B1
{do not allow answers purely about phase diff. e.g. with degrees or
used and no ref to }
(ii) recall of formula = ax/D C1
correct substitution for a, and D: e.g. = (4.86 10–7 2)/0.5 10–3 C1
x = 1.94 10–3 m (1.9 or 1.944) A1
(iii) central white fringe B1
other fringes are coloured (WTTE: e.g. allow spectrum formed) B1
[7]
7. diagram showing
laser/light source placed directly behind double slit AND screen placed
in front of slits B1
{single slit NOT required; no labelling required}
(i) D: allow any value between 30cm and 10m B1
(ii) a: allow any value between 0.1mm and 2mm B1
[3]
8. (a) (i) evidence of good practice: i.e distance for nx measured e.g.
5x = 18mm C1
x = 3.6 mm (OR 3.5 OR 3.7) A1
{x = 3.4, 3.8, 3.9, 4.0, or 4 mm, implying is directly
measured, and score 1 mark)
(ii) for O path difference = 0 B1
for A path difference = 3() B1
for B path difference = 1.5() B1
(b) recall of = ax/D OR =D/a OR B1
is smaller for blue light (than red light) hence is SMALLER (WTTE) B1
[7]
9. Maximum of 2 marks for correctly identifying the 4 errors OR stating the 2
correct notes:
i.e. errors in notes 1, 2, 3, and 6 (shown anywhere) B2
{5 or 6 or 2 or 1 notes nominated scores ZERO, 4 correct scores 2, 3 correct scores 1}
Valid corrections score 1 mark each: do not allow “NOT” corrections apart
from note 3
Note 1: In longitudinal waves vibrations are parallel to wave direction (WTTE) B1
{OR in transverse waves vibrations are perpendicular to wave
direction (WTTE}
Note 2 light (or any of the em waves) can travel through a vacuum (WTTE) B1
{allow sound/longitudinal waves cannot travel thro’ a vacuum}
Note 3: waves carry energy/disturbance (not displacement or info)
from…. (WTTE) B1
{allow “waves do not carry the medium” and “the medium carries
the waves from…….”}
Note 6: wavelength = distance from crest to crest/trough to trough/max
to max (WTTE) B1
[6]
Thursday, May 13, 2010
y12 questions on moments
Questions on moments
1. (i) pressure = force / area B1
(ii) moment = force multiplied by the perpendicular distance
(from the line of action of the force) to the pivot B1
[2]
2. (a) The net force acting on the object must be zero B1
The net moment (about any point) must also be zero B1
(b) Taking moments about A, we have
Sum of clockwise moments = sum of anticlockwise moments C1
(0.25 × 200) + (5.0 × 9.81 × 0.4) = 0.8F C1
F = 87 (N) A1
(c) These forces are opposite but not equal in magnitude. B1
[6]
3. (i) Moment is the force the perpendicular distance
from (the line of action of) the force to the pivot/point
(missing perpendicular 1, missing from the force to B2
the pivot / point 1)
(ii) Torque of a couple: one of the forces x B1
perpendicular distance between (the lines of action of) the forces
[3]
4. (i) 1 3600 1.0 = X 2.5 C2
one mark for one correct moment, one mark for the
second correct moment and equated to first moment A0
2 X = 1440 (N) C1
Y = 3600 – 1440 or 3600 1.5 = Y 2.5 A1
= 2160 (N) B1
(ii) Not a couple as forces are not equal B1
and not in opposite directions / the forces are in the
same direction C1
(iii) P = F / A B1
= 1440 / 2.3 10–2 B1
= 62609 (6.3 104)
unit Pa or N m–2
[9]
5. (i) 1 The (distribution of the) mass of the lawn mower is
not uniform B1
2. One correct moment about A stated
B 110 or 350 20 B1
B = (350 20) / 110 (moments equated) B1
B = 63.6 (N) A0
3. A = 350 63.6 = 286(.4) (N) A1
(ii) A goes down and B goes up B1
Turning effect of B is less / B needs greater force to
produce the same moment / if distance goes down
force needs to go up (to maintain the same turning effect) B1
[6]
6. (a) (i) • F × 25 sin15 / F × 0.25 sin 15 for one moment. (1)
• 450 × 40 cos 30 / 450 × 0.4 cos 30 for the other moment. (1)
• moments equated or stated, even if not correct.
[Do not accept forces resolved vertically] (1)
• Answer F = 2409 (N). (1) 4
(ii) • Answer F = 951 (N). 1
(b) • Link large force (2409N) with small angle (30°) /
The more nearly horizontal / the smaller the angle with the horizontal
your back is, the greater the force needed (from the muscles). (1)
• the force is large because the anti-clockwise moment is large (1)
• the anti-clockwise moment is large because the perpendicular
distance to the pivot is large. (1)
(First 3 points + any one of the following:) (1)
• consequence, eg tendon ‘goes’, etc.
• (Therefore) keep your back as vertical / upright as possible,
• … with the load close to your body …
• … and bend your knees / use leg muscles to do some of the lifting.
• ..back is strong in compression / weak in shear, etc. 4
[9]
7. (a) (i) force drawn vertically upwards at plunger B1
force drawn vertically at H B1
(ii) 20 500 / force on Plunger 120 (one correct moment stated) B1
Plunger force 120 = (20 500) B1
Plunger force = 83(.3) (N) A0
(b) (i) pressure = force / area
= 83 / 4 10–3 C1
= 20800 (Pa) A1
(ii) decrease area of plunger / decrease distance H to plunger /
increase F / increase length of arm B2
MAX 2
[8]
8. (i) W vertically down at G B1
Force at O vertical B1
(ii) V × 0.9 × cos60 = W × 0.35 × cos60 B1
V = (25 × 0.35) / 0.9 B1
= 9.7(22) (N) A0
(iii) total force is zero stated or implied / 25 – 9.7 C1
force at hinge = 15.3 (N) A1
(or may take moments about G or V)
[6]
1. (i) pressure = force / area B1
(ii) moment = force multiplied by the perpendicular distance
(from the line of action of the force) to the pivot B1
[2]
2. (a) The net force acting on the object must be zero B1
The net moment (about any point) must also be zero B1
(b) Taking moments about A, we have
Sum of clockwise moments = sum of anticlockwise moments C1
(0.25 × 200) + (5.0 × 9.81 × 0.4) = 0.8F C1
F = 87 (N) A1
(c) These forces are opposite but not equal in magnitude. B1
[6]
3. (i) Moment is the force the perpendicular distance
from (the line of action of) the force to the pivot/point
(missing perpendicular 1, missing from the force to B2
the pivot / point 1)
(ii) Torque of a couple: one of the forces x B1
perpendicular distance between (the lines of action of) the forces
[3]
4. (i) 1 3600 1.0 = X 2.5 C2
one mark for one correct moment, one mark for the
second correct moment and equated to first moment A0
2 X = 1440 (N) C1
Y = 3600 – 1440 or 3600 1.5 = Y 2.5 A1
= 2160 (N) B1
(ii) Not a couple as forces are not equal B1
and not in opposite directions / the forces are in the
same direction C1
(iii) P = F / A B1
= 1440 / 2.3 10–2 B1
= 62609 (6.3 104)
unit Pa or N m–2
[9]
5. (i) 1 The (distribution of the) mass of the lawn mower is
not uniform B1
2. One correct moment about A stated
B 110 or 350 20 B1
B = (350 20) / 110 (moments equated) B1
B = 63.6 (N) A0
3. A = 350 63.6 = 286(.4) (N) A1
(ii) A goes down and B goes up B1
Turning effect of B is less / B needs greater force to
produce the same moment / if distance goes down
force needs to go up (to maintain the same turning effect) B1
[6]
6. (a) (i) • F × 25 sin15 / F × 0.25 sin 15 for one moment. (1)
• 450 × 40 cos 30 / 450 × 0.4 cos 30 for the other moment. (1)
• moments equated or stated, even if not correct.
[Do not accept forces resolved vertically] (1)
• Answer F = 2409 (N). (1) 4
(ii) • Answer F = 951 (N). 1
(b) • Link large force (2409N) with small angle (30°) /
The more nearly horizontal / the smaller the angle with the horizontal
your back is, the greater the force needed (from the muscles). (1)
• the force is large because the anti-clockwise moment is large (1)
• the anti-clockwise moment is large because the perpendicular
distance to the pivot is large. (1)
(First 3 points + any one of the following:) (1)
• consequence, eg tendon ‘goes’, etc.
• (Therefore) keep your back as vertical / upright as possible,
• … with the load close to your body …
• … and bend your knees / use leg muscles to do some of the lifting.
• ..back is strong in compression / weak in shear, etc. 4
[9]
7. (a) (i) force drawn vertically upwards at plunger B1
force drawn vertically at H B1
(ii) 20 500 / force on Plunger 120 (one correct moment stated) B1
Plunger force 120 = (20 500) B1
Plunger force = 83(.3) (N) A0
(b) (i) pressure = force / area
= 83 / 4 10–3 C1
= 20800 (Pa) A1
(ii) decrease area of plunger / decrease distance H to plunger /
increase F / increase length of arm B2
MAX 2
[8]
8. (i) W vertically down at G B1
Force at O vertical B1
(ii) V × 0.9 × cos60 = W × 0.35 × cos60 B1
V = (25 × 0.35) / 0.9 B1
= 9.7(22) (N) A0
(iii) total force is zero stated or implied / 25 – 9.7 C1
force at hinge = 15.3 (N) A1
(or may take moments about G or V)
[6]
Wednesday, May 12, 2010
Y12 Photons
Y12 Questions on Photons
1. 3.9 eV = 3.9 × 1.6 × 10–19 J (= 6.24 × 10–19 J) (1)
λ = hc/E = 6.63 × 10–34 × 3.0 × 108 / 3.9 × 1.6 × 10–19 (= 320 × 10–9 m) (1) 2
[2]
2. (i) 1. E = hf / / f = 7.5 1017 (Hz) C1
(‘E = hf’ can be secured in (i))
/ E = 6.63 10–34 7.5 1017 C1
energy = 4.97 10–16 (J) 5.0 10–16 (J) (Allow 1 sf answer here) A1
2. (Possible ecf)
energy = 3.1 103 (eV) B1
(ii) The answer to (c)(i)1. and 1.4 (W) are used to determine the rate
of photons C1
(Possible ecf) C1
number = 2.8 1015 (s–1) (If 3100 eV is used, then allow 2/3 for 4.5 10–4) A1
[7]
3. (Allow any subject) C1
C1
v = 1.43 103 1.4 103 (ms–1) A1
[3]
4. (i) Visible (light) B1
(ii) work function = 1.9 1.6 10–19 M1
work function = 3.04 10–19 (J) 3.0 10–19 (J) A0
(iii) 1. E = hf / C1
E =
E = 3.9 10–19 (J) A1
2. hf = + KE(max) / hf = + ½ mv2
(Allow E = + ½ mv2 if E is qualified in (iii)1.) C1
3.9 10–19 = 3.0 10–19 + KE(max) / 3.9 10–19 = 3.04 10–19 + KE(max) C1
KE = 9.0 10–20 (J) / KE = 8.6 10–20 (J) (Possible ecf) A1
(iv) No change (to maximum KE of electron) B1
Each photon has same energy (but there are fewer photons) B1
(v) number of photons = ( 2.05 1017) (Possible ecf) C1
number of electrons = 0.07
number of electrons = 1.44 1016 (s–1) 1.4 × 1016 (s–1) A1
[11]
5.
(Four correct: 3 marks, three correct: 2 marks, two correct: 1 mark) B3
[3]
6. (i) particle / particulate / quantum / photon B1
(ii) wave B1
[2]
7. Maximum of three from points 1 to 6: B1 3
1. Photon mentioned (e.g.: photons interact with the surface electrons)
2. Energy is conserved (between the photon and the electron / in the interaction)
3. hf = KE(max)
4. A single photon interacts with a single electron / It is a one-to-one interaction
5. Electron is removed when photon energy is greater than / equal to the
work function
(energy) / (Allow ora)
6. Electron removed when frequency is greater than / equal to the threshold
frequency (Allow ora)
7. (Visible) light has lower frequency than the threshold frequency / Energy of
(visible) light photon is less than the work function (energy) (ora with uv) B1
8. Greater intensity of (visible) light means more photons (per unit time) /
energy of a photon remains the same B1
QWC Spelling, punctuation and grammar B1
Organisation B1
[7]
8. (a) quantum of energy / radiation / packet of energy B1
(b) (i) f = E/h = 5.60 × 10–19 /6.63 × 10–34 C1
f = 8.45 × 1014 (Hz) A1
(ii) 1 minimum energy to release an electron from the surface (of the metal) B1
2 5.60 × 10–19 – 4.80 × 10–19 ( = 8.0 × 10–20 J) B1
(iii) 8.0 × 10–20 = ½(9.1 × 10–31)v2 M1
giving v = 4.2 × 105 (m s–1) A1
(c) (i) Correct selection of: λ = h/p or λ = h/mv M1
where all symbols are defined A1
(ii) λ = 6.6 × 10–34 /(9.1 × 10–31 × 4.2 × 105) C1
λ = 1.7 × 10–9 (m) A1
[11]
9. (a) one (or more) electrons removed (or added) to an atom 1
(b) E = hf = hc/λ together with knowledge of symbol meaning (1)
= (1)
= 8.36 × 10–19 (J) (1) 3
(c) frequency of UV is greater than frequency of light
OR alternative statement in terms of wavelength.
so photon energy of visible light is less than photon energy of UV (1)
PLUS one of the idea of conservation of energy
it is not possible for a low energy photon to give a high energy photon
this is a one to one process (1) 2
(d) E = V/d and power of 10 correct for d (1)
= 30/0.00020 = 150 000 (1)
V m–1 (1) 3
[9]
10. (a) Maximum of five marks
Up to four from:
/ M1
All symbols (, h, m and v or p) defined A1
Electrons travel / move / propagate (through space) as a wave B1
Electrons are diffracted / ‘spread out’ M1
by the atoms / spacing between the atoms A1
The electrons are diffracted when their wavelength is less than or
comparable or same as size of atoms / gap between the atoms B1
Up to two from:
When the speed of electrons is increased) the rings ‘get smaller’ B1
(At greater speed of electrons) the wavelength is shorter B1
(At greater speed of electrons) there is less diffraction B1
QWC Organisation B1
Spelling, punctuation & grammar B1
(b) Electrons have mass / momentum / charge / can be ‘accelerated’ B1
[8]
11. (i) The minimum frequency needed to free an electron
(from the surface of a metal) B1
(ii)1 Line extended intersects (the f axis at) this value / At this frequency, Ek = 0 B1
(ii)2 ( = ) h 5.0 1014 / ( = ) 6.63 10–34 5.0 1014 C1
work function energy = 3.3 10–19 J A1
(iii)1 1 The gradient / slope of the line is the same B1
The gradient is equal to h / independent of the metal B1
(iii)2 The line is shifted to the right B1
The threshold frequency is greater (AW) B1
[8]
12. (a) Any five from: B1 × 5
1. Photoelectric (effect) mentioned
2. Photon(s) mentioned in correct context / E = hf
3. One-to-one ‘interaction’ between photon & electron
4. Surface electrons are involved
5. Electron released / photoelectric (effect) when photon
energy > / = work function (energy)
6. Electrons emitted / photoelectric (effect) when
frequency > / = threshold frequency
7. Energy is conserved (in the ‘interaction’ between photon and electron)
8. Reference to Einstein’s equation: hf = + KE(max)
[QWC: Spelling and Grammar]
(b) (i) 1. (energy of photon = 2.2 + 0.3) B1
2.5 (eV) B1
2. (energy =) 2.5 1.6 10–19 (Possible ecf from (b)(i)1.) C1
4.0 10–19 (J) (Allow 1 sf answer) A1
(ii) (f =) (Possible ecf) C1
(f = )
(f =) 6.03 1014 6.0 1014 (Hz) (Allow 6 1014) A1
(c) Each photon has more energy / There are fewer photons (in B1
a given time because intensity is the same)
Smaller current B1
[13]
13. (i) 1. The minimum frequency (of radiation \ waves) needed for electrons
to be released (from the metal surface) \ for photoelectric effect B1
2. Its temperature increases \ gets warm \ ‘heats up’ B1
(ii) E = 2.2 1.9 (= 4.1) C1
E = 4.1 1.6 10–19 = 6.56 10–19(J) C1
(Allow this mark for correct conversion of either 1.9 eV or 2.2 eV to joules)
C1
= 3.03 10–7 3.0 10–7(m)
(Allow 1 sf answer) A1
(Allow 3/4 marks for = 4.85 10–26 m when eV is not converted to joules)
[6]
1. 3.9 eV = 3.9 × 1.6 × 10–19 J (= 6.24 × 10–19 J) (1)
λ = hc/E = 6.63 × 10–34 × 3.0 × 108 / 3.9 × 1.6 × 10–19 (= 320 × 10–9 m) (1) 2
[2]
2. (i) 1. E = hf / / f = 7.5 1017 (Hz) C1
(‘E = hf’ can be secured in (i))
/ E = 6.63 10–34 7.5 1017 C1
energy = 4.97 10–16 (J) 5.0 10–16 (J) (Allow 1 sf answer here) A1
2. (Possible ecf)
energy = 3.1 103 (eV) B1
(ii) The answer to (c)(i)1. and 1.4 (W) are used to determine the rate
of photons C1
(Possible ecf) C1
number = 2.8 1015 (s–1) (If 3100 eV is used, then allow 2/3 for 4.5 10–4) A1
[7]
3. (Allow any subject) C1
C1
v = 1.43 103 1.4 103 (ms–1) A1
[3]
4. (i) Visible (light) B1
(ii) work function = 1.9 1.6 10–19 M1
work function = 3.04 10–19 (J) 3.0 10–19 (J) A0
(iii) 1. E = hf / C1
E =
E = 3.9 10–19 (J) A1
2. hf = + KE(max) / hf = + ½ mv2
(Allow E = + ½ mv2 if E is qualified in (iii)1.) C1
3.9 10–19 = 3.0 10–19 + KE(max) / 3.9 10–19 = 3.04 10–19 + KE(max) C1
KE = 9.0 10–20 (J) / KE = 8.6 10–20 (J) (Possible ecf) A1
(iv) No change (to maximum KE of electron) B1
Each photon has same energy (but there are fewer photons) B1
(v) number of photons = ( 2.05 1017) (Possible ecf) C1
number of electrons = 0.07
number of electrons = 1.44 1016 (s–1) 1.4 × 1016 (s–1) A1
[11]
5.
(Four correct: 3 marks, three correct: 2 marks, two correct: 1 mark) B3
[3]
6. (i) particle / particulate / quantum / photon B1
(ii) wave B1
[2]
7. Maximum of three from points 1 to 6: B1 3
1. Photon mentioned (e.g.: photons interact with the surface electrons)
2. Energy is conserved (between the photon and the electron / in the interaction)
3. hf = KE(max)
4. A single photon interacts with a single electron / It is a one-to-one interaction
5. Electron is removed when photon energy is greater than / equal to the
work function
(energy) / (Allow ora)
6. Electron removed when frequency is greater than / equal to the threshold
frequency (Allow ora)
7. (Visible) light has lower frequency than the threshold frequency / Energy of
(visible) light photon is less than the work function (energy) (ora with uv) B1
8. Greater intensity of (visible) light means more photons (per unit time) /
energy of a photon remains the same B1
QWC Spelling, punctuation and grammar B1
Organisation B1
[7]
8. (a) quantum of energy / radiation / packet of energy B1
(b) (i) f = E/h = 5.60 × 10–19 /6.63 × 10–34 C1
f = 8.45 × 1014 (Hz) A1
(ii) 1 minimum energy to release an electron from the surface (of the metal) B1
2 5.60 × 10–19 – 4.80 × 10–19 ( = 8.0 × 10–20 J) B1
(iii) 8.0 × 10–20 = ½(9.1 × 10–31)v2 M1
giving v = 4.2 × 105 (m s–1) A1
(c) (i) Correct selection of: λ = h/p or λ = h/mv M1
where all symbols are defined A1
(ii) λ = 6.6 × 10–34 /(9.1 × 10–31 × 4.2 × 105) C1
λ = 1.7 × 10–9 (m) A1
[11]
9. (a) one (or more) electrons removed (or added) to an atom 1
(b) E = hf = hc/λ together with knowledge of symbol meaning (1)
= (1)
= 8.36 × 10–19 (J) (1) 3
(c) frequency of UV is greater than frequency of light
OR alternative statement in terms of wavelength.
so photon energy of visible light is less than photon energy of UV (1)
PLUS one of the idea of conservation of energy
it is not possible for a low energy photon to give a high energy photon
this is a one to one process (1) 2
(d) E = V/d and power of 10 correct for d (1)
= 30/0.00020 = 150 000 (1)
V m–1 (1) 3
[9]
10. (a) Maximum of five marks
Up to four from:
/ M1
All symbols (, h, m and v or p) defined A1
Electrons travel / move / propagate (through space) as a wave B1
Electrons are diffracted / ‘spread out’ M1
by the atoms / spacing between the atoms A1
The electrons are diffracted when their wavelength is less than or
comparable or same as size of atoms / gap between the atoms B1
Up to two from:
When the speed of electrons is increased) the rings ‘get smaller’ B1
(At greater speed of electrons) the wavelength is shorter B1
(At greater speed of electrons) there is less diffraction B1
QWC Organisation B1
Spelling, punctuation & grammar B1
(b) Electrons have mass / momentum / charge / can be ‘accelerated’ B1
[8]
11. (i) The minimum frequency needed to free an electron
(from the surface of a metal) B1
(ii)1 Line extended intersects (the f axis at) this value / At this frequency, Ek = 0 B1
(ii)2 ( = ) h 5.0 1014 / ( = ) 6.63 10–34 5.0 1014 C1
work function energy = 3.3 10–19 J A1
(iii)1 1 The gradient / slope of the line is the same B1
The gradient is equal to h / independent of the metal B1
(iii)2 The line is shifted to the right B1
The threshold frequency is greater (AW) B1
[8]
12. (a) Any five from: B1 × 5
1. Photoelectric (effect) mentioned
2. Photon(s) mentioned in correct context / E = hf
3. One-to-one ‘interaction’ between photon & electron
4. Surface electrons are involved
5. Electron released / photoelectric (effect) when photon
energy > / = work function (energy)
6. Electrons emitted / photoelectric (effect) when
frequency > / = threshold frequency
7. Energy is conserved (in the ‘interaction’ between photon and electron)
8. Reference to Einstein’s equation: hf = + KE(max)
[QWC: Spelling and Grammar]
(b) (i) 1. (energy of photon = 2.2 + 0.3) B1
2.5 (eV) B1
2. (energy =) 2.5 1.6 10–19 (Possible ecf from (b)(i)1.) C1
4.0 10–19 (J) (Allow 1 sf answer) A1
(ii) (f =) (Possible ecf) C1
(f = )
(f =) 6.03 1014 6.0 1014 (Hz) (Allow 6 1014) A1
(c) Each photon has more energy / There are fewer photons (in B1
a given time because intensity is the same)
Smaller current B1
[13]
13. (i) 1. The minimum frequency (of radiation \ waves) needed for electrons
to be released (from the metal surface) \ for photoelectric effect B1
2. Its temperature increases \ gets warm \ ‘heats up’ B1
(ii) E = 2.2 1.9 (= 4.1) C1
E = 4.1 1.6 10–19 = 6.56 10–19(J) C1
(Allow this mark for correct conversion of either 1.9 eV or 2.2 eV to joules)
C1
= 3.03 10–7 3.0 10–7(m)
(Allow 1 sf answer) A1
(Allow 3/4 marks for = 4.85 10–26 m when eV is not converted to joules)
[6]
Tuesday, May 11, 2010
Y13 Stars
Questions on Stars
1. any 4 from:
end of H burning/red giant/supergiant (1)
onset of He fusion/fusion of heavier nuclei (1)
gravitational collapse of core (1)
supernova explosion/ star explodes (1)
suitable mass limit (chanderasekha limit 1.4M) (1)
supported against gavity by neutron gas pressure/ ref to
Fermi pressure (1)
internal structure protons and electrons combined/ very
thin atmosphere/ metallic crust (1) 4
[4]
2. Any 5 from
red shift data for galaxies (accept stars) 1
calculate velocity from red shift 1
galaxies/ stars receding from Earth 1
distance data for galaxies/ stars 1
velocity α distance / v/r = constant / v-r graph straight line 1
universe began at a single point 1
[5]
3. (a) Any two
stars rotate around galactic centre 1
star with velocity component towards Earth 1
reference to motion/shape of galaxy 1
or other valid points eg blue shift
(b) Ho = 75/ 3 × 1019 s–1 1
t ≈ 1/ 2.5 × 10–18 1
t ≈ 4 × 1017 s 1
[5]
4. Hydrogen atoms/particles (1)
Collapse under gravity/ decrease of gpe (1)
Increase in kinetic energy/ temperature (1)
Fusion of protons (1)
Energy released/ ref. to E = ∆mc2 (1)
[5]
5. Any 6 from
Nuclear/hydrogen burning ends (1)
Mass > Chandrasekhar limit (1)
Expanding gas/planetary nebular/red giant (1)
Gravitational collapse /ref. to burning He or higher metals (1)
Correct ref. to (Fermi) pressure/ radiation pressure (1)
(must have ref. to pressure or force from radiation.)
Neutron star (neutron by itself, not enough) (1)
Correct reference to Schwarzschild radius/
allow mass> 3M/ allow ref. critical radius (1)
Black Hole (1) 6
[6]
6. Ho2 = (1 × 10–26 × 8 × π × 6.67 × 10–11) / 3 C1
Ho = 2.36 × 10–18 s–1 A1
[2]
7. (i) v/c = ∆λ / λ (1)
∆λ = 656.3 × 10–9 × 6.1 / 3 × 108 (ignore minus sign) (1)
∆λ = 1.33 × 10–14 m (1) 3
(ii) Graph: any 4 points plotted correctly (1)
all correct (1) 2
(iii) graph: draw curve, reasonable attempt (1) 1
(iv) Either point where star moves perpendicular to line of sight (1) 1
(v) time = 72 h ± (1)h (ecf read value from their graph ± 1 h) (1) 1
(vi) r = 3√(6.7 × 10–11 × 4 × 1030 × [72 × 3600]2/ 4π2) ecf (1)
r = 7.70 × 109 m ecf . (1) 2
(use of t = 72h 1/2)
[10]
1. any 4 from:
end of H burning/red giant/supergiant (1)
onset of He fusion/fusion of heavier nuclei (1)
gravitational collapse of core (1)
supernova explosion/ star explodes (1)
suitable mass limit (chanderasekha limit 1.4M) (1)
supported against gavity by neutron gas pressure/ ref to
Fermi pressure (1)
internal structure protons and electrons combined/ very
thin atmosphere/ metallic crust (1) 4
[4]
2. Any 5 from
red shift data for galaxies (accept stars) 1
calculate velocity from red shift 1
galaxies/ stars receding from Earth 1
distance data for galaxies/ stars 1
velocity α distance / v/r = constant / v-r graph straight line 1
universe began at a single point 1
[5]
3. (a) Any two
stars rotate around galactic centre 1
star with velocity component towards Earth 1
reference to motion/shape of galaxy 1
or other valid points eg blue shift
(b) Ho = 75/ 3 × 1019 s–1 1
t ≈ 1/ 2.5 × 10–18 1
t ≈ 4 × 1017 s 1
[5]
4. Hydrogen atoms/particles (1)
Collapse under gravity/ decrease of gpe (1)
Increase in kinetic energy/ temperature (1)
Fusion of protons (1)
Energy released/ ref. to E = ∆mc2 (1)
[5]
5. Any 6 from
Nuclear/hydrogen burning ends (1)
Mass > Chandrasekhar limit (1)
Expanding gas/planetary nebular/red giant (1)
Gravitational collapse /ref. to burning He or higher metals (1)
Correct ref. to (Fermi) pressure/ radiation pressure (1)
(must have ref. to pressure or force from radiation.)
Neutron star (neutron by itself, not enough) (1)
Correct reference to Schwarzschild radius/
allow mass> 3M/ allow ref. critical radius (1)
Black Hole (1) 6
[6]
6. Ho2 = (1 × 10–26 × 8 × π × 6.67 × 10–11) / 3 C1
Ho = 2.36 × 10–18 s–1 A1
[2]
7. (i) v/c = ∆λ / λ (1)
∆λ = 656.3 × 10–9 × 6.1 / 3 × 108 (ignore minus sign) (1)
∆λ = 1.33 × 10–14 m (1) 3
(ii) Graph: any 4 points plotted correctly (1)
all correct (1) 2
(iii) graph: draw curve, reasonable attempt (1) 1
(iv) Either point where star moves perpendicular to line of sight (1) 1
(v) time = 72 h ± (1)h (ecf read value from their graph ± 1 h) (1) 1
(vi) r = 3√(6.7 × 10–11 × 4 × 1030 × [72 × 3600]2/ 4π2) ecf (1)
r = 7.70 × 109 m ecf . (1) 2
(use of t = 72h 1/2)
[10]
y12 Springs
Questions on springs
1. (a) The extension of a spring is directly proportional to the applied force M1
as long as the elastic limit is not exceeded) A1
(b) (i) Correct pair of values read from the graph
force constant = 12/0.080 C1
force constant = 150 (N m–1) A1
(ii) extension, x = × 80 (= 133.33) (mm) C1
(E = ½ Fx)
energy = ½ × 2/ × 133.33 × 10–3
energy = 1.33 (J) A1
(iii) The spring has not exceeded its elastic limit B1
(iv) (elastic potential energy = kinetic energy)
M1
m and k are constant, therefore x v. M1
[9]
2. (a) Young modulus = stress/strain
(As long as elastic limit is not exceeded) B1
(b) Strain has no units because it is the ratio of two lengths. B1
[2]
3. (a) A brittle material does not have a plastic region / it breaks at its elastic limit. B1
(b) Ultimate tensile strength is breaking stress for a material B1
Materials can be chosen / tested to prevent collapse of the bridge B1
[3]
4. (i) 1. Measure the time t for N oscillations. M1
frequency f = N/t A1
2. Measure the amplitude A of the oscillations using the ruler. M1
maximum speed is calculated using: vmax = (2πf )A A1
(ii) The maximum speed is doubled B1
because the frequency is the same and vmax A B1
(iii) F = (–) kx and F = ma
Therefore ma = (–) kx M2
ω2 = k / m
T = M1
[9]
5. (a) One reading from the graph e.g. 1.0 N causes 7 mm C1
Hence 5.0 (N) causes 35 0.5 (mm) A1
(allow one mark for 35 1 (mm)
(b) (i) Force on each spring is 2.5 (N) C1
extension = 17.5 (mm) allow 18 (mm) or reading from graph A1
[allow ecf from (a)]
(ii) strain energy = area under graph / ½ F e C1
= 2 0.5 2.5 17.5 10–3
= 0.044 (J) A1
[allow ecf from (b)(i)]
(c) E = stress / strain C1
Stress = force / area and strain = extension / length C1
extension = (F L) / (A E)
= (5 0.4) / (2 10–7 2 1011)
= 5.(0) 10–5 (m) A1
(d) strain energy is larger in the spring B1
extension is (very much larger) (for the same force) for the spring B1
[11]
6. (i) 1 Elastic as returns to original length (when load is removed) B1
2 Hooke’s law is obeyed as force is proportional to the extension B1
Example of values given in support from table B1
(ii) Measure (original) length with a (metre) rule / tape B1
Suitable method for measuring the extension e.g.
levelling micrometer and comparison wire or fixed
scale plus vernier or travelling microscope and marker / pointer B1
(iii) E = stress / strain C1
= (25 1.72) / (1.8 10–7 1.20 10–3) C1
= 1.99 1011 (Pa) A1
[8]
7. (i) Density = mass / volume B1
Area length = mass / density
Area = (2.0 10–3) / (7800 0.5) or 2.56 10–7 / 0.5 B1
= 5.1(3) 10–7 m2 A0
(ii) E = (F l) / (A e) / stress = F / A (1.6 108 and strain
= e / l (8 10–4) C1
F = (E A e) / l
= (2 1011 5.1 10–7 4.0 10–4) / 0.5 C1
=82 (N) (81.6) A1
(iii) Diameter for D is half G hence area is ¼ of G
Extension is 4x greater
Tension required is the same = 82 (N) A1
(iv) The extension is proportional to the force / Hooke’s B1
law (OWTE)
[7]
8. (i) Graph through origin with (short) linear section then reducing gradient. 1
(ii) Straight section - elastic; (1)
Curved section - plastic. (1) 2
[3]
9. (a) P.E. at top = 80 × 9.8(1) × 150 = 118 000 (J) (1)
K.E. at bottom and at top = 0 (1)
Elastic P.E. at top = 0, at bottom = P.E. at top for ecf = 118 000 J (1) 3
(b) 24 N m–1 × 100 m = 2400 N 1
(c) elastic P.E. is area under F-x graph (1)
graph is a straight line so energy is area of triangle (1)
elastic P.E. = ½ × kx × x = (½kx2) (1) 2
(d) loss of P.E. = 100 × 9.8(1) × 150 = 147 000 J (1)
gain of elastic P.E. = ½ × 26.7 × 1052 = 147 000 J (1) 2
(e) idea that a given (unit) extension for a shorter rope requires a greater force 1
[9]
10. (a) (i) Stress = force / area C1
force = stress area
= 180 106 1.5 10–4
= 27000 (N) A1
(ii) Y M = stress / strain C1
= 180 106 / 1.2 10–3 or using the gradient C1
= 1.5 1011 N m–2 A1
(b) brittle
elastic/ graph shown up to elastic limit
obeys Hooke’s law / force α extension / stress α strain
no plastic region B3
MAX 3
[8]
1. (a) The extension of a spring is directly proportional to the applied force M1
as long as the elastic limit is not exceeded) A1
(b) (i) Correct pair of values read from the graph
force constant = 12/0.080 C1
force constant = 150 (N m–1) A1
(ii) extension, x = × 80 (= 133.33) (mm) C1
(E = ½ Fx)
energy = ½ × 2/ × 133.33 × 10–3
energy = 1.33 (J) A1
(iii) The spring has not exceeded its elastic limit B1
(iv) (elastic potential energy = kinetic energy)
M1
m and k are constant, therefore x v. M1
[9]
2. (a) Young modulus = stress/strain
(As long as elastic limit is not exceeded) B1
(b) Strain has no units because it is the ratio of two lengths. B1
[2]
3. (a) A brittle material does not have a plastic region / it breaks at its elastic limit. B1
(b) Ultimate tensile strength is breaking stress for a material B1
Materials can be chosen / tested to prevent collapse of the bridge B1
[3]
4. (i) 1. Measure the time t for N oscillations. M1
frequency f = N/t A1
2. Measure the amplitude A of the oscillations using the ruler. M1
maximum speed is calculated using: vmax = (2πf )A A1
(ii) The maximum speed is doubled B1
because the frequency is the same and vmax A B1
(iii) F = (–) kx and F = ma
Therefore ma = (–) kx M2
ω2 = k / m
T = M1
[9]
5. (a) One reading from the graph e.g. 1.0 N causes 7 mm C1
Hence 5.0 (N) causes 35 0.5 (mm) A1
(allow one mark for 35 1 (mm)
(b) (i) Force on each spring is 2.5 (N) C1
extension = 17.5 (mm) allow 18 (mm) or reading from graph A1
[allow ecf from (a)]
(ii) strain energy = area under graph / ½ F e C1
= 2 0.5 2.5 17.5 10–3
= 0.044 (J) A1
[allow ecf from (b)(i)]
(c) E = stress / strain C1
Stress = force / area and strain = extension / length C1
extension = (F L) / (A E)
= (5 0.4) / (2 10–7 2 1011)
= 5.(0) 10–5 (m) A1
(d) strain energy is larger in the spring B1
extension is (very much larger) (for the same force) for the spring B1
[11]
6. (i) 1 Elastic as returns to original length (when load is removed) B1
2 Hooke’s law is obeyed as force is proportional to the extension B1
Example of values given in support from table B1
(ii) Measure (original) length with a (metre) rule / tape B1
Suitable method for measuring the extension e.g.
levelling micrometer and comparison wire or fixed
scale plus vernier or travelling microscope and marker / pointer B1
(iii) E = stress / strain C1
= (25 1.72) / (1.8 10–7 1.20 10–3) C1
= 1.99 1011 (Pa) A1
[8]
7. (i) Density = mass / volume B1
Area length = mass / density
Area = (2.0 10–3) / (7800 0.5) or 2.56 10–7 / 0.5 B1
= 5.1(3) 10–7 m2 A0
(ii) E = (F l) / (A e) / stress = F / A (1.6 108 and strain
= e / l (8 10–4) C1
F = (E A e) / l
= (2 1011 5.1 10–7 4.0 10–4) / 0.5 C1
=82 (N) (81.6) A1
(iii) Diameter for D is half G hence area is ¼ of G
Extension is 4x greater
Tension required is the same = 82 (N) A1
(iv) The extension is proportional to the force / Hooke’s B1
law (OWTE)
[7]
8. (i) Graph through origin with (short) linear section then reducing gradient. 1
(ii) Straight section - elastic; (1)
Curved section - plastic. (1) 2
[3]
9. (a) P.E. at top = 80 × 9.8(1) × 150 = 118 000 (J) (1)
K.E. at bottom and at top = 0 (1)
Elastic P.E. at top = 0, at bottom = P.E. at top for ecf = 118 000 J (1) 3
(b) 24 N m–1 × 100 m = 2400 N 1
(c) elastic P.E. is area under F-x graph (1)
graph is a straight line so energy is area of triangle (1)
elastic P.E. = ½ × kx × x = (½kx2) (1) 2
(d) loss of P.E. = 100 × 9.8(1) × 150 = 147 000 J (1)
gain of elastic P.E. = ½ × 26.7 × 1052 = 147 000 J (1) 2
(e) idea that a given (unit) extension for a shorter rope requires a greater force 1
[9]
10. (a) (i) Stress = force / area C1
force = stress area
= 180 106 1.5 10–4
= 27000 (N) A1
(ii) Y M = stress / strain C1
= 180 106 / 1.2 10–3 or using the gradient C1
= 1.5 1011 N m–2 A1
(b) brittle
elastic/ graph shown up to elastic limit
obeys Hooke’s law / force α extension / stress α strain
no plastic region B3
MAX 3
[8]
Monday, May 10, 2010
y12
Questions on electricity
1. (i) M marked at the end of the graph B1
(ii) current is 5 (A) and p.d is 6 (V) C1
P = VI \ p = 6.0 5.0
(Allow p = I2 R or p = V2 \ R) C1
power = 30 (W) A1
(iii) 1. VL = 1.0 (V) (From the I/V graph) \ RL = 1.0/2.0 or 0.5 () M1
VR = 1.2 2.0 \ RT = 1.2 0.5 M1
V = 1.0 2.4 \ V = 1.7 2.0 A1
voltmeter reading = 3.4 (V) A0
2. Vr = 4.5 3.4 (= 1.1 V) \ 4.5 = 2.0r 3.4 (Possible ecf) C1
r = 0.55 () (1.05 scores 0/2 since the lamp is ignored) A1
[9]
2. (i) p.d across 1.5 k resistor = 5.0 1.2 = 3.8 (V) B1
(ii) C1
C1
R = 474 () 470 () A1
(Using 3.8 V instead of 1.2 V gives 4.75 k - allow 2/3)
[4]
3. (i) Any four from:
1. The resistance of the thermistor decreases (as temperature is increased) B1
2. The total resistance (of circuit) decreases B1
3. The voltmeter reading increases B1
4. Explanation of 3. above in terms of ‘sharing voltage’
/ B1
5. The current increases / ammeter reading increases B1
6. Explanation of current increase in terms of B1
(Allow ecf for statements 3. and 5. if statement 1. is incorrect maximum
score of 2/4)
(ii) C1
C1
R = 467 () 470 () A1
(When 1.4 V and 3.6 V are interchanged, then R = 3.1 103 () can score2/3)
(Calculation of total circuit resistance of 1.67 103 () can score 2/3)
(Use of scores 0/3)
[7]
4. (i) (NTC) thermistor B1
(ii) Resistance decreases when temperature is increased. (ora) B1
(Allow correct credit for a PTC thermistor)
(iii) 1 I = (0.032-0.006 =) 0.026 (A) B1
2 (V200 = 0.026 200 =) 5.2 (V) / (V720 = 0.006 700 =) 4.2 (V) C1
E = 5.2 – 4.2 (Allow E = 4.2 – 5.2) C1
E = 1.0 (V) (Allow 1 sf answer) A1
(9.4 (V) scores 1/3)
[6]
5. (i) The resistance of LDR/circuit changes (as light intensity changes) B1
When blade blocks light, resistance of LDR/circuit is large(r) (ora) B1
Correct statement about p.d (Possible ecf) B1
(ii) 1. (V = 5.0 – 3.0)
2.0 (V) (Allow 1 sf answer) B1
2. V = I = 2.0/2200 / 9.1 10–4 (A) C1
(3.0 = ) (R = 3.0 / 9.1 10–4)
R = 3300 () R = 3300 () Possible ecf A1
(For VLDR = 2.0 V, R = 1.47 k. This scores 1/2)
(If 3.5 V given in (ii)1., then R = 940 . This scores 2/2)
[6]
1. (i) M marked at the end of the graph B1
(ii) current is 5 (A) and p.d is 6 (V) C1
P = VI \ p = 6.0 5.0
(Allow p = I2 R or p = V2 \ R) C1
power = 30 (W) A1
(iii) 1. VL = 1.0 (V) (From the I/V graph) \ RL = 1.0/2.0 or 0.5 () M1
VR = 1.2 2.0 \ RT = 1.2 0.5 M1
V = 1.0 2.4 \ V = 1.7 2.0 A1
voltmeter reading = 3.4 (V) A0
2. Vr = 4.5 3.4 (= 1.1 V) \ 4.5 = 2.0r 3.4 (Possible ecf) C1
r = 0.55 () (1.05 scores 0/2 since the lamp is ignored) A1
[9]
2. (i) p.d across 1.5 k resistor = 5.0 1.2 = 3.8 (V) B1
(ii) C1
C1
R = 474 () 470 () A1
(Using 3.8 V instead of 1.2 V gives 4.75 k - allow 2/3)
[4]
3. (i) Any four from:
1. The resistance of the thermistor decreases (as temperature is increased) B1
2. The total resistance (of circuit) decreases B1
3. The voltmeter reading increases B1
4. Explanation of 3. above in terms of ‘sharing voltage’
/ B1
5. The current increases / ammeter reading increases B1
6. Explanation of current increase in terms of B1
(Allow ecf for statements 3. and 5. if statement 1. is incorrect maximum
score of 2/4)
(ii) C1
C1
R = 467 () 470 () A1
(When 1.4 V and 3.6 V are interchanged, then R = 3.1 103 () can score2/3)
(Calculation of total circuit resistance of 1.67 103 () can score 2/3)
(Use of scores 0/3)
[7]
4. (i) (NTC) thermistor B1
(ii) Resistance decreases when temperature is increased. (ora) B1
(Allow correct credit for a PTC thermistor)
(iii) 1 I = (0.032-0.006 =) 0.026 (A) B1
2 (V200 = 0.026 200 =) 5.2 (V) / (V720 = 0.006 700 =) 4.2 (V) C1
E = 5.2 – 4.2 (Allow E = 4.2 – 5.2) C1
E = 1.0 (V) (Allow 1 sf answer) A1
(9.4 (V) scores 1/3)
[6]
5. (i) The resistance of LDR/circuit changes (as light intensity changes) B1
When blade blocks light, resistance of LDR/circuit is large(r) (ora) B1
Correct statement about p.d (Possible ecf) B1
(ii) 1. (V = 5.0 – 3.0)
2.0 (V) (Allow 1 sf answer) B1
2. V = I = 2.0/2200 / 9.1 10–4 (A) C1
(3.0 = ) (R = 3.0 / 9.1 10–4)
R = 3300 () R = 3300 () Possible ecf A1
(For VLDR = 2.0 V, R = 1.47 k. This scores 1/2)
(If 3.5 V given in (ii)1., then R = 940 . This scores 2/2)
[6]
y11 Triple
Y11 Triple questions on Med Physics
1. (a) A dot 3 squares up from centre line; 1
B any two from
4 squares vertically
negative direction line; 1
(b) any two from
6V
time base off
alternating/a.c.;; 2
allow 12V battery/d.c. for 1 mark
[4]
2. (a) minimum amount of energy to maintain life (per day)(OWTTE); 1
Allow:
• is the amount of energy expended while at rest
• is the number of calories you’d burn if you stayed
in bed all day
• number of calories needed to do nothing all day and
remain alive
• minimum amount of energy needed by the body to maintain
life
Reject heart rate
(b) take more exercise;
increase muscle to fat ratio;
eat a balanced diet;
if 4 ticks, deduct 1 mark from the total
if 5 ticks, then 0 marks 3
(c) (i) heart and other organs shown correctly; 1
(ii) the heart; 1
[6]
3. (a) (i) breathing, respiration, kidney action; 1
(ii) 200 × 2;
400 J; 2
(iii) energy transferred is equal to work done;
400 × 10 / 4200 = 1; 2
(iv) Any 2 of:
1. to reduce the effect of interference due to movement
2. to reduce the effect of electrical noise
3. back-up strategy
4. to help eliminate the reading due to non-pulsatile
(veinous or capillary) blood; 2
(b) (i) dangerous to the volunteer / unethical / OWTTE; 1
(ii) unreliable;
has been extrapolated / is a guess 2
[10]
4. all correct = 3 marks
2 correct = 2 marks
1 correct = 1 mark
action Number?
two gamma rays are produced 4
the radioactive isotope emits positrons 2 given
gamma rays are detected 5
a computer puts the images together 6
positrons annihilate electrons 3 given
a 3-D image is produced 7 given
the patient is injected with a radioactive isotope 1
[3]
5. (a) One from
• action potential (caused by heart muscle)
• p.d. / voltage (of the heart) with respect to time
• electrical activity (of the heart); 1
Accept:
• p.d./ voltage across the skin caused by heart activity
Ignore: heart beat/pulse
• heart rate/ pulse rate
ignore heart action
• heart activity
(b)
P = atria contract;
QRS = ventricles contract;
T = recovery wave; 2
all three correct = 2 marks
any one correct = 1 mark
(c) in order
allow descriptions in terms of
• the frequency or period of the heart beats
• specific section of the wave being too long/short
heart beats too slow/eq;
heart beats too fast/eq;
heart beat is irregular/eq; 3
for 3rd mark accept abnormal or (fast but) missing a beat
Ignore random
[6]
6. (a) KE = 1.6 x 10-19 x 25 x 103;;
[Evidence of 103 needed – deduct 1 mark if not used 103] 2
(b) A calculation to include:
1. K.E. = ½ x m x v2;
2. 4 x 10-15 = 0.5 x 9.1 x 10-31 x v2;
3. v = 9.4 x 107 ; 3
(c) Any four valid points, for example,
1. the electron will fall;
2. gravitational field is acting on it;
3. moving so fast;
4. (X/Y) plates used to compensate / position electron;
5. time to reach screen is very small / distance to screen short;
6. distance fallen will be negligible; 4
plus one communication mark for presenting relevant information in a 1
form that suits its purpose
[10]
7. (a) waves closer in glass;
emerging waves parallel to incident waves;
are the same wavelength as the incident beam;
3
(b) Any two of:
some reflected back into the water;
because angle greater than critical angle / 41° – 45° ;
these rays are totally internally reflected ; 2
[5]
8. (a) inserted down throat into/for internal examination of stomach; 1
(b) total internal reflection; 1
(c) angle 75 < x < 80; 1
(d) flexible/can bend easier etc. 1
[4]
9. (a) (i) change of speed (of light); 1
Allow change of direction /bending when qualified, e.g.
bends towards/away the normal, bends when it changes medium, bends when it enters a different medium
Ignore light bends / change of direction without qualification
(ii) correct angle at first reflection within set tolerance;
reasonable attempt to show continuous TIR along the fibre; 2
Allow:
• tolerance range is vertically between the first pt of impact,
and the end of the leader line
• straight by eye
(b) (i) mechanism is (T)IR;
Ignore endoscope
any one from
• idea of passage of light along a fibre
Allow tube for fibre
• idea of reflection inside the body; 2
Allow look for ans that describe mechanism not purpose or description
(ii) any valid suggestion; 1
Allow:
eg
more operations (possible)
idea of cosmetically better (scarring, hole size)
reducing pain
less anaesthetic/stay awake
less bleeding
reduced infection risk
fewer complications
faster recovery time
Ignore:
any implication of money
unqualified ‘safety’
unqualified ‘quicker’
[6]
10. calc. of distance (in m or in cm); 3.3(m)
330(cm)
Accept no alternatives
check the diagram for the calc of distance method
Reject incorrect units for one mark
e.g. N/m
substitution;
3.3 × 805
Acceptable Substitutions:
Distance
(m) Ans Distance
(cm) Ans
3.3
(or ecf) 2656 330
(or ecf) 265650
0.9 725 90 72450
1.4 1127 140 112700
1.8 1449 180 144900
1.9 1530 190 152950
2.4 1932 240 193200
2.8 2254 280 225400
Unit: J or
Nm Ncm
(Allow ecf into answers for 3.3m or 330cm)
answer in correct units; 2 660 J
2656.5(J)
2.656 kJ
ecf from substitution, as shown in above table. Answer plus correct unit.
[3]
11. (a) (i) A - heater filament
B - X deflecting plates
C - fluorescent screen
D - Y deflecting plates
E - accelerating anode 4
All five correct - 4 marks
four/three correct - 3 marks
two correct - 2 marks
one correct - 1 mark
(ii) An explanation to include:
1. electrons would not reach the screen;
2. undergo many collisions with air particles; 2
(b) A suggestion to include any two valid reasons, for example:
1. vacuum not good enough;
2. not a high enough voltage across the plates;
3. plates too far apart;
4. plates not long enough;
5. charged particles did not leave the region of the plates; 2
(c) Any two examples of where cathode rays are used, for example:
1. TV;
2. computers;
3. electron microscope; 2
[Allow CRO]
[10]
1. (a) A dot 3 squares up from centre line; 1
B any two from
4 squares vertically
negative direction line; 1
(b) any two from
6V
time base off
alternating/a.c.;; 2
allow 12V battery/d.c. for 1 mark
[4]
2. (a) minimum amount of energy to maintain life (per day)(OWTTE); 1
Allow:
• is the amount of energy expended while at rest
• is the number of calories you’d burn if you stayed
in bed all day
• number of calories needed to do nothing all day and
remain alive
• minimum amount of energy needed by the body to maintain
life
Reject heart rate
(b) take more exercise;
increase muscle to fat ratio;
eat a balanced diet;
if 4 ticks, deduct 1 mark from the total
if 5 ticks, then 0 marks 3
(c) (i) heart and other organs shown correctly; 1
(ii) the heart; 1
[6]
3. (a) (i) breathing, respiration, kidney action; 1
(ii) 200 × 2;
400 J; 2
(iii) energy transferred is equal to work done;
400 × 10 / 4200 = 1; 2
(iv) Any 2 of:
1. to reduce the effect of interference due to movement
2. to reduce the effect of electrical noise
3. back-up strategy
4. to help eliminate the reading due to non-pulsatile
(veinous or capillary) blood; 2
(b) (i) dangerous to the volunteer / unethical / OWTTE; 1
(ii) unreliable;
has been extrapolated / is a guess 2
[10]
4. all correct = 3 marks
2 correct = 2 marks
1 correct = 1 mark
action Number?
two gamma rays are produced 4
the radioactive isotope emits positrons 2 given
gamma rays are detected 5
a computer puts the images together 6
positrons annihilate electrons 3 given
a 3-D image is produced 7 given
the patient is injected with a radioactive isotope 1
[3]
5. (a) One from
• action potential (caused by heart muscle)
• p.d. / voltage (of the heart) with respect to time
• electrical activity (of the heart); 1
Accept:
• p.d./ voltage across the skin caused by heart activity
Ignore: heart beat/pulse
• heart rate/ pulse rate
ignore heart action
• heart activity
(b)
P = atria contract;
QRS = ventricles contract;
T = recovery wave; 2
all three correct = 2 marks
any one correct = 1 mark
(c) in order
allow descriptions in terms of
• the frequency or period of the heart beats
• specific section of the wave being too long/short
heart beats too slow/eq;
heart beats too fast/eq;
heart beat is irregular/eq; 3
for 3rd mark accept abnormal or (fast but) missing a beat
Ignore random
[6]
6. (a) KE = 1.6 x 10-19 x 25 x 103;;
[Evidence of 103 needed – deduct 1 mark if not used 103] 2
(b) A calculation to include:
1. K.E. = ½ x m x v2;
2. 4 x 10-15 = 0.5 x 9.1 x 10-31 x v2;
3. v = 9.4 x 107 ; 3
(c) Any four valid points, for example,
1. the electron will fall;
2. gravitational field is acting on it;
3. moving so fast;
4. (X/Y) plates used to compensate / position electron;
5. time to reach screen is very small / distance to screen short;
6. distance fallen will be negligible; 4
plus one communication mark for presenting relevant information in a 1
form that suits its purpose
[10]
7. (a) waves closer in glass;
emerging waves parallel to incident waves;
are the same wavelength as the incident beam;
3
(b) Any two of:
some reflected back into the water;
because angle greater than critical angle / 41° – 45° ;
these rays are totally internally reflected ; 2
[5]
8. (a) inserted down throat into/for internal examination of stomach; 1
(b) total internal reflection; 1
(c) angle 75 < x < 80; 1
(d) flexible/can bend easier etc. 1
[4]
9. (a) (i) change of speed (of light); 1
Allow change of direction /bending when qualified, e.g.
bends towards/away the normal, bends when it changes medium, bends when it enters a different medium
Ignore light bends / change of direction without qualification
(ii) correct angle at first reflection within set tolerance;
reasonable attempt to show continuous TIR along the fibre; 2
Allow:
• tolerance range is vertically between the first pt of impact,
and the end of the leader line
• straight by eye
(b) (i) mechanism is (T)IR;
Ignore endoscope
any one from
• idea of passage of light along a fibre
Allow tube for fibre
• idea of reflection inside the body; 2
Allow look for ans that describe mechanism not purpose or description
(ii) any valid suggestion; 1
Allow:
eg
more operations (possible)
idea of cosmetically better (scarring, hole size)
reducing pain
less anaesthetic/stay awake
less bleeding
reduced infection risk
fewer complications
faster recovery time
Ignore:
any implication of money
unqualified ‘safety’
unqualified ‘quicker’
[6]
10. calc. of distance (in m or in cm); 3.3(m)
330(cm)
Accept no alternatives
check the diagram for the calc of distance method
Reject incorrect units for one mark
e.g. N/m
substitution;
3.3 × 805
Acceptable Substitutions:
Distance
(m) Ans Distance
(cm) Ans
3.3
(or ecf) 2656 330
(or ecf) 265650
0.9 725 90 72450
1.4 1127 140 112700
1.8 1449 180 144900
1.9 1530 190 152950
2.4 1932 240 193200
2.8 2254 280 225400
Unit: J or
Nm Ncm
(Allow ecf into answers for 3.3m or 330cm)
answer in correct units; 2 660 J
2656.5(J)
2.656 kJ
ecf from substitution, as shown in above table. Answer plus correct unit.
[3]
11. (a) (i) A - heater filament
B - X deflecting plates
C - fluorescent screen
D - Y deflecting plates
E - accelerating anode 4
All five correct - 4 marks
four/three correct - 3 marks
two correct - 2 marks
one correct - 1 mark
(ii) An explanation to include:
1. electrons would not reach the screen;
2. undergo many collisions with air particles; 2
(b) A suggestion to include any two valid reasons, for example:
1. vacuum not good enough;
2. not a high enough voltage across the plates;
3. plates too far apart;
4. plates not long enough;
5. charged particles did not leave the region of the plates; 2
(c) Any two examples of where cathode rays are used, for example:
1. TV;
2. computers;
3. electron microscope; 2
[Allow CRO]
[10]
Y11 Triple
Medical Physics Answers
1. (a) A dot 3 squares up from centre line; 1
B any two from
4 squares vertically
negative direction line; 1
(b) any two from
6V
time base off
alternating/a.c.;; 2
allow 12V battery/d.c. for 1 mark
[4]
2. (a) minimum amount of energy to maintain life (per day)(OWTTE); 1
Allow:
• is the amount of energy expended while at rest
• is the number of calories you’d burn if you stayed
in bed all day
• number of calories needed to do nothing all day and
remain alive
• minimum amount of energy needed by the body to maintain
life
Reject heart rate
(b) take more exercise;
increase muscle to fat ratio;
eat a balanced diet;
if 4 ticks, deduct 1 mark from the total
if 5 ticks, then 0 marks 3
(c) (i) heart and other organs shown correctly; 1
(ii) the heart; 1
[6]
3. (a) (i) breathing, respiration, kidney action; 1
(ii) 200 × 2;
400 J; 2
(iii) energy transferred is equal to work done;
400 × 10 / 4200 = 1; 2
(iv) Any 2 of:
1. to reduce the effect of interference due to movement
2. to reduce the effect of electrical noise
3. back-up strategy
4. to help eliminate the reading due to non-pulsatile
(veinous or capillary) blood; 2
(b) (i) dangerous to the volunteer / unethical / OWTTE; 1
(ii) unreliable;
has been extrapolated / is a guess 2
[10]
4. all correct = 3 marks
2 correct = 2 marks
1 correct = 1 mark
action Number?
two gamma rays are produced 4
the radioactive isotope emits positrons 2 given
gamma rays are detected 5
a computer puts the images together 6
positrons annihilate electrons 3 given
a 3-D image is produced 7 given
the patient is injected with a radioactive isotope 1
[3]
5. (a) One from
• action potential (caused by heart muscle)
• p.d. / voltage (of the heart) with respect to time
• electrical activity (of the heart); 1
Accept:
• p.d./ voltage across the skin caused by heart activity
Ignore: heart beat/pulse
• heart rate/ pulse rate
ignore heart action
• heart activity
(b)
P = atria contract;
QRS = ventricles contract;
T = recovery wave; 2
all three correct = 2 marks
any one correct = 1 mark
(c) in order
allow descriptions in terms of
• the frequency or period of the heart beats
• specific section of the wave being too long/short
heart beats too slow/eq;
heart beats too fast/eq;
heart beat is irregular/eq; 3
for 3rd mark accept abnormal or (fast but) missing a beat
Ignore random
[6]
6. (a) KE = 1.6 x 10-19 x 25 x 103;;
[Evidence of 103 needed – deduct 1 mark if not used 103] 2
(b) A calculation to include:
1. K.E. = ½ x m x v2;
2. 4 x 10-15 = 0.5 x 9.1 x 10-31 x v2;
3. v = 9.4 x 107 ; 3
(c) Any four valid points, for example,
1. the electron will fall;
2. gravitational field is acting on it;
3. moving so fast;
4. (X/Y) plates used to compensate / position electron;
5. time to reach screen is very small / distance to screen short;
6. distance fallen will be negligible; 4
plus one communication mark for presenting relevant information in a 1
form that suits its purpose
[10]
7. (a) waves closer in glass;
emerging waves parallel to incident waves;
are the same wavelength as the incident beam;
3
(b) Any two of:
some reflected back into the water;
because angle greater than critical angle / 41° – 45° ;
these rays are totally internally reflected ; 2
[5]
8. (a) inserted down throat into/for internal examination of stomach; 1
(b) total internal reflection; 1
(c) angle 75 < x < 80; 1
(d) flexible/can bend easier etc. 1
[4]
9. (a) (i) change of speed (of light); 1
Allow change of direction /bending when qualified, e.g.
bends towards/away the normal, bends when it changes medium, bends when it enters a different medium
Ignore light bends / change of direction without qualification
(ii) correct angle at first reflection within set tolerance;
reasonable attempt to show continuous TIR along the fibre; 2
Allow:
• tolerance range is vertically between the first pt of impact,
and the end of the leader line
• straight by eye
(b) (i) mechanism is (T)IR;
Ignore endoscope
any one from
• idea of passage of light along a fibre
Allow tube for fibre
• idea of reflection inside the body; 2
Allow look for ans that describe mechanism not purpose or description
(ii) any valid suggestion; 1
Allow:
eg
more operations (possible)
idea of cosmetically better (scarring, hole size)
reducing pain
less anaesthetic/stay awake
less bleeding
reduced infection risk
fewer complications
faster recovery time
Ignore:
any implication of money
unqualified ‘safety’
unqualified ‘quicker’
[6]
10. calc. of distance (in m or in cm); 3.3(m)
330(cm)
Accept no alternatives
check the diagram for the calc of distance method
Reject incorrect units for one mark
e.g. N/m
substitution;
3.3 × 805
Acceptable Substitutions:
Distance
(m) Ans Distance
(cm) Ans
3.3
(or ecf) 2656 330
(or ecf) 265650
0.9 725 90 72450
1.4 1127 140 112700
1.8 1449 180 144900
1.9 1530 190 152950
2.4 1932 240 193200
2.8 2254 280 225400
Unit: J or
Nm Ncm
(Allow ecf into answers for 3.3m or 330cm)
answer in correct units; 2 660 J
2656.5(J)
2.656 kJ
ecf from substitution, as shown in above table. Answer plus correct unit.
[3]
11. (a) (i) A - heater filament
B - X deflecting plates
C - fluorescent screen
D - Y deflecting plates
E - accelerating anode 4
All five correct - 4 marks
four/three correct - 3 marks
two correct - 2 marks
one correct - 1 mark
(ii) An explanation to include:
1. electrons would not reach the screen;
2. undergo many collisions with air particles; 2
(b) A suggestion to include any two valid reasons, for example:
1. vacuum not good enough;
2. not a high enough voltage across the plates;
3. plates too far apart;
4. plates not long enough;
5. charged particles did not leave the region of the plates; 2
(c) Any two examples of where cathode rays are used, for example:
1. TV;
2. computers;
3. electron microscope; 2
[Allow CRO]
[10]
1. (a) A dot 3 squares up from centre line; 1
B any two from
4 squares vertically
negative direction line; 1
(b) any two from
6V
time base off
alternating/a.c.;; 2
allow 12V battery/d.c. for 1 mark
[4]
2. (a) minimum amount of energy to maintain life (per day)(OWTTE); 1
Allow:
• is the amount of energy expended while at rest
• is the number of calories you’d burn if you stayed
in bed all day
• number of calories needed to do nothing all day and
remain alive
• minimum amount of energy needed by the body to maintain
life
Reject heart rate
(b) take more exercise;
increase muscle to fat ratio;
eat a balanced diet;
if 4 ticks, deduct 1 mark from the total
if 5 ticks, then 0 marks 3
(c) (i) heart and other organs shown correctly; 1
(ii) the heart; 1
[6]
3. (a) (i) breathing, respiration, kidney action; 1
(ii) 200 × 2;
400 J; 2
(iii) energy transferred is equal to work done;
400 × 10 / 4200 = 1; 2
(iv) Any 2 of:
1. to reduce the effect of interference due to movement
2. to reduce the effect of electrical noise
3. back-up strategy
4. to help eliminate the reading due to non-pulsatile
(veinous or capillary) blood; 2
(b) (i) dangerous to the volunteer / unethical / OWTTE; 1
(ii) unreliable;
has been extrapolated / is a guess 2
[10]
4. all correct = 3 marks
2 correct = 2 marks
1 correct = 1 mark
action Number?
two gamma rays are produced 4
the radioactive isotope emits positrons 2 given
gamma rays are detected 5
a computer puts the images together 6
positrons annihilate electrons 3 given
a 3-D image is produced 7 given
the patient is injected with a radioactive isotope 1
[3]
5. (a) One from
• action potential (caused by heart muscle)
• p.d. / voltage (of the heart) with respect to time
• electrical activity (of the heart); 1
Accept:
• p.d./ voltage across the skin caused by heart activity
Ignore: heart beat/pulse
• heart rate/ pulse rate
ignore heart action
• heart activity
(b)
P = atria contract;
QRS = ventricles contract;
T = recovery wave; 2
all three correct = 2 marks
any one correct = 1 mark
(c) in order
allow descriptions in terms of
• the frequency or period of the heart beats
• specific section of the wave being too long/short
heart beats too slow/eq;
heart beats too fast/eq;
heart beat is irregular/eq; 3
for 3rd mark accept abnormal or (fast but) missing a beat
Ignore random
[6]
6. (a) KE = 1.6 x 10-19 x 25 x 103;;
[Evidence of 103 needed – deduct 1 mark if not used 103] 2
(b) A calculation to include:
1. K.E. = ½ x m x v2;
2. 4 x 10-15 = 0.5 x 9.1 x 10-31 x v2;
3. v = 9.4 x 107 ; 3
(c) Any four valid points, for example,
1. the electron will fall;
2. gravitational field is acting on it;
3. moving so fast;
4. (X/Y) plates used to compensate / position electron;
5. time to reach screen is very small / distance to screen short;
6. distance fallen will be negligible; 4
plus one communication mark for presenting relevant information in a 1
form that suits its purpose
[10]
7. (a) waves closer in glass;
emerging waves parallel to incident waves;
are the same wavelength as the incident beam;
3
(b) Any two of:
some reflected back into the water;
because angle greater than critical angle / 41° – 45° ;
these rays are totally internally reflected ; 2
[5]
8. (a) inserted down throat into/for internal examination of stomach; 1
(b) total internal reflection; 1
(c) angle 75 < x < 80; 1
(d) flexible/can bend easier etc. 1
[4]
9. (a) (i) change of speed (of light); 1
Allow change of direction /bending when qualified, e.g.
bends towards/away the normal, bends when it changes medium, bends when it enters a different medium
Ignore light bends / change of direction without qualification
(ii) correct angle at first reflection within set tolerance;
reasonable attempt to show continuous TIR along the fibre; 2
Allow:
• tolerance range is vertically between the first pt of impact,
and the end of the leader line
• straight by eye
(b) (i) mechanism is (T)IR;
Ignore endoscope
any one from
• idea of passage of light along a fibre
Allow tube for fibre
• idea of reflection inside the body; 2
Allow look for ans that describe mechanism not purpose or description
(ii) any valid suggestion; 1
Allow:
eg
more operations (possible)
idea of cosmetically better (scarring, hole size)
reducing pain
less anaesthetic/stay awake
less bleeding
reduced infection risk
fewer complications
faster recovery time
Ignore:
any implication of money
unqualified ‘safety’
unqualified ‘quicker’
[6]
10. calc. of distance (in m or in cm); 3.3(m)
330(cm)
Accept no alternatives
check the diagram for the calc of distance method
Reject incorrect units for one mark
e.g. N/m
substitution;
3.3 × 805
Acceptable Substitutions:
Distance
(m) Ans Distance
(cm) Ans
3.3
(or ecf) 2656 330
(or ecf) 265650
0.9 725 90 72450
1.4 1127 140 112700
1.8 1449 180 144900
1.9 1530 190 152950
2.4 1932 240 193200
2.8 2254 280 225400
Unit: J or
Nm Ncm
(Allow ecf into answers for 3.3m or 330cm)
answer in correct units; 2 660 J
2656.5(J)
2.656 kJ
ecf from substitution, as shown in above table. Answer plus correct unit.
[3]
11. (a) (i) A - heater filament
B - X deflecting plates
C - fluorescent screen
D - Y deflecting plates
E - accelerating anode 4
All five correct - 4 marks
four/three correct - 3 marks
two correct - 2 marks
one correct - 1 mark
(ii) An explanation to include:
1. electrons would not reach the screen;
2. undergo many collisions with air particles; 2
(b) A suggestion to include any two valid reasons, for example:
1. vacuum not good enough;
2. not a high enough voltage across the plates;
3. plates too far apart;
4. plates not long enough;
5. charged particles did not leave the region of the plates; 2
(c) Any two examples of where cathode rays are used, for example:
1. TV;
2. computers;
3. electron microscope; 2
[Allow CRO]
[10]
Y13
Questions on electromagnetism
1. magnetic flux = BA 1
meanings of B and A, i.e. flux density or field strength and area to it 1
magnetic flux linkage refers to the flux linking/passing through a coil; 1
and equals N × flux where N is the number of turns (of the coil) 1
Faraday’s law: induced e.m.f./voltage is proportional to rate of change of flux
linkage through it /correct mathematical formulation/AW 1
Lenz’s law: the direction of the induced e.m.f./voltage is such as to
oppose the motion/change that produced it 1
relationship of Lenz’s law to conservation of energy or other valid
explanation/discussion/description 2
max 5 marks
quality of written communication 2
[7]
2. (a) (i) F is towards ‘open’ end of tube; using Fleming’s L.H. rule 2
(ii) F = BIw 1
(iii) F = 0.15 × 800 × 0.0025; = 3.0 (N) 2
(b) (i) A voltage is induced across moving metal as it cuts lines of flux/AW; (1)
voltage is proportional to flux change per second/AW; (1)
the flux change per second is Bwv / is proportional to the area of
metal moving through the field per second / is proportional to v (1)
or Faraday’s law fully stated; with reasonable attempt to; (2)
relate flux linkage per second proportionally to speed (1) 3
(ii) flux (linkage) doubles; so using Faraday’s law V doubles/AW 2
[10]
3. sine or cosine wave of regular period and amplitude (1)
V doubles when the speed v of rotation of the coil doubles; (1)
when v doubles the rate of change of flux linking the coil doubles; (1)
the frequency of the a.c. signal doubles/period halves/AW (1)
V doubles when the number n of turns on the coil doubles; (1)
when n doubles there is twice as much flux linking the coil/AW; (1)
the frequency/period of the signal is unchanged; (1)
without iron core flux linking coil is much less/flux would spread in all
directions/flux not channelled through low reluctance path/AW (1)
amplitude of output voltage is smaller (1)
actually is tiny/negligible/mV rather than V max 7
Quality of Written Communication 2
[9]
4. (i) I = V/R = 12/50 (1)
= 0.24 A (1) 2
(ii) Power in primary = power in secondary / IpVp = IsVs (1)
Ip = 0.24 × 12 / 230 = 0.0125 A (1) 2
[4]
5. (a) (i) BA / = 0.05 × 0.05 × 0.026; = 6.5 × 10–5; Wb/T m2 3
(ii) BA sin 45°/BAcos 45° = 4.6 × 10–5 Wb ecf (a)i 1
(iii) 0 1
(b) (i) a point where curve crosses t-axis 1
(ii) voltage is proportional to the rate of change of flux linking the coil; 1
rate of flux change is zero/very small when the flux linking the
coil is a maximum 1
(iii) sinusoidal curve; of double the amplitude; and half the period 3
[11]
6. (a) (i) equally spaced horizontal parallel lines from plate to plate (1)
arrows towards cathode (1) 2
(ii) ½ mv2 = qV; v = √(2eV/m) = √(2 × 1.6 × 10–19 × 7000/9.1 × 10–31) so (1)
v = 4.96 × 107 (m s–1) (1) 2
(b) (i) arrow perpendicular to path towards centre of arc (1) 1
(ii) out of paper/upwards;using Fleming’s LH rule (for conventional
current) (2) 2
(iii) mv2/r; = Bqv; r = mv/Bq = ;= 9.4 × 10–2
(m) 4
(c) change magnitude of current in coils to change field; (1)
change field to change deflection; (1)
reverse field/current to change deflection from up to down (1) max 2 2
[13]
1. magnetic flux = BA 1
meanings of B and A, i.e. flux density or field strength and area to it 1
magnetic flux linkage refers to the flux linking/passing through a coil; 1
and equals N × flux where N is the number of turns (of the coil) 1
Faraday’s law: induced e.m.f./voltage is proportional to rate of change of flux
linkage through it /correct mathematical formulation/AW 1
Lenz’s law: the direction of the induced e.m.f./voltage is such as to
oppose the motion/change that produced it 1
relationship of Lenz’s law to conservation of energy or other valid
explanation/discussion/description 2
max 5 marks
quality of written communication 2
[7]
2. (a) (i) F is towards ‘open’ end of tube; using Fleming’s L.H. rule 2
(ii) F = BIw 1
(iii) F = 0.15 × 800 × 0.0025; = 3.0 (N) 2
(b) (i) A voltage is induced across moving metal as it cuts lines of flux/AW; (1)
voltage is proportional to flux change per second/AW; (1)
the flux change per second is Bwv / is proportional to the area of
metal moving through the field per second / is proportional to v (1)
or Faraday’s law fully stated; with reasonable attempt to; (2)
relate flux linkage per second proportionally to speed (1) 3
(ii) flux (linkage) doubles; so using Faraday’s law V doubles/AW 2
[10]
3. sine or cosine wave of regular period and amplitude (1)
V doubles when the speed v of rotation of the coil doubles; (1)
when v doubles the rate of change of flux linking the coil doubles; (1)
the frequency of the a.c. signal doubles/period halves/AW (1)
V doubles when the number n of turns on the coil doubles; (1)
when n doubles there is twice as much flux linking the coil/AW; (1)
the frequency/period of the signal is unchanged; (1)
without iron core flux linking coil is much less/flux would spread in all
directions/flux not channelled through low reluctance path/AW (1)
amplitude of output voltage is smaller (1)
actually is tiny/negligible/mV rather than V max 7
Quality of Written Communication 2
[9]
4. (i) I = V/R = 12/50 (1)
= 0.24 A (1) 2
(ii) Power in primary = power in secondary / IpVp = IsVs (1)
Ip = 0.24 × 12 / 230 = 0.0125 A (1) 2
[4]
5. (a) (i) BA / = 0.05 × 0.05 × 0.026; = 6.5 × 10–5; Wb/T m2 3
(ii) BA sin 45°/BAcos 45° = 4.6 × 10–5 Wb ecf (a)i 1
(iii) 0 1
(b) (i) a point where curve crosses t-axis 1
(ii) voltage is proportional to the rate of change of flux linking the coil; 1
rate of flux change is zero/very small when the flux linking the
coil is a maximum 1
(iii) sinusoidal curve; of double the amplitude; and half the period 3
[11]
6. (a) (i) equally spaced horizontal parallel lines from plate to plate (1)
arrows towards cathode (1) 2
(ii) ½ mv2 = qV; v = √(2eV/m) = √(2 × 1.6 × 10–19 × 7000/9.1 × 10–31) so (1)
v = 4.96 × 107 (m s–1) (1) 2
(b) (i) arrow perpendicular to path towards centre of arc (1) 1
(ii) out of paper/upwards;using Fleming’s LH rule (for conventional
current) (2) 2
(iii) mv2/r; = Bqv; r = mv/Bq = ;= 9.4 × 10–2
(m) 4
(c) change magnitude of current in coils to change field; (1)
change field to change deflection; (1)
reverse field/current to change deflection from up to down (1) max 2 2
[13]
Friday, May 07, 2010
Y11 triple
Questions on Gases
1. (a) (i) particles collide with/hit/bounce off (container) wall/eq; 1
(ii) An explanation to include:
1. greater volume/space/room;
2. fewer/less collisions (with container); 2
(b) (i) increases/eq; 1
(ii) increases/eq; 1
[5]
2. (a) (i) points plotted correctly;;
[Allow 1 mark for 4 correct]
straight line; 3
(ii) pressure increases with temperature; 1
(iii) line back tracked;
260 10 C; 2
(iv) absolute zero/ 0 K; 1
(b) An explanation to include:
• tyres get hot/increase in temperature;
• air expands/more collisions/air particles inside move more
/have more KE/OWTTE; 2
[9]
3. (a) An explanation to include:
1. moving gas particles;
2. hitting container walls;
[Allow bouncing off the sides for 2 marks]
[Ignore vibrating/collisions with other particles] 2
(b) increases;
increases;
stays the same;
stays the same; 4
(c) (i) increases;
in proportion/linearly/steady rate/OWTTE; 2
(ii) correctly indicated – intercept with horizontal axis; 1
(iii) zero/minimum; 1
[10]
4. (a) (i) An explanation to include:
• particles moving;
• collisions with walls/hitting walls;
• force produced;
plus 1 communication mark for ensuring that spelling,
punctuation and grammar are accurate, so that the meaning
is clear; 4
(ii) An explanation to include:
• more collisions;
• more particles present; 2
(b) (I) An explanation to include:
• not moving/all gases in solid state;
• no K.E.; 2
(ii) kelvin; 1
(iii) (+)373(K) 1
[10]
5. A calculation to show:
1/2. ;;
[1 mark for conversion to K and 1 mark for correct substitution]
3. P2 = 120 478 (Pa); 3
[Allow ;
P2 = 400 000 (Pa); 2 marks max]
[3]
6. (a) (i) An explanation to include two from:
1. molecules are moving / have (kinetic) energy;
2. colliding with container walls;
3. each collision produces a force; 2
plus one communication mark for using a suitable structure and
style of writing 1
(ii) An explanation to include:
1. molecules moving faster / more (kinetic) energy
/ moving more;
2. colliding more often with the walls / harder collisions
/ more force; 2
(b) (i) all points plotted correctly;; [Deduct 1 mark for every two errors]
straight line of best fit; 3
[Systematic error in vertical scale – max 2 marks]
(ii) straight line extended to pressure axis;
pressure of 104.5 ± 0.3 (kPa) / value from their graph; 2
[Bald, correct answer scores 2 marks]
(iii) molecules still have energy (to move) / molecules still moving; 1
(iv) –273°C / 0 K / absolute zero; 1
[12]
7. (a) (i) A description to include:
1. particles moving;
2. in all directions/randomly /
or implied by description (each other / walls); 2
[Arrows on diagram acceptable]
(ii) An explanation to include:
1. particles hit/collide with container walls;
2. producing a force; 2
(iii) pressure would increase/get bigger/larger; 1
(b) statements 2, 3 and 4 ticked ; ; ;
[If more than 3 ticked then deduct 1 mark for each error] 3
(c) An explanation to include two from:
1. temperature of air in tyre increases / hot / hotter;
2. particles hit more often/hit harder;
3. particles moving faster / more energy; 2
plus one communication mark for presenting relevant information in a
form that suits its purpose 1
[11]
8. (a) random;
collide;
force;
decreases;
decreases; 5
(b) (i) A description to include:
1. increases;
2. uniformly / equally / steadily / in a straight line /
positive correlation / OWTTE; 2
(ii) X on temperature axis between 250 and 300 K /
allow if on line in right place; 1
(iii) zero; 1
(iv) not moving; 1
[Reject mainly still]
[10]
1. (a) (i) particles collide with/hit/bounce off (container) wall/eq; 1
(ii) An explanation to include:
1. greater volume/space/room;
2. fewer/less collisions (with container); 2
(b) (i) increases/eq; 1
(ii) increases/eq; 1
[5]
2. (a) (i) points plotted correctly;;
[Allow 1 mark for 4 correct]
straight line; 3
(ii) pressure increases with temperature; 1
(iii) line back tracked;
260 10 C; 2
(iv) absolute zero/ 0 K; 1
(b) An explanation to include:
• tyres get hot/increase in temperature;
• air expands/more collisions/air particles inside move more
/have more KE/OWTTE; 2
[9]
3. (a) An explanation to include:
1. moving gas particles;
2. hitting container walls;
[Allow bouncing off the sides for 2 marks]
[Ignore vibrating/collisions with other particles] 2
(b) increases;
increases;
stays the same;
stays the same; 4
(c) (i) increases;
in proportion/linearly/steady rate/OWTTE; 2
(ii) correctly indicated – intercept with horizontal axis; 1
(iii) zero/minimum; 1
[10]
4. (a) (i) An explanation to include:
• particles moving;
• collisions with walls/hitting walls;
• force produced;
plus 1 communication mark for ensuring that spelling,
punctuation and grammar are accurate, so that the meaning
is clear; 4
(ii) An explanation to include:
• more collisions;
• more particles present; 2
(b) (I) An explanation to include:
• not moving/all gases in solid state;
• no K.E.; 2
(ii) kelvin; 1
(iii) (+)373(K) 1
[10]
5. A calculation to show:
1/2. ;;
[1 mark for conversion to K and 1 mark for correct substitution]
3. P2 = 120 478 (Pa); 3
[Allow ;
P2 = 400 000 (Pa); 2 marks max]
[3]
6. (a) (i) An explanation to include two from:
1. molecules are moving / have (kinetic) energy;
2. colliding with container walls;
3. each collision produces a force; 2
plus one communication mark for using a suitable structure and
style of writing 1
(ii) An explanation to include:
1. molecules moving faster / more (kinetic) energy
/ moving more;
2. colliding more often with the walls / harder collisions
/ more force; 2
(b) (i) all points plotted correctly;; [Deduct 1 mark for every two errors]
straight line of best fit; 3
[Systematic error in vertical scale – max 2 marks]
(ii) straight line extended to pressure axis;
pressure of 104.5 ± 0.3 (kPa) / value from their graph; 2
[Bald, correct answer scores 2 marks]
(iii) molecules still have energy (to move) / molecules still moving; 1
(iv) –273°C / 0 K / absolute zero; 1
[12]
7. (a) (i) A description to include:
1. particles moving;
2. in all directions/randomly /
or implied by description (each other / walls); 2
[Arrows on diagram acceptable]
(ii) An explanation to include:
1. particles hit/collide with container walls;
2. producing a force; 2
(iii) pressure would increase/get bigger/larger; 1
(b) statements 2, 3 and 4 ticked ; ; ;
[If more than 3 ticked then deduct 1 mark for each error] 3
(c) An explanation to include two from:
1. temperature of air in tyre increases / hot / hotter;
2. particles hit more often/hit harder;
3. particles moving faster / more energy; 2
plus one communication mark for presenting relevant information in a
form that suits its purpose 1
[11]
8. (a) random;
collide;
force;
decreases;
decreases; 5
(b) (i) A description to include:
1. increases;
2. uniformly / equally / steadily / in a straight line /
positive correlation / OWTTE; 2
(ii) X on temperature axis between 250 and 300 K /
allow if on line in right place; 1
(iii) zero; 1
(iv) not moving; 1
[Reject mainly still]
[10]
Thursday, May 06, 2010
y11 triple
Questions on Nuclear physics
1. (i) thermionic emission; 1
(ii) A description to include three from:
1. heat in filament (releases electrons);
2. reference to 50 kV supply;
3. KE (due to electric field);
4. wave energy/energy of X-rays/heat; 3
(iii) (50 kV) power supply; 1
[Reject heater filament]
(iv) e × V;
= 1.6 × 10–19 × 50 × 1000 = 8 × 10–15 (J);
[Bald correct answer scores 2 marks]
[Allow 50 KeV for 2 marks] 2
[7]
2. (a) 0 and – 1 for ;
234 and 91 for Pa; 2
(b) An explanation to include:
1. gamma ray a wave;
2. no mass/protons; 2
[4]
3. (a) atomic mass of alpha is 4/alpha is 2n + 2p; 1
(b) (i) look at ranges/absorption/path in E/B fields/ionising ability;
connection between energy and above; 2
(ii) gamma rays take away the excess energy; 1
(iii) different energy levels (in the nucleus); 1
[5]
4. (i) An explanation to include two from:
1. high speed/energy;
2. electrons/negative particles;
3. emitted from nucleus;
4. a neutron splits;
[Accept a neutron splits into a proton and an electron
for two marks] 2
(ii) 2
[4]
5. (a) An explanation to include:
• cannot be divided / broken down further;
• electrons; 2
(b) (i) A suggestion to include:
• neutrons have no charge/cannot be deflected by E/B fields 2
• (difficult to detect) travel though matter easily;
(ii) protons and neutrons contain 3 quarks;
different combinations of up and down quarks; 2
[6]
6. (i) u – up quark;
d – down (quark);
n – neutron;
P – proton;
+ – positron; 5
(ii) Any two from:
1. quarks; 2
2. positron;
3. neutrino;
[Allow ecf from part (a)(i) for
[Deduct 1 mark for each incorrect answer after two]
[7]
7. (a) (i) same proton number, different neutron number / OWTTE; 1
(ii) A description to include:
1. N/Z = 1 (until Z = 20) for lower proton number / initially;
2. it increases (with Z); 2
(b) (i) A description to include:
1. proton / loses proton / gains a neutron; 2
2. decays into neutron + +; 2
[Accept p n + + for two marks]
[Accept up to down quark answer]
(ii) for Ta N/Z = / 1. 41;
for Hf N/Z = / 1. 44; 2
[Accept a qualitative answer which is correct]
(iii) closer to (stable) curve / line of stability / energy has been emitted; 1
(c) An explanation to include:
1. nucleus of excess energy / nucleus undergoes rearrangement /
nucleus left in excited state / nuclear energy levels; 2
2. it takes away / gets rid of / loses energy as ;
[If word nucleus does not appear – 1 mark max]
[10]
8. (a) A calculation to include:
1. 236 – 234 / 144 – 142;
[Allow 1 mark if 143–142 or 235–234 = 1n]
2. = 2; 2
(b) proton number increases by 1;
mass number stays the same; 2
[4]
9. (a) An explanation to include:
cannot be divided/ broken down further;
electrons; 2
(b) (i) A suggestion to include:
neutrons have no charge/cannot be deflected by E/M
fields; 2
(difficult to detect) travel though matter easily;
(ii) protons and neutrons contain 3 quarks;
different combinations of up and down quarks; 2
[6]
10. (a) X marked on intersection of 19-21; 1
(b) reduce the number of neutrons/increase the number of protons/ 1
- decay/ emissions
(c) Any three from:
an electron/negative charge is emitted/-; reject electron shells
number of neutrons decreases/becomes 20;
number of protons increases/becomes 20;
sensible reference to the line of stability;
(equalises number of protons and neutrons to 20 = 2marks)
(equalises protons and neutrons = l mark) 3
(loss of neutrons and gain in protons = 1 mark)
change of quark; down quark to up quark;
(d) increases (by 1);
unchanged; 2
(e) by emitting a gamma ray/electromagnetic ray/wave/radiation; 1
[8]
1. (i) thermionic emission; 1
(ii) A description to include three from:
1. heat in filament (releases electrons);
2. reference to 50 kV supply;
3. KE (due to electric field);
4. wave energy/energy of X-rays/heat; 3
(iii) (50 kV) power supply; 1
[Reject heater filament]
(iv) e × V;
= 1.6 × 10–19 × 50 × 1000 = 8 × 10–15 (J);
[Bald correct answer scores 2 marks]
[Allow 50 KeV for 2 marks] 2
[7]
2. (a) 0 and – 1 for ;
234 and 91 for Pa; 2
(b) An explanation to include:
1. gamma ray a wave;
2. no mass/protons; 2
[4]
3. (a) atomic mass of alpha is 4/alpha is 2n + 2p; 1
(b) (i) look at ranges/absorption/path in E/B fields/ionising ability;
connection between energy and above; 2
(ii) gamma rays take away the excess energy; 1
(iii) different energy levels (in the nucleus); 1
[5]
4. (i) An explanation to include two from:
1. high speed/energy;
2. electrons/negative particles;
3. emitted from nucleus;
4. a neutron splits;
[Accept a neutron splits into a proton and an electron
for two marks] 2
(ii) 2
[4]
5. (a) An explanation to include:
• cannot be divided / broken down further;
• electrons; 2
(b) (i) A suggestion to include:
• neutrons have no charge/cannot be deflected by E/B fields 2
• (difficult to detect) travel though matter easily;
(ii) protons and neutrons contain 3 quarks;
different combinations of up and down quarks; 2
[6]
6. (i) u – up quark;
d – down (quark);
n – neutron;
P – proton;
+ – positron; 5
(ii) Any two from:
1. quarks; 2
2. positron;
3. neutrino;
[Allow ecf from part (a)(i) for
[Deduct 1 mark for each incorrect answer after two]
[7]
7. (a) (i) same proton number, different neutron number / OWTTE; 1
(ii) A description to include:
1. N/Z = 1 (until Z = 20) for lower proton number / initially;
2. it increases (with Z); 2
(b) (i) A description to include:
1. proton / loses proton / gains a neutron; 2
2. decays into neutron + +; 2
[Accept p n + + for two marks]
[Accept up to down quark answer]
(ii) for Ta N/Z = / 1. 41;
for Hf N/Z = / 1. 44; 2
[Accept a qualitative answer which is correct]
(iii) closer to (stable) curve / line of stability / energy has been emitted; 1
(c) An explanation to include:
1. nucleus of excess energy / nucleus undergoes rearrangement /
nucleus left in excited state / nuclear energy levels; 2
2. it takes away / gets rid of / loses energy as ;
[If word nucleus does not appear – 1 mark max]
[10]
8. (a) A calculation to include:
1. 236 – 234 / 144 – 142;
[Allow 1 mark if 143–142 or 235–234 = 1n]
2. = 2; 2
(b) proton number increases by 1;
mass number stays the same; 2
[4]
9. (a) An explanation to include:
cannot be divided/ broken down further;
electrons; 2
(b) (i) A suggestion to include:
neutrons have no charge/cannot be deflected by E/M
fields; 2
(difficult to detect) travel though matter easily;
(ii) protons and neutrons contain 3 quarks;
different combinations of up and down quarks; 2
[6]
10. (a) X marked on intersection of 19-21; 1
(b) reduce the number of neutrons/increase the number of protons/ 1
- decay/ emissions
(c) Any three from:
an electron/negative charge is emitted/-; reject electron shells
number of neutrons decreases/becomes 20;
number of protons increases/becomes 20;
sensible reference to the line of stability;
(equalises number of protons and neutrons to 20 = 2marks)
(equalises protons and neutrons = l mark) 3
(loss of neutrons and gain in protons = 1 mark)
change of quark; down quark to up quark;
(d) increases (by 1);
unchanged; 2
(e) by emitting a gamma ray/electromagnetic ray/wave/radiation; 1
[8]
Subscribe to:
Posts (Atom)