2007 Physics Intermediate 2 Finalised Marking Instructions
Scottish Qualifications Authority 2007 The information in this publication may be reprod...
Scottish Qualifications Authority 2007 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purposes written permission must be obtained from the Assessment Materials Team, Dalkeith. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre's responsibility to obtain the necessary copyright clearance. SQA's Assessment Materials Team at Dalkeith may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.
Physics – Marking Issues The current in a resistor is 1·5 amperes when the potential difference across it is 7·5 volts. Calculate the resistance of the resistor. Answers V=IR 7·5=1·5R R=5·0 Ω
Mark + Comment (½) (½) (1)
Issue Ideal answer
2.
5·0 Ω
(2) Correct answer
GMI 1
3.
5·0
(1½) Unit missing
GMI 2 (a)
4.
4·0 Ω
(0) No evidence/wrong answer
GMI 1
(0) No final answer
GMI 1
(1½) Arithmetic error
GMI 7
(½) Formula only
GMI 4 and 1
(½) Formula only
GMI 4 and 1
(1) Formula + subs/No final answer
GMI 4 and 1
1.
5.
Ω
V
R=
7.
R=
8.
R=
9.
R=
10.
R=
11.
R=
12.
R=
13.
R=
14.
V = IR 7·5 = 1·5 × R R = 0·2 Ω
15.
V = IR I 1·5 R= = = 0·2 Ω V 7·5
I V I V I V
=
7·5 = 4·0 Ω 1·5
6.
= 4·0 Ω
=
Ω
=
7·5 = 1·5
=
7·5 = 4·0 1·5
(1) Formula + substitution
GMI 2 (a) and 7
=
1·5 = 5·0 Ω 7·5
(½) Formula but wrong substitution
GMI 5
V 75 = = 5·0 Ω I 1·5
(½) Formula but wrong substitution
GMI 5
I 7·5 = = 5·0 Ω V 1·5
(0) Wrong formula
GMI 5
(1½) Arithmetic error
GMI 7
(½) Formula only
GMI 20
I V I V I
Ω
Page 2
2007 Physics Intermediate 2 Marking scheme Section A
1.
E
11.
D
2.
B
12.
C
3.
E
13.
B
4.
B
14.
A
5.
A
15.
C
6.
B
16.
D
7.
C
17.
C
8.
D
18.
D
9.
B
19.
E
10.
D
20.
C
Page 3
2007 Physics Intermediate 2 Sample Answer and Mark Allocation 21.
(a)
(b)
(i)
Notes
Marks
d=vt 3.2 = v × 20 v = 0·16 m/s
(½) (½) (1)
2
(ii) W = m g W = 60 × 10 W = 600 N
(½) (½) (1)
2
(iii) EP = m g h EP = 60 × 10 × 3.2 EP = 1920 J
(½) (½) (1)
2
(i) EK = EP 1920 = ½ × 60 × v2 v = 8 m/s
(½) (½) (1)
2
(ii) (actual speed) less air resistance during fall or not all EP changes to EK
(1) (1)
2 Total 10
Page 4
Sample Answer and Mark Allocation 22.
(a)
(b)
Notes
measure a distance using a tape/rule/trundle wheel measure time to travel this distance with stopwatch/clock use formula distance = average speed × time to calculate average speed
(½) (½) (½) (½)
(i) a = v – u t
(½)
a=0–8 2.5
(½)
a = -3·2 m/s2
(1)
(ii) speed (m/s) 8
0
(1)
axes
(½)
shape
(½)
values
(½) (½)
3
Marks
3
2
5.5 time (s)
(iii) distance = area under graph = (8 × 3) + (½ × 2·5 × 8) = 34m
(½) (½) (1)
2
2
Total 9
Page 5
Sample Answer and Mark Allocation 23.
(a)
(b)
(c)
(i)
Notes
Marks
EH = c m ∆T EH = 4180 × 10 × 80 EH = 3·34 × 106 J
(½) (½) (1)
2
(ii) E = P t 3·34 × 106 = 2·5 × 103 × t t = 1340 s
(½) (½) (1)
2
(iii) not all EH used to heat water OR EH lost to surroundings
(1)
1
P=IV 2·5 × 103 = I × 230 I = 10·9 A
(½) (½) (1)
2
EH = 1 m EH = 22·6 × 105 × 1·2 EH = 2·71 × 106 J
(½) (½) (1)
2 Total 9
Page 6
Sample Answer and Mark Allocation 24.
Notes
Marks
(a)
Transformers only work on a.c.
(1)
1
(b)
I = 150/3 = 50 mA P=IV P = 50 × 10-3 × 4·8 P = 0·24 W
(1) (½) (½) (1)
3
n s Vs = n p Vp
(½)
50 4⋅8 = 1000 V2
(½)
V2 = 96 V
(1)
(c)
(d) smaller (in a step down transformer), voltage steps down, current steps up or similar
2
(1) (1)
2 Total 8
Page 7
Sample Answer and Mark Allocation 25.
(a)
(b)
(c)
(i)
Notes
IR = infrared
(1)
Marks
1
(ii) both arrive at the same time both travel at the same speed (or speed of light or 3 × 108 m/s)
(1) (1)
2
Q=It Q = 3 × 2 × 60 × 60 Q = 21 600 C
(½) (½) (1)
2
VR = 8 – 2 = 6 V V=IR 6 = 15 × 10-3 × R R = 400 Ω
(1) (½) (½) (1)
3 Total 8
Page 8
Sample Answer and Mark Allocation 26.
Notes
1
(a)
thermistor
(b)
as temperature drops, voltage across thermistor rises or resistance of thermistor rises (1) when voltage goes above certain level MOSFET switches on (1) relay switch closes (and heater circuit is completed) (1)
3
to set the temperature at which the heater is switched on
1
(c)
(1)
Marks
(1)
Total 5
Page 9
Sample Answer and Mark Allocation 27.
(a)
(b)
Notes
Marks
(i)
refraction
(1)
1
(ii)
reflection
(1)
1
(iii)
red
(1)
1
(1) (1)
2
two forces: air resistance and weight balanced
Total 5
Page 10
Sample Answer and Mark Allocation 28.
(a)
(b)
Marks
(i)
(waveform) Q
(1)
1
(ii)
(waveform) Q
(1)
1
(i)
v=fλ 340 = 2 × 103 × λ λ = 0·17 m
(½) (½) (1)
2
d=vt 20·4 = 340 × t t = 0·06 s
(½) (½) (1)
2
(1) (1)
2
(ii)
(c)
Notes
(wavelength) decreased speed of sound slower
Total 8
Page 11
Sample Answer and Mark Allocation 29.
(a)
(b)
(c)
Notes
Marks
E=Dm E = 3 × 50 × 10-6 × 6 E = 9 × 10-4 J
(½) (½) (1)
2
lead absorbs X-rays or lead shields leg from X-rays
(1)
1
type of radiation or organ/type of tissue
(1)
1
Total 4
Page 12
Sample Answer and Mark Allocation 30.
(a)
(i)
(ii)
(iii)
(b)
(i)
(ii)
Notes
Marks
loss or gain of electrons from atom or molecule
(1)
1
alpha greatest ionisation (density)
(1) (1)
2
source Y long half-life but short range
(1) (1)
2
V =IR 9 = 30 × 10-3 × R R = 300 Ω
(½) (½) (1)
2
electrical to sound
(1)
1 Total 8
Page 13
Sample Answer and Mark Allocation 31.
(a)
(b)
(c)
Notes
Marks
cosmic rays radon gas or other correct answers
(1) (1)
2
N =At 4 = A × 10 A = 0·4 Bq
(½) (½) (1)
2
168 --- 84 in 4 minutes or 120 --- 60 or other pair of values half-life = 4 minutes
