Grade 12 Physical Science Paper 1 Memorandum (June)
[Pages:6]Physical Sciences
1 NSC ? Memorandum
June2014 Common Test
PHYSICAL SCIENCES P1 MEMORANDUM COMMON TEST JUNE 2014
NATIONAL SENIOR CERTIFICATE
GRADE 12
N.B. This memorandum consists of 6 pages including this page.
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Physical Sciences
SECTION A
QUESTION 1 1.1 C 1.2 D 1.3 C 1.4 A 1.5 A 1.6 B 1.7 A
2 NSC ? Memorandum
QUESTION 2
2.1
FN
Fapp
Ff
FN ? normal force Fg ? weight Ff ? friction Fapp ? applied force
Fg
(FN must be less than Fg)
2.2
Fnet = ma
29 cos36o + (-Ff) = 20 x 0
Ff = 23,46 N
2.3 Increase
2.4 FN increases Co-efficient of kinetic friction remains constant
From the equation fk = k N Friction directly proportional to normal force
June2014 Common Test
(7 x 2) [10]
(4) (3) (1) (3) [11]
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QUESTION 3
3 NSC ? Memorandum
3.1
0,57 s
3.2 vf = vi + at 0 = vi + (-9,8)(0,57) = 5,586 m.s-1
y = vit + ?at2 = (5,586)(0,57) + ?(-9,8)(0,57)2 = 1,59201 m
y = 4 x y1 = 4 x 1,59201 = 6,37 m
3.3 Inelastic ? The ball does not bounce to the original height.
3.4 y (m)
h
1 4
h
June2014 Common Test
(2)
(8) (2)
1,76
2,9 t (s)
Checklist
Marks
Criteria for graph
y and y1 correctly shown
1,76 and 2,9 correctly shown
Correct shape from 0 to 1,76 s
Correct shape from 1,76 to 2,9 s
Axes correctly labelled
(5)
[17]
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4 NSC ? Memorandum
June2014 Common Test
QUESTION 4
4.1 The total linear momentum of an isolated system remains constant in magnitude
and direction.
(2)
4.2
Total p before = Total p after
mvi + mvi = mvf + mvf
(75 x 3) + (2 x ?5) = (77) vf
vf
= 2,79 m.s-1 to the right
(4)
[6]
QUESTION 5
5.1 No.
The crate experiences a frictional force/non-conservative force.
The work done by the frictional force/non-conservative force changes the
mechanical energy of the crate.
(3)
5.2 A force for which the work done in moving an object between two points is
independent of the part taken.
(2)
5.3 A
E B
(5)
D C
5.4 Fg// = mgsin
= 120 x 9,8 x
= 176,4 N
(3)
5.5 O J force acts at right angles to the direction of motion. (Cos 90? = 0)
(2)
5.6 The net/total work done on an object is equal to the object's change in kinetic
energy.
(2)
5.7.1
W Ff = Ffxcos
= 50 x 10 x cos180?
= - 500 J
(3)
5.7.2
W Fg// = Fg//xcos
= 176,4 x 10 x cos 0?
= 1764 J
(2)
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June2014 Common Test
5.8
Wnet = K
WA + WB + WC + WD + WE = K
(0) + (- 500) + 1764 + 0 + WE = 0
WE = - 1264 J
(4)
OR
Wnc
= K + U
WB + WE
= 0 + mghf - mghi
- 500 + WE = 0 - (120)(9,8)(1,5)
WE = - 1264 J
5.9
WE = Fxcos
- 1264 = Fx cos180?
F = 126,4 N
P = Fv
= 126,4 x 1,25
= 158 W
(6)
[32]
QUESTION 6
6.1 fL
= v vL v vS
fs
580 = 340 + 0 x fs
340 ? 37,5
fs = 516,03 Hz
(4)
6.2 The apparent change in the frequency of a wave when there is relative motion
between the source of the wave and an observer.
(2)
6.3 No
An increase in the speed of the source producing the sound waves results in a greater
number of complete waves reaching the listener per second.
The apparent frequency increases and becomes greater than 580 Hz, the detector
cannot detect frequencies greater than 580 Hz.
(3)
6.4 Equal to.
There is no relative motion between the source and the listener.
(2)
6.5 6.5.1 change in pitch
(1)
6.5.2 change in colour
(1)
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June2014 Common Test
6.6 The Doppler flow meter uses the Doppler effect to measure the speed of blood
flowing through a person's blood vessels.
Abnormal blood speeds could be an indicator of a health risk.
(3)
[16]
QUESTION 7
7.1 The badge is referring to a blue shift caused by the Doppler Effect. For the badge to appear blue the car must be travelling at an extremely high speed. (2)
7.2 The claim made by the badge is not possible.
The car will have to travel at a speed comparable to the speed of light.
This is too large for the car.
(2)
[4]
TOTAL SECTION B: 86 GRAND TOTAL: 100
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