Download CMock exam IV paper 2

F6 Physics Mock Examination Suggested Answers (14-15)
1. (a) Energy removed when the tea is cooled down to 0 °C
= mc ΔT = 0.5 × 4000 × (90 – 0) = 180 000 J
Mass of ice needed = E / ℓf = 180 000 / (3.34 × 105) = 0.539 kg
(b) More ice should be needed.
(c)
(1M)
(1A)
(1A)
1M for starting from (all ice cubes have melted), 1A for approaching room temperature
2. (a) Temperature (of gas)
1M
Mass of air/gas Or number of atoms/molecules/moles of air/gas
1M
(b) Assumption: idea that volume occupied by trapped air proportional to length of air in tube
(i.e. volume = cross-sectional area × length)
1M
PL = a constant [accept PV = a constant] Or (If P doubled, L halved)
1A
At least 2 pairs of P, L values correctly read from graph
1A for each example
Examples:
P = 450 kPa, L = 10 cm,
PL=450 × 10 = 4500
P = 100 kPa, L = 45 cm,
PL=100 × 45 = 4500
P = 300 kPa, L = 15 cm,
PL=300 × 15 = 4500
Readings show that pL = 4500 (kPa cm) [± 100 kPa cm]
Or Readings show that p doubles when L is halved
PV 450  103  0.1  7.5  105
(c) n 

RT
8.31  (20  273)
(1M for n = PV/RT)
= 0.001386 mol
(1M)
23
No. of molecules = 0.001386 x 6.02 x 10 = 8.34 x 10
3. (a) Area = 4 x 20 / 2 = 40 Ns
20
(1A)
(1M for 4 x 20/2, 1A for 40 Ns)
Area under graph represents change of momentum or impulse. (1M)
(b) Change in velocity = change of momentum / mass (mv = Ft) (1M)
= 40 / 2.5 = 16 m s-1
(1A)
(c) The acceleration increases (uniformly) between 0 to 2s,
(1M, no need to have uniformly)
then decreases (uniformly) between 2s to 4s.
(1M for decrease, no mark for deceleration)
(d)
(1M for
1M for
)
GM
R2
6.67  1011  5.7  1026
g
 10.56 N kg-1
(6.0  107 )2
4. (a) g 
(b)
GMm mv2

r2
r
GM
v2 
r
v
(1M for formula g 
GM
)
R2
(1A)
(1M for either
GM
GMm mv2
or v 2 
)

2
r
r
r
GM
6.67  1011  5.7  1026

 8470 m s-1 (1M for substitution of Saturn’s mass, not Rhea, 1A)
r
5.3  108
5. (a) Light from two sources cannot have constant phase difference (or never be coherent).
(1M)
Only light waves from a single source have constant phase difference (or are coherent). (1M)
(b) (i) At D, path difference =
1

2
(1A)
(ii) At B, path difference = 
(1A)
(c) (i) Fringe separation s = 8 mm / 4 = 2 mm
(ii) s 
D
a
or  
(1A)
3
as 0.4  10  2  103

 5.33  10 7 m
D
1.5
(1M+1A)
(d) Light waves from the two slits are two coherent sources because they come from a single source. At places
where the light waves arrive in phase, constructive interference takes place and a bright fringe is formed. At
places where the light waves arrive exactly out of phase, destructive interference takes place, a dark fringe is
formed. (1M+1M+1M for coherent, constructive interference and destructive interference)
6. (a) (i)
air
water
1A for shorter wavelength, 1A for bending towards normal
sound waves
(ii)
air
water
1A for bending away from normal
(iii) Sound travels faster in water than in air while light travels faster in air than in water. 1A for each
light ray
(b) Light waves appear to go straight without spreading after passing the gap. (1M)
Sound waves spread after passing the gap.
(1M)
This is because the wavelength of visible light is much smaller than the gap width while the wavelength of
sound waves is comparable to the gap width.
(1M)
7. Connect a signal generator to a vibration generator. Tie one end of a thread to the vibrator and fix the thread to
a fixed end such as a stand as shown. (1A)
1A for connecting sig.gen. to vibration gen., and connecting the elastic string to vibration gen. and the stand
Gradually increase the vibrator frequency from zero. (1M)
When the frequency of the vibrator is slowly increased from zero, at first no stationary wave is produced. (1M)
When the frequency is increased to fo, a transverse stationary wave with one loop is formed. (1M)
When the frequency is further increased, it is found that at frequencies 2 fo, 3fo , ... , stationary waves with 2
loops, 3 loops, ...etc. are formed. (1M)
This shows that stationary waves are formed only at particular frequencies.
8. (a) Electric field pointing downwards.
(b) E = V/d = 160 / 0.05 = 3200 V m-1 (1M+1A)
(c) Electric force = QE = 1.6 x 10-19 x 3200
= 5.12 x 10-16 N
(d) Between the plates, there is an acceleration/ force on the electron directed upward.
(1M)
With an initial constant horizontal velocity, the resulting path is parabolic between the plates.(1M)
Outside the plates, there is no electric field, and no force acts on the electron, because of inertia, the
electron travels in a straight line.
(1M)
(e) Increase in KE = QV
1 2
mv  QV
2
=>
1
 9.1  10 31  (1.2  107 ) 2  1.6  1019V
2
V = 409.5 V (~410V)
(1A)
9. (a) Flux linkage = N = NBA cos 
= 200 x 3 x 10-2 x 2 x 10-4 = 0.0012 Wb (1.2 x 10-3 Wb)
(b) (i)   N
(1M for formula, 1M for substitution)

t
(1M for NBA cos )
(1A)(Accept 0.0012 Tm2)
(1M for formula)
 = NBA cos 0o – NBA cos 90o = 0.0012 Wb
t = 1/4 period = (1/4) (1/2) = 1/8 s
(1M for correct 1/8 s)
N
(1A)
 0.0012

 0.0096V
t
1/ 8
(9.6mV)
(ii) The induced emf depends on the rate of change of flux linkage or the rate at which the coil cuts the
magnetic field lines.
(1M)
When  = 0o or 180o or 360o, the plane of coil is parallel is perpendicular to the field lines, its motion does
not cut any magnetic field lines, so the induced emf is zero.
(1M)
When  = 900, or 270o, the plane of coil is parallel to the field lines, its motion cuts the field lines at the
fastest rate (or the rate of change of flux linkage is the greatest), thus the induced emf is a maximum. (1M)
(iii) The peaks would be of smaller amplitude (or the max. emf reduced) (1M)
because the rate of change of flux linkage is smaller (or rate of cutting field lines is less) (1M)
ln 2
k
2
ln 2
ln 2
k

 5.81  10 8 s 1
t1/ 2
138  24  3600
10. (a) (i) t 1 
(1M for formula, 1A)
(ii) Activity A = kN = 5.81 x 10-8 x 2.54 x 1015 = 1.48 x 108 Bq
(iii)
N  N 0e
 kt
 2.54  10 e
15
Fraction decayed =
(b)

ln 2
 21
138
=2.286 x 1015
2.54  1015  2.286  1015
 0.1
2.54  1015
(1M+1A)
(1M for formula)
(1M for numerator +1A)
Alpha particles are not able to penetrate the skin (from outside the body).
(1A)
If it is taken into the body, damage may be made to cells or DNA due to strong ionizing power (1A)
(c)
210
84
4
Po 206
82 Pb 2 
Top line correct 1A
MC
1-5 A C C C D
21-25 A C A C A
6-10
Bottom line correct 1A
BCCAB
26-30 C C A D C
11-15 C A B D C
31-33 A C B
16-20 D B A B D