Post on 21-Apr-2015
– 4 – N09/4/PHYSI/HP2/ENG/TZ0/XX/M+
SECTION A
A1. (a) both error bars of 15ms drawn correctly; [1]
(b) a straight line cannot be drawn through the
error bars; Accept the error bar comment with a
straight line drawn on graph.
that goes through the origin; [2]
(c) (i) 2 2500(m s ) ; [1]
(ii)
2
22
v v
v v;
2 2 527 2
27v ;
2 2 2( )300(m s )v or 2 2( )270(m s ) ; [3] or percentage error/uncertainty in (18.5 )19%v ;
percentage of error/uncertainty in 2 37 %v ;
absolute error2 2( )300(m s ) or 2 2( )270(m s ) ;
Answer must be to one or two significant figures.
(d)
2 3 2 2/ 10 m sv
3.0
2.5
2.0
1.5
1.0
0.5
0
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5
T / N use of gradient triangle over at least half of line;
gradient 640( 40) ;
2k to give 640 25( 1)k ;
unit of k is 1 12 2kg m or
121ms N ; [4]
Do not penalize omission of factor of 1000 for missing y-axis label if already
penalized in (c). Treat as ecf.
2 3 2 22.3 10 m sv
( ) 3.6 NT
– 3 – SPEC/4/PHYSI/HP2/ENG/TZ0/XX/M
A2. (a) Intensity
Distance along screen
general shape; relative position of secondary maxima / relative heights of secondary maxima; [2] Award [1 max] if not touching x-axis. (b) A B
maximum of B coincides with first minimum of A; [1]
(c) 7
62
1.22 1.22 5 10 2.4 10 rad25 10b
λθ−
−−
× ×= = = ××
;
168.1 10x=
×;
to give 112.0 10 mx = × ; [3] A3. (a) polarized light is light in which the (electric) field vector vibrates in one plane only /
OWTTE; the liquid crystal changes the plane in which (electric) field vector rotates; [2] (b) (i) nothing / whole area black; since the optical axes of 1P and 2P are at right angles / OWTTE; [2] (ii) since the liquid crystal rotates the plane of polarisation light is now
transmitted by 2P / OWTTE; the electric field across the parts of the liquid crystal in the shape of the
electrode on G no longer rotates the plane of polarization; the field of view of the observer will now contain a black area
corresponding to the shape of the electrode on G / OWTTE; [3]
– 3 – SPEC/4/PHYSI/HP2/ENG/TZ0/XX/M
A2. (a) Intensity
Distance along screen
general shape; relative position of secondary maxima / relative heights of secondary maxima; [2] Award [1 max] if not touching x-axis. (b) A B
maximum of B coincides with first minimum of A; [1]
(c) 7
62
1.22 1.22 5 10 2.4 10 rad25 10b
λθ−
−−
× ×= = = ××
;
168.1 10x=
×;
to give 112.0 10 mx = × ; [3] A3. (a) polarized light is light in which the (electric) field vector vibrates in one plane only /
OWTTE; the liquid crystal changes the plane in which (electric) field vector rotates; [2] (b) (i) nothing / whole area black; since the optical axes of 1P and 2P are at right angles / OWTTE; [2] (ii) since the liquid crystal rotates the plane of polarisation light is now
transmitted by 2P / OWTTE; the electric field across the parts of the liquid crystal in the shape of the
electrode on G no longer rotates the plane of polarization; the field of view of the observer will now contain a black area
corresponding to the shape of the electrode on G / OWTTE; [3]
– 6 – SPEC/4/PHYSI/HP2/ENG/TZ0/XX/M
SECTION B B1. Part 1 Simple harmonic motion and the greenhouse effect (a) the force acting/accelerating (on the body) is directed towards equilibrium (position); and is proportional to its/the bodies displacement from equilibrium; [2] (b) (i) 101.5 10 m−× ; [1] (ii) 121.1 10 sT −= × ;
12
11.1 10
f −⎛ ⎞= ⎜ ⎟×⎝ ⎠
;
139.1 10 Hz= × [2] (iii) 14 1(2 ) 5.7 10 (rad );fω −= π = × ( )2 2 27 2 20 2 281 1
max 02 2 1.7 10 (1.5) 10 (5.7) 10E m xω − −= = × × × × × × ;
186.2 10 J−= × [2] (c) (i) 2 2 26 27
p(4 ) 40 83 10 1.7 10k f m −= π = × × × × ;
1560 N m−≈ [1] (ii) use of F = kx and F = ma;
to give 10
20 227
560 1.5 10 5.0 10 ms1.7 10
a−
−−
× ×= = ××
; [2]
(d) (i) infra red radiation radiated from Earth will be absorbed by greenhouse gases; and so increase the temperature of the atmosphere/Earth; [2] (ii) the natural frequency of oscillation (of a methane molecule) is equal to
139.1 10 Hz× ; because of resonance the molecule will readily absorb radiation of this
frequency; [2]
– 12 – N09/4/PHYSI/HP2/ENG/TZ0/XX/M+
B3. Part 1 Simple harmonic motion
(a) 1. acceleration proportional to displacement from equilibrium/centre (of motion)
/mean position;
2. acceleration directed to equilibrium/centre/mean position; [2]
(b) (i) 2
d; [1]
(ii) sine/cosine curve shape reasonable; [1]
Do not allow semi-circle for half sine curve.
(iii) period labelled;
amplitude labelled; [2]
(c) (i) v 2a f seen/used;
13.3ms ; [2]
(ii) acceleration 2 24a f seen/used;
3 29.2 10 ms ; [2]
(d) cosine with the same period;
negative cosine;
Accept any amplitude. velocity
time
[2] Amplitudes need not be the same.
(e) (i) (a situation in which) a (resistive) force opposes the motion / the amplitude
decays with time; [1]
(ii) energy lost to surroundings / air resistance / frictional force is acting on the
fork; [1]
– 14 – M09/4/PHYSI/HP2/ENG/TZ1/XX/M+
B3. Part 1 Simple harmonic motion and waves
(a) is proportional to the displacement/distance (of the particle) from its equilibrium
position;
is directed towards the equilibrium position; [2]
(b) (i) overall correct shape;
with max of 0.06 J at x and zero at x ; [2]
(ii) max 0
2 2 2
K 4E mf x
;
from the graph maxK JE ;
and 0 mx ;
maxK
2
04
Ef
mx
;
to give Hzf [4] or
maxK
2
0
;E
kx
2
0.06;
0.05
= 48;
use of k
fm
;
= 2.0 Hz
(c) (i) the energy of the wave is propagated in a direction at right angles;
to the direction of oscillation of the particles; [2]
(ii) m ; [1]
– 15 – M09/4/PHYSI/HP2/ENG/TZ1/XX/M+
(d) (i) use of 22 1
1
sin sinv
v ;
2
1
v
v ;
this marking point is not necessary to award full credit.
1
2 sin 0.75 49 ; [3]
(ii)
any two lines as shown bending in the correct direction; [1]
– 12 – M09/4/PHYSI/HP2/ENG/TZ2/XX/M+
B3. Part 1 Simple harmonic motion and waves
(a) displacement is proportional to acceleration / vice versa;
because graph is straight-line through origin;
displacement and acceleration in opposite directions / acceleration always directed
towards origin;
because negative gradient; [4]
(b) use of 2 ( )a
x ;
2
3
2900
0.60 10
;
2 f ;
3
1 2900
2 0.60 10f
;
(to give 350Hz)f [4]
(c) (i) transfer of energy by means of vibrations/oscillations;
vibrations all in one direction parallel to direction of energy transfer; [2]
(ii) 330
350 or use of c f ;
0.94m ; [2]
Award [2] for bald correct answer.
– 13 – M09/4/PHYSI/HP2/ENG/TZ2/XX/M+
Part 2 Diffraction of light
(a) (i) spreading out of light;
beyond that predicted by the geometric pattern / by the obstacle shape / OWTTE; [2]
(ii) diagram:
central symmetrical maximum;
at least one secondary maximum on each side with smaller height no more
than one third height of central maximum;
minima drawn to zero; (i.e. sitting on x-axis) [3]
(iii) 9
3
620 10
0.4 10b
;
9
3
2.0 1.9 620 10(2 )
0.4 10w D
;
5.9mmw ; [3]
Award [3] for bald correct answer.
(b) (i) the images can be seen separately; [1]
(ii) diffraction occurs (at the aperture/iris of the eye);
each lamp gives rise to a diffraction pattern (at the back of the eye/on the
retina);
(for distant lamps) the two diffraction patterns overlap;
so that patterns cannot be distinguished / OWTTE; [4]
– 7 – N08/4/PHYSI/HP2/ENG/TZ0/XX/M+
SECTION B
B1. Part 1 Wave motion
(a) (i) 1.5 mm; [1]
(ii) 8.0 cm; [1]
(iii) distance travelled in 0.20 s is 3.2 cm;
so speed is 2
13.2 100.16ms
0.20
; [2]
(iv) 2
0.162.0Hz
8.0 10f
; [1]
(b) travelling waves transfer energy (standing waves do not);
travelling waves have a constant amplitude (standing waves do not);
standing waves have points that always have zero displacement (travelling waves
do not);
the phase of a travelling wave constantly changes (but in standing waves points in
between consecutive nodes have constant phase); [2 max]
(c) (i) it is the speed of energy transfer/rate/speed at which wavefronts move forward; [1]
(ii) a standing wave is formed from the superposition of two travelling waves;
wave speed refers to the speed of the travelling waves; [2]
(d) (i) the oscillating string collides with the air molecules surrounding it;
creating a pressure/longitudinal wave; [2]
(ii) wavelength of wave on string is 2 0.80 1.6m ;
frequency is then 240
150Hz1.6
;
sound has the same frequency and so wavelength is 340
2.3m150
; [3]
Award [1 max] for those using a wavelength of 0.80 m obtaining a wavelength
of 1.1 m in air. Accept alternative derivations that use a ratio and do not
calculate the frequency explicitly.
– 13 – M08/4/PHYSI/HP2/ENG/TZ2/XX/M
B3. Part 1 Wave phenomena (a) (i) C shown where graph line cuts x-axis; [1] (ii) time period 0.30 ms; =
use of v fλ= and 1 fT
= or vTλ
= ;
; [3] 3380 0.30 10 0.11mλ −= × × = ECF if time period misread. (b) (i) superposition of two waves / OWTTE; of same frequency and amplitude travelling in opposite directions; [2] (ii) stationary/standing wave is set up in the tube; heaps form at the (displacement) nodes / powder pushed away from antinodes; [2] (iii) wavelength ; (2 9.3 )18.6cm= × = speed 1(1800 0.186 ) 330ms−= × = ; [2] ECF if value of wavelength wrong. (c) heaps further apart means longer wavelength;
hence speed increases (as temperature rises);⎧⎨⎩
Do not award if there is no reasoning orreasoning is fallacious or misleading. [2]
(d) (i) when two waves meet; resultant displacement found by summing individual displacements; to give maximum displacement / displacement greater than that of an individual
wave; [3] (ii) line in correct position, labelled C; [1] (iii) line in correct position, labelled D; [1]
(e) use of axD
λ = and ; 24.0 10 ma −= ×
24.0 10 1.2
1.5λ
−× ×= ;
; [3] 23.2 10 mλ −= × ECF if value of “a” wrong [2 max].
– 11 – M08/4/PHYSI/HP2/ENG/TZ1/XX/M+
B2. Part 1 Waves (a) Transverse the particles (of the medium) vibrate at right angles; to the direction of energy transfer; Longitudinal the particles (of the medium) vibrate in the same direction as the direction of
energy transfer; [3] (b) (i) time period = 0.13 s;
1 Hz ;0.13
fT1⎛ ⎞= = = 7.7 (±0.3) ⎜ ⎟
⎝ ⎠ [2]
Award full marks for bald correct answer. (ii) 8 mm; [1]
(c) ;vf
λ =
157.7
;
1.95 cm 2.0 cmλ = ≈ [2] (d) start at (−1.2 →−2.0) on y-axis; sine curve of amplitude 8 mm; wavelength 2 cm; [3]
(e) use of 1 1
2 2
sinsin
vv
θθ
=
22 1
1
sin sin ;vv
θ θ=
sin 3020=
15to give 2 ;θ °= 42
angle °= 48 ; [3]
– 12 – M08/4/PHYSI/HP2/ENG/TZ1/XX/M+
(f) (i) each slit acts as a point source of waves; waves from these sources interfere; because of the principle of superposition; where a trough of one wave meets a crest of another the resultant displacement
will be minimum / waves arrives with opposite phase/completely out of phase / so destructive interference occurs; [4]
(ii) ;dsDλ
=
9.4 ;2.0
18×=
= 85 / 90 cm; [3] Award full marks for bald correct answer. (g) (i) the phase difference between them is constant; [1] (ii) fringes of equal intensity / intensity reducing from centre fringe; and equal width; [2] [max 1] if fringes do not touch axis. B2. Part 2 Magnetic fields (a) effective current in each side of coil = 20I;
each wire needs to produce 1 ;4
B
0
π ;rBIµ
2=
π 0.6 7.0 10 ;π 10 80
−5
−7
2 × × ×=
4 × ×
A= 2.6 [4]
Accept correct substitution for I into2
IBr
µ0=π
to show that 10B T−5= 7.0×
(b) plane vertical; plane normal to Earth field; [2]