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M13/4/PHYSI/HP2/ENG/TZ2/XX/M




MARKSCHEME


May 2013



PHYSICS



Higher Level



Paper 2




17 pages
–2– M13/4/PHYSI/HP2/ENG/TZ2/XX/M




This markscheme is confidential and for the exclusive use of
examiners in this examination session.

It is the property of the International Baccalaureate and must not
be reproduced or distributed to any other person without the
authorization of the IB Assessment Centre.
–3– M13/4/PHYSI/HP2/ENG/TZ2/XX/M



Subject Details: Physics HL Paper 2 Markscheme
Mark Allocation

Candidates are required to answer ALL questions in Section A [45 marks] and TWO questions in
Section B [2 25 marks]. Maximum total = [95 marks].


1. A markscheme often has more marking points than the total allows. This is intentional.

2. Each marking point has a separate line and the end is shown by means of a semicolon (;).

3. An alternative answer or wording is indicated in the markscheme by a slash (/). Either wording can
be accepted.

4. Words in brackets ( ) in the markscheme are not necessary to gain the mark.

5. Words that are underlined are essential for the mark.

6. The order of marking points does not have to be as in the markscheme, unless stated otherwise.

7. If the candidate’s answer has the same “meaning” or can be clearly interpreted as being of
equivalent significance, detail and validity as that in the markscheme then award the mark.
Where this point is considered to be particularly relevant in a question it is emphasized by
OWTTE (or words to that effect).

8. Remember that many candidates are writing in a second language. Effective communication is
more important than grammatical accuracy.

9. Occasionally, a part of a question may require an answer that is required for subsequent
marking points. If an error is made in the first marking point then it should be penalized. However, if
the incorrect answer is used correctly in subsequent marking points then follow through marks
should be awarded. When marking indicate this by adding ECF (error carried forward) on the script.

10. Do not penalize candidates for errors in units or significant figures, unless it is specifically referred
to in the markscheme.
–4– M13/4/PHYSI/HP2/ENG/TZ2/XX/M


SECTION A

A1. (a) 15


10
V/V

5


0
0 5 10 15 20 25 30
t/s
(i) smooth curve;
that passes through all error bars; [2]

(ii) correctly identifies three points from own graph;
correctly processes these three points using exponential/half life/constant ratio/
relationship;
to conclude that decay is exponential;
within uncertainty; [4]


(b) (i) evaluates a gradient over a minimum of 5 s to give an initial ⎫
⎪ (allow ECF
⎛ 12 ⎞ ⎬
rate for example, ⎜ = ⎟ 1.3 V s –1 for graph above;
⎝ 9.5 ⎠ ⎪ from the graph)

V s –1 ; [2]
Clear evidence of calculation of gradient must be seen.

(ii) obtains evidenced answer for t intercept; [1]
–5– M13/4/PHYSI/HP2/ENG/TZ2/XX/M


⎛ 10 ⎞
(c) C =⎜ = ⎟ 1.0 (±0.2) × 10 –6 Ω –1 s/F ; [1]
⎝ 10 × 10 6

Award [0] for absence of 106 unless unit is in terms of MΩ.



A2. (a) (i) s = 12.5/12.6 m; [1]

(ii) v = 2 gs or gt; (allow any use of suvat equations)
; [2]
Award [2] for a bald correct answer.



(b) v




0
0 t1 t2 t

straight line to water surface;

clear decrease after hitting surface;

constant non-zero speed reached smaller than ⎫ (speed must be less than
maximum; ⎬ maximum velocity) [3]

–6– M13/4/PHYSI/HP2/ENG/TZ2/XX/M


A3. (a) internal energy:
the sum of the potential and the (random) kinetic energy of the molecules/particles
of a substance;
thermal energy:
the (non-mechanical) transfer of energy between two different bodies as a result
of a temperature difference between them; [2]

(b) (i) ( ΔU ) = 0.25 × 4.2 ×103 × 27 ( = 2.835 ×10 J ) ;
4


= 2.8 × 104 J ; [2]

(ii) energy transfer = [300 ×120] – [2.8 ×104 ] = 7.65 ×103 J ;
7.650 ×103
rate of transfer = = 64 W ; [2]
120

Allow ECF from (b)(i).




A4. (a) (i) nuclides/atom/element/nucleus/nuclei that have the same proton number/
same element but different nucleon/neutron numbers / OWTTE; [1]

(ii) the time taken for the activity (of a radioactive sample) to decrease by half /
the time taken for half the (initial) number of radioactive nuclei/atoms/mass
to decay; [1]

(b) (i) 2; [1]

(ii) (mass difference = ) 7.0160 – [3.0161 + 4.0026] = (–) 2.7 × 10 –3 u ;
(energy required = ) (–) 2.7 × 10 –3 × 931.5 or 2.5151MeV ;
( ≈ 2.5 MeV) [2]
Allow unit conversions via mass and mc2.

(c) 3
1 H → 23 He + β− + v
β − or −1
0
e or e− or electron or beta particle;
0
v or v or antineutrino;
0 [2]
Allow answers in any order.
–7– M13/4/PHYSI/HP2/ENG/TZ2/XX/M


A5. (a) the work done per unit charge;
in bringing a positive test / positive point / positive small charge
⎫ (accept use of a “reference point where potential
from infinity to the point; ⎬ [3]
⎭ is zero”)

(b) (i) Q = 4πε 0 rV or krV ;
1
= × 10 −9 × 0.080 × 300 or 2.67 nC ;
9
( ≈ 2.7 nC) [2]


Q
(ii) E=
4πε 0 r 2
⎛ 9 × 109 × 2.7 × 10−9 ⎞
=⎜ = ⎟ 950 ;
⎝ 0.162 ⎠
−1 −1
940 / 950 N C or V m ; [2]

(iii) 950 V m –1 ; [1]
Allow ECF from (b)(ii).



A6. (a) (at constant speed) forces are balanced;
force is (electro) magnetic;
due to current in rod;
which arises from induced emf (from motion); [4]


(b) (Lenz’s law states) the direction of an induced emf/current is such as to oppose
the change to which it is due;
in this situation the induced current (in the rod) is in such a direction that the
magnetic force on the rod opposes the force F; [2]

⎛E ⎞
(c) Bv = ⎜ = ⎟ 0.015 Vm −2 ;
⎝l ⎠
F = ( Bev = ) 2.4 ×10 –21 N ; [2]
Award [2] for a bald correct answer.
–8– M13/4/PHYSI/HP2/ENG/TZ2/XX/M


SECTION B

B1. Part 1 Electric charge and electric circuits

(a) the force between two (point) charges;
is inversely proportional to the square of their separation and (directly)
proportional to (the product of) their magnitudes; [2]
Allow [2] for equation with F, Q and, r defined.

⎛ q q ⎞ 9 × 109 × [1.6 ×10−19 ]2
(b) (i) F = ⎜k 1 22 =⎟ ;
⎝ r ⎠ 4 ×10−20
= 5.8 ×10 –9 N ; [2]


(b)(i )
(ii) −19
or 3.6 ×1010 N C –1 or V m –1 ;
1.6 × 10
(directed) away from the proton; [2]
Allow ECF from (b)(i).


−11 −27
⎛ m ⎞ 6.67 × 10 × 1.673 × 10
(iii) H = ⎜G 2 =⎟ −20
= 2.8 × 10 –18 N kg –1 ;
⎝ r ⎠ 4 × 10
−18
H 2.8 × 10
= or 7.8 ×10 –29 C kg –1 ;
E 3.6 × 10 10


(≈ 10−28 C kg –1 ) [2]
Allow ECF from (b)(i).

(iv) 3.4 V; [1]
–9– M13/4/PHYSI/HP2/ENG/TZ2/XX/M


(c) (i) power supplied per unit current / energy supplied per unit charge / work
done per unit charge; [1]

5.1× 10−19
(ii) energy supplied per coulomb = or 3.19 V ;
1.6 × 10−19
( ≈ 3.2 V) [1]

⎛ 4.0 ×10−19 ⎞
(iii) pd across 5.0 Ω resistor = ⎜ −19
= ⎟ 2.5 V;
⎝ 1.6 ×10 ⎠
pd across r = (3.2 − 2.5 = ) 0.70 V;

and

either
⎛ 2.5 ⎞
current in circuit = ⎜ = ⎟ 0.5 A;
⎝ 5.0 ⎠

⎛ 0.70 ⎞
resistance of r = ⎜ = ⎟ 1.4 Ω;
⎝ 0.50 ⎠
or
0.70
resistance of r = × 5.0 ;
2.5
= 1.4 Ω;
or
3.2 = 0.5( R + r ) ;
resistance of r = 1.4 Ω; [4]
Award [4] for alternative working.
– 10 – M13/4/PHYSI/HP2/ENG/TZ2/XX/M


Part 2 Thermodynamic cycle

(a) a real gas can be liquefied (whereas an ideal gas cannot);
⎫ (do not accept “obeys equation
a real gas does not always obey (the equation ⎪
⎬ of state” without statement of
of state) PV = nRT (whereas an ideal gas does); ⎪
⎭ equation)
a real gas does not obey Boyle’s law for all values of pressure;
a real gas does not obey Charles’s law for all values of temperature; [2 max]
Accept discussion only of macroscopic effects.

(b) (i) for an adiabatic change there is no (non-mechanical)
transfer of (heat/thermal) energy (to or from the gas);
so the work done on or by the gas is equal to the (magnitude of the) change
in internal energy of the gas; [2]

(ii) for an isothermal change there is no change in
internal energy;
so that the work done on or by the gas is equal to
the energy (to or from the gas);
[2]
(c) CD;
(isothermal) compression / volume decreases;
in which work is done on the gas;
if there is to be no increase in internal energy the gas must lose energy to the
surroundings / OWTTE; [4]
– 11 – M13/4/PHYSI/HP2/ENG/TZ2/XX/M


B2. Part 1 Power production and the greenhouse effect

(a) energy output in a year = ( 4.0 ×109 × 3.2 ×107 = ) 1.28 ×1017 J
⎛ 1.28 ×1017 ⎞
energy input = ⎜ = ⎟ 3.2 ×1017 J ;
⎝ 0.4 ⎠
⎛ 3.2 ×10 17

mass of coal = ⎜ = ⎟ 1.3 ×1010 kg ; [3]
⎝ 2.4 ×10 ⎠
7


Allow approach using power output.

or
 4.0 
power required from coal = 
 =
 10GW ;
 0.4
⎛ 10 ×109 ⎞
mass of coal required each second = ⎜ = ⎟ 417 kg ;
⎝ 24 ×10 ⎠
6


mass of coal required each year = (417 × 3.2 ×107 =) 1.33 ×1010 kg
Allow alternative working leading to correct answer.


(b) advantage:
(nuclear power) does not produce carbon dioxide;
therefore it does not add to the greenhouse effect/global warming;
or
energy density of U-235 (fuel) is very high / small mass is required;
fuel is likely to last a long time/easier to transport / OWTTE;

disadvantage:
waste products (of U-235 fuel) are radioactive;
no safe method of disposal / long half-life; (do not accept “lasts a long time”) [4]
or
allows development of nuclear weapons;
mention of plutonium/uranium enrichment/dirty bomb;
or
accidents are potentially catastrophic;
leading to widespread mutations/cancers/contamination/other named effect;
or
power plant is more expensive;
plausible reason for the expense for example safety/complex plant/
decommissioning / OWTTE;
– 12 – M13/4/PHYSI/HP2/ENG/TZ2/XX/M


(c) power output of a turbine = 0.3 × 12 ρ Av3 = 0.3 × 0.5 ×1.2 × 3.14 × [42]2 × [12]3 ( = 1723kW) ;
4 ×109
number of turbines needed = (= 2322) ;
1.723 ×106
area needed = 2322 × 5.0 × 10 4 ;
= 1.2 × 108 m 2 ; [4]

(d) look for these main points:
the surface of Earth re-radiates the Sun’s radiation;
greenhouse gases (in atmosphere) readily absorb infrared;
mention of resonance;
the absorbed radiation is re-emitted (by atmosphere) in all directions;
(some of) which reaches the Earth and further heats the surface; [3 max]
Award [1 max] for responses along the lines that greenhouse gases trap infrared
radiation.

(e) total absorbed radiation = total emitted radiation = 238 W m –2 ;
1
⎡ 238 ⎤ 4
temperature of Earth = ⎢ = 255 K; [2]
⎣ 5.67 ×10 ⎥⎦
−8
– 13 – M13/4/PHYSI/HP2/ENG/TZ2/XX/M


Part 2 Optical resolution and polarization

(a) intensity



S2 S1




S1 / S 2 shows a central maximum flanked by at least one subsidiary maximum on
each side;
intensity of first maximum is less than one-third of the intensity of the central
maximum; (judge by eye)
first minimum of S1 / S 2 coincides with central maximum of S 2 / S1 ;

3.5 × 106
(b) angle (subtended at eye for Moon to be resolved) = ;
d
⎛ λ ⎞ 1.22 × 4.2 ×10 −7
= ⎜1.22 = ⎟ ;
⎝ b ⎠ 4.0 ×10−3
3.5 ×106 × 4.0 ×10−3
d= or 2.7 ×1010 m ;
1.22 × 4.2 ×10−7
(≈ 3 ×1010 m) [3]

(c) (i) the (electric/magnetic) vector/direction of electric field (of the light) vibrates
only in one plane; [1]
Accept a labelled diagram.

(ii) φ = tan −1 n = 54D or 55D or 0.95 rad;
( )
angle to surface = 90 – 55 = 36D or 35D or 0.62 rad; [2]
Award [2] for a bald correct answer.
– 14 – M13/4/PHYSI/HP2/ENG/TZ2/XX/M


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