2016 HL P2 EoY12 Exam MS. Question 1.

– 6 – M12/4/PHYSI/SP2/ENG/TZ2/XX/M

SECTION A A1. (a)

f / Hz

70

60

50

40

30

20

10

0

0 5 10 15 20 25 h / m smooth curve as above; (judge by eye) [1] Do not allow point-to-point curve. Do not allow curve to “curl round” at low or high h. Single “non-hairy” line only is acceptable. (b) choice of points separated by ( 7.5)h' t e.g. [6.0, 37] [15, 6.0]; recognize fh constant for an inverse relation; calculates fh correctly for both points; state that two calculated numbers are not equal (therefore not inverse); [4] Award [3 max] if data points are not on line. Award [3 max] if data points are too close together ( 7.5)h' t . Award [2 max] if both of above. (c) (i) a straight-line that goes through all the error bars; and drawn through the origin; (allow 1

2r square) [2] (ii)

read-off of suitable point(s) on line separated by at least half of drawn line;

­®¯ (allow implicit use of origin)

calculation of gradient to give 31.5( 0.2) 10r u ; 1 2s m or 2Hz m ; [3] (d) the relation might not hold/extrapolate for larger values of h / outside range of

experiment / values would be close to origin and with large (percentage experimental) error / girders of this height could buckle under their own weight / OWTTE; [1]

Question 2.

– 6 – M07/4/PHYS/SP2/ENG/TZ1/XX/M+

SECTION A A1. (a) correct positioning of all three error bars; correct length of error bars – at least 3 mm long, even if only two are shown; [2] (b) a smooth curve through the data points within 2 mm of each data point; [1] Award [0] for points joined by straight-line segments. (c) line of best-fit must be a curve; in order to pass through the error bars; [2] Award [0] for plain “curve” without attempt at an explanation. (d) (i) 20( 2) C; [1] (ii) large enough triangle-hypotenuse 6 cm, from which to get slope; correct calculation of slope at 50cmx to give 1( )1.05 0.25 C cm ; for a more accurate calculation in the range 1( )1.05 0.15 C cm ; [3]

(e) realization that 50 10

50 10

R R

x x

if rate is proportional to temperature gradient;

substitution to get ;W2505.181.1

43

50

10

1050 x

x

RR

with a comment about the agreement of the result with the given value of 50R ; [3] A2. (a) (i) initial momentum 500 6 3000 Ns; final momentum 500 ( 1) 700 5 3000 Ns ; [2] (working must be shown to award marks) Allow approach that shows equal and opposite momentum changes. (ii) initial kinetic energy 1

2 500 36 9000J ; final kinetic energy 1 1

2 2500 1 700 25 9000J ; [2] (working must be shown to award marks) (b) impulse change of momentum 700 5 3500 Ns ; duration of collision 2.0s ;

to give 3500 1800 N2.0

F ; [3]

Accept force in the range 1700 N to 1800 N even with three significant figures.

Question 3.

– 13 – M12/4/PHYSI/HP2/ENG/TZ1/XX/M

Part 2 Projectile motion (a) (i) zero; [1]

(ii) horizontal: any horizontal line not on t-axis (accept lines above or below t-axis); vertical: any diagonal line starting at origin (accept positive or negative gradients); [2]

horizontal component vertical component vx

t

vy

t

(b) (i) 2 21 1110 10 ;2 2y ys a t t u u

4.690 4.7s;t | [2] (ii) 5.0 4.690;x xs u t u 23m;xs [2] (c) lower maximum height; lower horizontal range;

asymmetrical with horizontal range before maximum height more than horizontal range after maximum height; [3]

Question 4.

– 8 – M06/4/PHYSI/SP2/ENG/TZ1/XX/M+

A2. (a) there are no positions; the lamp is effectively in series with 100 kΩ no matter what the position of S; this means that the pd across it will always be close to zero (very small) / never

reach 6 V; Or there are no positions; the resistance of the filament is much smaller that 100 kΩ ; so (nearly) all the potential of the battery appears across the variable resistance; [3] Award [1] max for correct answer with no argument or incorrect argument.

Anthropomorphic answers such “battery has a lot of resistance to overcome” score [1] max. Must mention that voltmeter is effectively in series with battery to get full marks.

(b) VIR

= ;

12

0.80 A15

= = ; [2]

(c)

A

V

S12 V

correct position of ammeter; correct position of voltmeter (either to the right or left of the lamp); [2] A3. (a) towards the centre of Earth; [1] (b) gravity/gravitational/centripetal; [1] (c) 424 3600 8.64 10 sT = × = × ;

substitute into 2 2 7

2 2 8

4 4π 4.2 10(8.64) 10

RaTπ × ×

= =×

;

20.23ms−= ; [3]

Question 5.

– 12 – M13/4/PHYSI/SP2/ENG/TZ2/XX/M

Part 2 Thermal concepts

(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) ( ) ( )3 40.25 4.2 10 27 2.835 10 JUΔ = × × × = × ;

4

2.8 10 J= × ; [2]

(ii) energy transfer4 3

300 120 – 2.835 10 7.65 10 J[ ] [ ]= × × = × ;

rate of transfer

37.650 10

64 W120

×= = ; [2]

Allow ECF from (b)(i).

(c) (i) total energy supplied to water ( ) 5500 300 – 500 64 1.18 10 J= × × = × ;

specific latent heat

5

6 –11.18 102.4 10 J kg

0.05

Qm

⎛ ⎞×= = = ×⎜ ⎟⎝ ⎠

; [2]

Award [1 max] for 500 answer to (b)(ii)

0.05

× .

(ii) all the thermal energy is used to separate the molecules/break the bonds

between molecules;

and not to increase their (average) kinetic energy;

average kinetic energy is a measure of the temperature (of the water); [3]

Question 6. – 10 – M07/4/PHYS/SP2/ENG/TZ1/XX/M+

B2. Part 1 Waves on a string (a) (i) wavelength 3.0cm ; [1] (ii) period 0.25ms ; hence frequency 4000Hz ; [2] (Bald answer 4000 Hz scores [2])

(iii) 13

0.03 0.03 4000 120ms0.25 10

c ; [1]

Watch ECF from (i) and (ii) (b) (i) correct labelling of amplitude of 2.0 mm; [1] (Any line from equilibrium to crest or trough) (ii) cosine wave from x = 0 ; period constant throughout ; [2] (c) (i) pulse of similar shape and size; and inverted; [2] Accept pulse that is of similar width but smaller amplitude. (ii) the string pulls on the wall and so the wall pulls in the opposite direction on

the string by Newton’s third law; the wall pushes on the string creating an inverted pulse; [2] (d) (i) the oscillating left end creates a travelling wave to the right ; which gets reflected by the fixed end; at any one time there are two waves on the string travelling in opposite

directions whose displacements/amplitudes are added (creating the standing wave); [3]

(ii) c f

4 4.0m3L ;

hence 120 30Hz4.0

f ; [2]

Use ECF for wave speed from (a)(iii).

Question 7. – 9 – M12/4/PHYSI/HP2/ENG/TZ1/XX/M

A4. (a) (i) 330100 ;330 120s

vf fv u

§ ·ª º ª ºc u¨ ¸« » « »¨ ¸ ¬ ¼¬ ¼© ¹

157Hz;f c [2]

(ii) 330 120100 ;330

ov uf fv

§ · ª º ª ºc u¨ ¸« » « »¬ ¼ ¬ ¼© ¹

136Hz;f c [2] (b) Measured wavelength Measured wavespeed Moving source less than O0 equal to v0; Moving observer equal to O0; greater than v0; [3] A5. (a) general cos2 shape; zero intensity at 90˚ and maximum at both 0˚ and 180˚; [2]

transmitted intensity I

90 180 angle /T

(b) solution placed between two crossed polarizer; plane of polarization rotated by an amount that depends on concentration;

rotation of second polarizer/intensity of transmitted light is a measure of concentration; or solution placed between two polarizers (at any orientation); second polarizer rotated until light intensity reaches zero/maximum; rotation of second polarizer is a measure of concentration; [3]

0

Question 8.

– 13 – M07/4/PHYSI/HP2/ENG/TZ2/XX/M+

Part 2 Friction (a) nature of the surfaces; normal reaction; relative motion of the surfaces; [2 max] (b) friction is the frictional force between an object and a surface / two surfaces; static friction is (the frictional force) when the object/surfaces are at rest; dynamic friction is(the frictional force) when the object is sliding / one of the

surfaces is sliding / moving with respect to the other; some additional comment e.g. friction varies from zero to maximum / maximum

value of static friction always greater than kinetic friction; [3 max] Award [1 max] for an answer such as “friction force on an object at rest and friction

force on a moving object”. Some appreciation that it is friction between two surfaces is required.

(c) 7.2 0.6012s ; [1]

(d) it will accelerate; since the coefficient of dynamic friction is less than coefficient of static friction; therefore, frictional force acting is less than 7.2 N / a net force greater than zero

acting on the block; [3] Award [0] for a bald statement or incorrect reasoning.

Question 9.

Question 10.

– 8 – N12/4/PHYSI/HP2/ENG/TZ0/XX/M

A3. (a) θ–7

–4

6.2 104.5 10

×=

×(= –31.38 10× );

distance (= –31.38 10 3.4× × =4.68)≈4.7mm; [2] (b) (i) in order to be (just) resolved the first minimum of diffraction pattern (of one

image) coincides with the central maximum of the other (image) / OWTTE; [1] (ii) criterion specifies > 4.7 mm in this case / clear use of answer to (a) as

distance;

4.7 6.03.4

! "× =$ %& '

8.3 mm; [2]

Award [1 max] if factor of 1.22 used. (c) for white light: central maximum white, laser central maximum is monochromatic; white light fringes/lines will be coloured; blue diffracted least / OWTTE; [2 max] A4. (a) providing the temperature/physical conditions are constant and pd∝ current; [1] or providing the temperature/physical conditions are constant and the resistance

is constant; (b) (i) current for one lamp=1.5 A;

131.5

=8.67;

so 8; [3] Must show working for full credit. Allow any suitable method. (ii) 4.0 Ω; [1] (iii) resistance of incorrect lamp=16 Ω;

total resistance of “correct” lamps in parallel=1.3 Ω or 1 1 1 1 116 4 4 4R

= + + + ;

total resistance=1.2 Ω; [3]

Question 11. – 11 – M15/4/PHYSI/HP2/ENG/TZ1/XX/M

Part 2 Vibrations and waves

(e) (periodic) motion in which acceleration/restoring force is proportional to thedisplacement from a fixed point;directed towards the fixed point / in the opposite direction to the displacement; [2]

(f) (i) ( fω = π = π×2 2 125 )0 7854 rad –1s ;

xω= − = −a 20 0( 78542 ×0.85×10–3 =) (−)5.2×104 ms−2 ; [2]

(ii) correct substitution into xω=TE m 2 2102 irrespective of powers of 10;

0.14 to 0.15 J; [2]

(g) (i) 0.264 m; [1]

(ii) longitudinal;progressive / propagate (through the air) / travels with constant speed(through the air);series of compressions and rarefactions / high and low (air) pressure; [3]

(h) (i) S leads L / idea that the phase of L is the phase of S minus an angle;18π4

period / 1 10–4 s / 0.1 m s;

/ 0.79 rad / 45 degrees; [3]

(ii) agreement at all zero displacements;maxima and minimum at correct times;constant amplitude of 1.60 mm; [3]