Q1(a) net force

Correct. You stated that there is no net force acting on the body.

no net force acting on the body

Q1(a) net torque

Correct. You also stated there is no net torque acting on the body.

no net torque

Q1(b) horizontal

Correct horizontal resolution: $T_1 \cos 20^\circ = T_2 \cos 10^\circ$.

$T_1 \cos 20^\circ = T_2 \cos 10^\circ$

Q1(b) vertical

Correct vertical resolution equalling 700 N.

$T_1 \sin 20^\circ + T_2 \sin 10^\circ = 700$

Q1(b) T1

Correct. $T_1 = 1380\,\mathrm{N}$.

$T_1 = 1380$

Q1(b) T2

Correct. $T_2 = 1320\,\mathrm{N}$.

$T_2 = 1320$

Q1(c) T value

Your T = 185 N is too low. With moments about the base, T is about 277 to 280 N. Note: 150 N is the given wire tension, not the support-cable answer.

$T = 185\,\mathrm{N}$

What to do next: Redo with moments about the base; expect T around 280 N (2 s.f.).

Q1(c) need moments

You balanced horizontal forces. Take moments about the base of the pole instead.

$T_{\mathrm{cable},x} = T_{\mathrm{wire},x}$

What to do next: Pivot at the base so the ground reaction has zero moment, then equate moments of the wire and cable.

Q1(c) wrong equation

This force-balance equation is not the moment equation about the base.

$T \sin 53.13^\circ = 150 \cos 10^\circ$

What to do next: Moment of wire: 150 cos10 x 1.8. Moment of cable: T cos(theta to vertical) x 1.2, or use horizontal components with heights as levers.

Q2(a)(i) definition

Correct. Oscillations are perpendicular to the direction of wave travel.

vibration of particles perpendicular to direction of wave travel

Q2(a)(i) example

Accepted. Water waves are a valid transverse example.

water wave

Q2(a)(ii) polarisation

Correct idea: oscillations limited to one plane.

limiting of a wave to one plane

Q2(b)(i) Malus 1

Malus law is $I = I_0 \cos^2 \theta$, not $I_0^2 \cos \theta$. After the first filter use $I_0 \cos^2 30^\circ$.

$I_1 = I_0^2 \cos 30^\circ$

What to do next: $Write I1 = I_0 \cos^{2} 30 = (3/4) I_0.$

Q2(b)(i) filter angle

The angle between the two transmission axes is 30 degrees (60 minus 30), not another 30 to the vertical alone.

second filter … $\cos 30^\circ$

What to do next: $Use \cos^{2} 30 for the second filter relative to the first axis.$

Q2(b)(i) IT

$I_T$ should be about $0.56\,I_0$ (or $\frac{9}{16} I_0$), not a power of $I_0$.

$\dfrac{3\sqrt{3}}{8} I_0^4$

What to do next: $I_T = I_0 \cos^{2} 30 \cos^{2} 30 = (3/4)*(3/4) I_0 = 9/16 I_0.$

Q2(b)(ii) setup

Use intensity proportional to amplitude squared: amplitude ratio $= \sqrt{I_T/I_0}$.

$\left(\tfrac{1}{2} A\right)\cos 30^\circ$

What to do next: $From I_T/I_0 = 9/16, amplitude ratio = 3/4 = 0.75.$

Q2(b)(ii) ratio

Your ratio $\sqrt{3}/4$ is too small. The correct amplitude ratio is about $0.75$.

$\dfrac{\sqrt{3}}{4}$

What to do next: $sqrt(9/16) = 0.75.$

Q2(b)(iii) maxima

Correct. Intensity maxima at $0^\circ$, $180^\circ$ and $360^\circ$ at $I_0$.

$peaks at I_0$

Q2(b)(iii) zeros

Correct. Intensity is zero at 90 and 270 degrees.

zeros at 90 and 270

Q2(c)(i) longitudinal

Correct. Sound is longitudinal, so it cannot be polarised.

longitudinal wave

Q2(c)(ii) period

Correct. You used T = 1/f = 1/740 s.

$\dfrac{1}{740}$

Q2(c)(ii) time-base

Correct. Time-base setting $= 0.20\,\mathrm{ms\,cm^{-1}}$.

0.20

Q3(a) definition

Correct. Work done per unit positive charge from infinity to the point.

work done per unit positive charge ... from infinity

Q3(b)(i) V formula

Correct potential formula.

$V = \dfrac{Q}{4\pi\varepsilon_0 r}$

Q3(b)(i) Z=54

Correct. You showed there are 54 protons.

54 protons

Q3(b)(i) data

Correct. You used a point from the graph ($V = 780\,\mathrm{V}$ at $r = 1.0 \times 10^{-10}\,\mathrm{m}$).

$780 ... 1.0 \times 10^{-10}$

Q3(b)(ii) point charges

Correct. Sizes are negligible compared with the separation.

radius ... negligible compared to the separation

Q3(b)(iii) force

Correct use of Coulomb force.

$F_E = \dfrac{Qq}{4\pi\varepsilon_0 r^2}$

Q3(b)(iii) value

Correct. $a = 1.86 \times 10^{16}\,\mathrm{m\,s^{-2}}$.

$1.86 \times 10^{16}$

Q3(b)(iii) a=F/m

Correct. Acceleration from a = F/m.

$a = F_E/m$

Q3(b)(iv) conservation

Correct. Gain in kinetic energy equals loss in electric potential energy.

loss in EPE $=$ gain in KE

Q3(b)(iv) zero at infinity

Correct. Electric potential energy at infinity is zero.

$EPE_f = 0$

Q3(b)(iv) value

0.072 J is far too large. You need $U = Qq/(4\pi \varepsilon_0 r)$, which is about $6.2 \times 10^{-19}\,\mathrm{J}$.

0.072 J

What to do next: $Include both charges Q = 54e and q = e in the potential energy.$

Q4(a)(iii) path

Correct. The slower proton curves downward toward plate Q.

curved path toward Q

Q4(a)(i) Fm

Correct. By Fleming's left-hand rule, the magnetic force on the proton is upwards.

magnetic force ... upwards

Q4(a)(i) polarity

Correct. Plate P is positive so the electric force is downwards.

Plate P is positive

Q4(a)(ii) eV to J

Correct. You converted 64 eV using $1.6 \times 10^{-19}$.

$64 x 1.6 \times 10^{-19}$

Q4(a)(ii) v setup

Correct approach: find v from kinetic energy.

$v = sqrt(2KE/m)$

Q4(a)(ii) wrong mass

You used the electron mass $9.11 \times 10^{-31}\,\mathrm{kg}$. For a proton use about $1.67 \times 10^{-27}\,\mathrm{kg}$.

$9.11 \times 10^{-31}$

What to do next: Recalculate v with proton mass; v is about 1.11 x 10^5 m s^-1.

Q4(a)(ii) equate

Correct. You equated Bqv = qE so E = Bv.

$E = Bv$

Q4(a)(ii) E value

Your E is too large because v was wrong. With the proton mass, $E$ is about $4980\,\mathrm{N\,C^{-1}}$.

$2.13 \times 10^{5}$

What to do next: E = Bv with the corrected proton speed.

Q4(b) F/L formula

You need F/L = (mu0 IX IY)/(2 pi d).

$F = BIL$ only

What to do next: Combine B and F = BIL to get force per unit length.

Q4(b) blank B

No working for the field due to wire Y at X.

blank

What to do next: $B = (mu0 IY)/(2 \pi d) into the page at X.$

Q4(b) numerical

No numerical answer. $F/L$ is about $7.7 \times 10^{-5}\,\mathrm{N\,m^{-1}}$.

blank answer line

What to do next: $Substitute IX=5.5 A, IY=8.4 A, d=0.12 m.$

Q4(b) direction

Direction is blank. Parallel currents attract, so the force on X is toward Y (to the right).

blank direction

What to do next: State toward wire Y / to the right.

Q5(a) emf

Correct idea: work done by the source per unit charge round a complete circuit.

work done by the source in driving unit charges

Q5(b)(i) Req

Correct. S and T in parallel give $120\,\Omega$.

120

Q5(b)(i) open circuit

Good. You recognised that when the switch is open, E = 12.0 V.

$when open: E = 12.0$

Q5(b)(i) E

When the switch is open the voltmeter reads the emf, so E = 12.0 V, not 6.6 V.

$E = 6.6 V$

What to do next: $Set E = 12.0 V from the open-circuit reading.$

Q5(b)(i) answers

Both final values on the answer lines are incorrect (need $E = 12.0\,\mathrm{V}$ and $r$ about $13.3\,\Omega$).

$6.6 V and 540 \Omega$

What to do next: Rewrite the answer lines with the corrected pair.

Q5(b)(i) r

$r = 540\,\Omega$ is wrong. Use the potential divider: $10.8 = 12.0 \times 120/(120+r)$, giving $r$ about $13.3\,\Omega$.

$r = 540$

What to do next: $I = V/R = 10.8/120 when closed, then r = (E-V)/I.$

Q5(b)(ii) current

Current through r when closed is about 0.090 A (for example 10.8/120), not 0.01 A.

0.01

What to do next: I = E/(R+r) or I = V/R with the correct E and r.

Q5(b)(ii) energy

$19.8\,\mathrm{J}$ is wrong. Energy dissipated in the internal resistance is $I^2 r t$ (about $32\,\mathrm{J}$ with correct values).

19.8 J

What to do next: Use I^2 r t with I about 0.090 A, r about 13.3 ohm, t = 300 s.

Q5(b)(iii) NTC

Correct. An NTC thermistor's resistance falls as temperature rises.

resistance decrease

Q5(b)(iii) current up

Correct. Total current increases when total resistance falls.

total current ... increases

Q5(b)(iii) V falls

The terminal pd falls, not rises. Larger current means a larger Ir drop, so V = E - Ir decreases.

voltmeter reading increases

What to do next: End with: voltmeter reading decreases.

Q6(a) masses

Correct idea: Solar System sources are not massive enough compared with merging black holes.

solar system lacks strong gravitational sources

Q6(b)(i) dL

You used the fractional change alone. Multiply by the tube length 4.0 km: $\Delta L$ is about $4 \times 10^{-18}\,\mathrm{m}$.

$2.4 \times 10^{-21} m$

What to do next: $delta L = (delta L / L) x 4.0 \times 10^{3} m.$

Q6(b)(ii) too small

Correct. The length change is far too small compared with a wavelength-scale path difference.

relatively small

Q6(b)(ii) reflections

Correct idea: many reflections build up a usable path / phase difference.

many reflections to change the phase difference

Q6(b)(iii) zero amp

Also say the two beams have equal amplitude, so complete destructive interference gives zero resultant amplitude.

destructive interference ... zero amplitude

What to do next: Equal amplitude + antiphase => resultant amplitude zero.

Q6(b)(iii) antiphase

With no gravitational wave the path lengths are equal, and the beams meet with phase difference pi (antiphase), not path difference pi/2.

path difference is $\pi/2$, $\lambda/2$

What to do next: $State equal paths and phase difference of \pi radians.$

Q6(c)(i) photon energy

Correct. Photon energy E = hc/lambda.

$E photon = hc/\lambda$

Q6(c)(i) value

Correct. N = $5.35 \times 10^{20}$ s^-1.

$5.35 \times 10^{20}$

Q6(c)(i) N from P

Correct. $N$ per second $= P / E_{\mathrm{photon}}$.

$N/t = P \lambda / hc$

Q6(c)(ii) dp=2p

Correct. On reflection, change in momentum is 2p.

2p

Q6(c)(ii) momentum

Correct. Photon momentum p = h/lambda.

$p = h/\lambda$

Q6(c)(ii) force

Correct. Average force about $6.7 \times 10^{-7}$ N.

$6.7 \times 10^{-7}$

Q6(c)(iii) absorbed

Correct. Absorbed photons give a smaller momentum change, so the average force decreases.

force ... will decrease

Q6(d)(i) mass defect

Correct. Mass defect is 3 solar masses.

$62 - (29+36)$

Q6(d)(i) energy

Correct. Energy released = $5.4 \times 10^{47}$ J.

$5.4 \times 10^{47} J$

Q6(d)(ii) point source

Correct idea: treat the source as a point and use the inverse-square law.

$I proportional to 1/r^{2}$

Q6(d)(ii) power

Correct. Power = E/t = $2.7 \times 10^{49}$ W.

$2.7 \times 10^{49}$

Q6(d)(ii) intensity formula

Write $I = P / (4\pi r^2)$ with $r$ in metres.

$P/A = I$

What to do next: $Convert 1.3 \times 10^{9} light-years to metres, then divide P by 4 \pi r^{2}.$

Q6(d)(ii) blank I

No final intensity. With the correct steps, $I$ is about $0.014\,\mathrm{W\,m^{-2}}$.

blank

What to do next: $Finish I = P/(4 \pi r^{2}) numerically.$

Q6(e)(i) time

0.015 s does not match the chirp peaks well. Read the time between the two largest peaks on Fig 6.2 (about 0.007 to 0.012 s).

0.015

What to do next: Compare the Washington and Louisiana peak times carefully.

Q6(e)(ii) speed method

Correct method: $\text{speed} = \text{distance} / \text{time}$ using your time interval.

$2.00 \times 10^{8}$

What to do next: With a better delta t near 0.01 s you get a speed closer to c.

Q6(e)(iii) path

The issue is not a medium slowing the wave. Gravity bends / lengthens the actual path compared with the straight surface distance.

travelling through a medium

What to do next: Say the true path is longer because of gravitational fields of Earth and the Solar System.

Q6(f) advantage 1

Being closer to sources is not a valid advantage here. Better points: much less vibration from human activity, easier vacuum, or less dust in the beam path.

closer to the sources

What to do next: Pick two from: less vibration, easier vacuum without pumps, less dust.

Q6(f) advantage 2

Gravitational waves do not need a higher speed in space for this mark. Use a scheme advantage such as reduced vibration or easier vacuum.

absence of a medium ... higher speeds

What to do next: Replace with reduced seismic noise or vacuum quality.