Calculate the Carnot efficiency of the nuclear power plant.

Explain, with a reason, why a real nuclear power plant operating between the stated temperatures cannot reach the efficiency calculated in (a).

The nuclear power plant works at 71.0% of the Carnot efficiency. The power produced is 1.33 GW. Calculate how much waste thermal energy is released per hour.

Discuss the production of waste heat by the power plant with reference to the first law and the second law of thermodynamics.

correct conversion to K «622 K cold, 885 K hot»

${\eta }_{\mathrm{C}\mathrm{a}\mathrm{r}\mathrm{n}\mathrm{o}\mathrm{t}}=1-\frac{T_{\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{d}}}{T_{\mathrm{h}\mathrm{o}\mathrm{t}}}=1-\frac{622}{885}=0.297$ or 29.7%

Award [1 max] if temperatures are not converted to K, giving result 0.430.

the Carnot efficiency is the maximum possible

the Carnot cycle is theoretical/reversible/impossible/infinitely slow

energy losses to surroundings «friction, electrical losses, heat losses, sound energy»

OWTTE

0.71 × 0.297 = 0.211

Allow solution utilizing wasted power «78.9%».

1.33/0.211 × 0.789 = 4.97 «GW»

4.97 GW × 3600 = 1.79 × 10¹³ «J»  

Award [2 max] if 71% used as the overall efficiency giving an answer of 1.96 × 10¹² J.

Award [3] for bald correct answer.

Watch for ECF from (a).

Law 1: net thermal energy flow is Q_IN–Q_OUT

Q_OUT refers to “waste heat”  

Law 1: Q_IN–Q_OUT = ∆Q=∆W as ∆U is zero

Law 2: does not forbid Q_OUT=0

Law 2: no power plant can cover 100% of Q_IN into work

Law 2: total entropy must increase so some Q must enter surroundings  

OWTTE

The velocity of the falling object is 1.89 m s^–1 at 3.98 s. Calculate the average angular acceleration of the flywheel.

Show that the torque acting on the flywheel is about 0.3 Nm.

(i) Calculate the tension in the string.

(ii) Determine the mass m of the falling object.

ALTERNATIVE 1

${\omega }_{\text{final}}=\frac{v}{r}=31.5$ «rad s^–1»

«$\omega ={\omega }_o+\alpha t$ so» $\alpha =\frac{\omega }{t}=\frac{31.5}{3.98}=7.91$ «rad s^–2»

ALTERNATIVE 2
$a=\frac{1.89}{3.98}=0.4749$ «m s^–2»

$\alpha =\frac{a}{r}=\frac{0.4749}{0.060}=7.91$ «rad s^–2»

Award [1 max] for r = 0.24 mm used giving $\alpha$ = 1.98 «rad s^–2».

$\mathrm{\Gamma }=\frac{1}{2}\mathrm{M}{\mathrm{R}}^2\alpha =\frac{1}{2}\times 1.22\times {0.240}^2\times 7.91$

= 0.278 «Nm»

At least two significant figures required for MP2, as question is a “Show”.

i

$F_T=\frac{\mathrm{\Gamma }}{r}$

$F_T=4.63$ «N»

Allow 5 «N» if Γ=  0.3 Νm is used.

ii

$F_T=mg-ma$ so $m=\frac{4.63}{9.81-0.475}$
m = 0.496 «kg»

Allow ECF

Write down an expression, in terms of M, v and R, for the angular momentum of the system about the vertical axis just before the collision.

Just after the collision the system begins to rotate about the vertical axis with angular velocity ω. Show that the angular momentum of the system is equal to $\frac{4}{3}M{R}^2\omega$.

Hence, show that $\omega =\frac{v}{4R}$.

Determine in terms of M and v the energy lost during the collision.

Show that the angular deceleration of the system is 0.043 rad$$s^–2.

Calculate the number of revolutions made by the system before it comes to rest.

$\frac{M}{3}vR$

[1 mark]

evidence of use of: $L=I\omega =\left(M{R}^2+\frac{M}{3}{R}^2\right)\omega$

[1 mark]

evidence of use of conservation of angular momentum, $\frac{MvR}{3}=\frac{4}{3}M{R}^2\omega$

«rearranging to get $\omega =\frac{v}{4R}$»

[1 mark]

initial KE = $\frac{M{v}^2}{6}$

final KE = $\frac{M{v}^2}{24}$

energy loss = $\frac{M{v}^2}{8}$

[3 marks]

$\alpha$ «= $\frac{3}{4}\frac{\mathrm{\Gamma }}{M{R}^2}$» = $\frac{3}{4}\frac{0.01}{0.7\times {0.5}^2}$

«to give $\alpha$ = 0.04286 rad$$s⁻²»

Working OR answer to at least 3 SF must be shown

[1 mark]

$\theta =\frac{{\omega }_i^2}{2\alpha }$ «from ${\omega }_f^2={\omega }_i^2+2\alpha \theta$»

$\theta$ «$=\frac{v^2}{32R^2\alpha }=\frac{{2.1}^2}{32\times {0.5}^2\times 0.043}$» = 12.8 OR 12.9 «rad»

number of rotations «= $\frac{12.9}{2\pi }$» = 2.0 revolutions

[3 marks]

State what is meant by an adiabatic process.

Identify the two isothermal processes.

Determine the temperature of the gas at A.

The volume at B is 2.30 × 10^–3$$m³. Determine the pressure at B.

Show that $P_B{V}_B^{\frac{5}{3}}=nR{T}_C{V}_C^{\frac{2}{3}}$

The volume at C is 2.90 × 10^–3$$m³. Calculate the temperature at C.

State a reason why a Carnot cycle is of little use for a practical heat engine.

«a process in which there is» no thermal energy transferred between the system and the surroundings

[1 mark]

A to B AND C to D

[1 mark]

$T=\frac{PV}{nR}$

$T\left(=\frac{512\times {10}^3\times 1.20\times {10}^{-3}}{0.150\times 8.31}\right)\approx 493$ «K»

The first mark is for rearranging.

[2 marks]

$P_B=\frac{P_a{V}_A}{V_B}$

$P_B=267\text{ KPa}$

The first mark is for rearranging.

[2 marks]

«B to C adiabatic so» $P_{\text{B}}{V}_{\text{B}}^{\frac{5}{3}}=P_{\text{C}}{V}_{\text{C}}^{\frac{5}{3}}$ AND P_CV_C = nRT_C «combining to get result»

It is essential to see these 2 relations to award the mark.

[1 mark]

$T_{\text{C}}=\left(\frac{P_{\text{B}}{V}_{\text{B}}^{\frac{5}{3}}}{nR}\right)V_{\text{C}}^{\frac{-2}{3}}$

$T_{\text{C}}=$ «$\left(\frac{267\times {10}^3\times {\left(2.30\times {10}^{-3}\right)}^{\frac{5}{3}}}{0.150\times 8.31}\right){\left(2.90\times {10}^{-3}\right)}^{\frac{-2}{3}}$» = 422 «K»

[2 marks]

the isothermal processes would have to be conducted very slowly / OWTTE

[1 mark]

Show that the final angular velocity of the bar is about $3 \mathrm{rad} {\mathrm{s}}^{-1}$.

Draw the variation with time $t$ of the angular displacement $\theta$ of the bar during the acceleration.

Calculate the torque acting on the bar while it is accelerating.

The torque is removed. The bar comes to rest in $30$ complete rotations with constant angular deceleration. Determine the time taken for the bar to come to rest.

${{\omega }_f}^2=0+2\times 0.110\times 6\times 2\mathrm{\pi }$  ✓

${\omega }_f=2.88 «\text{rad} {\text{s}}^{-1}»$ ✓

Other methods are possible.
Answer 3 given so look for correct working
At least 2 sig figs for MP2.

concave up from origin ✓

 $\mathrm{\Gamma }=«\mathrm{I} \mathrm{\alpha } \mathrm{so} \mathrm{\Gamma }=0.110\times 0.0216=» 2.38\times {10}^{-3} «\mathrm{N} \mathrm{m}»$ ✓

$\alpha =\frac{2.9^2}{2\times 2\mathrm{\pi }\times 30}=$ OR $-0.022 «\mathrm{rad} {\mathrm{s}}^{-2}»$ ✓

$t «=\frac{{\omega }_f-{\omega }_i}{\alpha }=\frac{-2.9}{-0.0220}»=130«\mathrm{s}»$✓

Other methods are possible.

Allow $\mathit{131} \mathrm{s}$ if $\mathit{2}\mathit{.}\mathit{88}$ used

Allow $\mathit{126} \mathrm{s}$ if $\mathit{3}$ used

Award [2] marks for a bald correct answer

Deduce the linear acceleration of the centre of mass of the probe.

Calculate the resultant torque about the axis of the probe.

The forces act for 2.00 s. Show that the final angular speed of the probe is about 16 rad$$s^–1.

Determine the final angular speed of the probe–satellite system.

Calculate the loss of rotational kinetic energy due to the linking of the probe with the satellite.

zero

[1 mark]

the torque of each force is 9.60 x 10³ x 6.0 = 5.76 x 10⁴ «Nm»

so the net torque is 2 x 5.76 x 10⁴ = 1.15 x 10⁵ «Nm»

Allow a one-step solution.

[2 marks]

the angular acceleration is given by $\frac{1.15\times {10}^5}{1.44\times {10}^4}$ «= 8.0 s^–2»

ω = «$\alpha$t = 8.0 x 2.00 =» 16 «s^–1»

[2 marks]

1.44 x 10⁴ x 16.0 = (1.44 x 10⁴ + 4.80 x 10³) x ω

ω = 12.0 «s^–1»

Allow ECF from (b).

[2 marks]

initial KE $\frac{1}{2}\times 1.44\times {10}^4\times {16.0}^2=1.843\times {10}^6$ «J»

final KE $\frac{1}{2}\times \left(1.44\times {10}^4+4.80\times {10}^3\right)\times {12.0}^2=1.382\times {10}^6$ «J»

loss of KE = 4.6 x 10⁵ «J»

Allow ECF from part (c)(i).

[3 marks]

Identify the mechanism leading stars to produce the light they emit.

Outline why the light detected from Jupiter and Vega have a similar brightness, according to an observer on Earth.

Outline what is meant by a constellation.

Outline how the stellar parallax angle is measured.

Show that the distance to Vega from Earth is about 25 ly.

«nuclear» fusion

Do not accept “burning’’

brightness depends on luminosity and distance/b = $\frac{L}{4\pi d^2}$

Vega is much further away but has a larger luminosity

Accept answer in terms of Jupiter for MP2

a group of stars forming a pattern on the sky AND not necessarily close in distance to each other

OWTTE

the star’s position is observed at two times, six months apart, relative to distant stars

parallax angle is half the angle of shift

Answers may be given in diagram form, so allow the marking points if clearly drawn

$\frac{1}{0.13}$ = 7.7 «pc»

so d = 7.7 x 3.26 = 25.1 «ly»

Explain what is meant by proper length.

Draw a spacetime diagram for this situation according to an observer at rest relative to the tunnel.

Calculate the velocity of the train, according to an observer at rest relative to the tunnel, at which the train fits the tunnel.

For an observer on the train, it is the tunnel that is moving and therefore will appear length contracted. This seems to contradict the observation made by the observer at rest to the tunnel, creating a paradox. Explain how this paradox is resolved. You may refer to your spacetime diagram in (b).

the length of an object in its rest frame

OR

the length of an object measured when at rest relative to the observer

world lines for front and back of tunnel parallel to ct axis

world lines for front and back of train

which are parallel to ct′ axis

realizes that gamma = 1.25

0.6c

ALTERNATIVE 1

indicates the two simultaneous events for t frame

marks on the diagram the different times «for both spacetime points» on the ct′ axis «shown as Δt′ on each diagram»

ALTERNATIVE 2: (no diagram reference)

the two events occur at different points in space

statement that the two events are not simultaneous in the t′ frame

Complete the diagram, with a Newtonian mounting, continuing the two rays to show how they pass through the eyepiece.

When the Earth-Moon distance is 363 300 km, the Moon is observed using the telescope. The mean radius of the Moon is 1737 km. Determine the focal length of the mirror used in this telescope when the diameter of the Moon’s image formed by the main mirror is 1.20 cm.

The final image of the Moon is observed through the eyepiece. The focal length of the eyepiece is 5.0 cm. Calculate the magnification of the telescope.

The Hubble Space reflecting telescope has a Cassegrain mounting. Outline the main optical difference between a Cassegrain mounting and a Newtonian mounting.

plane mirror to the left of principal focus tilted anti-clockwise

two rays which would go through the principal focus

two rays cross between mirror and eyepiece AND passing through the eyepiece

eg:

$\frac{2\times 1737}{363300}=\frac{0.0120}{f}$

f = 1.25 «m»

Allow ECF if factor of 2 omitted answer is 2.5m

M = $\frac{1.25}{0.05}$ = 25

parabolic/convex mirror instead of flat mirror

eyepiece/image axis same as mirror

Justify why the thermal energy supplied during the expansion AB is 416 J.

Show that the temperature of the gas at C is 386 K.

Show that the thermal energy removed from the gas for the change BC is approximately 330 J.

Determine the efficiency of the heat engine.

State and explain at which point in the cycle ABCA the entropy of the gas is the largest.

ΔU = 0 so Q = ΔU + W = 0 + 416 = 416 «J»

Answer given, mark is for the proof.

[1 mark]

ALTERNATIVE 1

use $p{V}^{\frac{5}{3}}=c$ to get $T{V}^{\frac{2}{3}}=c$

hence $T_{\text{C}}=T_{\text{A}}{\left(\frac{V_{\text{A}}}{V_{\text{C}}}\right)}^{\frac{2}{3}}=612\times {0.5}^{\frac{2}{3}}=385.54$

«T_C ≈ 386K»

 ALTERNATIVE 2

$P_{\text{C}}{V_{\text{C}}}^{\gamma }=P_{\text{A}}{V_{\text{A}}}^{\gamma }$ giving P_C = 1.26 x 10⁶ «Pa»

$\frac{P_{\text{C}}{V}_{\text{C}}}{T_{\text{C}}}=\frac{P_{\text{A}}{V}_{\text{A}}}{T_{\text{A}}}$ giving $T_{\text{C}}=1.26\times \frac{612}{2}=385.54$ «K»

«T_C ≈ 386K»

Answer of 386K is given. Look carefully for correct working if answers are to 3 SF.

There are other methods:

Allow use of P_B = 2 x 10⁶ «Pa» and $\frac{P}{T}$ is constant for BC.

Allow use of n = 0.118 and T_C = $\frac{P_{\text{C}}{V}_{\text{C}}}{nR}$

[2 marks] 

$Q=\mathrm{\Delta }U+W=\frac{3}{2}\frac{P_{\text{A}}{V}_{\text{A}}}{T_{\text{A}}}\mathrm{\Delta }T+0$

$Q=\frac{3}{2}\times \frac{4.00\times {10}^6\times 1.50\times {10}^{-4}}{612}\times \left(386-612\right)$

«–332 J»

Answer of 330 J given in the question.
Look for correct working or more than 2 SF.

[2 marks]

$\text{e}=\frac{Q_{\text{in}}-Q_{\text{out}}}{Q_{i\text{n}}}=\frac{412-332}{416}$

e = 0.20

Allow $\frac{416-330}{416}$.

Allow e = 0.21.

[2 marks]

entropy is largest at B

entropy increases from A to B because T = constant but volume increases so more disorder or ΔS = $\frac{Q}{T}$ and Q > 0 so ΔS > 0

entropy is constant along CA because it is adiabatic, Q = 0 and so ΔS = 0
OR
entropy decreases along BC since energy has been removed, ΔQ < 0 so ΔS < 0

[3 marks]

A solid sphere of radius $r$ and mass $m$ is released from rest and rolls down a slope, without slipping. The vertical height of the slope is $h$. The moment of inertia $I$ of this sphere about an axis through its centre is $\frac{2}{5}m{r}^2$.

Show that the linear velocity $v$ of the sphere as it leaves the slope is $\sqrt{\frac{10gh}{7}}$.

conservation of rotational and linear energy

OR
$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$ ✓

using $I=\frac{2}{5}m{r}^2$ AND $\omega =\frac{v}{r}$ ✓

with correct manipulation to find the requested relationship ✓

The diagram shows two methods of pedalling a bicycle using a force F.

In method 1 the pedal is always horizontal to the ground. A student claims that method 2 is better because the pedal is always parallel to the crank arm. Explain why method 2 is more effective.

in method 1 the perpendicular distance varies from 0 to a maximum value, in method 2 this distance is constant at the maximum value
OR
angle between F and r is 90° in method 2 and less in method 1
OR
Γ = F × perpendicular distance

perpendicular distance/ torque is greater in method 2

Show that the angular acceleration of the merry-go-round is 0.2 rad s^–2.

Calculate, for the merry-go-round after one revolution, the angular speed. 

Calculate, for the merry-go-round after one revolution, the angular momentum.

Calculate the new angular speed of the rotating system.

Explain why the angular speed will increase.

Calculate the work done by the child in moving from the edge to the centre.

Γ «= Fr = 50 × 2» = 100 «Nm»

α « $=\frac{\mathrm{\Gamma }}{I}=\frac{100}{450}$ » =0.22 «rads^–2»

Final value to at least 2 sig figs, OR clear working with substitution required for mark.

[2 marks]

«${\omega }_t^2-{\omega }_0^2=2\alpha \mathrm{\Delta }\theta$»

«${\omega }_t^2-0=2\times 0.22\times 2\pi$»

${\omega }_t=1.7$ «rads^–1»

Accept BCA, values in the range: 1.57 to 1.70.

[1 mark]

«L = Iω = 450 × 1.66»

= 750 «kgm² rads^–1»

Accept BCA, values in the range: 710 to 780.

[1 mark]

«I = 450 + mr²»

I «= 450 + 30 × 2²» = 570 «kgm²»

«L = 570 × ω = 747»

ω = 1.3 «rads^–1»

Watch for ECF from (a) and (b).

Accept BCA, values in the range: 1.25 to 1.35.

[2 marks]

moment of inertia will decrease

angular momentum will be constant «as the system is isolated»

«so the angular speed will increase»

[2 marks]

ω_t = 1.66 from bi AND W = ΔE_k

W = $\frac{1}{2}$ × 450 × 1.66² – $\frac{1}{2}$ × 570 × 1.31² = 131 «J»

ECF from 8bi

Accept BCA, value depends on the answers in previous questions.

[2 marks]

On the diagram, draw and label the forces acting on the hoop.

Show that the linear acceleration a of the hoop is given by the equation shown.

a = $\frac{g\times \sin q}{2}$

Calculate the acceleration of the hoop when θ = 20°. Assume that the hoop continues to roll without slipping.

State the relationship between the force of friction and the angle of the incline.

The angle of the incline is slowly increased from zero. Determine the angle, in terms of the coefficient of friction, at which the hoop will begin to slip.

weight, normal reaction and friction in correct direction

correct points of application for at least two correct forces

Labelled on diagram.

Allow different wording and symbols

Ignore relative lengths

ALTERNATIVE 1

ma = mg sin θ – F_f

I$\alpha$ = F_f x r

OR

mr $\alpha$ = F_f

$\alpha$ = $\frac{a}{r}$

ma = mg sin θ – mr $\frac{a}{r}$ → 2a = g sin θ

Can be in any order

No mark for re-writing given answer

Accept answers using the parallel axis theorem (with I = 2mr²) only if clear and explicit mention that the only torque is from the weight

Answer given look for correct working

ALTERNATIVE 2

mgh = $\frac{1}{2}$Iω² + $\frac{1}{2}$ mv²

substituting ω = $\frac{v}{r}$ «giving v = $\sqrt{gh}$»

correct use of a kinematic equation

use of trigonometry to relate displacement and height «s = h sin θ»

For alternative 2, MP3 and MP4 can only be awarded if the previous marking points are present

1.68 «ms^–2»

ALTERNATIVE 1

N = mg cos θ

F_f ≤ μmg cos θ

ALTERNATIVE 2

F_f = ma «from 7(b)»

so F_f = $\frac{mg\sin \theta }{2}$

F_f = μmg cos θ

$\frac{mg\sin \theta }{2}$ = mg sin θ – μmg cos θ

OR

mg $\frac{\sin \theta }{2}$ = μmg cos θ

algebraic manipulation to reach tan θ = 2μ

Show that the volume of the gas at the end of the adiabatic expansion is approximately 5.3 x 10^–3 m³.

Using the axes, sketch the three-step cycle.

The initial temperature of the gas is 290 K. Calculate the temperature of the gas at the start of the adiabatic expansion.

Using your sketched graph in (b), identify the feature that shows that net work is done by the gas in this three-step cycle.

$500000\times {\left(2\times {10}^{-3}\right)}^{\frac{5}{3}}=100000\times V^{\frac{5}{3}}$

V = 5.3 x 10^–3 «m³»

Look carefully for correct use of pV^γ = constant

correct vertical and horizontal lines

curve between B and C

Allow tolerance ±1 square for A, B and C

Allow ECF for MP2

Points do not need to be labelled for marking points to be awarded

use of PV = nRT OR use of $\frac{P}{T}$ = constant

T = «5 x 290 =» 1450 «K»

area enclosed

work is done by the gas during expansion

OR

work is done on the gas during compression

the area under the expansion is greater than the area under the compression

Explain the direction in which the person-turntable system starts to rotate.

Explain the changes to the rotational kinetic energy in the person-turntable system.

«person rotates» anticlockwise ✓

the person gains angular momentum «in the opposite direction to the new wheel motion» ✓

so that the total angular momentum is conserved ✓

OWTTE

Award [1 max] for a bald statement of conservation of angular momentum

the rotational kinetic energy has increased ✓

energy is provided by the person doing work «flipping the wheel» ✓

OWTTE

Show that at C the pressure is 1.00 × 10⁶Pa.

Show that at C the temperature is 254 K.

Show that the thermal energy transferred from the gas during the change B → C is 238 J.

The work done by the gas from A → B is 288 J. Calculate the efficiency of the cycle.

State, without calculation, during which change (A → B, B → C or C → A) the entropy of the gas decreases.

ALTERNATIVE 1:

$P_c=P_B=\frac{P_A{V}_A}{V_B}$ ✔

= $\frac{2.8\times {10}^6\times 1\times {10}^{-4}}{2.8\times {10}^{-4}}$ «= 1.00 × 10⁶Pa» ✔

ALTERNATIVE 2:

$2.8\times {10}^6\times {1.00}^{\frac{5}{3}}=P_c\times {1.85}^{\frac{5}{3}}$ ✔

$P_c=2.8\times {10}^6\times \frac{{1.00}^{\frac{5}{3}}}{{1.85}^{\frac{5}{3}}}$ «= 1.00 × 10⁶Pa» ✔

ALTERNATIVE 1:

Since T_B = T_A then T_c = $\frac{V_c{T}_B}{V_B}$ ✔

 = $\frac{1.85\times 385}{2.8}$ «= 254.4K» ✔

ALTERNATIVE 2:

$\frac{2.80\times 1.00}{385}=\frac{1.00\times 1.85}{T_c}$ «K»✔

$T_c=385\times \frac{1.00\times 1.85}{2.80}$ «= 254.4K» ✔

work done = «pΔV = 1.00 × 10⁶ × (1.85 × 10⁻⁴ − 2.80 × 10⁻⁴ =» −95 «J» ✔

change in internal energy = «$\frac{3}{2}$pΔV = −$\frac{3}{2}$ × 95 =» −142.5 «J» ✔

Q = −95 − 142.5 ✔

«−238 J»

Allow positive values.

net work is 288 −238 = 50 «J» ✔

efficiency = «$\frac{288-238}{288}$ =» 0.17 ✔

along B→C ✔

Calculate the force the support exerts on the rod.

Calculate, in rad s^–2, the initial angular acceleration $\alpha$ of the rod.

After time t the rod makes an angle θ with the horizontal. Outline why the equation $\theta =\frac{1}{2}\alpha t^2$ cannot be used to find the time it takes θ to become $\frac{\pi }{2}$ (that is for the rod to become vertical for the first time).

At the instant the rod becomes vertical show that the angular speed is ω = 2.43 rad s^–1.

At the instant the rod becomes vertical calculate the angular momentum of the rod.

taking torques about the pivot R × 4.00 = 36.0 × 2.5 ✔

R = 22.5 «N» ✔

36.0 × 2.50 = 30.6 × $\alpha$ ✔

$\alpha$ = 2.94 «rad s^–2» ✔

the equation can be applied only when the angular acceleration is constant ✔

any reasonable argument that explains torque is not constant, giving non constant acceleration ✔

«from conservation of energy» Change in GPE = Change in rotational KE ✔

$W\frac{L}{2}=\frac{1}{2}I{\omega }^2$ ✔

$\omega =\sqrt{\frac{36.0\times 5.00}{30.6}}$ ✔

«ω = 2.4254 rad s^–1»

L = 30.6 × 2.43 = 74.4 «J s» ✔

The moment of inertia of the wheel is 1.3 × 10^–4 kg m². Outline what is meant by the moment of inertia.

In moving from point A to point B, the centre of mass of the wheel falls through a vertical distance of 0.36 m. Show that the translational speed of the wheel is about 1 m s^–1 after its displacement.

Determine the angular velocity of the wheel at B.

Describe the effect of F on the linear speed of the wheel.

Describe the effect of F on the angular speed of the wheel.

an object’s resistance to change in rotational motion

OR

equivalent of mass in rotational equations

OWTTE

[1 mark]

ΔKE + Δrotational KE = ΔGPE

OR

$\frac{1}{2}$mv² + $\frac{1}{2}$I$\frac{v^2}{r^2}$ = mgh

$\frac{1}{2}$ × 0.250 × v² + $\frac{1}{2}$ × 1.3 × 10^–4 × $\frac{v^2}{1.44\times {10}^{-4}}$ = 0.250 × 9.81 × 0.36

v = 1.2 «m s^–1»

[3 marks]

ω «= $\frac{1.2}{0.012}$» = 100 «rad s^–1»

[1 mark]

force in direction of motion

so linear speed increases

[2 marks]

force gives rise to anticlockwise/opposing torque on

wheel ✓ so angular speed decreases ✓

OWTTE

[2 marks]

Show that the final volume of the gas is about 53 m³.

Calculate, in J, the work done by the gas during this expansion.

Determine the thermal energy which enters the gas during this expansion.

Sketch, on the pV diagram, the complete cycle of changes for the gas, labelling the changes clearly. The expansion shown in (a) and (b) is drawn for you.

Outline the change in entropy of the gas during the cooling at constant volume.

There are various equivalent versions of the second law of thermodynamics. Outline the benefit gained by having alternative forms of a law.

ALTERNATIVE 1

«Using $\frac{V_1}{T_1}=\frac{V_2}{T_2}$»

V₂ = $\frac{47.1\times (273+19)}{(273-12)}$

V₂ = 52.7 «m³»

ALTERNATIVE 2

«Using PV = nRT»

V = $\frac{243\times 8.31\times (273+19)}{11.2\times {10}^3}$

V = 52.6 «m³»

[2 marks]

W «= PΔV» = 11.2 × 10³ × (52.7 – 47.1)

W = 62.7 × 10³ «J»

Accept 66.1 × 10³ J if 53 used

Accept 61.6 × 10³ J if 52.6 used

[2 marks]

ΔU «= $\frac{3}{2}$nRΔT» = 1.5 × 243 × 8.31 × (19 – (–12)) = 9.39 × 10⁴

Q «= ΔU + W» = 9.39 × 10⁴ + 6.27 × 10⁴

Q = 1.57 × 10⁵ «J»

Accept 1.60 × 10⁵ if 66.1 × 10³ J used

Accept 1.55 × 10⁵ if 61.6 × 10³ J used

[3 marks]

concave curve from RHS of present line to point above LHS of present line

vertical line from previous curve to the beginning

[2 marks]

energy is removed from the gas and so entropy decreases

OR

temperature decreases «at constant volume (less disorder)» so entropy decreases

OWTTE

[1 mark]

different paradigms/ways of thinking/modelling/views

allows testing in different ways

laws can be applied different situations

OWTTE

[1 mark]

Show that the work done on the gas for the isothermal process C→A is approximately 440 J.

Calculate the change in internal energy of the gas for the process A→B.

Calculate the temperature at A if the temperature at B is −40°C.

Determine, using the first law of thermodynamics, the total thermal energy transferred to the building during the processes C→A and A→B.

Suggest why this cycle is not a suitable model for a working heat pump.

evidence of work done equals area between AC and the Volume axis ✓

reasonable method to estimate area giving a value 425 to 450 J ✓

Answer 440 J is given, check for valid working.

Examples of acceptable methods for MP2:

- estimates 17 to18 small squares x 25 J per square = 425 to 450 J.
- 250 J for area below BC plus a triangle of dimensions 5 × 3, 3 × 5, or 4 × 4 small square edges giving 250 J + 187.5 J or 250 J + 200 J.

Accurate integration value is 438 J - if method seen award [2].

«use of $U=\frac{3}{2}nRT$ and $pV=nRT$ to give»

$\mathrm{\Delta }U=\frac{3}{2}\mathrm{\Delta }pV$ ✔

«$=\frac{3}{2}\times -2.5\times {10}^5\times 1\times {10}^{-3}$»

=«–»375«J»  ✔

Another method is possible: eg realisation that ΔU for BC has same magnitude, so ΔU = 3/2 PΔV.

T_A = 816«K» OR  543«°C»✔

for CA ΔU = 0 so Q = W = −440 «J» ✔

for AB W = 0 so Q = ΔU = −375 «J» ✔

815 «J» transferred to the building ✔

Must use the first law of thermodynamics for MP1 and MP2.

the temperature changes in the cycle are too large ✔

the cycle takes too long «because it contains an isothermal stage» ✔

energy/power output would be too small ✔

At SL, Correct answers were rare and very few candidates used the fact the work done was area under the curve, and even fewer could estimate this area. At HL, the question was better answered. Candidates used a range of methods to estimate the area including counting the squares, approximating the area using geometrical shapes and on a few occasions using integral calculus.

Not very many candidates seem to know the generalised formula ΔU =1.5(P2V2 -P1V1) however many correct answers were seen.

The temperature at A was found correctly by most candidates.

The main problem here was deciding whether each Q was positive or negative. But the question was quite well answered.

Because the question was about a heat pump rather than a heat engine very few answers were correct. Only a very small number of candidates mentioned the fact that the isothermal change would take an impracticably long time.

Show that the pressure at B is about 130 kPa.

Calculate the ratio $\frac{V_{\mathrm{A}}}{V_{\mathrm{C}}}$.

determine the thermal energy removed from the system.

explain why the entropy of the gas decreases.

state and explain whether the second law of thermodynamics is violated.

$P_{\mathrm{B}}=\frac{250\times {10}^3}{1.5^{{\displaystyle \frac{5}{3}}}} «\mathrm{from} P_{\mathrm{B}}{\left(1.5 V_{\mathrm{A}}\right)}^{\frac{5}{3}}=250\times {10}^3\times {V_{\mathrm{A}}}^{\frac{5}{3}}»$ ✔

= 127 kPa ✔

$«127\times {10}^3\times 1.5 V_{\mathrm{A}}=250\times {10}^3 V_{\mathrm{C}}»$

1.31 ✔

ALTERNATIVE 1

work done $∆W=«-»250\times {10}^3\times 1.5\times {10}^{-3}=«-»375 «\mathrm{J}»$ ✔

change in internal energy $∆U=\frac{3}{2}\times 0.300\times 8.31\times \left(-150\right)=«-»561 «\mathrm{J}»$
OR
$∆U=\frac{3}{2}P∆V=\frac{3}{2}\times 375=«-»563 «\mathrm{J}»$ ✔

thermal energy removed $∆Q=375+561=936 «\mathrm{J}»$
OR
$∆Q=375+563=938 «\mathrm{J}»$ ✔

ALTERNATIVE 2

$∆Q=«nCp∆T=»\frac{5}{2}\times nRT$ ✔

thermal energy removed $∆Q=0.300\times 2.5\times 8.31\times 150$ ✔

$=935 «\mathrm{J}»$ ✔

ALTERNATIVE 1

«from b(i)» $∆Q$ is negative ✔

$∆S=\frac{∆Q}{T}$ AND $∆S$ is negative ✔

ALTERNATIVE 2

T and/or V decreases ✔

less disorder/more order «so S decreases» ✔

ALTERNATIVE 3

T decreases ✔

$∆S=K\times \ln \left(\frac{T2}{T1}\right)<0$ ✔

NOTE: Answer given, look for a valid reason that S decreases.

not violated ✔

the entropy of the surroundings must have increased
OR
the overall entropy of the system and the surroundings is the same or increased ✔

Show that the pressure at B is about 5 × 10⁵ Pa.

For the process BC, calculate, in J, the work done by the gas.

For the process BC, calculate, in J, the change in the internal energy of the gas.

For the process BC, calculate, in J, the thermal energy transferred to the gas.

Explain, without any calculation, why the pressure after this change would belower if the process was isothermal.

Determine, without any calculation, whether the net work done by the engine during one full cycle would increase or decrease.

Outline why an efficiency calculation is important for an engineer designing a heat engine.

«$p_1{V}_1^{\frac{5}{3}}=p_2{V}_2^{\frac{5}{3}}$»

$1.1\times {10}^5\times 5^{\frac{5}{3}}=p_2\times 2^{\frac{5}{3}}$

p₂ «= $\frac{1.1\times {10}^5\times 5^{\frac{5}{3}}}{{2.5}^{\frac{5}{3}}}$» = 5.066 × 10⁵ «Pa»

Volume may be in litres or m³.

Value to at least 2 sig figs, OR clear working with substitution required for mark.

[2 marks]

«W = pΔV»

«= 5.07 × 10⁵ × (5 × 10^–3 – 2 × 10^–3)»

= 1.52 × 10³ «J»

Award [0] if POT mistake.

[1 mark]

ΔU = $\frac{3}{2}$pΔV = $\frac{3}{2}$5.07 × 10⁵ × 3 × 10^–3 = 2.28 × 10^–3 «J»

Accept alternative solution via T_c.

[1 mark]

Q «= (1.5 + 2.28) × 10³ =» 3.80 × 10³ «J»

Watch for ECF from (b)(i) and (b)(ii).

[1 mark]

for isothermal process, PV = constant / ideal gas laws mentioned

since V_C > V_B, P_C must be smaller than P_B

[2 marks]

the area enclosed in the graph would be smaller

so the net work done would decrease

Award MP2 only if MP1 is awarded.

[2 marks]

to reduce energy loss; increase engine performance; improve mpg etc

Allow any sensible answer.

[1 mark]

State the torque provided by the force W about the axis of the flywheel.

Identify the physical quantity represented by the area under the graph.

Show that the angular velocity of the flywheel at t = 5.00 s is 200 rad s^–1.

Calculate the maximum tension in the string.

The flywheel is in translational equilibrium. Distinguish between translational equilibrium and rotational equilibrium.

At t = 5.00 s the flywheel is spinning with angular velocity 200 rad s^–1. The support bearings exert a constant frictional torque on the axle. The flywheel comes to rest after 8.00 × 10³ revolutions. Calculate the magnitude of the frictional torque exerted on the flywheel.

zero ✔

«change in» angular momentum ✔

NOTE: Allow angular impulse.

use of L = lω = area under graph = 1.80 «kg m²s^–1» ✔

rearranges «to give ω= area/I»  1.80 = 0.5 × 5.00 × 0.060² × ω ✔

«to get ω = 200 rad s^–1 »

$«\frac{0.40}{0.012}=»33.3 \mathrm{N}$ ✔

translational equilibrium is when the sum of all the forces on a body is zero ✔

rotational equilibrium is when the sum of all the torques on a body is zero ✔

ALTERNATIVE 1

$0={200}^2+2\times \alpha \times 2\mathrm{\pi }\times 8000$ ✔

$\alpha =«-»0.398 «\mathrm{rad} s^{-2}»$ ✔

torque = $\alpha I=0.398\times \left(0.5\times 5.00\times 0.{060}^2\right)=3.58\times {10}^{-3} «\mathrm{N} \mathrm{m}»$ ✔

ALTERNATIVE 2

change in kinetic energy $=«-»0.5\times \left(0.5\times 5.00\times 0.{060}^2\right)\times {200}^2=«-»180 «\mathrm{J}»$ ✔

identifies work done = change in KE ✔

torque = $\frac{W}{\theta }=\frac{180}{2\mathrm{\pi }\times 8000}=3.58\times {10}^{-3} «\mathrm{N} \mathrm{m}»$✔

Calculate the work done during the compression.

Calculate the work done during the increase in pressure.

Calculate the pressure following this process.

Outline how an approximate adiabatic change can be achieved.

«–» $3\times {10}^3$ «$\mathrm{J}$» ✓

$0$ «$\mathrm{J}$» ✓

OWTTE

use of $P{V}^{\frac{5}{3}}$ is constant «$4.0\times {10}^5\times {\left(2.0\times {10}^{-2}\right)}^{\frac{5}{3}}=P_2\times {\left(5.0\times {10}^{-2}\right)}^{\frac{5}{3}}$» ✓

$P_2=8.7\times {10}^{4 }«\mathrm{Pa}»$ OR $87$ «$\mathrm{kPa}$» ✓

Award [2] marks for a bald correct answer

adiabatic means no transfer of heat in or out of the system ✓

should be fast ✓

«can be slow if» the system is insulated ✓

OWTTE

Outline why the normal force acting on the ladder at the point of contact with the wall is equal to the frictional force F between the ladder and the ground.

Calculate F.

The coefficient of friction between the ladder and the ground is 0.400. Determine whether the ladder will slip.

«translational equilibrium demands that the» resultant force in the horizontal direction must be zero✔

«hence N_W = F»

Equality of forces is given, look for reason why.

«clockwise moments = anticlockwise moments»

50 × 2cos 60 = N_W × 4sin 60 ✔

«$N_{\text{W}}=F=\frac{50\times 2\cos 60}{4\sin 60}$»

F = 14.4«N» ✔

maximum friction force = «0.4 × 50N» = 20«N» ✔

14.4 < 20 AND so will not slip ✔

Many candidates stated that the resultant of all forces must be zero but failed to mention the fact that horizontal forces must balance in this particular question.

Very few candidates could take moments about any point and correct answers were rare both at SL and HL.

The question about the slipping of the ladder was poorly answered. The fact that the normal reaction on the floor was 50N was not known to many.

Show that the total kinetic energy E_k of the sphere when it rolls, without slipping, at speed v is $E_{\text{K}}=\frac{7}{10}m{v}^2$.

A solid sphere of mass 1.5 kg is rolling, without slipping, on a horizontal surface with a speed of 0.50 m s^-1. The sphere then rolls, without slipping, down a ramp to reach a horizontal surface that is 45 cm lower.

Calculate the speed of the sphere at the bottom of the ramp.

E_k = E_k linear + E_k rotational

OR

$E_{\text{k}}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$  ✔

$=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2\times {\left(\frac{v}{r}\right)}^2$  ✔

«$=\frac{7}{10}m{v}^2$»

Answer is given in the question so check working is correct at each stage.

Initial $E_K=\frac{7}{10}\times 1.50\times {0.5}^2$ «=0.26J»  ✔

Final $E_K=0.26+1.5\times 9.81\times 0.45$ «=6.88J»  ✔

$v=«\sqrt{\frac{10}{7}\times \frac{6.88}{1.5}}=»2.56$ «m s^–1»  ✔

Other solution methods are possible.

The derivation of the formula for the total kinetic energy of a rolling ball was well answered.

Although there were many correct answers, many candidates forgot to include the initial kinetic energy of the ball at the top of the ramp. The process followed to obtain the answer was too often poorly presented, candidates are encouraged to explain what is being calculated rather than just writing numbers.