State the direction of the resultant force on the ball.

On the diagram, construct an arrow of the correct length to represent the weight of the ball.

Show that the magnitude of the net force F on the ball is given by the following equation.

                                          $$F=\frac{mg}{\tan \theta }$$

The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.

Outline whether this ball can move on a horizontal circular path of radius equal to the radius of the bowl.

Outline why the ball will perform simple harmonic oscillations about the equilibrium position.

Show that the period of oscillation of the ball is about 6 s.

The amplitude of oscillation is 0.12 m. On the axes, draw a graph to show the variation with time t of the velocity v of the ball during one period.

A second identical ball is placed at the bottom of the bowl and the first ball is displaced so that its height from the horizontal is equal to 8.0 m.

The first ball is released and eventually strikes the second ball. The two balls remain in contact. Determine, in m, the maximum height reached by the two balls.

towards the centre «of the circle» / horizontally to the right

Do not accept towards the centre of the bowl

[1 mark]

downward vertical arrow of any length

arrow of correct length

Judge the length of the vertical arrow by eye. The construction lines are not required. A label is not required

eg: 

[2 marks]

ALTERNATIVE 1

F = N cos θ

mg = N sin θ

dividing/substituting to get result

ALTERNATIVE 2

right angle triangle drawn with F, N and W/mg labelled

angle correctly labelled and arrows on forces in correct directions

correct use of trigonometry leading to the required relationship

tan θ = $\frac{\text{O}}{A}=\frac{mg}{F}$

[3 marks]

$\frac{mg}{\tan \theta }$ = m$\frac{v^2}{r}$

r = R cos θ

v = $\sqrt{\frac{gR{\cos }^2\theta }{\sin \theta }}/\sqrt{\frac{gR\cos \theta }{\tan \theta }}/\sqrt{\frac{9.81\times 8.0\cos 22}{\tan 22}}$

v = 13.4/13 «ms ^–1»

Award [4] for a bald correct answer 

Award [3] for an answer of 13.9/14 «ms ^–1». MP2 omitted

[4 marks]

there is no force to balance the weight/N is horizontal

so no / it is not possible

Must see correct justification to award MP2

[2 marks]

the «restoring» force/acceleration is proportional to displacement

Direction is not required

[1 mark]

ω = «$\sqrt{\frac{g}{R}}$» = $\sqrt{\frac{9.81}{8.0}}$ «= 1.107 s^–1»

T = «$\frac{2\pi }{\omega }$ = $\frac{2\pi }{1.107}$ =» 5.7 «s»

Allow use of or g = 9.8 or 10

Award [0] for a substitution into T = 2π$\sqrt{\frac{I}{g}}$

[2 marks]

sine graph

correct amplitude «0.13 m s^–1»

correct period and only 1 period shown

Accept ± sine for shape of the graph. Accept 5.7 s or 6.0 s for the correct period.

Amplitude should be correct to ±$\frac{1}{2}$ square for MP2

eg: v /m s^–1   

[3 marks]

speed before collision v = «$\sqrt{2gR}$ =» 12.5 «ms^–1»

«from conservation of momentum» common speed after collision is $\frac{1}{2}$ initial speed «v_c = $\frac{12.5}{2}$ = 6.25 ms^–1»

h = «$\frac{{v_c}^2}{2g}=\frac{{6.25}^2}{2\times 9.81}$» 2.0 «m»

Allow 12.5 from incorrect use of kinematics equations

Award [3] for a bald correct answer

Award [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.

Allow ECF from MP1

Allow ECF from MP2

[3 marks]

The charge per unit area on the surface of the wall is σ. It can be shown that the electric field strength E due to the charge on the wall is given by the equation

$E=\frac{\sigma }{2{\epsilon }_0}$.

Demonstrate that the units of the quantities in this equation are consistent.

The thread makes an angle of 30° with the vertical wall. The ball has a mass of 0.025 kg.

Determine the horizontal force that acts on the ball.

The charge on the ball is 1.2 × 10⁻⁶C. Determine σ.

The thread breaks. Explain the initial subsequent motion of the ball.

Calculate the charge on Q. State your answer to an appropriate number of significant figures.

Outline, without calculation, whether or not the electric potential at P is zero.

identifies units of $\sigma$ as ${\text{C\hspace{0.05em}m}}^{-2}$ ✓

$\frac{C}{m^2}\times \frac{N{m}^2}{C^2}$ seen and reduced to ${\text{N\hspace{0.05em}C}}^{-1}$ ✓

Accept any analysis (eg dimensional) that yields answer correctly

horizontal force $F$ on ball $=T \sin 30$ ✓

$T=\frac{mg}{\cos 30}$ ✓

$F «=mg \tan 30 = 0.025\times 9.8 \times \tan 30» = 0.14 «\text{N}»$ ✓

Allow g = 10 N kg⁻¹

Award [3] marks for a bald correct answer.

Award [1max] for an answer of zero, interpreting that the horizontal force refers to the horizontal component of the net force.

$E=\frac{0.14}{1.2\times {10}^{-6}}«=1.2\times {10}^5»$ ✓

$\sigma =«\frac{2\times 8.85\times {10}^{-12}\times 0.14}{1.2\times {10}^{-6}}»=2.1\times {10}^{-6} «{\text{C\hspace{0.05em}m}}^{-2}»$ ✓

Allow ECF from the calculated F in (b)(i)

Award [2] for a bald correct answer.

horizontal/repulsive force and vertical force/pull of gravity act on the ball ✓

so ball has constant acceleration/constant net force ✓

motion is in a straight line ✓

at 30° to vertical away from wall/along original line of thread ✓

$\frac{Q}{0.{22}^2}=\frac{1.2\times {10}^{-6}}{0.{18}^2}$ ✓

$«+»1.8\times {10}^{-6} «\text{C}»$✓

2sf ✓

Do not award MP2 if charge is negative

Any answer given to 2 sig figs scores MP3

work must be done to move a «positive» charge from infinity to P «as both charges are positive»
OR
reference to both potentials positive and added
OR
identifies field as gradient of potential and with zero value ✓

therefore, point P is at a positive / non-zero potential ✓

Award [0] for bald answer that P has non-zero potential

Sketch, on the diagram, the variation of displacement of the air molecules with distance along the pipe when t = $\frac{3}{4f}$.

An air molecule is situated at point X in the pipe at t = 0. Describe the motion of this air molecule during one complete cycle of the standing wave beginning from t = 0.

The speed of sound c for longitudinal waves in air is given by

$c=\sqrt{\frac{K}{\rho }}$

where ρ is the density of the air and K is a constant.

A student measures f to be 120 Hz when the length of the pipe is 1.4 m. The density of the air in the pipe is 1.3 kg m^–3. Determine the value of K for air. State your answer with the appropriate fundamental (SI) unit.

Demonstrate, using a second ray, that the image appears to come from the position indicated.

Outline why the observer detects a series of increases and decreases in the intensity of the received signal as the boat moves along the line XY.

horizontal line shown in centre of pipe ✔

«air molecule» moves to the right and then back to the left ✔

returns to X/original position ✔

wavelength = 2 × 1.4 «= 2.8 m» ✔

c = «f λ =» 120 × 2.8 «= 340 m s⁻¹» ✔

K = «ρc² = 1.3 × 340² =» 1.5 × 10⁵ ✔

kg m^–1s^–2 ✔

construction showing formation of image ✔

Another straight line/ray from image through the wall with line/ray from intersection at wall back to transmitter. Reflected ray must intersect boat.

interference pattern is observed

OR

interference/superposition mentioned ✔

maximum when two waves occur in phase/path difference is nλ

OR

minimum when two waves occur 180° out of phase/path difference is (n + ½)λ ✔

Each rod is to have a resistance no greater than 0.10 Ω. Calculate, in m, the minimum radius of each rod. Give your answer to an appropriate number of significant figures.

Calculate the maximum number of lamps that can be connected between the rods. Neglect the resistance of the rods.

One advantage of this system is that if one lamp fails then the other lamps in the circuit remain lit. Outline one other electrical advantage of this system compared to one in which the lamps are connected in series.

Outline how eddy currents reduce transformer efficiency.

Determine the peak current in the primary coil when operating with the maximum number of lamps.

ALTERNATIVE 1:

$r=\sqrt{\frac{\rho l}{\pi \text{R}}}$ OR $\sqrt{\frac{7.2\times {10}^{-7}\times 12.5}{\pi \times 0.1}}$ ✔

r = 5.352 × 10⁻³ ✔

5.4 × 10⁻³ «m» ✔

For MP2 accept any SF

For MP3 accept only 2 SF

For MP3 accept ANY answer given to 2 SF

ALTERNATIVE 2:

$A=\frac{7.2\times {10}^{-7}\times 12.5}{0.1}$ ✔

r = 5.352 × 10⁻³ ✔

5.4 × 10⁻³ «m» ✔

For MP2 accept any SF

For MP3 accept only 2 SF

For MP3 accept ANY answer given to 2 SF

current in lamp = $\frac{5}{24}$ «= 0.21» «A»

OR

n = 24 × $\frac{8}{5}$ ✔

so «38.4 and therefore» 38 lamps ✔

Do not award ECF from MP1

when adding more lamps in parallel the brightness stays the same ✔

when adding more lamps in parallel the pd across each remains the same/at the operating value/24 V ✔

when adding more lamps in parallel the current through each remains the same ✔

lamps can be controlled independently ✔

the pd across each bulb is larger in parallel ✔

the current in each bulb is greater in parallel ✔

lamps will be brighter in parallel than in series ✔

In parallel the pd across the lamps will be the operating value/24 V ✔

Accept converse arguments for adding lamps in series:

when adding more lamps in series the brightness decreases

when adding more lamps in series the pd decreases

when adding more lamps in series the current decreases

lamps can’t be controlled independently

the pd across each bulb is smaller in series

the current in each bulb is smaller in series

in series the pd across the lamps will less than the operating value/24 V

Do not accept statements that only compare the overall resistance of the combination of bulbs.

«as flux linkage change occurs in core, induced emfs appear so» current is induced ✔

induced currents give rise to resistive forces ✔

eddy currents cause thermal energy losses «in conducting core» ✔

power dissipated by eddy currents is drawn from the primary coil/reduces power delivered to the secondary ✔

power = 190 OR 192 «W» ✔

required power $=190\times \frac{100}{95}$ «200 or 202 W» ✔

so $\frac{200}{240}=0.83$ OR 0.84 «A rms» ✔

peak current = «$0.83\times \sqrt{2}$ OR $0.84\times \sqrt{2}$» = 1.2/1.3 «A» ✔

Write down the maximum magnitude of the rate of change of flux linked with the coil.

State the fundamental SI unit for your answer to (a)(i).

Explain why the graph becomes negative.

Part of the graph is above the t-axis and part is below. Outline why the areas between the t-axis and the curve for these two parts are likely to be the same.

Predict the changes to the graph when the magnet is dropped from a lower height above the coil.

«−» 5.0 «mV»  OR  5.0 × 10⁻³ «V» ✓

Accept 5.1

kg m²A⁻¹s⁻³ ✓

ALTERNATIVE 1

Flux linkage is represented by magnetic field lines through the coil ✓

when magnet has passed through the coil / is moving away ✓

flux «linkage» is decreasing ✓

suitable comment that it is the opposite when above ✓

when the magnet goes through the midpoint the induced emf is zero ✓

ALTERNATIVE 2

reference to / states Lenz’s law ✓

when magnet has passed through the coil / is moving away ✓

«coil attracts outgoing S pole so» induced field is downwards ✓

before «coil repels incoming N pole so» induced field is upwards
OR
induced field has reversed ✓

when the magnet goes through the midpoint the induced emf is zero ✓

OWTTE

area represents the total change in flux «linkage» ✓

the change in flux is the same going in and out ✓

«when magnet is approaching» flux increases to a maximum ✓

«when magnet is receding» flux decreases to zero ✓

«so areas must be the same»

magnet moves slower ✓

overall time «for interaction» will be longer ✓

peaks will be smaller ✓

areas will be the same as before ✓

Allow a graphical interpretation for MP2 as “graph more spread out”

ai
Well answered, with wrong answers stating 8 for the difference or 3 without realising that the sign does not matter.

aii
Very few candidates managed to get the correct fundamental SI unit for V. All kinds of errors were observed, from power errors to the use of C as a fundamental unit instead of A.

bi) Most scored best by marking using an alternative method introduced to the markscheme in standardisation. There were some confused and vague comments. Clear, concise answers were rare.

bii) It was common to see conservation of energy invoked here with suggestions that energy was the area under the graph. Many candidates described the shapes to explain why the areas were the same rather than talking about the physics e.g. one peak is short and fat and the other is tall and thin so they balance out.

c) A surprising number didn't pick up on the fact that the magnet would be moving slower. As a result, they discussed everything happening sooner, i.e. the interaction with the magnet and the coil, and that led onto things happening quicker so peaks being bigger.

Calculate, for the surface of $\text{Io}$, the gravitational field strength g_Io due to the mass of $\text{Io}$. State an appropriate unit for your answer.

Show that the $\frac{\mathrm{gravitational} \mathrm{potential} \mathrm{due} \mathrm{to} \mathrm{Jupiter} \mathrm{at} \mathrm{the} \mathrm{orbit} \mathrm{of} \mathrm{Io}}{ \mathrm{gravitational} \mathrm{potential} \mathrm{due} \mathrm{to} \mathrm{Io} \mathrm{at} \mathrm{the} \mathrm{surface} \mathrm{of} \mathrm{Io}}$ is about 80.

Outline, using (b)(i), why it is not correct to use the equation $\sqrt{\frac{2G\times \text{mass of Io}}{\text{radius of Io}}}$ to calculate the speed required for the spacecraft to reach infinity from the surface of $\text{Io}$.

An engineer needs to move a space probe of mass 3600 kg from Ganymede to Callisto. Calculate the energy required to move the probe from the orbital radius of Ganymede to the orbital radius of Callisto. Ignore the mass of the moons in your calculation. 

$«\frac{GM}{r^2}=\frac{6.67\times {10}^{-11}\times 8.9\times {10}^{22}}{{\left(1.8\times {10}^6\right)}^2}=»1.8$ ✓

N kg⁻¹  OR  m s⁻²  ✓

$\frac{1.9\times {10}^{27}}{4.9\times {10}^8}$  AND  $\frac{8.9\times {10}^{22}}{1.8\times {10}^6}$ seen ✓

$«\frac{1.9\times {10}^{27}\times 1.8\times {10}^6}{4.9\times {10}^8\times 8.9\times {10}^{22}}=»78$  ✓

For MP1, potentials can be seen individually or as a ratio.

«this is the escape speed for $\text{Io}$ alone but» gravitational potential / field of Jupiter must be taken into account  ✓

OWTTE

$-G{M}_{\text{Jupiter}}\left(\frac{1}{1.88\times {10}^9}-\frac{1}{1.06\times {10}^9}\right)=«5.21\times {10}^7 {\text{J\hspace{0.05em}kg}}^{-1}»$  ✓

« multiplies by 3600 kg to get » 1.9 × 10¹¹«J» ✓

Award [2] marks if factor of ½ used, taking into account orbital kinetic energies, leading to a final answer of 9.4 x 10¹⁰«J».

Allow ECF from MP1

Award [2] marks for a bald correct answer.

Calculate the average force exerted by the racquet on the ball.

Calculate the average power delivered to the ball during the impact.

Calculate the time it takes the tennis ball to reach the net.

Show that the tennis ball passes over the net.

Determine the speed of the tennis ball as it strikes the ground.

A student models the bounce of the tennis ball to predict the angle θ at which the ball leaves a surface of clay and a surface of grass.

The model assumes

• during contact with the surface the ball slides.
• the sliding time is the same for both surfaces.
• the sliding frictional force is greater for clay than grass.
• the normal reaction force is the same for both surfaces.

Predict for the student’s model, without calculation, whether θ is greater for a clay surface or for a grass surface.

$F=\frac{\mathrm{\Delta }mv}{\mathrm{\Delta }t}/m\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}/\frac{0.058\times 64.0}{25\times {10}^{-3}}$  ✔

$F$ = 148 «N»≈150«N»  ✔

ALTERNATIVE 1

$P=\frac{\frac{1}{2}m{v}^2}{t}/\frac{\frac{1}{2}\times 0.058\times {64.0}^2}{25\times {10}^{-3}}$  ✔

$P=4700/4800$«$\text{W}$»  ✔

ALTERNATIVE 2

$P=\text{average}Fv\text{ / 148}\times \frac{64.0}{2}$  ✔

$P=4700/4800$«$\text{W}$»  ✔

horizontal component of velocity is $64.0\times \cos 7^{\circ }=63.52$«$\text{m }{\text{s}}^{-1}$»   ✔ 

$t=$«$\frac{11.9}{63.52}$»$\text{0}\text{.187/0}\text{.19}$«$\text{s}$»  ✔

ALTERNATIVE 1

u_y=64sin7/7.80«ms^–1»  ✔

decrease in height = 7.80 × 0.187 + $\frac{1}{2}$ × 9.81 × 0.187² / 1.63«m»  ✔

final height = «2.80 – 1.63» = 1.1/1.2«m»  ✔

«higher than net so goes over»

ALTERNATIVE 2

vertical distance to fall to net «=2.80 – 0.91» = 1.89«m»  ✔

time to fall this distance found using «$1.89=7.8t+\frac{1}{2}\times 9.81\times t^2$»

$t$ = 0.21«s»  ✔

0.21«s» > 0.187«s»   ✔

«reaches the net before it has fallen far enough so goes over»

ALTERNATIVE 1

Initial KE + PE = final KE /

$\frac{1}{2}\times 0.058\times {64}^2+0.058\times 9.81\times 2.80=\frac{1}{2}\times 0.058\times v^2$  ✔

$v=64.4$«$\text{m }{\text{s}}^{-1}$»  ✔

ALTERNATIVE 2

$v_v=$«$\sqrt{{7.8}^2+2\times 9.81\times 2.8}$» = 10.8«$\text{m }{\text{s}}^{-1}$»  ✔

«$v=\sqrt{{63.5}^2+{10.8}^2}$»

$v=64.4$«$\text{m }{\text{s}}^{-1}$»  ✔

so horizontal velocity component at lift off for clay is smaller ✔

normal force is the same so vertical component of velocity is the same ✔

so bounce angle on clay is greater ✔

At both HL and SL many candidates scored both marks for correctly answering this. A straightforward start to the paper. For those not gaining both marks it was possible to gain some credit for calculating either the change in momentum or the acceleration. At SL some used 64 ms-1 as a value for a and continued to use this value over the next few parts to the question.

This was well answered although a significant number of candidates approached it using P = Fv but forgot to divide v by 2 to calculated the average velocity. This scored one mark out of 2.

This question scored well at HL but less so at SL. One common mistake was to calculate the direct distance to the top of the net and assume that the ball travelled that distance with constant speed. At SL particularly, another was to consider the motion only when the ball is in contact with the racquet.

There were a number of approaches students could take to answer this and examiners saw examples of them all. One approach taken was to calculate the time taken to fall the distance to the top of the net and to compare this with the time calculated in bi) for the ball to reach the net. This approach, which is shown in the mark scheme, required solving a quadratic in t which is beyond the mathematical requirements of the syllabus. This mathematical technique was only required if using this approach and not required if, for example, calculating heights.

A common mistake was to forget that the ball has a vertical acceleration. Examiners were able to award credit/ECF for correct parts of an otherwise flawed method.

This proved difficult for candidates at both HL and SL. Many managed to calculate the final vertical component of the velocity of the ball.

As the command term in this question is ‘predict’ a bald answer of clay was acceptable for one mark. This was a testing question that candidates found demanding but there were some very well-reasoned answers. The most common incorrect answer involved suggesting that the greater frictional force on the clay court left the ball with less kinetic energy and so a smaller angle. At SL many gained the answer that the angle on clay would be greater with the argument that frictional force is greater and so the distance the ball slides is less.

Identify a property of electrons demonstrated by this experiment.

Show that the energy E of each electron in the beam is about 7 × 10⁻¹¹J.

The de Broglie wavelength for an electron is given by $\frac{hc}{E}$. Show that the diameter of an oxygen-16 nucleus is about 4 fm.

Estimate, using the result in (a)(iii), the volume of a tin-118 $\left({}_{50}{}^{118}\text{Sn}\right)$ nucleus. State your answer to an appropriate number of significant figures.

wave properties ✓

Accept reference to diffraction or interference.

440 x 10⁶ x 1.6 x 10^-19  OR  7.0 × 10^-11 «J» ✓

$\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{7\times {10}^{-11}}$  OR  $\frac{1.24\times {10}^{-6}}{440\times {10}^6}$  OR  2.8 × 10^-15«m» seen ✓

read off graph as 46° ✓

«Use of $D=\frac{\lambda }{\sin \theta }$=» 3.9 × 10^-15 m ✓

Accept an angle between 45 and 47 degrees.

Allow ECF from MP2

ALTERNATIVE 1

use of $R\propto A^{\frac{1}{3}}$   OR  $V\propto A$ ✓

volume of $\text{Sn}=\frac{4}{3}\pi \left(\frac{A_{Sn}}{A_O}\right)r_O^{ 3}$ or equivalent working ✓

2.3 to 2.5 × 10^-43«m³»✓

answer to 1 or 2sf ✓

ALTERNATIVE 2

use of $R=R_{\text{o}}\times A^{\frac{1}{3}}$ ✓

volume of $\text{Sn}=\frac{4}{3}\pi R^3$  OR  5.9 x 10^-15 seen ✓

8.5 × 10^-43 «m³»✓

answer to 1 or 2sf ✓

Although the question expects candidates to work from the oxygen radius found, allow ALT 2 working from the Fermi radius.

MP4 is for any answer stated to 1 or 2 significant figures.

ai) Well answered.

aii) Well answered.

aiii) This was generally well done but quite a few attempted the small angle approximation. Probably worth a mention in the report.

b) Most gained credit from the first alternative solution, trying to use the data as the question intended. There were the inevitable slips and calculator mistakes. Most got the fourth mark.