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 centre of the ball, still carrying a charge of $1.2\times {10}^{-6} \text{C}$, is now placed $0.40 \text{m}$ from a point charge Q. The charge on the ball acts as a point charge at the centre of the ball.

P is the point on the line joining the charges where the electric field strength is zero.
The distance PQ is $0.22 \text{m}$.

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

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 the 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.

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

$«+»1.8\times {10}^{-6} «\text{C}»$ ✓

2sf ✓

Do not award MP2 if charge is negative

Any answer given to 2 sig figs scores MP3

Calculate the thermal energy transferred from the sample during the first $30$ minutes.

Estimate the specific heat capacity of the oil in its liquid phase. State an appropriate unit for your answer.

The sample begins to freeze during the thermal energy transfer. Explain, in terms of the molecular model of matter, why the temperature of the sample remains constant during freezing.

Calculate the mass of the oil that remains unfrozen after $60$ minutes.

$«15\times 30\times 60»=27000 «\mathrm{J}»$ ✓

$27\times {10}^3=0.32\times c\times \left(290-250\right)$ OR $2100$ ✓

$\mathrm{J} {\mathrm{kg}}^{-1} {\mathrm{K}}^{-1}$ OR $\mathrm{J} {\mathrm{kg}}^{-1} { }^0{\mathrm{C}}^{-1}$ ✓

Allow any appropriate unit that is $\frac{energy}{mass\times termperature}$

«intermolecular» bonds are formed during freezing ✓

bond-forming process releases energy
OR
«intermolecular» PE decreases «and the difference is transferred as heat» ✓

«average random» KE of the molecules does not decrease/change ✓

temperature is related to «average» KE of the molecules «hence unchanged» ✓

To award MP3 or MP4 molecules/particles/atoms must be mentioned.

mass of frozen oil $«=\frac{27\times {10}^3}{130\times {10}^3}»=0.21 «\mathrm{kg}»$ ✓

unfrozen mass $«=0.32-0.21»=0.11 «\mathrm{kg}»$ ✓

Show that the time taken for the ball to reach the surface of the table is about 0.2 s.

Sketch, on the axes, a graph showing the variation with time of the vertical component of velocity v_v of the ball until it reaches the table surface. Take g to be +10 m s⁻².

The net is stretched across the middle of the table. The table has a length of 2.74 m and the net has a height of 15.0 cm.

Show that the ball will go over the net.

Determine the kinetic energy of the ball immediately after the bounce.

Player B intercepts the ball when it is at its peak height. Player B holds a paddle (racket) stationary and vertical. The ball is in contact with the paddle for 0.010 s. Assume the collision is elastic.

Calculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures.

t = «$\sqrt{\frac{2d}{g}}$=» 0.22 «s»
OR

t = $\sqrt{\frac{2\times 0.24}{9.8}}$  ✓ 

Answer to 2 or more significant figures or formula with variables replaced by correct values.

increasing straight line from zero up to 0.2 s in x-axis ✓

with gradient = 10 ✓

ALTERNATIVE 1 

$t=\frac{1.37}{12}=$«0.114 s» ✓

$y=\frac{1}{2}\times 10\times 0.{114}^2=0.065$ m ✓

so (0.24 − 0.065) = 0.175 > 0.15  OR  0.065 < (0.24 − 0.15) «so it goes over the net» ✓

ALTERNATIVE 2

«0.24 − 0.15 = 0.09 = $\frac{1}{2}\times 10\times t^2$ so» t = 0.134 s ✓

0.134 × 12 = 1.6 m ✓

1.6 > 1.37 «so ball passed the net already»  ✓

Allow use of g = 9.8.

ALTERNATIVE 1 

KE = $\frac{1}{2}$mv² + mgh = $\frac{1}{2}$0.0027 ×10.5² + 0.0027 × 9.8 × 0.18 ✓

0.15 «J» ✓

ALTERNATIVE 2

Use of v_x = 10.5 AND v_y = 1.88 to get v = «$\sqrt{10.5^2 + 1.{88}^2}$» = 10.67 «m s⁻¹» ✓

KE = $\frac{1}{2}$ × 0.0027 × 10.67² = 0.15 «J»  ✓

$\text{Δ}v = 21$ «m s⁻¹» ✓

$F=\frac{0.0027 \times 21}{0.01}$

OR

5.67 «N» ✓

any answer to 2 significant figures «N» ✓

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, in kg m^–1 s^–2, the value of K for air.

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.

«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 ✔

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 ✔

Outline two differences between the momentum of the box and the momentum of the load at the same instant.

The vertical acceleration of the load downwards is 2.4 m s⁻².

Calculate the tension in the string.

Show that the speed of the load when it hits the floor is about 2.1 m s⁻¹.

The radius of the pulley is 2.5 cm. Calculate the angular speed of rotation of the pulley as the load hits the floor. State your answer to an appropriate number of significant figures.

After the load has hit the floor, the box travels a further 0.35 m along the ramp before coming to rest. Determine the average frictional force between the box and the surface of the ramp.

The student then makes the ramp horizontal and applies a constant horizontal force to the box. The force is just large enough to start the box moving. The force continues to be applied after the box begins to move.

Explain, with reference to the frictional force acting, why the box accelerates once it has started to move. 

direction of motion is different / OWTTE ✓

mv / magnitude of momentum is different «even though v the same» ✓

use of ma = mg − T «3.5 x 2.4 = 3.5g − T »

OR

T = 3.5(g − 2.4) ✓

26 «N» ✓

Accept 27 N from g = 10 m s⁻²

proper use of kinematic equation ✓

$\sqrt{\left(2\times 2.4\times 0.95\right)}=2.14$ «m s⁻¹» ✓

Must see either the substituted values OR a value for v to at least three s.f. for MP2.

use of $\omega =\frac{v}{r}$ to give 84 «rad s⁻¹»

OR

$\omega =2.1/0.025$ to give 84 «rad s⁻¹» ✓

quoted to 2sf only✓

ALTERNATIVE 1

«$v^2=u^2+2as\Rightarrow 0=2.1^2-2a\times 0.35$» leading to a = 6.3 «m s^-2»

OR

« $x=1/2\left(u+v\right)t$ » leading to t = 0.33 « s » ✓

F_net = « $ma=1.5\times 6.3$ = » 9.45 «N» ✓

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

friction force = net force – weight down ramp = 2.1 «N» ✓

ALTERNATIVE 2

kinetic energy initial = work done to stop 0.5 x 1.5 x (2.1)² = F_NET x 0.35 ✓

F_net = 9.45 «N» ✓

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

friction force = net force – weight down ramp = 2.1 «N» ✓

Accept 1.95 N from g = 10 m s^-2.
Accept 2.42 N from u = 2.14 m s^-1.

static coefficient of friction > dynamic/kinetic coefficient of friction / μ_s > μ_k ✓

«therefore» force of dynamic/kinetic friction will be less than the force of static friction ✓

there will be a net / unbalanced forward force once in motion «which results in acceleration»

OR

reference to net F = ma ✓

Many students recognized the vector nature of momentum implied in the question, although some focused on the forces acting on each object rather than discussing the momentum.

Some students simply calculated the net force acting on the load and did not recognize that this was not the tension force. Many set up a net force equation but had the direction of the forces backwards. This generally resulted from sloppy problem solving.

This was a "show that" questions, so examiners were looking for a clear equation leading to a clear substitution of values leading to an answer that had more significant digits than the given answer. Most candidates successfully selected the correct equation and showed a proper substitution. Some candidates started with an energy approach that needed modification as it clearly led to an incorrect solution. These responses did not receive full marks.

This SL only question was generally well done. Despite some power of 10 errors, many candidates correctly reported final answer to 2 sf.

Candidates struggled with this question. Very few drew a clear free-body diagram and many simply calculated the acceleration of the box from the given information and used this to calculate the net force on the box, confusing this with the frictional force.

This was an "explain" question, so examiners were looking for a clear line of discussion starting with a comparison of the coefficients of friction, leading to a comparison of the relative magnitudes of the forces of friction and ultimately the rise of a net force leading to an acceleration. Many candidates recognized that this was a question about the comparison between static and kinetic/dynamic friction but did not clearly specify which they were referring to in their responses. Some candidates clearly did not read the stem carefully as they referred to the mass being on an incline.

Show that the time taken for the battery to discharge is about 3 × 10³ s.

Deduce that the average power output of the battery is about 240 W.

Friction and air resistance act on the bicycle and the girl when they move. Assume that all the energy is transferred from the battery to the electric motor. Determine the total average resistive force that acts on the bicycle and the girl.

Calculate the component of weight for the bicycle and girl acting down the slope.

The battery continues to give an output power of 240 W. Assume that the resistive forces are the same as in (a)(iii).

Calculate the maximum speed of the bicycle and the girl up the slope.

On another journey up the slope, the girl carries an additional mass. Explain whether carrying this mass will change the maximum distance that the bicycle can travel along the slope.

Determine the internal resistance of the battery.

Calculate the emf of one cell.

Calculate the internal resistance of one cell.

time taken $\frac{2.0\times {10}^4}{7}$«= 2860 s» = 2900«s» ✔

Must see at least two s.f.

use of E = qV OR energy = 4.3 × 10³ × 16 «= 6.88 × 10⁵ J» ✔

power = 241 «W» ✔

Accept 229 W − 241 W depending on the exact value of t used from ai.

Must see at least three s.f.

use of power = force × speed OR force × distance = power × time ✔

«34N» ✔

Award [2] for a bald correct answer.

Accept 34 N – 36 N.

66 g sin(3°) = 34 «N» ✔

total force 34 + 34 = 68 «N» ✔
3.5 «ms^-1»✔

If you suspect that the incorrect reference in this question caused confusion for a particular candidate, please refer the response to the PE.

Look for ECF from aiii and bi.

Accept 3.4 − 3.5 «ms^-1».

Award [0] for solutions involving use of KE.

Award [0] for v = 7 ms^-1.

Award [2] for a bald correct answer.

«maximum» distance will decrease OWTTE ✔

because opposing/resistive force has increased
OR
because more energy is transferred to GPE
OR
because velocity has decreased
OR
increased mass means more work required «to move up the hill» ✔

V dropped across battery OR R_circuit = 1.85 Ω ✔

so internal resistance = $\frac{4.0}{6.5}$ = 0.62«Ω» ✔

For MP1 allow use of internal resistance equations that leads to 16V − 12V (=4V).

Award [2] for a bald correct answer.

$\frac{16}{5}$ = 3.2 «V» ✔

ALTERNATIVE 1:

2.5r = 0.62 ✔

r = 0.25 «Ω» ✔

ALTERNATIVE 2:

$\frac{0.62}{5}$ = 0.124 «Ω» ✔

r = 2(0.124)= 0.248 «Ω» ✔

Allow ECF from (d) and/or e(i).

This question was generally well answered. Candidates should be reminded on questions where a given value is being calculated that they should include an unrounded answer. This whole question set was a blend of electricity and mechanics concepts, and it was clear that some candidates struggled with applying the correct concepts in the various sub-questions.

Many candidates struggled with this question. They either simply calculated the weight, used the cosine rather than the sine function, or failed to multiply by the acceleration due to gravity. Candidates need to be able to apply free-body diagram skills in a variety of “real world” situations.

This question was well answered in general, with the vast majority of candidates specifying that the maximum distance would decrease. This is an “explain” command term, so the examiners were looking for a detailed reason why the distance would decrease for the second marking point. Unfortunately, some candidates simply wrote that because the mass increased so did the weight without making it clear why this would change the maximum distance.

The player’s foot is in contact with the ball for 55 ms. Calculate the average force that acts on the ball due to the football player.

The ball leaves the ground at an angle of 22°. The horizontal distance from the initial position of the edge of the ball to the wall is 11 m. Calculate the time taken for the ball to reach the wall.

The top of the wall is 2.4 m above the ground. Deduce whether the ball will hit the wall.

In practice, air resistance affects the ball. Outline the effect that air resistance has on the vertical acceleration of the ball. Take the direction of the acceleration due to gravity to be positive.

The player kicks the ball again. It rolls along the ground without sliding with a horizontal velocity of $1.40 {\text{m\hspace{0.05em}s}}^{-1}$. The radius of the ball is $0.11 \text{m}$. Calculate the angular velocity of the ball. State an appropriate SI unit for your answer.

$\text{Δ}p=0.45\times 19 \mathit{\text{OR }}a =\frac{19}{0.055}$ ✓ 

$«=F=\frac{0.45\times 19}{0.055}»160 «\text{N}»$ ✓

Allow [2] marks for a bald correct answer.

Allow ECF for MP2 if 19 sin22 OR 19 cos22 used.

$\text{horizontal speed =} 19\times \cos 22 «=17.6{\text{\hspace{0.05em}m\hspace{0.05em}s}}^{-1}»$ ✓ 

$\text{time}=«\frac{\text{distance}}{\text{speed}}=\frac{11}{19 \cos 22}=» 0.62 «\text{s}»$ ✓

Allow ECF for MP2

$\text{initial vertical speed}=19\times \sin 22 «= 7.1 {\text{m\hspace{0.05em}s}}^{-1}»$ ✓ 

$«7.12\times 0.624-0.5\times 9.81\times 0.{624}^2=» 2.5 «\text{m}»$ ✓

ball does not hit wall OR 2.5 «m» > 2.4 «m» ✓

Allow ECF from (b)(i) and from MP1

Allow g = 10 m s⁻²

air resistance opposes «direction of» motion
OR
air resistance opposes velocity ✓

on the way up «vertical» acceleration is increased OR greater than g ✓

on the way down «vertical» acceleration is decreased OR smaller than g ✓

Allow deceleration/acceleration but meaning must be clear

$13 «\text{rad}» {\text{s}}^{-1}$✓

Unit must be seen for mark

Accept Hz

Accept $4 \pi «\text{rad}» {\text{s}}^{-1}$

The glider reaches its launch speed of 27.0 m s^–1 after accelerating for 11.0 s. Assume that the glider moves horizontally until it leaves the ground. Calculate the total distance travelled by the glider before it leaves the ground.

The glider and pilot have a total mass of 492 kg. During the acceleration the glider is subject to an average resistive force of 160 N. Determine the average tension in the cable as the glider accelerates.

The cable is pulled by an electric motor. The motor has an overall efficiency of 23 %. Determine the average power input to the motor.

The cable is wound onto a cylinder of diameter 1.2 m. Calculate the angular velocity of the cylinder at the instant when the glider has a speed of 27 m s^–1. Include an appropriate unit for your answer.

After takeoff the cable is released and the unpowered glider moves horizontally at constant speed. The wings of the glider provide a lift force. The diagram shows the lift force acting on the glider and the direction of motion of the glider.

Draw the forces acting on the glider to complete the free-body diagram. The dotted lines show the horizontal and vertical directions.

Explain, using appropriate laws of motion, how the forces acting on the glider maintain it in level flight.

At a particular instant in the flight the glider is losing 1.00 m of vertical height for every 6.00 m that it goes forward horizontally. At this instant, the horizontal speed of the glider is 12.5 m s^–1. Calculate the velocity of the glider. Give your answer to an appropriate number of significant figures.

correct use of kinematic equation/equations

148.5 or 149 or 150 «m»

Substitution(s) must be correct.

a = $\frac{27}{11}$ or 2.45 «m s^–2»

F – 160 = 492 × 2.45

1370 «N»

Could be seen in part (a).
Award [0] for solution that uses a = 9.81 m s^–2

ALTERNATIVE 1

«work done to launch glider» = 1370 x 149 «= 204 kJ»

«work done by motor» $=\frac{204\times 100}{23}$

«power input to motor» $=\frac{204\times 100}{23}\times \frac{1}{11}=80$ or 80.4 or 81 k«W»

ALTERNATIVE 2

use of average speed 13.5 m s^–1

«useful power output» =  force x average speed «= 1370 x 13.5»

power input = «$1370\times 13.5\times \frac{100}{23}=$» 80 or 80.4 or 81 k«W»

ALTERNATIVE 3

work required from motor = KE + work done against friction «$=0.5\times 492\times {27}^2+\left(160\times 148.5\right)$» = 204 «kJ»

«energy input» $=\frac{\text{work required from motor}\times 100}{23}$

power input $=\frac{883000}{11}=80.3$ k«W»

Award [2 max] for an answer of 160 k«W».

$\omega =$ «$\frac{v}{r}=$» $\frac{27}{0.6}=45$

rad s^–1

Do not accept Hz.
Award [1 max] if unit is missing.

drag correctly labelled and in correct direction

weight correctly labelled and in correct direction AND no other incorrect force shown

Award [1 max] if forces do not touch the dot, but are otherwise OK.

name Newton's first law

vertical/all forces are in equilibrium/balanced/add to zero
OR
vertical component of lift mentioned

as equal to weight

any speed and any direction quoted together as the answer

quotes their answer(s) to 3 significant figures

speed = 12.7 m s^–1 or direction = 9.46^º or 0.165 rad «below the horizontal» or gradient of $-\frac{1}{6}$

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.

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]

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]

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.

The 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«$\text{N}$»≈150«$\text{N}$»  ✔

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

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

$P=4700/4800«\text{W}$»  ✔

horizontal component of velocity is 64.0 × cos7° = 63.52 «ms⁻¹» ✔

$t=«\frac{11.9}{63.52}=»0.187/0.19«\text{s}$»   ✔

Do not award BCA. Check working.

Do not award ECF from using 64 m s^-1.

ALTERNATIVE 1

u_y = 64 sin7/7.80 «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 + $\frac{1}{2}$ × 9.81 ×t²»

t = 0.21 «s»✔

0.21 «s» > 0.187 «s» ✔

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

Other alternatives are possible

ALTERNATIVE 1

Initial KE + PE = final KE /

$\frac{1}{2}$ × 0.058 × 64² + 0.058 × 9.81 × 2.80 = $\frac{1}{2}$ × 0.058 × v² ✔

v = 64.4 «ms⁻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.

Estimate the power input to the heating element. State an appropriate unit for your answer.

Outline whether your answer to (a) is likely to overestimate or underestimate the power input.

Discuss, with reference to the molecules in the liquid, the difference between milk at 11 °C and milk at 84 °C.

State how energy is transferred from the inside of the metal pipe to the outside of the metal pipe.

The missing section of insulation is 0.56 m long and the external radius of the pipe is 0.067 m. The emissivity of the pipe surface is 0.40. Determine the energy lost every second from the pipe surface. Ignore any absorption of radiation by the pipe surface.

Describe one other method by which significant amounts of energy can be transferred from the pipe to the surroundings.

energy required for milk entering in 1 s = mass x specific heat x 73 ✓

16 kW OR 16000 W ✓

MP1 is for substitution into mcΔT regardless of power of ten.

Allow any correct unit of power (such as J s^-1 OR kJ s^-1) if paired with an answer to the correct power of 10 for MP2.

Underestimate / more energy or power required ✓

because energy transferred as heat / thermal energy is lost «to surroundings or electrical components» ✓

Do not allow general term “energy” or “power” for MP2.

the temperature has increased so the internal energy / « average » KE «of the molecules» has increased OR temperature is proportional to average KE «of the molecules». ✓

«therefore» the «average» speed of the molecules or particles is higher OR more frequent collisions « between molecules » OR spacing between molecules has increased OR average force of collisions is higher OR intermolecular forces are less OR intermolecular bonds break and reform at a higher rate OR molecules are vibrating faster. ✓

conduction/conducting/conductor «through metal» ✓

use of $P=e\sigma A{T}^4$ where T = 357 K ✓

use of $A=2\pi r l$ « = 0.236 m²» ✓

P = 87 «W» ✓

Allow 85 – 89 W for MP3.

Allow ECF for MP3.

convection «is likely to be a significant loss» ✓

«due to reduction in density of air near pipe surface» hot air rises «and is replaced by cooler air from elsewhere»

OR

«due to» conduction «of heat or thermal energy» from pipe to air ✓

Most candidates recognized that this was a specific heat question and set up a proper calculation, but many struggled to match their answer to an appropriate unit. A common mistake was to leave the answer in some form of an energy unit and others did not match the power of ten of the unit to their answer (e.g. 16 W).

Many candidates recognized that this was an underestimate of the total energy but failed to provide an adequate reason. Many gave generic responses (such as "some power will be lost"/not 100% efficient) without discussing the specific form of energy lost (e.g. heat energy).

This was generally well answered. Most HL candidates linked the increase in temperature to the increase in the kinetic energy of the molecules and were able to come up with a consequence of this change (such as the molecules moving faster). SL candidates tended to focus more on consequences, often neglecting to mention the change in KE.

Many candidates recognized that heat transfer by conduction was the correct response. This was a "state" question, so candidates were not required to go beyond this.

Candidates at both levels were able to recognize that this was a blackbody radiation question. One common mistake candidates made was not calculating the area of a cylinder properly. It is important to remind candidates that they are expected to know how to calculate areas and volumes for basic geometric shapes. Other common errors included the use of T in Celsius and neglecting to raise T ^4. Examiners awarded a large number of ECF marks for candidates who clearly showed work but made these fundamental errors.

A few candidates recognized that convection was the third source of heat loss, although few managed to describe the mechanism of convection properly for MP2. Some candidates did not read the question carefully and instead wrote about methods to increase the rate of heat loss (such as removing more insulation or decreasing the temperature of the environment).

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.

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» ✔

ALTERNATIVE 2:

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

r = 5.352 × 10⁻³ ✔

5.4 × 10⁻³ «m» ✔

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

OR

n = 24 × $\frac{8}{5}$ ✔

so «38.4 and therefore» 38 lamps ✔

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.