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 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» ✓

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.

Outline why a force acts on the airboat due to the fan blade.

Show that a mass of about 240 kg of air moves through the fan every second.

Show that the tension in the rope is about 5 kN.

Estimate the distance the airboat travels to reach its maximum speed.

Deduce the mass of the airboat.

The fan is rotating at 120 revolutions every minute. Calculate the centripetal acceleration of the tip of a fan blade.

ALTERNATIVE 1

there is a force «by the fan» on the air / air is accelerated «to the rear» ✓

by Newton 3 ✓

there is an «equal and» opposite force on the boat ✓

ALTERNATIVE 2

air gains momentum «backward» ✓

by conservation of momentum / force is rate of change in momentum ✓

boat gains momentum in the opposite direction ✓

Accept a reference to Newton’s third law, e.g. N’3, or any correct statement of it for MP2 in ALT 1.

Allow any reasonable choice of object where the force of the air is acting on, e.g., fan or blades.

$\pi R^2$ OR «mass of air through system per unit time =» $Av\rho$ seen ✓

244 «kg s⁻¹» ✓

Accept use of Energy of air per second = 0.5 ρΑv³ = 0.5 mv² for MP1.

«force = Momentum change per sec = $A{v}^2\rho$ = » 244 x 20 OR 4.9 «kN» ✓

Allow use of 240

recognition that area under the graph is distance covered ✓

«Distance =» 480 - 560 «m» ✓

Accept graphical evidence or calculation of correct geometric areas for MP1.

MP2 is numerical value within range.

calculation of acceleration as gradient at t = 0 «= 1 m s^-2» ✓

use of F=ma OR $\frac{4900}{1}$ seen ✓

4900 «kg» ✓

MP1 can be shown on the graph.

Allow an acceleration in the range 1 – 1.1 for MP2 and consistent answer for MP3

Allow ECF from MP1.

Allow use of average acceleration = $\frac{18}{40}$

or assumption of constant force to obtain 11000 «kg» for [2]

Allow use of 4800 or 5000 for MP2

ALTERNATE 1

« ω = » 4$\pi$ rad s⁻¹ ✓

« a = r ω²= » 280 « m s⁻² » ✓

ALTERNATE 2

« $v=\frac{2\pi r}{T}$ » = 22.6 m s⁻¹ ✓

« $a=\frac{v^2}{r}$»= 280 « m s⁻² » ✓

Allow ECF from MP1 for wrong ω (120 gives 2.6 x 10⁴ « m s⁻² »)

Allow ECF from MP1 for wrong T (2 s gives 18 « m s⁻² »)

The majority succeeded in making use of Newton's third law to explain the force on the boat. The question was quite well answered but sequencing of answers was not always ideal. There were some confusions about the air hitting the bank and bouncing off to hit the boat. A small number thought that the wind blowing the fan caused the force on the boat.

bi) This was generally well answered with candidates either starting from the wind turbine formula given in the data booklet or with the mass of the air being found using $\rho Av$.

1bii) Well answered by most candidates. Some creative work to end up with 240 was found in scripts.

1ci) Many candidates gained credit here for recognising that the resistive force eventually equalled the drag force and most were able to go on to link this to e.g. zero acceleration. Some had not read the question properly and assumed that the rope was still tied. There was one group of answers that stated something along the lines of "as there is no rope there is nothing to stop the boat so it can go at max speed.

1cii) A slight majority did not realise that they had to find the area under the velocity-time graph, trying equations of motion for non-linear acceleration. Those that attempted to calculate the area under the graph always succeeded in answering within the range.

1ciii) Use of the average gradient was common here for the acceleration. However, there also were answers that attempted to calculate the mass via a kinetic energy calculation that made all sorts of incorrect assumptions. Use of average acceleration taken from the gradient of the secant was also common.

Draw the free-body diagram for the sledge at the position shown on the snow slope.

After leaving the snow slope, the girl on the sledge moves over a horizontal region of snow. Explain, with reference to the physical origin of the forces, why the vertical forces on the girl must be in equilibrium as she moves over the horizontal region.

When the sledge is moving on the horizontal region of the snow, the girl jumps off the sledge. The girl has no horizontal velocity after the jump. The velocity of the sledge immediately after the girl jumps off is 4.2 m s^–1. The mass of the girl is 55 kg and the mass of the sledge is 5.5 kg. Calculate the speed of the sledge immediately before the girl jumps from it.

The girl chooses to jump so that she lands on loosely-packed snow rather than frozen ice. Outline why she chooses to land on the snow.

Show that the acceleration of the sledge is about –2 m s^–2.

Calculate the distance along the slope at which the sledge stops moving. Assume that the coefficient of dynamic friction is constant.

The coefficient of static friction between the sledge and the snow is 0.14. Outline, with a calculation, the subsequent motion of the sledge. 

arrow vertically downwards labelled weight «of sledge and/or girl»/W/mg/gravitational force/F_g/F_{gravitational} AND arrow perpendicular to the snow slope labelled reaction force/R/normal contact force/N/F_N

friction force/F/f acting up slope «perpendicular to reaction force»

Do not allow G/g/“gravity”.

Do not award MP1 if a “driving force” is included.

Allow components of weight if correctly labelled.

Ignore point of application or shape of object.

Ignore “air resistance”.

Ignore any reference to “push of feet on sledge”.

Do not award MP2 for forces on sledge on horizontal ground

The arrows should contact the object

gravitational force/weight from the Earth «downwards»

reaction force from the sledge/snow/ground «upwards»

no vertical acceleration/remains in contact with the ground/does not move vertically as there is no resultant vertical force

Allow naming of forces as in (a)

Allow vertical forces are balanced/equal in magnitude/cancel out

mention of conservation of momentum

OR

5.5 x 4.2 = (55 + 5.5) «v»

0.38 «m s^–1»

Allow p=p′ or other algebraically equivalent statement

Award [0] for answers based on energy

same change in momentum/impulse

the time taken «to stop» would be greater «with the snow»

$F=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}$ therefore F is smaller «with the snow»

OR

force is proportional to rate of change of momentum therefore F is smaller «with the snow»

Allow reverse argument for ice

«friction force down slope» = μmg cos(6.5) = «5.9 N»

«component of weight down slope» = mg sin(6.5) «= 6.1 N»

«so a = $\frac{F}{m}$» acceleration = $\frac{12}{5.5}$ = 2.2 «m s^–2»

Ignore negative signs

Allow use of g = 10 m s^–2

correct use of kinematics equation

distance = 4.4 or 4.0 «m»

Alternative 2

KE lost=work done against friction + GPE

distance = 4.4 or 4.0 «m»

Allow ECF from (e)(i)

Allow [1 max] for GPE missing leading to 8.2 «m»

calculates a maximum value for the frictional force = «μR=» 7.5 «N»

sledge will not move as the maximum static friction force is greater than the component of weight down the slope

Allow correct conclusion from incorrect MP1

Allow 7.5 > 6.1 so will not move

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 $4 \pi «\text{rad}» {\text{s}}^{-1}$

Calculate the speed of the ball as it leaves the racket.

Show that the average force exerted on the ball by the racket is about 50 N.

Determine, with reference to the work done by the average force, the horizontal distance travelled by the ball while it was in contact with the racket.

Draw a graph to show the variation with t of the horizontal speed v of the ball while it was in contact with the racket. Numbers are not required on the axes.

links 0.84 to Δp ✔

$v=$«$\frac{0.84}{5.8\times {10}^{-2}}=$» 14.5 «m s^–1»✔

NOTE: Award [2] for bald correct answer

use of Δt = «(28 – 12) × 10^–3 =» 16 × 10^–3 «s» ✔

$\overline{F}$ =«$\frac{∆p}{∆t}=$» $\frac{0.84}{16\times {10}^{-3}}$ OR  53 «N» ✔

NOTE: Accept a time interval from 14 to 16 ms
Allow ECF from incorrect time interval

E_k = $\frac{1}{2}$ × 5.8 × 10^–2 × 14.5² ✔

E_k = W ✔

s = «$\frac{W}{F}=\frac{{\displaystyle \frac{1}{2}\times 5.8\times {10}^{-2}\times 14.5^2}}{53}=$» 0.12 « m » ✔

Allow ECF from (a) and (b)

Allow ECF from MP1

Award [2] max for a calculation without reference to work done, eg: average velocity × time

graph must show increasing speed from an initial of zero all the time ✔
overall correct curvature ✔

Show that the electric field strength due to the point charge at the position of the electron is 3.4 × 10⁸ N C^–1.

Calculate the magnitude of the initial acceleration of the electron.

Describe the subsequent motion of the electron.

$E=\frac{k\times q}{r^2}$ ✔

$E=\frac{8.99\times {10}^9\times 6.0\times {10}^{-3}}{0.4^2}$  OR  $E=3.37\times {10}^8 «\mathrm{N} {\mathrm{C}}^{-1}»$ ✔

NOTE: Ignore any negative sign.

$F=q\times E$ OR  $F=1.6\times {10}^{-19}\times 3.4\times {10}^8=5.4\times {10}^{-11} «\mathrm{N}»$ ✔

$a=«\frac{5.4\times {10}^{-11}}{9.1\times {10}^{-31}}=» 5.9\times {10}^{19}« \mathrm{m} {\mathrm{s}}^{-2}»$✔

NOTE: Ignore any negative sign.
Award [1] for a calculation leading to $a=« \mathrm{m} {\mathrm{s}}^{-2}»$
Award [2] for bald correct answer

the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔

NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force as attractive so concludes that the acceleration will increase

The molar mass of water is 18 g mol⁻¹. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas.

State one assumption of the kinetic model of an ideal gas.

Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer.

Explain why the temperature of water remains at 100 °C during this time.

The heater is removed and a mass of 0.30 kg of pasta at −10 °C is added to the boiling water.

Determine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible.

Specific heat capacity of pasta = 1.8 kJ kg⁻¹ K⁻¹
Specific heat capacity of water = 4.2 kJ kg⁻¹ K⁻¹

Show that each resistor has a resistance of about 30 Ω.

Calculate the power transferred by the heater when both switches are closed.

E_k = « $\frac{3}{2}(1.38\times {10}^{-23})\left(373\right)$» = $7.7\times {10}^{-21}$ «J» ✓

v = «$\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 6.02\times {10}^{23}\times 373}{0.018}}$» = 720 «m s⁻¹» ✓

particles can be considered points «without dimensions» ✓

no intermolecular forces/no forces between particles «except during collisions»✓

the volume of a particle is negligible compared to volume of gas ✓

collisions between particles are elastic ✓

time between particle collisions are greater than time of collision ✓

no intermolecular PE/no PE between particles  ✓

Accept reference to atoms/molecules for “particle”

«mL = P t» so «$L=\frac{1600\times 200}{0.14}$» = 2.3 x 10⁶ «J kg^-1» ✓

J kg⁻¹ ✓

«all» of the energy added is used to increase the «intermolecular» potential energy of the particles/break «intermolecular» bonds/OWTTE ✓

Accept reference to atoms/molecules for “particle” 

use of mcΔT ✓

0.86 × 4200 × (100 – T) = 0.3 × 1800 × (T +10) ✓

T_eq = 85.69«°C» ≅ 86«°C» ✓

Accept T_eq in Kelvin (359 K).

$P=\frac{v^2}{R} \mathrm{so} \frac{{220}^2}{1600} \mathrm{so} R=30.25$ «Ω» ✓

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

use of parallel resistors addition so R_eq = 15 «Ω» ✓

P = 3200 «W» ✓

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.

With the door open the air in the refrigerator is initially at the same temperature and pressure as the air in the kitchen. Calculate the number of molecules of air in the refrigerator.

Determine the pressure of the air inside the refrigerator.

The door of the refrigerator has an area of 0.72 m². Show that the minimum force needed to open the refrigerator door is about 4 kN.

Comment on the magnitude of the force in (b)(ii).

$N=\frac{pV}{kT}$ OR $N=\frac{1.0\times {10}^5\times 0.36}{1.38\times {10}^{-23}\times 295}$ ✔

$N=8.8\times {10}^{24}$ ✔

NOTE: Allow [1 max] for substitution with T in Celsius.
Allow [1 max] for a final answer of n = 14.7 or 15
Award [2] for bald correct answer.

use of $\frac{p}{T}$ = constant  OR  $p=\frac{nRT}{V}$  OR  $\frac{NkT}{V}$ ✔
$p=9.4\times {10}^4$« Pa »✔

NOTE: Allow ECF from (a)
Award [2] for bald correct answer

$F=A\times ∆p$ ✔

$F=0.72\times \left(1.0-0.94\right)\times {10}^5$OR 4.3 × 10³ « N »✔

NOTE: Allow ECF from (b)(i)
Allow ECF from MP1

force is «very» large ✔

there must be a mechanism that makes this force smaller
OR
assumption used to calculate the force/pressure is unrealistic ✔

At position B the rope starts to extend. Calculate the speed of the block at position B.

Determine the magnitude of the average resultant force acting on the block between B and C.

Sketch on the diagram the average resultant force acting on the block between B and C. The arrow on the diagram represents the weight of the block.

Calculate the magnitude of the average force exerted by the rope on the block between B and C.

between A and B.

between B and C.

The length reached by the rope at C is 77.4 m. Suggest how energy considerations could be used to determine the elastic constant of the rope.

use of conservation of energy

OR

v² = u² + 2as

v = «$\sqrt{2\times 60.0\times 9.81}$» = 34.3 «ms^–1»

[2 marks]

use of impulse F_ave × Δt = Δp

OR

use of F = ma with average acceleration

OR

F = $\frac{80.0\times 34.3}{0.759}$

3620«N»

Allow ECF from (a).

[2 marks]

upwards

clearly longer than weight

For second marking point allow ECF from (b)(i) providing line is upwards.

[2 marks]

3620 + 80.0 × 9.81

4400 «N»

Allow ECF from (b)(i).

[2 marks]

(loss in) gravitational potential energy (of block) into kinetic energy (of block)

Must see names of energy (gravitational potential energy and kinetic energy) – Allow for reasonable variations of terminology (eg energy of motion for KE).

[1 mark]

(loss in) gravitational potential and kinetic energy of block into elastic potential energy of rope

See note for 1(c)(i) for naming convention.

Must see either the block or the rope (or both) mentioned in connection with the appropriate energies.

[1 mark]

k can be determined using EPE = $\frac{1}{2}$kx²

correct statement or equation showing

GPE at A = EPE at C

OR

(GPE + KE) at B = EPE at C

Candidate must clearly indicate the energy associated with either position A or B for MP2.

[2 marks]

Calculate the speed of the combined masses immediately after the collision.

Show that the collision is inelastic.

Describe the changes in gravitational potential energy of the oscillating system from t = 0 as it oscillates through one cycle of its motion.

0.40 «m s⁻¹» ✔

initial energy 24 mJ and final energy 12 mJ ✔

energy is lost/unequal /change in energy is 12 mJ ✔

inelastic collisions occur when energy is lost ✔

maximum GPE at extremes, minimum in centre ✔

Candidates fell into some broad categories on this question. This was a “show that” question, so there was an expectation of a mathematical argument. Many were able to successfully show that the initial and final kinetic energies were different and connect this to the concept of inelastic collisions. Some candidates tried to connect conservation of momentum unsuccessfully, and some simply wrote an extended response about the nature of inelastic collisions and noted that the bobs stuck together without any calculations. This approach was awarded zero marks.

This straightforward question had surprisingly poorly answers. Candidate answers tended to be overly vague, such as “as the bob went higher the GPE increased and as it fell the GPE decreased.” Candidates needed to specify when GPE would be at maximum and minimum values. Some candidates mistakenly assumed that at t=0 the pendulum bob was at maximum height despite being told otherwise in the question stem.

Determine the initial acceleration of the spacecraft.

Estimate the maximum speed of the spacecraft.

Outline why scientists sometimes use estimates in making calculations.

Outline why the ions are likely to spread out.

Explain what effect, if any, this spreading of the ions has on the acceleration of the spacecraft.

Outline what is meant by the gravitational field strength at a point.

Newton’s law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting a spherical planet of uniform density.

change in momentum each second = 6.6 × 10⁻⁶ × 5.2 × 10⁴ «= 3.4 × 10⁻¹kg m s⁻¹» ✔

acceleration = «$\frac{3.4\times {10}^{-1}}{740}$ =» 4.6 × 10⁻⁴ «m s⁻²» ✔

ALTERNATIVE 1:

(considering the acceleration of the spacecraft)

time for acceleration = $\frac{30}{6.6\times {10}^{-6}}$ = «4.6 × 10⁶» «s» ✔

max speed = «answer to (a) × 4.6 × 10⁶ =» 2.1 × 10³ «m s⁻¹» ✔

ALTERNATIVE 2:

(considering the conservation of momentum)

(momentum of 30 kg of fuel ions = change of momentum of spacecraft)

30 × 5.2 × 10⁴ = 710 × max speed ✔

max speed = 2.2 × 10³ «m s⁻¹» ✔

problem may be too complicated for exact treatment ✔

to make equations/calculations simpler ✔

when precision of the calculations is not important ✔

some quantities in the problem may not be known exactly ✔

ions have same (sign of) charge ✔

ions repel each other ✔

the forces between the ions do not affect the force on the spacecraft. ✔

there is no effect on the acceleration of the spacecraft. ✔

force per unit mass ✔

acting on a small/test/point mass «placed at the point in the field» ✔

satellite has a much smaller mass/diameter/size than the planet «so approximates to a point mass» ✔ 

Explain why the path of the proton is a circle.

Show that the radius of the path is about 6 cm.

Calculate the time for one complete revolution.

Explain why the kinetic energy of the proton is constant.

magnetic force is to the left «at the instant shown»
OR
explains a rule to determine the direction of the magnetic force ✔

force is perpendicular to velocity/«direction of» motion
OR
force is constant in magnitude ✔

force is centripetal/towards the centre ✔

NOTE: Accept reference to acceleration instead of force

$qvB=\frac{m{v}^2}{R}$✔

$R=\frac{1.67\times {10}^{-27}\times 2.0\times {10}^6}{1.6\times {10}^{-19}\times 0.35}$ OR 0.060 « m »

NOTE: Award MP2 for full replacement or correct answer to at least 2 significant figures

$T=\frac{2\mathrm{\pi }R}{v}$✔

$T=«\frac{2\mathrm{\pi }\times 0.06}{2.0\times {10}^6}=1.9\times {10}^{-7}$ « s » ✔

NOTE: Award [2] for bald correct answer

ALTERNATIVE 1
work done by force is change in kinetic energy ✔
work done is zero/force perpendicular to velocity ✔

NOTE: Award [2] for a reference to work done is zero hence E_k remains constant

ALTERNATIVE 2
proton moves at constant speed ✔
kinetic energy depends on speed ✔

NOTE: Accept mention of speed or velocity indistinctly in MP2

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 value of the resultant force on the aircraft when hovering.

Outline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved.

Determine $v$. State your answer to an appropriate number of significant figures.

The package and string are now released and fall to the ground. The lift force on the aircraft remains unchanged. Calculate the initial acceleration of the aircraft.

zero ✓

Blades exert a downward force on the air ✓

air exerts an equal and opposite force on the blades «by Newton’s third law»
OR
air exerts a reaction force on the blades «by Newton’s third law» ✓

Downward direction required for MP1.

«lift force/change of momentum in one second» $=1.7v$ ✓

$1.7v=\left(0.95+0.45\right)\times 9.81$ ✓

$v=8.1«{\mathrm{ms}}^{-1}»$ AND answer expressed to $2$ sf only ✓

Allow $\mathit{8}\mathit{.}\mathit{2}$ from $g\mathit{=}\mathit{10}\mathit{ }m{s}^{\mathit{-}\mathit{2}}$.

vertical force = lift force – weight OR $=0.45\times 9.81$ OR $=4.4 «\mathrm{N}»$✓

acceleration$=\frac{0.45\times 9.81}{0.95}=4.6 «{\mathrm{ms}}^{-2}»$✓

Determine H.

Label the time and velocity graph, using the letter M, the point where the ball reaches the maximum rebound height.

State the acceleration of the ball at the maximum rebound height.

Draw, on the axes, a graph to show the variation with time of the height of the ball from the instant it rebounds from the floor until the instant it reaches the maximum rebound height. No numbers are required on the axes.

Estimate the loss in the mechanical energy of the ball as a result of the collision with the floor.

Determine the average force exerted on the floor by the ball.

Suggest why the momentum of the ball was not conserved during the collision with the floor.

H = «$\frac{1}{2}$gt² =» 4.9 «m» ✓

Accept other methods as area from graph, alternative kinematics equations or conservation of mechanical energy.
Award [1] for a bald correct answer in the range 4.9 - 5.1.
Award [0] if time used is different than 1.0 s.

M at 1.6 s ✓

«g =» 9.80 «ms⁻²» ✓

Accept 9.81, 10 or a plain “g”.
Ignore sign if provided.

concave down parabola as shown «with non-zero initial slope and zero final slope» ✓

Award [1] mark if curve starts from a positive time value.
Award [0] if the final slope is negative.

« loss of KE is $\frac{1}{2}\times 0.25\times \left(9.8^2-5^2\right)=$» $8.9$«J» ✓

Award [1] mark for an answer in the range 8.7 - 9.5.

$∆p=0.250\times \left(9.8+5.0\right)$ ✓

F_net = «$\frac{∆p}{∆t}=\frac{3.7}{0.1}=$» $37$ «N» ✓

N $=37+0.250\times 9.8=39.5$ «N» ✓

Allow ECF for MP2 and MP3.

there is an external force acting on the ball

OR

some momentum is transferred to the floor ✓

Allow references to impulse instead of force.
Do not award references to energy.

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]

Draw and label the free-body diagram for the person.

The person must not slide down the wall. Show that the minimum angular velocity $\omega$ of the cylinder for this situation is

$\omega =\sqrt{\frac{g}{\mu R}}$

where $\mu$ is the coefficient of static friction between the person and the cylinder.

The coefficient of static friction between the person and the cylinder is $0.40$. The radius of the cylinder is $3.5 \mathrm{m}$. The cylinder makes $28$ revolutions per minute. Deduce whether the person will slide down the inner surface of the cylinder.

arrow downwards labelled weight/W/$mg$ and arrow upwards labelled friction/$F$ ✓

arrow horizontally to the left labelled «normal» reaction/$N$ ✓

Ignore point of application of the forces but do not allow arrows that do not touch the object.

Do not allow horizontal force to be labelled ‘centripetal’ or $R$.

See $F=\mu N$ AND $N=mR{\omega }^2$ ✓

«substituting for N» $\mu m{\omega }^{2}R=mg$ ✓

ALTERNATIVE 1

minimum required angular velocity $«=\sqrt{\frac{9.81}{0.40\times 3.5}}»=2.6«\mathrm{rad} {\mathrm{s}}^{-1}»$ ✓

actual angular velocity $«=\frac{2\mathrm{\pi }}{\left({\displaystyle \frac{60}{28}}\right)}»=2.9«\mathrm{rad} {\mathrm{s}}^{-1}»$✓

actual angular velocity is greater than the minimum, so the person does not slide ✓

ALTERNATIVE 2

Minimum friction force $=mg=«9.81 \mathrm{m}»$ ✓

Actual friction force $«=\mu mR{\omega }^2=0.40 \mathrm{m}\times 3.5{\left(2\mathrm{\pi }\frac{28}{60}\right)}^2»=12.0 \mathrm{m}$ ✓

Actual friction force is greater than the minimum frictional force so the person does not slide ✓

Allow $\mathit{2}\mathit{.}\mathit{7}$ from $g\mathit{=}\mathit{10}\mathit{ }m{s}^{\mathit{-}\mathit{2}}$.

State what is meant by the binding energy of a nucleus.

Draw, on the axes, a graph to show the variation with nucleon number $A$ of the binding energy per nucleon, $\frac{\text{BE}}{A}$. Numbers are not required on the vertical axis.

Identify, with a cross, on the graph in (a)(ii), the region of greatest stability.

Show that the energy released in this decay is about 6 MeV.

The plutonium nucleus is at rest when it decays.

Calculate the ratio $\frac{\text{kinetic energy of alpha particle}}{\text{kinetic energy of uranium}}$.

the energy needed to «completely» separate the nucleons of a nucleus

OR

the energy released when a nucleus is assembled from its constituent nucleons ✓

Accept reference to protons AND neutrons.

curve rising to a maximum between 50 and 100 ✓

curve continued and decreasing ✓

Ignore starting point.

Ignore maximum at alpha particle

At a point on the peak of their graph ✓

correct mass numbers for uranium (234) and alpha (4) ✓

$234\times 7.600+4\times 7.074-238\times 7.568$ «MeV» ✓

energy released 5.51 «MeV» ✓

Ignore any negative sign.

«$\frac{K{E}_{\alpha }}{K{E}_U}=$»$\frac{\frac{p^2}{2m_{\alpha }}}{\frac{p^2}{2m_U}}$  OR  $\frac{m_U}{m_{\alpha }}$ ✓

«$\frac{234}{4}=$» $58.5$ ✓

Award [2] marks for a bald correct answer.

Accept $\frac{117}{2}$ for MP2.

Label with arrows on the diagram the magnetic force F on the proton. 

Label with arrows on the velocity vector v of the proton.

The speed of the proton is 2.16 × 10⁶ m s^-1 and the magnetic field strength is 0.042 T. For this proton, determine, in m, the radius of the circular path. Give your answer to an appropriate number of significant figures.

F towards centre ✔

v tangent to circle and in the direction shown in the diagram ✔

«$qvB=\frac{m{v}^2}{R}\Rightarrow »R=\frac{mv}{qB}/\frac{1.673\times {10}^{-27}\times 2.16\times {10}^6}{1.60\times {10}^{-19}\times 0.042}$  ✔

R = 0.538 «m»✔

R = 0.54 «m» ✔

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

This was generally well answered although usually to 3 sf. Common mistakes were to substitute 0.042 for F and 1 for q. Also some candidates tried to answer in terms of electric fields.

From A to B, 24 % of the gravitational potential energy transferred to kinetic energy. Show that the velocity at B is 12 m s^–1.

Some of the gravitational potential energy transferred into internal energy of the skis, slightly increasing their temperature. Distinguish between internal energy and temperature.

The dot on the following diagram represents the skier as she passes point B.
Draw and label the vertical forces acting on the skier.

The hill at point B has a circular shape with a radius of 20 m. Determine whether the skier will lose contact with the ground at point B.

The skier reaches point C with a speed of 8.2 m s^–1. She stops after a distance of 24 m at point D.

Determine the coefficient of dynamic friction between the base of the skis and the snow. Assume that the frictional force is constant and that air resistance can be neglected.

Calculate the impulse required from the net to stop the skier and state an appropriate unit for your answer.

Explain, with reference to change in momentum, why a flexible safety net is less likely to harm the skier than a rigid barrier.

$\frac{1}{2}{v}^2=0.24\text{gh}$

$v=11.9$ «m s^–1»

Award GPE lost = 65 × 9.81 × 30 = «19130 J»

Must see the 11.9 value for MP2, not simply 12.

Allow g = 9.8 ms^–2.

internal energy is the total KE «and PE» of the molecules/particles/atoms in an object

temperature is a measure of the average KE of the molecules/particles/atoms

Award [1 max] if there is no mention of molecules/particles/atoms.

arrow vertically downwards from dot labelled weight/W/mg/gravitational force/F_g/F_{gravitational} AND arrow vertically upwards from dot labelled reaction force/R/normal contact force/N/F_N

W > R

Do not allow gravity.
Do not award MP1 if additional ‘centripetal’ force arrow is added.
Arrows must connect to dot.
Ignore any horizontal arrow labelled friction.
Judge by eye for MP2. Arrows do not have to be correctly labelled or connect to dot for MP2.

ALTERNATIVE 1
recognition that centripetal force is required / $\frac{m{v}^2}{r}$ seen

= 468 «N»

W/640 N (weight) is larger than the centripetal force required, so the skier does not lose contact with the ground

ALTERNATIVE 2

recognition that centripetal acceleration is required / $\frac{v^2}{r}$ seen

a = 7.2 «ms^–2»

g is larger than the centripetal acceleration required, so the skier does not lose contact with the ground

ALTERNATIVE 3

recognition that to lose contact with the ground centripetal force ≥ weight

calculation that v ≥ 14 «ms^–1»

comment that 12 «ms^–1» is less than 14 «ms^–1» so the skier does not lose contact with the ground

ALTERNATIVE 4

recognition that centripetal force is required / $\frac{m{v}^2}{r}$ seen

calculation that reaction force = 172 «N»

reaction force > 0 so the skier does not lose contact with the ground

Do not award a mark for the bald statement that the skier does not lose contact with the ground.

ALTERNATIVE 1
0 = 8.2² + 2 × a × 24 therefore a = «−»1.40 «m s⁻²»

friction force = ma = 65 × 1.4 = 91 «N»

coefficient of friction = $\frac{91}{65\times 9.81}$ = 0.14

ALTERNATIVE 2
KE = $\frac{1}{2}$mv² = 0.5 x 65 x 8.2² = 2185 «J»

friction force = KE/distance = 2185/24 = 91 «N»

coefficient of friction = $\frac{91}{65\times 9.81}$ = 0.14

Allow ECF from MP1.

«76 × 9.6»= 730
Ns OR kg ms^–1

safety net extends stopping time

F = $\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}$ therefore F is smaller «with safety net»

OR

force is proportional to rate of change of momentum therefore F is smaller «with safety net»

Accept reverse argument.

Determine the magnitude of the average decelerating force that the ground exerts on the egg.

Explain why the egg is likely to break when dropped onto concrete from the same height.

ALTERNATIVE 1:

initial momentum = mv = $\sqrt{2\times 0.058\times 0.63}$ «= 0.27 kg m s⁻¹»

OR

mv = $0.058\times \sqrt{2\times 9.81\times 1.1}$ «= 0.27 kg m s⁻¹» ✔

force = «$\frac{\text{change in momentum}}{\text{time}}$ =» $\frac{0.27}{0.055}$ ✔

4.9 «N» ✔

F − mg = 4.9 so F= 5.5 «N» ✔

ALTERNATIVE 2:

«E_k = $\frac{1}{2}$mv² = 0.63 J» v = 4.7 m s⁻¹ ✔

acceleration = «$\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$ =» $\frac{4.7}{55\times {10}^{-3}}$ = «85 m s⁻²» ✔

4.9 «N» ✔

F − mg = 4.9 so F= 5.5 «N» ✔

ALTERNATIVE 1:

concrete reduces the stopping time/distance ✔

impulse/change in momentum same so force greater

OR

work done same so force greater ✔

ALTERNATIVE 2:

concrete reduces the stopping time ✔

deceleration is greater so force is greater ✔

Allow reverse argument for grass.

Radioactive decay is said to be “random” and “spontaneous”. Outline what is meant by each of these terms.

Random: 

Spontaneous:

Calculate the binding energy per nucleon for uranium-238.

Calculate the ratio $\frac{\mathrm{kinetic} \mathrm{energy} \mathrm{of} \mathrm{alpha} \mathrm{particle}}{\mathrm{kinetic} \mathrm{energy} \mathrm{of} \mathrm{thorium} \mathrm{nucleus}}$.

random:
it cannot be predicted which nucleus will decay
OR
it cannot be predicted when a nucleus will decay ✔

NOTE: OWTTE

spontaneous:
the decay cannot be influenced/modified in any way ✔

NOTE: OWTTE

234 × 7.6  OR  4 × 7.07 ✔

BE_U =« 234 × 7.6 + 4 × 7.07 – 4.27 =» $1802$ « MeV » ✔

$\frac{B{E}_{\mathrm{U}}}{A}=«\frac{1802}{238}=» 7.57$ « MeV » ✔

NOTE: Allow ECF from MP2
Award [3] for bald correct answer
Allow conversion to J, final answer is 1.2 × 10^–12

states or applies conservation of momentum ✔

ratio is «$\frac{E_{\mathrm{k}\alpha }}{E_{\mathrm{kTh}}}=\frac{{\displaystyle \frac{p^2}{2m_{\alpha }}}}{{\displaystyle \frac{p^2}{2m_{\mathrm{Th}}}}}=\frac{234}{4}$» 58.5 ✔

NOTE: Award [2] for bald correct answer

The work done to move a particle of charge 0.25 μC from one point in an electric field to another is 4.5 μJ. Calculate the magnitude of the potential difference between the two points.

Determine the force on Q at the instant it is released.

Describe the motion of Q after release.

On the diagram draw an arrow to show the direction of the magnetic field at Q due to wire X alone.

Determine the magnitude and direction of the resultant magnetic field at Q.

«$V=\frac{4.5}{0.25}=$» 18 «V» ✓

$F=\frac{8.99\times {10}^9\times 68\times {10}^{-6}\times 0.25\times {10}^{-6}}{0.{48}^2}$ ✓

$F=0.66$ «N» ✓

Award [2] marks for a bald correct answer.

Allow symbolic k in substitutions for MP1.

Do not allow ECF from incorrect or not squared distance.

Q moves to the right/away from P «along a straight line»

OR

Q is repelled from P ✓

with increasing speed/Q accelerates ✓

acceleration decreases ✓

arrow of any length as shown ✓

«using components or Pythagoras to get» B = 21 «mT» ✓

directed «horizontally» to the right ✓

If no unit seen, assume mT.

A black body is on the Moon’s surface at point A. Show that the maximum temperature that this body can reach is 400 K. Assume that the Earth and the Moon are the same distance from the Sun.

Another black body is on the Moon’s surface at point B.

Outline, without calculation, why the aximum temperature of the black body at point B is less than at point A.

The albedo of the Earth’s atmosphere is 0.28. Outline why the maximum temperature of a black body on the Earth when the Sun is overhead is less than that at point A on the Moon.

Outline why a force acts on the Moon.

Outline why this force does no work on the Moon.

T = ${\left(\frac{1360}{\sigma }\right)}^{0.25}$   ✔

390 «K» ✔

Must see 1360 (from data booklet) used for MP1.

Must see at least 2 s.f.

energy/Power/Intensity lower at B ✔

connection made between energy/power/intensity and temperature of blackbody ✔

(28 %) of sun’s energy is scattered/reflected by earth’s atmosphere OR only 72 % of incident energy gets absorbed by blackbody ✔

Must be clear that the energy is being scattered by the atmosphere.

Award [0] for simple definition of “albedo”.

gravitational attraction/force/field «of the planet/Moon» ✔

Do not accept “gravity”.

the force/field and the velocity/displacement are at 90° to each other OR there is no change in GPE of the moon ✔

Award [0] for any mention of no net force on the satellite.

Do not accept acceleration is perpendicular to velocity.

Many candidates struggled with this question. A significant portion attempted to apply Wein’s Law and simply stated that a particular wavelength was the peak and then used that to determine the temperature. Some did use the solar constant from the data booklet and were able to calculate the correct temperature. As part of their preparation for the exam candidates should thoroughly review the data booklet and be aware of what constants are given there. As with all “show that” questions candidates should be reminded to include an unrounded answer.

This is question is another example of candidates not thinking beyond the obvious in the question. Many simply said that point B is farther away, or that it is at an angle. Some used vague terms like “the sunlight is more spread out” rather than using proper physics terms. Few candidates connected the lower intensity at B with the lower temperature of the blackbody.

This question was assessing the understanding of the concept of albedo. Many candidates were able to connect that an albedo of 0.28 meant that 28 % of the incident energy from the sun was being reflected or scattered by the atmosphere before reaching the black body.

This was generally well answered, although some candidates simply used the vague term “gravity” rather than specifying that it is a gravitational force or a gravitational field. Candidates need to be reminded about using proper physics terms and not more general, “every day” terms on the exam.

Some candidates connected the idea that the gravitational force is perpendicular to the velocity (and hence the displacement) for the mark. It was also allowed to discuss that there is no change in gravitational potential energy, so therefore no work was being done. It was not acceptable to simply state that the net displacement over one full orbit is zero. Unfortunately, some candidates suggested that there is no net force on the moon so there is no work done, or that the moon is so much smaller so no work could be done on it.

Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs.

The slits are separated by 1.5 mm and the laser light has a wavelength of 6.3 x 10^–7 m. The slits are 5.0 m from the train track. Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum.

Estimate the speed of the train.

In another experiment the student replaces the light sensor with a sound sensor. The train travels away from a loudspeaker that is emitting sound waves of constant amplitude and frequency towards a reflecting barrier.

The sound sensor gives a graph of the variation of output voltage with time along the track that is similar in shape to the graph shown in the resource. Explain how this effect arises.

«light» superposes/interferes

pattern consists of «intensity» maxima and minima
OR
consisting of constructive and destructive «interference»

voltage peaks correspond to interference maxima

«$s=\frac{\lambda D}{d}=\frac{6.3\times {10}^{-7}\times 5.0}{1.5\times {10}^{-3}}=$» 2.1 x 10^–3 «m» 

If no unit assume m.
Correct answer only.

correct read-off from graph of 25 m s

v = «$\frac{x}{t}=\frac{2.1\times {10}^{-3}}{25\times {10}^{-3}}=$» 8.4 x 10^–2 «m s^–1»

Allow ECF from (b)(i)

ALTERNATIVE 1

«reflection at barrier» leads to two waves travelling in opposite directions

mention of formation of standing wave

maximum corresponds to antinode/maximum displacement «of air molecules»
OR
complete cancellation at node position

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