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

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 $4 \pi «\text{rad}» {\text{s}}^{-1}$

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

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.

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.

(i) Define gravitational field strength.

(ii) State the SI unit for gravitational field strength.

A planet orbits the Sun in a circular orbit with orbital period T and orbital radius R. The mass of the Sun is M.

(i) Show that $T=\sqrt{\frac{4{\pi }^2{R}^3}{GM}}$.

(ii) The Earth’s orbit around the Sun is almost circular with radius 1.5×10¹¹ m. Estimate the mass of the Sun.

(i)  «gravitational» force per unit mass on a «small or test» mass 

(ii)  N kg^–1

Award mark if N kg^-1 is seen, treating any further work as neutral.
Do not accept bald m s^–2

i
clear evidence that v in $v^2=\frac{4{\pi }^2{R}^2}{T^2}$ is equated to orbital speed $\sqrt{\frac{GM}{R}}$
OR
clear evidence that centripetal force is equated to gravitational force
OR
clear evidence that a in $a=\frac{v^2}{R}$ etc is equated to g in $g=\frac{GM}{R^2}$ with consistent use of symbols
Minimum is a statement that $\sqrt{\frac{GM}{R}}$ is the orbital speed which is then used in $v=\frac{2\pi R}{T}$
Minimum is F_c = F_g ignore any signs.
Minimum is g = a.

substitutes and re-arranges to obtain result
Allow any legitimate method not identified here.
Do not allow spurious methods involving equations of shm etc

$\ll T=\sqrt{\frac{4{\pi }^{2}R}{\left(\frac{GM}{R^2}\right)}}=\sqrt{\frac{4{\pi }^2{R}^3}{GM}}\gg$

ii
«T = 365 × 24 × 60 × 60 = 3.15 × 10⁷ s»

$M=\ll \frac{4{\pi }^2{R}^3}{G{T}^2}=\gg =\frac{4\times {3.14}^2\times {\left(1.5\times {10}^{11}\right)}^3}{6.67\times {10}^{-11}\times {\left(3.15\times {10}^7\right)}^2}$
2×10³⁰«kg»

Allow use of 3.16 x 10⁷ s for year length (quoted elsewhere in paper).
Condone error in power of ten in MP1.
Award [1 max] if incorrect time used (24 h is sometimes seen, leading to 2.66 x 10³⁵ kg).
Units are not required, but if not given assume kg and mark POT accordingly if power wrong.
Award [2] for a bald correct answer.
No sf penalty here.

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 $-\frac{1}{6}$

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.

Explain why a centripetal force is needed for the planet to be in a circular orbit.

State the nature of this centripetal force.

Determine the gravitational field of the planet.

The following data are given:

Mass of planet            $=8.0\times {10}^{24}$ kg
Radius of the planet    $=9.1\times {10}^6$ m.

«circular motion» involves a changing velocity ✓

«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓

there must be a force/acceleration towards centre/star ✓

without a centripetal force the planet will move in a straight line ✓

gravitational force/force of gravity ✓

use of $\frac{GM}{R^2}$ ✓

6.4 «Nkg⁻¹ or ms⁻²» ✓

State the direction of the magnetic field.

Calculate, in N, the magnitude of the magnetic force acting on the electron.

Explain why the electron moves at constant speed.

Explain why the electron moves on a circular path.

out of the page plane / ⊙

Do not accept just “up” or “outwards”.

[1 mark]

1.60 × 10^–19 × 6.8 × 10⁵ × 8.5 = 9.2 × 10^–13 «N»

[1 mark]

the magnetic force does not do work on the electron hence does not change the electron’s kinetic energy

OR

the magnetic force/acceleration is at right angles to velocity

[1 mark]

the velocity of the electron is at right angles to the magnetic field

(therefore) there is a centripetal acceleration / force acting on the charge

OWTTE

[2 marks]

Show that the intensity of the solar radiation at the location of Titan is 16 W m⁻²

Titan has an atmosphere of nitrogen. The albedo of the atmosphere is 0.22. The surface of Titan may be assumed to be a black body. Explain why the average intensity of solar radiation absorbed by the whole surface of Titan is 3.1 W m⁻²

Show that the equilibrium surface temperature of Titan is about 90 K.

The orbital radius of Titan around Saturn is $R$ and the period of revolution is $T$.

Show that $T^2=\frac{4{\mathrm{\pi }}^2{R}^{ 3}}{GM}$ where $M$ is the mass of Saturn.

The orbital radius of Titan around Saturn is 1.2 × 10⁹m and the orbital period is 15.9 days. Estimate the mass of Saturn.

incident intensity $\frac{1360}{9.3^2}$ OR $15.7\approx 16$ «W m⁻²» ✓

Allow the use of 1400 for the solar constant.

exposed surface is ¼ of the total surface ✓

absorbed intensity = (1−0.22) × incident intensity ✓

0.78 × 0.25 × 15.7  OR  3.07 «W m⁻²» ✓

Allow 3.06 from rounding and 3.12 if they use 16 W m⁻².

σT ⁴ = 3.07

OR

T = 86 «K» ✓

correct equating of gravitational force / acceleration to centripetal force / acceleration ✓

correct rearrangement to reach the expression given ✓

Allow use of $\sqrt{\frac{GM}{R}}=\frac{2\mathrm{\pi }R}{T}$ for MP1.

$T=15.9\times 24\times 3600$ «s» ✓

$M=\frac{4{\mathrm{\pi }}^2{\left(1.2\times {10}^9\right)}^3}{6.67\times {10}^{-11}\times {\left(15.9\times 24\times 3600\right)}^2}=5.4\times {10}^{26}$«kg» ✓

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

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» ✔ 

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]

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.

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.

Determine the orbital period for the satellite.

Mass of Earth = 6.0 x 10²⁴ kg

Determine the mean temperature of the Earth.

Suggest how the difference between λ_S and λ_E helps to account for the greenhouse effect.

Not all scientists agree that global warming is caused by the activities of man.

Outline how scientists try to ensure agreement on a scientific issue.

$\frac{m{v}^2}{r}=G\frac{Mm}{r^2}$

leading to T² = $\frac{4{\pi }^2{r}^3}{GM}$

T = 5320 «s»

Alternative 2

«$v=\sqrt{\frac{G{M}_E}{r}}$» = $\sqrt{\frac{6.67\times {10}^{-11}\times 6.0\times {10}^{24}}{6600\times {10}^3}}$ or 7800 «ms^–1»

distance = 2$\pi$r = 2$\pi$ x 6600 x 10³ «m» or 4.15 x 10⁷ «m»

«T = $\frac{d}{v}=\frac{4.15\times {10}^7}{7800}$» = 5300 «s»

Accept use of ω instead of v

T = «$\frac{2.90\times {10}^{-3}}{{\lambda }_{\text{max}}}=$» $\frac{2.90\times {10}^{-3}}{10.1\times {10}^{-6}}$

= 287 «K» or 14 «°C»

Award [0] for any use of wavelength from Sun

Do not accept 287 °C

wavelength of radiation from the Sun is shorter than that emitted from Earth «and is not absorbed by the atmosphere»

infrared radiation emitted from Earth is absorbed by greenhouse gases in the atmosphere

this radiation is re-emitted in all directions «including back to Earth»

peer review

international collaboration

full details of experiments published so that experiments can repeated

[Max 1 Mark]