a1, a2,a3 YEE [109 marks]

1. [Maximum mark: 4] SPM.2.SL.TZ0.3

A satellite moves around Earth in a circular orbit.

(a) Draw an arrow on the diagram to represent the direction of the

acceleration of the satellite. [1]

Markscheme

arrow normal to the orbit towards the Earth ✓

(b) The following data are given:

Mass of Earth, M = 5.97 × 1024 kg

Radius of Earth, R = 6.37 × 106 m
Orbital period of the satellite, T = 5.62 × 103 s

(b.i) Kepler’s Third Law of orbital motion states that T 2 = kr 3

where k is a constant and r is the orbital radius of the satellite.

4π
2

Show that k =
GM
. [1]

Markscheme
 2πr GM
use of v orbital =
T
AND either v orbital = √
r
or
2
mv
orbital

r
=
GM m

r
2
correctly manipulated ✓

4π
2

«to yield T 2 = (
GM
)r
3
»

Allow use of ω.

(b.ii) Determine the height of the satellite above the Earth’s surface. [2]

Markscheme

−11 24 2
2 3 6.67×10 ×5.97×10 ×(5620)
r = √
3 GM T

4π
2
= √
4π
2
✓

= 6. 83 × 10
6
«m»

6 6 5
height = «6. 83 × 10 − 6. 37 × 10 =» 4. 6 × 10 «m» ✓

2. [Maximum mark: 4] SPM.2.SL.TZ0.1

A block of mass 45 kg is placed on a horizontal table. There is no friction between
the block and the table.

An object of mass 15 kg is placed on top of the block.

A force F acts on the block so that it accelerates. The acceleration of the object
and the acceleration of the block are the same so that they do not move relative
to each other.
 The coefficient of static friction between the block and the object is 0.60.
(a) State the nature and direction of the force that accelerates the
15 kg object. [1]

Markscheme

static friction force «between blocks»

AND

directed to the right ✓

(b) Determine the largest magnitude of F for which the block and
the object do not move relative to each other. [3]

Markscheme

F = 60a ✓

Ff = 0.6 × 15 × 9.8 «= 88.2 N» ✓

⇒ F = 350 «N» ✓
F
88. 2 = 15 ×
60

Allow use of a = 0.6g leading to 353 N.

3. [Maximum mark: 20] EXE.2.SL.TZ0.2

An Alpine village uses an electric tram system to transport visitors from a lower
station up to an upper station at the village. The length of the tramline is 3.0 km
and the gradient of the tramline is a constant 10°.

The tram has a weight of 5.0 × 104 N and can carry a maximum of 75 passengers
of average weight 710 N.

The energy is supplied to each tram through a single overhead cable with a
resistance per unit length of 0.024 Ω km−1. The tram rails are used for the return
path of the current. The return path and the connections from the cable to the
electric motor in the tram have negligible resistance.

The power supply maintains a constant emf of 500 V between the rails and the
cable at the upper station.

Assume that the current through the motor is constant at 600 A and that the
motor efficiency is always 0.90 for the entire range of voltages available to the
tram.
(a) A tram is just leaving the lower railway station.

Determine, as the train leaves the lower station,

(a.i) the pd across the motor of the tram, [2]

Markscheme

Resistance of cable = 0.072 Ω ✓

Pd is (500 − 0.072 × 600) = 457 V ✓

(a.ii) the mechanical power output of the motor. [2]

Markscheme

Power input = 457 × 600 = 274 kW ✓

Power output = 0.9 × 274 = 247 kW ✓

(b) Discuss the variation in the power output of the motor with
distance from the lower station. [2]

Markscheme

The pd across the motor increases as the tram travels up the track ✓

(As the current is constant), the power output also rises ✓

(c) The total friction in the system acting on the tram is equivalent
to an opposing force of 750 N.

For one particular journey, the tram is full of passengers.

Estimate the maximum speed v of the tram as it leaves the lower

station. [4]

Markscheme

Total weight of tram = 75 × 710 + 5 × 104 = 1.03 × 105 N ✓

Total force down track = 750 + 1.03 × 105 sin (10) = 1.87 × 104 N ✓

Use of P= F × v ✓

(v = 247 000 ÷ 1.87 × 104)= 13 m s−1 ✓

(d) The tram travels at v throughout the journey. Two trams are
available so that one is returning to the lower station on
another line while the other is travelling to the village. The
journeys take the same time.

It takes 1.5 minutes to unload and 1.5 minutes to load each

tram. Ignore the time taken to accelerate the tram at the
beginning and end of the journey.

Estimate the maximum number of passengers that can be

carried up to the village in one hour. [4]

Markscheme

Time for run = s/v = 3000 ÷ 13.2 = 227 s ✓

3 minutes loading = 180 s

So one trip = 407 s ✓

And there are 3600/407 trips per hour = 8.84 ✓

So 8 complete trips with 75 = 600 passengers ✓

(e) There are eight wheels on each tram with a brake system for
each wheel. A pair of brake pads clamp firmly onto an annulus
made of steel.

The train comes to rest from speed v. Ignore the energy

transferred to the brake pads and the change in the
gravitational potential energy of the tram during the braking.

Calculate the temperature change in each steel annulus as the

tram comes to rest.

Data for this question

 The inner radius of the annulus is 0.40 m and the outer radius is
0.50 m.

The thickness of the annulus is 25 mm.

The density of the steel is 7860 kg m−3 [4]

The specific heat capacity of the steel is 420 J kg−1 K−1

Markscheme

Work leading to volume = 7.1 x 10−3 m3 ✓

Work leading to mass of steel = 55 .8 kg ✓

Kinetic Energy transferred per annulus =

1 2
mv 1 1.03×10
5
2 2
= × × 13
8 16 9.81

= 110 kJ ✓
5
Ek 1.1×10
ΔT = = = 4. 7 K ✓
mc 55.8×420

(f ) The speed of the tram is measured by detecting a beam of

microwaves of wavelength 2.8 cm reflected from the rear of the
tram as it moves away from the station. Predict the change in
wavelength of the microwaves at the stationary microwave
detector in the station. [2]

Markscheme

Δλ v
Use of λ
≈
c
✓

1.2 nm ✓

4. [Maximum mark: 13] 24N.2.SL.TZ0.1

(a) In a “loop-the-loop” toy, a car of mass 0.12 kg is released from
rest. The initial position of the car is 45 cm above level ground.
The radius of the circular loop is 15 cm. The car reaches the top
of the loop at position P. Frictional and air resistance forces are
negligible.

(a.i) Show that the speed of the car at P is 1.7 m s−1. [2]

Markscheme

1
mg × 0. 15 =
2
mv
2
✔

−1
v = √ 2 × 9. 81 × 0. 15 OR 1. 72 «m s » ✔

Award MP1 for recognition that KE of car at P is GPE lost.

Do not award MP1 for answers based on suvat equations. MP2 can still be awarded for a correct answer.

(a.ii) Determine the normal force exerted by the loop on the car at P. [3]

Markscheme

N + mg =
mv

r
✔
 2

✔
1.72
N = 0. 12 × ( − 9. 81)
0.15

N = 1. 2 «N» ✔

Allow 1.1 or 1.2 depending on g and rounding of v.

Award [0] for answers based on N = W.

(a.iii) State why the car stays in contact with the loop. [1]

Markscheme

ALTERNATIVE 1
the normal force is greater than zero ✔

ALTERNATIVE 2
−1 AND actual
minimum speed at P = √ 9. 81 × 0. 15 = 1. 2 «m s »

speed is greater ✔

Do not accept a statement of the value of N e.g. N = 1.2 «N».

(b) At point A the car collides with a block of mass 0.18 kg and
sticks to it. After the collision, the car and the block move
together with speed 1.2 m s−1.
(b.i) Calculate the speed of the car just before it collides with the
block. [2]

Markscheme

ALTERNATIVE 1
mg × 0. 45 =
1

2
mv
2
✔

−1
v =«√ 2 × 9. 81 × 0. 45» = 3. 0 «m s »✔

ALTERNATIVE 2
0.12 × v = (0.12 + 0.18) × 1.2 ✔

v= 3.0 «m s−1» ✔

Do not award BCA.

For ALT1 award MP1 for recognition that KE of car at P is GPE lost.

For ALT1 do not award MP1 for answers based on suvat equations. MP2 can still be awarded for a correct
answer.

(b.ii) The surface from A to B is rough and the combined car and
block come to rest at B. The distance AB is 0.20 m. Determine the
rate of change of momentum of the combined car and block
from A to B. [3]

Markscheme

ALTERNATIVE 1
rate of change of momentum is the net force ✔

Fd =
1

2
(m 1 + m 2 )u
2
✔
2

F =«
0.30×1.2

2×0.20
=» 1. 08 ≈ 1. 1 «N» ✔
 ALTERNATIVE 2
rate of change of momentum is the net force ✔
2 2
−2
a =« =»3. 6 «m s »✔
u 1.2
=
2d 2×0.20

F = «ma = 0.30 × 3.6 =»1.08 ≈ 1.1 «N» ✔

ALTERNATIVE 3
t =
2×0.2

1.2
OR 0. 333 «s» ✔

Δp= (0.18 + 0.12) × 1.2 OR 0.36 «Ns» ✔

Δp
F «= Δt
=
0.36

0.333
=» 1. 1 «N» ✔

Do not award BCA.

MP1 can be awarded if an equation stated implies that force is rate of change of momentum or the final
answer is clearly a force.

(b.iii) Calculate the dynamic coefficient of friction between the rough

surface and the combined car and block. [2]

Markscheme

μ =«
F

(m 1 +m 2 )g
=»
1.08

0.30×9.81
✔

μ= 0.367 ≈ 0.37 ✔

Allow ECF from (bii).

5. [Maximum mark: 9] 24N.2.SL.TZ0.3

(a.i) State what is meant by a transverse wave. [1]
 Markscheme

the displacement/oscillation is at right angles/perpendicular to the

direction of energy transfer/propagation ✔

Accept vibration/movement of particles for oscillation.

A transverse wave is travelling to the right on a string. The graph shows, at one
particular time, the variation of the displacement y with distance x along the
string. Two points, P and Q, on the string have been marked.

(a.ii) Explain whether P or Q has the greater acceleration. [2]

Markscheme

P and Q are performing SHM ✔

«in SHM» acceleration is proportional to displacement so P ✔
acceleration is a maximum at P AND zero/minimum at Q ✔
 MP2 can be expressed as an equation.

(a.iii) The speed of the wave is 62 m s−1. Calculate the period of the
wave giving your answer to the correct number of significant
figures. [2]

Markscheme

wavelength is 1.20 «m» ✔

T =« » 1. 9 × 10 «s» ✔
λ 1.20 −2 −2
= = 1. 936 × 10
c 62

For MP2 the final answer must be to 2 s.f.

Allow ECF from MP1 provided the answer is to 2 s.f.

(a.iv) Calculate the average speed of P during one complete

oscillation. [2]

Markscheme

distance travelled is 4 × 0.04 OR 0.16 «m» ✔

v =«
0.16
=» 8. 3 «m s
−1» ✔
−2
1.936×10

Allow ECF from (aiii).

Allow 8.3 or 8.4 «m s −1».

(b) The string is now stretched between an oscillator and a fixed

point. When the oscillator is set to a frequency of 120 Hz the
standing wave shown in the diagram is observed on the string.
 Draw the standing wave that would be observed on the same
string when the oscillator is set to a frequency of 180 Hz.

[1]

Markscheme

Accept either the solid or dashed line alone, or 2 solid lines.

(c) Physicists and engineers study simple harmonic oscillations

even though most oscillations are not simple harmonic.
Suggest why this is a useful thing to do. [1]

Markscheme

SHM helps understand wave motion

OR
Real/complex systems resemble SHM for small oscillations
OR
Complex oscillations can be modelled as a combination of SHM / reference
to Fourier analysis
OR
If friction is small, SHM can be a good approximation to the real system /
small damping ✔
6. [Maximum mark: 8] 24N.2.SL.TZ0.102
A small ball is released from rest at time t = 0 in front of a vertical ruler. A multi-
flash photograph is taken of the ball at t = 0 and every 0.050 s from then on.

The distance s fallen by the ball is related to the acceleration g of the ball and t by
gt2.
1
s= 2

(a) Determine g using the photograph. [3]

Markscheme
 Read at least two points correctly and consistently ✓

Use 1

2
⋅ g ⋅ t
2
with a length interval from two non-consecutive points
OR for two (or more) length intervals, using consistent time intervals ✓

Correct calculation of g ✓

Award [2] max if they use one single length interval of consecutive points.

Award [1] max if they miss to subtract the initial point in their length interval or if they use inconsistent
time intervals.

Do not penalize significant figures in the final answer.

(b) The percentage uncertainty in t is ± 5 %.

Estimate the absolute uncertainty in your value of g. [3]

Markscheme

Use 10% for delta t2 ✓

Estimate an uncertainty for s using an absolute of 0.5 OR 1 mm AND

propagate it to 1/their length interval OR 2/their length interval
respectively ✓

Add both relative or percentage uncertainties AND calculate the absolute

uncertainty ✓

Allow ECF for MP3.

Do not penalize significant figures in the final answer.

(c) Suggest a suitable value for the maximum duration of the flash. [2]
 Markscheme

Flash must be so short that blurring is within the absolute uncertainty /

error of determination of s ✓

(Final) speed is around 2 m s−1 and error is about 1 mm so (maximum)

duration around 1

2000
s✓

Award [1] max if they argue that images might merge and state a maximum of 0.025 s (or less).

Award [1] max if they use the uncertainty of time and state a time of (2 x) 0.05 x 0.05 s = 0.005 s (or less).

Award MP2 if they state an interval of 0.001 s or less.

7. [Maximum mark: 6] 24M.2.SL.TZ1.1

A stationary ball is hanging from a light string. A pellet from an air rifle is
travelling horizontally and becomes embedded in the ball. The velocity of the
pellet when it strikes the ball is 160 m s−1.

The following data are given.

Mass of the ball = 250 g

Mass of the pellet = 2.0 g

(a) Calculate the speed of the ball and the pellet immediately after
the impact. [2]

Markscheme

(2.0)(160) = (250 + 2.0)v ✔

(2.0)(160)
v =« 250 + 2.0
=»1.3 «m s−1» ✔

Award [1 max] for 1.28 m/s (mass of pellet neglected)

Award [2] for BCA

(b) Suggest why the combined kinetic energy of the ball and the
pellet after the impact is less than the initial kinetic energy of
the pellet. [2]

Markscheme

«Work is done by contact forces» to /penetrate/deform/squash/change

shape of the ball / the interaction causes deformation of the ball. ✔

«this requires energy transfer» from kinetic to other forms e.g. PE of

deformation/heat/internal ✔

Allow ‘embedded’.

Do not allow ‘inelastic collision’

MP2 requires at least one other appropriate energy form to be mentioned having been transferred from KE.
NOT ‘sound energy’

The ball with the embedded pellet rises to a maximum vertical height h.
(c) Draw and label the free-body diagram for the ball at height h.

[2]

Markscheme

arrow along the string direction line labelled tension / T ✔

arrow downwards of approximately correct length labelled weight / W / mg

✔

Allow FT or T for tension in MP1

 Allow Fg, FW, mg, or W for Weight in MP2

Do not allow “gravity” for weight in MP2

Do not allow Fc for tension in MP1

Ignore any additional forces.”

8. [Maximum mark: 5] 24M.2.SL.TZ1.2

A student throws a ball towards a wall. The ball is released from a point 1.8 m
above the ground and 8.0 m from the wall. The initial velocity of the ball makes
an angle of 48° with the horizontal. Air resistance is negligible.
The diagram shows the initial path of the ball. P is a point on the path.

(a) Draw, on the diagram, an arrow to show

(a.i) the velocity of the ball at P. Label this arrow v. [1]

Markscheme

arrow tangent to the path in the correct direction ✔

If the line when produced backwards goes below the curve – no mark.

Arrows not beginning at P score [0]

(a.ii) the acceleration of the ball at P. Label this arrow a. [1]

Markscheme

arrow vertically downwards ✔

The ball takes 1.3 s to reach the wall. The initial speed of the ball is 9.2 m s−1.

(b) Determine the height above the ground at which the ball hits
the wall. [3]

Markscheme

initial vertical velocity vy = 9.2 sin 48° «6.8 m s−1» ✔

h = 1.8 + 9.2 sin 48° × 1.3 − 1

2
× 9.8 × 1.32 ✔

2.4 «m» ✔

Award [2] for h=0.6 m (candidates have not taken the initial height of 1.8 m into account).

Award [2 max] for h=19 m

Award 3 for BCA

9. [Maximum mark: 10] 24M.2.SL.TZ2.1
A ball of mass 2.7 g is released from rest from a height of 28 m above horizontal
ground.

(a) Show that in the absence of air resistance the ball impacts the
ground with a speed of about 23 m s−1. [1]

Markscheme

v = ≪√2gh = ≫ √2 × 9. 81 × 28 OR 23.4 «m s−1» ✔

Starting point can be suvat equations or conservation of energy.

Answer must be seen to at least 3 s.f. or correct substitution shown.
Allow 23.7 for use of g = 10

(b) An air resistance force F acts on the ball. F can be modeled by F =

kv2 where v is the speed and k is a constant.

(b.i) Determine the unit of k in terms of fundamental units. [2]

Markscheme

−2
kg m s
[k] = m
N
2
s
−2
= m
2
s
−2
✔

[k] = kgm
−1
✔
 kg m s −2 OR m2 s −2 seen for MP1

(b.ii) Describe how the ball reaches terminal speed. [2]

Markscheme

ALTERNATE 1
Resistance/drag force/friction increases with speed ✔
Until it becomes equal to the weight ✔
Net force/acceleration is then zero «and so speed is constant» ✔

ALTERNATE 2
Resistance force increases with speed ✔
Until GPE lost no longer converted to KE ✔
But to thermal energy/work done against resistive force ✔

(c) The graph shows the variation with time t of the speed v of the
ball from the instant it is released until it impacts the ground.

(c.i) State the value of the area under the curve. [1]
 Markscheme

28 «m» ✔

Accept range 25 to 31 «m» for those who counted squares

Do not allow displacement or distance travelled. A value is required.

(c.ii) Determine k. [2]

Markscheme

Alternate 1

mg = kv
2 with v = 9.5 m s−1 ✔

−3

k= « =»2.9 × 10−4 ✔
mg

v
2
= 2.7×10

9.5
2
×9.81

Alternate 2

Determination of acceleration from gradient of a tangent at any point

other than t = 0 ✔

Use of mg − kv2 = ma to find k ✔

For example:
Gradient at t = 1 s is 3.9 «m s−2» and speed is 7.4 «m s−1»
−3
mg−ma 2.7×10 ×(9.81 − 3.9)
k=« v
2
= 7.4
2
=»2.9 × 10−4

Allow 3.0 × 10−4 for use of g = 10.

Allow ECF for MP2 from incorrect read off from graph.
Ignore any units on k.
 Do not award MP1 for simply calculating the gradient and equating it to k. For MP1 there needs to be a
recognition that the gradient is the acceleration.
Range for Alternate 2
(2.5 to 3.3) × 10−4

(d) The ball rebounds from the ground with speed 7.8 m s−1. The
ball is in contact with the ground for a time T. The average
resultant force on the ball during this time is 1.1 N.

Determine T. [2]

Markscheme

ALTERNATE 1

Δp = «2.7 × 10−3 × (7.8 + 9.5) =» 0.0467 «N s» ✔

Δp
«1.1 = so T = =» 0.042 «s» ✔
0.0467

Δt 1.1

ALTERNATE 2

a=« F

m
=» 1.1

2.7 × 10
−5
= 407 «m s −2» ✔

T = « 9.5407
+ 7.8
=» 0.042 «s» ✔

Watch for ECF from incorrect value of v in cii).

Award [1] for t = 0.076 «s » using an impact speed of 23 m s −1.

10. [Maximum mark: 6] 24M.2.SL.TZ2.5

A probe of mass m has landed on the equator of a rotating asteroid of mass M
and radius R.
The asteroid rotates with angular speed ω.
(a) By drawing a free-body diagram for the probe, show that the
normal force, N, on the probe from the asteroid is given by N = m(
GM

R
2
−ω2R).

[2]

Markscheme

GM m

R
2
OR mg N

Arrow for N and correct labelling ✔

−N = mω2R result follows ✔

GM m
2
R
 N arrow must be shorter than weight.
Award [0] if there are extra forces or incorrect length arrows in the diagram.
On diagram allow Weight, W, Fg, gravitational force but not gravity.
Allow rotated diagrams if the surface of the asteroid is shown.
For MP2 the algebra must follow from a diagram with correct length arrows.

(b) Another probe orbits the Sun.

(b.i) The distance between the probe and the Sun is 4 times the
distance between the Earth and the Sun. Show that the
intensity of the solar radiation at the surface of the probe is 85
W m−2. [2]

Markscheme

Selection of solar constant, 1360 seen ✔

«Intensity is proportional to d−2 so»

P

2
4π(4d )

« I

1360
= P
»
2
4πd

I = 1360

4
2
=«85.0 W m−2» ✔

Allow ECF for MP2 from incorrect solar constant.

(b.ii) Estimate the equilibrium temperature of the probe assuming it

behaves as a black body. [2]

Markscheme

σT4 = 85 OR T = √ ✔
4 85
−8
5.67 × 10
 T = 196.7 2 ≈ 2.0 × 102 «K» ✔

Award [2] for 140 K, if factor of ¼ is included

11. [Maximum mark: 12] 23N.2.SL.TZ1.1

A space probe of mass 95 kg is designed to land on the surface of an asteroid.
The gravitational field strength g of the asteroid at its surface is 2.7 × 10−3 m s−2.

(a) The radius r of the asteroid is 230 km. Calculate the mass of the
asteroid. [2]

Markscheme

2
−3 5
2
gr 2.7×10 ×(2.3×10 )
M = « G
= » −11
✔
6.67×10

« kg » ✔
18
2. 1 × 10

(b) The probe is carried to the asteroid on board a spacecraft.

Calculate the weight of the probe when close to the surface of

the asteroid. [1]

Markscheme

0.25−0.26 «N» ✔
(c.i) As the probe approaches the surface of the asteroid, a rocket
engine is fired to slow its descent. Explain how the engine
changes the speed of the probe. [3]

Markscheme

ALTERNATIVE 1
the engine exerts an upward/opposing force «on the probe» ✔
«upward» force is greater than weight/grav force OR there is an upward
resultant/net force ✔
« by NII » this causes deceleration/reduction in speed ✔

ALTERNATIVE 2
the engine/probe exerts a force on the fuel molecules/gas ✔
«by NIII» an equal and opposite force acts on the engine/probe ✔
« by NII » this causes deceleration/reduction in speed ✔

ALTERNATIVE 3
engine causes change in momentum to fuel molecules/gas ✔
«by conservation of momentum» the probe has an equal and opposite
change in momentum ✔
this results in deceleration/reduction in speed ✔

Marks may only be awarded from one alternative.

Examiners should determine which alternative provides the most marks.
MP3 must have a reduction in speed not just a change in speed

(c.ii) A constant force of 12.0 N is exerted by the rocket engine.

Determine the time for which the rocket must fire to reduce the
speed of the probe from 0.64 m s−1 to zero. State your answer
to an appropriate number of significant figures. [4]

Markscheme

ALTERNATIVE 1
 net force on probe=12−0.26=«11.7» «N» ✔

change in momentum=0.64×95=« 60.8 Ns » ✔

time=« 60.8/11.7»=5.2 to 5.3 « s » ✔

any answer to 2 s.f. ✔

ALTERNATIVE 2

net force on probe=12−0.26=« 11.7 » «N» ✔

acceleration «=F/m »=11.7/95 «=0.12 » ✔

time=« 0.64/0.12 »=5.2 to 5.3 « s » ✔

any answer to 2 s.f. ✔

Allow ECF from 1b

(d) As the probe lands, a small stone resting on a rock on the

asteroid’s surface is projected horizontally from the top of the
rock. The horizontal speed of the stone is 34 m s−1 from a
height of 1.9 m above the surface of the asteroid.

Estimate the horizontal distance from the stone’s point of

projection along the line AB at which the stone lands. Ignore
the curvature of the asteroid. [2]
 Markscheme

time to reach surface = «√ »37. 5 «s» ✔

2×1.9
−3
=
2.7×10

distance travelled horizontally=« 34 × 37.5 »=1300 «m» ✔

Award [2] for BCA

(Allow unrounded correct answers e.g. 1275)
Check units match power of ten e.g. 1.3 km scores both marks
Award [1 max] for 21 «m» (g taken as 9.81)
Watch for ECF from incorrect t.

12. [Maximum mark: 12] 23N.2.SL.TZ2.1

A space probe of mass 95 kg is designed to land on the surface of an asteroid.
The gravitational field strength g of the asteroid at its surface is 2.7 × 10−3 m s−2.

(a) The radius r of the asteroid is 230 km. Calculate the mass of the
asteroid. [2]

Markscheme

2
2 −3 5
gr 2.7×10 ×(2.3×10 )
M = « G
= » −11
✔
6.67×10

18
2. 1 × 10 « kg » ✔

(b) The probe is carried to the asteroid on board a spacecraft.

 Calculate the weight of the probe when close to the surface of [1]
the asteroid.

Markscheme

0.25−0.26 «N» ✔

(c.i) As the probe approaches the surface of the asteroid, a rocket

engine is fired to slow its descent. Explain how the engine
changes the speed of the probe. [3]

Markscheme

ALTERNATIVE 1
the engine exerts an upward/opposing force «on the probe» ✔
«upward» force is greater than weight/grav force OR there is an upward
resultant/net force ✔
« by NII » this causes deceleration/reduction in speed ✔

ALTERNATIVE 2
the engine/probe exerts a force on the fuel molecules/gas ✔
«by NIII» an equal and opposite force acts on the engine/probe ✔
« by NII » this causes deceleration/reduction in speed ✔

ALTERNATIVE 3
engine causes change in momentum to fuel molecules/gas ✔
«by conservation of momentum» the probe has an equal and opposite
change in momentum ✔
this results in deceleration/reduction in speed ✔

Marks may only be awarded from one alternative.

Examiners should determine which alternative provides the most marks.
MP3 must have a reduction in speed not just a change in speed

(c.ii) A constant force of 12.0 N is exerted by the rocket engine.

Determine the time for which the rocket must fire to reduce the
 speed of the probe from 0.64 m s−1 to zero. State your answer
to an appropriate number of significant figures. [4]

Markscheme

ALTERNATIVE 1

net force on probe=12−0.26=«11.7» «N» ✔

change in momentum=0.64×95=« 60.8 Ns » ✔

time=« 60.8/11.7»=5.2 to 5.3 « s » ✔

any answer to 2 s.f. ✔

ALTERNATIVE 2

net force on probe=12−0.26=« 11.7 » «N» ✔

acceleration «=F/m »=11.7/95 «=0.12 » ✔

time=« 0.64/0.12 »=5.2 to 5.3 « s » ✔

any answer to 2 s.f. ✔

Allow ECF from 1b

(d) As the probe lands, a small stone resting on a rock on the

asteroid’s surface is projected horizontally from the top of the
rock. The horizontal speed of the stone is 34 m s−1 from a
height of 1.9 m above the surface of the asteroid.
 [2]

Estimate the horizontal distance from the stone’s point of

projection along the line AB at which the stone lands. Ignore
the curvature of the asteroid.

Markscheme

time to reach surface = «√ »37. 5 «s» ✔

2×1.9
−3
=
2.7×10

distance travelled horizontally=« 34 × 37.5 »=1300 «m» ✔

Award [2] for BCA

(Allow unrounded correct answers e.g. 1275)
Check units match power of ten e.g. 1.3 km scores both marks
Award [1 max] for 21 «m» (g taken as 9.81)
Watch for ECF from incorrect t.

© International Baccalaureate Organization, 2025