2.

1 QP+MS IB [168 marks]

1. A large stone is dropped from a tall building. What is correct about the [1 mark]
speed of the stone after 1 s?
A. It is decreasing at increasing rate.
B. It is decreasing at decreasing rate.
C. It is increasing at increasing rate.
D. It is increasing at decreasing rate.

Markscheme
D

Examiners report
[N/A]

2. The graph shows how the position of an object varies with time in the [1 mark]
interval from 0 to 3 s.

At which point does the instantaneous speed of the object equal its average speed
over the interval from 0 to 3 s?

Markscheme
C
 Examiners report
[N/A]

3. The minute hand of a clock hanging on a vertical wall has length [1 mark]
L = 30 cm.

The minute hand is observed pointing at 12 and then again 30 minutes later when
the minute hand is pointing at 6.
What is the average velocity and average speed of point P on the minute hand
during this time interval?

Markscheme
A

Examiners report
[N/A]
4. A projectile is launched at an angle above the horizontal with a horizontal [1 mark]
component of velocity Vh and a vertical component of velocity Vv . Air
resistance is negligible. Which graphs show the variation with time of Vh and of Vv
?

Markscheme
D

Examiners report
[N/A]
 Two players are playing table tennis. Player A hits the ball at a height of 0.24 m
above the edge of the table, measured from the top of the table to the bottom of
the ball. The initial speed of the ball is 12.0 m s−1 horizontally. Assume that air
resistance is negligible.

5a. Show that the time taken for the ball to reach the surface of the table is [1 mark]
about 0.2 s.

Markscheme
t = «√ 2gd =» 0.22 «s»
OR

t = √ 2×0.24
9.8
✓

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

correct values.

Examiners report
[N/A]
5b. Sketch, on the axes, a graph showing the variation with time of the [2 marks]
vertical component of velocity vv of the ball until it reaches the table
surface. Take g to be +10 m s−2.

Markscheme
increasing straight line from zero up to 0.2 s in x-axis ✓
with gradient = 10 ✓

Examiners report
[N/A]
5c. The net is stretched across the middle of the table. The table has a [3 marks]
length of 2.74 m and the net has a height of 15.0 cm.
Show that the ball will go over the net.

Markscheme
ALTERNATIVE 1
1.37
t= 12 =«0.114 s» ✓
1
y= 2 × 10 × 0. 1142 = 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 = 12 × 10 × t2 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.

Examiners report
[N/A]
 The ball bounces and then reaches a peak height of 0.18 m above the table with a
horizontal speed of 10.5 m s−1. The mass of the ball is 2.7 g.

5d. Determine the kinetic energy of the ball immediately after the bounce. [2 marks]

Markscheme
ALTERNATIVE 1
KE = 12 mv2 + mgh = 12 0.0027 ×10.52 + 0.0027 × 9.8 × 0.18 ✓
0.15 «J» ✓
ALTERNATIVE 2
Use of vx = 10.5 AND vy = 1.88 to get v = «√10. 52 + 1. 882 » = 10.67 «m 
s−1» ✓
KE = 12 × 0.0027 × 10.672 = 0.15 «J» ✓

Examiners report
[N/A]
5e. Player B intercepts the ball when it is at its peak height. Player B holds a [3 marks]
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.

Markscheme
Δv = 21 «m s−1» ✓
0.0027×21
F= 0.01
OR
5.67 «N» ✓
any answer to 2 significant figures «N» ✓

Examiners report
[N/A]
 A football player kicks a stationary ball of mass 0.45 kg towards a wall. The initial
speed of the ball after the kick is 19 m s−1 and the ball does not rotate. Air
resistance is negligible and there is no wind.

6a. The player’s foot is in contact with the ball for 55 ms. Calculate the [2 marks]
average force that acts on the ball due to the football player.

Markscheme
19
Δp = 0. 45 × 19ORa = 0.055
✓
0.45×19
«=F = 0.055
»160«N» ✓
Allow [2] marks for a bald correct answer.
Allow ECF for MP2 if 19 sin22 OR 19 cos22 used.

Examiners report
[N/A]
6b. The ball leaves the ground at an angle of 22°. The horizontal distance [2 marks]
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.

Markscheme
horizontal speed =19 × cos 22« = 17. 6 m s−1 » ✓
11
time = « distance
speed
= 19cos22
= »0. 62 «s» ✓
Allow ECF for MP2

Examiners report
[N/A]
6c. The top of the wall is 2.4 m above the ground. Deduce whether the ball [3 marks]
will hit the wall.

Markscheme
initial vertical speed = 19 × sin 22« = 7. 1 m s−1 » ✓
«7. 12 × 0. 624 − 0. 5 × 9. 81 × 0. 6242 = »2. 5 «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

Examiners report
[N/A]
6d. In practice, air resistance affects the ball. Outline the effect that air [2 marks]
resistance has on the vertical acceleration of the ball. Take the direction
of the acceleration due to gravity to be positive.

Markscheme
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

Examiners report
[N/A]

6e. The player kicks the ball again. It rolls along the ground without sliding [1 mark]
−1
with a horizontal velocity of 1. 40 m s . The radius of the ball is 0. 11 m .
Calculate the angular velocity of the ball. State an appropriate SI unit for your
answer.
Markscheme
13 «rad» s−1 ✓

Unit must be seen for mark

Accept Hz
Accept 4π «rad» s−1

Examiners report
[N/A]
 A vertical wall carries a uniform positive charge on its surface. This produces a
uniform horizontal electric field perpendicular to the wall. A small, positively-
charged ball is suspended in equilibrium from the vertical wall by a thread of
negligible mass.

7a. The charge per unit area on the surface of the wall is σ. It can be shown [2 marks]
that the electric field strength E due to the charge on the wall is given by
the equation
σ
E= 2ε 0
.

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

Markscheme
identifies units of σ as Cm−2 ✓
Nm2
C
m2
× C2
seen and reduced to N C−1 ✓
Accept any analysis (eg dimensional) that yields answer correctly

Examiners report
[N/A]
7b. The thread makes an angle of 30° with the vertical wall. The ball has a [3 marks]
mass of 0.025 kg.
Determine the horizontal force that acts on the ball.

Markscheme
horizontal force F on ball = T sin 30 ✓
mg
T= cos30
✓

F « = mg tan 30 = 0. 025 × 9. 8 × tan 30» = 0. 14«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.

Examiners report
[N/A]

−6 
7c. The charge on the ball is 1.2 × 10−6 C. Determine σ. [2 marks]

Markscheme
0.14
E= « = 1. 2 × 105 » ✓
1.2×10−6
−12
σ = « 2×8.85×10 ×0.14
» = 2. 1 × 10−6 «Cm−2 » ✓
1.2×10−6

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

Award [2] for a bald correct answer.

Examiners report
[N/A]
7d. The thread breaks. Explain the initial subsequent motion of the ball. [3 marks]

Markscheme
horizontal/repulsive force and vertical force/pull of gravity act on the ball ✓
so ball has constant acceleration/constant net force ✓
motion is in a straight line ✓
at 30° to vertical away from wall/along original line of thread ✓

Examiners report
[N/A]

−6 
 The centre of the ball, still carrying a charge of 1.2 × 10−6 C, is now placed 0.40 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 m.

7e. Calculate the charge on Q. State your answer to an appropriate number [3 marks]
of significant figures.

Markscheme
Q 1.2×10−6
2 = ✓
0.22 0.182
−6
« + »1. 8 × 10 «C»✓
2sf ✓

Do not award MP2 if charge is negative

Any answer given to 2 sig figs scores MP3

Examiners report
[N/A]
7f. Outline, without calculation, whether or not the electric potential at P is [2 marks]
zero.

Markscheme
work must be done to move a «positive» charge from infinity to P «as both
charges are positive»
OR
reference to both potentials positive and added
OR
identifies field as gradient of potential and with zero value ✓
therefore, point P is at a positive / non-zero potential ✓

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

Examiners report
[N/A]

8. A horizontal force F acts on a sphere. A horizontal resistive force kv2 acts [1 mark]
on the sphere where v is the speed of the sphere and k is a constant. What is the
terminal velocity of the sphere?

A. √ Fk

B. k
F
C. F
k

D. √ Fk
 Markscheme
D

Examiners report
[N/A]

9. P and Q leave the same point, travelling in the same direction. The [1 mark]
graphs show the variation with time t of velocity v for both P and Q.

What is the distance between P and Q when t = 8. 0 s?

A. 20 m
B. 40 m
C. 60 m
D. 120 m

Markscheme
B

Examiners report
[N/A]
10. A balloon rises at a steady vertical velocity of 10 m s−1 . An object is [1 mark]
dropped from the balloon at a height of 40 m above the ground. Air resistance is
negligible. What is the time taken for the object to hit the ground?
A. 10 s
B. 5 s
C. 4 s
D. 2 s

Markscheme
C

Examiners report
Even though over half the candidates are choosing the correct response it has
a low discrimination index. Many are choosing D indicating that they forgot to
take the velocity upward as negative.

11. The variation with time t of the acceleration a of an object is shown. [1 mark]

What is the change in velocity of the object from t = 0 to t = 6 s?

A. 6 m s–1
B. 8 m s–1
C. 10 m s–1
D. 14 m s–1

Markscheme
C
 Examiners report
[N/A]

12. A ball falls from rest in the absence of air resistance. The position of the [1 mark]
centre of the ball is determined at one-second intervals from the instant
at which it is released. What are the distances, in metres, travelled by the centre
of the ball during each second for the first 4.0 s of the motion?
A. 5, 10, 15, 20
B. 5, 15, 25, 35
C. 5, 20, 45, 80
D. 5, 25, 70, 150

Markscheme
B

Examiners report
[N/A]

13. An object is thrown from a cliff at an angle to the horizontal. The ground [1 mark]
below the cliff is horizontal.
Three quantities are known about this motion.
I. The horizontal component of the initial velocity of the object
II. The initial angle between the velocity of the object and the horizontal
III. The height of the cliff
What are the quantities that must be known in order to determine the horizontal
distance from the point of projection to the point at which the object hits the
ground?

A. I and II only
B. I and III only
C. II and III only
D. I, II and III
 Markscheme
D

Examiners report
[N/A]

The graph shows the variation with time t of the horizontal force F exerted on a
tennis ball by a racket.

The tennis ball was stationary at the instant when it was hit. The mass of the
tennis ball is 5.8 × 10–2 kg. The area under the curve is 0.84 N s.

14a. Calculate the speed of the ball as it leaves the racket. [2 marks]
 Markscheme
links 0.84 to Δ p ✔
0.84
v =« =» 14.5 «m s–1»✔
5.8×10−2
NOTE: Award [2] for bald correct answer

Examiners report
[N/A]

14b. Show that the average force exerted on the ball by the racket is about [2 marks]
50 N.

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

¯¯¯ =« Δp =»
F 0.84
OR 53 «N» ✔
Δt 16×10−3
NOTE: Accept a time interval from 14 to 16 ms
Allow ECF from incorrect time interval

Examiners report
[N/A]
14c. Determine, with reference to the work done by the average force, the [3 marks]
horizontal distance travelled by the ball while it was in contact with the
racket.

Markscheme
Ek = 12 × 5.8 × 10–2 × 14.52 ✔
Ek =W ✔
1
×5.8×10−2×14.52
s = «W = 2
=» 0.12 « m » ✔
F 53

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

Examiners report
[N/A]
14d. Draw a graph to show the variation with t of the horizontal speed v of [2 marks]
the ball while it was in contact with the racket. Numbers are not
required on the axes.

Markscheme

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

Examiners report
[N/A]

5
 The air in a kitchen has pressure 1.0 × 105 Pa and temperature 22°C. A
refrigerator of internal volume 0.36 m3 is installed in the kitchen.

15a. With the door open the air in the refrigerator is initially at the same [2 marks]
temperature and pressure as the air in the kitchen. Calculate the
number of molecules of air in the refrigerator.

Markscheme
pV 1.0×105×0.36
N= OR N= ✔
kT 1.38×10−23×295

N = 8. 8 × 1024 ✔
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.

Examiners report
[N/A]

The refrigerator door is closed. The air in the refrigerator is cooled to 5.0°C and
the number of air molecules in the refrigerator stays the same.

15b. Determine the pressure of the air inside the refrigerator. [2 marks]
 Markscheme
p nR T
use of = constant OR p= V
OR NkT
V
✔
T
4
p = 9. 4 × 10 « Pa »✔
NOTE: Allow ECF from (a)
Award [2] for bald correct answer

Examiners report
[N/A]

15c. The door of the refrigerator has an area of 0.72 m2. Show that the [2 marks]
minimum force needed to open the refrigerator door is about 4 kN.

Markscheme
F = A × Δp ✔
F = 0. 72 ×(1. 0 − 0. 94)×105 OR 4.3 × 103 « N »✔
NOTE: Allow ECF from (b)(i)
Allow ECF from MP1

Examiners report
[N/A]
15d. Comment on the magnitude of the force in (b)(ii). [2 marks]

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

Examiners report
[N/A]

An electron is placed at a distance of 0.40 m from a fixed point charge of –6.0 mC.

16a. Show that the electric field strength due to the point charge at the [2 marks]
position of the electron is 3.4 × 108 N C–1.

Markscheme
k×q
E= r2
✔
8.99×109×6.0×10−3
E= OR E = 3. 37 × 108 «N C−1 » ✔
0.42
NOTE: Ignore any negative sign.
 Examiners report
[N/A]

16b. Calculate the magnitude of the initial acceleration of the electron. [2 marks]

Markscheme
F = q × E OR F = 1. 6 × 10−19 × 3. 4 × 108 = 5. 4 × 10−11 «N» ✔
−11
a = « 5.4×10−31 = » 5. 9 × 1019 « m s−2 »✔
9.1×10
NOTE: Ignore any negative sign.
Award [1] for a calculation leading to a = « m s−2 »
Award [2] for bald correct answer

Examiners report
[N/A]
16c. Describe the subsequent motion of the electron. [3 marks]

Markscheme
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

Examiners report
[N/A]
17. A sky diver is falling at terminal speed when she opens her parachute. [1 mark]
What are the direction of her velocity vector and the direction of her
acceleration vector before she reaches the new terminal speed?

Markscheme
C

Examiners report
[N/A]

18. A stone is thrown downwards from the edge of a cliff with a speed of 5.0 [1 mark]
m s–1. It hits the ground 2.0 s later. What is the height of the cliff?
A. 20 m
B. 30 m
C. 40 m
D. 50 m
 Markscheme
B

Examiners report
This question was well answered by the majority of candidates and had a high
discrimination index.

19. A ball is thrown upwards at an angle to the horizontal. Air resistance is [1 mark]
negligible. Which statement about the motion of the ball is correct?
A. The acceleration of the ball changes during its flight.
B. The velocity of the ball changes during its flight.
C. The acceleration of the ball is zero at the highest point.
D. The velocity of the ball is zero at the highest point.

Markscheme
B

Examiners report
Candidate responses were divided between responses B (correct), D, and to a
lesser extent, C. Many candidates appeared to focus on vertical velocity only
or confused vertical velocity and acceleration values. This question had the
highest discrimination index, suggesting that it would be a useful question for
class discussion.
20. The graph shows the variation of velocity of a body with time along a [1 mark]
straight line.

What is correct for this graph?

A. The maximum acceleration is at P.
B. The average acceleration of the body is given by the area enclosed by the
graph and time axis.
C. The maximum displacement is at Q.
D. The total displacement of the body is given by the area enclosed by the graph
and time axis.

Markscheme
D

Examiners report
[N/A]
21. A sports car is accelerated from 0 to 100 km per hour in 3 s. What is the [1 mark]
acceleration of the car?
A. 0.1 g
B. 0.3 g
C. 0.9 g
D. 3 g

Markscheme
C

Examiners report
Response D was the most common (but incorrect) response, with candidates
neglecting to convert km/h to m/s.

22. A girl throws an object horizontally at time t = 0. Air resistance can be [1 mark]
ignored. At t = 0.50 s the object travels horizontally a distance x in
metres while it falls vertically through a distance y in metres.
What is the initial velocity of the object and the vertical distance fallen at t = 1.0
s?
 Markscheme
D

Examiners report
The correct response (D) was the most common selection by a minority of
candidates, with incorrect responses being roughly equally distributed among
the remaining options. This question has one of the highest discrimination
indexes.

23. A boy throws a ball horizontally at a speed of 15 m s-1 from the top of a [1 mark]
cliff that is 80 m above the surface of the sea. Air resistance is negligible.
What is the distance from the bottom of the cliff to the point where the ball lands
in the sea?
A. 45 m
B. 60 m
C. 80 m
D. 240 m

Markscheme
B

Examiners report
[N/A]
 A student strikes a tennis ball that is initially at rest so that it leaves the racquet
at a speed of 64 m s–1. The ball has a mass of 0.058 kg and the contact between
the ball and the racquet lasts for 25 ms.

24a. Calculate the average force exerted by the racquet on the ball. [2 marks]

Markscheme
F= Δ mv /m Δ v / 0.058×64.0 ✔
Δt Δt 25×10−3
F = 148«N»≈150«N» ✔

Examiners report
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.

24b. Calculate the average power delivered to the ball during the impact. [2 marks]
 Markscheme
ALTERNATIVE 1
1 2 1 2
2 mv 2 ×0.058×64.0
P= t
/ −3 ✔
25×10
P = 4700/4800«W» ✔
ALTERNATIVE 2
64.0
P = averageF v/148 × 2 ✔

P = 4700/4800«W» ✔

Examiners report
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.

The student strikes the tennis ball at point P. The tennis ball is initially directed at
an angle of 7.00° to the horizontal.

The following data are available.

Height of P = 2.80 m
Distance of student from net = 11.9 m
Height of net = 0.910 m
Initial speed of tennis ball = 64 m s-1

24c. Calculate the time it takes the tennis ball to reach the net. [2 marks]
 Markscheme
horizontal component of velocity is 64.0 × cos7° = 63.52 «ms−1» ✔
11.9
t = « 63.52 = »0.187/0.19«s» ✔
Do not award BCA. Check working.
Do not award ECF from using 64 m s-1.

Examiners report
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.

24d. Show that the tennis ball passes over the net. [3 marks]
 Markscheme
ALTERNATIVE 1
uy = 64 sin7/7.80 «ms−1»✔
decrease in height = 7.80 × 0.187 + 12 × 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 + 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

Examiners report
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.

24e. Determine the speed of the tennis ball as it strikes the ground. [2 marks]
 Markscheme
ALTERNATIVE 1
Initial KE + PE = final KE /
1 2
2 × 0.058 × 64 + 0.058 × 9.81 × 2.80 =
1
2 × 0.058 × v2 ✔
v = 64.4 «ms−1» ✔
ALTERNATIVE 2
vv = «√7.82 + 2 × 9.81 × 2.8» = 10.8«ms−1 » ✔
« v = √63.52 + 10.82 »
v = 64.4«ms−1 » ✔

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

24f. The student models the bounce of the tennis ball to predict the angle θ [3 marks]
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.
 Markscheme
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 ✔

Examiners report
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.

A student strikes a tennis ball that is initially at rest so that it leaves the racquet
at a speed of 64 m s–1. The ball has a mass of 0.058 kg and the contact between
the ball and the racquet lasts for 25 ms.

25a. Calculate the average force exerted by the racquet on the ball. [2 marks]

Markscheme
F= Δ mv /m Δ v / 0.058×64.0 ✔
Δt Δt 25×10−3
F = 148 «N»≈150«N» ✔
 Examiners report
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.

25b. Calculate the average power delivered to the ball during the impact. [2 marks]

Markscheme
ALTERNATIVE 1
1 2 1 2
2 mv 2 ×0.058×64.0
P= t
/ −3 ✔
25×10
P = 4700/4800«W» ✔
ALTERNATIVE 2
64.0
P = averageF v/ 148 × 2 ✔

P = 4700/4800«W»✔

Examiners report
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.
 The student strikes the tennis ball at point P. The tennis ball is initially directed at
an angle of 7.00° to the horizontal.

The following data are available.

Height of P = 2.80 m
Distance of student from net = 11.9 m
Height of net = 0.910 m
Initial speed of tennis ball = 64 m s-1

25c. Calculate the time it takes the tennis ball to reach the net. [2 marks]

Markscheme
horizontal component of velocity is 64.0 × cos 7∘ = 63.52«ms−1 » ✔
t =« 63.52
11.9
»0.187/0.19 «s» ✔

Examiners report
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.
25d. Show that the tennis ball passes over the net. [3 marks]

Markscheme
ALTERNATIVE 1
uy=64sin7/7.80«ms –1» ✔
decrease in height = 7.80 × 0.187 + 12 × 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» ✔
1
time to fall this distance found using «1.89 = 7.8 t + 2 × 9.81 × t 2»

t = 0.21«s» ✔

0.21«s» > 0.187«s» ✔

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

25e. Determine the speed of the tennis ball as it strikes the ground. [2 marks]

Markscheme
ALTERNATIVE 1
Initial KE + PE = final KE /
1
2 × 0.058 × 642 + 0.058 × 9.81 × 2.80 = 1
2 × 0.058 × v2 ✔
v = 64.4«ms−1 » ✔

ALTERNATIVE 2
vv =«√7.82 + 2 × 9.81 × 2.8» = 10.8«ms−1» ✔
«v = √63.52 + 10.82»

v = 64.4«ms−1» ✔
 Examiners report
This proved difficult for candidates at both HL and SL. Many managed to
calculate the final vertical component of the velocity of the ball.

25f. A student models the bounce of the tennis ball to predict the angle θ at [3 marks]
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.

Markscheme
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 ✔
 Examiners report
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.

26. A truck has an initial speed of 20 m s–1. It decelerates at 4.0 m s–2. What is [1 mark]
the distance taken by the truck to stop?

A. 2.5 m
B. 5.0 m
C. 50 m
D. 100 m

Markscheme
C

Examiners report
[N/A]
27. A projectile is fired at an angle to the horizontal. Air resistance is [1 mark]
negligible. The path of the projectile is shown.

Which gives the magnitude of the horizontal component and the magnitude of
the vertical component of the velocity of the projectile between O and P?

Markscheme
A

Examiners report
[N/A]
28. A runner starts from rest and accelerates at a constant rate throughout a [1 mark]
race. Which graph shows the variation of speed v of the runner with
distance travelled s?

Markscheme
A

Examiners report
[N/A]
29. A projectile is fired at an angle to the horizontal. The path of the projectile [1 mark]
is shown.

Which gives the magnitude of the horizontal component and the magnitude of
the vertical component of the velocity of the projectile between O and P?

Markscheme
A

Examiners report
[N/A]
 Ion-thrust engines can power spacecraft. In this type of engine, ions are created in
a chamber and expelled from the spacecraft. The spacecraft is in outer space
when the propulsion system is turned on. The spacecraft starts from rest.

The mass of ions ejected each second is 6.6 × 10–6 kg and the speed of each ion is
5.2 × 104 m s–1. The initial total mass of the spacecraft and its fuel is 740 kg.
Assume that the ions travel away from the spacecraft parallel to its direction of
motion.

30a. Determine the initial acceleration of the spacecraft. [2 marks]

Markscheme
change in momentum each second = 6.6 × 10−6 × 5.2 × 104 «= 3.4 ×
10−1 kg m s−1» ✔
3.4×10−1
acceleration = « 740
=» 4.6 × 10−4 «m s−2» ✔

Examiners report
[N/A]
 An initial mass of 60 kg of fuel is in the spacecraft for a journey to a planet. Half of
the fuel will be required to slow down the spacecraft before arrival at the
destination planet.

30b. Estimate the maximum speed of the spacecraft. [2 marks]

Markscheme
ALTERNATIVE 1:
(considering the acceleration of the spacecraft)
30
time for acceleration = = «4.6 × 106» «s» ✔
6.6×10−6
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» ✔

Examiners report
[N/A]
30c. Outline why scientists sometimes use estimates in making calculations. [1 mark]

Markscheme
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 ✔

Examiners report
[N/A]
 In practice, the ions leave the spacecraft at a range of angles as shown.

30d. Outline why the ions are likely to spread out. [2 marks]

Markscheme
ions have same (sign of) charge ✔
ions repel each other ✔

Examiners report
[N/A]
30e. Explain what effect, if any, this spreading of the ions has on the [2 marks]
acceleration of the spacecraft.

Markscheme
the forces between the ions do not affect the force on the spacecraft. ✔
there is no effect on the acceleration of the spacecraft. ✔

Examiners report
[N/A]

On arrival at the planet, the spacecraft goes into orbit as it comes into the
gravitational field of the planet.

30f. Outline what is meant by the gravitational field strength at a point. [2 marks]

Markscheme
force per unit mass ✔
acting on a small/test/point mass «placed at the point in the field» ✔
 Examiners report
[N/A]

30g. Newton’s law of gravitation applies to point masses. Suggest why the law[1 mark]
can be applied to a satellite orbiting a spherical planet of uniform
density.

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

Examiners report
[N/A]
 Ion-thrust engines can power spacecraft. In this type of engine, ions are created in
a chamber and expelled from the spacecraft. The spacecraft is in outer space
when the propulsion system is turned on. The spacecraft starts from rest.

The mass of ions ejected each second is 6.6 × 10–6 kg and the speed of each ion is
5.2 × 104 m s–1. The initial total mass of the spacecraft and its fuel is 740 kg.
Assume that the ions travel away from the spacecraft parallel to its direction of
motion.

31a. Determine the initial acceleration of the spacecraft. [2 marks]

Markscheme
change in momentum each second = 6.6 × 10−6 × 5.2 × 104 «= 3.4 ×
10−1 kg m s−1» ✔
3.4×10−1
acceleration = « 740
=» 4.6 × 10−4 «m s−2» ✔

Examiners report
[N/A]
 An initial mass of 60 kg of fuel is in the spacecraft for a journey to a planet. Half of
the fuel will be required to slow down the spacecraft before arrival at the
destination planet.

31b. (i) Estimate the maximum speed of the spacecraft. [3 marks]

(ii) Outline why the answer to (i) is an estimate.
 Markscheme
(i) ALTERNATIVE 1:
(considering the acceleration of the spacecraft)
30
time for acceleration = = «4.6 × 106» «s» ✔
6.6×10−6
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» ✔

(ii) as fuel is consumed total mass changes/decreases so acceleration

changes/increases
OR
external forces (such as gravitational) can act on the spacecraft so
acceleration isn’t constant ✔

Examiners report
[N/A]

31c. Outline why scientists sometimes use estimates in making calculations. [1 mark]
 Markscheme
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 ✔

Examiners report
[N/A]

In practice, the ions leave the spacecraft at a range of angles as shown.

31d. Outline why the ions are likely to spread out. [2 marks]

Markscheme
ions have same (sign of) charge ✔
ions repel each other ✔
 Examiners report
[N/A]

31e. Explain what effect, if any, this spreading of the ions has on the [2 marks]
acceleration of the spacecraft.

Markscheme
the forces between the ions do not affect the force on the spacecraft. ✔
there is no effect on the acceleration of the spacecraft. ✔

Examiners report
[N/A]

32. An object is projected vertically upwards at time t = 0. Air resistance is [1 mark]

negligible. The object passes the same point above its starting position at
times 2 s and 8 s.
If g = 10 m s–2, what is the initial speed of the object?
A. 50
B. 30
C. 25
D. 4
 Markscheme
A

Examiners report
[N/A]

33. The distances between successive positions of a moving car, measured at [1 mark]
equal time intervals, are shown.

The car moves with

A. acceleration that increases linearly with time.
B. acceleration that increases non-linearly with time.
C. constant speed.
D. constant acceleration.

Markscheme
D

Examiners report
[N/A]
34. A boy runs along a straight horizontal track. The graph shows how his [1 mark]
speed v varies with time t.

After 15 s the boy has run 50 m. What is his instantaneous speed and his average
speed when t = 15 s?

Markscheme
C

Examiners report
[N/A]

35. Two balls X and Y with the same diameter are fired horizontally with the [1 mark]
same initial momentum from the same height above the ground. The
mass of X is greater than the mass of Y. Air resistance is negligible.
What is correct about the horizontal distances travelled by X and Y and the times
taken by X and Y to reach the ground?
 Markscheme
C

Examiners report
[N/A]

36. A parachutist of total mass 70 kg is falling vertically through the air at a [1 mark]
constant speed of 8 m s–1.
What is the total upward force acting on the parachutist?
A. 0N
B. 70 N
C. 560 N
D. 700 N

Markscheme
D

Examiners report
[N/A]

37. A ball starts from rest and moves horizontally. Six positions of the ball are [1 mark]
shown at time intervals of 1.0 ms. The horizontal distance between X, the
initial position, and Y, the final position, is 0.050 m.

What is the average acceleration of the ball between X and Y?

A. 2000 m s–2
B. 4000 m s–2
C. 5000 m s–2
D. 8000 m s–2
 Markscheme
B

Examiners report
[N/A]

An elastic climbing rope is tested by fixing one end of the rope to the top of a
crane. The other end of the rope is connected to a block which is initially at
position A. The block is released from rest. The mass of the rope is negligible.

The unextended length of the rope is 60.0 m. From position A to position B, the
block falls freely.

38a. At position B the rope starts to extend. Calculate the speed of the block [2 marks]
at position B.
 Markscheme
use of conservation of energy
OR
v2 = u2 + 2as
v = «√2 × 60.0 × 9.81 » = 34.3 «ms–1»
[2 marks]

Examiners report
[N/A]

At position C the speed of the block reaches zero. The time taken for the block to
fall between B and C is 0.759 s. The mass of the block is 80.0 kg.

38b. Determine the magnitude of the average resultant force acting on the [2 marks]
block between B and C.
 Markscheme
use of impulse Fave × Δt = Δp
OR
use of F = ma with average acceleration
OR
F = 80.0×34.3
0.759
3620«N»
Allow ECF from (a).
[2 marks]

Examiners report
[N/A]

38c. Sketch on the diagram the average resultant force acting on the block [2 marks]
between B and C. The arrow on the diagram represents the weight of
the block.
 Markscheme
upwards
clearly longer than weight
For second marking point allow ECF from (b)(i) providing line is upwards.
[2 marks]

Examiners report
[N/A]

38d. Calculate the magnitude of the average force exerted by the rope on [2 marks]
the block between B and C.

Markscheme
3620 + 80.0 × 9.81
4400 «N»
Allow ECF from (b)(i).
[2 marks]

Examiners report
[N/A]
 For the rope and block, describe the energy changes that take place

38e. between A and B. [1 mark]

Markscheme
(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]

Examiners report
[N/A]

38f. between B and C. [1 mark]

Markscheme
(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]
 Examiners report
[N/A]

38g. The length reached by the rope at C is 77.4 m. Suggest how energy [2 marks]
considerations could be used to determine the elastic constant of the
rope.

Markscheme
k can be determined using EPE = 12 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]

Examiners report
[N/A]
39a. At position B the rope starts to extend. Calculate the speed of the block [2 marks]
at position B.

Markscheme
use of conservation of energy
OR
v2 = u2 + 2as
v = «√2 × 60.0 × 9.81 » = 34.3 «ms–1»
[2 marks]

Examiners report
[N/A]

39b. Determine the magnitude of the average resultant force acting on the [2 marks]
block between B and C.
 Markscheme
use of impulse Fave × Δt = Δp
OR
use of F = ma with average acceleration
OR
F = 80.0×34.3
0.759
3620«N»
Allow ECF from (a).
[2 marks]

Examiners report
[N/A]

39c. Sketch on the diagram the average resultant force acting on the block [2 marks]
between B and C. The arrow on the diagram represents the weight of
the block.
 Markscheme
upwards
clearly longer than weight
For second marking point allow ECF from (b)(i) providing line is upwards.
[2 marks]

Examiners report
[N/A]

39d. Calculate the magnitude of the average force exerted by the rope on [2 marks]
the block between B and C.

Markscheme
3620 + 80.0 × 9.81
4400 «N»
Allow ECF from (b)(i).
[2 marks]

Examiners report
[N/A]
39e. between A and B. [1 mark]

Markscheme
(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]

Examiners report
[N/A]

39f. between B and C. [1 mark]

Markscheme
(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]

Examiners report
[N/A]
39g. The length reached by the rope at C is 77.4 m. Suggest how energy [2 marks]
considerations could be used to determine the elastic constant of the
rope.

Markscheme
k can be determined using EPE = 12 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]

Examiners report
[N/A]
 An elastic climbing rope is tested by fixing one end of the rope to the top of a
crane. The other end of the rope is connected to a block which is initially at
position A. The block is released from rest. The mass of the rope is negligible.

The unextended length of the rope is 60.0 m. From position A to position B, the
block falls freely.

In another test, the block hangs in equilibrium at the end of the same elastic rope.
The elastic constant of the rope is 400 Nm –1. The block is pulled 3.50 m vertically
below the equilibrium position and is then released from rest.

39h. Calculate the time taken for the block to return to the equilibrium [2 marks]
position for the first time.

Markscheme
T = 2π√ 80.0
400
= 2.81 «s»

time = T4 = 0.702 «s»

Award [0] for kinematic solutions that assume a constant acceleration.

[2 marks]

Examiners report
[N/A]
39i. Calculate the speed of the block as it passes the equilibrium position. [2 marks]

Markscheme
ALTERNATIVE 1
2π
ω = 2.81 = 2.24 «rad s–1»

v = 2.24 × 3.50 = 7.84 «ms–1»

ALTERNATIVE 2
1 2 1 2 1 2 1 2
2 kx = 2 mv OR 2 400 × 3.5 = 2 80v
v = 7.84 «ms–1»
Award [0] for kinematic solutions that assume a constant acceleration.
Allow ECF for T from (e)(i).
[2 marks]

Examiners report
[N/A]

40. The variation of the displacement of an object with time is shown on a [1 mark]
graph. What does the area under the graph represent?
A. No physical quantity
B. Velocity
C. Acceleration
D. Impulse
 Markscheme
A

Examiners report
[N/A]

41. An object is thrown upwards. The graph shows the variation with time t of [1 mark]
the velocity v of the object.

What is the total displacement at a time of 1.5 s, measured from the point of
release?
A. 0 m
B. 1.25 m
C. 2.50 m
D. 3.75 m

Markscheme
B

Examiners report
[N/A]
 42. An object is released from a stationary hot air balloon at height h above [1 mark]
the ground.
An identical object is released at height h above the ground from another balloon
that is rising at constant speed. Air resistance is negligible. What does not
increase for the object released from the rising balloon?
A. The distance through which it falls
B. The time taken for it to reach the ground
C. The speed with which it reaches the ground
D. Its acceleration

Markscheme
D

Examiners report
[N/A]

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