Physics Educators’ Network of Trinidad and Tobago
Docendo Discimus By Teaching we Learn
63 Scotland Drive, Cocoyea, San Fernando.
Phone: 795-5633.
Email: kamlamatthews@hotmail.com
PENTT EXAMINATIONS REVIEW COMMITTEE
Report on CAPE Unit 1 Physics Paper 2 2022
Dear Sir/Madam,
Please find attached a copy of the 2022 Examination Review document for CAPE Unit 1
Physics, Paper 2, from the Physics Educators Network of Trinidad and Tobago (PENTT):
The committee compared solutions and made general and specific comments, as shown in the
document.
Suggested Solutions/ Specific Comments
Question 1
1(a) CAPE 2017 Syllabus Unit 1 Module 1 Topic 4- Effects of forces. Specific Objective 4.2:
Explain the nature, cause and effects of resistive forces; 4.3: Use concept of terminal velocity to
solve problems involving motion through a fluid.
1(a) Define EACH of the following terms:
Comment: Syllabus has “Explain” not “Define”- so any reasonable ‘explanation’ should be
accepted.
Answer: (i) Drag force: For a (body/object) moving through a fluid, the drag force is the :(Force which opposes its motion) OR (Resistive force on it) OR (Retarding force on it).
(ii)Terminal Velocity: For a (body/ object) moving through a fluid, the terminal
velocity is its:- (constant velocity) OR (velocity when the [Net/ Resultant] force on it is zero) OR
(velocity when it is in equilibrium) OR (velocity at which its acceleration is zero).
(iii) Uniform motion: Motion with:- (constant velocity) OR (constant speed in a
straight line) OR (constant change of distance in a straight line with time) OR (zero
acceleration).
1(b) CAPE 2017 Syllabus Unit 1 Module 1 Topic 4- Effects of forces. Specific Objective 4.3: Use
concept of terminal velocity to solve problems involving motion through a fluid. Include Stoke’s
law for viscous drag = 6πɳrv.
1(b) The following equation is used to express the drag force, 𝐅𝐃 , experienced by a moving
spherical object: 𝐅𝐃 = 6πɳrv
1(b) (i) State the meaning of EACH term on the right-hand side of the equation above.
Comment: By “Term”, did they mean “Quantity” or did they want 6π as well.
Answer: r = radius of (sphere/object, spherical object)
ɳ = (viscosity/ coefficient of viscosity) of fluid
v = velocity of (sphere/ object/spherical object) OR Terminal velocity when
(sphere/object) is in (dynamic) equilibrium.
1(b) (ii) CAPE 2017 Syllabus Unit 1 Module 1 Topic 4- Effects of Forces, Specific Objective 4.2
Resistive forces/ Drag Forces, Specific Objective 4.3 Terminal Velocity.
1(b) (ii) State ONE condition under which the equation in (b) applies for the motion of a
spherical object.
Comment: “Condition” is NOT mentioned in S.O. 4.2 or 4.3, so any reasonable answer should
be accepted
Answer: Object is: (spherical) or (smooth) or (rigid) or (small) or (small radius) or (flow is
streamlined) or (there is no turbulence) or viscosity is constant or At Terminal Velocity: (the
object must be in equilibrium) or (net force acting on object is zero).
1
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
1(c) CAPE 2017 Syllabus Unit 1 Module 1 Topic 4- Effects of forces. Specific Objective 4.3: Use
concept of terminal velocity to solve problems involving motion through a fluid. Include Stroke’s
law for viscous drag = 6πɳrv.
1(c) (i) On the grid provided in Figure 1 on page 7, plot a graph of velocity, v, against time,
t. Draw a smooth curve through the points.
1(c) (ii) Explain the shape of the graph in (c)(i)
Answer: The 3 Points can be found from any of Following:(Gradient/Acceleration) of sphere:
(i) is constant from t = 0 s to about 0.2 s or is maximum at t = 0 s.
(ii) decreases (with time) or (from = 0.2 to 2.8 s)
(iii) is zero at t = 2.8 s
(iv) remains zero from t = 2.8 to 3.0 s.
(Velocity) of sphere:
(i) is zero at t = 0 s
(ii) increases (with time) or increases (from t = 0 to 2.8 s)
(iii) is maximum at t = 2.8 s.
(iv) remains (maximum) or (constant) or (terminal velocity) from t = 2.8 to 3.0 s
2
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
1 (c) (iii) Use the graph in (c) (i) to determine the terminal velocity, vt , of the metal sphere
in ms-1.
Answer: vt = 4.10 𝐜𝐦 𝐬−𝟏 or 4.10 x 𝟏𝟎 𝟐 𝐦 𝐬 𝟏 or 0.0410 𝐦 𝐬 𝟏
1 (d) (i) Determine the units of k (in 𝒗𝒕 =
𝒎𝒈
𝟔𝝅𝒌𝒓
)
Comment: Question asked for “Units”, not “Base Units”.
Answer:
OR 𝑘 ≡
𝑘 =
≡
( )(
)
(
)(
)
( )(
)
= kg m-1s-1 (Base Units)
= Nm-2s (Units)
OR 𝑘 ≡ Pa.s (Units)
1(d) (ii) Given that m = 5 x 𝟏𝟎 𝟑 kg and r = 1 x 𝟏𝟎 𝟑 m, determine the value of k for
glycerin at 30 0C.
=(
Answer: 𝑘 =
(𝟓 𝐱 𝟏𝟎
)(𝟏 𝐱
)( .
)
)( .
)
= 63.47 or 63.46 (4 s.f.) or 63.5 (3s.f.)
or 63 (2 s.f.)
1(e) Explain how the terminal velocity will be affected if
1(e) (i) a sphere of the same mass but twice the radius is used
Answer: EITHER: 𝑣 𝛼
OR (
=
=
=
⸫ If r is doubled, v is halved or (v = 2.05 𝑐𝑚𝑠
⸫ v2 =
)
v )
1(e) (ii) a sphere of a different metal is used.
Comment: Question is ambiguous. There is no mention of what is kept constant, mass or radius.
Answer: Different metal ⸫ different density (𝜌 =
=
⸫𝜌𝛼
)
(i)
If radius (r) constant 𝜌 𝛼 m, but v 𝛼 m ⸫ v 𝛼 𝜌
If 𝜌 increases, v increases. If 𝝆 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐞𝐬, 𝐯𝐭 decreases.
(ii)
If mass (m) constant ⸫ 𝜌 𝛼
, but ⸫ v 𝛼
⸫ v 𝛼 (𝜌)
If 𝜌 increases, v increases. If 𝝆 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐞𝐬, 𝐯𝐭 decreases.
1(f) CAPE 2017 Syllabus Unit 1 Module 1 Topic 4- Effects of Forces. 4.1 Explain the origin of
the upthrust acting on a body wholly or partially immersed in a fluid.
1(f) According to Archimedes’ principle, when a body is immersed in a fluid it experiences
an upthrust. Explain why the upthrust is neglected in the calculation on terminal velocity
in this experiment.
Comment: ‘Why Upthrust is neglected’ is NOT mentioned in S.O. 4.1, so any reasonable answer
should be accepted.
Answer: Either:- Upthrust (U) depends on density of Fluid (𝜌 ), but Weight (W) depends on
density of sphere (𝜌 ).
𝝆𝑭 ≪ 𝝆𝑺
so
U≪W
⸫ U can be ignored.
Or At terminal velocity (v ), F is large compared to Upthrust (F ≫ U) ⸫ U can be ignored
Or Upthrust (U) is already included in the constant (k) given in the formula.
3
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
Question 2
2(a) CAPE 2017 Syllabus Unit 1 Module 2 Topic 2- Properties of Waves, Specific Objective 2.2,
Differentiate between Transverse and Longitudinal Mechanical waves.
2(a) (i) State ONE similarity and ONE difference between a transverse wave and a
longitudinal wave.
2 (a)(i) Answer: Similarity:
(1) Both deal with (vibration /oscillation/ displacement/ disturbance) of particles.
(2) Both can be progressive or stationary
(3) Both have a /wavelength/amplitude/speed/frequency or v = f λ
(4) Both have f of the source
(5) If progressive: Energy transferred from source to areas around it.
If stationary: There is no net transfer of energy in space.
(6) Both have maximum displacement in one direction (e.g. crest or centre of
compression) and opposite direction (e.g. trough or centre of rarefaction).
Difference:
(1) Direction of (vibration /oscillation/displacement/disturbance) of particles with respect
to direction of (travel/propagation) of the wave is (i) perpendicular for transverse
(ii) (Parallel/ to and fro along) for longitudinal.
(2) Transverse waves can be polarised but longitudinal waves cannot.
2(a) (ii) Give ONE example of a transverse wave and ONE example of a longitudinal wave.
2 (a)(ii) Answer: Transverse: Any e.m. radiation, water (in ripple tank), slinky or string moved
up and down, s-seismic waves.
Longitudinal: sound, slinky moved to and fro, p-seismic waves.
2 (b) CAPE 2017 Syllabus Unit 1 Module 2 Topic 2- Properties of waves, Specific Objective 2.11
Discuss application of sound waves to musical instruments
2 (b) Explain how a stringed instrument, such as a guitar, produces a sound.
Answer: When plucked:
(i)
(The string vibrates) or (a standing/stationary wave is created in the string)
(ii)
The (vibrating) string pushes the air (molecules)
(iii) The air molecules vibrate, (passing on the sound/ producing a sound/ heard as sound/
passing on a series of compressions and rarefactions).
2 (c) CAPE 2017 Syllabus Unit 1 Module 2 Topic 2- Properties of waves, Specific Objective 2.9.
Explain properties of stationary waves and perform related calculations. For example: waves on
strings.
2 (c) (i) At a certain setting on the signal generator, a standing wave with three antinodes is
set up on the string. In the space below, sketch and label a diagram to show how the string
would look when this standing wave is set up.
Comment: MELDE’S EXPERIMENT:
(1) In the longitudinal mode, the vibrator is vibrating in the Horizontal plane along the
direction of the string, so nodes are formed at both the vibrator and pulley.
4
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
(2) In the transverse mode, the vibrator is vibrating in the horizontal plane perpendicular to
the direction of the string, so an antinode is formed at the vibrator and a node at the
pulley.
(3) Most experiments give the instructions to ignore the loops at vibrator and pulley and
measure the distance between the rest of the nodes or antinodes.
(4) Ignoring the loops at the end also applies to sound waves reflected off a wall.
(5) If the standing wave questions in (08/2, 10/5, 17/15, 22/2) had ignored the loops at both
ends and had given a distance between the rest, there would not have been a wide range
of answers for those questions.
Answer:
L=n
L = (n- ) or (2n – 1)
2 (c) (ii) Write an equation for the wavelength when the string has n antinodes and use this
equation to calculate the wavelength of the wave sketched in (c) (i).
Answer:
L=n
⸫
λ=
In (c)(i) L = 2.76 m, n = 3 ⸫ λ =
( )( .
)
= 1.84 m
n=3
L = (n- )
⸫λ=
or (2n – 1)
⸫λ=
(
(
)
)
λ=
( )( .
)
= 2.21 m
( )( . )
) ]
λ = [(
= 2.21 m
n=3
5
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
2 (c) (iii) CAPE 2017 Syllabus Unit 1 Module 2 Topic 2- Properties of waves, Specific Objective
2.5. Use the equation v = f λ to solve problems involving wave motion.
2 (c) (iii) Using the equation in (c)(ii), show that the relationship between the frequency of
the vibrator and the number of antinodes, n, is given by: f = (𝐯/𝟐𝒍)n
Where v is the wave velocity and l is the length of the string.
v=fλ ⸫f=
Answer:
λ=
λ=
⸫ f=
(
OR λ = (
⸫ f = (𝒗/𝟐𝒍)n
⸫ f=
)
⸫ f = (𝑣/2𝑙) (n - )
(
)
)
⸫ f=
⸫ f = (𝑣/2𝑙) (2n - 1)
(
)
2 (d) (i) On the grid provided in Figure 3 on page 15, plot a graph of f against n. Draw the
line of best fit through the points.
6
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
2 (d) (ii) Using the graph in (d)(i), determine the velocity of the waves on the string
Comment: Answers depend on Scale used and Best fit
Answer: f = (𝒗/𝟐𝒍)n
y = (m) x
For the graph of f vs n, gradient (m) =(
From graph above: (f starting at 0): m =
) and v = 2Lm = (2)(2.76) (m)
=
= 85.7 Hz
v = 2Lm = (2) (2.76)(85.7) = 473 ms-1
From graph using (f with broken scale at 50): m =
Range: (85.0-87.5) Hz
Range: (469-483) ms-1
=
= 92.4 Hz
v = 2Lm = (2) (2.76)(92.4) = 510 ms-1
2 (e) (i) CAPE 2017 Syllabus Unit 1 Module 2 Topic 2- Properties of waves, Specific Objective
2.3 Differentiate between transverse and longitudinal mechanical waves.
Explanation of the movement of particles in the medium of transmission and the
energy of the waves.
2 (e) An explosion occurs at one end of a pier of length 125m. The sound reaches the other
end of the pier by travelling through three media: air, water and a slender
handrail of solid steel.
2 (e) (i) List the order of the media through which the sound arrives at the other end of the
pier. Give ONE reason for your response.
Comment: “ List the order of the media through which the sound arrives” is not clear.
It should be: “List the order, first to last or list the order, fastest first”
Answer: Either: Order: First to arrive, down to last to arrive: Steel, Water, Air.
Or: Order: Last to arrive, down to first to arrive: Air, Water, Steel.
Reason: (1) Sound travels by vibration of (particles/molecules), so the closer the
(particles/molecules) or the denser the medium, the faster the sound travels
OR (2) 𝝆𝒔𝒕𝒆𝒆𝒍 ˃ 𝝆𝒘𝒂𝒕𝒆𝒓 ˃ 𝝆𝒂𝒊𝒓
OR (3) separation of particles/molecules of steel ˂ water ˂ air
2 (e) (ii) Calculate the time taken for the sound waves to travel through the water to reach
the other end of the pier, given that the speed of sound in water is 1482ms-1.
Answer: v =
t=
=
^
= 0.0843 s (3sf) or 0.084 s (2 sf)
t = 0.0843 s (3sf) or 0.084 s (2 sf)
7
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
Question 3
3 (a) CAPE 2017 Syllabus Unit 1 Module 3 Topic 6-Mechanical Properties of materials.
Specific Objective: 6.9. Definitions of stress, strain.
3 (a) Define EACH of the following terms, and state the unit in which it is measured:
(i)
Stress (ii) Strain
3 (a)(i) (ii)Answer:
(i) Stress = (Force per unit cross-sectional Area) or (Force per unit Area) or (Force per unit Area
normal to Force) or (Force acting normally per unit Area). Unit = Pascal or Pa or Nm-2
(ii) Strain = (Extension per unit length) or
length) or
or (change in length divided by original
or (Deformation per unit length) or (fractional change in length). No unit.
3 (b) CAPE 2017 Syllabus Unit 1 Module 3 Topic 6-Mechanical Properties of materials,
Specific Objective 6.8. Discuss stretching of wire in terms of Load extension….Hooke’s Law
Specific Objective 6.10. Perform experiments…Young’s Modulus of wire.
Specific Objective 6.11. Perform experiments based on F-e graphs for.. ductile material
Specific Objective 6.12. Deduce the strain energy in a deformed material from a force-extension
graph.
Specific Objective 6.13. Distinguish between elastic and inelastic deformations of a material
3 (b)(i) Describe the behavior of the metal wire in EACH of the following segments of the
graph:
Comment: S.O. 6.8 asks for ‘discussion ‘of stretching of wire in terms of F-e graphs, NOT
Stress-Strain Graphs, so a variety of words can be used. Also the graph did not differentiate
between proportional and elastic limit.
3 (b)(i) Answer:
AB: (Proportional Region/Deformation) or (Hooke’s Law obeyed) or (Force proportional to
extension) or (stress proportional to strain) or (elastic region) or (when force/stress is removed,
it returns to [original length/ zero strain/ zero extension].
BC: [(Inelastic/ Plastic) (Region/ Deformation)] or (Hooke’s Law NOT Obeyed) or (Force not
proportional to extension) or (stress not proportional to strain) or ( Proportional/ Elastic Limit
exceeded) or (small increase in force/stress causes large increase in length/strain).
CD: Permanent (deformation/extension/ strain) when (force/ stress) removed or Same (Young’s
Modulus (E)/ Force constant (k)/ Gradient) for CD as for AB.
3 (b) (ii) Explain what is meant by the ‘elastic limit’ of a wire.
Comment: “Definition” of elastic limit is NOT in S.O. 6.8-6.12 so any reasonable answer should
be allowed.
3 (b) (ii) Answer:
Furthest point or maximum (stress/force) or max( strain/ extension/deformation) at which it
can return to its (original length/ zero strain/ zero extension) when (force/ stress) is removed
Or [end of Elastic and beginning of Plastic (Region / Deformation]
Or Point beyond which it will have a permanent (deformation/ extension) when
(force/stress) is removed.
Or Point beyond which it undergoes plastic deformation.
8
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
3 (b) (iii) If EACH major interval on the vertical axis of the graph in Figure 4 represents
150 MPa, determine the elastic limit of the wire.
3 (b) (iii) Answer: Taking Elastic limit = Proportional Limit (Point B) = 450 MPa
3 (b) (iv) The wire can be described as a ductile material. State TWO properties of ductile
materials.
Comment: S.O.6.11 asks for experiment for F-e graph for ductile material. It does NOT ask for
properties of ductile materials, so any reasonable answer should be accepted.
3 (b) (iv) Answer:
Ductile Materials(1) Have both elastic and plastic deformation.
(2) Have a large (plastic region/deformation) before (breaking/fracturing)
(3) Have a large (extension/strain/deformation) before (breaking/fracturing)
(4) Can be stretched into a wire without breaking/ can be hammered thin.
3 (b) (v) In terms of the wire, what does the interval AD on the graph paper represent?
3 (b) (v) Answer: Permanent (Strain/extension/deformation/stretch/displacement)
3 (b) (vi) State how the energy per unit volume stored in the wire during the loading
process can be determined from the graph.
Comment: S.O. 6.12 asks for (“Energy from F-e graph”, not (Energy per unit volume from a
stress-strain graph).
3 (b) (vi) Answer: From: Area under graph/ area under loading graph/ area under ABC.
3 (c) (i) State what is meant by the term ‘high tensile strength’.
Comment: Meaning of ‘high tensile strength’ NOT in S.O. 6.8 – 6.12, so any sensible answer
should be allowed.
3 (c) (i) Answer: Large (force/stress) (it can withstand/ applied) before it (breaks/ fractures).
3 (c) (ii) Spiders use silk to build webs to catch insects. Using the graph in Figure 5, explain
how the properties of spider silk make it more suitable than silkworm silk for
building webs to catch insects.
3 (c) (ii) Answer: Spider silk:
(1) Can take a larger (stress/force) for same (strain/extension)
(2) Can take a larger (stress/force) before (breaking/ fracturing)
(3) Has a higher tensile (strength/ stress)
(4) Has a greater (Young’s Modulus/ Gradient/ Force constant)
[where FORCE = force of struggling insect]
3 (c) (iii) Using the graph in Figure 5, determine the MAXIMUM amount of energy that
can be stored by silkworm silk before it fractures.
Comment: Question 3c (iii) asks for Energy, NOT Energy/Volume, which is the area under
graph. You can’t get Energy from Energy/Volume, since length is not given.
3 (c) (iii) Answer:
Area of 1 𝑐𝑚 block = (20 x 10 ) (0.025) = 0.5 x 10
Range for Maximum Energy/Volume = Area under silk worm graph
= (15 – 16) 𝑐𝑚 blocks = (15 – 16)(0.5 x 106) = (7.5 – 8.0) x 𝟏𝟎𝟔 J 𝐦 𝟑 or J/𝐦𝟑
9
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
3 (c) (iv) Using the graph in Figure 5, determine the Young’s Modulus of spider silk for
small stresses.
3(c)(iv) Answer:
Young’s Modulus (E) =
(up to Elastic Limit)
E = Gradient of straight line through origin
E=
=
(
)
= 1.43 x 10 Pa OR E =
.
=
(
)
.
= 1.44 x 10 Pa
Range for E: (1.35-1.49) 𝟏𝟎𝟗 Pa
3 (c) (v) An insect flies into a spider’s web and becomes attached to a single thread. This
creates a tension of 580 μN in the thread. The thread extends by approximately 3% of its
original length. Calculate the radius of a single thread of spider silk.
3 (c) (v) Answer:
Strain =
Stress =
.
Strain =
= 0.03
=
Young’s Modulus (E) =
E=
(
(
(e = 3% of L0 = 0.03 L0)
)
)( .
)
⸫r=
=
(
)
( )( .
(
)
( )( )( .
)
but A = πr2
)
/
When E = 1.43 x 109 Pa, r = 2.07 x 10-6m
Range: for E = (1.35-1.49) 10 Pa ,
r = (2.03 – 2.14) 𝟏𝟎 𝟔 m
General comments:
Drawing many Graphs is very time consuming.
Suggestion on limiting the number of Graphs which have to be Drawn in the Examination.
One Question could have a Graph to Daw and the students would be required to draw the Best
Fit, determine a Readout or Gradient and perform Calculations using it.
One Question could have an already plotted Graph to Interpret.
One Question could have points already plotted and the students would be required to draw the
Best Fit, determine a Readout or Gradient and perform Calculations using it.
Or the Question could have the Graph already Drawn and the students would be required to draw
the Best Fit, determine a Readout or Gradient and perform Calculations using it.
Or one Question could have a Graph to Sketch
10
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
PENTT Examination Review Committee 2022
 Aarti Dowlath: Teacher at St Augustine Girls’ High School
[PENTT District Committee Coordinator for St George East]
 Allison Ramsey: Teacher at Holy Faith Convent, Couva
 Anthony Ramdath: Teacher at St George’s College
 Bernadette Harrychandsingh-Hassranah: Teacher at Couva Government Secondary
[PENTT District Committee Coordinator for Caroni]
 Charlene Ramkissoon-Vidale : Teacher at El Dorado West Secondary
[PENTT Vice president & District Committee Coordinator for St George East]
 Cherisse Crichlow: Teacher at St. Joseph Convent – St. Joseph
 Cheryl-Ann Ramadhar: Retired [formerly Teacher at St. Mary’s College, Vice Principal at
Chaguanas North Secondary]
 Cheryl Ramlogan: Teacher at St. Joseph’s Convent, San Fernando
[PENTT District Committee Coordinator for Victoria]
 Daryll Williams: Teacher at St Benedict’s College
 Davatee Maharaj: Teacher at Vishnu Boys’ Hindu College
 Dianne Dilchan-Dass: Teacher at Tabaquite Secondary
[PENTT District Committee Coordinator for SEED]
 Hana Subratti: Teacher at Tableland Secondary
[PENTT District Committee Coordinator for SEED]
 Hollis Sankar: Retired Curriculum Coordinator MOETT
 Kamla Matthews: Retired Teacher [formerly at Holy Faith Convent, Couva]
[PENTT President, Exam Review Coordinator/Editor]
 Karen Dilraj: Teacher at Couva Government Secondary
 Kubair Rampersad: Teacher at San Fernando West Secondary
 Larissa Mohammed: Teacher at Hillview College
[PENTT District Committee Coordinator for St George East]
 Nadira Nandlal: Teacher at Naparima College
[PENTT Vice president & District Committee Coordinator for Victoria]
 Natalie Cummings-Medina: Teacher at St. Joseph Convent – St. Joseph
 Natasha Lewis- Dos Santos: Teacher at Queen’s Royal College
[PENTT District Committee Coordinator for POS]
 Neil Garibsingh: Teacher at San Fernando Central Secondary
[PENTT District Committee Coordinator for Victoria]
 Nigel Superville: Teacher at Roxborough Secondary, Tobago
[PENTT District Committee Coordinator for Tobago]
 Nirmala Ramnarine: Teacher at Shiva Boys’ Hindu College
[PENTT District Committee Coordinator for St Patrick]
 Nirmala Singh: Teacher at Northeastern College
[PENTT District Committee Coordinator for NEED]
 Nisha Ramlal: Teacher at St. Joseph’s Convent, San Fernando
 Rhona Edwards-Cato: Teacher at Queen’s Royal College
[PENTT District Committee Coordinator for POS]
 Samlal Mannie: Retired Teacher [formerly at Carapichaima East Secondary]
 Sanjeev Maharaj: Curriculum Officer [formerly Teacher at Hillview College]
[PENTT Vice president & District Committee Coordinator for St George East]
 Sarah Shah Dookran: Teacher at Waterloo High
[PENTT District Committee Coordinator for Caroni]
 Suraj Gopaul: Teacher at Shiva Boys’ Hindu College
[PENTT District Committee Coordinator for St Patrick]
 Vernessa Sankar: Teacher at Chaguanas North Secondary
[PENTT District Committee Coordinator for Caroni]
 Vishalli Tancoo: Teacher at Naparima Girls’ High School
[PENTT District Committee Coordinators for Victoria]


PENTT Exam Review 2022 sent to:
Allister Ramrattan: MOETT Curriculum Officer
Anna Singh: MOETT Director, Curriculum, Planning and Development Division.
11
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
Physics Educators’ Network of Trinidad and Tobago
Docendo Discimus By Teaching we Learn
63 Scotland Drive, Cocoyea, San Fernando.
Phone: 795-5633.
Email: kamlamatthews@hotmail.com
PENTT EXAMINATIONS REVIEW COMMITTEE
Report on CAPE Unit 1 Physics Paper 2 2021
Dear Sir/Madam,
Please find attached a copy of the 2021 Examination Review document for CAPE Unit 1
Physics, Paper 2, from the Physics Educators Network of Trinidad and Tobago (PENTT):
The committee compared solutions and made general and specific comments, as shown in the
document.
Suggested Solutions/ Specific Comments
Question 1
1(a) CAPE 2017 Syllabus Unit 1 Module 1 Topic 3- Motions Specific Objective 3.2: Use graphs
to represent displacement, speed, velocity and acceleration in a single dimension.
1(a) Plot a graph of velocity versus time. Draw the best smooth curve through the points
Comment: Best smooth curve implies bell shaped curve.
1
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
1(b) Describe the motion….explaining clearly…
Comment: Definition of Motion- mathematically described in terms of distance/displacement,
speed/velocity, acceleration and time”.
Comment: Forces were NOT ASKED FOR until 1(c), so the points about Forces shown
below are unnecessary in 1 (b).
0-22: Drag/Air Resistance/Viscous force (A) < Weight(W) ⸫ Resultant Force F = W – A (down)
20-40: A α v ⸫ A increases. When A = W, F = 0, so acc = 0.
40-45: When parachute opens, large upward force (A) = (Lift for Square parachute or Drag for
Round parachute). F = A – W ⸫ deceleration.
45-60: When A = W, F = 0, so acc = 0.
60-62: On landing, upward force of ground on girl (R) > W, F = R – W, ⸫ deceleration.
Answer:
Time, t/s
Description of motion
0-20
(i) Velocity increases
(ii) (m = decreasing and positive), so (decreasing) acceleration
20-40
(i) Velocity constant/ Terminal Velocity
(ii) m = 0, so zero acceleration
40-45
(i) Velocity decreases
(ii) (m = increasing and negative), so (increasing) deceleration
45-60
(i) Velocity constant (lower)/ (lower) terminal velocity
(ii) (m = 0), so zero acceleration
60-62
(i) Velocity decreases to zero/ comes to rest.
(ii) ( m = increasing and negative), so (increasing) deceleration
Range of Teachers’ Answers for 1(b):
Some teachers mentioned combinations of any of the following points:
(i) Motion- Velocity and acceleration.
(ii) Parachute open or closed.
(iii)Forces.
Sample answer for Q1(b) received from a member.
t/s
Description of motion
0 - 20
The parachutist is falling with the parachute closed, initially with acceleration g. Drag
forces increase with velocity until the net force on the body is zero. At this point terminal
velocity is reached
20 - 40
The parachutist continues to fall with terminal velocity with the parachute still closed.
40 - 45
The parachute opens and there is a sudden increase in drag force, giving rise to a
deceleration, thus reducing the velocity.
45 - 60
With a decrease in velocity, drag force also decreases, until net force is again zero. Hence
constant velocity ( new terminal velocity)
60 - 62
This is the interval the parachutist hits the ground and velocity reduces to zero in 2
seconds.
2
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
1(c) CAPE 2017 Syllabus Unit 1 Module 1 Topic 4- Effects of Forces, Specific Objective 4.2
Resistive forces/ Drag Forces, Specific Objective 4.3 Terminal Velocity.
1(c) Identify the 2 major forces acting on the parachutist during the time intervals 20-40 s and
45-60 s and state the relationship between the forces.
Answer: 20-40s: (i) Weight/Pull of Gravity (W)
(ii) Drag/ Air Resistance/ Viscous Force (A)
45-60s: (i) Weight/ Pull of Gravity (W)
(ii) Lift (for square parachute)/ Drag (for Round parachute)/Air Resistance (A)
Relationship: either A= W (Magnitude)
or A = -W [Magnitude and direction]
or A-W = 0
1(d) CAPE 2017 Syllabus Unit 1 Module 1 Topic 3- Motions
1(d) (i) Explain why the car is accelerating although its speed remains constant.
Comment: No Specific Objective. NOT on the Syllabus.
Answer: Speed is constant,
but direction is changing.
Velocity has magnitude (speed) and direction
(or velocity is a vector),
so velocity is changing.
Acceleration is the rate of change of velocity, ⸫ it accelerates.
1(d) (ii) CAPE 2017 Syllabus Unit 1 Module 1 Topic 3- Motion, Specific Objective 3.20 a =
1(d) (ii) For the car and driver calculate the centripetal acceleration
Answer: a =
. = 64ms-2
1(d) (iii) CAPE 2017 Syllabus Unit 1 Module 1 Topic 3- Motion, Specific Objective 3.21, Use the
equations of circular motion to solve problems….Banking
1(d) (iii) For the car and driver calculate the normal reaction force (R)
Answer:
RH = R sin θ = ma ……(i)
RV = R cos θ = mg ……(ii)
Either: R =
=
=
(
)(
)
R = 5.68 x 104 N (or 56811 N)
3
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
OR: R =
=
=
(
)( .
)
R = 1.04 x 104 N (or 10377 N)
OR: R = 𝑅 + 𝑅 = (𝑚𝑎) + (𝑚𝑔)
R = (680 𝑥 64) + (680 𝑥 9.81)
R = 4.40 x 104 N (or 44028 N)
Comment: (i) ÷ (ii)
𝒂
𝟔𝟒
Tan θ = 𝒈 = 𝟗.𝟖𝟏 = 𝟔. 𝟑𝟐
θ = 81.3⁰
But θ was given as θ = 50⁰ in the question
If θ = 81.3⁰ was used, then all the methods would give the SAME answer.
𝒎𝒂
(𝟔𝟖𝟎)(𝟔𝟒)
𝒎𝒈
(𝟔𝟖𝟎)(𝟗.𝟖𝟏)
R = 𝒔𝒊𝒏 𝟖𝟏.𝟑 = 𝒔𝒊𝒏 𝟖𝟏.𝟑 = 𝟒𝟒𝟎𝟐𝟕 = 4.4 x 104 N,
R = 𝒄𝒐𝒔 𝟖𝟏.𝟑 =
𝒄𝒐𝒔 𝟖𝟏.𝟑
= 𝟒𝟒𝟏𝟎𝟏 = 4.4 x 104 N,
R = 𝑹𝟐𝑽 + 𝑹𝟐𝑯 = 4.4 x 104 N
1(e)(i) CAPE 2017 Syllabus Unit 1 Module 1 Topic 3- Motions, Specific Objective 3.26, discuss
the motion of geostationary satellites and their applications.
1 (e)(i)Define the term ‘geostationary satellite’
Answer: One orbiting at a Fixed Height above the Equator with the Same Period of Rotation or
the Same Angular Velocity.
1(e) (ii) CAPE 2017 Syllabus Unit 1 Module 1 Topic 3- Motions, Specific Objective 3.20,
F = 𝑚𝜔 𝑟, Specific Objective 3.21, F =G
1(e) (ii) State the physical characteristics of the earth which determine the fixed orbital radius,
r, for a geostationary satellite.
Answer:
F=G
= m ω r,
Or F = G
= m ω r = m0 (2𝜋/T)2 r
r=(
)
r3 = Gm1T2/4𝜋2
r depends on mass of Earth (m1) and angular velocity of Earth (ꞷ) OR Period of Rotation of
Earth (T =
).
Comment: Physical Characteristics means ‘Aspects Physically Apparent’
1(e) (iii) CAPE 2017 Syllabus Unit 1 Module 1 Topic 3- Motions, Specific Objective 3.26,
discuss the motion of geostationary satellites and their applications.
1(e) (iii) State two applications of geostationary satellites.
Answer:
Communications
Weather Forcasting
Spotting Pollution
Spying
4
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
Question 2
2(a) CAPE 2017 Syllabus Unit 1 Module 2 Topic 2- Properties of Waves, Specific Objective 2.1,
Use the following terms: displacement, Amplitude, Period, frequency, velocity in relation to the
behaviour of waves.
2(a) Define (i) Amplitude (ii) Period
2 (a)(i) Answer: Amplitude is the maximum displacement from the equilibrium position.
2 (a)(ii) Answer: Period is the time for one complete oscillation/ cycle/ vibration.
2 (b) CAPE 2017 Syllabus Unit 1 Module 2 Topic 2- Harmonic Motion, Specific Objective 1.2
Recall the conditions necessary for simple harmonic motion.
2 (b) State the 2 conditions necessary for simple harmonic motion.
Answer: (i) Acceleration of particle is proportional to its distance away from a fixed point.
(ii) Acceleration of particle is directed towards the fixed point.
2 (c) (i) CAPE 2017 Syllabus Unit 1 Module 2 Topic 1- Harmonic Motion, Specific Objective 1.4
Period of mass on spring, T = 2𝜋
2 (c) (i) Calculate the mass of the wooden block
Answer:
T = 2𝜋
T = 8/20 = 0.4 s
, T2 =
m = T2 k/4 𝜋2 = (0.4)2 (26.5)/ 4 𝜋2
m = 0.107 kg
2 (c) (ii) CAPE 2017 Syllabus Unit 1 Module 2 Topic1- Harmonic Motion, Specific
Objective1.1Use the equations of simple harmonic motion to solve problems T=
2 (c) (ii) Calculate the angular frequency of the oscillations
Answer: Either
T = 0.4s
Or f = 20/8 = 2.5 Hz
T=
ꞷ=
ꞷ = 2πf = 15.7 rad s-1
ꞷ= .
ꞷ = 15.7 rad s-1
2 (c) (iii) CAPE 2017 Syllabus Unit 1 Module 2 Topic 1- Harmonic Motion, Specific
Objective1.1Use the equations of simple harmonic motion to solve problems 𝑣 = 𝜔𝐴
2 (c) (iii) Calculate the maximum velocity of the wooden block.
Answer: 𝑣 = 𝜔𝑥
or 𝑣 = 𝜔𝐴
𝑣 = (15.7)(0.02),
𝑣 = 0.314 ms-1
A or 𝑥 = 2cm = 0.02m
2 (d) (i) CAPE 2017 Syllabus Unit 1 Module 2 Topic 1- Harmonic Motion, Specific Objective1.9
Describe damped oscillations and represent such motion graphically.
2 (d) (i)The amplitude of the oscillations… decreases..by 0.2 cm with each oscillation.
…sketch the first 5 oscillations.
5
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
2 (d) (i) Answer: T = 1 osc = 0.4s
5 osc. = 5 x 0.4 = 2.0s
2 (d) (ii) CAPE 2017 Syllabus Unit 1 Module 2 Topic 1- Harmonic Motion, Specific
Objective1.10 Explain how damping is achieved in some real-life examples
2 (d) (ii) Identify the phenomenon depicted by the waveform sketched in (d) (i) and state 2 real
life situation in which this phenomenon occurs
Answer: Phenomena: Damping/Light Damping/ Underdamping
2 Situations: Pendulum, Mass on a spring, Swing, Hammock, any naturally vibrating body.
2 (e) (i) CAPE 2017 Syllabus Unit 1 Module 2 Topic 3- Physics of the Ear and Eye, Specific
Objective3.2 Threshold of Hearing
2 (e) (i) Define the term Threshold of Hearing and state its value…
Answer: Minimum intensity for audibility at a particular frequency.
(At a frequency f=1000Hz), the threshold intensity I0 = 10-12 Wm-2.
2 (e) (ii) CAPE 2017 Syllabus Unit 1 Module 2 Topic 3- Physics of the Ear and Eye, Specific
Objective3.2 Intensity Level = 10 log
2 (e) (ii) Write an equation to show the relationship between sound intensity level and the
threshold of hearing I0
Answer: Intensity level (IL) = 10 log
.
2 (e) (iii) Determine the intensity in Wm-2 for an average sound intensity level of 90dB
Answer:
Either:
Or:
Intensity level (IL) = 10 log
Intensity level (IL) = 10 log
90 = 10 log
90 = 10 log
9 = log
9 = log
9 = log I – log I0
9 = log I – log 10-12
9 = log I – (- 12)
log 109 = log
109 =
6
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
9 = log I + 12
Log I = 9 - 12 = -3
I = 10-3 Wm-2 or 0.001
Wm-2
I = 109 x 10-12
I = 10-3 Wm-2 or 0.001
Wm-2
2 (e) (iv) CAPE 2017 Syllabus Unit 1 Module 2 Topic 3- Physics of the Ear and Eye, Specific
Objective3.1Consider Intensity.
Comment: The Syllabus did NOT mention the definition of Intensity as Power/Area,
⸫ NO SPECIFIC OBJECTIVE
2 (e) (iv) …for 3 hours. Determine the amount of energy incident his eardrum, if its surface
area is 0.5 cm2
Answer: I =
=
E = IAt where I = 10-3 Wm-2
A = 0.5 cm2 = (0.5 x 10-4) m2
t = 3 hrs = (3 x 3600)s = 10800s
E = 5.4 x 10-4 J
7
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
Question 3
3 (a) CAPE 2017 Syllabus Unit 1 Module 3 Topic 4-The Kinetic Theory of Gases Specific
Objective 4.4, Basic assumptions of the kinetic theory of gases.
3 (a)
State THREE assumptions of the kinetic theory of ideal gases
Answer:
Number- A sample of gas contains a very large number of molecules
Same- The molecules are identical
Motion- The molecules move randomly
Time- Duration of collision is negligible compared with time between collisions
Force- Force of Attraction between neighbouring molecules is negligible
Volume- Volume occupied by molecules themselves is negligible compared with the
volume occupied by the gas [or The separation between molecule > size of the molecule]
Elastic-The collisions of the molecule (with each other and walls of container are
perfectly elastic)
Temperature- The average kinetic energy of the molecule is proportional to temperature
in Kelvin.
3 (b) (i) CAPE 2017 Syllabus Unit 1 Module 3 Topic 4-The Kinetic Theory of Gases Specific
Objective 4.7, Ek =
𝑘𝑇.
3 (b) (i) Calculate the average translational kinetic energy of ONE molecule of Ne gas
Answer:
𝐸 =
𝑘𝑇,
T = 270C = (27 + 273)K = 300K
= ( )(1.38 𝑥 10
)(300)
𝐸 = 6.21 x 10-21 J
3 (b) (ii) CAPE 2017 Syllabus Unit 1 Module 3 Topic 4-The Kinetic Theory of Gases Specific
Objective 4.3, PV =nR𝑇, PV = NkT
3 (b) (ii) Calculate the number of molecules of Ne gas in the metal sphere
Answer:
Either
PV = NkT
Or
PV = nRT
N=
n=
N=( .
.
)(
( .
n=
)
N = 1.69 x 1023 molecules
( .
)(
)
)
n = 0.281 moles
N = n x NA
N = (0.281)(6.02 x 1023)
N = 1.69 x 1023 molecules
3 (b) (iii) CAPE 2017 Syllabus Unit 1 Module 3 Topic 4-The Kinetic Theory of Gases Specific
Objective 4.6, PV =
𝑁𝑚𝑐 , Specific Objective 4.7, rms speed- 𝑐 , Specific Objective 4.2,
1mole = NA molecules.
3 (b) (iii) Calculate the rms speed of the Ne molecules
Answer: Mass of one molecule (m) =
,
=
-26
m = 3.35 x 10
(
.
( .
)
)
kg
8
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
Either
Or
PV =
𝐸 = 𝑚𝑐
𝑁𝑚𝑐
𝑐 =
𝑐 =
( )
( .
𝑐 =( .
)
)( .
𝑐 =
)
𝑐 = 3.71 x 105
( .
)
( .
)
cr = 𝑐 = √3.7 𝑥 10 = 609 ms-1
cr = 𝑐 = √3.7 𝑥 10 = 609 ms-1
Or
Or
PV = 𝑛 𝑀 𝑐
𝐸 =
𝑐 =
( )
𝑐 = ( .
𝑚𝑐 =
.
)
cr = 𝑐 = √3.7 𝑥 10 = 609 ms-1
3𝑘𝑇
𝑚
𝑐
=
=
(3)(1.38 𝑥 10 )(300)
3.35 𝑥 10
.
)(
𝑘𝑇
cr = 𝑐 = √3.7 𝑥 10 = 609 ms-1
3 (c) CAPE 2017 Syllabus Unit 1 Module 3 Topic –Heat Transfer, Specific Objective 3.7,solve
problems using Stefan’s Equation
3 (c) (i) Complete row 3 of table….
Comment: Mistake in unit for T
3 (c) (i) Answer:
T4/K4 9.60
(x 109)
(3sf)
10.9
12.3
13.8
15.5
17.4
3 (c) (ii)…plot a graph of P versus T4. Draw the line of best fit …
9
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
3(c)(ii)
3(c)(ii) CAPE 2017 Syllabus Unit 1 Module 3 Topic –Heat Transfer, Specific Objective 3.7,solve
problems using Stefan’s Equation, Black Body Radiation, P = σAT4
3(c)(ii) Determine the gradient of the graph above and hence deduce the surface area of the
metal sphere
Answer:
(
)
Slope m =
= (
m= .
= 20.3 𝑥 10 𝑊
P = σAT4
A=
. )
.
m = 2.03 x 10 WK-4
Range of Teachers’ Answers for slope (m) = (1.90 to 2.03)10-8 WK-4
A=
.
.
.
A=
A = 0.358 m2
Range of Teachers’ Answers for A = (0.335 to 0.358) m2
10
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
3(d) CAPE 2017 Syllabus Unit 1 Module 3 Topic –Heat Transfer, Specific Objective 3.1,
Mechanism of thermal conduction
3(d) Describe the mechanism of thermal conduction in metallic conductors
Answer:
(i) Lattice Atom Vibration:
Atoms vibrate about fixed positions in the lattice.
When heated, atoms vibrate faster, transmitting energy to neighbouring atoms.
(Spread of energy is slow).
(ii) Free Electron Diffusion/Movement:
Metals have free or valence electrons (which wander randomly through the metal).
When heated, free electrons move to cool end and collide with the ions in the lattice,
giving them energy.
(Spread of energy is faster).
CAPE - General Comments from Teachers
 The students prepared for a wide Range of Specific Objectives under each listed Topic.
 Only a narrow range of Specific Objectives was examined under each listed Topic.
 Examining a wide range of Specific Objectives allows the students to capitalise on their
individual strengths.
 Examining a narrow range of Specific Objectives means that some students, who are weak in
those areas, will not stand a chance.
 Some parts of Questions were not on the Syllabus. (i.e. did not have a Specific Objective)
 Students can be very distressed when they see parts of Questions which are not on the
Syllabus.
PENTT Examination Review Committee 2021
 Aarti Dowlath: Teacher at St Augustine Girls’ High School
[PENTT District Committee Coordinator for St George East]
 Allison Ramsey: Teacher at Holy Faith Convent, Couva
 Anthony Ramdath: Teacher at St George’s College
 Bernadette Harrychandsingh-Hassranah: Teacher at Couva Government Secondary
[PENTT District Committee Coordinator for Caroni]
 Charlene Ramkissoon-Vidale : Teacher at El Dorado West Secondary
[PENTT Vice president & District Committee Coordinator for St George East]
 Cherisse Crichlow: Teacher at St. Joseph Convent – St. Joseph
 Cheryl- Ann Ramadhar : Retired Teacher [formerly Teacher at St. Mary’s College, Vice
Principal at Chaguanas North Secondary]
 Cheryl Ramlogan: Teacher at St. Joseph’s Convent, San Fernando
[PENTT District Committee Coordinator for Victoria]
 Daryll Williams: Teacher at St Benedict’s College
 Davatee Maharaj: Teacher at Vishnu Boys’ Hindu College
 Dianne Dilchan-Dass: Teacher at Tabaquite Secondary
[PENTT District Committee Coordinator for SEED]
 Hana Subratti: Teacher at Tableland Secondary
[PENTT District Committee Coordinator for SEED]
 Hollis Sankar: Retired Curriculum Coordinator MOETT
 James Tudor: Teacher at Barataria South Secondary
 Kamla Matthews: Retired Teacher [formerly at Holy Faith Convent, Couva]
[PENTT President, Exam Review Coordinator/Editor]
 Karen Dilraj: Teacher at Couva Government Secondary
11
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
 Kubair Rampersad: Teacher at San Fernando West Secondary
 Larissa Mohammed: Teacher at Hillview College
[PENTT District Committee Coordinator for St George East]
 Melissa Jagroop: Teacher at Toco Secondary
 Nadira Nandlal: Teacher at Naparima College
[PENTT Vice president & District Committee Coordinator for Victoria]
 Natalie Cummings-Medina: Teacher at St. Joseph Convent – St. Joseph
 Natasha Lewis- Dos Santos: Teacher at Queen’s Royal College
[PENTT District Committee Coordinator for POS]
 Neil Garibsingh: Teacher at San Fernando Central Secondary
[PENTT District Committee Coordinator for Victoria]
 Nigel Superville: Teacher at Roxborough Secondary, Tobago
[PENTT District Committee Coordinator for Tobago]
 Nirmala Ramnarine: Teacher at Shiva Boys’ Hindu College
[PENTT District Committee Coordinator for St Patrick]
 Nirmala Singh: Teacher at Northeastern College
[PENTT District Committee Coordinator for NEED]
 Nisha Ramlal: Teacher at St. Joseph’s Convent, San Fernando
 Rhona Edwards-Cato: Teacher at Queen’s Royal College
[PENTT District Committee Coordinator for POS]
 Samlal Mannie: Retired Teacher [formerly at Carapichaima East Secondary]
 Sanjeev Maharaj: Curriculum Officer [formerly Teacher at Hillview College]
[PENTT Vice president & District Committee Coordinator for St George East]
 Sarah Shah Dookran: Teacher at Waterloo High
[PENTT District Committee Coordinator for Caroni]
 Suraj Gopaul: Teacher at Shiva Boys’ Hindu College
[PENTT District Committee Coordinator for St Patrick]
 Vernessa Sankar: Teacher at Chaguanas North Secondary
[PENTT District Committee Coordinator for Caroni]
 Vishalli Tancoo: Teacher at Naparima Girls’ High School
[PENTT District Committee Coordinators for Victoria]


PENTT Exam Review 2021 sent to:
Allister Ramrattan: MOETT Curriculum Officer
Anna Singh: MOETT Director, Curriculum, Planning and Development Division.
12
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
Physics Educators’ Network of Trinidad and Tobago
Docendo Discimus By Teaching we Learn
63 Scotland Drive, Cocoyea, San Fernando.
Phone: 795-5633.
Email: kamlamatthews@hotmail.com
PENTT EXAMINATIONS REVIEW COMMITTEE
Report on CAPE Unit 1 Physics Paper 2 2019
Dear Sir/Madam,
Please find attached a copy of the 2019 Examination Review document for CAPE Unit 1
Physics Paper 2, from the Physics Educators Network of Trinidad and Tobago (PENTT):
The committee compared solutions and made general and specific comments, as shown in the
document.
Suggested Solutions/ Specific Comments
Question 1
1(a)-(c) CAPE 2017 Syllabus Unit 1 Module 1 Topic3-Motions, Specific Objective 3.5: Use
equations of motion…. Specific Objective 3.7: Show that projectile motion is parabolic.
1(a) (i) x or xH or sH or sX = u cos𝜃 t
2
1(a) (ii) y or xV or sV or sY = u sin𝜃t – ½ gt
1(a) (iii) Comment: Straightforward proof.
1(b) Table Headings:
Comment:
 Since a Quantity is defined as the product of a number and a unit, then a number is
the Quotient of a Quantity and a unit, i.e. Quantity/Unit.
 We would like to suggest that table headings and graphical axes should be
labelled Quantity/Unit. E.g. 𝛉/℃ and H/m
1(b) (i) – Sin2 𝛉 to 3 significant figures.
0.117
0.250
0.413
Sin2 𝛉
0.587
0.750
0.883
Or Sin2 𝛉 to 2 significant figures.
0.12
0.25
Sin2 𝛉
0.59
0.75
0.88
0.41
1(b) (iii) - Gradient (m):
Comment: Gradient (m) [depended on best fit]. There was a range of Best Fit Graphs.
Teachers’ answers for Gradient (m) ranged from 11.4 to 11.6 m.
1(b) (iv) – Initial Velocity (u):
Answer:
Substituting for gradient [m = H/sin2 𝜃] into v2 = u2 + 2ax [where v = 0 and x = H],
gives u2 = 2gm
Teachers’ answers for (u) ranged from 15.0 to 15.1 m/s
1(c) Height of object:
Answer:
Either using the equations from a(i) and a(ii) or the equation from a(iii).
Teachers’ answers for Height ranged from 4.3 to 4.5 m.
1(d): CAPE 2017 Syllabus Unit 1 Module 1 Topic 3-Motions, Specific Objective 3.2:
Use graphs to represent…velocity…
The object falls vertically to the ground and bounces twice before coming to rest. Sketch a
velocity-time graph of its motion.
Possible Answers: Note: Rebound Time may have been taken as Instantaneous [Vertical dotted
line] or Non- Instantaneous [Non vertical dotted line].
1
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
N.B. This is highly unlikely. If the student feels that the upward moving projectile causes the
object to move up before falling down, he will get the one of the graphs below.
Question 2
2(a)-(c) CAPE 2017 Syllabus Unit 1 Module 2 Topic 3-Physics of the Eye & Ear, Specific
Objective 3.5: Solve problems using Lens Formulae. Power in D…. Specific Objective 3.7(d):
Short Sight. Calculation of Power of correcting lens. Specific Objective 3.8: Discuss how defects
of the eye can be corrected.
2(a)-(c) A grandmother.. is unable to read her newspaper clearly, unless she holds it …15cm
from her eye. Normal near point..is 25cm.
Comment:
The Real Near point is the closest distance on which the eye can focus and the Real Far point is
the farthest distance on which the eye can focus.
Normally, calculations involving Short Sight have the person’s Real Near Point as the Normal
Near Point of 25cm. Then the person’s Real Far Point is given and we are asked to find the
Power of the Corrective Lens needed to view an object at the Normal Far Point of infinity.
The question did not state the object position. Was it at the Normal Near Point of 25cm or at the
Normal Far Point of infinity?
2(a) (i) Identify ..eye defect.
Answer: Short Sight or Myopia or Near sight.
2(a) (ii) Name the type of lens…to correct eye defect…
Answer: Diverging
2(b) (i) Sketch…showing how… eye forms the image without her corrective lens.
Comment: The question did not state the object position. Was it at 15cm, 25cm or ∞?
Possible Answers:
2(b) (ii) Sketch…showing how… eye forms the image with her corrective lens.
Comment:
The question did not state the object position. Was it at 25cm or ∞?
2
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
Possible Answers:
2(c) Calculate the Power of the lens needed to correct her eye defect.
Possible Answers:
Assuming v = Image Distance = Real Near Point = - 15cm = - 0.15 m
And u = Object Distance = Normal Near Point = 25cm = 0.25m, P = - 2.7 D
Or Assuming v = Image Distance = Real Far Point = - 15cm = - 0.15 m
And u = Object Distance = 𝐍𝐨𝐫𝐦𝐚𝐥 𝐅𝐚𝐫 𝐏𝐨𝐢𝐧𝐭 = ∞, P = - 6.7 D
2(d) A magnifying glass has a focal length of 6.4cm. An object of height 1.5cm is placed a
distance of 2.5cm away from the optical centre of the lens. By scale drawing, determine the
Image height and Image Distance.
Comment: Drawing to scale is not on the CAPE 2017 Syllabus.
It is on the CSEC 2013 Syllabus S.O.5.5 and is a standard CSEC type question.
A graph page could have been given for this part.
Teachers’ answers: Image height = 2.4 to 2.7cm and Image Distance = 4.0 to 4.5cm
2(e) CAPE 2017 Syllabus Unit 1 Module 2 Topic 3-Physics of the Eye & Ear, Specific Objective
3.9: Discuss the formation of focused images in the simple camera.
With the aid of a diagram, show how an image
is formed in a simple camera.
Comment: Straightforward CSEC skill.
2(f)(i) Define the term ‘refraction'
Comment: ‘Define ‘refraction’ is not on the CAPE 2017 Syllabus or the CSEC 2013 Syllabus.
Possible Answer: The [bending/change of direction] of [a wave/ray/light] as it passes from
one medium to another, due to its change of speed.
2(f)(ii) CAPE 2017 Syllabus Unit 1 Module 2 Topic 2- Properties of Waves, Specific Objectives
2.25-2.26: Explain Total Internal Reflection. Determine the value of Critical Angle.
A light ray hits a prism …Determine by calculation whether the ray will pass through PR.
Answer: 𝒄 = 41.1° , but ̂ = 45°. ̂ > 𝒄, ∴Total Internal Reflection at PR.
∴ Ray will not pass through PR.
2(f)(iii) On the diagram in Fig.3, draw the passage
of the ray of light through the glass prism.
Comment: Straightforward CSEC skill.
Question 3
3 Table Headings: Comment:
 Since a Quantity is defined as the product of a number and a unit, then a number is
the Quotient of a Quantity and a unit, i.e. Quantity/Unit.
 We would like to suggest that table headings and graphical axes should be
labelled Quantity/Unit. E.g. 𝐭/℃ and emf/mV
3(b)(i) Using the calibration curve plotted…determine the temperatures when the
thermocouple reads 51.0 mV
3
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
Comment: Read-off [depended on best fit curve]. There was a range of Best Fit Curves.
Teachers’ answers for Read-off ranged from t = 21 to 24 ℃ and 112 to 116 ℃
3(b)(ii) With reference to the answer in (b)(i) and the calibration curve..state why this
thermocouple is NOT the best option for use as a thermometer.
Possible Answers: Same emf at 2 different temperatures/ For same emf 2 possible values of
Temperature/Non Linear [scale]
3(c) CAPE 2017 Syllabus Unit 1 Module 3 Topic1- Design and use of Thermometers, Specific
Objective 1.3: Discuss the Advantages… of these Thermometers. Give typical situations
where…thermometers will be best suited…
State 2 Advantages of a thermocouple and give 2 corresponding situations where it will be the
best choice of thermometer.
Possible Answers:
Advantage
Situation
Remote Sensing
Kiln/Furnace/Stove/Toaster/Car
Engine/Remote Locations
Very High Temperatures
Kiln/Furnace/Stove/Toaster/Car Engine
Wide Temperature Range
Kiln/Furnace/Stove/Toaster/Car Engine
Fast Response/Quick Acting
Rapidly Varying Temperatures/Exhaust
Gases from Engines
Measures Surface Temperature
Any Surface
Measures Temperature of Small volumes
Any Small Volume
Measures Temperature at a point
Centre of a flame
Durable/Robust
Industry
3(d)(i) CAPE 2017 Syllabus Unit 1 Module 3 Topic 4- The kinetic Theory of Gases, Specific
Objective 4.3: Use… pV = nRT Module 3 Topic 5- First law of Thermodynamics, Specific
Objective 5.4 Draw graphs…different types of systems…isobaric, isochoric, adiabatic.
…sketch a P-V graph of the gas engine cycle, clearly labelling EACH stage and its
corresponding Pressure and Volume.
Comment: There is a contradiction between the Pressure and Volume given for stage C
and the Statement that an Adiabatic Expansion occurs from B to C.
∆U = Q + W. Adiabatic ∴ Q = 0, ∆U = W. Expansion ∴ - W, so ∆U = - W
i.e. When work is done by the gas in expanding adiabatically,
the Internal Energy Decreases and the Temperature Drops.
But in the question, the Temperature increased from B to C.
[TB = 800K. Using pV = nRT, TC =1393K].
So B to C was not an Adiabatic Expansion.
It could have been another type of expansion. [See second graph below].
Since the Data given for the stage B to C was conflicting, could credit be given to students who
drew a rectangle?
Answer: Using pV = nRT, PB = [6.9 or 6.91] x 105 Pa
4
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
3(d)(ii) CAPE 2017 Syllabus Unit 1 Module 3 Specific Objective 4.3: Use… pV = nRT
Calculate the initial Temperature of the gas (Stage A)
Answer: Using pV = nRT or PA/TA = PB/TB, TA = 694 or 695 K.
3(d)(iii) CAPE 2017 Syllabus Unit 1 Module 3- Either, Specific Objective 5.6: Solve problems
𝟑
involving or the First Law of Thermodynamics & Specific Objective 4.8: EK = 𝟐 n RT or
Specific Objective 4.3: Use… pV = nRT Or, Specific Objective 5.1:EH = n Cv ∆θ
Determine the heat energy supplied to the gas…from stage A to stage B
Answer:
𝟑
Either: Using 1st Law: QV = ∆U = 𝟐 V ∆P = 6.83 J = 6.8 J
𝟑
Or: Using 1st Law: QV = ∆U = 𝟐 n R ∆T = 6.87 J = 6.9 J
𝟑
Or: Using Definition of CV: QV = n CV ∆T = 𝟐 n R ∆T =6.87 J = 6.9 J
3(e) CAPE 2017 Syllabus Unit 1 Module 3 Topic 2- Thermal Properties, Specific Objective 2.1:
Express the Internal Energy of a System as the Sum of the Kinetic and Potential Energies
associated with the molecules of the System.
Define the Internal Energy of a Gas.
Comment:
The Syllabus gave Internal Energy of a System, not a Gas.
Possible Answers:
The Sum of the Kinetic and Potential Energies of the molecules [of the system]
Or, The Sum of the Kinetic Energies of the molecules [of the gas]
3(f) CAPE 2017 Syllabus Unit 1 Module 3 Topic 5- First law of Thermodynamics, Specific
Objective 5.5: Express the First Law of Thermodynamics: change in internal energy, the heat
supplied to the system and the work done on the system; ∆U = ∆Q + ∆W
State the First Law of Thermodynamics, clearly defining each term.
Answer:
∆U = ∆Q + ∆W or
∆U = Q + W
where
∆U = increase/change in internal energy [of the system]
∆Q or Q = heat supplied to the system or heat gained by the system
∆W or W = work done on the system.
Or
Q=U+W
where
Q = heat supplied to the system or heat gained by the system
U = increase/change in internal energy
W = work done by the system.
3(g) With reference to the First Law of Thermodynamics, calculate the change in internal
energy of the gas…from stage B to stage C.
Answer: Based on the information given: Adiabatic Q = 0, ∆U = W =37 J or -37J
Comment:
Contradiction in the question: For Adiabatic expansion ∴ ∆U =-W = -37 J i.e. When work is
done by the gas in expanding adiabatically, the Internal Energy Decreases and the Temperature
Drops. But in the question, the Temperature increased ∴ ∆U = +37 J?
3(h)…going from stage D back to stage A. State what happens to the volume, temperature and
pressure of the gas…
Answer:
VA = 5.0 x 10-5 m3, VD = 7.0 x 10-5 m3. VA < VD ∴ Volume decreases.
TA = 694 K. Using pV = nRT, TD =972K. TA < TD ∴ Temperature decreases.
PA = PD = 6.9 x 105 Pa. ∴ Pressure remains constant.
Or The Volume, Temperature and Pressure all return to their original values
5
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
General Comments
There are 5 Topics in Module 1, 3 Topics in Module 2 and 6 Topics in Module 3.
This Paper tested 1 Topic in Module 1, 2 Topics in Module 2 and 2 or 3? Topics in Module 3.
We believe that the Syllabus is too wide for such a limited range of Topics to be tested.
General Comments on Graphs
Question 1 - 1 Graph to Plot and 1 Graph to Sketch
Question 3 - 1 Graph to Plot and 1 Graph to Sketch
We would like to suggest that only one of the 3 Questions should test Graphical and Data
Analysis skills [XS skill].
This would allow the examination of a wider range of Topics [which would focus on KC and
UK skills].
Whilst the Paper 1 would test more Specific Objectives and hence a greater width of the
Syllabus, the Paper 2 should test the Mastery of a topic and its Specific Objectives.
General Comments from a variety of Teachers







The Questions could have been set to test most of the topics in each Module.
I am disturbed by the very small range of Topics examined.
Syllabus too wide for such limited testing.
No spread of Syllabus Coverage.
Too narrow a scope!
Better clustering of topics required.
Examining a wide range of Topics allows the students to capitalise on their individual
strengths. Examining a narrow range of Topics means that some students, who were weak in
those areas, did not stand a chance.
 The students prepared for a wide Range of Topics. It was unfair to test them on one third of
the Topics.
PENTT Examination Review Committee 2019
2019 U1 Review Contributors:
 Allison Ramsey: Teacher at Holy Faith Convent, Couva
 Anthony Ramdath: Teacher at St George’s College
 Bernadette Harrychandsingh-Hassranah: Teacher at Couva Government Secondary
[PENTT District Committee Coordinator for Caroni]
 Cheryl- Ann Ramadhar : Vice Principal at Chaguanas North Secondary
[formerly Teacher at St. Mary’s College]
 Cheryl Ramlogan: Teacher at St. Joseph’s Convent, San Fernando
[PENTT District Committee Coordinator for Victoria]
 Davatee Maharaj: Teacher at Vishnu Boys’ Hindu College
 Kamla Matthews: Retired Teacher [formerly at Holy Faith Convent, Couva]
[PENTT President, Exam Review Coordinator/Editor, U1Exam. Review Collator]
 Natasha Lewis- Dos Santos: Teacher at Queen’s Royal College
[PENTT District Committee Coordinator for POS]
 Nirmala Ramnarine: Teacher at Shiva Boys’ Hindu College
[PENTT District Committee Coordinator for St Patrick]
 Rhona Edwards-Cato: Teacher at Queen’s Royal College
[PENTT District Committee Coordinator for POS]
 Samlal Mannie: Retired Teacher [formerly at Carapichaima East Secondary]
 Vishalli Tancoo: Teacher at Naparima Girls’ High School
[PENTT District Committee Coordinators for Victoria]
2019 Review Committee Members - Feedback & Comments:
 Aarti Dowlath: Teacher at St Augustine Girls’ High School
[PENTT District Committee Coordinator for St George East]
6
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
 Allison Ramsey: Teacher at Holy Faith Convent, Couva
 Anthony Ramdath: Teacher at St George’s College
 Bernadette Harrychandsingh-Hassranah: Teacher at Couva Government Secondary
[PENTT District Committee Coordinator for Caroni]
 Charlene Ramkissoon-Vidale : Teacher at El Dorado West Secondary
[PENTT Vice president & District Committee Coordinator for St George East]
 Cherisse Crichlow: Teacher at St. Joseph Convent – St. Joseph
 Cheryl- Ann Ramadhar : Vice Principal at Chaguanas North Secondary
[formerly Teacher at St. Mary’s College]
 Cheryl Ramlogan: Teacher at St. Joseph’s Convent, San Fernando
[PENTT District Committee Coordinator for Victoria]
 Collin Samaru: Teacher at Presentation College, Chaguanas
 Daryll Williams: Teacher at St Benedict’s College
 Davatee Maharaj: Teacher at Vishnu Boys’ Hindu College
 Dianne Dilchan-Dass: Teacher at Tabaquite Secondary
[PENTT District Committee Coordinator for SEED]
 Hana Subratti: Teacher at Tableland Secondary
[PENTT District Committee Coordinator for SEED]
 Hollis Sankar: Retired Curriculum Coordinator MOETT
 James Tudor: Teacher at Barataria South Secondary
 Kamla Matthews: Retired Teacher [formerly at Holy Faith Convent, Couva]
[PENTT President, Exam Review Coordinator]
 Karen Dilraj: Teacher at Couva Government Secondary
 Larissa Mohammed: Teacher at Hillview College
[PENTT District Committee Coordinator for St George East]
 Melissa Jagroop: Teacher at Toco Secondary
 Nadira Nandlal: Teacher at Naparima College
[PENTT Vice president & District Committee Coordinator for Victoria]
 Natalie Cummings-Medina: Teacher at St. Joseph Convent – St. Joseph
 Natasha Lewis- Dos Santos: Teacher at Queen’s Royal College
[PENTT District Committee Coordinator for POS]
 Neil Garibsingh: Teacher at San Fernando Central Secondary
[PENTT District Committee Coordinator for Victoria]
 Nigel Superville: Teacher at Roxborough Secondary, Tobago
[PENTT District Committee Coordinator for Tobago]
 Nirmala Ramnarine: Teacher at Shiva Boys’ Hindu College
[PENTT District Committee Coordinator for St Patrick]
 Nirmala Singh: Teacher at Northeastern College
[PENTT District Committee Coordinator for NEED]
 Nisha Ramlal: Teacher at St. Joseph’s Convent, San Fernando
 Rhona Edwards-Cato: Teacher at Queen’s Royal College
[PENTT District Committee Coordinator for POS]
 Samlal Mannie: Retired Teacher [formerly at Carapichaima East Secondary]
 Sanjeev Maharaj: Curriculum Officer [formerly Teacher at Hillview College]
[PENTT Vice president & District Committee Coordinator for St George East]
 Sarah Shah Dookran: Teacher at Waterloo High
[PENTT District Committee Coordinator for Caroni]
 Suraj Gopaul: Teacher at Shiva Boys’ Hindu College
[PENTT District Committee Coordinator for St Patrick]
 Vernessa Sankar: Teacher at Chaguanas North Secondary
[PENTT District Committee Coordinator for Caroni]
 Vishalli Tancoo: Teacher at Naparima Girls’ High School
[PENTT District Committee Coordinators for Victoria]
PENTT Exam Review 2019 sent to:


Allister Ramrattan: MOETT Curriculum Officer
Anna Singh: MOETT Curriculum Officer
7
With the compliments of the Physics Educators Network of Trinidad and Tobago (PENTT)
02138020/CAPE/KMS 2018
C A R I B B E A N
E X A M I N A T I O N S
C O U N C I L
CARIBBEAN ADVANCED PROFICIENCY EXAMINATIONS®
PHYSICS
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
MAY/JUNE 2018
-202138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
Question 1.
S.O [3.3, 3.8, 3.15]
KC
(a)
“Newton’s first law of motion” states that a body
will continue in its state of rest or uniform motion/constant
velocity in a straight line (1),
unless a resultant/unbalanced/net external force (1)
makes it behave differently.
UK
XS
2
1 mark for each underlined point
(b)
(i)
(ii)
(iii)
Scale (linear & more than ½ of page)
– 1 mark
Axes (quantities & units & corr.orientation) - 1 mark
Line of best fit (any reasonable line)
-1 mark
Plot points (6 to 7 correct)– 2 marks
(5 to 6 correct) – (1)
(< 5 correct) = (0)
5
Maximum height - Point at which the line cuts the Xaxis
= 3.0 s +/- 5% (1)
1
At maximum height, v = 0 m/s.
V2 = u2 + 2 as (where u = 25 m/s) (1)
therefore, 0 = 252 + 2 × (-9.8) × s (1)
and hence s = 31.9 m
(1)
(iv)
3
P.E = mgh
(1)
= 0.02 × 9.8 × 31.9 = 6.25 J (1)
2
mv2
Alternative solution: ½
= K.E = P.E (1)
=0.5 x 0.02 x 25 x 25 = 6.25 J (1)
(v)
The value of P.E. calculated would be smaller (1)
because some energy would be converted to heat energy
(1) OR any other reasonable explanation.
Total 15 Marks
2
2
7
6
-302138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
Question 2.
S.O [2.5, 2.10, 3.5]
KC
(a)
(i)
Loudness of a sound is related to the intensity and
hence the energy of the sound wave. (1)
(ii)
Pitch of a sound is determined by the frequency (or
wavelength) of the sound.(1)
UK
XS
3
(iii) The quality of a sound is determined by the relative
strengths of the overtones. (1)
(b)
f/Hz
V/m3
1/√𝑽
225
250 × 10-6
63.2
255
200 ×
10-6
70.7
150 ×
10-6
81.6
125 ×
10-6
89.4
100 ×
10-6
100.0
400
80 ×
10-6
111.8
445
65 × 10-6
124.0
290
320
350
(i) Completed Column 3 (All 7 correct) - 2 marks
(5 to 6 correct) – 1 mark
(< 5 correct) – 0 mark
2
(ii) Scale (linear & more than ½ page)
– 1 mark
Axes (quantities & units & corr. Orientation) - 1
mark
Line of best fit - 1 mark
Plot points (5 to 7 correct)
– 1 mark
< 5 correct
- 0 marks
[Insist on Unit for quantities on the graph]
4
(iii) Gradient = (y2 – y1) / (x2 – x1)
1
(read-off values (1 XS)
= (434 – 212) / (120-61)
(1)
substitution (1)
= 3.8 Hz m1.5
(1)
(answer (1)) (accept 3.2 – 4.0)
(c)
c/2 
Slope =
A
L
(correct
2
(1) rearranging formula
to extract slope
c =
3.8  2
4
2.50  10
5.8  102
& substn)
(1) rearrange to make c the subj.
3
-402138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
c = 361.6 m/s
(1) (accept 340.0m/s to 385m/s)
(calculation)
Total 15 Marks
Question 2. (continued)
3
7
5
-502138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
Question 3.
S.O [6.7, 6.8, 6.9, 6.10]
-602138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
KC
UK
XS
(a)
6
(must have a Vernier to measure the
very small
extensions)





Measure the diameter of the wire and hence determine its
Cross Sectional Area, A. (1)
Measure the length, l0, of the wire. (1)
Vary the weights and note the corresponding extensions
(1). .
Plot a graph of weight versus extension and determine its
gradient (1)
Young’s modulus = (Gradient × l/a) (1).
1 mark for diagram
1 mark for each step up to 5 marks
(b)
(i)
Best fit line through the Origin (1)
1
(ii)
Energy density = area under the graph
determine area)
(1 for method to
Accept (1.54 to 1.86) x 105 J m-3
(1 actual
determination of area)
= (1.54 to 1.86 x 10-4 × J mm-3
(1 Unit
conversion)
(c)
(i)
(ii)
Stress / Strain(steel) = 2 × 1011;
(1)
Stress / Strain(copper) = 1.2 × 1011
Strain(steel) / Strain(copper) = 1.2 × 1011 / 2 × 1011 = 0.6
(1)
(not accepting 5:3 ratio for ratio of Strain
(steel)/Strain (copper))
esteel / ecopper = 0.6 (Eqn1) (1) recog. ratio of strains
= ratio of extns.
3
2
3
-702138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
esteel + ecopper = 0.0003 (Eqn 2)
Substn. eqn 1 into eqn 2
0.6ecopper + ecopper = 0.0003
(1)
Therefore, ecopper = 0.0003 / 1.6 = 0.00019m
(1)
Total 15 marks
8
7
Question 4.
S.O [4.2, 4.4, 4.5]
KC
(a)






Frictional forces(FF) act to oppose the tendency of a body
to move
FF act to slow down a body that is in motion (dynamic
friction)
[FF oppose forces which tend to move/accelerate a body] –
2 points above included in this statement.
It is a force which acts between two surfaces in contact
Its magnitude is proportional to the normal forces acting
on the 2 surfaces
Accept example/examples for 1 mark
(NATURE) Any 3 will be accepted (3marks)

It is due to the roughness of surfaces which are in contact
(1 mark), (CAUSE).

A minimum force is needed to cause a body at rest to begin
moving (1)
Friction causes the slowing downing of a body already in
motion
(1)
Friction causes the K.E of a moving body to be converted
into heat energy(1)
Can induce static electricity (1)
(EFFECT) (2 marks)



3
1
2
UK
XS
-802138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
(b)
(b)
(i)
(ii)

Weight, W acting downwards at midpoint of ladder AND
length of ladder. (1)

Frictional force, F at base of ladder acting along
the ground towards the wall AND base at 3.2 m away
from wall. (1)

Reaction force, S, at base of ladder acting upwards
also 3.2 m away from wall. (1)

Reaction force, R, at top of ladder acting away from
the wall.(1)
(For ease of marking)
4 Forces shown (3 marks); 2 distances shown (1 mark)
– subtract a mark for any missing.
4
The total/net torque about A is zero
(1) Recog.
moments has to be take OR statement of PoM
Therefore, (F × 4.5) + (245 × 1.6) = S × 3.2
4.5 F + 392 = 3.2 S
(1) Moments correctly taken
But S = 245 N (no resultant vertical/upward forces =
downward forces) (1)
Therefore, 4.5F + 392 = 784
(1)
i.e. F = 87.1 N
(1)
Total 15 marks
5
6
9
KC
UK
-
Question 5
S.O. [1.1, 1.2, 1.3, 1.4]
XS
-902138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
(a)
Simple harmonic motion refers to the motion of a particle
whose acceleration is always:
 directed towards a fixed point and
(1)
 directly proportional to its distance from that
point.(1)
(b)
a = -ω2 𝑥
(2)
mark for the minus sign)
(c)
The total energy (sum of P.E. and K.E) is constant OR total
energy of the oscillator is being interchanged between kinetic
and potential.
(1)
P.E. maximum; K.E. =0; on the ends of the oscillation.
(1)
K.E. maximum; P.E = 0; at the midpoint of oscillation.
(1)
(1 mark for the equation AND 1
2
2
3
Accept also (for 2 marks)
Non-zero P.E. and K.E. at all points between both ends and the
midpoint of oscillation.
(d)
(i)
(ii)
Maximum speed occurs when x = 0
i.e. Vmax = ωA = 0.785 × 5.0 = 3.9m/s
(1)
1
Maximum Kinetic Energy =
mv2 = 0.5 × 2.0 × (3.9)2
2
(1)
= 15.2 J
(e)
(i)
1
2
(1)
T = 2π/v
OR T = (2πr)/v
(1)
(v = rω → ω = v/r)
Therefore T =
2 r

=
2  3.14  2
= 2.5 seconds (1)subst;
5
3
(1)ans.
Alternatively, T = 2π/ω (1); Calculation of ω (1); Calculation of T
(1)
(ii)
T = 2π √ l / g 
= 2 × 3.14 ×
= 2.8 seconds
(2 / 9.8)
2
(1) substn.
(1) ans.
Total 15 marks
8
7
-
-1002138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
Question 6
S.O [4.1, 4.3, 5.5, 5.6]
KC
(a)
(b)
PV
=
(1)
n = PV / RT = (1.03 × 105 × 2.80 × 10-4) / (8.31 × 305)
=
1.14
×
10-2
(1)
(i)



(ii)



(c)
(d)
(e)
nRT
(1)
moles.
3
First law of thermodynamics: ∆U = ∆Q + ∆W (accept
any form as long as what follows is consistent.
(1)
Since it’s an adiabatic compression ∆Q = 0 and ∆W
is positive the work done on the gas is positive.
(1)
Hence ∆U is increasing & since temp. of a gas and
its internal energy are directly proportional, the
temperature rises
(1).
3
When a gas is compressed, the molecules of the gas
collide with greater frequency. (1)
As the piston moves down, molecules hitting the
piston rebound with a greater velocity and hence
their K.E. increases since K.E. = ½ mv2.
(1)
The temperature of a gas is directly related to the
average K.E. of the molecules, so if K.E. increases,
the temperature will increase. (1)
3
(P1V1 / T1) = (P2V2 / T2)
P1 = 1.03 × 105 Pa; V1 = 2.80 × 10-4 m3; T1 = 305 K
substn.
P2 = ? ; V2 = 3.4 × 10-5 m3; T2 = 795 K
Therefore P2 = (P1V1 / T1) × T2/V2
So P2 = 2.22 × 106 Pa
(1) Ans
UK
(1)
2
∆U = ∆Q + ∆W
Since no heat is added to the system, ∆Q = 0
So, ∆U = ∆W = 85 J
(1)
1
E = n Cv ∆θ
(1)
∆θ = 795 – 305 = 490 K
n = 1.14 × 10-2 moles
(1)
Substn. of all values
E = 85 J
So Cv = E/n ∆θ
= 85 / (1.14 × 10-2 x 490) = 15.23 J mol-1 K-1 (1) ans.
3
XS
-1102138020/CAPE/KMS 2018
PHYSICS
UNIT 1 — PAPER 02
KEY AND MARK SCHEME
Total 15 marks
6
9
-
02138020/Cr8rytB 2017
CARIBBEAN
EXAUTNATTONS
COT'NCrL
CARIBBEAI.I N)VANCED PROFICIENCY EXAI,IINATIONS.
PHTSICS
T'NIT1-PAPER02
I,IARK SCHEI,IE
HAY/.,I,NE 2017
-2-
o28eo2o/cuEls 20L7
PEYSICS
T'NTT1-PAPER02
}IARK SCITEME
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I
-3-
02138020/CrPrlB 20L7
PEYSICS
I,IIITl.PAPER02
I.IARK SCHEME
Orestion 1 (continued)
rc gr
(a)
Graph:
Labelled Axes correct orientationGood Scale (i.e. no multiples of
etc.) more than 50t of grid area -
correctly Plotted Points - . (2) r3
(-1 for orctr tacorarct plotl
Three straight Lines -
(b)
(c)
(d)
(1, I3
(u r8
13
1
3 or odd nuubers
I
(1) rs
2
[5 urfr,
(1) Uf
Block accelerates dorn the plane.
Block collides uith uall and rebounds. (U l[[
(U ttrt
Block decelerates up the plane.
(tC or zturr dva fa {rcr;J.lttlo - rc nrtr)
[3 rrlr,
I
ISI
I
1
I
tllt
(i) Acceleration = Gradient of initial slope (first
(1, l[
segroent of graph) (SOI)
(U u
= S=3.4ms-2
Ll{r15
(ii1 tength of incline - Area under the graph (U G
(Ullf
=0.5xL.20x4
- 2.1O m Correct ansuer (U fll
G
I€ngth of incline - Eguation of notion
lilust use acceleration as
calculated in Part c(i)
e.c.t. tril
ansuer from calculation
Correct
,
catcutatton above
(F.
A.)
S.
.
._)
veloclt.les are dlfferent (in both rnagmLtude and
(l E,
Block loses energy during collision (1 rc,
direction) (sol)
lotrl' 15 hr}t
1
I
1
1
1
I
1
2
8
5
-{-
o2138020/CUAnG 2017
PETSTCS
T'NTTl-PAPER02
I,IARK SCHEME
Ouestion 2
g.o [2.15, 2.16, 2.181
rc t[[ I8
(a)
tct5
allc
Laaa Lilatar
(acctc,
r.}--
-c====----=--
O.
D
(1 rrrk for rreagnat of -rl pior of qrpntnr) lld
(1 rart for nlrtlrr dtrtru D f.sgt er cqlrd
to
rt,t t)
a
t{ith the apparatus setup as shoun above, ensure that
the laser bean passes un-deviated through the narror
xazor slit to strike the screen (U.
a
A spreading pattern of bright and dark fringes wiII
be observed on the screen (1) which indlcates that
the light waves bend around the edges of the slit (1)
and hence disproves the partlculate aature of light
(or supports the uave nature of light). (1,
16
rrrtr, El
6
-5-
o2L38020lcI,E!/S 20L7
PEYSICS
T'NTT1-PAPER02
I,IARX SCEEI|E
Orestion 2 (continqgQ
rc
(b)
ott
P
(1)
I
8@ro
I
D
3ctu
[1 D* rcl
[1 srl
From diagram: tan g = ylD and sin 0 = A/a
tltren D >> a
1
lrfl
1
For sruall O; sin 0 - tan e (U
I
ThereforeY=AD/a (11
1
[2 rrrLr, El
(c) (i)
y-
3.6
4
[1 nrl,
= 0.9 mt - 0.9 x 10-qm
^va
-'.4--
tEl
0.9x10-3x0.55x10-3
0.8
d
IBI - robotihtton
[1 urt, Bl - .Drrr8
fringes: l.e. 3 tEw
f - 3.6 m /3 - 1.2 m or 1.2 r10-r u.
.... I = yalD - 8.1 rl0-? r
to the 4e dlaL tringe [a - 0 to 41: f .". * qr.N
I = valD - ?.0 rtO-? r
1
1
[1 rst,
- 6.3 x !Q-? I
1
x8
-6-
02138020/CrDrlS 20L7
PEYSICS
T'NTT1-PAPER02
I,iARK SCHEME
Question 2 (continued)
(c) (ii)
1 rrrl.qurl
1 rert dor
tat E ttl,
of poetr
[2 rerlr, Ilf,l
rc tf, I8
2
lotd 15 rertr
3
6
6
02138020/CrDt/:E 201?
PETSICS
UNTTl-PAPER02
I,IARK SCHEME
Question 3
8.O [5.t, 5.5, 5.6t
rc
(a)
69.= I + tl
OR AO
- AU + An (in this case An Ls work
done br)
(1 rc)
1
(1 rc)
I
(AU = changeiniruoalcoetp
I Q = t catruppllcdtosystem
I W=worhdoneonslrstm
lll
3 cor:*t
12
rrstr, El
T'f,
18
-8-
02138020/Cu!lB 20Lt
PEYSICS
T'NIT1-PAPER02
T,IARK SCHEME
Ouestion 3 (continued)
(b) (i)
a.
at
o
o
lllll
tarra rllrl taall rrrr lrrll
o
rrtlr!rltt
rrrrr
lllll
llrar
!tlIr
o
rllltlltlr
aatrt
Etrlrrola, P(lfoo Da,
t
o
-
t
o
C'
rrlar aaal2
atall ttlat rlarl talll llraa a!arr
rlaaa ltaat trlla lrrrl
traat
trrlt ltrrt ttlll
raal tlltl
lllrr rrrrr !lill
trrlt tttrl I l lM tlttl
tlllr
llt!t
trlll
Itttl trrrt rllrl rtara Itatr
trlal rrlrl lllrt
lflllrllla
ttrtt
tltl!
!trat rttlt rrrrr
rarrl lllll
rtlal lltlt:ltrr
rlrlttlllt
tll la rrlat lllrl
tttll
lll:t lllll
ttttr
ltara
lllll
atlat lrall tltlt
tllla
allal tllta laral ratlr rrraa lrrat
llatr ttalt tllll
lrtla Italr llrll
rllal Itltr tltaa atrll aalrl trrta llaaa atata
rat!a !ltlt
aa lat aaala lalla lala l ttart lllaa IIIII
rlttt !!!rl llrlr
all l t t rlla ltlll
tltll
tttll
!lralttta!
llrra
!!rlt rrltl Ill ll lttrl rllltlatrt
rtrtl
rrlrl
lllrl
Iatrr
Itrlt
arrll
rtttt
:trfl
llllr
rlaal
lllrl
!rrrt
latar rr rrt lalll Irrrr rrrrt llral ltltl
Itltr
trtrl ltl!a lrrrl talrl lrlrl lrlrt Itrlt
TTIII atart !rrll
rltlt rlrrr
lrlll
ltllr trra!
ttitr aEa
alall
ll-ra aaatt tt!tt latrl rrrtl rlna trait rrrtr
rltlt rarrr lalra tIIrr rrrtt lllat lat!l
rEn tlttt tEn rltra anu alalt rlnr laatl
rtraa
taala trrtr
r-!at datt alttr rr-al alataED
rrlal
rllll
ttttt
llrll
III'I
tlltt
!rrrl
trlat
rtrar
rtrtr luat rltll ltltt rrttl f tur lrral allll Ittrt rttrl ltltr
rrrrt tltll
lrttt ttlla tlrll
Itlll
ltara rrart tllrr T"II rlllt
lllll
tllta tIrI
lrrlr tlrrt rrrll raall allrt trrtt tirrl
IttIt
latlt rrrll aiaat
llrrr rrrtt ltlta trttl
tlrrl lrrll llltt
lrltt
aaraa llll!
!lIll
ttrlt !!!ll
lralr
tltla !laaa
rrlrr rrlrt !latr ltltl
laarr ratll llllt
raaar rllla llatt aaatr laall tlrlr
laaat llaaa laaal ttllt
tltar ataaa aaaaa rltta talaa ltata airar
lttll
arrat
tllat ltrrr !arIt Ilral llrlt
rrltl
lrlal rrrrl
lattt IIIII
trEra !lart ttalt lrltr !rll!
lrltr
attrllllta
rrlll Ittll
llrrr lrrrt IlllI
ratlt
ttltl
!laIl
lttlr tlllr
lrltl
I lll I Irrrl
o Itllt
arrrl arata rrtlr
lrrtl
!ratt trllr llITII'ltl tl:tl
rlatl
tllrr rallI
tltlt
arrl!
l:Ira
ttlll
lllll
tlrla
I"II
at-tttrtrl
:taat
rrarr III'I
taarl tlrrr
rarra
raatr
rrarl
araaa
atrra
o
rrial
rarra tarar a"I'Itara r raal aa arl rrlar'---.:alIl
alalr
rrraa
Irata .II"
rarra
aalaa
arrar
i-_,-. llllt
lltll
tllrl
o ttrlt ltttl allll llrll Ittll
!aarl rIlL
tlrrl
Ittlt rrrtt ltltt rrrll rrlrl!
III!'
raaa lrrr r I rr aa a)- -- - alrrr tttrr laaar Itar!
arrrl
trttt IIIT' lllrl
trrll rrtla alral ttrtt
Ialat alrtt llllt
rrata tllrl l!tal
llrrl
rrrli
lrttr
rrr-rrllr ltlrt ltrrl
I'ITI
!rlrr irlli
ItrIr rrIIl lllll
-aaal
llall aElt
lllaa taaaa !rlll
ralll taa-l Iula lattr rulr lalta aaala lMl alaaa. rl-fl
ttait llllt
ltlll
tEtr
aErI rrrrtl![lattrl
trlat lEat talaa ltaat ltlMar---rrl arrll
Itttl rEaa aarll tlala aral Irtt
lrrll lalll ftttt laat! llatl !rrrl
IIIII lIIrr !aatl rtttt
!lrlr !rlIl
Itrrr ltltl
t!!tt trrtl llltl
lurl tltlt
!ttll
llrll
arllt rrtrt
lllll
tttal IITII llarr rrrrl llr--.
lEar rrrll
IIIII rrtrl
r:Itr rrrrr lllll
trrrt Ertl tarr trtll
rrrlt ,--rtt !!rrr tllrl rEll
rnll ltrrr
I'III
rtllr rraa tallr rElr tIT!'
tlltl
rtrll rllal tlttt
trllr llllt
ltlrt laarl Itttl larat aaarl arrll attta
larrr tllll
tt.U rar[ lllll
trrrr tattt trraa attaa
taara a, -ar tlltr alraa rllllllaaa
trrll trtrr rratt
ttlaa lalaa rllal llaaa !IIII
rrrrl tttlt ltrrt
talat tlttt arlra rtIIl aarlr
t!alI ttatt'.altt
aal!a trtlt
!lttr llltl
ttrtt !lall lllll
rtrrl ltttl
!lttl
tltrt
rrrllllrrl
rarltlatr!
rltta
II'IT tarta rrtlt !llu.l!aIl
lrrrt rrrlt tlalt rrrrr rlrll
lrrlt atall tltll
tllla rrara llarr tllrr
rIIII l,- ,aaaaaaa rlltl
rlall larlt
ITIII ltlaa
ilrrr trrra
talaa
latll !lllt
rlalt
tlrlllalra
lllatttalt
alaa
laall
lalar
llIlT
arttt
rltlr
Ilatt alall lllrl llrrl
tatll
rtall
,aatt
llrlt
trtra
ttltl
rrtrt
tarrl
zlall rlrll lrlll
tlarr aaall .IIII
!llrr lalta tltrt arra! rtatt
trtll tlaft
tllll
ttrrr rrrl! rllll
rtltr lltrt a)af / llllt
llrar t!!ll
llttr rrrt! llll!
rllrl
llrtl
!IIII
rrrrr
lllll
!rlll
rull arlar
ltt"l
llrll
trltl lIIll llraa II!TI Itttr rtlt!
C' lrlrr
lraE lrlrl l!aal rlnr r!all lurt
!rrrl frrll rrrtr
aaf aa lrlrr
MH trrlr U
ll!4, tlrrr ITIII
l!!!t trtlt
lallr laarr ralrl rr!!r tarn llEt
aaaar lrttl llltl
/IT tlall
lEtl tllll
llrla lratl fsal IIIII
C) tlrla
a-tal aarar
trttt tarrr aaan llrat IIIII
rr aat llarl lrall
taalt
rr-ta
taat! EI!'
aaaat
lEat
!rtra
utra ltltr IITII
o
lrrrt tlrlt lIlt
trtll a taa tlttMrrlrrlll rrart
trrll tatla llrI
trtt! llrll
lttlt
rIIII ttrrl
t' Ettl
llaat !rarl tlrrr lrtll
ITtD '.rrar Ittrl
!ltlt
lrrar
tllrt lrlll
atrra lltlt
rlttl !rral tttrl
lllll
ttall trrar llall
,atal aalll
latr! ltrla
trrtt !tlll
ltrrl rtrtt IITII rlllt lllll
atrrr
ttttl
laarr ttlla ataat lrlrl lltll
ralla lllll
llltt
latd alrat trtl! laala aallr Itarr
ralattaaat
trtII taarl lalll tlatl alaal alaaa
traaa lttal
taaar ralla ralll laara aartt llaaa trr, ,la taaaa
lrlll
rlrtl rraal llarl trrrl llrtt lltlltlltl lrr!l
rrlll
llrrl IIITI
tttlt lllrr lla
Itllr llrtt tlalr lrraa
lrrlt ltlrr rrrlr llrla ltlta lratl ltlrt
tr!rr lrtlt
ltltr ttttl lllrr !tt ll rllll
tatl! ttrtr tlllt
ttllt
laarl
trrrlallll
ttlll
Itltl
tlltt !rrlr
llt!! lllll
ltrl! II'II
al /ll lltll
ttrrl trarr raalr fllrr
Itltl
rtrt! rralt atrra
rraar tllll
llllt
ttrlr ltlra tltll
rr art !trlt
llrtr lltaa trllr lrarl f!art ltllr
.T'II
II'I'
tlata rlrra
Ilaar rrtat ItalI lrllr alllataatl
/aar rlrll
latta rlraa rrrll iiala rratl tnat IT altt
lrrl! !lltt ralar
ttllt trllr lrltt
ltrll
!rtlr llll!
ltlll
llllr
tarta trttr
ltrrt lallt tlltl tll!t
tltll
tllrl
l!lrl
rltaa lrtrl rrrar rrrll ltrrl
tllrr tltll
!IIII latta rlrtl Irlll
rllll
lrlll Ilrll
llltl
llltl
lt![
llIUlrl[
rErr rarll llfll
IE'T
lltlt
atarr allll
attll
lltar rrtll llnl
lllll
ltttl ErtI
ataat ltad allu aarar ltarr Iaaal tr.] lEal lrlEllu traaa
lEat
ttlaa iiaal laall Errl taalr Itr', trrtt
o(
II'E
latrt
tran
rrrra
lrara
aattl
aarat
rtal
,
llaal
lIIal
!aaar
laraa
aaaa
arala
aara!
laaal
iaara
P
TIE' tlrlt rrllr
rrlli ltrll
lttI
larlt IIII'
lrtrt
lnll
I'II! rllal
rl![
tt[r
lalll rrrrr lrall rllr,r
,ltlr
ttllr !alrt rlrrl
larrr ralll llIlI
tltll
ttttl I'III
rtr rt trrrr lnll
rrrn lll!! llrrr tllll
C' Iaarr
larlr Itrtl
ttrl! ltltllttll
tatlt latlr rlllr
IIIII
Itta! rtala
rll rr tI.II
trtrr rrttl lll!! Irarl llrrt
TIIII
rrtlt lallalratr
rllll
lallr
laaat
allra
lrrrr
tlrlt
trr
aalrt
ltlll
rrllt
lr
trrra rrtlt trlla ltrrr
o
attta
lrtla Itral ar-la
la
ltl
tarrl
rrtaa
attlt
liiia
tallr
atarttlla!
Irara tt ul!
lllallrrlr
IIII'
ITIT:IITII
C' laraararlt
rrrrr
llrrr
ttttt
trtlt
ttrtl
llrlr
tlttr
tttlt I rllr ttlaf lIlll
tttrl tttlr I llal lllrl
lllal
IIIT!
rllta tllar rrrrr
ttltl
!alrl
L
lrtat
Itrtt
lllll
IIITT
rlatt f ,aaa tllrr
rllat tlllt
lrrlt llttl
a
trarl
trrll
lrtll
IIIII
ttllt
lrrrr
tllrl aaiaa tlral llraa attaa !lrtt
tllll
tltrr !tlaI
tttll
IlIll
lrtll
rtrr. rrrll rllll
'.lttt
alrll
tlitt
larrr
llttt
lal aa aaaat aaall lalrl artta aall r laalt lllaa lllar ratat
IIIII
allta
alttt aattl trata lrrla talat latll
ttr tarlt latlr tarar trlla lllrr aaiaa
tlatr lrrrr aallI
l!al!
rlart
rlrll TTIIT lrltl
lllll
lrlrr trltl rrarr tlrat rrrrr
llll
l t[!t
lllta
lllll
lllll
tllra tttll
lrrtl ttart IIIIT ralrr IIIII
ltllt
II'II ttrtt aarrt tltrr tllra
trU I
trarl
trlll ltrll
lrtal llltt
ITTE Htl
tl:rl
lmt
EIII
tllLl
IITII
tltrl lEn lltaa rDrr utrl rlllr
tlllr
ft!at rrtll lrlll
IITII llarl
rllll
tltE
llur
lllll
!ITII atra! ftrrt ltrrt lttta
tlrrr
tulr rrrrl ttttt !trrr lrr tl
ttlar t-taa
laaar raaal lEll taaaa lattt
tutl
taial tlrar raaaa
tarai taara rttlt lt-tl tat at ataaa aaaat
tttll
lltrr ltrlr llrat lttll
tllrl
!ltDlllm
tttrr !lrtl Ittlr lrtlt
lrlat
lrrar rtall rllla lErr lI'/ll
ratll
IITII llltl
lalll !ttrl
!aatr llaat !tIIt !ttla allrr
taa!a Illar tl ata aalat !aaar ltaat
ItItr rtlll
llttr !ltll
llttt
rrtrt
!lart Itrrl tlrtr rrrll altlr
lrlll
r, ltl IIIII
!lrrr
IIT'I
,la:t rrrla
llltllllrr
llrll
tiall
tlllr
tttat ralat
II'II rrtla talli altra llllr
r,,lta ll
lallrrlrrl
lltar rarla llllaltaaa
r rr tll lrlra aaaar lrart !ralr atala
tarlt alra! tarll rlala llll
tttt al
ttlrt
tlrlr ilaaa lllal laara I, ttl!
tltrl rttlr rttlt
ttrrr rlll! laalt lrall
lltll
Itala alfrr Ilatt t!!tl
II
rllll
talra
lllrt ltltt
!tttt
lrltt taatt lllll
tt ttt rrrll tllll
lattt rlltt tlrrl llltl
Itlt llttt
!taal
tlrl r tr ll! lrlll tlrrr
arlll tlarl llrtl
ttrtr rlrtl lllal llall
latrl larll tltla llllt
ltlt IIIII ltlla
llltl
trrrl
latat ltllt
tarlt
Itta! tltal lalrt
tlltt
!tllMal!
ttllt
tllra
ITIII
llttt
altt
lrltt
ttlaa
ItlrlttllT
llalrrrrr!
C' lalrt taaaa tttlt
taaaa attta aaaar
raarl lllll
I'III
alrra
llatt
taalr
trrat
ttllt
allat
/tr!a
lEta
larrt
o trrrr rrttt ttala !rrrr rtatt rElr ttarl rlltl
!lrtl
rrfat Itltl rllr
tltrt trrrr :rlll nltl
arlrl lrtll
ttlrl
llra ratal tEtl
rrrra
rIEl
trln
lrtrl
tlrra
rrltt
laatr
/
rrlll
tllrl
rrlll
llta
llltl
lttar
aaall
!utt
rrrfl
latll
trtrt
Irrla trtlt
rrrtt
tlEi
ltlrr
rtlat
ttall
r
rrrat
lrl[
Itllt
lrrrt lrrrl lrala lrrr
rlrlt
tlut
tltar IIT'T larll
llrrr
llrrr
lrlrt
!larl
rEII
!atar
lrlrl
l!rll
lrllt rraar rllrl Ear a rirll
llall
tatal
tlttr raaat
lllrl
tarrt
tlttrtlarr
IIIII
lalaa
altlt tlaaa
tlttt llrr rtI rtilt
,ttrr raaaa
ltlla
r.TII
rllll
lllll
lrttl
laarr
larrl
tIrIt
rrrll
I!m llrrl
rrtlt llllt
rrttr tlllt
tE
[lrt
!T'II
aarlr tata!
rlttl t:r!r arrar rrrrl
tt!H arlat lrtta
al rrrlt ttrl! Ittrl
ltltlrllla
atlllltr
!llrl
rrtlt alatr
l!aal
rrrlt rrtlr tllrr llrll
rartt Illtt
lllli
trrla
trlll lllat ttI l a lll lllr !tlaI
ttatt
Irrar
ltlal
llatl
trlr!
Ittrl rtlrr tlaal tlllt
tlrttlllrr
trtttltattrlalrlrr
tarat
lll
tatlt lllrr
tlalr trtaa allal alrla
lallt
rr rarttlllll
ttlaraarraaalllaar
lllta
ltt lttII tllla tll Il llrtl
rIIIT ltllr
ttral
lll tr!atltrll
lltl!
allll
talat
aa
o
ttlla
alttt rrlrl lrlll
aaatt tttll
tllll
lr lra IITII TTIT' IIIII rrrtl lrlrr tlrlt.IT tatlt
car
llllt
rrltl tlrll
alrrt
llttr ttaaa rltlt
!lrtl rrlll
rtatl
ll
ttl
rrrltlallt
lltll
lttlt
taala
tlttt ataat attla llrll
rrrrarrraaltllltt
aaa ,ataa aaatl tlaar rtaat taarr
l.a
r-rra aalra
larat ,llla lrrrl atlt! alrrr
rallr aataa trllt lt raa rataa ltaat
J
a
lttII araal lrrrl rllll
lttll
alrrr IITII
rarrl rr ,rrr IITII IIIII
tlltr
lrtrr
tltrt llttt lttrl !rlll llln
tatl lflrl
rtrH rarrt talll a' rall laltt llatl
rrall
tttE arlll lnrl
rrtlt !tarl ltr!t ftlfl
lr!a taatl tltrl tttrt
lalal r7 llE
ttrtr tarat
I'TI'
lrttt lllra rltta lrltr aatll trlll
larrl lrlll
lrrla lllll
C' laara
lrrlt tlata tttrt ltral
Eatt aaaaa larll ! /taa aaalr llrlr iraaa ltatl tlrlt rtala
o
ttrll llrrr rrlll llrll
aratt Itall ltrll
r lrrr ltrlt
trltr
lrrra
tratt
talll
ltrrr
!ratt
It lrrrrraraaaallll
llrrr rrlrl lllII
rrttl ararr !trll !llll
llra rlltl tltrt lllll
I'III
lttrt ralrr llll!
rtrll ttttt
Mrt IIIII llllr
I rlll
trarl tlarr
Itttt
trttr lllll
aarrl
rtrra laaar rrlat rrlll
lalaalalllarlat,atla
tttar taata atarr rtlaa IIIII itaat
rrrll lllat aalla rrart lrllr lrrlt araat
atllr attla taatt
rrrrr rrrrr trtla rrrat trara aalar tltal ttata
lllr!
trrrl rlrrt lltlt
rrlrl tltrl trrll frlll
rrltt rlrla Itrlr rlrta
llrlalrlltllrlll
lrlal tattllllrt
trtlt
rrrrr rrlrl lllrrtrrrrtllll
trtll rrlra Itlrl lrrat tllrrlrrtl
rrrtrrrlrr
tltrr tralt raltt lIlIr tltlr
rrtrr rltrl llIIr llal!
rrlll alrrr !llal lralr
artll llrlr allrl
tlllt tlltrtlttt
ltlta tttll
tlllt
lllrr
tlttl
lalll
ttltl
lltll
lttll
tllra
rrrlt
rtrtr
tttta
rrrlt
rtllr
allrt
ltttt
lllll
rrrrr rrart ttttt
riall
llatt
rlrra
Itrar
tlata
tlllr
aalaa
tlrt!
trrlr
rltlt
lrlal
alrll Ittll !tara T!IIt
trraa Iart ttlll
talal rltrt ttttr latlr l[tl
tlttt .IITI trrtt tnn ttltr
l allra
rrtll rlrla trrrr tltlt nlu trtrl tlllt
!lrat
rtttr
ttlar
lrlrt
!lall
rttrl
lrrat
lrrrr
llrlr rartt rrrtt ttrat larrl lllal rrrrr arrrr lttal !Ett
trrar ralrt ttrlt tattt rrtlt tttrr lttta rillt
!tlal
rlrrl lratl tlrrr ttltt I'III ltrtt
rrrrr rarrr tlttl
lrlai lllf
tattt
llall lrrra lrlat tlala ltllt
ratll tllla laaaa lrlta t-aat IITII atlat rritt
I taaaatrral
tltra ltalt aatat ataaa
trlt! taara ltlla trlll
Itralrrrarttaal
alara trt!t aalar ttarl tarlt traar laalr
rlrrr raarl allll lrrtt
rllll
rrllt rtrrr tllla rtlll
ltr!a tltll
trlrt !tll I l!!tt ttalt altlr l!ala
I lltlt
tlalt taalr
alrrr rrllt llll I lrrrarllrl
lrtta
ltllt
tlllt
ttllltllrl
llll
I llllr
IIIfI lltrt rITIT tlr!!
ltrlt
tarrrrrrrt
ttlll
tllrl
lrtal lrrar rrrll
IIIIIITIIT
trtrt
arrtt latrl ltlrr rTIII
ltt!a alall
!tt ta tr lrr ralll tlarl llrartrtra
rrrrr rrllt
rr-ar tltl
arar !I lrrtr
lllll
talrtlltla
aaIlt tlall tllll
ttlat
o
aatra t trtt aaatl !aa tt llrltlalltallla
tt
a trll
aatll
alr
tlalr
tatlt
I taaaa ltatt
lt
tl2lllll
tltra
C' I tarrr rratr !tllt
ltla!
alrli !alrlarlla
la r lalI rrlll attlt ltllr
tttat
IIIII
tal
rtrtl
lrrtr
II'II
lttlt
rrrr llttl
lrrrrrrrrl
O I lrrrt ltttl
rrltl fli ll t rttt rrrll ttlar r!rrrl
IttIt lalrt Itllt
lalalllrll
!lIll
lllll
lltrr lltrl
rrllt ttl
ftlat
rlllt
I tllrt
llllt
lltl!
!tara
lrtrr Itarl
ttrar !rala llrrr rairr raaaa ltlal
tataa llrrr
lll
talir
IiIII
' /rtrr rrtaa ltrtt
llllt
tarra rutr
I
1
i
l
l
I
t
I
-
8t
c
e
L
I
L
I
I
I
I
I
I
I
I
I
I
-9-
o2138020/CrElB 20L7
PEYSICS
UNTTl-PT.PER02
I,TARK SCHEHE
Orestion 3 (continued)
rc
(b)
(i)
Labelled tures (1 fS1
Good Scale (no ruultiples of 3 etc. >50t of grid)
(1 rc)
(1
Points !l8, lll 6 Potat conrct
Best smooth cunre (1 18)
[l rrlc,
(ii) ' Irork done -
675 t
(iii)
30 i,
50 J
(2 lfr)
(1 tll oofYl
[3 rrrtr,
1
I
1
I8l
1
2
Gl
W - nRT
n--
18
1
[ff."]ff:'::T] .tlrTn"' suitabre 'nethod
= 6?5 t
T'f,
1
Ptt
RT
(1 Ur,
= 0.16 moles (1 u0 (can use any point
from table or graPh)
#
1
[Using {.00 x 10s Pa and 0.001 n3 from Table 2l
[2 rrrtr,
(iv)
OBI
AU=e+If
1
Since expansion is isothermal, (1 rcl
T : constant =) AU - 0
1
i. No change in internal energy (1 Uf)
[2rrrLl, 1rc+tUtrl
(v)
au-o+w
0 -AU-n
1
(1 url
-Q-(-675J)
- +675 J (1 t[r)
[2 rertr,.tnrl
f .E....If b(tl) or (tv) taorarct th.a .cf
lotel 15 trrlr
I
3
8
I
-10-
o2t3ao2olcLP!,ls 20t7
PHYSICS
I'NTT1-PAPER02
I,IARK SCHEI.IE
Ouestion I
g.o. 11.3, 2.L, 2.2, 3.5, 5.U
rc
(a)
(it The SI base unit of vo1rrc is
u I9
m3. (11
[1 Dst, El
1
(ii)
v
t
tPrr
*1
Rearrange to make C the subJect of the formula:-
C=
rPr|t
w7
1
(1 ur)
Substitution for nnits of eacb quantity:Unit of C =
.rg n-r-2
I
s ms
T
1
(1 u$
- kg m-1 s-1
(1 ur)
(drra.loo.l raellnir eccqrt$lo)
(rarrrr PSA b. iD brr uatt l
(b)
[3 nstr,
(i) tfork done by a force is the force x @
. (1'
Unit: Nm or J or kgm23-2 (1)
(il)
1
I
[2 usts, El
Corq)onent of reight ' 840 x 9.8 x sin(7.0)
(1 I8l - Go88.Gt trig
= 1003.2 N I 1003 N
(1 E, -
I[Il
1
1
lurcttoa
robott'totto t t^otr.a
1
[2 rrrtr,
Gl
-11-
o2l,3ao2olctDus 20L7
PEYSICS
IINITl-PAPER02
}RRT SCHEI,TE
Ouestion a (contlnued)
rc 0r
(c)
(i) Resultant force acting on car during deceleration (1 t[t,
({500n-1003)N=3rt97N
Fnet = ma :.
. = *=X=42ms-2
18
1
(1 E'
1
[2 rrrtr, ltrtl
(ii).rr2-u2+2as
(1 !C)
.0-202+21_4.21 s
s - 47.6 m (1 ltr3,
(iii)
1
J2 rrrls,
I
E,ttfl
F* = lrmo2
= 0.5 x 8rl0 x lQ2 - 1.7 x 103J Subatttutt6a t aaaf,at
(168000 J)
[1 lrt, IEI
(iv1 tfork done
I
1
: ffi; : f;:**
o -2.1 , 10s J SrtiEtto
t .Dtnr
rfgo)
(ao prorlQr for aot breflg rgntlrr
[1 ru*, url
(v) Ouantitles are not equal because the initial potential
energy of the car ras not taken inEo account.
[1 rrL, UlEl
totel 15 urte
1
a
11
0
a
-t2-
ozl3i,lo2olcrEI,lB 20L7
PEYSICS
UNTTl-PAPER02
I,IARK SCHEME
Ouestion 5
8.O. 12.5, 2.12, 3.2, 3.31
(al
(t) Ttrreshold of hearing refers to the intensity level
rhere a rmd 1l lurt rodtblo. (1 rc)
rhereas threshold of pain refers to the intensity
Ievel beyond at rhich rouad Dotag to pro&o
dteodort or erla. OR the maximrrm intensity that can
be heard sithout pain. (1 rc)
(ii)
a
a
[2 urtr,
The human ear responds to a ride range of
1
frequencies.' G rc)
The response of the hunan ear to sound intensity
is logarithmic. (1 rc,
r (dB) = to roe(il
1
To determine Intensity in tl m-z today
-> 115 = ,o rre(il(l tE,
=> 115
11.5=IgI-lgIo
11.5-lgI-Ig(10-12)
11.5-19I-(-12)
11.5 - 19 I + 12 (1 flll
11.5 -12 - Ig I
-0.5 - 19 I (1 Ur)
lO-0.5 - J
0.316 *D-2 = r (1 url
I
Et
(1 I[)
- 'o b.(il(l rB'
-) 11.r - b8G) (1 url
1
El
[2 trrt't,
(iii)
rc ur I8
I
-) 11.r - bsG) (1 ur)
.:.lgll.s = 3.16x 10u = (l)fr rrf
... f = (316 x t011xro) (1
1
1
1
1
url
- 3.16x 1011 xl x 10-u
i
= 3.16 x l0-tWm-2(1 1p0
1
I
[5 rrtr,
I[tl
a
-13-
o2l38U20lCrD!/B 20L7
PETSICS
UNITl-PA.PER02
I,TARK SCHEME
Ouestion 5 (continued)
rc
(b)
(i)
Ilaves going toward the uaII Lnterfere rlth waves
reflected frou the uall and nlni"n occur rhere there
is destructive interference bptreen incident and
reflected raves. (1 rc)
l{inina are not zero intensity because the reflected
waves have srnaller arylitudes than the incident ra\res.
. (1 rcl
[2 rrrtc,
(ii)
I8
I
I
El
1
{ loops betueen raII and speaker -> there are 4 |
between the wall and the speaker.
v-fh
TE
1
(1 E)
1
* i=0.2S. => l=15m
f =1, =#=zztnz
Otbor Eoastbl,o SolutLos:
(1 url
(1 rB'
[3 nrlr,
t[[l
-ti
4i internodal loops - 22 xiT -3m ...4_i
m
f = v/l
f=2558r
si -3m *r.-3,
f = v/It
Eor v
= 340 ms-l
f-283E:
lotel 15 rrrtr
6
9
0
o
-1tl-
o2l,38020lgtg!lB 20L7
PEYSICS
T'NIT1-PAPER02
I,IARK SCHEME
Ouestion 6
8.O [2.3, 2.1, 2.7, 2.11t
rc ur I8
(al
(i)
is the bret
The heat capacity of a body or 6lct
rnrlq, required to produce uaf.t tqtoretun rtro (lf,
or lt) in that body or obJect. (1 rc,
Unit: iI K_r or iI oC-r (1 rc)
[2 urtr,
El
(ii) Specific heat capacity is the b..t.m!9I7 reguired to
produce unlt tqlrnton
rtro (lf or lt) in tnLt rlr
(Ug) of a substance. (1 rcl
Unit: J kg.r 6-t 08 iI kg-r og-r
(iii)
(1 rcl
[2 lllIr,
Et
pbuo/rtrtr chry. rl,tbmt r tqrrt 5-1 cn'-gr (rt
oortrat tarntonD. (1 E)
(b)
(i)
[2 rr:tr,
1
I
El
Q=lVt=mcAO or Cl= nu60 or P = CIt = nDAO/t (or any
(1 Uf,
other equivalent eguation)
For flow of 5 g s-l -) 149=(0.00S)(+X5) + L... Eg'n I
(1 ur)
oR 140 = (SX+XS)
For flow of 10 g s-r => ZSO=(OO1)(OXS) + L... Eg'n 2
(1 ul)
oR ZSO = (r0Xc.)(5)
Snbtracting Eq'n I from Eq'n 2t- (1 ltrl,
=> 110 = 0.025cr oR 110 =25t4
110
:. tb= G
OR
1
1
Specific latent heat is tbe b.lt osgy requlred to
cause ualt rlrr (tlg) of a substance to undergo a
Unit: J kg-r (1 rc)
1
1
1
1
I
2
110
c-= E
= l.(fQ I lrg-r X-l on = {.{f g-r 1-r
(1 E-uol.tl
(1 f-Goaa.cturrrr)
[6 rr:kr, Ilttl
(ii)
The heat loss to the environnent remains constant (or
any other equivalent resPonse).
[1 r.tt, Irfl
1
a,
-15-
02138020/CrPtls 20L7
PEYSICli
T'NIT1-PAPER02
t,tARK SCHEI{E
Question 5 (continued)
rc ut I8
(iii)
a
a
o
The calculation does not involve the heat
capacities of the various Parts of the apParatus
and there ls no need to knor thelr values.
the inputs can be adJusted to nininlze heat loss
1
to the environnent.
Arry other equivalent resPonse.
Any one advantage
[1 rrrt,
El
(1v) Hg in glasa thernooeter or any other equlvalent (1,
1
(accePt resistance, themocouple, Iab thermometer)
[1 rert', fl
lot l 15 rrslr
7
I
0
02138020/CAPE/MS 2016
CARIBBEAN
EXAMINATlONS
COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATIONS@
PHYSICS
UNITl_PAPER02
FINAI, MARK SCHEME
MAY/JUNE 2016
'z-
o2t3ao2o/cAPE/Ms 2016
PHYS ICS
UNITl_PAPER02
MARK SCHEME
Question 1
S.O:3.1,3.2,3.3,3.8
t,K
(a)
(i)
a
5
(ii
(b)
)
(i)
KC
xs
Acceleration is the rate of change of velocity
oR a : dvldt oR change of velocity divided by
elapsed ti:me
Unit: m s-'
Both definition and unit correct (1)
The Newton's Second Law:
. External Unba-l-anced force ,/ (resultant) (1)
. Produces an accel-erat'ion (rate of change of
momentum) directly proportionaf to the magnitude
of the unbalanced force in the di'rection of the
unbalanced force (1)
. E=ma (1)
Graph - See Page 3
AI1 7 Points correct
Plot:
5-6Pointscorrect
Lalce]. (1)
Sca].e (1)
Best ].ine (1)
t2t
(1)
1
(ii)
(iii)
(iv)
The reaction time of the driver
(1)
Strai-ght line feature beyond 0'5s OR bet 0.5 & 3.5s
non-zero giadient
(1)
oR The slope of the graph between 0'5 & 3. 5s
OR constant
Distance travelled : A-rea under graph (1) (1)
(Area of rectangle) + (Area of triangle)
i
: (15 , 0.5) + (l' 15 ' 3) (1)
1
SOI
2
4
: 7.5 + 22.5 : 30 m (1)
[acce1>t al.so area of TraPezium]
Eota]. 15 marks
6
4
5
-3-
o2t38O2O/capn/us 2Ore
PHYSICS
UNIT1-PAPER02
MARK SCHEME
Questi-on 1 cont'd
ISca]-e
il-l-:
I
r
- on x-axis 4 cm q 1.O sec
on y-axis 2 96=2m-r
I :r
l.
I.ll'.1
:
i :l
ti i
(1)
Labef
(f)
Scale
Plot
{2)
Best line (1)
16
;14
5
d
0)
,
!)
8Lz
IO
8
:\
\
6
4
i--,
\i
\:
0.o
o.o
r.o
2.O
3.O
Tine (s)
rl
4.0
-4-
o2L3gO2O/cAPE/Ms 2016
PHYSICS
UNITl-PAPER02
MARK SCHEME
Question 2
S.o: 2.3, 2.5,2.7,2.9
uI(
(a)
(i)
7.6 mm (Read off
(ii )
1B0o oR rI radians (accept antiphase or
liii
(b)
(c)
xs
1
equi val ent ) (1)
1
V = f). (1) SOI
1
:15x0.8:12ms-1(1)
1
between X and Y = 0o oR 0
radians (accept "in Phase") (1)
(i)
Phase difference
(j-i)
iii
Antinode - Point of maximum
displacement/amPlitude (1 )
Nod; - point of zero displacement/amplitude
: 2 (1)
Number of antinodes in Figure 3
(i)
Graph - See Page 5
(
KC
from Figure 1) (1)
Plot
(1)
A11 points correct (2)
6 -.7 points correct (1)
Lalce]- (1)
Scale (1); Best line(1)
(ii )
Gradient = (Change in wavel-ength) / (Change rn
wave speed)
(0.4s - o.oo) / G7 .s - 0.00)
o.oL2
(1)
Frequency : L / (gradi-ent of Plot)
o .oL2
= L/
oa
2 a *
-
(1)
5
4
Total 15 rnarks 6
ysl
s
ect ana
(Accept al so a pl ot of v vs )' al ong with consistent corr
q
}2L3AO2O /CAPE,/MS 201 6
PHYS ICS
UNITl-PAPER02
MARK SCHEME
Questi-on 2 conL'd
liil:
- Scale:
Labcl
r - axis 2 cm =5 m s-r
- y - axis 4 cm =O.1m l:
..: :1.:....:l:.. I
I
:l
;.,.1
o.5
,-.1.--l t-
(I )
Plot a1I Points eortect (2) Best Line ( 1 )
--
-_:1
I
r ; ,,(
i
)
i
/
o.4
i
t
E
Ii
t-:- _ -
1l
tn
c
o
o
I
6
o.3
/
f'
o.2
/
{ {
/
I
L
{
I
.,1:
o.1
)
- -,
i
. i.,
-f-
-f
o
5
10
t5
20
2s
wave speed v/m s-l
30
3s
40
-5-
o 2L38O2O /cAPE/Msl 201 6
PHYSTCS
UNITl_PAPER02
MARK SCHEME
Question 3
S.O: 6.6, 6-9, 6.10, 6.11
t,K
(a)
xs
Extension is directly proportional to the applied
force oR F = -kx (1) (accePt F = kx)
(i)
Provided that the elas ti-clProPortional
not exceeded (1)
(second mark cErll only be awarded if
correct)
(
KC
l-irn-it is
first
is
Hang a known mass from Lhe spring and measure the
ext.ension. (1)
ii)
Record mass and corresponding extension (1)
Repeat for 4 (rinim-rm) additional
masses (1)
use
PloL a graph of weight versus extension and
(1)
k
constant'.
spring
tf'"
the graph to determit'L
(accept aver age of f i ve( or mre) det ermi nat i ons )
Hang unknown werght from the spring and measure
.oti"=po*ding extension, x (1)
(b)
(i)
}vq
Determj-ne the welght of the rock using
mass(1)
the
and hence deternrine
: Area under graph
work done in extending spring
(1) sor
1
area
(nethod f or
: a.rea of A + area of rectangle + area between
of
curved Portion under AB and uPPermost side
squares
)
(c
:ount
AB
recLangle under
+(Lx B) + counting of squares (1)
:(a,rbxh)
sor
Calcu]-a
tl
the area
O.O29, 33) + (0 016 , 33) + 49 squares
(1)
1
1
sor
converting squares to q+ergy^
(5 x o.oo5)/25 : o 000lJ
1
=
(1)
"qlr"r"
So 49 squares = 49 x 0'0001 = o. o4eJ
Final cal culation and answer
: 0.48 J + 0.53 J +0.049J=1
i^7.d
(1)
06 + 0.02J
(any suitable,/acctrrate method used to arrive at
ttre final ans!iler within ttre tolerance
range will earn 4 rtarks)
(ii
6
Inel-astrc deformation:
1
1
-'7 -
02L38020 /CAPE,/MS 2O1 6
PHYSICS
UNITl-PAPER02
MARK SCHEME
a
applied Load > elastic Li-rtrit (1)
Therefore causes per:manent deformation (1)
Total 15 marks
5
4
6
-B-
o2L38O 20 /CAPE,/MS 2016
PHYS ICS
UNIT1_PAPER02
MARK SCHEME
Questi-on 4
UK
(a)
dr (1)
Upward force, on l-ower surface (1)
(mrst rention pressure in order to get both of the
first 2 rmrks)
Upthrust is the upward force minus the dovmward
force (1)
L{hen upthrust is less than weight of obiect; obiect
will sink (1)
trdhen upthrust is egual to the weight of ob3ect' the
oblect will float (1)
(b)
KC
Downward force, on upper surface due to pressure at
(i)
Weight of water collected : 0'25 N (1)
(ii )
Mass of HzO : 0 - 025 kg
1000 kglm3
5w:
l'. vo1 . of H2o = m= 0.025:
0 - 000025 m3
t
Volume
stone)
1
(1)
1000
of H2O : volume of stone : o.ooo025 m3 (1)
Density of stone :
(mass
5
of sLone) /
3
(volume of
(0.1), / (0. 000025)
= 4000 kqm "
(1)
(c) (i)
(ri )
(rii)
same in
The vol-ume of the liqurd displaced is the
and
water
both cases (when ornament is immersed in
(1)
when immersed in corn oil) '
than
Upthrust on ornament due to corn oil is less
(1)
that due to water
OR corn oi-I is fess dense than water
N
wt. of corn o-i1 displaced : (2'7 '1'87)N = 0'23
.. mass of corn oi-I displaced = 0'023 kq (1)
: (80 - 50) cm'
Volume of corn oil displaced
= 30 cm3 : 3Ox1O-6 m3
(1)
1
1
I
1
xs
-9-
o2t3ao 20 /cAPE/Msi 2O1 6
PHYSICS
UNITl-PAPER02
MARK SCHE},IE
.'. Densi'ty of (r) corn oil :
Mass of corn oil : 0.023 = '761 kg m '
30"10-u
Vof . of corn oi-],
(comect units in order to obtain both rnarks)
Alternative
2
Of r C
wei-ght of corn oj-1 d-isplaced = (2'7
1.87) = 0-23N
(1)
:
Volume of corn oil displaced = (80 - 50) cm3
(1)
10-6 m3
3]
[accept ans. below if presented in kgcm
(lieight of corn oil)
DensitY of corn oi:l
g)
x
of corn oil
'76'1
:
kgm-3
(1) ans
mark)
unit
earn
(unit must be stated to
S.O: 4.L, 4-2
30x
/ (VoI'
(1) unit
10
5
- 10-
o2l38O2O /CAPE,/MS 2O1 6
PHYSICS
UNITl_PAPER02
MARK SCHEME
Question 5
(rnathod rm:st be practically
feasible to earn ANY of
the first 5 m^arks)
Any suitable experiment to prove Snell's T'aw
ExamPle: (2t
(a)
UK
KC
Diagram
1. A clearly labelled diagram showing (i) glass
b1ock, (ii) norma;I, (iii) incident ray' (iv)
refracted ray (v) emergent ray, (vi)angIe of
.incidence, i, (vii) angle of refraction, r '
Sheet
of,
PaPef
Normal- - -
2
(subtract 1 mark for everl'rnissing conqponent)
(sr:btract 1 mark for conqglete J-isting of
apparatus but no t{i agram)
l2l
Procedure
trace'
ray
obtaining
of
method
Credible
Credible method of obtaining i and r.
4
Indicate at least 4 repet'itions of 2 &' 3
(subtract 1 mark for onrission of any steps)
5
(1)
Precaution
accuracy
the
Any sensible step taken Lo improve
of the resul-ts.
Anal vsis & Interpretation
5. Pfot a graph of sin i versus sin r
(b)
(i)
'7
(1)
't. A straight
line passing through the ori:gin
verifies Snell's Law :
(1)
(sin i/sin r) = constant
medium into
parti-cular
one
from
for lrght travelling
another (1)
sPeed of light in bl-ock A : Speed of light iq air
Refractrve index of A
2
3r1o8 : L.67 x 1o-8 m'-t (1)
1.8
XS
- 11-
o2t3so2o /cAPE,/Ms 201 6
PHYSICS
UNIT.l _ PAPER 02
MARK SCHEME
Questi-on 5 cont'd
(j-i)
(1) fonmrla
Sin 72 : 1-B
Sin 0
Sin 0 = sin (72) /l.B : 0.53
(1) substn &
= O = 30o
ans
(ful I mrks f or exact ans\4er : 31 9 degrees; even if
appr ox' deni ed
(1) for:rtr]-a
Sin O. : rr2/\1
(1)su.bstn
& ans
= 1.40 .'. 0" = 51"
1-BO
(Accept exact ans \,1er : 51. 05 de gr ees )
At AC boundary 0 i (angle of incidence) : 22 + O
a
)
(iii)
(iv)
(22 + 30"\
2
1
= 52o
(1)
angle of 51" at AC I
52' is greater than the critical
j-s
internalJ-y
totally
light
and therefore the
(1)
AC.
at
reflected
Tota1 15 marks I
7
-72-
02L3AO2O/cAPE/Ms 2016
PHYS ICS
UNITl_PAPER02
MARK SCHEME
Question 6
S.O: 3-2| 3.3t 3.1
t,K
(a)
short wavelength radiation (mainly visible)
passes through the earth atmosphere (1)
KC
enters and
This j-s absorbed bY the earth (land c sea) and wa-rms
them up (1)
The warm land and sea then re-radiates outwards ' (1)
is longer wavelength IR
radiation
The re-radiated
radiation (1)
A portion of the outgoing IR radiation is trapped' by
the greenhouse gases in the atmosphere' (1)
6
This traPPed radiation causes the temPerature of the
earLh's atmosPhere to j-ncrease' (1)
(b)
Area of glass window : L x B
(2. B
(1)
2.0):
4.8 m2
1
2
KA AE
t
(1) ,
Ax
0.03 m (1) (unit conversion)
AX
=-0
96
)
x 4-B x (35 - 26) /0-03 (1)
1382 J s-t (1)
(c)
Heat inPut into room to cause 3"c rise in
temperature: 2
AT (1)
5.7 x 10s, 3
C
1.?'106J(1)
Time required for this temperature rl-se
1 . 7,106
1382
(1)
: 1,230 seconds (1)
= 20.5 minutes
2
xs
02138020/CAPE/MS 2015
C A R I B B E A N
E X A M I N A T I O N S
C O U N C I L
CARIBBEAN ADVANCED PROFICIENCY EXAMINATIONS ®
PHYSICS
UNIT 1 – Paper 02
MARK SCHEME
2015
- 2 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Question 1
KC
(a) Newton’s law of universal gravitation states that any two bodies
attract each other with a force that is directly proportional to
the product of their masses (1) and inversely proportional to the
square of their distances apart. (1)
(both conditions for 2 marks, any one wrong 0 marks)
(b)
(i)
Radius =
G M m
Weight = F =
(1) Formula
2
r
6.79  106
2
6.67  1011  6.42  1023  1.4
=
6 2
(3.39  10 )
= 3.39 × 106
= 5.20 N
(ii)
UK XS
2
3
(1) substn + ans
(1)
G M m
(1) Formula
r
6.67  1011  6.42  1023  1.4
3.39  106
=
Gravitational P.E =
= -1.77 × 107 J
(1) substn + ans
2
OR
mgh = Wr
= 5.20 x 3.39 x 106
= 1.76 x 107 J (not penalizing for not having negative
sign)
(c)
(i) Plot of graph
(ii)
Gradient = GME m
From graph Gradient =
- scale(both axes must be correct)(1)
(convenient, even counting, more
than ½, correct way)
- line of best fit
(1)
- points all EIGHT correct
(2)
(-1 mark for any incorrect point)
∴ m =
Gradient
GM E
4
(1)
(1.3  0.3)  105
(23  5)  1015
= 5.56 × 1018 N m–2 (±0.56 x 1018)
(1)Calculating Gradient
(gradient not given if not in range EXCEPTION: give if axes
wrong way and 1/gradient is in range)
4
- 3 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Question 1 cont’d
∴ m =
5.56  1018
6.67  1011  5.6  1024
= 1.5 × 104 kg (1) substn + ans
(1)
(ECF for mass with wrong gradient)
Specific Objective(s): 3.21, 5.4, 5.5
Total 15 marks
2
9
4
- 4 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
This point
be
should
middle of
square
is off,
the
in
1cm
the
- 5 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Question 2
KC UK XS
(a)
(i) SHM is the motion of a particle whose acceleration is always:


(ii)
Directed towards a fixed point (restoring force)
(1)
Directly proportional to its distance from that point (1)
(equation accepted with correctly defined variables)
Ek =
2
1
mw2Xo and w = 2πf
2
∴ Ek =
2
(1) formula
1
× 5.8 × 10–3 × (2π × 4.5)2 × (3 × 10–3)2
2
= 2.1 × 10–5 J (2.08 x 10-5 J)
3
(1) substn + ans (1)
(iii) Increasing amplitude with frequency remaining constant means
that acceleration of the plate increases. (1)
At some
point, the acceleration becomes greater than g (1) and the
cube loses contact with the plate.
(iv) a = (-) w2Xo and g = acceleration
⇒ 9.81 = -(2πf)2 Xo
2
(1)
(1) formula
w = 2f
= -4π2f2 Xo
∴ Xo =
(b)
3
9.81
9.81
=
= 1.2 × 10-2 m (1.23 x 10-2 m)
2 2
4  (3.14)2  (4.5)2
4 f
substn + ans (1)
(i) Plot of graph:
Scale (1)
Line (1)
All 7 correct points (2)
(-1 for any incorrect plot)
4
(ii) Angular frequency, ω = 2 πf
f ≈12.7 Hz(from graph)
= 2× 3.14 × 12.7
± 0.1 Hz
= 79.8 rads-1
(1) substn + ans
1
Specific Objective(s): 1.1, 1.2, 1.3, 1.6
Total 15 marks
3
8
4
- 6 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Question 2 cont’d
- 7 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
- 8 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Question 3
KC
UK XS
(a) Two reasons why different temperatures may be recorded:


(b)
Thermometers are usually calibrated assuming a linear variation
of thermometric property with temperature. (1)
Neither property varies linearly with temperature. (1)
or any other valid reason; e.g. construction of the thermometer
(Note that the assumption stated in the first part of the
markscheme may not be explicitly stated in the response and the
candidate should not be penalized)
2
(i) Absolute zero of temperature – Temperature at which atoms
have minimum/zero energy (1)(volume of gas goes to zero,
temperature below which they cannot go etc.)
2
(ii) The absolute thermodynanic scale of temperature does not
depend on the property of any substance (1).
(c)
(i) 50 oC = (50.00 + 273.15) K = 323.15 K
(1)
1
(ii) Change in temperature = 30.00 K
(1)
1
(d) Consider a resistance thermometer (T) (or any other suitable
thermometer)
(1)
Consider the thermometric property for the choice of thermometer(1)
Let R0 = resistance of wire at ice point.
This resistance is
measured with the wire in a mixture of pure melting ice and water.
(1)
Let R100 = resistance of wire at steam point. This resistance is
measured with the wire in steam above water boiling at standard
atmospheric pressure.
(1)
Then the
5
unknown temperature /°C can be determined from
𝑅 –𝑅
 = [  0 ] × 100
(1)
𝑅100 – 𝑅0
(e)
(i) Using Rt = [R0/(1 + t2)]:
When t = 100 °C, we find that R100 = 0.667R0 or R0=1.5 R100
When t = 80 °C we find that R80 = 0.758R0 or R0=1.32 R80
1
1
- 9 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
(ii) When the mercury thermometer reads 80 °C the temperature as
recorded by the resistance thermometer is
 = [(0.758R0 – R0)/(0.667R0 – R0)] × 100
= [(0.758 – 1)/(0.667 – 1)] x 100
= 72.7 °C
2
(1)
(1)
(1)
Specific Objective(s): 1.1, 1.4, 1.5
Total 15 marks
4
5
6
- 10 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Question 4
KC UK XS
(a) Ax = A cos 
Ay = A sin 
(b)
(1)
(1)
2
Angle
and
distance
labeled
correctly
20o, 145 m
(1)
35o, 105 m
(1)
Resultant from beginning to end
with arrows in correct direction
(1)
(i)
3
(ii) Ay = 145 cos 20
= 136.3 m
By = -105 sin 35
= -60.2 m
Cy = Ay + By = 76.1 m
(iii) Ax = 1455 sin 20
= 49.6 m
Bx = 105 cos 35
= 86 m
Cx = Ax + Bx = 135.6 m
(1)
2
(1)
(1)
1
1
(1)
Cx2  Cy2 (1) = (135.6)2  (75.8)2
= 155.3 m
OR
C2 = A2 + B2 – 2AB Cos C
= 1452 + 1052 – 2 x 145 x 105 Cos 75
= 155.3 m
C =
(1) substn + ans
(1)
(1)
(1)
3
Angle the resultant makes with the x-axis is
Cy
76.1
 = tan-1
= tan-1
= 29.3o
(1) substn + ans
Cx
135.6
(c)
2s
t = 

 a 
x
 2 [L] 
in Dimensions ⟹ [T] = 
2 
[L][T] 
x
∴ [T] = [2 [T]2]x
LHS = RHS if x =
1
2
Specific Objective(s): 1.3, 2.6
UNITS
LHS = S
 m 
RHS =  2 
 ms 
= (s+2)x
∴ 2x = 1
1
x =
2
x
(1)
1
(1)
2
(1)
- 11 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Total 15 marks
5
10
0
- 12 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Question 5
KC UK XS
(a)
4
(1)(correct symbol must be used for the mark to be
awarded)
(maximum of 2 marks awarded if no diagrams are used)
(b)
Converging
lens
can also be drawn
in this manner
3
(1)
(2)
(3)
(c)
Diagram must show a converging lens as magnifying glass (1)
Position of object between lens and focal point (1)
Diagram construction to show image (1)
(i) Lens is converging ⟹ f = +18 cm
Object is real ⟹ u = +12 cm
1
1
1


V
u
f (1)
1
1
1
1
1
1



=
=
V
f
u
36 (1)
18 12
4
∴ V = -36 cm. (1)
The image is in front of the lens. (1)
(ii) The –ve sign means the image is virtual. (1)
(1)
(1)
V
36

 -3; the image is upright. (1)
Magnification =
u
12
Specific Objective(s): 3.6, 3.8, 3.9
(iii)
1
3
- 13 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Total 15 marks
7
8
Question 6
KC
UK XS
(a) Young’s modulus is a measure of the stiffness of an elastic
stress
material. It is the ratio of stress to strain or Y=
(1).
strain
Y
=
f
eA
(2)or
2
𝜎
𝜀
(1 mk for this equation if not clearly
defined)
(1) any of the formula
(b) In elastic deformatic, Hooke’s Law is obeyed (1) and atoms undergo
small displacement. The energy stored in the material is fully
recovered (1) when the load is removed.
In inelastic deformation, Hooke’s Law is not obeyed(1) and the
original shape is not recovered. (1) Some energy is converted to
heat and not recovered when the load is removed. (1)
(c)
E =
5
stress
stress
⟹ strain =
E
strain
(1)
1.5  10
1.8  1010
8
=
Strain = 0.0083
If
= 0.47 m ⟹
e
= 0.0083
⟹ e = 0.0083 ×
= 0.0083 × 0.47
= 0.0039 m = 3.9 mm
Penalize for units or incorrect conversion – 1mk
3
(1)
(1)
0
- 14 02138020/CAPE/MS 2015
PHYSICS
UNIT 1 – PAPER 02
MARK SCHEME
Question 6 cont’d
KC
(d)
(i)
E =
F
eA
But strain =
Manipulation ⇒ F =
e
UK XS
= 0.0010
EeA
(1)
2
∴ F = 2.4 × 1011 × 0.001 × 1.3 × 10-4
F = 3.12 × 104 N
(1)
(ii) 8 – Tonne = 8000 kg; F = 78480 N (1)
= 8 m
E =
F
eA
∴ e =
78480  8
= 0.020 m (1mark)
2.4  1011  1.3  104
Rearanging ⇒ e =
F
EA
(1)
3
Penalize for lack of units here -1mk
= 2.0 cm
Specific Objective(s): 6.6, 6.7, 6.11, 6.12
Total 15 marks
7
8
Unit 1 Paper 2 – 2014
Question 1
(a)
vH = v cos θ
= 300 cos 40
= 230 m/s
1 mk
s = vt
= 230 x 39
= 8970 m
1 mk
(i)
Picture 1
1 mk
(ii)
Picture 3 or 4
1 mk
(iii)
EP loss = mgΔh
= 2 x 0.7
= 1.4 J
1 mk
(i)
(ii)
(b)
1 mk
1 mk
1 mk
(c)
Scale – 1 mk
(d)
Points – 2 mks
Line – 1 mk
gradient = Δy/Δx
1.0 – 0.21
= 107 – 0.35
1 mk
= 0.59 (± 0.03)
1 mk
Hrebound = 0.59 Hbefore
1 mk
Total 15 marks
Question 2
Unit 1 Paper 2 – 2014
(a)
(b)
(i)
20 Hz to 20 000 Hz
(ii)
use audio frequency generator attached to ear phone
adjust to low frequencies and find threshold
adjust to high frequencies and find threshold
repeat for other ear and compare
v = f and
 v = /T

f = 1/T
1 mk
1 mk
1 mk
1 mk
g
T2

T2
=
=
2
g
1 mk
2
g
2
2
T= √
T= √
1 mk
g
2×3.14×0.8
1 mk
9.8
= 0.716 s
(d)
1 mk
= √
T
2
(c)
1mk
1 mk
(i)
Radio waves
IR
1
Visible
2
UV
3
X-ray
4
Decreasing wavelength
(ii)
1
2
3
4
10-2 → 10-3
10-6 → 10-7
10-7 → 10-8
10-8 → 10-9
3 mks (-1 each one wrong)
Total 15 marks
Unit 1 Paper 2 – 2014
Question 3
(a)
(b)
(c)
(d)
(e)
Volume V
(mm3)
498
534
618
660
720
2 mks
(i)
Axes labels /scale – 2 mks
Plots – 1 mk
(ii)
1 mk
1 mk
Extrapolation of line
Read off of value
275oC (± 25)
Line – 1 mk
No
1 mk
Conditions are not ideal. Gases are real therefore will always have volume.
70 mm length → 420 mm3
1 mk
Read of θ from graph
T = θ + 273 = 218 K
1 mk
1 mk
(-55 oC)
(i)
pV = nRT
(ii)
V  T at constant pressure
𝑉1
𝑇1
OR
=
1 mk
𝑉2
𝑇2
Calculate volume
Convert of Kelvin
Correct calculation
1 mk
1 mk
1 mk
1 mk
1 mk
Total 15 marks
Unit 1 Paper 2 – 2014
Question 4
(a)
(i)
change in direction = change in velocity as velocity is a vector quantity
change in velocity = acceleration
1 mk
(ii)
a = ms-2
v2
r
=
=
1 mk
(ms−1 )2
m
m2 s2
m
= ms-2
(b)
1 mk
v2 = u2 + 2as
= 0 + 2 x 6 x 150
= 1800 m2s-2
s = ut + ½at2
t = 7.07 s
v = u + at
v = 42.4 ms -1
OR
2 mks
F = mv2/r
1 mk
= 500 x 1800/200
= 4500 N towards the centre of the circle
(c)
1 mk
1 mk
1 mk
1 mk
(i)
T+W
T
W
(ii)
(iii)
At the lowest point
Tension is largest at that point
1 mk
1 mk
T = m2r + mg
1 mk
2
20 = 0.5 x 0.5 x  + 0.5 x 9.8
2 = (20 – 4.9)/0.25
1 mk
 = 7.8 rad/s
1 mk
Total 15 marks
Unit 1 Paper 2 – 2014
Question 5
(a)
waves are coherent if they have the same frequency and constant phase difference.
(b)
(i)
Diagram
2 mks
Let P be the position of the 1st bright fringe, so that BP – AP = 
BAN  sinθ = /a
PMO  tanθ = y/D
From
From
1 mk
When θ is small; D >> a; sinθ = tanθ

𝑦
 =
1 mk
𝑎
𝑦 =
𝐷
𝐷
1 mk
𝑎
Alternative derivation using Pythagoras’ Theorem can be found here:
http://www.a-levelphysicstutor.com/wav-light-inter.php
or
http://www.tutorvista.com/content/physics/physics-iv/optics/youngs-double-slit.php
(it is a bit long)
(ii)
y = 1.4 mm,
D = 0.5 m,
𝑎𝑦
a = 0.2 mm
 = 𝐷
=
0.2 ×10−3 ×1.4 ×10−3
1 mk
0.5
= 5.6 x 10-7 m
= 560 nm
1 mk
1 mk
Unit 1 Paper 2 – 2014
(c)
(i)
For first order spectrum n = 1
θ = 27.7/2 = 13.85o
d sinθ = n
𝑛
5.89 ×10−7
d =
=
𝑠𝑖𝑛𝜃
OR
sin 13.85
-6
d = 2.46 x 10 m
No. of rulings =
1
𝑑
=
𝑠𝑖𝑛𝜃
𝑛
=
sin 13.85
5.89 ×10−7
1
2.46 ×10−6
1 mk
1 mk
𝑑 𝑠𝑖𝑛𝜃

2.46 ×10−6
=
5.89 ×10−7
1 mk
= 4.17
 no. of spectra = 4
1 mk
1 mk
1 mk
-1
p = 4.06 x 10 m
= 406 mm-1
For maximum number of spectra: sinθ  1
n=
p=
5
1 mk
= 406504 rulings /m
= 406 rulings /mm
(ii)
θ = 27.7/2 = 13.85o
sinθ = np
1 mk
1 mk
1 mk
Total 15 marks
1 mk
Unit 1 Paper 2 – 2014
Question 6
(a)
For conduction, convection and radiation the marks are allotted as follows:


Appropriate and correct identification of heat transfer process in the solar water heater
It’s influence on the construction the solar water heater
1 mk
(6 marks)
Greenhouse effect description
1 mk
Conduction
Takes place between the solar collector and the copper pipes.
Copper pipes used because copper is a very good conductor of heat
Convection
Convection currents in the air above the collector plates can cause heat loss
The collector plate is covered by the glass which reduces air currents
Radiation
Radiation from the sun is absorbed by the collector plate
The collector plate is painted black to increase the efficiency of absorption.
Greenhouse effect
Short wave radiation from the sun passes through the glass cover and is absorbed
Re-emitted long wave radiation cannot escape through the glass trapping the heat energy
(b)
(i)
30
Brick
10
Foam
15
Brick
25
-5
Graph showing three distinct regions
1 mk
Slope of middle region much steeper than end regions
Two end regions having same gradient
1 mk
1 mk
1 mk
Unit 1 Paper 2 – 2014
(ii)
Equivalent brick thickness
k1
k2
=
x2 =
x1
1 mk
x2
0.05 ×0.48
0.016
= 1.5 m or 150 cm
(iii)
1 mk
Total equivalent brick = 1.7 m
Δθ = 35oC
P = kA Δθ/x
1 mk
=
1 mk
0.48 ×1 ×35
= 9.9 W
1.7
1 mk
Total 15 marks
June 2013
Unit 1 PaPEI 2
(D
Velocitv
,/ms-l'
Time
t/s
0
1.0
0
9.8
19.6
2.0
3.0
4.0
29.4
39.2
tu,,
lmk
45
40
i
35
E
5
o l- -.
0
2
time/s
Scale I mk
Plot - 2 mks
-
Bestfit- I mk
(iii)
Height = *." *der graph
h=7zbxh
: lzx4x39
=78m
lmk
lmk
4
!
I
(b) (i)
lmk
lmk
V; = V; COS0
vyi = vi sino
(ii)
x = vxi t: vi coso. t
lmk
Y=vrit-%af
lmk
= (v; sin0)t - '/zgt'
(ll1)
t = ----i"'rri cos0
I =uisind
xi
1
x!
tane x,' - 2oz=9
.x2
cosz=b
This is the equation of a parabole
4 mks
!
')
2
(a)
similarities - both carry energy, both have waveleagth,
both can be refracted,
reflected d.iffracted and
I mk
interfered.
differences
- transverse - direction of propagation perpendicular to
displacement of particles, no compressions o.
rarefactions
longitudinal - direction ofpropagation parallel
to displacement of
particles, compressions and
I mk
rarefactions
examples
-
transverse - e_m. waves
lmk
lmk
longitudinal _ sormd waves
(b)
speed ofsound at 0.C = 340 m/s
speedofsoundarlooc=(0.61 x 10)+340=346.1
m/s
I mk
time to hear echo = 2 x tirne ior
sourtd to travel from shooter to wall
.'.tilne. for sonnd to trave! = S.3.4:4.i5
! _rs
Distance between shooter and wall :
vxt
= 346.1 x 4.t5
=1436m
(")
t=
tzo
zo
second.s
0.36
0.41
0.47
0.53
0.59
I mark each column
t second.s
1
0. l8
0.21
0.24
0.27
0.30
tmk
I
120
)
i
I
I
100
I
j
I
I
i
I
80
I
I
{
i
(,
o 60
E
!
40
0
0.18
o.27
o.27
o.24
t /2 seconds
Scale on both axes
I mk each
Line ofbest fit
Points
lmk
lmk
(iii)
sfleed ofsound: gradient of graph
Large gradient tiaogle
v=333m/s
I mk
imk
lmk
0.3
t-
3
Parts (a) and (b) cou.ld not have been
(")
done because the equation at the top was
r.m.s. c l/{m
r.m.s., : kdmn
r.m.s.o = Ptr/*
*"-_J_xfio
/lrLro
,lmn
lmk
k
=*=fi=t.ot
(d) prV: nrRTr
2 mks
pV = n2RT2
I rn-k
t -n-7.
tL2
lmk
prl,
1.33x104 x1_12 x273
-
1,o1x1os xz23
4 m.ks
:0.18 moles
(e)
Area under graph
lmk
: [(6+qn x \ + [(4+2.95)t2 x 2
l
=5 +
6.95
=11.95x105 J
(D
1.05 mj
lmk
lmk
I mk for line
I mk for answer
incorrect
4.
(a)
For a system in equilibrium, the sum of the anticlockwise moments about a point
must be equal to the sumof the clockwise moments about the same point.
2 mks
Diagram
lmk
Explanation
2 mks (must show equality of moments with person of 50
kg sitting at least twice as far away from pivot as person of 100 kg)
(b)
(i)
For a system ofcolliding bodies, thc total momentum befor collision is
equal to the total momentum after
1 mk
collision
(ii)
a)
Betbre collision
mrx4-m-lx3
After collision
m1 x -1.5 + m: x 5.5 (1)
(r)
,'.4m, + 1.5 mr : 5.5m2 + 3m2
5.5 mr
: 8.5 mz
m1/m2:8.5/5.5
= l-55
b)
m1
Imk
: l 55 m2
.'. m2
- 5.48 kg
lmk
Total E6 before collision = Yz m1v12 + y2 .r4t,v22
: kx8.5 x42 + Y"x5.4E x (-3)2
:92.7 J
lmk
Total EK after collision
(c)
Imk
: t/z m1y12 + lz mzvz2 I mk
: %x 8.5 x(-1.5)'? + %x 5.48 x (5.5)2
=92.45J
tmk
Yes, it was an elastic collision- The total EK wa.s conservd in the collision.
lmk
5.
(a)
f : 1200 H4 length of wire:0.4 m
From diagram L:21,
lmk
lmk
0.4 m = 21.
1"=0.2m
v : Il" = 1200 x 0.2 =240 nrls
(b)
(r)
If tension is doubled, speed increases by a tactor
(ii)
If length is doubled, speed remains constant because
depend on
(c)
1mk
L
amplitude = 5xl0{ m
speed does not
(ii)
Zn0t =740
imk
l mk
)y=2n/i40:0.0085 m
lmk
(iii) 2fi=251300
r mk
f = 2513O0/2n : 3gggi.64 Hz
speed ofwave: fl,
:39995.64 x 0.0085
lmk
lmk
= 339.96 m/s
(d)
I mk
I mk
(i)
(iv)
of J2
any application of sound wave
lmk
1 mk
Sound waves ofthis type are
used.in medicine for diagrrostic purposes
lmk
High frequency pulses are dir
p,,r.""o"o*t"o-";;;;;"# j:;",f;"Iilffi
By scanning the ultrasonic waves acr6c< rh. r.^,.r.,
I.i;;H::H,H.;
^-r -,-^
I mk
,}";T"H",
qen.craredfr omvarious""J::ffi'"*Hfl?.il1t:";ffi
'" "',
the inner anatomy
*
6
Sketch - 1 mk
Label - I mk
h
Force on base due to weight of liquid : Ahpg
Pressure = F/A
: hpg
(b)
(i)
Derivation - 3 mks
p=hpc
= 2500 x l.04xl05 x 9.B
:2.-s48 x i07 pa
orhrtitr rtinn 1 +L-
atrswer 1 mk
(iD
imk
E = stress/strain
69xt0e = 2.548x10%hain
srrain : 2.548x rcl rcgxl}e
= 3.69 x 10{ m
= 0.369 mm
(c) (D
II lll.l!.
-t-
A - proporrional limir
lmk
- limit beyond which extension is not proportional
B - breaking point
(iD
lEk
- point where wire breaks I mk
Area under graph
= Yzb xh
= y2x 0.72x SOnIO-3
:0.018 J
to force applied
1mk
lmk
lmk
UNIT 1_ PAPER 02
MARK SCHEME
20ll
Ouestion 1
UK
(a)
(i)
Displacement: Distance moved in a Particular direcEion (1,
vector
(ii)
Rate
of
(1) vector
change of
displacement at given
vector/scalar
ilcorrect-2narks
3coEEect-1mark
(iii)
At a certain tsime, rate of change of velocity
(iv1
Kinetic energy: energy due Lo motion (1) scalar
(b) (i)
e.g. (3.0, 261 (1)
time
6
(1,
a) 26 ::"
26
=
n
b)
(1)
3
" (,leJ = . ln 0
(i)
1.5
40 =:o
2
(3)"
n
40
Any tswo accura!6 read-offg
(1)
40 = x" (4)'
(1)
(11
(4)r'5
1
ro= 5
(iii)
XS
vector
e.!r. (4.0,40)
(ii)
KC
Take gradient t = 30 s,
]. e
(11
,=#=T=3.8ms-1 (1)
l.-
t
( to)
5 (1)
3
l
,LF i( [, ]
f
_i
(f0.2 m s-1)
Module 1
Specific
U
Olrj ectives : , 1
2,3.t, 3.2. 3.3
(l i ri l.r l
tr
4
6
5
UNITl-PAPER02
MARK SCHEME
I
Ouestion 2
UK
(a)
(b)
T=2n
(subtEact 1 nark if axes revers€d)
5
a) At max amPlitude, t = 12.5-HZ (Ll
(1,
w = Znf - 78.5 rad/s
2fl
b) T = - = 0.080 s
(c)
XS
1
(1)
Points (2)
ScaIe (2!
Line (1)
(i )
(ii)
lE
{r
KC
6. os
0.08 = 2n l-k'rr
{
2
1
l:
(1)
(1)
3
K.rt : 308.4 N,/m (1)
k=Y = r54 N/m (1)
(d)
(i)
Amplitude
<-
Mass rvithout plasticine
2
Mass with plastlcine
Frequencg Hz
sttap€ (tr} -ti.-----+ reduced anplituda
{ resonant f,requency decrease (11 ]
(ii)
Tuning circuit in radio
oB
Acoustical resonance
(1t
(1)
wind or strlnd iinstruments
1
oR darping of car suspension
Module 2
specLfic objectives: L-1, 1.7, L.g
1:,'',
6
{
5
L
UNITl-PAPER02
MARK SCHEME
Question 3
UK
Q+
AU=
(a)
Iteat added
to gas
internal energy
(b)
2
t{ork done
M€anings ( 1)
on gas
(i) Points (2)
Line (1)
scale (1)
(ii)
XS
EorEuIa (1)
sl
i
change in
KC
4
(1)
tlork done - area under curve
i) count squares
ii) use Erapeziun rule etc
Students may
(2t
3
Work done by gas = 544 J
(iii)
PV = nRT
PV
RT
(1)
2
400
8.31{ ' 300 = 0. 16 mol
(1'
(1'
(iv) Isothermal. process (1)
AU=0
1
1
(v) au = 0, Q +W =0
(1)
_
(_644
J}
=
2
=o0 : 544 .l
(1'
Module 3
Specific Objectives:
rr
4.t, s.4, 5.5, 5.6
1
7
Fr:{
'
f
7
a
EnrJrer
UNITl-PAPER02
MARK SCHEME
ouestion {
UK
(a) Rate of change of momentum of a body is proportionaL to the
applied force, and takes place in the direction of the force.
(AIBo acc€Pt lrE = Da" stataE€ntl
Fd
Fo
Fd
(b) (i)
KC
1
or
1
.30N
mg
(ii) Medium that exerts a reEarding force on a mass moving
1
through thaE mediun whlch depends on Ehe veloci ty.
(c)
(i)
e'i, .:pv
m
(11
:2s00 x 0.002
=5kg (1)
(ii)
(iii)
Mg - Fb = 30
(1)
Fu=m9-30=5(9.8)
- 19 N (1)
30
2
30 - r" = 16 (1)
Initially.
30 30
m5
(iv)
2
v = O = E6:0
(1)
[If
FxEr = Ea uc6d'
only 2 Darksl
3
G ms-z (1)
30 - 5 V: rn a (1)
at. terminal veLocity,
30=5vt
vt = 6ms-l (1t
a
0 (1) (or €qui]'ibrius Fo = 30N)
3
XS
a
UNIT 1-
PAPER 02
MARK SCHEME
Question 5
UK
(i) Diffractj-on: spreading of wave as iL passes through
aperture or around an obstacle (11
(a)
KC
an
2
Refraction: Bending as wave moves from one medium to
another (1)
(ii)
rf aperture width >> wavelength of wave + little
diffraction (1)
or no
2
If aperture widt h = wavelength of wave a appreciable
diffraction (1)
OR DIAGN.EMSi
(iii)
(b)
lilaves from diffraction
at apertures (1) interfere
constructively at certain angJ.es which are differenE for
each colour (1)
(d =
1
;-;-;l6E = 2.2 x 10-6 m)
1 (or use sin9 : nh p)
A (ultraviolet ) : 2.2 x 10-6 sin 11.8"
: 450 nn
d sine = n}r; n
)\ (red)
(c) At g
n
10'6 sin 54.80
t2)
4; 4th order violet
1o-e
light (1)
3
l
3,' 3'd dldei lred Iighc
3rd order of red 1i9ht coincides with 4'h order of violet
Modul€ 2
5
(1)
10-6 sln 5{ .0o
450 , lO-e
599
12t
599 nm
(1t
2.2
(1)
2.2 , lO-5 sin 15.8"
54 .8"
2.2
2
I
r,/a
1ight.
1
a] I
t:atfjl{
Specific Objectives : 2.r9 , Z .L7
;, 1 rl, r,
I
1
XS
UNIT 1.
PAPER 02
MARK SCHEME
Ouestion 5
UK
(a)
(i)
KC
Choose two experimental fixed points and measure the
(1)
values of the thermometric Properties at these reference
AIl other temPeratures can by read by linear
points.
interpolation (1!
e
xo
x
x 100"c
xroo- xo
3
(or awarded in b (i) )
(1)
(1)
(ii)
The prope-rty is measured ats.the Erip1e Point of vrater and
(1)
scaled to ..the unit for each kelvin.
2-t3-t6
1
The lower fixed
2
point is 0 K (absolut6 zero)
(iii)
Mercury expands linearly wiEh increase in temperature.
Herctrry does not "wet" glass, andr so responds guickly to
temperature change.
(any one of the above or otl.er
1
reasonab].e answer)
(b) (i)
e
Re
R"
100
Rroo- Ro
950 - 3750
215 - 3750
x 100
e = -19 .2"C
(ii)
(c)
(i)
2
(1)
(1)
Different thermometric properties will not be linear on the
mercury scale. (OR Dercury tlr€rEgEeter calibrated on a
(1)
CelEius scal€)
1
(1)
P
IV = 10 r 120
12 00 w
2
(1)
I
XS
UNIT 1_ PAPER 02
MARK SCHEME
uestion 6 cont'd
UK
KC
(ii) O=mcAe
Om
P=-=-CAe
tt
(1)
(1)
ko 1 min x
4200 x Ae
1200 = 0.3 -i x --mtn ou s
A6 = 57.1"C
et - 20" = 57.1"c
or = 77 .loc (1)
(iii)
3
Results taken at steady state i.e. apparaEus has a constant
temperat.ure. (1)
Module 3
gpecl,f J.c Olrjectiwes: 1.1
1
L.3, L.4, 1.5, 2.11
'_t.
,,t.
7
;
t
,ll
I
xs
2
O?'L3BOZO / CAEEIIIS 2 0l.:.
PHYSTCS
UNIT ]. - PAPER 02
i.,.
F--.
F4rFv,€
KC
*l*,i
(1)
(a) (i)
Eil
1
EE
(1)
(ii)
.mg - Fd = ma
2
mg-bv"=na
(1)
o-Pr'="
-m
(b) (i)
lEt
g
0
0.89
1.s6
z .44
3 .52
4.80
Is _Y-
:_-.oO
- '-:0-a)
. 40 9'
(
-0 .051
0.193
0.387
0.547
0.681
1. Dark fot
1.18
1.30
1.40
1.48
3
1. s4
eacb correct
;":?
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PlrYsrcs
UNfT 1- PAPER 02
i'ilix scE.c
^1,arri
1
^-,
j
^^hr
*l*rl
(ii )
(iii
sqales - 1. II|E E]E
5 e'ccuaate pLogr - 3 roErkE
5 acquEEtse lr1ots - 2 aan:ke
4 e,cclrrate D].ots - 1 Ea:rlc
<4 accuraue D].otE - O
(1)
)
m=-=2.05
xz- xt
2
(1)
n = 2 ( integerc )
{iv)
bV.: (a = 0)
=
.mE
-b
(1)
mE
a
1
(=J
8.5 x _2.-gq
0.251
3
1
)'
55.4 ms-1
(a)
(1)
Su.bject ObJ ect.Lwes: u4! 1.3. L.6, 3.].5, 4.2, 4.3
1
I
5
4
OZ'!3AAZA / C;EE /lltst 20j.l
PI{YSICS
PAPER 02
UNIT 1-
MERT, SCHE-!,E
li
€ i::.:-
:
KC
(a) (i)
(ii)
(b) (i)
UK
xs
( 1)
heard, except
As t'he water leve1 fa1ls, feeble sounds are
or Loud
obtairled
are
where certain length A of air column
sound heard when L = i.
2
1
Name: resonance
Frequeircy of
fork, f lHzt
256
2A8
of 0.781 0.695
Lengttr
string,
320
3 84
450
512
o.625
o .52r
o .444
0,391
L.50
L .92
2 .25
2 :56
L
7
(m)
r/ I (m
(ii)
L .24
)
! .44-
7
5go
5()0
I
450
ET
,r
4a
35
I
P
3()
r tl
I
o5
rrtl
L
,d,1-5
Ulco.gtb ar
I
l I
3
t t tJ-
z-s
I
3 Earks
(-1 for eacb LDcoffisct Dlots)
rII
5
32a3 e o2o,lclgE /uts 20La
PEYSICS
ITNIT ]. - PAPER 02
MARK SCI:EME
e: ioD 2 cont ' d
KC
t = 4l .7
.iiil
1
1
2
1
(1)
(1)
from giral>h, f = 480 Hz
iv)
xs
c:I,j
ll- = 2.4 m-'
(
'rIK
(1)
= 200 ms-l (14 ms-l)
(1)
x2- tl
(1)
(w)
(1)
1E
J-
sradieDE =;
1
2
F
1lu
3
(1)
(199)2 = 1l
4U
100
\\* -_ _-------.---..-4 x (199 )'
= 5.31 x 1o-{ kg m-l
(t 0.15 kg m-1)
VA,PL
Srrcc:LtrLc Objccliwes: u2: 2.9
(1)
3
7
5
6
O2]--ae o2O / cAPF./Bs zALi-
PBYS]CS
UNTT 1- PAPER 02
ILA3I'SeHEr"E
ouestion 3
-li
xc lur lxs
I
(a)
tso cbErge 1kg- of solid
ouan-.itY of therma.I energry required
ternperature
safte
JubsEance t.o 1kg of tiquid ac Che
(b) (i)
MelEing PoinE 272 K
(1)
(1)
Boiling Points 430 K
(1)
(ii)
Gradients P
GrEdient Q
z7z - ?30
2
21.0 K min-l (1) t+ 0.51
430 - z7z
z7-a3
4
l-L.3 K min- I (1) t r 1. oI
(1)
a
(iii)
ri-ses more s1owly
c for liguid i-s larger: ttte temP.
(1)
(iv)
€ = Ig c A T
(1)
(1)
P=-=mc
AT
AE
1- x '1-O' = 2
x c x 11.3
c
1x105
22.6
44oo ,, kg-l K-l
(v) Q = ro!!
1x1O5al!=2xL1
Lf = 5.5 x 105 'f kg-l
tB, P1
s:rccif ic objectives s ldi3.2.5' 2-7,2'e,z'LL
(1)
(1)
(1)
(1)
(1)
-7
AZLZSA2O /CzEE/fig 2OLL
PHTSICS
UNIT 1- PAPER 02
irrlR.i( SCliEviE
€iuest--i cl
{
l*"1*l*"1
(a) (i)
(II'
K E statseroerrt, 12 wz
(2)
P E stsatsements mgh
(2)
4
K E aI-ways posiEive because v2 is positiwe
PE could be neEaEive
2
ToEaIEnerg-y=KE+PE
(b) (i )
* mwz = m9 (hr - hz)
v2 = 2s (hl - h2)
(1)
2s(\ - h,)
2 x 9.8 x !2
= .E1g
= I5-J
(ii)
(1)
m-s _
y=uts+*at2
(1)
7
u=O, so:
20 20
9.8
s
E = 1.43 s
(1)
EI
(1)
(iii ) x=vrt
(1)
= 15.3 x (1.43)n
= 21.9 m
(c)
Neglect air resistsance
on track
No friction
(1)
]
2
,fL.P2
rEI
B
D213SA2A / cE. Ey'Ms 20L'L
PHYSICS
UNTT 1- PAPER 02
lvi;l.n-t SCi{EilE
Sres LIcll
EI\
(a) (i)
(ii)
refraction
artd d.iagran
2
diffraction
and diagram
2
(1)
" diffraclion at each slit
. act as soujrceg of wawes (1)
o w}.ich interfere
(1)
(b) (i )
d
(ii)
Ysooo *
tso give rise
UK
3
to specEra (1)
(1)
B)
1.557 x 1o-5 Iri
a
(1)
d Sino = nr\
et = Sin-'-zA?
2 x 0. 5563 r:D (1)
1.567 po )
(1)
51. 9"
Sin-1
Oz
Sin-1
Sin-1
5
2|,2
cL
2 x 0.4851 }lt! (1)
1. 567 Pm
(1)
(iii)
AnEu].ar separatsion = 61
O2
1
(1)
D. P2
sE>eeific object:tvas t
7
a
xs
I
02138C2 0 /CAPEIfiS 203.L
PHYSTCS
UI TTI-PAPEF02
tval1li sciiiEsA'
,i-re!-:::.::
KC
(a) P v
(b)
(ii)
(c)
(1)
1.03 x 105 x2.9 x 1o-{
8.31 x 315
RT
(i)
(L)
DRT
PV
n
(1)
AU=Q+ITI
Pzvz
T1
I2
3
(1)
1. L4 x 1O-2 moI
(1)
whenQ=0,AU=w
gas
and
tshe
T}.e gas .is cornpressed so work is done on tshe
(1)
interllal energr!, illcreases i,e- ttle temp rises
Mowing tshe piston dordn trits ttre molecules so they move
fastser (in a sma].].er space) (1) Movirrg fas-.er means tsrrey
l.ave Illore kinetic energry (1) i.e. the teloPerature rises (1)
Plvr
D.
UK
3
3
(1)
1.03 x 1os x2.9 x Lo-a x 1go
2
3,5 x 1o-s x 315
(a)
= 2.l-4 x 105 Pa
(d) AU=Q+vl
(1)
AU = 0 + 90,r = 90.I
(1)
(e) AU=nc,AT
90 = O.Ol-L4 x c.x (790 - 315)
( a)"
3
(1)
L6,5 .T nol.-l k-l
s9€cif,:Lc objectives ! ,!3.2.2,
1
4.L, 5.a, 5.2,5.5,
5.6
5
9
xs
Unit 1 2010
Question 1
(a)
(i)
5
- +--
4
..i --+r -
---:---i--;-.--.:-
'--i--- -i"'
-i
r
3
2
1
E
0
J
5--:
-1
-2
3
.Y-.,
- --'
-4
13 accurate points
2 mks
< 13 accurate points
1mk
Lines
2 mks
(ii)
1''
tls
Uniform acceleration
Collision Ball rebounds
Ball changes direction and comes to rest
(iii) (a)
v-u
1mk
t
4.2-O:
1
3.5 ms-'
1.2-0
(b)
1mk
1mk
1mk
lmk
Area under line = %b xh
:Yzxl.2x4.2
:2.5m
I mk
1 mk
5
(c)
m(u-u)
| =ma:'t
1mk
:- o.a(-2.+-+.2)
1mk
L.25-7.2
: -91.2 N
(iv)
Inelastic collision
Loses kinetic energy
1mk
1mk
Total 15 marks
Question 2
(a)
(i)
Red
Violet
White
Violet
I rnk
Red
(ii)
Refraction l mk
Red
Correct colour dispersion 1 mk
White
Violet
(iiD
1mk
(b) (i)
01
sin 01
0z
sin 02
31.00
0.s 15
48.3
0.746
41.9
0.667
75.20
0.967
43.6
0.690
90.00
I
Tolerance + 2 in final digit
1 mk for read-off
I mk for calculation
(6 mks)
1mk
Critical angle of glass 43.6o
(ii)
55o > critical angle .'. total intemal reflection
(iii) gradient read off
substitution
answer 1.45 t 0.01
lmk
1mk
1mk
1mk
Total 15 marks
a
Question 3
(a)
(i)
Load
Load
Load
Extension
Extension
Steel wire
GIass
Extension
Polymeric material
(1 mk each)
(ii)
(b)
Stress: force per unit area
lmk
Strain: extension/original length
1 mk
(i)
Extension, ALlm
(ii)
0.012
0.006
0
0.019
0.028
0.050
lmk
0.6
0.5
0.4
.. _.. ---_: _ _.L^-: . ,.
,1---,-.-.. .--i
a!
I
!t(!
0.3
!
o.2
0.1
0
0
0.01
0.02
0.03
extension / m
Axes/Scale 1 mk
6 accurate points 2 mks
(< 6 accurate points) 1 rnk
1 rnk
Line
0.04
0.05
0.05
(iii)
r
E
(iv)
mgL
AAL
lmk
: gradienr (+ )
1mk
0.2-0
..
gradient=#:16.7kgm-l
1mk
0.012 -0
E: gradient (+ )
E_
1.67x 5x1o-2 x9.8 :1.82
3x10-2x 1.5x10-3
(1 mk)
1mk
x 106 Nm-2
(1 mk)
Total 15 marks
Question 4
(a)
(i)
(ii)
(b)
(i)
:0
I rnk
I torque:0
lmk
Tcos26o=W
T=150lcos26o
lmk
lmk
T= 167N
1mk
F:Tsin26o
lmk
F:73 N
1mk
X forces
fJ* : 1.6 cos 20o :1.5 rns-l
Uy: 1.6 sin 20o:0.55 ms'l
lmk
lmk
(iD y: ut +'Aat2
1mk
Y:-2'2m
lmk
-2.2:0.55t- 4.9t2
1mk
t: 0.73 s
lmk
OR
(iiD
time up
x2
time down
1mk
1mk
add
1mk
lmk
x:uxt
: 1.5 x 0.73
1mk
:1.1 m
lmk
Total 15 marks
Question 5
(a)
(i)
Threshold of hearing: Intensity level where a sound becomes audible I mk
Threshold of pain: Intensity level beyond which sounds become unbearable
lmk
(b)
(ii)
Ear responds to a wide range of frequency
(iii)
0
: 10 16910 I
lmk
(iv) 0:10 log,oj4
1.0 x i-o-12
lmk
:96 dB
lmk
lo
lmk
(v)
1 mk
Threshold of hearing increases in intensity
Audible frequency range decreases, especially at high end of frequency
(i)
Waves going towards wall interfere with waves reflected
Minimum occurs where there is destructive interference
Not zero because the reflecting wave has a smaller amplitude
(ii)
Three loops in standing wave so h : 1.5 m
lmk
1mk
lmk
lmk
lmk
f = vllu
:330/1.5
:220 Hz
1 mk
1 mk
OR
Four loops in standing wave so h : 1.125 m
f : vl)"
1 mk
= 330/1.125
(iii)
=293H2
lmk
)":2 m
1 mk
Maximum is at 50 cm from wall
imk
lmk
Total 15 marks
Question 6
(a)
(i)
The e-m waves cause atoms in the cold body to vibrate faster and so get hotter
lmk
(ii)
1 mk
Radiation from the sun passes through the COz
1 mk
It is absorbed at the surface and causes warming
The warmed surface radiates at a longer wavelength than the radiation from the
I mk
sun
This radiation cannot escape because it cannot pass through the greenhouse gases
1mk
(b)
(i)
H--kA- AT
A,X
1 mk (may or may not have the minus sign)
: 80.4 x 4.6 x6+7- q1o
1mk
:277 kW
lmk
4xL0 '
(ii) P= oAQf -rt)
= 5.67 x 10-8 x 4.6 x (9204 - 3034)
: 185 kW
Conversion from
oC
to K
1mk
1mk
lmk
lmk
(tf P = o A If used.... max2out of 4 ...P : 187 kW)
(iiD 185 kW of the power conducted through stove walls from interior is radiated
1 mk
away by the exterior surface
Thus (277 - 185) kW = 92 kW is transferred by convection and conduction into
2 mks
the surrounding
air
Total 15 marks
i
)-\.-\_ i,-
\
*J ---
L
Question 1
(a)
ib)
(i)
F=Ap(momentum)
t
(ii)
impulse = F . t (change in mornentum = force x time)
ri)
Wr = L8 kg x 9.8 Nkg'r
: 17.6 N
I mk
(r r)
Wr: 0.40 kg x 9.8 N kg-r = 3.9 N
1 mli
(1il)
r= i.s-_U.4 =5.6s
I mk
I mk
I mk
0.25
(c) (i)
ii)
(
2 mks fminus 1 for each error]
plotling points
curve
1 mli
iine
intercept
I mk
1mk
15
Sbaded area
E
o
U
A
tf.-rrrrr
ry+4
t . 0.85 6ec --r-Fr-t
lll
1o
-
o
k
5
0
ri
ll
00
1.oo 2.oo
t 0.05 s
3.00
--rl=l. I I _T.lIL
!
4.00
5.00
6,.00.
Tlme E/E
Imk
(iii)
t = 0.85
(i")
shaded area
(v)
change in momentum = area ofshaded part
lmk
=13Ns
2 mks
(any method may be used to find area
il'tvlrt,jt ir;c,a trscil 501-s (1 tlr[:))
Total l5 marl6
)
Question 2
(a)
(i)
Periodic motion
Motion is repetitive or cyclic in nature
Simpie harmonic motion
(r r)
/,
Acceleration proportionai to dispiacement
1mli.
Acceleration opposition to dispiacement
I nrk
L\
nhi";.
., n^la.,h^,,r,n,-
l^."
:
).1:;: : :::l: ::t :h: ::l:
3
Divisions of i 0 cm or less
4
Pole is stuck in the sea bed with the zero marking coinciding
with the sea bed
5
Choose a calm day
6
Take reading every half hour or hourly
7
Observe for 24 hours
[any 5 points I ml< each]
(c)
(i)
Minimumdepth=4m
(ii)
:I hour
11.5
: 9 + 5 sin (1.45 x 10-at)
sin-l (
)
: 1.45 x 10at
L2
I mk
t1
f"
-tir 'l -,
lmk
= 18055 sec
.- r 02 irorrls
I ml(
,r
1.{9 x lO-.
c-r:{
L.ts x1o-'
'l-.t6ll .' cr:
(iii)
lrnh
Length of time : 4 hours
t rri
Total 15 marks
Question 3
(a) t= h-ho rloo
l mlr
= 105 - -50 x 100
1mk
: 57.4oC
1mk
720
-
-50
(b) To agree with old Celsius scale
(r)
I mk
i P,, x i (-t' Pa r 4.3i'i i 3,480
Pr./Pr.
0.6590 I 0.6592
2.57'7
I Rt'
L000
0.6s93
0.65 95
0 6598
(ii) A real gas rs used
OR
Readings are taken for different masses of gas in the bulb
(c)
0
E.G 1o-'
, -,,'5.59
6599 1 -,Y
5 1O-r
MO
x0 + l.I:.tx
0 .5s997
ttl :.0.00023191
.5 E
,R, 'i:r :t 0.194795
-|*:l.i{it#-
E
tllr
EIE
Scale/ Axis
2 mks
Points
2 mks
Best fit line
I mk
(-l for every pt. wrong)
Question 3 cont'd
(d) lntercepr = 0.660
l
(e) T:0.660 x273 16
l mk
I
I
: 180.28 K
I
I mk
I
I
ll.l
U= l6U.l6* lt).lJ
: -92 8't "C
I mk
I
Toia I l5 rn a rlr :\
l
Question 4
(a) (i)
(ij)
Energy: capacity ro do work
I mk
Kinetic energy: energy possessed by virtue of motion
1 rnli
Potential energy: energy possessed by vifiue ofposition
lmk
v2 = u2 + 2as
1 mk
1or an obJect starting at rest u = 0 ms I
a:- 2r
1 mk
1
rlli -
c -_ rr^s - /
LJ\
-
)s
\
(b) (i)
?r
I nrk
/2 nt
mgh = |4 mv2
" = #cE:
1 mk
2 x 9.81 x10
v = l4 ms'l
(
ii)
-1.
1mk
V=COI
14
o:- r :-=l.4rad
r0
sl
I mk (vatu e)
(r-) (i)
1 mk (unit)
.T
mg
(ij) T -.g
I - nrI
I nik
S.r
= 7s(#+ e.e)
I nrk
= 2205 N
lmk
Total 15 nrar
Question 5
(a) (i)
Accommodation: The ability ofthe eye to change its rens porver so that it can focus
near and distant ob..lects onto r.he retina.
I mk
Depth offocus: The range of image disrances that can be accepted by the retina
as
a focussed image.
lmk
(iD
(b) (i)
Long-sightedness or hvpermetroDia
1 rnk
1 mli
R e.,.,s
r)
(ignore absent arrows on rays)
,.1
luuar
I
I
I rnk
I rnk
Correct lens - couvex
Correct rays from the lens
(c) (i)
.;
6 = loo = 56
t.!
",r,
'l nr li
u :25 cm
l
11'
uvf
I mk
--J---_
1
L
25
+
1
50
lmk
v : - -50 c.nr
Near point 50 cm lron hcr
cyes
1 mlr
Qu
tion 5 cont'd
(c)
(i1) u=+40cm
f:50cm
1mk
111
vfu
11
50
I mk
40
r':-200cm
Near point 200 cm lrom her eyes
(iii) u:25 cm
v:-200cm
t
11
---+
f25
-
200
I mk
f : 28.6 cm
P=3.5D
lmk
Total l5 mar
Question 6
(a) (i)
(alsoacceptQ=AU+W) (ignoreabsence ofA)
^U=Q+w
1
chanse in
work done ON
heat added
to the system
intemal
energy (1)
the system
(1)
0)
(1i) C, - No work done in constant volume process
Co
I nrk
- ii-,;i",1es wL-,t k,jo,re itr r,otrstatrI pressule pru\,ass
L.-C,-R
or
(b) (D pV:nRr
Il:
1mk
Cp=C.-R
1 mli
1mk
DV
read offvalues from any isotherm
-RT
line
1 mk
1 mk
n = 0.25 moles
(ii) a) Q - nC,AT = 0.:5 x; R x 250 = 779 J
I mk srrbstitution antl itnswer
1 mk equatio-n
b) o - nc^Af
tz = 0.25 x:R x 250 = 1300 J
I ml< substitution 1 ml< altstver
(also accepr nC,A'I' + PAV)
(iii) W=pAV =2x10sx(0.005-0.0026) OR 2xlO5 x(0.0052-0.0026)
= 480J
I mk equation
:520J
I mk substitution and lnswer
Also accept W = 1300 -779 = 521 J
'l'olrti I5 t:ra l'li
.r-=.)
I 'cg4a' '7*r Ltl P L
-4-
tAr n t\rtti
SECTION A
,
AttemPt ALL questions'
You MUST write your answers
l.
(a)
in this answer booklet'
data in Table'
Usine electronic timing the following
l' w'as obtained for a steel
batl bearing
taltiriefromrest. Usethedatatoll:8"ffi;;;;lghttinegraphlono""t)t".[Tfr:ri
the acceleration or rree fall'
;;;;;;r'r'
TABLE 1
Time, r/s
12 mm
!2 ms
0.400
0.281
c. a l.g o
0.600
0.342
o.t lYo
0.800
0.414
o't?l*
1.000
0.456
o ')0
1.200
0.500
o.L{e-o
I.400
0.534
o- ?8f )^
Working for determination of
5
SD
t'z/i
Distance, Y/m
*s t-
t
W
$
g'
Sn ?W
-r
5
T+ rtf,
\ - >oo-
o
z
t+
o'&$f o
lP +'l @
1-b w
'lI clo ,r4l t J
+
.7^
[3 marks]
GO ON TO THE NEXT PAGE
-5-
ffi
I-a
ttffl1ir.]
i't
f
.t
l-3
l-'t-
f'r
,-o
!1
L8
_.1
t
i
.6
'Y-
.,]
o
GO ON TO THF NFXT PAGF
-6u. pitive)
to show how the
donur,
on the axes below sketch graphs (using y,
tne ralling steel bearing vary with
acceleration, a, velociry, ,, ,"a"'ilpro"*"ti
"f
(b)
time
v
v
t
lo,.I
Using g: 9.8 m s-'z rather than the val
oft"rlt has fallen a distance of 0-90 tn'
(.)
\,
l,
2- lL
t
[3 marksl
t
t
ue from (a), find the velocity of
the steel bearing
,1)
A
+ )--,G-{
o +
r[r'r)C.r")
|1-6 .r
v'
+'^
I
,vt
c
[3 marks]
Total 15 marks
GO ON TO THE NEXT PAGE
(a)
2.
(D
What is meant by the term 'diffractioni?
.*0; j,d-
s
Ur'v..ar\l-a
t 8^-.],\
.-31
o( c-+e.*l .9*.*-l{
[1 mark ]
Plane waves incident on small aperhfes are represented in Figure I below' Show
on the diagram, the resulting waves after passage through the slits'
(
I
I
:
I
X
I
/-
T
a-t->
(-^
)
lfi Anr
l^/ewr.-!
I
,
tr^+
,--d}
Figure 1
[2 markst
(iiD
Mark on Figure l, TWO places where constuctive interference occurs and TWO
places wheie destructive- interference occurs. [Use a circle O for constructive
'interference
[2 marksl
and an X for destructive interference.]
(i")
In the space below draw wave diagrams to show how TWO waves could superpose
[2 marksl
to proiuce constructive and destructive interference'
l
.'.,. ',,y
a.
t
t
I
A
A
t-
C.*ilil^-EJ
\l\r9.\,$---k ;
IE
;\,,
l-,^-o
\--
\"-s'h/
\Xf.^r.r-s aD".".r-*
GO ON TO THE NEXT PAGE
,a!
'-/'
!
t'
I-.'
Two vertical dipole antennas, A and B, ernit coherent signalS from a radio station
broadcasting at i frequency of 9Q0_kllz as shown in Figure 2. The antennas are spaced
600 m apart.
(b)
R
& >6aot,,.
I
.a
-
J/-
&
P
,9
: 3 .otro "
B.
.ootr o -..--<
q
Figure 2
- 535 ,v\ .
(i)
A maximum intensity of the signal is received at P some distance away on the
line perpendicular to AB as shown and the next maximum is detected at R on a
line at an angle 0. Calculate the value of 0.
\
*L:S.'
-il
\A
,&-..v
l-F 3]]
+
)3. ? o
$*,n=t
4qp
= o' l-t
S
' [5 marks]
(iD
Explain why at Q, a poir,.t midway between P and R, the signal received is very
weak.
P* q lQ,t'
.r-^*r+ c,"-*
S As,"V-;no
.34
[1 mark ]
At what angle to the mid-line would the next minimum of the interferetrce pattern
(iii)
occur?
,e':". -t\.r .*.ilt,.- ,-,r.\
&
J}Al"(
YL
-t^^ g
r
l-<-+3 a )
5 a-p
F
," I ?Rnrn/aApF ?noR
E6
v
o
.*.r- c.r*"*,t*1 J* U
o . a] ),r-
z 6 *--t O "r
[2 marks]
Total 15 marks
GO ON TO THE NEXT PAGE
-93'
(o)
(i)
Explain what is meant by EACH of the following tenns
'Stress':
g'kl,rx j
cJ-
Srr-'.- .
-! Srp-r-l
,t*A e-ya-r,l - b-f".; *-L
'z-
F
N (v\
o{ /-_
]
[1 mark ]
'Strain':
9 [./o^--- ;
\R{ f.-{-
-$q . o-v -;..i
SOr"-;
t sq*
Lrf\^.
+-
JL
-a-
I mark ]
(it)
Figure_3_ sbows a graph ofapplied force F, against
extension, L,l, for along steel
wire. Mark on the graph the region where the wire obeys Hooke,s law.
{rn"^ oJ" + *-(rze
f'orce F
A
& r+r\U_t J.**- 6'.t
_o-_a\ -
o
Extension A/
Figure 3
[1 mark ]
2211802o/c. AP.F ) ooe.
GO ON TO THE NEXT PAGE
-10(iii)
ln the space below, sketch graphs to show how
if the rnaterial is
the giaph in Figure 3 will change
(b) Rubber
(a) Glass
r
Vx*l;,^J
?
LErP
\'l"e'f;.-
["
'"-ry
tfu-*t')
al-
LL
,
[2 marks]
(b)
I m long and 2 x l0-3 m in diameter
Figure 4 shows a 0.5 kg mass on a steel wire
a vertical circle' The mass is at its lowest
iv?rrl -.arfr. : z x td' Pa) Leiig whLtedm in
the wire is 7.2 x
tni. p"i"i i. 38 s-r. The breaking stress of
"t
108 N m'.
ilil?;li;;"to"ity
r-.r^ezs .-
s.*t:* -\ Fro.. ! . r._*tr_
lm
=t[o,.o* JL>
-i-to
M
3-l*
I
<
*1
Figure 4
(i)
On Figure 4 show the forces acting on
the mass'
[2 marks]
GO ON TO THE NEXT PAGE
- 11-
(iD
At the LOWEST point, calculate
a)
..
the tensiou in the wire
fl* ,t^-"*-,,',.f- 9c-.* -f - *.X ? TWV
g.
r-
*t +
T
tvr-V
vla
,-,,-(I.er+
:-
fe+ N.
b)
Ce
,,5)
v-Y-y
'
*-4
t rr)
[2 marksl
the stress experienced by the wire
S,Vr.s =.-
+
*= ?, r"
n[rrr;])
7zt
3.,y Fro-
)'tv rro (
C
J,7/+ro I (u,
[2 marksl
c)
the extension of the wire.
Y -
g
Ykvn
g" SFo.^,.
L
a')i +r0 r
,dt'*^J
Yh'o=->
)
f { o).
_'r
1.,>-yF{o-r_l
_.'
l'l6 T-(o - n^
c/ l.l[
(c)
vr,,rwu .
Explain whether or not the wire breaks at this point.
t
s I-i J-5)
\r_ r^t..a *.../.t .164 lw"*
[3 marks]
b+-<-a..r-c- Y-e-J
&^"-'." J J"* -tQ.*^- *fu k^-J
t+
B'3) +ro € P^ 1 +.> +( 08 /n
31r,.-r[1 mark ]
Total 15 marks -
\4,'
3xT):1 ,
,? I ?enrn/aapF
GO ON TO THE NEXT
ltu
Y
Page 1
Question
write on both sides ofthe paper and start each answer on a new page
Do not
Do not
write
in this
margin.
rryrite
in this
rnargin.
(.t '
.jfu
=(fu
1r)-"'.(-
3**-d^
ho-t*,'....*-' -W.-r
r*-f.4*-fr +, ,trtto
turl...--[
#.-*-\
"-&J
dR-,r^".*-ui .
F\v-t[
EJ
; +q*.g,
.*r"&-$
l"-Aoh, q
$*
<frs-, ff-._"--,
-ur-Jio;
. Lhr
t' .,*- ffk r-rt-"&
rt
-]."*-l-
*{*"1
-tt.a
rfu
6-l f4.r--^{
f}l*
\b)
Qra .-s.,.,.,
tLr-ru
\,')
-rtnnro
\.,J
tN-4-p-4.t
N
tt
r--t
,{tJt
^rrJJ"
*[.-u-. A-"-.L-
-rfu
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o-a-,.^,-
--Lf- -
-(^-* "^,---!. V^e
k,a 97.,:.-.a
+. \tu
h-r+ ,nrt
Y*
cr-,4-
I
.h--4" ,*;ft" k,r* c-l^->." Slr.
$--q,\,-{ r-4 + 'tt { qe-r-tr---{- \I=-l^{ /
k Yt ( ftrx-*--(
L *-fu
,h"-,< 6rre
)
I
+
I
:
j
E
.L
F*--. Ls r.t+ c,oa,,-i-b- -
[r..*X
fltr.
-'n-
Question .
\k .
write on both sides ofthe paier and star[ each answer on a new page
Page 2
Do not
Do not
write
in this
write
rnargin.
(.t
in this
h4L
Pt
margin.
Pt
L
{,t *
Vo
9'f [rs-:
1
L,t-
,5Q
t_
P
.6 rt ..{srD T,
+-
st
+
5 Irz
r
a
5'7 + 3oo 'l-
.L
2
-- &L
v
0.of 3I
e 'ofB?x-
4. r1r\
fh
t
1.q
t'6 yllo
A
-\ e.ot-6 F-{oA o/r.*l
fDL
0''.",. n .'
L
"Tu ) Y'(o El"a
I
Question
write ou both sides ofthe paper and start each auswer on a new page
Page 3
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write
in this
write
iu this
margin
\i,
la
e'-ir'"1
margin.
Q,r
>t
(
^
o.=to-eF
T(.a w\ry(r&- 3
*..-u
9^--*-tU-- ur-evr-1,'!-JtL
(vt (i). +,
A+A
ru
0.A
..-4
f-({-'"r-l- *JL
c"-U{"'L
{.2
_ \,^.
? w. vr,r
ft+ )'- > Br..,r,u.,.
t
% Ao**+u
-3
I 1t-ro
3 tt;=
P f r-ro
-7
91re-]
'23 l+z
= to-6:
l3.o- to . (
te.+ tt + .
1+=r
L.,.0;.
L-,
Qrrestion. . ...!.1
Page 4
write on both sides ofthe paper and start each answer on a new page
Do rot
Do not
write
wlite
in this
margin
in this
nrargin
w Qt
I/**
I
o
,l*^/l 8rl
u"1*' :l1u
I 1
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I
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b
#'
w
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*/'fT"
I
0,5{b
/^ft-
TEST CODE 22738020
FORM TP 2008246
MAY/JI.INE 2OO8
CAIIIBI]EAN EXAMINATIONS
COUNCTL
ADVANCED FROF,ICIENCY
EXAMINATION
PHYSICS - UNIT I - PAPER
Insert for euestiou 5 (b)
Registr-ation Nuruber ...........
Cerrtre Number .............
Candidate,s Number
rr#
/^ ft
a,r
&,-
I
7
{
--rf
t.
"?
{t
02
r.q
\
- 15 from
marksl
[2
when the wavelenf ltlr is changed
(i)
By how much does the frequency decrease
3 mm to 8 mm?
(ii)
With water of the depth used in the experimbnt' give. bv
ress than 2'0 mm is
iican be shown that the sp eeg of
#;;;i;;;i;;velength
n
::fB+', 'S;**
-.
+c
k
-,3i;
2,"
where k and c are constants and v is measured
I
")
Take readings from the graph and
in mm s-r'
draw a suitable table of valueii#"*z
b)
the insert'
Using the graph paper provided on
test the relationshiP.
c)
Use your graph to determine
-of
.:
go
lc-
,-
3
Total 15 marks
F(D
=_)
0'rt
-Ft o
f'7r1t
o
--
5^'Yf
[10 marks]
['
the value of
(r."- (--r-)
rf,,s?{_,-
plot a staight line graph to
7
!
-tlo 3
I
tww
-r5
':i
;:
c
1
ii
GO ON TO THE NEXT PAGE
I
6
Pase 5
Questioll
write o[ both sides ofthe paper and start each ans]ver on a new page
Do not
write
in this
nrargin.
Do not
\ryrite
in this
(41- ) .
,<4
P
ct)
,rf,.r", $1, )
Sr,r*
1+tu .--lw*-L'zlJ
I
@).
.,r,.rct
Lc,-a
€) ,rp -*ro)-
?q
rbp
&
'tL*u'".'1
?o
,{- w
c+- r
b"-L\.'- F
J,r<!-a
margin
f
Lh
o' )-2 .F(o
9.o w
0i')' TL 1^-.t,--r r) \fu
- Jo
Ra
,hV ( . ,,) Crr)
to 11
>>
=- >)- fn
*it L t- tI (D--0,)
o.j7)-
-)
tr
.
;.
\
)'
b
Question
write on both sides ofthe paper and start each answer on a new page
Do not
wl ite
in this
margilr.
Page 6
l
Do not
(v
writ€
in this
. margin.
t
1
.L-
X
Gi ,n.*L"-u 1 {lq*-t [o-*.j, T*"
c- ts Cr-*N+ (t
t*
)
\(w
TI
> Soslc
L , z6-g k-.
f-b" -l-r o--P xe(lo3*->6,Pe )
7">+ -Frof yS.6rp,f-E+ >
.5ft w
s lpJ.
v !!v
?>
11v.r I I
(a) (i)
4 .5
4
l
.1
.-!
3-5
3
I
-5
E
2
1-5
-t
'i
--,
I
1
I
0.5
-1
0
0
0_5
L
z
1.5
2.5
3
tls
.d'cis
(1)
Sca.le
(i)
Plotting
(?)
(1)
Best fit
(ii)
Acceieration of sphere decreases to zero as time progresses [or any other logicai, correct
(1)
explanationsl
Vt = 4.12 ms-l
(1)
(iii) a = v2 - vt = 2.45 ms-2 (+ 0.15)
12-tl
(1)
(1)
(b)
(i)
k= me
(1)
,.)
(1)
ms '. m
(iD
k
=
(5 x l0-1 (9.8)
(1)
6xx 7 xTOa x4.12
= u.6Jkgm's'
(iii)
(1)
Tertrinal velocity decreases by a factor of2
Total 15 marks
(2)
QUESTiON 2
(a)
Standrng wave ( stationar,*.' wave set up by pluct<lng the
strieg) (i )
Vibrating stiing acts on air moiecules and transmits sound wave
&)
A
A
(1)
A
(i)
(1)
(iir
-L
1,
2x2.16 = 1.84m
='lt).=
^
J
1.:v/f (1)
(iii) L: n-1" (1)
2
hence L: nv
2f
f:v
n
2L
(1)
(1)
(cj
80
l_.
I
i
i
1
10
.,,.i
-)
--J
1
l
:l
I
60
I
ti
I
l
..... ,- . .,-t- -: -- -- :
I
L
I
I
i
50
i
I
-L-.
l
..1
_
AO
i
I,
-r--i
___--.1
___l
30
i
l
I
l
I
I
-t
20
-.-,]- i'-"--
10
I
i
0
-."----t-,--.-
0
1
z
3
.'-----l---'--' -r--'-6
45
n
Axis
(l )
Scale
(1 )
Plotting
Best fit
(2 )
(1 )
Calculations
Gradient = 8.6 + 0.2
Hz
(1)
v:2L x gradient
(1)
v = 48 ms-l
(i)
Total 15 marks
I
'7
9
QUESTIOI{ 3
a
Thenmometen
.tdvamtaEe
Drsadvamtage
not verv accurate
sirnDle to use
cheap
Liquid in glass thenaometer
ortable
quick responding
Thermocouple
temperature must be
calculated ftom e.m.f.
awkward and bulky to use
cannot be used for rapidly
^L-tures
accurate
Constant volume gas
thermometer
verv accurate
(l mark each for a total of 6 marks)
O) 6)
Adiust heisht af tube to fixeC pornt
(iD a)
r1)
h6 - ice point - place ice and pure water in the beaker in equilibrium, adjust tube
to fixed point and read of mercury
b)
height
(1)
- steam point - heat pure water until boiling, adjust tube to fixed point and
read of mercury
(i)
h1e6
height
c)
ht - boiling point of Iiquid - place liquid in beaker, heat until boiling, adjust tube
to fixed point and read of mercury
d)
(c)
(1)
Iabel correct height on diagram
height
(1)
(1)
t: 100 (16.8 - 5)
20_5
= 78.7'C
(ii)
Pr-:P"*+hpg
(1)
(1)
: Pg (hdu,, + h)
Total 15 marks
If hpg: 13600 x 9.8 x 0168
= 13600 x 9.8 x 0.928
(1)
= 724lcPa
(1)
: 22.4kPa
(1 mark onty)
QUESTTON 4
ia) (i)
(ii)
Diie:rion changes (i)
a: r2lt
(b)
cha:rgiag (i)
(1)
Centripetal force perpendicular to dtection ofmotion
(ir
hence accelerating.
(1)
Towards centre of circle
(iiD
so velocitrr is
(1)
I
(l)
mg
(ir)
a)
T cos 300 = mg
(1)
T : mg/ cos 30o
(i)
T=1x9.8
cos 30
- I 1.-r 1\
b)
R=0.25m
T sin 30o = mfiR
(1)
(1)
(1)
).b) = I xv'
tan JQo = y211t
(1)
025
OR
v2 = 1.41 m2s-2
v : 1.19 ms'r
v2 = rg tan 30o
v = 1.19ms-l
(1)
(1)
(")
(i)
nass moves as projectile Lr parabollc paih
(ii)
(l )
1.5 ='/zx9.8xt2
(1)
t = 0.55s
r1)
Total 1 5 marks
(1)
QL,ESTION 5
G)
(,)
Each slit becomes a souice cfwave by
diffraction
(1)
D
/)
Theses waves are superposed. 'tr4ieie crest meets
and
crest there is constructive interference
'
( 1)
*"ets trough ( out of
phase)
ma)(imurn intensiti. cancettrng occur, *h"r. "r".t
+
+
(ir)
(1)
Blue has the shortest wavelength
(1)
from sin 0 : nVd, blue has the smallest 0
(1)
(1)
B is yellow, C is red
(ii1)
(b)
(1)
Paths from slits on each side of the central beam
(1)
colours are il Phase
n=dsin0
(1)
sin0= I
T
d=1/6x10s = 1.67x10-6
n:1.67xi0-6x1 = 2.8
SgO * tO'
max. n is .'. 2 orders
(1)
(1)
are the same' No path difference
(1)
"'
all
(ii)
0r :2x589xiC-'
(1)
0r = 45.Co (44.9'7")
(l)
i.67 r 10{
H.
= 2x590x10-e
rc7
"
10.6
(1)
0r = 45.1o (45.07)
Separation
: 0. 10'
(1)
QUESTICN 6
(a)
(i)
AU: Q +W
AU --- inclease in internal energy
Q - heat supplied to the gas
W --- work done on the gas
(3)
(iD
At constaat volume W : 0
(i
)
therefore the heat supplied (Q, = AU
: nc"AT) is
useC to increase the iLrternal energv of the gas
(l)
At constant pressure W = pAV
-Jierefcre the heat supplied (Q" = AU + pAV
: ncpaT) is useci to <io work anci increase i.'." ii,teri..l energ;"' Sc fcr
more heat is required, hence cp>
(b) (0
DrYrTr
(1)
(1) OR Y7=PAV
W: area = 1.01 x 105 x 0.0225
: 2.27 W
(ii)
cv
(1)
2plYr
-T-
Tz: 2Tt : 546K
(1)
(iii) Qr-z = ncvAT (1)
LT = 546 -273:273 K
= 1x3izx 8.314x273
: 3402.9 t
(1)
Q2-3 : ncrAT (1)
= lx5lzx 8.314 x 546
LT = t092 - 546 = 5a6 K
= 11343.2 J
Q total = 14.7 kJ
(1)
+-]1e sa:ne
AT a'lot
(n
)
eifrciency : \tr
2.27
74.7
: i5% or 0.15
Total 15 marks
(1)
(1)
'-"4
PETSTCS
uult 01 - PTDER, 02
taiB gcEDC 200?
i.
OrestLon !.
(a)
IE at
tr - i19.2 t 0.4 co (1)
2
D - 1 .g2 t O.O{ cm (1,
(b,
wgfgh 10 spheres (1,
by.gstrrr"ttng fraction of division
9?Btn
Dlvide !0lg
by l0 (1,
(c)
8D3
' " ?
3
(1'
' 3.706 x 1o-5 m'
,=+:-#*F - 8.203 x ro, ks p-r (1,
I emdr ln m - lt
(1t
a
t errqr in D3 - 6.25t (1,
t eror ln dens!.ty - Z.2St (1)
l
:
Actualierror - 0.595 x 10! kg n-l
i
Flnalltf p - (8.2 t 0.6, x 103 kg 6-r (1)
Spoclttc Obtlcttrrr: tGdrle 1: 6.1, 6.5
1
I
6
ut t!
01 - DIDIA 02
tasB Serot 2oo?
i.
OrestLon 2
(al
G I9
1
-{v
1
63.2 or 63.3
70.7
81.6 or 81.?
89. {
100
LL2
L24
Graph
Scales (1,
Plot 12,
Bcct llnc (1,
(b) Grad - 3.6 *, O.2
(cl
Oathod 1 rnd rco-rcfr U
"-2tf ,G gradient (1,
a
rE
3
€. $.
5.9 r
2.55 x 1
- 3{.0 n s-r
x 3.6 (1'
(1,
gtaMa GJootltrr: tGdnlo 2t 1.1, 5.6, S.9
a
6
PETAECS
UIrlE 01 - PIPER 02
,\
tllna scEEtG 2007
aa0
120
t00
r80
i60
a
u,
C)
I
a0
l,
It
C'
20
,o
to
;o
120 0
0
sa
PEISICg
urr! 01 - PrDln 02
raRr SCElc 200?
i.i
Oueation 3
u; IS
(a) Scales (11
Plor l2l
I
Good curve (1,
(b) e.g. n.D. - 55 squares x (2.5 x O.S
xtO-i, U (1,
(c) (1)
2
20 N
I
(1r, v - I!
- gradlent x !
(1'
GradLent - 9.3 x l0! tf n-r
(1)
Ar(
fr
A
3
9.3 x 10-3 x2
- 9.3 x 10ro N n-r (1,
2 x tO'1
(t 0.2 x l0ro)
{nat!I,a OJoctirrr: tGcblo 3: 6.?1 6.t, 6.10, 6.11
I
6
PEISres
UlrlT 01 - PrPtR 02
toRr scEA 2007
OuestLon I
tf,
BC
(a) (i,
scalar iquantltler can be expressed in terns of
I
thelr qagnltudc (f,
a
e.g. naber density, temperature, etc (ll
.l
Vectorg have uagmitude and dlrectlon (1'
c.g.
velocltyr dlsplaceucnt (1'
(il)
r oqrnrat
y coqloot
1{.8
31.7
20.8
-t2
2A N
I
-14.1
-11. 1
B
+21.5
+5.6
r{sco
I
35N
:
2{N
!
I
I
(1 rst
for orch :or conrot)
Rr
9:gllllgl"oryon"nrs
to the no{itlve x-axis 22.2 N tL, at 1{.6o (t}
(b, (f) thc resut
force nust be zero (1, p + O - If (U
The resul
L
PxIr-ll
2
(tl, A
B
2
a
torgue uust be zero (ft
(1'
f,r oN (1'
Px6 - 200 x 1.5 (1,
P -50N
(1,
F - Pl= s0 N (11
6
i
c
I
Coublnlng E and R the resultant is 206 N
angle of 7
gncl,ttc 0bloatlrrr:
to the ground (1, 1)
1: 1.2, 1.3, a.l, a.s
at an
I
L2
6
02138020letPRltrg.lzl
PEISrcs
urlr 01 - PIPEB 02
i,?
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PElttrcs
a
Wlr 01 - PrDtR 02
toRr trcEDG 200?
OuestLon S
tc
(1,
(1,
(1)
_,.'
DerivatioD
E1 r px - nai - * *-,,,1 - 1rrr,
,gT2---
(a) (t)
(1r)
Urr
3
I
Eofuula asEuues g Ls constant or only applles
to
anill distances.
1
I
(lii) .1"{"g p_rlnciple in equation
(Lops of Ep ls equal t-o ttre gain
ln E1) ..
1
1
*-iru2+msh
-n
2
1
1
(1'
lF - 1,!4 + 2 x 343 - g30
(1)
=
v-29 ms-l (1)
Galrt in % - 1200 I 35 - 42 OOO;,
(b) (f)
(tl,
powdr -
*,
=
{2000 r
24
(i11) Eff r Porer outPut
poyer input
(c)
input - 1?50 I
(1)
1750 f,
I
(lt
2
(1,
3
H .(U - 2sgo w (1)
Vcloel,Qr (dora l,t +)
i,r poritlvr)
I
I
3
-m
|F - )nn,+mx9.Bx35
2
v
2
v
(11) t-,i. + 1a tz
35
-
I
,L.9
t - zl.zs
C
(1,
t
3
t2 (u
(1,
qrat,ttc obJoct{ra: tlodulo 1: S.2, S.3, S.5,
5.6, S.7r 5.g
I
t2
DEtStcS
tnrl8 01 . Df,DIR 02
la8f scrc
Ouestton 6
(al
G,
Can be constructed
the same
fron
2007
i.;
progrcsslve raves
and .anpll tude
the same
t
of
r travelllng rtth
(u, There is no vlbratlon
.g Ar node (1'
B ls an antinode
rlth larger anplltude
than
C ls in antlphese
(111, pltch
rlth B (ft
- frequancy (1,
Ioudness _ lntcnslty
(tt (or eiryllhdcl
euallty _ ovGrtone structure
(1,
(b)
(r, c - fl =r l. g - rro uzll'^ ^a i6o-Al;'0.3{u
6,
(r,
(11, p- .,OdB,
- (10 dB)
(1'
I x l0'2 f, u-'
(r,
x l0'rz r n'2
- .ioo as (1,
(111, Threshold
of paln: 120 - l0 t9
r - 10ux I x l0-u
. i 1.0 nu-2
(1,
#
(1'
C (1,
PEESTCS
omr 01 - PIPIR 02
tGRf, gcsEtc 2oO?
ChrestLon 6 rpe"t ,d)
(b) (lil)
EC
I
dconr ,d)
I
Ftom lnverse law
I - i
= r, - rr4
t1,
f2 -
urs
6
x 1o-2 x 2to2- { rr (1}
(1,
I
i
I
nehce
I
xz- 2.0n (1,
$tecttta ObJofErrr: Io&rlo 2t l.t, 1.5, l.9r a.!0, a.1l
I
u2
v--.rcv3u I lj,LEE.l ,trE' I SVV t
PETSICS
UNIT 01 -
TPER 02
tOnA sCM
200? i,i
ChrcstLon ?
(al
(r)
Refracti.ve Lndex ls the ratlo of the spced of
ltEht Ln vaguo (or alr) to the speed ln the
E
I'E
t
nedl.un.
'
(11) Diagrara directloo tovards the oornal (1,
ravelength decrcases (U
(1U) Derlvatlon of lar
(1v) FrequencY ts lnvariant
.
tl1 Ir - nz la
2
3
(1,
2
(11
2
(b, (1) csrd' '=*i=d
- 1.99 r 10'n s-r (U
1.51
cbtr ' '=*l=d ' 1'94 x 1o'n g-t (u
1.55
(111 Red. 1.51 sl'n 30: 1 x sln 02 (1,
I
Q. {9.0" (1}
Blue 1.55 sln 30 ' 1 x sr.n 0l
Oar
- 50.8o (1,
DlspersLoo - 1.8c (1,
6
(c) Eormrla ntr ' a sln 0 (1,
ctrange
a ' (1200 x 10lr-r ' 8.33 x 10-?m [or cquivaleat(11
of foruulal
Red (uslng flrst order)
sln0-
678x10-.'- (1)
!-'
8.33 r 10-'
- 0.81t1
0 - 5l.lo (1!
Blue slnilar1Y 0tro.' 30.40 (1'
Dlspcrsion 24' (U
L.9, 1.a, 1.10, 3.3, 3.8,
{notflc obJoctlvrts tlodd. 2z 3.9
8
u2
(
PETSICS
Intr!8 01 - PrPEn 02
taRr scEtc 200?
i.i
Orestlon 8
(e) (1)
rc ur
GE rs theraoneter needed
in order to calibrate
ot her thermometers(U. Thi;-i" u"".o".-it-grrr."
re a$l1es on the.Ldeal gas (ther:nodynanic) z---l
gc ale (11 .
(i$
e.g. A thermocouple could be used (11 .
sna1l size rould be useful (1i.'
al
The
2
2
2
(iut
2
arn glrr rryr trro (21 .
(b)
(it
I
R -i Ro elA
5
i
icel pol.nt - 273k (1)
stei1n polnt - 373 k (1)
I
' 'ilT# - 48'r I ssl2?3
iffi
\273 3n)
"(t-
i
i
i
B=3.9{x1O3k
(it1
Tm'5- - 3.97 x 1o-3 fl
f,or
" ' ir.9?
_ 3.94 x tO,
97x
3.9,! x ld
2200
.9? x 1.0-3
-
I
50C
(1)
x 1O-3 et'e' r tdrr
2200
t-
(1,
?360 (l
i
t
(u
1)
ln 48 1 =
!
-3t373
(1,
T
3
(1)
= 298 k (1)
PEtstcs
)
utsls ol - DltPlR 02
tasr SCEIIG
i't
'OO7
Orestton t (Contrd)
E G
(b, (cont,d,
'
g=
(Iil) (a; 100
Qu -
3
R-
R t.-
- Rr-
g, = 2200-zsso
153 - 7350,
(1,
71.6oC (1,
(b) Reslstance ngt linear utth
SD.cfl& Objoctirrr: Uodulo 3: L.l,
teuperature.
1.2,1.3, t.t, l'.s
I
I
u
a
PEIgICS
urr8 01 - PIPER 02
urRB scEuE 200?
''i
I
Orestion 9
!
i
I
(a)
i
(1,
BC
fn motion and colll.de with
e)
the ralls (1,
Force because
.r,"ngJ-"i
ioi"n.*
Pressure is sum
"ror roices-alJio.o (1,
I
i
i
uf
6
by area (t)
More coltiilons per second
so greater (1,
average force.
b)
c
Mole-cules move faster (1,
Hlt harder and nore
often (1,
(lit Ur ing pV - nRT
lreTz
P2 '
P1 (1,
rLTr
t!
frpryera ture conversions\
l2r9e,
I (1'
(lto,c' - 300k
- 32ok
Pz
)
i' 6.7 x 32O
illoox{.5x10s
(r,
= 11.5 x 105 pa (1,
(b) (r,
Au- Q+w
(1,
Au. change in Lnternal energy
Oi. ::::9Y
bY heatins-'(u
'dq"9 by
'
WF energy
added
doing
I
2
,o"[
(ril a)
b)
Q- n q Ae (1,
- 6.2 x L2.5 x 25
- 1940 iI (1,
Au-Q-19{0U (1)
c)
Au-e+r
(1'
8
19110- 3200 + w (U
w = -L260 ;I
work done by the gas
= 1260 .r (11
Q= n Cp a0 (t,
d)
-P= 6.23200
,.ZS - 20.6 ,, nol-l k-r (U
co
gnettl,c 6Joati,vor: ttodtrlo
i
3: {._1, {.3, S.1, S.2, S.3,
5.S, s.c'-'',
I
L2