CSEC Physics
Prep for Success
Answer Series 2
Complete
Part 1 – Physical Quantities and Units, Density, Moments
Part 2 – Graphs, Hooke’s Law
Part 3 – Velocity, Acceleration, Equations of Motion, Kinetic Energy
Part 4 – Specific Heat Capacity, Latent Heat
Part 5 – Boyle’s Law, Pressure Law
Part 6 – Rays of Light, Reflection, Refraction
Join the next series
Get weekly answers in your email
CLICK HERE
CXCPhysics.com
CSEC Physics
Prep for Success
Series 2
Part one
Physical Quantities and Units
Density
Moments
Get more study material at
CXCPhysics.com
Jun 2012 P2. Q2
Physical quantities and Units
Density
1.
a) Complete the table below which relates the physical quantities and their derived
SI units.
Physical Quantities
Derived SI Unit
Area
m2
Volume
m3
Densit
Kgm-3
(3 marks)
State two fundamental quantities and their corresponding SI Units
Fundamental Quantity
SI Units
Lengt
m
Temperature
0
C
Any of the following 3 will be correct as well
Time
s
Current
A
Mass
kg
(4 marks)
b)
i. Calculate the density of gasoline which occupies 150 cm 3 of space and has a mass
of 102 g.
m
v
102 g
ρ=
150 cm3
−3
ρ=0.68 gcm
ρ=
There’s no need to convert the units grams (g) and centimeter cube (cm3) to the SI
Units. It’s sufficient to work with the units as given.
(3 marks)
Get more study material at CXCPhysics.com
ii. If 325 g of mercury of density 13.6 gcm-3 occupies a certain space, determine the
volume of the space.
m
ρ=
v
m
v= ρ
325 g
v=
13.6 gcm−3
3
v=23.9cm
(3 marks)
iii. Calculate the relative density of gasoline using the density of mercury as the base
quantity for the comparison.
ρ
ρ REL = ρgasoline
mercury
3
0.68 gcm
ρ REL =
13.6 gcm3
ρ REL =0.05
Take note that the units cancel and so relative density has not units
(2 marks)
Jan 2014 Paper 2. Q 2
Physical Quantities and Units
Moments
2.
a)
i. The figure below shows some physical quantities and their units. Draw an
arrow from each quantity to its unit.
(3 marks)
Get more study material at CXCPhysics.com
ii. Complete the table below to show the names of three forces and situations in
which each force acts.
Force
Situation
1. Gravit
A javelin falling in the air
2. Upthrust
A body in a uid
3. Magnec
Between two magnec poles
b) Patrick and Patricia showed their class a ‘magical’ balancing act. They both sat
on one side of a uniform, 6.0 m plank. Patrick sat 1.0 m from the pivot located
0.5 m from the center of the plank. Patricia at 0.5 m from the pivot, on the same
side as Patrick.
The figure below shows the ‘magical’ balancing act. X represents the pivot point.
i. Label as W, the point on the plank, where the weight acts.
(1 mark)
W
$e weight is at te cen&r of te plank. It is 0.5 m +om te pivot. It has , be
loca&d on te right of te pivot because it is te weight tat provides te
coun&rbalance Pa.ick and Pa.icia.
ii. Calculate the weight of the plank
From te principle of moments
sum of te clockwise moments = sum of te an-clockwise moments
W ×0.5 m=(400 N ×0.5 m)+(500 N×1 m)
Get more study material at CXCPhysics.com
(4 marks)
W ×0.5 m=200 Nm+500 Nm
700 Nm
W=
0.5 m
W =1400 N
iii. If Patrick sat alone on the plank, calculate where he would have to sit, in
order to maintain his ‘magical’ balancing act. The pivot remains as before.
If Pa.ick alone sat on te plank and it balanced, te clockwise moment crea&d
by te weight of te plank wi4 be equal , te an-clockwise moment crea&d by
his weight.
Let y be te unknown distance he has , sit +om te pivot for te plank ,
balance
sum of te clockwise moments = sum of te an-clockwise moments
1400 N ×0.5 m=500 N × y
700 Nm=500 N × y
700 Nm
=y
500 N
y=1.4 m
(3 marks)
Get more study material at CXCPhysics.com
CSEC Physics
Prep for Success
Series 2
Part two
Graphs
Hooke’s Law
Get more study material at
CXCPhysics.com
Jan 2010 P3/2. Q2
Graphs
Hooke’s Law
Two fourth form Physics students were asked to work together to investigate the relationship
between extension, e, and the stretching force, F, for a spiral spring. Four sets of their results are
shown in the figure below.
(a) Use the diagrams shown above to complete Table 1.
Mass (m) /g
Stretching force (F) /N Scale Reading /mm
Extension (e) /mm
0
0
10.0
0.0
200
2
12.1
2.1
400
4
14.3
4.3
600
6
16.3
6.3
800
8
18.4
8.4
1000
10
21.4
11.4
1200
12
25.2
15.2
(4 marks)
Get more study material at CXCPhysics.com
(b) On graph paper, plot a graph of extension, (e) against stretching force, (F).
(8 marks)
Get more study material at CXCPhysics.com
(c) Identify the part of the graph where Hooke’s law is NOT obeyed. Label the beginning and
end of the part as A and B respectively.
( 1 mark)
Get more study material at CXCPhysics.com
(d) Calculate the gradient of the straight line portion of the graph
y 2− y 1
x 2−x 1
8.4−0.4 mm
gradient=
8.0−0.4 N
8.0 mm
gradient=
7.6 N
−1
gradient =1.05 mmN
gradient =
−1
gradient=1.1 mmN 2 sig. fig .
Get more study material at CXCPhysics.com
(4 marks)
(e) Use the graph to find the mass that would give an extension of 5.0 mm.
(Acceleration due to gravity = 10.0 ms-2).
Use
e graph nd
Find y = 5 mm on
e force
at produces an exnsion of 5 mm
e y-axis and read
e corresponding value on
e x-axis.
%is graph reads 4.8N
Using W= mg, where W is
gravi-, we can nd
W =mg
W
m=
g
4.8 N
m=
10.0 ms−2
m=0.48 kg
e weight (force), m is
e mass
e mass and g is accelera,on due at produces a force of 4.8 N
Get more study material at CXCPhysics.com
(2 marks)
(f) Identify ONE possible source of error and ONE precaution to be taken in conducting this
experiment.
One precau,on is take
e readings of
e leng
of
e spring at eye level avoid errors
due para0ax.
One source of error could arise 1om
accura a degree of half of
e measuring ins2ument,
e ruler, as it can only be
e sma0est division. %e sma0est division on
mm and answers can only be given accuraly 0.05 mm, hence
Get more study material at CXCPhysics.com
e ruler is 0.1
is is a source of error.
(2 marks)
CSEC Physics
Prep for Success
Series 2
Part three
Velocity
Acceleration
Equations of Motion
Kinetic Energy
CLICK HERE
Get this and more in your email
CXCPhysics.com
June 2010 Paper 2 Question 2
Graphs
Hooke’s Law
(a) Define each of the following terms
i. Velocity
e veloci
is
e ra of change of displacement wi
me. It is measured in ms-1.
(2 marks)
ii. Acceleration
e ra of change of veloci
wi
me. It is measured in ms -2.
(2 marks)
iii. Linear momentum
e product of
e mass and
e veloci
of a body. It is measured in kgms -1.
(2 marks)
(b) In 2008, at the Beijing Olympics, Usain Bolt of Jamaica reclaimed his title as the world’s fastest
man. He completed the 100 m final in a world record time of 9.69 s. He accelerated uniformly
from rest for the first 6.5 seconds, covering 60 m before coasting at maximum speed to finish.
i. Calculate his average speed for the first 6.5 s.
d=60 m
t=6.5 s
average speed=
total distance
total time
average speed=
60 m
6.5 s
−1
average speed=9.23ms
−1
average speed=9ms correct to 1 sig fig
(2 marks)
Get more study material at CXCPhysics.com
ii. What was his maximum speed?
He got # maximum speed afr running for 6.5 s
veloci ; v, )nal veloci
erefore ga ering values for u, inial
afr 6.5 s, i.e. maximum speed; t, me and s, displacement
u=0
v=?
t=6.5 s
s=60 m
Using the third equation of motion s=(
60 m=(
u+v
)t
2
0+ v
)6.5 s
2
2×60=(0+v )6.5
120=(6.5×0)+(6.5×v)
120=(6.5×v)
v=
120
6.5
−1
v=18.46 ms
−1
v=18 ms correct to 2 sig. fig .
( 2 marks)
iii. What was his acceleration during the first 6.5 s?
Using the same values from the previous question
u=0
−1
v=18 ms
t=6.5 s
Using equation of motion v=u+at , we can rearrange this to get, firstly
v−u=at then, making a the subject we get
a=
(v−u)
t
a=
(18 ms −0)
6.5 s
−1
−2
a=2.77ms
Get more study material at CXCPhysics.com
−2
a=2.8 ms correct to 2 sig . fig .
(2 marks)
(a) What major form of energy did Bolt possess when he crossed the finish line?
Kinetic energy
( 1 mark)
(b) Calculate the value of this energy of his mass was 86 kg.
1 2
KE= mv
2
m=86 kg
−1
v=18 ms
this is the same maximum speed he had when he finished the race. We
calculated this in question (b) iii. above
1
−1 2
KE= ×86 kg×(18 ms )
2
1
2
KE= ×86×(324)
2
KE=13, 932 J
KE=14 kJ correct to 2 sig . fig .
(2 marks)
Get more study material at CXCPhysics.com
CSEC Physics
Prep for Success
Series 2
Part Four
Specific Heat Capacity
Latent Heat
CLICK HERE
Get this and more in your email
CXCPhysics.com
June 2011 Paper 2 Question 2
Specific Heat capacity
Latent Heat
(a)
i. Complete the table by inserting the correct symbol and SI Unit which relate to the
quantity shown in column 1.
Quantity
Symbol
SI Unit
Specific Heat Capacity
c
J kg-1K-1
lv
J kg-1
Specific Latent Heat of
Vaporization
(3 marks)
ii. Define the term specific heat capacity of a substance
e specic heat capaci is e amount of heat at is required raise e
mperature of a mass of 1 kg by 1 degree.
(2 marks)
iii. Write the equation that relates specific heat capacity with heat capacity.
C =mc
(1 mark)
(b) A busy housewife left 25 g of ice in an open insulated container while she was answering
her cellphone. When she returned, the ice at 0 0C was converted to water at 3 0C.
i. Calculate the energy needed for the ice to totally melt and to reach its present
temperature. Assume no heat loss.
Assuming no heat loss e tal heat required has be calculad in two parts.
First, heat is needed change e ice at 0 0C war at 0 0C, is uses e lant heat
of )sion of ice.
E H =ml f
en, heat is needed change e mperature of e war at 0 0C war at 3 0C.
e specic heat capaci is used calcula is part.
E H =mc Δ Θ
Get more study material at CXCPhysics.com
erefore tal heat can be found using is equa+on
E H =(mlf )+(mc Δ Θ)
We must ensure at units for like quan++es are coherent (i.e e same). So since e
values for e specic heat capaci of war and e specic lant heat of )sion of ice
are given in J kg-1K-1 and J kg-1 we should convert e mass 0om 25 g kg.
25 g=(25÷1 000)kg
m=0.025 kg
erefore
−1
−1
−1
E H =(0.025 kg×340 000 J kg )+(0.025 kg×4 200 J kg K ×3)
E H =(8500 J )+(315 J )
E H =8 815 J
E H =8.8 kJ correct 2 sig g.
Specific heat capacity of water = 4 200 J kg -1K-1
Specific latent heat of fusion of ice = 340 000 J kg -1
(6 marks)
ii. If this melting and heating activity took place over 300 s, calculate the rate at which the
ice / water was receiving heat.
is ques+ons is e same as asking for how much energy is 3ansferred every second.
It can be calculad by dividing e tal energy received by e +me
rate=
total energy received
time
rate=
8800 J
300 s
rate=29.3 J / s
rate=30 J / s correct to 1 sig. fig.
(3 marks)
Get more study material at CXCPhysics.com
CSEC Physics
Prep for Success
Series 2
Part Five
Boyle’s Law
Pressure Law
CLICK HERE
Get this and more in your email
CXCPhysics.com
Jan 2008 Paper 2 Question 2
Boyle’s Law
Pressure Law
(a)
i. Boyle’s law states that the absolute pressure of a fixed mass of gas is inversely
proportional to the volume once the temperature doesn’t change
( 2 mark)
ii. If you were to carry out an experiment to investigate Boyle’s Law, this is the linear
graph that you would expect to get.
P
1/V
(4 marks)
(b)
i.
This is how you can use the kinetic theory to explain an increase in pressure in the
tires of a vehicle after driving at a high speed for a very long time. The Kinetic
theory states that all matter is made up of particles that are in constant motion. The
average kinetic energy of the particles is related to the temperature. The particles
collide off each other and the walls of the container and so create a pressure on the
walls. After the high speed chase, the friction between the tires and the ground
would’ve increased the heat transferred to the tires. The transfer of heat energy to the
particles increased their kinetic energy and therefore increased their motion. An
Get more study material at CXCPhysics.com
increase in motion results in more frequent collisions of the particles on the walls
and among themselves and this results in an increase in pressure.
(4 marks)
ii. Atmospheric Pressure is 100 KPa.
The pressure before the chase, P1 is 210 KPa over atmospheric pressure
P1 = 210 Kpa + 100 KPa = 310kPa
The pressure after the chase, P2 is 260 KPa over atmospheric pressure
P2 = 260 KPa+100KPa = 360 KPa
The temperature before the chase is 24 0C. When working with the Gas laws, it is the
Kelvin temperature that’s used. Therefore convert this to Kelvin
0
0
24 C +273 C=297 K this is T1
Summary
P1 = 310 KPa
P2 = 360 KPa
T1 = 297 K
T2 = ?
Assuming there’s no change in volume, the pressure law will apply. Equation for
pressure law is
P1 P2
=
T1 T 2
Make T2 subject
P1 T 2=P 2 T 1
T 2=
P2 T 1
P1
T 2=
360000×297
310 000
T2 = 344.9 K
T2 = 345 K correct to 2 sig fig
(5 marks)
Get more study material at CXCPhysics.com
CSEC Physics
Prep for Success
Series 2
Part Six (Final)
Rays of Light
Reflection
Refraction
CLICK HERE
Get this and more in your email
CXCPhysics.com
June 2008 Paper 2 Question 3
Light Rays
Reflection
Refraction
(a) Take note of the following features Z, X, Y, W, V, x and label them
X
Z
Y
^z ^y
w
^
W
c^
V
x
Z – incident ray
X – normal
Y – reflected ray
W – refracted ray
V – emergent ray
X – lateral displacement of Z
(6 marks)
(b)
i. If
^z is 600 then
^y is also 600.
(1 marks)
ii. The reason is that the law of reflection states that the angle of incidence is equal to
the angle of reflection. In this case
^z is the angle of incidence and ^y is the angle
of reflection. The law therefore says that they are both equal.
Get more study material at CXCPhysics.com
(1 mark)
iii. If the refractive index of this block is 1.5 then w
^ can be calculated using Snell’s
Law.
Snell’s Law states that the ratio of the angle of incidence, i^
to the angle of
refraction r^ is a constant. For light passing from air into glass, this constant is the
refractive index of the glass. Therefore we can write
n=
i^
^z
⇒ n=
r^
w
^
1.5=
^=
w
60
^
w
60
1.5
w
^ =40
^ =400
w
(4 marks)
iv. If we want to determine the value that angle w, w
^ must exceed in order for E to be
totally internally reflected, we must consider that angle w and angle c are equal.
(Geometry review: Interior angles are equal). Angle c is the critical angle of the
glass and when the critical angle is exceeded, the ray will be totally internally
reflected. The question wants us to find the critical angle of the glass.
We can apply the relationship between the critical angle of glass and its refractive
index.
n=
1
sin c
1.5=
1
1
Rearrange this to make sin c the subject we get sin c=
sin c
1.5
sin c=
1
= 0.67
1.5
−1
c=sin 0.67
−1
c=sin 0.67
c=42
0
(3 marks)
Get more study material at CXCPhysics.com