Dalziel High School
Higher Physics
Mechanics and Properties of
Matter
Unit 1
Dalziel High School
Crawford Street Motherwell, Lanarkshire ML1 3AG
T 01698 274900 F
jmonaghan@dalziel.n-lanark.sch.uk
Physics@dalziel
Dalziel High School
Mechanics
Scalars and vectors
VECTORS
VECTORS
Distance and Displacement.
Distance
is the total path length. It is fully described by magnitude (size) alone. Displacement is the
VECTORS
Distance
and Displacement.
direct length from a starting point to a finishing point. To describe displacement both magnitude and
Distance
is themust
total
path length. It is fully described by magnitude (size) alone.
Distance
and
Displacement.
direction
be given.
Displacement
is
the
direct
lengthIt isfrom
starting
point to (size)
a finishing
point. To describe
Distance is the total path length.
fully a
described
by magnitude
alone.
Displacement
is
the
direct
length
from
a
starting
point
to
a
finishing
point.
To
describe
displacement
Exampleboth magnitude and direction must be given.
displacement both magnitude and direction must be given.
A woman walks 3 km due North (000) and then 4 km due East (090). Find her
ExampleExample
A woman
a) distance
travelled
walks
3 km
dueNorth
North (000)
and and
then 4then
km due
A woman
walks
3 km
due
(000)
4 East
km (090).
due East (090).
Find her a) distance travelled
b)
displacement
i.e.
how
far
she
is
from
where
she
started ?
Find her a) distance
travelled
b) displacement
i.e. how far she is from where she started ?
b) displacement i.e. how far she is from where she started ?
4 km
B
3 km
3 km
4 km
B
C (Finish)
C (Finish)
Displacement
A (Start)
Displacement
Using a scale of 1cm: 1km draw an accurate scale diagram as shown above.
Distance
= ABscale
+ BC diagram
= 7 km as shown above.
Using a scale of 1cm:a)1km
drawtravelled
an accurate
b) Measuring AC = 5 cm.
Convert
the scale gives the
a) Distance travelled = AB
+ BCusing
= 7 km
A magnitude
(Start) of the displacement = 5 km
Use a protractor to check angle BAC = 53o that is 53o east of north.
b) MeasuringUsing
AC = 5acm.
Convert
using1km
the scale
the
of the
displacement
= 5 km
Displacement
= 5magnitude
km (053)
scale
of 1cm:
drawgives
an accurate
scale
diagram
as shown
above.
travelled
= AB
Usea)a Distance
protractor
check
angle
BAC+= BC
53o =
that7 iskm
53° east of north. Speedto
and
Velocity
b) Measuring
AC
= 5 cm.
These
quantities
are fundamentally different.
Displacement
= 5 two
km (053)
Convert using the scale gives the magnitude of the displacement = 5 km
displacement
that is
53o east
of north.
Average
velocity
=
Average speed =
time
time
Displacement = 5 km (053)
Speed and Velocity
distanceBAC = 53o
Use a protractor to check angle
These two quantities are fundamentally different.
Velocity has an associated direction, being the same as that of the displacement.
Speed Velocity
and
Velocity
The unit
for both
these quantities
is metres
per the
second,
m as
s-1. that of the displacement. The unit for both
has
an associated
direction,
being
same
thesequantities
quantities isare
meters
per second, m different.
s-1.
These two
fundamentally
Vectors and Scalars
A scalar quantity is completely defined by stating its magnitude.
A vector quantity is completely defined by stating its magnitude and direction.
distance
Examples
are givenspeed
below. =
Average velocity
Average
time
Vectors
Displacement
Velocity
Velocity has an associated direction,
being the
Acceleration
The unit for both these quantities
is metres per
Force
Impulse
Scalars
Distance
Speedas that of
same
Time m s-1.
second,
Mass
Energy
=
displacement
time
the displacement.
Vectors and Scalars
A scalar quantity is completely defined by stating its magnitude.
Higher Physics
1 quantity is completely defined by stating its magnitude and direction.
A Unit
vector
Examples are given below.
Vectors
Scalars
1
Dalziel High School
Vectors and Scalars
A scalar quantity is completely defined by stating its magnitude.
A vector quantity is completely defined by stating its magnitude and direction. Examples are given below.
Vectors
Scalar
Displacement
Distance
Velocity
Speed
Acceleration
Time
Force
Mass
Impulse
Energy
Addition of Vectors
Addition
of Vectors
When vectors
are being
added, their magnitude and direction must be taken into account.
This can beWhen
done
using
scale
diagram
and adding
themust
vectors
‘tip
tail’,This
then
vectors
are a
being
added,
their magnitude
and direction
be taken
into to
account.
can joining the
done using apoints.
scale diagram
adding
the vectors
‘tip to tail’,
then joining
the starting
tarting andbefinishing
Theandfinal
sum
is known
as the
resultant,
the and
single vector that
finishing points. The final sum is known as the resultant, the single vector that has the same effect as
has the same effect as the sum of the individuals.
the sum of the individuals.
Addition of Vectors
Example Example
When
vectors
are being added, their magnitude and direction must be taken into account.
Find the resultant force acting
point O.
Find the resultant
force actingThis
atatcan
point
O.using a scale diagram and adding the vectors ‘tip to tail’, then joining the
be done
starting and finishing points. The final sum is known as the resultant, the single vector that
has the same effect as the sum of the individuals.
8N
O
Example
0
Find the resultant force acting at point O.
45
5N
8N
O
45
0
1: Choose
a suitable
e.g.to
1 5cm
to 1 N.
Step
1: Choose
aStep
suitable
scale,
e.g.scale,
1 cm
1N N.
Step 2: Arrange arrows “tip to tail”.
Step 1:to
Choose
a suitable scale, e.g. 1 cm to 1 N.
Step
2: Arrange
arrows
“tip
tail”.
Step 3: Draw
Stepin2:resultant
Arrangevector,
arrowsmeasuring
“tip to tail”.its length and direction.
Step 3: Draw in resultantStep
vector,
measuring
itsmeasuring
lengthitsand
direction.
3: Draw in
resultant vector,
length
and direction.
(000)
(000)
A
8N
A
8N
B
B
450
5N
450
5N
1 cm to 1 N.
C
= 12.2 cm
Force = 12.2
= N12.2 N
Using a protractor, angle BAC measures 120
0
Bearing
=
+ 120 = 1020
Using a protractor, angle BAC measures 12° Bearing = 90+12 =90102°
Resultant Force
= 12.2 N at (102)
C
Resultant Force = 12.2 N at (102)
Vectors at right angles
AC
12.2thencm
If the vectors are at =
right angles
it may be easier to use Pythagoras to find the resultant
and
trigonometry
to
find
an
angle. N
Force
= 12.2
1cm to 1N.
1 cm to 1 N.
AC
AC = 12.2
cm Force
0
Addition of more than twoUsing
vectors a protractor, angle BAC measures 12
0
0
each vector
is placed “tip to tail” to the previous vector.
BearingUse a scale diagram=and ensure
900that
+ 12
= 102
The resultant vector is the vector from the starting point to the finishing point in magnitude
Resultant
Force
= 12.2 N at (102)
and direction.
Higher Physics Unit 1
Resultant of a number of forces
Vectors at right angles
The resultant of a number of forces is that single force which has the same effect, in both
magnitude
the sum of
individual
forces.
f the vectors are at right angles
thenand
it direction,
may beaseasier
totheuse
Pythagoras
to find the resultant
2
Dalziel High School
Vectors at right angles
If the vectors are at right angles then it may be easier to use Pythagoras to find the resultant and
trigonometry to find an angle.
Addition of more than two vectors
Use a scale diagram and ensure that each vector is placed “tip to tail” to the previous vector. The
resultant vector is the vector from the starting point to the finishing point in magnitude and direction.
Resultant of a number of forces
The resultant of a number of forces is that single force which has the same effect, in both magnitude
and direction, as the sum of the individual forces.
Rectangular
components
of of
a vector
Rectangular
components
a vector
Resolution
of a vector
into horizontal
vertical
Resolution
of a vector
into horizontal
and vertical and
components.
components.
Rectangular components of a vector
Any
vector
can
bevector
split
into
a horizontal
component
vcomponent
h and vertical
Any vector
v canvbe
split
into aup
horizontal
component
vh and vertical
vv. component vv.
Resolution
of aup
into
horizontal
and vertical
components.
Rectangular components of a vector
Any vector
v can be split up into a horizontal component vh and vertical component vv.
vv components.
= ?
Resolution
v of a vector into horizontal and vertical
Any vector v can be split up into a horizontal component
vv = ? vh and vertical component vv.
v
is equivalent to
is equivalent to
v
vv = ?
vh = ?
is equivalent to
vh = ?
vh = ?
v
v
v
v v vv
vh v
h
vv
vh
Example
A shell
isExample
fired from
a cannon as shown.
Example
Example
A fired
shell isfrom
firedafrom
a cannon
as shown.
A shell is
cannon
as shown.
Calculate
its a) horizontal
component
of velocity
Calculate
its
a)
horizontal
component
velocity
Calculate its a) horizontal
component
ofofvelocity
b)
vertical
component
of
velocity.
b)
vertical
component
of
velocity.
b) vertical component of velocity.
b)
vertical component of velocity.
A shellitsis fireda)from
a cannon
as shown.
Calculate
horizontal
component
of velocity
-1
a)cos
vh =60°
u = s-1
cos 60° = 25-1 m s
a) vh =v cosu
=50
25m
a) v ==50
v cos
u v=cos
cos50
60°
= 25 m s
h
-1
43
b)uv=
=50
v sin
=
50 =
sin 60° =
ms
a) vh = vvv cos
cosusin
60°
sinv 60°
u ==50
60° =2543mms-1s-1
b) vv = v b)
sinu ==50v sin
43 m s-1
b) vv = v sin u = 50 sin 60° = 43 m s
-1
-1
50mm
ms s-1
s-1
50
50
0
6060
0
60
So
So
0
50 m s -1
50 m s -1
So
50 m s -1
43.3 m s -1
is equivalent to
43.3 m s -1
43.3 m
is equivalent to
s -1
25 m is
s -1equivalent to
60 0
60 0
60 0
25 m s -1
25 m s -1
Higher Physics Unit 1
3
Dalziel High School
EQUATIONS OF MOTION
EQUATIONS OF MOTION
Acceleration
Acceleration is defined as the change in velocity per unit time. The unit is metre per second squared,
Acceleration
EQUATIONS
OF MOTION
m s-2.
Acceleration is defined as the change in velocity per unit time.
The unit is metre per second squared, m s-2.
Acceleration
v −u
where
t
Acceleration is defined as the change in velocity per unit time.
The unit is metre per second squared, m s-2.
v -u
a=
t
a=
v = final velocity
u = initial velocity
= time taken
v = finalt velocity
where
v -u
a = u = initial velocity, t = time utaken
= initial velocity
where v = final velocity,
t
t = time taken
Measuring acceleration
Acceleration is measured by determining the initial velocity, final velocity and time taken. A
Measuring
acceleration
Measuring
acceleration
double mask
which interrupts a light gate can provide the data to a microcomputer and give a
Acceleration
isismeasured
by
determining the initial velocity, final velocity and time taken. A
Acceleration
measured
by determining the initial velocity, final velocity and time taken. A double mask which interrupts a
direct reading
of acceleration.
double mask which interrupts a light gate can provide the data to a microcomputer and give a
light gate
canof
provide
the data to a microcomputer and give a direct reading of acceleration.
direct
reading
acceleration.
Acceleration-time and velocity-time graphs
Acceleration-time and
graphs
Acceleration-time
andvelocity-time
velocity-time
graphs
Constant velocity
Constant positive acceleration
(velocity
increasing)
Constant positive
acceleration
Constant velocity
(velocity increasing)
-1
v / m s -1
v/ms
v/ms
-1
v/ms
t/s
a / am/sm-2 s
Constant deceleration
Constant
negative acceleration
Constant deceleration
(velocity
decreasing)
Constant negative
acceleration
v/ms
(velocity decreasing)
-1
v/ms
t/s
t/s
-2
a / m sa -2/ m s
t/st / s
a / m s -2a / m s
acceleration is positive
t/s
-2
t/s
t/s
acceleration is positive
a=0
t/s
-2
t/s
a=0
t/s
-1
-1
t/s
acceleration is negative
acceleration is negative
Constant velocity and constant acceleration
Constant
and
constant
acceleration
Constantvelocity
velocity
and
constant
acceleration
The velocity time graph below illustrates these terms.
The
timetime
graph
below
illustrates
these terms.
Thevelocity
velocity
graph
below
illustrates
these terms.
OA
v / m s -1
v / m s -1
A
AB
B
A
B
BC
C
O
O
Higher Physics Unit 1
t/s
C
t/s
is constant acceleration,
OA
is constant
acceleration,
the acceleration
is positive.
is constant
the velocity,
acceleration is positive.
the acceleration
is zero.
AB
is constant
velocity,
is constant
deceleration,
the acceleration is zero.
the acceleration
is deceleration,
BC
is constant
negative.
the acceleration is
negative.
4
Dalziel High School
Equations of motion
v = u + at
s = displacement (m)
2
1
s
=
ut
+
at
2
Equations of motion
u = Initial velocity (m s-1)
where: v = final velocity (m s-1)
a = acceleration (m s-2)
u - initial velocity of object at time t = 0
v = u + 2as
t = time (s)
v - final velocity of object at time t
a - acceleration of object
t - time to accelerate from u to v
1
s = ut
+ atof2 motion apply providing:
These
equations
s - displacement in time t.
2
v2= u + at2
๏
where:
the motion is in a straight line
2
๏v
is uniform.
= the
u2 acceleration
+ 2as
When using the equations of motion, note:
These equations of motion apply providing:
thequantities
motion u,
is v,ins aand
straight
the
a are allline
vector quantities
๏
the acceleration is uniform.
๏
a positive direction must be chosen and quantities in the reverse direction must be given
negative
Whenausing
thesign
equations of motion, note:
the quantities u, v, s and a are all vector quantities
a deceleration will be negative, for movement in the positive direction.
๏
a positive direction must be chosen and quantities in the reverse direction must be given
a negative sign
a deceleration
be negative, for movement in the positive direction.
Derivation
of equationswill
of motion
Derivation
of equations
of motion
The
velocity - time
graph for an object
accelerating uniformly from u to v in time t is shown below.
The velocity - time graph for an object accelerating uniformly from u to v in time t is shown
below.
v / m s-1
v −u
v -u
a=
v
a =
t
t
A
u
Changing
thethe
subject
of theof
formula
gives: gives:
Changing
subject
the formula
v = u + at --------- [1]
B
v = u + at
--------- [1]
The displacement s in time t is equal to the area under the velocity time graph.
t/s
The displacement s in time t is equal to the area under the velocity time graph.
s=Area of triangle A + Area of triangle B
Area
Area
of triangle A + area of rectangle B
s= 21 (v=− u)t
+ ut
1
s = 2 (v-u)t + ut
but from equation 1,
v − u = at
so = at
but from equation 1,
v-u
s = 21 (at )t + ut
1 2
+ ut
s =s ut= + 21 at(at)t
----------[2)
2
s =
Higher Physics Unit 1
1
ut + at2
2
--------- [2]
5
Dalziel High School
Using
v = u + at
v 2 = (u + at )2
v 2 = u2 + 2uat + a 2t 2
v 2 = u2 + 2a(ut + 21 at 2 )
v 2 = u2 + 2aS
Projectile motion
A projectile has a combination of vertical and horizontal motions. Various experiments show that these horizontal and vertical
motions are totally independent of each other.
Closer study gives the following information about each component.
Horizontal: constant speed
Vertical: constant acceleration downward (due to gravity).
Example
An object is released from an aircraft traveling horizontally at 1000 m s-1. The object takes 40 s to reach the ground.
a) What is the horizontal distance travelled by the object?
b) What was the height of the aircraft when the object was released?
c) Calculate the vertical velocity of the object just before impact.
d) Find the resultant velocity of the object just before hitting the ground.
Before attempting the solution, you should divide your page into horizontal and vertical and enter appropriate information
given or known.
Horizontal
a)
Vertical
s=?
b)
u = 1000m s −1
s=?
v = 1000m s −1
u = 0m s −1
v=
a=0
t = 40s
a = 9.8m s −2
t = 40s
s = ut = 1000 × 40 = 40000m
s = ut + 21 at 2
s = 0 + 21 × 9.8 × 402
s = 7840m
c)
v = u + at
v = 0 + 9.8 × 40
v = 392m
v = 10002 + 3922
−1
d) v = 1074m s
tanx =
Resultant velocity is
392
= 21°
1000
Higher Physics Unit 1
6
Dalziel High School
NEWTON’S SECOND LAW, ENERGY AND POWER
Dynamics deals with the forces causing motion and the properties of the resulting moving system.
Newton’s 1st Law of Motion
Newton’s 1st law of Motion states that an object will remain at rest or travel with a constant speed in a straight line (constant
velocity) unless acted on by an unbalanced force.
Newton’s 2nd Law
Newton’s 2nd law of motion states that the acceleration of an object:
๏
varies directly as the unbalanced force applied if the mass is constant
๏
varies inversely as the mass if the unbalanced force is constant.
These can be combined to give
a∝
a=
Fun
m
kFun
m
kFun = ma
The unit of force, the newton is defined as the resultant force which will cause a mass of 1kg to have an acceleration of 1 m
s-2. Substituting in the above equation.
k × 1= 1× 1
k =1
Provided F is measured in newtons, the equation below applies.
Fun = ma
Free Body Diagrams
Some examples will have more than one force acting on an object. It is advisable to draw a diagram of the situation showing
the direction of all forces present acting through one point. These are known as free body diagrams.
Examples
1.
On take off, the thrust on a rocket of mass 8000 kg is 200,000 N. Find the acceleration of the rocket.
Resultant force = 200000 - 78,400 = 121,600 N
a=
Fun
m
=
121600
= 15.2m s−2
8000
Higher Physics Unit 1
7
F
121600
=
= 15.2 m s-2
m
8000
Dalziel High School
a =
2.
is standing
a set of bathroom
in a stationary
lift (ainnormal
everyday occurrence!).
The reading on
2.A woman
A woman
is on
standing
on a setscales
of bathroom
scales
a stationary
lift (a normal
the scales
is 500 N.occurrence!).
When she presses
thereading
ground floor
the liftisaccelerates
downwards
and the reading
everyday
The
on button,
the scales
500 N. When
she presses
the on
the scales
at
this
moment
is
450
N.
Find
the
acceleration
of
the
lift.
ground floor button, the lift accelerates downwards and the reading on the scales at
this moment is 450 N. Find the acceleration of the lift.
Weight = 500 N
W = 500 N
Force upwards = 500 N
F = 450 N
(reading on scales)
Lift accelerates downwards,
unbalanced force acts.
A ski tow pulls 2 skiers who are connected by a thin nylon rope along a frictionless surface. The tow uses a force of
W=F
Resultant Force = Weight - Force from floor
70 N and the skiers have masses of 60 kg and 80 kg. Find
=W-F
= the
500system
N
a) the acceleration of
= 500 – 450
3.Tension
Lift is stationary, forces balance
b) the tension in the rope.
= 50 N
Resultant Force
a =
m
50
50
= 1 m s-2
=
3.
a) Total
mass, m = 140kg
Tension
A ski tow pulls 2 skiers who are connected by a thin nylon rope along a frictionless
F
70
surface.
usessa-2 force of 70 N and the skiers have masses of 60 kg and 80 kg.
∴a = unThe
= tow
= 0.5m
m 140
Find a) the acceleration of the system
b) the tension in the rope.
b) Consider the 60kg skier alone.
Tension, T = ma = 60 × 0.5 = 30N
T
60 kg
Resolution of a Force
70 N
80kg
In the previous section, a vector was split into horizontal and vertical components. This can obviously apply to a force.
a)
Total mass, m = 140 kg
F
70
a =
=
= 0.5Fm=Fs-2sinθ
v
m
140
F
b)
θ
Higher Physics Unit 1
is equivalent
Consider
theto60 kg skier alone.
Tension, T = ma = 60 × 0.5 = 30 N
Fh =F cos θ
8
Resolution
of a Force
Dalziel High School
In the previous section, a vector was split into horizontal and vertical components.
This can obviously apply to a force.
Example
Fv = Fsin
F
A man pulls a garden roller of mass 100 kg with a force of 200 N acting
at 30° to the horizontal. If there is a frictional force of 100isNequivalent
between the to
roller and the ground, what is the
acceleration of the roller along the ground?
Fh = Fcos
Fh = F cos θ = 200 cos 30° =173.2 N
Example
Resultant Fh = 173.2 - Friction = 173.2 - 100 = 73.2 N
A
man pulls a garden roller of mass 100 kg with a force of 200 N acting
at 300 to the horizontal. If there is a frictional force of 100 N between the roller
Fun 73.2
and
what is−2 the acceleration of the roller along the ground?
a = the
=ground,
= 0.732ms
100
m
Fh = F cos = 200 cos 300 =173.2 N
Force Acting
a Plane
Resultant
FhDown
= 173.2
- Friction = 173.2 - 100 = 73.2 N
F
30 0
73.2
Ifaan=object is=placed on =
a slope
then
0.732
mitss-2weight acts vertically downwards. A certain component of this force will act down
Fh
m
100
Friction
the slope. The weight can be split into two components at right angles to each other.
100 N
Force Acting Down a Plane
If an object is placed on a slope then its weight acts vertically downwards. A certain
component of this force will act down the slope. The weight can be split into two
components at right angles to each other.
Reaction Force
mg sinθ
Component of weight down slope = mgsin
θ
mgsin
mg
mg
θ
Component of weight down slope = mgsinθ
Component perpendicular to slope =
mg cosθ
mgcos
mgcosθ
Component perpendicular to slope = mgcos
Example
Example
A wooden block of mass 2 kg is placed on a slope at 30° to the horizontal as shown. A
A wooden block of mass 2 kg is placed on a slope at 30° to the horizontal as shown. A frictional force of 4 N acts up the
frictional force of 4 N acts up the slope. The block slides down the slope for a distance
slope. The block slides down the slope for a distance of 3 m. Determine the speed of the block at the bottom of the slope.
of 3 m. Determine the speed of the block at the bottom of the slope.
3m
2 kg
Friction = 4 N
30 0
3 m 2 kg
Component
of weight acting down slope = mg sin30° = 2 × 9.8 × 0.5 = 9.8 N
Friction = 4 N 300
Resultant force down slope = 9.8 - friction = 9.8 - 4 = 5.8 N
Component of weight acting down slope = mg sin30° = 2 × 9.8 × 0.5 = 9.8 N
a = F/
= 5.8 / 2
= 2.9 m s-2
Resultant force down slope = 9.8 - friction = 9.8 - 4 = 5.8 N
Higher Physics Unit 1
v2 = u2 + 2as
= 0 + 2 × 2.9 × 3
= 17.4
9
Dalziel High School
a=
Fun
m
5.8
a=
2
a = 2.9ms−2
v 2 = u 2 + 2as
v 2 = 0 + 2 × 2.9 × 3
v 2 = 17.4
∴v = 4.2m s−1
Conservation of Energy
The total energy of a closed system must be conserved, although the energy may change its form.
The equations for calculating kinetic energy Ek, gravitational potential energy Ep and work done are given
below.
v = 4.2 m s-1
E p = mgh
Conservation
of Energy
E = 21 mv 2
Thek total
energy of a closed system must be conserved, although the energy may change its
W = Fd
form.
The equations for calculating kinetic energy Ek, gravitational potential energy Ep and
work
done
given
below. in joules J.
Energy
and are
work
are measured
1
Example E = mgh
work done = force × displacement
p
Ek = mv2
2 of 0.3 m. If its speed at the bottom is found to be 2 m s-1,
A trolley is released down a slope from a height
Energy
Find and work are measured in joules J.
a) the energy difference between the Ep at top and Ek at the bottom.
Example
A trolley is released down a slope from a height of 0.3 m. If its speed at the bottom is found
the2work
tob)be
m s-1done
, findby
a) friction
the energy difference between the Ep at top and Ek at the bottom.
b)
the
done by friction
c) the force of friction on work
the trolley
c) the force of friction on the trolley
1 kg
2m
h = 0.3 m
a) Ep at top = mgh = 1 × 9.8 × 0.3 = 2.94 J
1 = 2.94J
a) Ep at top = mgh1 = 1×9.8×0.3
×1×4=2J
Ek at bottom = mv2 =
2
2
2
1
1
Ek
at bottom=
mv == 0.94
× 1×J4 = 2J
Energy
difference
2
2
b) Work done by friction = energy difference (due to heat, sound) = 0.94 J
Energy difference = 0.94 J
c) Work done = Force of friction × d = 0.94 J
d=2m
b) Work done by friction = energy difference (due to heat, sound) = 0.94 J
0.94
= 0.47 N
2
Force of friction = 0.47 N
c) Work done=Force of friction
F ×= d
Power
Power is the rate of transformation of energy from one form to another.
energy
work done
F
=
=
Higher Physics time
Unit 1
time
P =
Power is measured in watts W.
displacement
= F
t
average velocity
10
Dalziel High School
W
d
0.94
F=
2
F = 0.47N
F=
∴ Force of friction = 0.47 N
Power
Power is the rate of transformation of energy from one form to another.
Power is measured in watts W.
P=
Energy Work done
F × displacement
=
=
= F × Average Velocity
Time
Time
time
MOMENTUM AND IMPULSE
The momentum of an object is given by:
Momentum = mass × velocity of the object.
Momentum is measured in kg m s −1
Note: momentum is a vector quantity. The direction of the momentum is the same as that of the velocity.
Conservation of Momentum
When two objects collide it can be shown that momentum is conserved provided there are no external forces applied to the
system.
For any collision:
Elastic and inelastic collisions
An elastic collision is one in which both kinetic energy and momentum are conserved. An inelastic collision
is one in which only momentum is conserved.
Example
a) A car of mass 1200 kg traveling at 10 m s-1 collides with a stationary car of mass 1000 kg. If the cars
lock together find their combined speed.
b) By comparing the kinetic energy before and after the collision, find out if the collision
is elastic or inelastic.
Higher Physics Unit 1
11
Dalziel High School
Draw a simple sketch of the cars before and after the collision.
Before
After
10ms-1
v=0
m=1200kg
m=1000kg
v=?
m=1200+1000kg
a) Total momentum of all objects before = Total momentum of all objects after.
m1u1 + m2u2 = m1v1 + m2 v 2
m1u1 + m2u2 = m1+2 v1+2
m1+2 since masses now stuck and are moving together.
(1200 × 10) + 0 = 2200v1+2
12000 = 2200v1+2
12000
2200
v1+22 = 5.5m s −1
E k = 21 mv
v1+2 =
b) Kinetic energy before
E k = 21 × 1200 × 102
E k = 60000J
E k = 21 mv 2
Kinetic energy after
E k = 21 × 2200 × 5.52
E k = 33275J
Kinetic energy is not the same, so the collision is inelastic
Vector nature of momentum
Remember momentum is a vector quantity, so direction is important. Since the collisions dealt with will act along the same
line, then the directions can be simplified by giving:
momentum to the right a positive sign and momentum to the left a negative sign.
Example
Find the unknown velocity below.
Higher Physics Unit 1
12
Dalziel High School
Before
4ms-1
After
2ms-1
v=?
m=8kg
m=6kg
m=8+6kg
Total momentum of all objects before = Total momentum of all objects after.
m1u1 + m2u2 = m1v1 + m2 v 2
m1u1 + m2u2 = m1+2 v1+2
(8 × 4) + (6 × −2) = 14v1+2
20 = 14v1+2
20
14
= 1.43m s −1
v1+2 =
v1+2
Trolleys will move to the right at a
1.43ms-1
since v is positive.
Explosions
A single stationary object may explode into two parts. The total initial momentum will be zero. Hence
the total final momentum will be zero. Notice that the kinetic energy increases in such a process.
Example
Two trolleys shown below, initially stationary, are exploded apart. Find the unknown velocity.
Before
After
3ms-1
m=2kg
v=?
m=1kg
Total momentum before = Total momentum after
m1u1 + m2u2 = m1v1 + m2 v 2
0 = (2 × (−3)) + (1× v 2 )
0 = −6 + v 2
v 2 = 6m s −1
Higher Physics Unit 1
13
Dalziel High School
Impulse
An object is accelerated by a force F for a time, t. The unbalanced force is given by:
F = ma =
Unbalanced force =
m(v − u) mv − mu
=
t
t
Change in momentum
= Rate of change of momentum
time
Ft=mv-mu
The term Ft is called the impulse and is equal to the change in momentum. Note: the unit of impulse,
Ns will be equivalent to kg m s-1.
The concept of impulse is useful in situations where the force is not constant and acts for a very short period of time. One
example of this is when a golf ball is hit by a club. During contact the
unbalanced force between the club and the ball varies with time as shown
below.
Since F is not constant the impulse (Ft) is equal to the area under the
graph. In any calculation involving impulse the unbalanced force calculated
is always the average force and the maximum force experienced would be
greater than the calculated average value.
Examples
1.In a snooker game, the cue ball, of mass 0.2 kg, is accelerated from the rest to a velocity
of 2 m s-1 by a force from the cue which lasts 50 ms. What size of force is exerted by the cue?
u=0
v = 2 m s-1 t = 50 ms = 0.05 s m = 0.2 kg
F=?
Ft = mv - mu
F × 0.05 = 0.2 × 2 F=8N
2.A tennis ball of mass 100 g, initially at rest, is hit by a racket. The racket is in contact with the ball for 20 ms and the force
of contact varies over this period as shown in the graph.
Determine the speed of the ball as it leaves the racket.
F /N
Impulse = Area under graph
= ½×20×10-3 ×400 = 4Ns
u=0
m = 100 g = 0.1 kg v=?
400
Ft = mv - mu = 0.1v
4 = 0.1 v
v = 40 m s-1
20 Higher Physics Unit 1
t / ms
14
Dalziel High School
3. A tennis ball of mass 0.1 kg travelling horizontally at 10 m s-1 is struck in the opposite direction by a tennis racket. The
tennis ball rebounds horizontally at 15 m s-1 and is in contact with the racket for 50 ms. Calculate the force exerted on the
ball by the racket.
m = 0.1 kg u = 10 m s-1 v = -15 m s-1 (opposite direction to u) t = 50 ms = 0.05 s
Ft = mv - mu
0.05 F = (0.1 × (-15)) - (0.1 × 10) = -1.5 - 1 = -2.5
∴F =
−2.5
= −50N (Negative indicates force in opposite direction to initial velocity)
0.05
Newton’s 3rd Law and Momentum
Newton’s 3rd law states that if an object A exerts a force (ACTION) on object B then object B will exert an
equal and opposite force (REACTION) on object A.
This law can be proved using the conservation of momentum.
Consider a jet engine expelling gases in an aircraft.
Let FA be the force on the aircraft by the gases
and Momentum
and FG beNewton’s
the force3rd
on Law
the gases
by the engine (aircraft).
Newton’s 3rd law states that if an object A exerts a force (ACTION) on object B then object
B will exert an equal and opposite force (REACTION) on object A.
This law can be proved using the conservation of momentum.
Consider a jet engine expelling gases in an aircraft.
Let FA be the force on the aircraft by the gases
and FG be the force on the gases by the engine (aircraft).
FA
FG
FA
FG
Let the positive direction be to the left (direction of Fa)
In a small time, let mG be the mass of the gas expelled and
mA be the mass of the aircraft.
total momentum before = total momentum after
O = mGvG + mA vA
mG vG = - mAvA [vG and vA in opposite directions]
(mG vG - O) = -(mA vA - O)
Change in momentum of gas = - (change in momentum of aircraft)
Changes in momentum of each object are equal in size but opposite in direction.
If forces act in time, t
Force = change in momentum
t
FG = (mG vG - O)
t
But
(mG vG - O)
FA = (mA vA - O)
t
=
(mA vA - O)
(uA = uG = O)
from above
since t is the same for the engine and gas.
FG = - FA
The forces acting are equal in size and opposite in direction.
Higher Physics Unit 1
15
Dalziel High School
DENSITY AND PRESSURE
Density
AND PRESSURE
The mass per unit volume of a substance is called the density, r.
(The symbol, ρ, is the Greek letter rho).
er unit volume of a substance is called the density, r.
ol, , is the Greek letter rho).
= density
per cubic metre, kg mρ = density in kilograms per cubic
metre, in
kg kilograms
m-3
= m
m = mass in kilograms, kg
m =V
mass in kilograms, kg V = volume in cubic metres, m3
V = volume in cubic metres, m3
DENSITY AND PRESSURE
Example
he density of a 10 kg block of DENSITY
carbon measuring
10 cm by 20 cm by 25 cm.
AND PRESSURE
Density
Calculate the density of a 10 kg block of carbon measuring 10 cm by 20 cm by 25 cm.
3
The
mass
per= unit
of a =
substance
ate volume, V,
in m
: V
0.1
×volume
0.2 × 0.25
0.005 m3is called the density, r.
Density
(The symbol, ,The
is the
Greek
letter
rho).of a substance is called the density, r.
mass
per unit
volume
= density in kilograms per cubic metre, kg m=
m
(The10
symbol, , is the Greek letter rho).
m
m = mass in=kilograms,
kg
density in kilograms
per cubic metre, kg mV
0.005=
3
V
3
-3
m
10 cm
V = volume
cubicin metres,
m kg
=
m in
= mass
kilograms,
m
= 2000 kg m
V
V = volume in cubic metres, m3
25 cm
Example
20 cm
Example
Calculate
the density
of a 10 kg block of carbon measuring 10 cm by 20 cm by 25 cm.
f Solids, Liquids
and Gases
Calculate
the density of a 10 kg block of carbon measuring 10 cm by 20 cm by 25 cm.
3 : V = 0.1 × 0.2 × 0.25 = 0.005 m3
calculate
volume,
in mrelative
ble opposite,First,
it can
be seen
thatV,the
Substance
Density
(kg m3-3)
calculate
volume,
V, in volume,
m3 : V V,
= 0.1
0.25
= 0.005
First, calculate
in m×3 :0.2
V =× 0.1
× 0.2
× 0.25m
= 0.005 m3
of the densitiesFirst,
of solids
and liquids
are
Ice
920
the relative magnitude of gases are smaller by
=
?
=
m
10
Water
1000
=?
=
m
10
1000.
m
=
10
kg
V
0.005
m
=
10
kg
V
0.005
Steam
0.9
d melts to a liquid, there is3 little relative 3
10 cm 10 cm
V = 0.005=m2000 kg m-3
=Aluminium
2000 kg m-3
V
=
0.005
m
2700
olume due to expansion. The densities of
solids have similar magnitudes.
Iron
7860
25 cm 25 cm
20 cm
uid evaporates to a gas, there is
a large of
relative
Perspex
1190 20 cm
Densities
Solids, Liquids
and Gases
Densities
of
Solids,
Liquids
and
Gases
olume due toDensities
the expansion
ofFrom
the material.
of Solids,
Liquids
Gases
theand
table
opposite,Ethanol
it can be seen that the relative 791
Substance
Density
(kg m-3)
From
the
table
opposite,
it
can
be
seen
that
the relative
e of a gas is approximately
1000
times
greater
-3
magnitude
thethat
densities
of solids
and liquids
are Substance
From the table opposite,
it can beof
seen
the relative
of the densities
of
Olive
oilmagnitude
915
(kg m ) 920
Ice Density
magnitude
thesimilar
densities
of
solids
and
liquids
ume of the same
mass
ofof
the
liquid
but
the
relative
magnitude
ofare
gases
are smaller
by
solids and
liquids
are solid
similar or
but
the
relative
magnitude
of gases
are smaller
by aIce
Vinegar
1050 Water
920 1000
a factor of
1000.
similar
but the relative
magnitude
of gases are smaller by
substance. factor
of 1000.
Oxygenthere is little relative Water
1.43 Steam
1000 0.9
When
a solidof
melts to a liquid,
factor ofthan
1000.
es of gases areasmaller
the
densities
Nitrogen
1.25 Aluminium
change
volume
due
tois
expansion.
The densities ofSteam
0.9 2700
When
aapproximately
solid
to 1000.
ainthere
liquid,
there
little
iquids by a factor
When
aof
solid
melts melts
to a liquid,
is little
relative
changerelative
in volume due to
liquids
and
solids
have
similar
magnitudes.
Iron
Aluminium
2700 7860
change in
to expansion.
The
densities
of
expansion.
Thevolume
densitiesdue
of liquids
and solids have
similar
magnitudes.
When a liquid evaporates to a gas, there is a large relative
Perspex
1190
liquids
andevaporates
solids
have
similar
7860
herefore, thatWhen
the aspacing
of the
particles
is amagnitudes.
must
be approximately
10
times
liquid
to a in
gas,
there
isgas
a large
relative
changeofinthe
volume
dueIron
to
change
volume
due
to
the
expansion
material.
Ethanol
791
When
a liquid
a large1000
relative
in a liquid or
solid.
Perspex
1190
the
expansion
of theevaporates
material.
The to
volume
of is
athere
gas
isis
approximately
1000
times
The volume
ofaa gas,
gas
approximately
times
greater
Olive oil
915
change
volume
due
themass
expansion
ofmass
theliquid
material.
than
volume
of the
same
of the
solid
orsubstance.
liquid
Ethanol Vinegar
791 1050
greater
thaninthe
volume
of
theto
same
of the
solid
or
form
of the
The volume of aform
gas of
is the
approximately
substance. 1000 times greater
Olive oil Oxygen
915 1.43
Gas
The
densities
of gasesThe
arethe
smaller
than
the
densities
of solids
liquids
by a factor
densities
ofmass
gases
smaller
than
the
densities
of
than
the volume
of
same
ofarethe
solid
orand
liquid
1050 1.25
solids and liquids by a factor of approximately 1000. Vinegar Nitrogen
ofform
approximately
1000.
of the substance.
10d
Oxygen
1.43
The densities of gases are smaller than the densities of
It follows, therefore, that10d
the spacing of the particles is
a gas must be approximately 1.25
10 times
Nitrogen
solids and liquids by a factor of approximately 1000.
d
greater than in a liquid or solid.
10d
It follows, therefore, that the spacing of the particles is a gas must be approximately 10 times greater than in a liquid or solid.
It follows, therefore, that the spacing of the particles is a gas must be approximately 10 times
occupied by each particle
Volume occupied by each particle
Gas
Solid or solid.
greater
than
in
a
liquid
= d3
= (10d) 3 = 1000 d 3
or
liquid
Gas
Solid
d
or
liquid
Higher Physics
Unit
1 Volume occupied by each particle
hanics and Properties
of Matter
(H)
= d3
d
10d
10d
10d
10d
Volume
2 occupied by each particle
10d
= (10d) 3 = 1000 d 3
10d
16
Nitrogen
solids and liquids by a factor of approximately 1000.
1.25
It follows, therefore, that the spacing of the particles is a gas must be approximately 10 times
a liquid or solid.
Dalziel
Highthan
School
greater
in
Gas
Solid
or
liquid
10d
10d
10d
d
Volume occupied by each particle
= d3
Volume occupied by each particle
= (10d) 3 = 1000 d 3
Physics: Mechanics and Properties of Matter (H)
Pressure
2
Pressure on a surface is defined as the force acting normal (perpendicular) to the surface.
p = pressure in pascals, Pa
F = normal force in newtons, N
A = area in square metres, m2
p=
F
A
1 pascal is equivalent to 1 newton per square metre; ie 1 Pa = 1 N m-2.
Example
Calculate the pressure exerted on the ground by a truck of mass 1600 kg if each wheel has an area of 0.02 m2 in contact
with the ground.
Total area A = 4 × 0.02 = 0.08 m2
Normal force F = weight of truck = mg = 1600 × 9.8 = 15680 N
Area = 0.02 m2
p=?
F = 15680 N A = 0.08 m2
Pressure In Fluids
F 15680 p = =
A 0.08
Higher Physics Unit 1
17
A
A = 0.08 m2
0.08
= 196,000 Pa or 196 kPa
Dalziel High School
Pressure
In196
Fluids
= 196,000
Pa or
kPa
Fluid
is a general
term which
and gases.
Anyand
equations
apply
to liquids atthat
rest apply
equally to
apply
to gases
Fluid
is a general
termdescribes
which liquids
describes
liquids
gases.thatAny
equations
liquids
at rest.
at rest
equally apply to gases at rest.
TheThe
pressure
at a point
a fluidin
at arest
of density
depth
h below the
surface,
is given the
by
pressure
at ainpoint
fluid
at restρ,of
density
, depth
h below
p=hρg
p=h
g
surface, is given by
p = pressure in pascals, Pa
h = depth in metres, m
= density of the fluid in kg m-3
g = gravitational field strength in N kg-1
p = pressure in pascals, Pa
h = depth in metres, m
ρ = density of the fluid in kg m-3
g = gravitational field strength in N kg-1
Example
Example
Calculate
the pressure
due to the
water
at a depth
in water.
Calculate
the pressure
due
to the
waterofat15amdepth
of
p=?
h = 15 m
-3
water = 1000 kg m
g = 9.8 N kg-1
15 m in water.
p=h g
= 15 × 1000 × 9.8
= 147000 Pa
surface
15 m
Buoyancy Force (Upthrust)
When a body is immersed in a fluid, it appears to “lose” weight. The body experiences an
force due to being immersed in the fluid. This upwards force is called an upthrust.
Buoyancyupwards
Force (Upthrust)
When a body is immersed in a fluid, it appears to “lose” weight. The body experiences an upwards force due to being
or buoyancy
canupthrust.
be explained
immersedThis
in the upthrust
fluid. This upwards
force is force
called an
in terms of the forces acting on the body
due to the pressure acting on each of the surfaces of the body.
This
upthrustMechanics
or buoyancy
can be of
explained
in terms of the forces
Physics:
andforce
Properties
Matter (H)
3
acting on the
body
due
to
the
pressure
acting
on
each
of
the
surfaces
of
Pressure on the top surface ptop = htop g
the body. Pressure on bottom surface p
h top
p top
bottom = hbottom g
Pressure on the top surface ptop = htopρg
The bottom surface of the body is at a greater depth
than
the surface
top surface,
Pressure on
bottom
pbottom =therefore
hbottomρg the pressure on the
bottom surface is greater than on the top surface.
The bottom
surface
of theinbody
is at
a greater
depth than
the top
This
results
a net
force
upwards
on the
body due
surface, therefore
the pressure
the bottom
surface
is greater
to the liquid.
Thisonupward
force
is called
thethan on
the top surface.
This results in a net force upwards on the body due to
upthrust.
h bottom
p bottom
the liquid. This upward force is called the upthrust.
Notice that the buoyancy force (upthrust) on an object depends on the difference in the pressure on the top and bottom of
Notice that the buoyancy force (upthrust) on an object depends on the difference in the
pressure on the top and bottom of the object. Hence the value of this buoyance force does
not depend on the depth of the object under the surface.
the object. Hence the value of this buoyance force does not depend on the depth of the object under the surface.
GAS LAWS
Kinetic Theory of Gases
The kinetic theory tries to explain the behaviour of gases using a model. The model
considers a gas to be composed of a large number of very small particles which are far apart
and which move randomly at high speeds, colliding elastically with everything they meet.
Higher Physics
Unit 1
Volume
The volume of a gas is taken as the volume of the container. The volume
occupied by the gas particles themselves is considered so small as to be
negligible.
18
Dalziel High School
GAS LAWS
Kinetic Theory of Gases
The kinetic theory tries to explain the behaviour of gases using a model. The model considers a gas to be composed of a
large number of very small particles which are far apart and which move randomly at high speeds, colliding elastically with
everything they meet.
The volume of a gas is taken as the volume of the container. The volume
Volume
occupied by the gas particles themselves is considered so small as to be
negligible.
Temperature
The temperature of a gas depends on the kinetic energy of the gas
particles. The faster the particles move, the greater their kinetic energy
and the higher the temperature.
Pressure
The pressure of a gas is caused by the particles colliding with the walls of
the container. The more frequent these collisions or the more violent these
collisions, the greater will be the pressure.
Relationship
Between
Pressure
and Volume
of a Gas of
Relationship
Between
Pressure
and Volume
a Gas
For a fixed mass of gas at a constant temperature, the pressure of a gas is inversely proportional to its volume.
For a fixed mass of gas at a constant temperature, the pressure of a gas is inversely
proportional to its volume.
p∝ 1
V
Graph
p × V = constant
p
p1 V1 = p2 V2
p
V
0
1
V
0
Example
Example
The pressure
of a gas enclosed in a cylinder by a piston changes from 80 kPa to 200 kPa.
If there is no change in temperature and the initial volume was 25 litres, calculate the new
The pressure of a gas enclosed in a cylinder by a piston changes from 80 kPa to 200 kPa. If there is no change in
volume.
temperature and the initial volume was 25 litres, calculate the new volume.
p11 =80kPa
= 80 kPa
p
p 1 V1 = p2 V2 p1 V1 =p2 V2
V
=
25
litres
80
V2
V11=25litres × 25 = 200 ×
80×25=200×V2
p2 = 200 kPa V2 = 10 litres
p
V2 = 10 litres
V22==200
? kPa V2 = ?
Relationship Between Pressure and Temperature of a Gas
If a graph is drawn of pressure against temperature in degrees celsius for a fixed mass of gas
at a constant volume, the graph is a straight line which does not pass through the origin.
When the graph is extended until the pressure reaches zero, it crosses the temperature axis at
-273 oC. This is true for all gases.
p
Higher Physics Unit 1
T / oC
-273
19
V1 = 25 litres 80 × 25 = 200 × V2
p2 = 200 kPa V2 = 10 litres
V2 =School
?
Dalziel High
Relationship
Betweenand
Pressure
and of
Temperature
Relationship
Between Pressure
Temperature
a Gas
of a Gas
If a graph
temperature
in degrees
celsius for
fixed masscelsius
of gas atfor
a constant
thegas
If isa drawn
graphofispressure
drawnagainst
of pressure
against
temperature
ina degrees
a fixedvolume,
mass of
graph isat
a straight
line
which
does
not
pass
through
the
origin.
When
the
graph
is
extended
until
the
pressure
reaches
a constant volume, the graph is a straight line which does not pass through the origin. zero,
it crosses
the temperature
at -273 ℃. until
This is the
true pressure
for all gases.
When
the graphaxis
is extended
reaches zero, it crosses the temperature axis at
-273 oC. This is true for all gases.
p
T / oC
-273
Kelvin Temperature Scale
-273oC is called absolute zero and is the zero on the kelvin temperature scale. At a
Kelvin Temperature
temperatureScale
of absolute zero, 0 K, all particle motion stops and this is therefore the lowest
-273℃ possible
is called absolute
zero and is the zero on the kelvin temperature scale. At a temperature of absolute zero, 0 K, all
temperature.
particle One
motion
stops andon
this
is therefore
the lowest possible
division
the
kelvin temperature
scaletemperature.
is the same size as one division on the celsius
One division
on the kelvin
temperature
scale is the same
size as one
division
on thein
celsius
temperature
scale, i.e.
temperature
scale,
i.e. temperature
differences
are
the same
kelvin
as in degrees
celsius,
temperature
are theincrease
same in kelvin
as in degrees
celsius,
a temperature increase
10°C
the same as a
e.g. differences
a temperature
of 10°C
is the same
ase.g.
a temperature
increaseof of
10isK.
Noteincrease
the unit
of K.
the kelvin scale is the kelvin, K, not degrees kelvin, °K!
temperature
of 10
Converting Temperatures Between °C and K
Note the unit of the kelvin scale is the kelvin, K, not degrees kelvin, °K!
Converting °C to K
add 273
Converting
Between °C andBetween
K
Converting
Temperatures
°C and K
Converting
KTemperatures
to °C
subtract 273
Converting
°C
to
K
add
273
If the graph of pressure against temperature is drawn using the kelvin temperature scale, zero
K is
to °C
273
Converting
tothe
K kelvin temperature
add 273 scale and the graph now goes through the
onConverting
the graph
thesubtract
zero°C
on
If
the
graph
of
pressure
against
temperature
is
drawn
using273
the kelvin temperature scale, zero on the graph is the zero on the
Converting
K
to
°C
subtract
origin.
p
If the graph
of the
pressure
against
temperature
is drawn using the kelvin temperature scale, zero
kelvin temperature
scale and
graph now
goes through
the origin.
on the graph is the zero on the kelvin temperature scale and the graph now goes through the
Physics:
of Matter (H)
5
origin.Mechanics and Properties
p
T/K
0
For a fixed mass of gas at a constant volume, the pressure of a gas is directly proportional to
its temperature measured in kelvin (K).
T/K
0
For a fixed mass of gas at a constant
volume, the pressure
p
p1 pof2 a gas is directly proportional to
constant
its temperature measured in kelvin (K).
T
T1 T2
p
T
Example
p
p1 it ispheated
Hydrogen in a sealed container at 27 °C has a pressure
of 1.8 × 105 Pa. If
to a
2
constant
For
a
fixed
mass
of
gas
at
a
constant
volume,
the
pressure
of
a
gas
is
directly
proportional
to
its
temperature
measured in
T
temperature of 77 °C, what will be its new pressure?
T1 T2
Example
kelvin (K).
p1 = 1.8 × Hydrogen
105 Pa in a sealed container at 27 °C has a pressure of 1.8 × 105 Pa. If it is heated to a
T1 = 27 °Ctemperature
= 300 K of 77 °C, what will be its new pressure?
p2 Example
=?
5
= 1.8
10
T2Hydrogen
= 77 °Cinp=1a350
K×
p2 =Pa
2.1 × 105 Pa
5
sealed
container
T1 = 27 °C = 300atK27 °C has a pressure of 1.8 × 10 Pa. If it is heated to a temperature of 77 °C, what will be
its new pressure?
p2 = ?
= 77 °C = 350 K p2 = 2.1 × 105 Pa
p1 =1.8×10T5 2Pa
p
T1 =27°C=300K
p2 = ?
T2=77°C=350K
Higher Physics Unit 1
T
p1
T1
=
p2
T2
p
1.8 × 10
= 2
300
350
p2 = 2.1× 105 Pa
5
20
Dalziel High School
Relationship
Between
Volume
and Temperature
Relationship
Between
Volume
and Temperature
of a Gas
of a Gas
If If
a graph
is drawn
of volume
against temperature,
in degrees celsius,
for a fixed
mass of for
gas aatfixed
a constant
a graph
is drawn
of volume
against temperature,
in degrees
celsius,
masspressure,
of gas the
at a isconstant
the graph
isthrough
a straight
line which
does
not
pass through
origin.
graph
a straightpressure,
line which does
not pass
the origin.
When the
graph
is extended
until thethe
volume
reaches zero,
Relationship Between Volume and Temperature of a Gas
When
the graph
is extended
the
reaches
zero, again it crosses the temperature
again
it crosses
the temperature
axis until
at -273
°C.volume
This is true
for all gases.
If
a
graph
is
drawn
of
volume
against
temperature,
in
degrees celsius, for a fixed mass of gas
axis at -273 °C. This is true for all gases.
Relationship
Between
Volume
and
Temperature
Gas through the origin.
at a constant pressure, the graph is a straight line which does of
nota pass
V
When theIfgraph
is extended
until
the volume
zero, again
it crosses
the temperature
a graph
is drawn of
volume
againstreaches
temperature,
in degrees
celsius,
for a fixed mass of gas
axis at -273
This pressure,
is true forthe
all graph
gases.is a straight line which does not pass through the origin.
at a°C.
constant
When the graph is extended until theVvolume reaches zero, again it crosses the temperature
axis at -273 °C. This is true for all gases.
T / oC
V
-273
o C temperature scale, the
If the graph of volume against temperature is drawn using the
T /kelvin
graph now goes through the origin.
-273
If the graph of volume against temperature is drawn using the kelvin temperature scale, the graph
T / o C now goes through the
If the graph of volume againstVtemperature
is
drawn
using
the
kelvin
temperature scale, the
origin.
-273
graph now goes through the origin.
If the graph of volume against temperature is drawn using the kelvin temperature scale, the
V the origin.
graph now goes through
0
V
T/K
For a fixed mass of gas at a constant pressure, the volume of a gas is directly proportional to
T/K
its temperature measured in kelvin0 (K).
For a fixed mass of gas at a constant
pressure, the volume
T / K proportional to
V
V1 ofVa2 gas is directly
0
V
T
=
constant
=
its temperature measured in kelvin (K).
Tthe
Tproportional
T at a the
1
2
For a fixed mass
at amass
constant
pressure,
volume pressure,
of a gas is directly
its is
temperature
in
Forofa gas
fixed
of gas
constant
volume
of a togas
directly measured
proportional
to
its
temperature
measured
in
kelvin
(K).
kelvin (K).
V
V1
V
= constant
= 2
Example V T
T1
T2
T
400 cm3 of air is at a temperature of 20 V
°C. At what temperature
the
volume be 500 cm3
V1will V
2
V∝ T
= constant
=
if the air pressure does
not change?
T
T2
T
1
Example
3 is at a temperature of 20 °C. At what temperature will the volume be 500 cm3
cm3 of air
V400
1 = 400 cm
500
400
V
V1
Example
Example
if1 the
air°C
pressure
does not change?
=
= 2
=
20
=
293
K
T
3
3
400 cm of air
is at
a temperature of 20 °C. At what temperature
will At
the
volume
be 500 cm
pressure
doesbe
not500 cm3
T23 if the
293
T2what
400
°C.
temperature
willairthe
volume
3 cm of air is at a temperature of 20 T
1
V
=
500
cm
2
change?
3 air pressure does not change?
the
= ?400ifcm
500
400 to temperature
V back
TV21=
T2 = 366 K = 93 °CV1(convert
= the question)
= 2 scale in
T1 = 20 °C = 293 K 3
T2
T2 293
V1 VT
1 =3 400 cm
= 21
500
V2 400
V1
V2 = 500Vcm
T
T
=
=
1
2
=
20
°C
=
293
K
T
3
1
V1T=2 400
= ?cm Gas Equation
T2 = 366 K = 93 °C (convertT1backTto
temperature
T2
293
Combined
2
400 500
V2 = 500 cm3
T1 = 20°C = 293K
=
scale in the question)
293 K T=
T2 = ?
T2 = 366
2 93 °C (convert back to temperature
3
combining
the above three relationships, Tthe
following
relationship
for the
pressure,
=
366K
V2By
= 500
cm
scale
in the
question)
2
Combined Gas Equation
volume
and
temperature
of
a
fixed
mass
of
gas
is
true
for
all
gases.
T2 = ?
Combined Gas Equation
three
following
the pressure,
By combining
relationships,
T2the
= 366
K = 93 °Crelationship
(convert back for
to temperature
scale in the
p theVabove
p
V
p
V
1
1
2
2
of a fixed mass
of gas is true
for=all
volume
and temperature
= constant
gases.
question)
By combining
the above three relationships,
relationship for the pressure,
T1 the following
T2
1
volume and temperature of a fixed mass of gas is true for all gases.
p
V
p 1 V1
p 2 V2
= constant
=
Combined Gas Equation
T1
T21
1
p
V
p1 V
p 2 V2
= constant
By combining the above three relationships,
the following relationship for the pressure,
= volume and temperature of a fixed
T1
T2
mass of gas is true for all gases. 1
Physics: Mechanics and Properties of Matter (H)
7
Physics: Mechanics and Properties of Matter (H)
7
Higher Physics Unit 1
21
Dalziel High School
pV
pV p V
= Constant ∴ 1 1 = 2 2
T
T1
T2
Example
A balloon contains 1.5 m3 of helium at a pressure of 100 kPa and at a temperature of 27 °C. If the pressure is increased to
250 kPa at a temperature of 127 °C, calculate the new volume of the balloon.
p1 = 100 kPa
V1 = 1.5 m3
T1 = 27°C = 300K
p2 = 250 kPa
V2 = ?
p1V1
T1
=
p2 V2
T2
100 × 1.5 250 × V2
=
300
400
V2 = 0.8m3
T2 = 127°C =400K
Gas Laws and the Kinetic Theory of Gases
Pressure - Volume (constant mass and temperature)
Consider a volume V of gas at a pressure p. If the volume of the container is reduced without a change in temperature, the
particles of the gas will hit the walls of the container more often (but not any harder as their average kinetic energy has not
changed). This will produce a larger force on the container walls. The area of the container walls will also reduce with
reduced volume.
As volume decreases, then the force increases and area decreases resulting in, from the definition of pressure, an increase in
pressure,
i.e. volume decreases hence pressure increases and vice versa.
Pressure - Temperature (constant mass and volume)
Consider a gas at a pressure p and temperature T. If the temperature of the gas is increased, the kinetic energy and hence
speed of the particles of the gas increases. The particles collide with the container walls more violently and more often. This
will produce a larger force on the container walls.
As temperature increases, then the force increases resulting in, from the definition of pressure, an increase in pressure,
i.e. temperature increases hence pressure increases and vice versa.
Volume - Temperature (constant mass and pressure)
Consider a volume V of gas at a temperature T. If the temperature of the gas is increased, the kinetic energy and hence
speed of the particles of the gas increases. If the volume was to remain constant, an increase in pressure would result as
explained above. If the pressure is to remain constant, then the volume of the gas must increase to increase the area of the
container walls that the increased force is acting on, i.e. volume decreases hence pressure increases and vice versa.
Higher Physics Unit 1
22