CIE A2 Physics (9702)
Key:
Blue → These are derivations that might get too mathematical for some. It is for a more comprehensive
understanding. Those who aren’t really interested in knowing more can ignore.
Red → Very very important stuff from past years, they’re basically marking schemes.
1
Physics
Contents
1 Circular Motion
3
2 Gravitational Field
6
3 Oscillation
13
4 Ideal Gases
16
5 Temperature
19
6 Thermal Properties
21
7 Communication
23
8 Electric Fields
33
9 Capacitance
38
10 Electronics
47
11 Magnetic Fields and Electromagnetism
53
12 Electromagnetic Induction
59
13 Alternating Current
62
14 Nuclear Physics
68
15 Quantum Physics
71
16 Medical Physics
78
2
1 Circular Motion
Concepts and Formulae Derivation
Centripetal force is the net force acted on an object towards its centre of rotation. The object is accelerating
due to this net force, but the speed of the object remains constant, the acceleration caused by the centripetal force
only affects the instantaneous direction of motion. We define the velocity of covering a certain angle by angular
velocity, ω.
∆θ
ω=
∆t
To calculate centripetal force, we can use Newton’s 2nd law, F⃗ = m⃗a, the acceleration towards the centre, ⃗a, is
given by
⃗v 2
⃗a =
= rω 2
r
It can be derived as follow
Derivation of centripetal acceleration.
(Source: Physics Wiki)
∆⃗v = ⃗v2 − ⃗v1
When θ → 0, ∆⃗v → 0 which means
⃗v1 ≈ ⃗v2 = ⃗v =⇒ ∆⃗v ≈ ⃗v θ
This yields the following relation
∆⃗v
⌢
=
AB
⃗v θ
rθ
=
⃗v
r
⌢
Since AB= ⃗v ∆t, we can deduce the following equation
∆⃗v
⃗v ∆t
=
⃗v
r
∆⃗v
=⇒
∆t
=
⃗v 2
r
= ⃗a
We can also multiply the angular velocity, ω, by the velocity, ⃗v , to get the acceleration, ⃗a, this is because the change
in angle causes the direction of velocity to change (the magnitude isn’t changing, ∆⃗v here just means the direction
of velocity is changing).
⃗v ∆θ ∆⃗v
ω⃗v =
=
= ⃗a
∆t
∆t
Therefore with acceleration figured out, we can find the centripetal force F⃗c using Newton’s 2nd law
F⃗c = m⃗a = m
3
⃗v 2
r
= mrω 2
Horizontal Rotation
T cos θ = mg
T sin θ = Fc = m
⃗v 2
r
= mrω 2
Vertical Rotation
If the speed of the objects moving in the circle path is constant, then
T1 = T2 − mg = T3 − mg cos θ = T4 + mg = T5 + mg cos θ = m
4
⃗v 2
r
Vehicle on a Banked Road
N = mg cos α
F⃗c = N sin α = m
⃗v 2
r
Since weight of the car is constant, therefore Fc , is constant, as a result if the velocity of the vehicle, ⃗v , is
increased, r, will so increase which implies that the car will move up the slope to increase the radius of circular
rotation so that the constant is not changed.
5
2 Gravitational Field
Concepts
In A-level syllabus, direction of forces in this chapter are usually disregarded, but in the notes here, directions
will be taken into account to provide better understanding of the mathematics behind it. For convention, all the
displacement is measured relative to the body that is creating the gravitational field that we are considering.
If an object has mass, M , it will create a gravitational field around it which expands to infinity. The strength
of the gravitational field generate by mass M , is denoted by ⃗g , which can be defined by
⃗g = −
GM
⃗r 2
where, G, is universal gravitational constant (≈ 6.67 × 10−11 m3 kg −1 s−2 ), and ⃗r is the displacement between the
centre of mass of object (provided that the it is a body with uniform density) that generates the gravitational
field and the point of measuring its strength. To understand why negative sign is present, we first have to define
where the positive direction of the displacement will be, for simplicity’s sake, we define the positive direction of
displacement to be anywhere away from the centre of mass of the body creating the gravitational field. Since the
direction gravitational field is towards the centre of body creating it, therefore the direction force is in opposite
direction to the direction of positive displacement, hence negative sign is needed to indicate the direction.
When two objects are placed into each others’ gravitational fields, both of them will experience a pulling force
from each other
蓓
⃗ then the strength of gravitational
Let the displacement between m1 and m2 from both reference point be R,
Gm2
field experienced at m1 due to m2 will be −
and the strength of gravitational field experienced at m2 due to
⃗ 2
R
Gm1
Gm2
Gm1
m1 will be −
, although −
< −
but since m1 > m2 , therefore the two forces experienced by both
⃗ 2
⃗ 2
⃗ 2
R
R
R
are in fact the same in magnitude.
|F⃗1 | = F1 =
Gm2
Gm1
m1 =
m = F2 = |F⃗2 |
2
⃗
⃗ 2 2
R
R
This is also known as Newton’s Law of Universal Gravitation.
Note that gravitational field strength is a vector, hence that means in a two body system they will be a point
between the two masses where the gravitational vector field add up to zero. But this does not implies that the point
in that particular space has no gravitational potential, this is because gravitational potential is a scalar quantity
hence they do not cancel out. Another definition of the gravitational field strength will also be derived in the next
section when we talk about gravitational potential.
In A-level syllabus we will only be concerning with the gravitational force and gravitational field strength inside
a perfectly spherical and uniform density object, usually a planet. The derivation of the following equation
6
is beyond A-level syllabus. During this derivation, we will avoid considering force as vectors to avoid vector
calculus, hence we will consider deriving using magnitude. We first will derive the magnitude gravitational force
inside a spherical planet then using it to derive the gravitational field strength inside the spherical planet. In order
to derive the equation for the gravitational force, we would first need to enter wacko mode. Let’s assume an object
P of mass m is inside a spherical planet with mass Mp and radius R, let r be the distance between the object and
the centre of mass of the planet. Let’s also only consider a thin slice of surface shell of mass dMs , and the width of
this surface shell will be Rdθ for relatively small dθ
Let’s say that this slice on the surface shell exerts a small force of dF on the object P , to figure what dF is we
first consider another small section within the slice with mass d2 Ms (section in red), labelling the force exerted on
P by this small section as d2 F , we can calculate the force exerted using gravitational force formula
Gm 2
d Ms
s2
d2 F =
if we sum all the d2 Ms along the slice we get the total mass of the slice which is dMs and the total force dF , in
other words when we integrate over this slice we get the force exerted by in
Z
d(dF ) =
Gm
Z
s2
d(dMs )
shell
dF =
Gm
dMs
s2
However, since there is partial cancellation due to the vector nature of the force in conjunction with the circular
band’s symmetry, the leftover component is
dFr =
Gm
s2
cos(φ)dMs
where dFr is the resultant small force exerted by the slice
To obtain the total force exerted by the whole outer spherical shell we would need to integrate again
Z
cos(φ)
dMs
s2
Fr = Gm
7
To evaluate this integral, we first need to do some variable changes. We first notice that the area of the outer
spherical shell is 4πR2 and the area of the thin slices (section in blue) is 2πR sin(θ)Rdθ = 2πR2 sin(θ)dθ this implies
that the mass of the thin shell dMs is given by
dMs =
1
2πR2 sin(θ)
Ms dθ = Ms sin(θ)dθ
2
4πR
2
where Ms is the mass of the outer spherical shell. This gives
Z
Fr = Gm
GmMs
cos(φ)
dMs =
2
s
2
Z
sin(θ) cos(φ)
dθ
s2
Next, we can see that by using law of cosine we get the following equations
cos(φ) =
r2 + s2 − R2
2rs
r + R2 − s2
2
cos(θ) =
2Rr
Differentiating the second equation implicitly we get (r, R are constants)
r2 + R2 − s2
d(cos(θ)) = d
!
2Rr
− 2s
ds
2Rr
s
ds
sin(θ)dθ =
Rr
− sin(θ)dθ =
Using the results obtain we can now continue with the derivation
Fr =
=
=
=
Z
GmMs
2
sin(θ) cos(φ)
dθ
s2
Z
GmMs
sin(θ)dθ cos(φ)
2
GmMs 1
2
GmMs
Rr
Z
2Rr
Z
s2
s cos(φ)
ds
s2
cos(φ)
ds
s
Inserting the expression of cos(φ) derived from law of cosine we get
Fr =
=
GmMs
2Rr
GmMs
Z
r 2 +s2 −R2
2rs
s
Z
1+
4Rr2
8
ds
r 2 − R2
s2
!
ds
To determine the bound of the integration, we may look at the original image, it’s clear to see that to find the
total force exerted by the outer spherical shell, our value s ranges from its minimum value R − r which is at the left
hand side of the planet, to its maximum value, R + r which is on the right hand side of the planet, integrating over
this 2 bounds is the same as summing all the force exerted by the thin slices of shell to give the total force exerted
by the outer spherical shell
Fr =
GmMs
R+r
Z
1+
4Rr2
R−r
=
GmMs
R+r
Z
4Rr2
d
ds
r 2 − R2
s2
"
s−
=
4Rr2
s−
r 2 − R2
s
ds
r 2 − R2
R−r
GmMs
!
#
s
!
ds
R+r
R−r
=0
Now, we have proven that the outer shell exerts no force on the object P , we can also proof that for the shell that
has radius R − δR by using the same method as above for all the infinitely thin slices of spherical shell enveloping
the inner crust until we reach the inner sphere of radius r.
The magnitude of gravitational force experienced by P is then just
GMp′ m
FG =
r2
where Mp′ is the mass enclosed in the sphere with radius r, since one of our assumptions is that the planet has
uniform density, this allows us to express the mass enclosed in the sphere, Mp′ in terms of r
Mp′ =
4πr3
9
3
ρ
where ρ is the density of the planet. Substituting this into FG yields
GMp′ m
FG =
r2
=
4Gmπr3
=⇒ F⃗G = −
=
r2
4Gmπρ⃗r
4Gmπρr
3
3
This means that inside the planet, the gravitational force, FG is directly proportional to displacement between the
object and the centre of mass of planet, ⃗r.
To obtain gravitational field strength inside planet we can divide the gravitational force experienced an object
by its mass. Therefore the gravitational field strength in a planet’s crust is given by
⃗g =
F⃗G
m
=−
4Gmπρ⃗r
3m
=−
4Gπρ⃗r
3
which is also directly proportional to the separation between the object and the centre of mass of the planet.
The graph for gravitational force FG and gravitational field strength g will graph of this general shape
where the green region representing the gravitational field strength or gravitational force in the planet or spherical
object and blue region is representing position outside the planet/ spherical object.
Gravitational Potential and Gravitational Potential Energy
GM
An object of mass M will generate a gravitational field with the magnitude strength equals to 2 at R distance
R
away from the mass M as discussed from previous section. Assume another object is at infinity which is out of
the influence of the gravity produced by mass M , then it is said to have zero gravitational potential energy. Now
let’s say the object is brought into the gravitational field and is d away from mass M from infinity, at that exact
position in space, it is said to have gravitational potential of ϕ, given by
ϕ=−
GM
d
The negative sign is due to the relative position from infinity(which is said to have zero gravitational potential
energy). Let the mass of the object be m, then the gravitational potential energy, GP E, of the object at this
10