Formula Sheet for Stage 6 Physics
Preliminary Course
v  f
H.S.C. Course - Core
Ep = - G

1 
I  
 d 

E
R
F  mg
v x  ux
2
2
2
y
V
I
3
2
L  L 1
v
a 
av
r
t
V
t
vu
t
 F  ma
1
Ek  mv 2
2
p  mv 

Impulse  Ft
mm
F  G 12 2
d
r
GM

T
4
0
2
v
c
2
2
m0
m
v2
1 2
c
t
t 
v
1
c
Vout
Vin
0
v
2
A0 
2
F = BI
V0
V+  V- 
Constants
F
II
k 1 2
l
d
Speed of light
in vacuum
c
3 X 108 ms-1
  Fd
Acceleration
due to gravity
g
9.81 ms-2
  nBIAcos 
Universal Gravitation G
6.67 X 10-11 Nm2 kg-2
Electric constant
k
9 X 109 NC-1m-1
Magnetic constant
k
2 X 10-7 Ns2C-2
Planck’s Constant
h
6.626 X 10-34 Js
Ryberg’s Constant
R
1.10 X 107 m-1
E  hf
mass of proton
mp
1.67 X 10-27 kg
c  f
mass of neutron
mn
1.68 X 10-27 kg
mass of electron
me
9.11 X 10-31 kg
v
n

v n
p
p
s
s
F  qvB sin 
3
2
4 2 r 3
GT
 1
1
1 
 R  2  2 

ni 
n f
h

mv
m1  m2 
r
GM

T
4
2
d
)
10
IA
 100(M B  M A ) / 5
IB
1
y  uy t  ay t 2
2
P=VI
av
M  m  5 log(
v  u  2ayy
2
y
Energy = VIt
a 
Ir Z2  Z1 

2
Io Z2  Z1 
x  u xt
F
q
v av  
Z = 
m1 m2
r
2
2
v1 sin i


v 2 sin r
H.S.C. Course - Options
E
V
d
charge on an electron e
1.602 X 10-19 C
How to Use the Formulas for Stage 6 Physics
Preliminary Course
Formula
v  f
Name
Wave
Equation
8.2.1
Comments
v= velocity (m/s)
f = frequency (hz)
 = wavelength (m)

1 
I  
 d 
Intensity Law
8.2.3
I = intensity (no units)
d = distance (m)
v
v
Snell’s Law
8.2.4
v1 = speed in first medium
v2 = speed in second medium
sin i = angle in first medium
sin r = angle in second medium
 Angles are always
measured to the normal
E = Electric field (N/C, V/m)
F = Force (N)
q = charge (C)
2
1

2
sin i
( 1 n 2 )
sin r

 F
E
q

Electric Field
8.3.2
Ohm’s Law
8.3.2
R = resistance ()
V = voltage (V)
I = current (A)
Energy = VIt
Electrical
Energy
8.3.4
Energy (J)
V = voltage (V)
I = current (A)
T = time (s)
P=VI
Electrical
Power
8.3.4
P = power (W)
V = voltage (V)
I = current (A)
R
V
I
Typical Problem
Calculate the wavelength
of a water wave travelling
at 3 m/s whose frequency
is 6 Hz.
A globe is viewed from 2
metres at a certain
brightness. How bright
does it appear 6 metres
away?
A light ray travelling at 3
X 108 m/s in air enters a
pool of water at 45 to the
normal. If it slows down
to 2.4 X 108 m/s, what is
its angle in the water?
What is the force exerted
on a 2 X 10-6 C charge
moving in an electric field
of size 5 X 103 V/m?
What is the current
through a 6 resistor
when a voltage of 18 V is
applied across it?
How much energy is
delivered to an electric
kettle if a current of 2A is
used for 3 minutes
plugged into the mains
(240V)?
What current is required
by a 2400W electric heater
plugged into the mains
(240V)?
Typical Answer
v 3
v  f       0. 5m
f 6


 I  1   d  3d , 1  1 thus it appears 9 times less
2

9
d2
d 

bright.
v
v
1

2
v sin i
sin i

sin r  2
sin r
v1
2.4  10 8 sin 45 

 38 
8
3  10

 F
E   F  qE  2  10 6 X 5  10 3  0.01N
q
R
V
V 18
 I    3A
I
R 6
Energy = VIt = 240 X 2 X 3 X 60 (sec)
= 86400 = 86.4 kJ
P  VI  I 
P 2400

 10 A
V
240
Formula
r
v av  
t
Name
Average
Velocity
8.4.1


V
a av 
t

Average
Acceleration
8.4.2
 F  ma

Newton’s
Second Law
8.4.2
1
Ek  mv 2
2
Kinetic
Energy
8.4.3
Momentum
8.4.4
p  mv
Impulse  Ft 
FG
m1m 2
d2
r
GM

T
4
3
2
2
Impulse
8.4.4
Universal
Gravitation
8.5.4
Comments
vav = average velocity (m/s)
r = distance covered (m)
t = time (s)

aav = average acceleration (ms-2)
v = change in velocity (m/s)
t = change in time (s)

 F = sum of all forces (N)
m = mass (kg)

a = acceleration (ms-2)
Ek = kinetic energy (J)
m= mass (kg)
v = speed (m/s)
P = momentum (Ns, kgm/s)
m = mass (kg)
v = velocity (m/s)
Impulse = change in momentum
(Ns, kgm/s)
F = force (N)
G = universal gravitation
constant (6.67 X 10-11 Nm2 kg-2)
m1 = mass of body 1 (kg)
m2 = mass of body 2 (kg)
d = separation between the two
bodies (m)
Kepler’s Third r = radius of motion (m)
Law
T = period of motion (s)
8.5.4
G = universal gravitation
constant (6.67 X 10-11 Nm2 kg-2)
M = mass of system (kg)
Typical Problem
How much distance is
covered by a car travelling
for 4 mins at an average
speed of 16 m/s?
A train accelerates from
12 m/s to 18 m/s in 12 s.
What is the value of this
acceleration?
Typical Answer
r
v av    r  v av  t  16  4  60  384m
t
A ball possesses 10 J of
energy. What is its mass if
it is moving at 2 m/s?
What is the momentum of
1400 kg car moving at 6
m/s?
A ball of mass 0.5 kg
travelling at 3 m/s hits
wall and bounces back at
the same speed. If it is in
contact with the wall for
0.1 s, what is the force
exerted by the wall?
What is the gravitational
force between the Earth
(m=6X1024 kg) and 2 kg
ball given that the radius
of the Earth is 5.8X106 m
Ek 
What is the period of a
1012 kg comet that orbits at
5 X 108 m?
r 3 GM

T 
T 2 4 2


V 18  12
a av 

 0.5ms  2
t
12
2E
1 2
2  10
mv  m  2 k  2  5kg
2
v
2


p  mv  1400  6  8400kgms1
Impulse  p f  p i  m(v f  v i )  0.5(2  (2))

Impulse
2
Impulse  Ft  F 

 20 N
t
0.1
F G
m1 m 2
d2

6.67 10 11 X 2 X 6 X 10 24
 19.6 N
(5.8 10 6 m) 2
4 2 r 3

GM
4 2 X 5 X 10 8
 17203s
6.67 X 10 11 X 1012
How to Use the Formulas for Stage 6 Physics
HSC Course (Core)
Formula
Name
Gravitational
Potential
Energy
9.2.1
mm
Ep = - G 1 2
r
F  mg
v x  ux
2
2
v  u  2ayy
2
y
2
y
x  u xt
1
y  uy t  ay t 2
2
r
GM

T
4
3
2
2
L  L 1
v
t 
v
0
t
0
v
1
c
2
2
v
c
2
2
Comments
Ep = Potential energy
G = universal gravitation constant
(6.67 X 10-11 Nm2 kg-2)
m1 = mass of body 1 (kg)
m2 = mass of body 2 (kg)
r = separation between the two
bodies from infinity to r (m)
Gravitational
F = force (N)
Force
M = mass (kg)
9.2.1
g = gravitational constant at the
surface of the Earth 9.81 ms-2
Newtons’ Laws ux = initial speed in x direction (m/s)
of Motion
vx = final speed in x direction (m/s)
9.2.2
uy = initial speed in y direction (m/s)
vy = final speed in y direction (m/s)
a = constant acceleration (ms-2)
x, y = change in displacement (m)
t = time (s)
Kepler’s third
r = radius of motion (m)
law
T = period of motion (s)
9.2.2
G = universal gravitation constant
(6.67 X 10-11 Nm2 kg-2)
M = mass of system (kg)
Relativistic
Lv = apparent length (m)
Length
Lo = “rest” length (m)
Contraction
v = relative velocity (m/s)
9.2.4
c = speed of light (3 X 108 m/s)
Relativistic time tv = apparent time (s)
dilation
to = “rest” time (s)
9.2.4
v = relative velocity (m/s)
c = speed of light (3 X 108 m/s)
Typical Problem
What is the
gravitational potential
energy between two
100 kg masses
through a distance of
2000 m?
What is the weight of
a 100 kg person?
What is the maximum
height of a projectile
launched at 45 to the
horizontal at 50 m/s?
Typical Answer
mm
6.67 X 10 11 X 100 X 100
Ep  G 1 2 
 3.3 X 10 12 J
r
2000


F  mg  100 X 9.81  98.1N
Let up be positive.
At max. height, vy=0, thus
0  u v2  2a g yy  u 2 sin 2 45   2 X 9.81yy
y 
What is the period of
a 1012 kg comet that
orbits at 5 X 108 m?
What is the apparent
length of a spaceship
of rest length 150m
travelling at 0.9c?
How much slower
does an astronaut
travelling at 0.9c
appear to an observer
“at rest”
u 2 sin 2 45  50 2 X sin 2 45 


2 X 9.81
19.6
r 3 GM

T 
T 2 4 2
Lv  L0
tv 
4 2 r 3

GM
4 2 X 5 X 10 8
 17203s
6.67 X 10 11 X 1012
(0.9c) 2
v2
1  2  150 1 
 65.4m
c
c2
t0
2
v
1 2
c

1
(0.9c)
1
c2
2

1
0.19
 2.29 times slower
Formula
F = BI
F
II
k 1 2
l
d
Name
Magnetic force
on a currentcarrying wire of
length l in a
magnetic field
9.3.1
Force per unit
length
9.3.1
  Fd
torque
9.3.1
  nBIAcos
torque on a coil
immersed in a
magnetic field
9.3.1
v
n

v n
Transformer
equation
9.3.4
p
p
s
s
Comments
F = force (N)
B= Magnetic Field (T)
I = current (A)
l = length (m)
Typical Problem
Calculate the force on
2m of wire carrying a
current of 4A in a
magnetic field of
0.1T.
Typical Answer
F  BI  0.1X 4 X 2  0.8N
F = force (N)
l = length (m) per unit
I1, I2 = two currents
parallel=repulsive,
antiparallel=attractive
d = separation of the two currents
(m)
k=magnetic constant (2 X 10-7 NC1 -1
m )
 =torque (Nm)
F =force (N)
d=distance (m)
What is the force per
unit length on two
wires, both carrying
10A, separated by a
distance of 3m?
I I
F
10 X 10
 k 1 2  2 X 10 7 X
 6.67 X 10 6 N / m
l
d
3
What is the torque on
a nut when a 0.6 m
spanner has a force of
80 N applied on it?
What is the torque on
a 0.20 m2 coil of 200
turns immersed in a
magnetic field of 0.2
T carrying a current
of 3 A?
  Fd  80 X 0.6  48Nm
 =torque (Nm)
n =number of turns of coil
B=magnetic field (T)
I = current (A)
A = area of coil immersed in
magnetic field (m2)
cos =angle between the coil and
the magnetic field
Vp = primary voltage (V)
Vs = secondary voltage (V)
Np = number of turns in the primary
coil
Ns = number of turns in the
secondary coil
A transformer is
required to step down
mains voltage (240V)
to 12 V. If the
primary coil has 960
turns, how many turns
are required in the
secondary coil?

  nBIAcos  200 X 0.2 X 3X 0.20  24Nm
vp
vs

np
ns
 n ss 
n p vs
vp

960 X 12
 48 turns.
240
Formula
F  qvB sin 
E
V
d
Name
Magnetic force
on a charge in a
magnetic field
9.4.1
Electric Field
9.4.1
E  hf
Energy of a
Photon
9.4.2
c  f
Wave Equation
9.4.2
Comments
F = force (N)
q = charge (C)
v = velocity (m/s)
B = magnetic field (T)
Sin  = angle between the velocity
and the magnetic field
E = Electric Field (N/C, V/m)
V = Voltage (V)
d = distance (m)
E = Energy (J)
h = Planck’s constant 6.626 X 10-34
Js
f = frequency (Hz)
c = speed of light 3 X 108 m/s
f = frequency (Hz)
 = wavelength (m)
Typical Problem
What is the force on
an electron travelling
at 105 m/s in a
magnetic field of 3 T
at an angle of 30 to
the field?
What is the electric
field between the
prongs of a mains
outlet (240V) if its
separation is 18 mm?
What is the energy of
a photon of yellow
light (f=5.1X1015 Hz)
?
What is the frequency
of yellow light given
that its wavelength is
590 nm?
Typical Answer
F  qvB sin   1.6 X 10 19 X 10 5 X 3 X sin 30   2.4 X 10 14 N
E
V
240

 1.33 X 10 5 V / m
3
d 18 X 10
E  hf  6.67 X 10 34 X 5.1X 1015  3.2 X 10 18 J
c  f  f 
v


3 X 10 8
 5.1X 1015 Hz
590 X 10 9
HSC Course (Options)
Formula
Z = 
Name
Acoustic
Impedance
9.6.1
I r Z 2  Z 1 

(  R)
I o Z 2  Z 1 2
2
M  m  5 log(
d
)
10
IA
 100(M B  M A ) / 5
IB
m1  m2 
4 2 r 3
GT
 1
1 
 R  2  2 

ni 
n f
1
Comments
Z = Acoustic impedance (kgm2s-1
=Rayls)
 = acoustic density (kgm)
 = speed of sound in medium
(m/s)
Reflection
Ii=initial intensity
Intensity
Io= output intensity
9.6.1
Z2 = acoustic impedance
(medium 1)
Z1 = acoustic impedance
(medium 2)
Astronomical
M = absolute magnitude
Distance
m = relative magnitude
9.7.4
d = distance in parsecs
5  5 100  5.02
Ratio of intensity IA,B = intensity of objects A and B
9.7.4
MA,B = absolute magnitude of A
&B
5  5 100  5.02
Kepler’s Third
r = radius of motion (m)
Law
T = period of motion (s)
9.7.5
G = universal gravitation constant
(6.67 X 10-11 Nm2 kg-2)
M1+M2 = total mass of system
(kg)
Ryberg’s
equation
9.8.1
 = wavelength (m)
ni,f = quantum states (shells)
R = Ryberg’s constant 1.10 X 107
m-1
Typical Problem
What is the acoustic
impedance of vaseline given
that its acoustic density is
0.003 kgm and the speed of
sound in vaseline is 650 m/s?
What is the reflection intensity
at the interface of air (acoustic
impedance = 0.5 kgm2s-1) and
vaseline (Z=1.95 kgm2s-1)?
How far away is a star that
appears m=4.5 on Earth while
its absolute magnitude is –3.4?
How much dimmer is Sirius A
(M=-4.3) compared to Echelon
(M=-2.1)?
Typical Answer
Z =   0.003 X 650  1.95kgm2 s 1
I r Z 2  Z 1 
(1.95  0.5) 2


 0.35
I o Z 2  Z 1 2 (1.95  0.5) 2
2
M  m  5 log(
1
1
d
)  d  10
10
mM
5.02
4.5  ( 3.4 )
5.02
d  10
 143 par sec s
IA
 100(M B  M A ) / 5
IB
Sirius A and B orbit each other T = 61 X 365.25 X 24 X 60 X 60
every 61 years. What is the
= 1925013600 s
radius of this orbit if Sirius A
GT (m1  m 2
4 2 r 3
27
m1  m 2 
r  3
has a mass of 10 kg and
GT
4 2
Sirius B is 1029 kg?
6.67 X 10 11 X 1925013600(10 27  10 29 )
r 3
4 2
r = 6.9 X 109 m
What wavelength of light is
 1
1
1 
1 1
 R 2  2   1.1X 107 X ( 2  2 )
produced in hydrogen by a
n
4 1
ni 
transition between the first and 
 f
fourth orbitals (shells)?
1
 
 9.7 X 10 8 m
1
1
1.1X 10 7 ( 2  2 )
4 1
Formula
h

mv
Name
Wavelength of a
particle
9.8.2
Vout
Vin
Amplifier Gain
9.9.6
A0 
V0
V+  V- 
Open Gain Loop
9.9.6
Comments
 = wavelength (m)
h = Planck’s constant 6.626 X 1034
Js
m = mass of particle (kg)
v = speed of particle (m/s)
Vout = output voltage (V)
Vin = input voltage (V)
Ao = Amplifier gain
Vo = output voltage (V)
V+ = positive input voltage (V)
V- = negative input voltage (V)
Typical Problem
What is the wavelength
associated with an elephant
(m=1400 kg) moving at 4 m/s?
Typical Answer
h
6.626 X 10 34


 1.2 X 10 37 m
mv
1400 X 4
A BC547 transistor has a gain
of 120. What is the output
voltage if the input voltage is
60 mV?
A 741 op-amp has an output
voltage of 12 V when its V+ is
2.3 V and its V- is 1.2 V.
What is its gain in this
configuration?
hFE 
Vo
Vo  Vi hFE  120 X 60 X 10 3  0.72V
Vi
A0 
V0
12

 10.9
V+  V-  (2.3  1.2)