Step 2 : Decide which equations to use in the problem
We know the collision is inelastic and there was a denite change in shape of the
objects involved in the collision - there were two objects to start and after the collision
there was one big mass of me
steam to drive a steam generator as discussed above.
Hydroelectric power Water from a river is diverted to turn a water turbine to create electricity
similar to the principles of steam generation. The water is returned to the river after driving the
turbi
Blades
Nacelle -Vessel
containing all
moving parts
Detail of Nacelle Vessel (top view)
to blades
Shaft 1 (slow spin)
Shaft 2 (Fast spin)
Gears
Generator
Biomass
Biomass is anything organic i.e. plant or animal matter. It can be used in the place of coal a
Sun
Transmission
cables &
pylon
Heated oil stored
in insulated tank
for night use
Steam
pipes
Mirrored trough
reflector, reflects
sunlight onto black
pipes
Steam
Turbine
Generator
Water pipes
Heat exchanger transfers
Black pipes carrying oil heat from hot
Steam rushes through
Turbine, causing the
blades to spin (attached to
the shaft so shaft spins)
Generator
Shaft
Boiler
Magnet attached
to shaft
Steam
Water
Turbine blades
Heat
Coils
Electricity transmitted
through power cable
Coal
Power Station
Pylons sup
We already know m, g and vA , but what is hA ? Note that if we let hB = 0 then
hA = 0.5m as A is 0.5m above B. In problems you are always free to choose a line
corresponding to h = 0. In this example the most obvious choice is to make point B
correspond t
Essay 1 : Energy
Author: Asogan Moodaly
Asogan Moodaly received his Bachelor of Science degree (with honours) in Mechanical Engineering
from the University of Natal, Durban in South Africa. For his nal year design project he worked on
a 3-axis lament wind
Question: How much potential energy does a brick with a mass of 1kg gain if it is
lifted 4m.
Answer:
Step 1 : Analyse the question to determine what information is provided
The mass of the brick is m = 1kg
The height lifted is h = 4m
These are in the co
Kinetic Energy
Kinetic energy is the energy of motion that an object has. Objects moving in straight lines
possess translational kinetic energy, which we often abbreviate as E k .
The translational kinetic energy of an object is given by
Ek = 1 mv 2
2
Ek
These are relatively new technologies.
Liquid Fuels Liquid fuels are used mainly for transportation. Petrol and diesel are the most
common liquid fuels and are obtained from oil.
Sasol is the only company in the world that makes liquid fuels from coal; an
Essay 2 : Tiny, Violent Collisions
Author: Thomas D. Gutierrez
Tom Gutierrez received his Bachelor of Science and Master degrees in physics from San Jose
State University in his home town of San Jose, California. As a Masters student he helped work on a
l
9.1.2
Inelastic Collisions
Denition: An inelastic collision is a collision in which total momentum
is conserved but total kinetic energy is not conserved;
the kinetic energy is transformed into other kinds of energy.
So the total momentum before an inelas
total kinetic energy is not the same as before the collision.
, E
pAf
ter
kAf ter
So:
pBef ore
+
pm pa
EkBef ore
Ekm + Eka
=
pAf
ter
pAf ter
=
but
=
EkAf ter
=
EkAf ter
Worked Example 49
Inelastic Collision
Question: Lets consider the collision of two
Step 4 : Show the conservation of kinetic energy
We start by writing down that the kinetic energy before the collision KBefore is equal
to the kinetic energy after the collision KAfter
Before
After
KBefore
K1 + K 2
=
=
KAfter
K3 + K4
0 + K2
K2
=
=
(9.10)
After the collision:
2
, E
p3 k3
1
, E
p4 k4
Step 3 : Decide which equations to use in the problem
Since the collision is elastic, both momentum and kinetic energy are conserved in the
collision. So:
EkBef ore
=
EkAf ter
and
=
pBef ore
pAf
ter
and ) so
Before the balls collide, the total momentum of the system is equal to all the individual momenta
added together. The ball on the left has a momentum which we call 1 and the ball on the
p
, it means the total momentum before the collision is
right has a
We will have a look at the collision between two pool balls. Ball 1 is at rest and ball
2 is moving towards it with a speed of 2 [m.s1 ]. The mass of each ball is 0.3 [Kg].
After the balls collide elastically, ball 2 comes to a stop and ball 1 moves o. Wh
Chapter 9
Collisions and Explosions
In most physics courses questions about collisions and explosions occur and to solve these we
must use the ideas of momentum and energy; with a bit of mathematics of course! This section
allows you to pull the momentum
We know we need the mass and the speed to work out Ek and we are given both of
these quantities. We thus simply substitute them into the equation for E k :
Ek
=
=
=
=
1
mv 2
2
m
1
(900kg)(16.67 )2
2
s
2
kgm
125 000 2
s
125 000 J
Worked Example 44
Mixing U
IN THE PRESENCE OF FRICTION
Mechanical energy is not conserved
(The mechanical energy lost is equal
to the work done against friction)
U = Ubef ore Uaf ter = Work Done Against Friction
Worked Example 46
Using Mechanical Energy Conservation
Question: A 2kg
Potential Energy
If you lift an object you have to do work on it. This means that energy is transferred to
the object. But where is this energy? This energy is stored in the object and is called potential
energy. The reason it is called potential energy i
p
= f inal initial = m m
p
p
v
u
Worked Example 36
Change in Momemtum
Question: A rubber ball of mass 0.8kg is dropped and strikes the oor at a velocity
of 6 m.s1 . It bounces back with an initial velocity of 4 m.s1 . Calculate the change
in momentum of
the velocity of ball 2, ,
v2
all in the correct units!
Step 2 : Decide how to tackle the problem
What is being asked? We are asked to calculate the total momentum of the system. In this example our system consists of two balls. To nd the total momentum
F Res
t=0
u
m
t = t
F Res
m
v
Figure 7.3: An object under the action of a resultant force.
Law of Momentum: The applied resultant force acting
on an object is equal to the rate of change
of the objects momentum and this force
is in the direction of the c
Therefore,
T = 2.36 106 s.
Combining the distance travelled by the moon in an orbit and the time taken by the
moon to complete one orbit, we can determine the magnitude of the moons velocity
or speed,
v
=
=
=
Distance
time
C
T
1.02 103 m.s1 .
Step 4 : Fin
m1
u1
u2
m2
Figure 7.1: Before the collision.
7.4
What properties does momentum have?
You may at this stage be wondering why there is a need for introducing momentum. Remarkably
momentum is a conserved quantity. Within an isolated system the total momentu
v1
m1
m2
v2
Figure 7.2: After the collision.
p total bef ore = total af ter
p
m1 + m 2 = m 1 + m 2
u1
u2
v1
v2
m1
m2
: mass of object 1 (kg)
: mass of object 2 (kg)
u1
u2
: initial velocity of object 1 (m.s1 + direction)
: initial velocity of object 2 (
Step 2 : Decide how to tackle the problem
What is being asked? We are asked to calculate the momentum which is dened as
= m .
p
v
Thus, we need the mass and velocity of the ball but we have only its mass and the
magnitude of its velocity. In order to det
the moons mass,
the distance to the moon, and
the time for one orbit of the moon
with mass in the correct units but all other quantities in the incorrect units. The
units we require are
seconds (s) for time, and
metres (m) for distance
Step 2 : Decid