Download P4 revision

P4 revision
Forces and motion
Interaction pairs
 Forces are equal in
magnitude(size)
 Forces act in oppostie
directions
 Each force acts on a
different body.
Question
A boy kicks a ball in the park. Discuss the
situation.
Include a point about interaction pairs
Name the other force.
(2marks)
Moving objects
In this situation we still have our interaction pair. Having an
interaction pair does not mean that you will not get a
resultant force.
It does mean that you get friction, there are three types of
friction :
Friction:
 Friction between solid surfaces which are
gripping. Eg. Earths crust.
 Friction between solid surfaces which are
sliding past each other. Eg. Pieces of a car
engine.
 Friction or drag from from fluids(liquids or
gases)
Question
The overall force on a rocket at takeoff is
an upwards force. Explain this using the
idea of interaction pairs and resultant
forces.
(3marks)
Speed
Units
 In physics, units are really important. Although
you are used to using mph in everyday life in
physics we use m/s or metres per second.
Distance is measured in metres and time in
seconds.
 Look for tricky questions where you are given
distance in km or time in minutes or hours. You
might need to convert. Look at the units of the
answer for guidence.
Distance- Time graphs
C
D
B
A
Speed – Time graphs
Velocity- Time graphs
Velocity
(m/s)
B
A
C
Time (s)
D
E
Speed
Average speed: This is the distance
travelled in a given time.
Instantaneous speed: This is the average
speed but over a very small time. (used in
spee cameras)
Velocity: Speed in a given direction. Eg.
5m/s north. Velocity is a vector as it has a
length an a dircetion. Speed is scalar.
Techographs
Question...
Give two reasons why haulage businesses
fit their lorrys with tachographs. (2 marks)
Momentum
Again, you will need
to be careful with your
units here. You might
be given something in
g, but we use kg!
P= momentum
m= mass
v = velocity
(kgm/s)
(kg)
(m/s)
Conservation of momentum
The momentum before and interaction and
after an interaction is alwas the same. So
if it is zero before say, a car accident, it
must be zero afterwards. This is achieved
by a change of mass or velocity.
Example
If they then push away from each
other so they move in opposite
directions. The child in the pink
coat has a velocity of 2m/s what
is the velocity of the child in the
red coat?
Mass = 30kg
Mass = 20kg
We can take this a before. They are not
moving. Therfore their velocity is zero,
and so is their momentum.
mxv
+ mXv = p
(30)(0) + (20)(0) = 0kgm/s
After
mXv
+
mXv
(30)(2) + (20)v
60 = -20v
v = -3m/s
=
P
= 0 kgm/s
As the momentum
is conserved.
Change in momentum
If the resultant force on an object is not
zero, its momentum changes in the
direction of the force.
Change in momentum = Force X Time
Again be careful with units. Force should
be in newtons and time in seconds.
Car safety
If someone’s momentum changes very
quickly, the forces on the body will be very
large and more likely to cause injury.
We want the change in momentum to take
place over the largest time possible so the
force is minimal.
In a car we have many safety features to
try and do this.
Safety features
Crumple Zones
Airbags
Seat belts
Bike and motorcycle helmets
Question
A force of 500N acts on a car of mass
1500kg for half a minute. What is the
speed of te car after this time?
Clue you need to use both equations.
Work Done
Work done = Force applied x Distance
moved
Question: If a force of 1500N is applied to
a wall however the wall cannot be moved,
what is the work done?
Work done is measured in joules (J)
Work done
 Work done = Energy Transferred
 Energy is also measured in joules and this is a
really important equation. It will be on the cover
of your exam paper but it is easly forgotten and
often examined without being directly asked.
You might be asked for an assumption you
made or why the work done is not equal to the
energy transferred.
Kinetic Energy
KE= ½ m X v2
Energy is always conserved (it cannot be
created of destroyed) but it is often lost as
heat or sound. (if friction is causing a loss
this is heat too) So in reality the kinetic
energy is a little less than the energy
transferred.
Gravitational Potential Energy
GPE = m x g x h
 g is the gravitational
field strength and at
GCSE is rounded to
10N/kg.
GPE to KE
Falling objects convert GPE to kinetic
energy.
Kinetic energy = Potential energy
Gained
Lost
This is a commonly asked question
Example
A rolercoaster carriage has a weight of
5000N (a mass of 500kg) and is 20m
above its lowest point. Assuming the roller
coaster is stationary at its greatest height,
what is the speed of the carriage when it
reaches its lowest point?
(These questions are often asked in
stages)
Steps
 1. Write down what your are given to see what
equations to use.
 Mass = 500kg
 Weight (mass x g) = 5000N
 H = 20m
 Possible eqns: KE= ½ mv2
GPE = Wh
GPE lost = KE gained.
Step 2
So we can calculate GPE and then
because this is equal to KE gained, the
KE.
GPE = Wh
(5000)(20) = 100,000J
Therefore KE gained = 100,000J
Step 3
Ke= 100,000J
M= 500kg
V= ?
We need to rearrange our equation
KE = ½ mv2
2KE = mv2
2KE = v2
√ (2KE/m) = v
m
Answer
 √ (2KE/m) = v

√((2 x 100,000)/500) = v
 v = 20m/s
These are the hardest questions you get
asked.
Questions
A 600kg(6000N) rollercoaster carriage is
travelling at 40m/s. What is the maximum
height it could climb if all its kinetic energy
is transferred to gravitational potential
energy?
(this is the opposite)
A trolley is stationary at the top of a hill.
The trolley weighs 200N(mass 20kg) and
the hill is 50m high. Assuming all of its
GPE is converted into KE, how fast will the
trolley be going when it reaches the
bottom of the hill?