Christopher Chan Peak Dot Point Summary: Physics Module 2: Moving About 8.4.1 – Vehicles do not typically travel at a constant speed 8.4.1.1 – Identify that a typical journey involves speed changes Typical journeys involve various speed changes e.g.: - Acceleration/Deceleration - Coming to rest/Moving from rest Furthermore, most typical journeys begin at a speed of zero, meaning speed must change for the journey to take place 8.4.1.2 – Distinguish between the instantaneous and average speed of vehicles and other bodies Instantaneous speed – exact speed of a vehicle/body at a particular instant of time Average speed/velocity – indicated by the change in total distance/displacement divided by the time taken during a given time v avg = ∆s ∆t v avg → = ∆s → ∆t → E.g. A car travels 10km East and then 10km South in 20 minutes. Determine: a) The total distance and displacement travelled b) The average speed of velocity of the journey DIAGRAM
Transcript
Christopher Chan
Peak Dot Point Summary: Physics
Module 2: Moving About
8.4.1 – Vehicles do not typically travel at a constant speed
8.4.1.1 – Identify that a typical journey involves speed changes Typical journeys involve various speed changes e.g.:
- Acceleration/Deceleration- Coming to rest/Moving from rest
Furthermore, most typical journeys begin at a speed of zero, meaning speed must change for the journey to take place
8.4.1.2 – Distinguish between the instantaneous and average speed of vehicles and other bodies
Instantaneous speed – exact speed of a vehicle/body at a particular instant of time Average speed/velocity – indicated by the change in total distance/displacement
divided by the time taken during a given time
vavg=∆ s∆ t
vavg→=∆s→
∆ t→
E.g. A car travels 10km East and then 10km South in 20 minutes. Determine:a) The total distance and displacement travelledb) The average speed of velocity of the journey
DIAGRAM
a) Total distance = 20km
Total displacement = √102+102 = 14.14km (2dp) South-East
b) Average Speed = distancetime
= 2013
= 60km/hr
Average velocity = displacement
time
= 14.14km13
= 42.42km/hr South-East 8.4.1.3 – Distinguish between scalar and vector quantities in equations
Scalar quantities – only magnitude is specified Vector quantities – both magnitude AND direction must be specified Examples below:
In equations whether the values are scalar/vector must be taken into account E.g. If a question asks for the velocity of a moving body, DIRECTION as well as
magnitude must be given (such as in the question above) 8.4.1.4 – Compare instantaneous and average speed with instantaneous and average
velocity Instantaneous speed vs. Instantaneous velocity
- Instantaneous speed is a scalar quantity, instantaneous velocity a vector quantity (has direction)
- Both provide a value at an exact point in time Average speed vs. Average velocity
- Average speed is a scalar quantity, average velocity a vector quantity (has direction, again)
- Average speed is calculated using all the values in the journey, average velocity is only start to finish
8.4.1.5 – Define average velocity as:
vav=∆r∆ t
Average velocity is equal to the change in displacement divided by the change in time
Calculated by using starting and ending positions and time taken
E.g. What is the average velocity of a car that moved 40 km East and 80 km West in 2 hours?
Displacement = 40km West
Time taken = 2 hours
vav=∆r∆ t
∴ Average velocity=402
Average velocity=20km /hrWest
8.4.2 – An analysis of the external forces on vehicles helps to understand the effects of acceleration and deceleration
8.4.2.1 – Describe the motion of one body relative to another
The motion of a body can be described in different perspectives E.g. Car A moving at 40km/hr East would be moving at 40km/hr East to a stationary
observer. However another car B moving at 40km/hr West would perceive this car to be moving faster.
Exactly how fast does car A appear to be travelling relative to B?i .e V A /B=40km /hr East−40km /hrWest
¿40 km /hr East+40km /hr East
= 80km /hr East
Thus, the velocity of object A relative to object B is given by the equation:
V A /B=V A−V B
8.4.2.2 – Identify the usefulness of using vector diagrams to assist solving problems Vector diagrams can give a general idea of the overall displacement of a body in
many motion questions Vector diagrams are used to add vectors (Head to Tail), to find a resultant vector This can be used to find average velocity, total displacement, etc.
8.4.2.3 – Explain the need for a net external force to act in order to change the velocity of an object
Newton’s First Law of Motion (An object will remain at rest or moving at constant velocity unless acted upon an external unbalanced force)
Thus, according to this law the velocity of an object can only change when there is a net external force applied to the object
8.4.2.4 – Describe the actions that must be taken for a vehicle to change direction, speed up and slow down
According to Newton’s First Law, a net external force must be applied to a vehicle to change direction/speed
This can be seen in a typical driver’s journey, pressing on the accelerator/brake (providing thrust/driving force) or using the steering wheel to change speed/direction
On earth there are however other forces such as gravity, air resistance and friction which slow a car down
8.4.2.5 – Describe the typical effects of external forces on bodies including:- Friction between surfaces- Air resistance
External forces such a friction and air resistance usually oppose the motion of bodies E.g. friction between tyres of vehicles and roads and air resistance occurring when
vehicles are in motion Most real life journeys consist of these net external forces (and thus, it is hard to
observe Newton’s First Law of Motion being applied) 8.4.2.6 – Define average acceleration as:
aav=∆v∆ t
Therefore
aav=v−ut
Average acceleration is as stated above It is the change in velocity over the change in time (thus v-u on t) Average acceleration only takes into account the beginning and ending points
8.4.2.7 – Define the terms ‘mass’ and weight’ with reference to the effects of gravity Mass and weight are often confused Mass is the amount of matter in a body (does not change despite location) Weight is the force a body exerts at a particular area, i.e. on Earth this is shown by:
W = mg Where W = weight in newtons, m = mass and g = force due to gravity (9.8m/s/s) on
Earth
E.g. If a person has a mass of 50kg, calculate the weight of this person on the Earth (Take gravity to be 9.8m/s/s)
m = 50kg, g = 9.8m/s/s
W = mg
= 50*9.8
= 490N
8.4.2.8 – Outline the forces involved in causing a change in the velocity of a vehicle when:
- Coasting with no pressure on the accelerator
- Pressing on the accelerator
- Pressing on the brakes
- Passing over an icy patch on the road
- Climbing and descending hills
- Following a curve in the road
8.4.2.9 – Interpret Newton’s Second Law of Motion and relate it to the equation:
∑ F=ma Newton’s second law states that the net/imprest force is equal to the product of
mass and acceleration Since mass is assumed constant, simply put the acceleration of an object is directly
proportional to the force (both being in the same direction) The mass being involved is what makes this an equation
8.4.2.10 – Identify the net force in a wide variety of situations involving modes of transport and explain the consequences of the application of that net force in terms of Newton’s Second Law of Motion
The net force is the sum of all forces acting on a body The net force is in the direction of where the vehicle is travelling in transportation
(thrust/driving force overcomes friction/resistance forces) The net force is proportional to acceleration and depends on mass according to
Newton’s Second Law of Motion If acceleration is kept constant this force will depend on mass, thus heavier objects
will have more net force than lighter ones
8.4.3 – Moving vehicles have kinetic energy and energy transformations are an important aspect in understanding motion
8.4.3.1 – Identify that moving object possesses kinetic and that work done on that object can increase that energy
All moving objects have kinetic energy, this energy being quantified in the formula:
KE=12mv2
Where KE=Kinetic Energy ( Joules )m=mass (kg)
v=velocity (ms−1 )
E.g. An object of mass 100kg is moving at a constant speed of 10m/s. Calculate the kinetic energy of the body.
m = 100kg
v = 10m/s
KE = ?
KE=12mv2
KE = ½ (100) (10*10)
= 5000J or 5kJ
If work is applied to the object the velocity increases (thus increasing KE) KE is directly proportional to the square of the velocity of an object (mass is
constant) Work is defined through the equation:
W=FsWhere W = work, F = Net force, S = Displacement
E.g. Johnny pushes a box of trash 500m using a force of 50N. Calculate the work he has done.
s = 500m
F = 50N
W = ?
W=Fs
W = 500*50
= 25000J or 25kJ
8.4.3.2 – Describe the energy transformations that occur in collisions
There are two types of collisions: elastic and inelastic Elastic collisions are when two bodies collide then rebound off each other
- In elastic collisions KE of each body can change BUT it is conserved Inelastic collisions are when two bodies collide then stick together and move off as a
single body- In inelastic collisions KE IS NOT conserved, but rather transformed into
other types of energy (sound, etc.) The equation for the conservation of momentum is given in 8.4.4.3 For inelastic collisions this formula must be used:
m1u1+m2u2=(m1+m2 )vc 8.4.3.3 – Define the law of conservation of energy
The law of conservation states that ‘’In an isolated system, energy cannot be created nor destroyed but only transformed from one form to another”
This is similar to the law of conservation of mass, however energy can be transformed into different forms
E.g. when a car crashes into a tree, the kinetic energy does not just disappear, but is rather transformed into different forms of energy such as deformation, heat, light and sound
An example of energy transformations in car collisions (seen in deformation)
8.4.4 – Change of momentum relates to the forces acting on the vehicle or the driver
8.4.4.1 – Define momentum as:p=mv
The law of reflection states that momentum is equal to the product of mass and velocity
Momentum can be seen in real life, e.g. a person running, cars moving, the momentum of a tennis ball as it is hit
E.g. If a baseball of mass 500g is moving through the air at a velocity of 10m/s North, calculate the momentum of the ball.
m = 500g = 0.5kg (NOTE: always check unit conversion)
v = 10m/s
p = ?
p = mv
= (0.5)(10)
= 5kgm/s North
8.4.4.2 – Define impulse as the product of force and time Given by the formula
I=Ft
Impulse is defined as the product of force and time, i.e. the measurement of a force over a period of time
E.g. Calculate the impulse of a man pushing a box with a force of 500N over 20 seconds.
F = 500N
t = 20 seconds
I = ?
I = Ft
= (500)(20)
= 10000Ns
8.4.4.3 – Explain why momentum is conserved in collisions in terms of Newton’s Third Law of motion
According to Newton’s Third Law for every action there is an equal and opposite reaction (action-reaction forces)
In collisions this still applies, therefore the momentums before and after the collisions must be equal
This can be proven mathematically :
Given vehicle A and B are involved in a collision
F A=−FB [Newton' s3 rd Law ]
mAaA=−mBaB
Now ,v=u+at
a= v−ut
mA(v A−uAt
)=−mB(v B−uBt
)
mA v A−mAuA=−mB vB+mBuB
∴mAuA+mBuB=mA v A+mB v B
8.4.5 – Safety devices are utilised to reduce the effects of changing momentum
8.4.5.1 – Define the inertia of a vehicle as its tendency to remain in uniform motion or at rest
Inertia is the tendency for an object to remain in uniform motion or at rest E.g. if a car suddenly stops the passenger in the vehicle continues to go forward
(because of his/her inertia) The car in this sense may have forces slowing it down; friction, etc., however the
passenger does not and may be injured in such a situation This can also be seen in the images below where the coin remains at rest and drops
into the glass when the cardboard is flicked fast
8.4.5.2 – Discuss reasons why Newton’s First Law of Motion is not apparent in many real world situations
There are various reasons why Newton’s First Law of Motion is not apparent in real life situations:
- There are many external forces on Earth which can oppose a body’s motion- E.g. friction, air resistance, gravity and drag- Thus, the law is not apparent in real life situations- If there was no friction an object moving at uniform speed would maintain it
forever
8.4.5.3 – Assess the reason for the introduction of low speed zones in built-up areas and the addition of air bags and crumple zones to vehicles with respect to the concepts of impulse and momentum
Low speed zones in built-up areas (such as school zones) were introduced because it gives drivers more time to stop
In terms of physics this can be seen through the momentum formula: p = mv (the less velocity, the less momentum, thus the faster u can stop)
It can also be seen through the stopping distance formula:
s=mu2
2F(by decreasing initial velocity the stopping distance decreases significantly,
stopping distance proportional to square of initial velocity, assuming mass and friction are constants)
Where s = stopping distance, m = mass, u = initial velocity, F = friction
Air bags and crumple zones provide essential safety for drivers and reduction in possible damage of vehicles
Air bags and crumple zones increase time of collisions for vehicles as well as passengers
Since Impulse is given by: I = Ft, increasing the time of the collision therefore, decreases the force of the impact
8.4.5.4 – Evaluate the effectiveness of some safety features of motor vehicles There are various safety features of vehicles which can be evaluated:
Safety Feature Advantages DisadvantagesAir Bags Cushions head to prevent
impact on dashboard Reduces force acting on
occupant by increasing time
Can be activated unnecessarily Can suffocate children Rapid deployment of airbags
can injure unrestrained passengers
Crumple Zones Absorbs energy at collision by converting KE into thermal energy
Reduces damage/impact Can be at front and back for
more protection Increases distance over the
whole force is acting, thus decreasing average force
After destruction car will be unusable
Passengers may still continue to move at the collision
Seat Belts Prevents ejection from vehicle
Minimises body’s contact with interior of car, thus preventing injuries
Spread the force of impact over larger parts of the body, reducing severity of injuries
Can result in injuries due to the fibres of the seat belt causing harm to chest/abdominal regions
May cause serious damage to back
Can be uncomfortable
Head Restraints Can save lives by preventing severe neck injuries
Provides a force that acts against the passenger’s head in accords to inertia
