PHYSICS TUTORIAL NOTES

SYLLABUS TOPIC 8.4

These notes are meant as a guide only and are designed to focus your thoughts on the dot points mentioned in the syllabus. They give a very brief overview of the topic and should be used in conjunction with your class notes, your textbooks and your research from other areas such as the library and the Internet.

Notes compiled by:

CARESA EDUCATION SERVICES

“1. Vehicles do not typically travel at a constant speed” (syllabus)

“Students learn to identify that a typical journey involves speed changes” (syllabus)

Think of any typical journey you have made such as a bus ride to school. The vehicle starts from rest, accelerates, decelerates and stops. Since you start from rest, if you didn’t change speed you would stay where you were.

Distinguish between the instantaneous and average speed of vehicles and other bodies

Instantaneous speed refers to a speed at a particular instant of time; the smallest interval of time you can think of, say a trillionth of a second. You can determine the instantaneous speed of a car by looking at its speedometer. The average speed of a vehicle is determined by dividing the distance travelled by the time taken. To determine the average speed for a car trip set the trip meter to zero before starting out. Look at your watch as you begin your journey and again as you arrive at your destination. Determine the time taken for the trip. Look at the trip meter to determine the distance travelled. The average speed for the trip can be determined by dividing the distance travelled by the time taken.

Distinguish between scalar and vector quantities in equations

Vector quantities have magnitude and direction whereas scalar quantities have magnitude only. An example of a vector quantity would be force (which direction are you pushing?) and an example of a scalar quantity would be mass (no direction applies).

In the equation; F = ma           F and a are vectors while m is a scalar.

Common scalar quantities: distance, speed, time, mass, work,

Common vector quantities: displacement, velocity, acceleration, force,

Compare instantaneous and average speed with instantaneous and average velocity

First compare distance to displacement. Distance is the length travelled. It is a scalar and no direction is required. The approximate distance of Brisbane from Sydney for instance is 1000 km. However the displacement of Brisbane from Sydney is 1000 km north (direction approximate only). Displacement is the length from the starting point with direction specified. Suppose someone takes a trip from Brisbane to Sydney and back again. The distance travelled is 2000 km but the displacement (distance from the starting point) is zero.

Speed is the rate of change of distance i.e. speed = distance travelled/time taken.

Velocity is the rate of change of displacement i.e. velocity = final displacement/time taken

Instantaneous speed is the speed at a given instant whereas instantaneous velocity is speed at a given instant with direction specified. For instance if you look at the speedometer of a car you could say that its instantaneous speed was 50 km/h but if you also knew the direction that it was travelling (say west) you could say that its instantaneous velocity was 50 km/h west. Average speed is the distance travelled divided by the time taken whereas average velocity is the change in displacement divided by the time taken.

Define average velocity as vav = Dr/Dt

Average velocity is the change in displacement divided by the time taken. vav = Dr/Dt

Students plan, choose equipment or resources for, and perform a first-hand investigation to measure the average speed of an object or a vehicle

Maybe you can do an investigation such as Jill and Jack below.

Jill and Jack travelled from home to the athletic field for their school’s athletic carnival. They set the trip meter of the car to zero and Jack looked at his watch and noted the time as 8.23 a.m. Jill then drove Jack to the athletic field and as they arrived Jack noted that the time was 8.48 a.m. They noted that the reading on the trip meter was 12.5 km. They then calculated the average speed.

Time taken = 8.48 – 8.23 = 25 minutes = 25 x 60 = 1500 seconds

Distance = 12.5 km = 12500 metres

Average speed = 12500/ 1500  = 8.33 m/s

Jill then ran the 100 metres race in 13.4 seconds. Jack worked out that her average speed was 100/13.4 = 7.46 metres per second.

During a break they took a compass over to the track and noted that the 100m track was pointing north-east. Jill then said that her average velocity for the 100m was 7.46 metres per second north-east.

Jack then lined up for the 400m run. He ran on the inside lane so that 400m was exactly 1 lap of the track. His time for the run was 54.6 seconds.

Jill worked out that Jack’s average speed for the 400m was 400/54.6 = 7.326 m/s

They then joked that since Jack finished the race at the same point at which he started, his displacement was zero and so his average velocity for the race was zero.

Students solve problems and analyse information using the formula: vav = Dr/Dt

Where r = displacement

Q.        Jason walked 100m north and then 100m east in 3minutes and 20 seconds.

What was his   (i) displacement

(ii) average velocity.

A.        (i)         Displacement = straight line distance with direction specified.

r =  = = 141.4 m north-east

(ii)               time = (3 x 60) + 20 seconds = 200 seconds

vav = Dr/Dt

Average velocity = 141.4/200 = 0.707 ms-1 north-east

Students present information graphically of:

-          displacement vs time

-          velocity vs time

for objects with uniform and non-uniform velocity

Consider an object moving at a constant velocity of 2 m/s. A table of displacement and time for the first five seconds is shown below.

Table of displacement against time

 Time (s) Displacement (m) 0 0 1 2 2 4 3 6 4 8 5 10

A graph showing the motion of this object is shown below. Graph of displacement against time

There are a couple of things to note about the graph.

(i)                 Since the object is moving with uniform velocity the graph is a straight line.

(ii)               The rise of the graph represents the change in displacement Dr

The run of the graph represents the change in time Dt

The gradient of the graph = rise / run = Dr / Dt = velocity v

The gradient of a displacement/time graph is the velocity.

Now consider the case of an object that travels at 2 ms-1 for the first two seconds and then at 4 ms-1 for the next three seconds. A table of displacement against time is shown below.

Table of displacement against time

 Time (s) Displacement (m) 0 0 1 2 2 4 3 8 4 12 5 16

The graph for the object’s motion is shown below. Note that the graph bends at 2 seconds. This is because it changes velocity.

To find the velocity at 4 seconds we have to find the gradient of the second part of the graph.

Velocity = gradient = rise/run = (16 – 4) / (5 – 2) = 12/3 = 4 ms-1

Taking the total displacement (16m) and dividing by the total time (5s) will give the average velocity for the journey.

Average velocity = 16/5 = 3.2 ms-1

The displacement of an object that starts from rest can be determined from the formula; r = ½ at2 where r = displacement, a = acceleration and t = time.

The following displacement / time table represents an object starting from rest and accelerating at 2 ms-2.

Table of displacement against time

 Time (s) Displacement (m) 0 0 1 1 2 4 3 9 4 16 5 25

A graph of the object’s motion is shown below. Note that the displacement/time graph for an object that is accelerating is parabolic.

The following graph shows an object moving at a constant velocity of 2 ms-1 There are a couple of things to note about the graph.

Since the velocity is constant (no acceleration) the graph is horizontal (gradient = 0)

The area underneath the graph = rise x run = velocity x time = displacement.

The following graph shows an object starting from rest and accelerating at 2 ms-2. The graph is a straight line indicating that the acceleration is constant.

The gradient of the graph is rise/run = change in velocity / change in time

This is acceleration.

The gradient of a velocity/time graph is acceleration.

Since the acceleration is constant the average velocity can be found by vav = (v-u)/2

Where v = final velocity and u = initial velocity.

The displacement, r is given by; r = vav t

This is the area under the graph

The area under a velocity/time graph is displacement.

“2. An analysis of the external forces on vehicles helps to understand the effects of acceleration and deceleration” (syllabus)

Describe the motion of one body relative to another

A set of coordinates from which we take measurements is a frame of reference. While we normally take the Earth as our frame of reference and describe motion relative to the Earth it is just as valid to take a moving car as our frame of reference and describe motion relative to the car.

e.g. a car is travelling north at 50 km/h relative to a telegraph pole or the telegraph pole is travelling south at 50 km/h relative to the car.

As Galileo showed, the laws of mechanics are the same for a body at rest and one moving at constant velocity. This is known as Galilean relativity.

Q.1.     A car is moving at 60 km/h north and a utility is moving at 50 km/h south on a long straight road. What is the velocity of the (i) car relative to the utility (ii) the utility relative to the car?

A.1.     Suppose we start taking measurements when the car and utility pass each other. One hour later the car is 110 km north of the utility so the velocity of the car relative to the utility is 110 km/h north. Similarly the utility is 110 km south of the car so the velocity of the utility relative to the car is 110 km/h south.

Note that the velocities of the car and utility relative to each other are equal and opposite.

Q.2.     A coyote and a road-runner are running north at 50 km/h on a long straight road. They are side-by-side when the road-runner suddenly says “beep beep” and accelerates to 80 km/h. What is the new speed of (i) the road-runner relative to the coyote (ii) the coyote relative to the road-runner?

A.2.     One hour later the road-runner will be 30 km north of the coyote and the coyote will be 30 km south of the road-runner. Hence the velocity of the road-runner relative to the coyote is 30 km/h north and the velocity of the coyote relative to the road-runner is 30 km/h south.

Identify the usefulness of using vector diagrams to assist solving problems

Vector diagrams use arrows drawn to scale to represent vector quantities in both magnitude and direction.

Vector addition can be undertaken by joining the arrows, head to tail, and finding the resultant vector.

e.g. A river is flowing at 2 km/h and a canoeist is paddling downstream at a speed that would be 4 km/h in still water. What is the velocity of the canoeist relative to a person on the bank of the river?

The velocity of the river can be added to the velocity of the canoeist to give the velocity of the canoe relative to the bank. Velocity of canoe relative to bank = 6 km/hà

i/e. 6 km/h downstream.

e.g. A river is flowing at 2 km/h and a canoeist is paddling upstream at a speed that would be 4 km/h in still water. What is the velocity of the canoeist relative to a person on the bank of the river?

The velocity of the river can be added to the velocity of the canoeist to give the velocity of the canoe relative to the bank. Velocity of canoe relative to bank = 2 km/h ß

i.e. 2 km/h upstream.

Two Dimensions: Vectors can also be added and subtracted in two dimensions.

Consider the case of forces of 3N and 4N acting at right angles to each other. Intuitively we know that the resultant force will be an angle upwards and to the right.

To find the magnitude and direction of the resultant we construct a polygon of vectors. In this case, since there are only two vectors to be added our polygon will be a triangle.

Transpose one of the vectors, keeping the direction constant, so that the vectors are joined from the head of one to the tail of the other. Complete the polygon by joining the tail of the first vector to the head of the last vector. This represents the resultant in magnitude and direction.

In the above example the resultant would be 5N (by Pythagoras) at 53o to the 3N force and 37o to the 4N force.

Several vectors can be added by drawing a polygon where the vectors are represented in magnitude and direction by arrows. The resultant is found by transposing the vectors and joining the tail of one to the head of the previous one. The polygon is completed by joining the tail of the first vector to the head of the last vector. Note that the order in which we take the vectors doesn’t matter.

A car drives 6 km north then 4 km east, 8 km north, 6 km east, 2 km south and then 1 km west. What was its final displacement?

Draw the vectors in magnitude and direction so that they are joined from head to tail. The resultant can be found by accurate scale measurements or by calculating the north-south displacement and the east-west displacement and then using trigonometry to find the resultant. In this case the resultant is 15 km 37o east of north.

Note: Only give compass directions when they are given in the question.

Sometimes we want to resolve vectors into components i.e. consider a single vector as two separate vectors that could replace it.

An example would be resolving the velocity of a projectile fired at an angle q at velocity v into its vertical and horizontal components. Let the horizontal and vertical components be vx and vy respectively.

Sin q = = vy = v sin q

Cos q = = vx = v cos q

e.g. A stone is thrown at a speed of 20 ms-1 at an angle of 30o up from the horizontal. What are the horizontal and vertical components of its velocity?

vx = v cos q  = 20 cos 30 = 17.32 ms-1 horizontally

vy = v sin q = 20 sin 30 = 10 ms-1 up.

Explain the need for a net external force to act in order to change the velocity of an object

A body remains at rest or in uniform motion unless acted on by an external force. (Newton’s first law of motion) This is known as the law of inertia. The motion of a body will not change of its own accord; an external force i.e. from a source outside the body, is required.

Describe the actions that must be taken for a vehicle to change direction, speed up and slow down

The motion of a vehicle depends on the force between the tyres of the vehicle and the road.

To change direction the front wheels are turned so that they are aligned in the direction of desired travel. This causes the road to exert a frictional force on the tyres at right angles to the direction they are rotating i.e. at right angles to the front wheels. To speed up the driver presses the accelerator. This causes more fuel to be provided to the cylinders which causes them to oscillate faster which then, via the transmission, causes the wheels to rotate faster. As the wheels rotate faster there is an increase in static friction between the tyres and the road and this increases the reaction force to accelerate the car forward.

To gradually slow down the driver eases off the accelerator. This causes less fuel to be provided to the cylinders which causes them to oscillate slower which then, via the transmission, causes the wheels to rotate slower. As the wheels rotate slower there is a decrease in static friction between the tyres and the road and this decreases the reaction force to accelerate the car forward. While Newtons First Law says that a body will remain in uniform motion unless acted on by a force, the air resistance and friction in the moving parts of the car soon bring it to rest.

To slow down more effectively, the driver presses the brake pedal. This increases the friction between the brake pads and the drums or discs which slows the rotation of the wheels and in turn causes a forward force between the wheels and the road. The reaction force to this forward force causes the car to slow down and stop.

If too much pressure is applied to the brakes they completely stop the rotation of the wheels. This causes the wheels to slide (skid) over the road and since sliding friction is less than static friction the car travels a much greater distance before stopping.

Describe the typical effects of external forces on bodies including:

-          friction between surfaces

-          air resistance

When dealing with friction and cars there are different types of friction to consider.

(i)                 Static friction: This is the friction between the tyre and the road and provides the reaction force that causes the wheel to roll and the car to move. When it is removed (e.g. a car bogged in mud) the wheels rotate but because there is no reaction force between the tyres and the surface, the car will not move.

(ii)               Sliding friction is the friction between two surfaces that prevents one surface sliding over the other. It applies when a car is parked on a hill to prevent the car sliding down the hill. It also applies when hard braking stops the wheels from turning and the car skids to a stop.

(iii)             Movement friction is the friction that opposes the motion of a moving body and would include air resistance as well as the friction between moving parts of a car such as the pistons and cylinders.

When an object moves through a fluid (liquid or gas) it has to push the particles of fluid out of the way. When the fluid is air this is known as air resistance. Streamlining, i.e. sloping the shape of the car so that air flows over it, greatly reduces air resistance. Air resistance includes the movement of a car through still air as well as the movement of air against the motion of the vehicle (wind).

define average acceleration as: aav = therefore        aav = Acceleration is the rate of change of velocity hence the average acceleration is the total change in velocity divided by the time taken. i.e. aav = The change in velocity is the final velocity – the initial velocity. Dv = v – u where

Dv = change in velocity,        v = final velocity,        u = initial velocity

Hence aav = define the terms “mass” and “weight” with reference to the effects of gravity.

Mass is the amount of matter in a body whereas weight is the force due to gravity acting on that mass. Fweight = mg

The unit of mass is the kilogram and is defined as the mass of the International Prototype Kilogram. This is a cylinder of platinum-iridium alloy which is defined as having a mass of one kilogram exactly and is kept in the International Bureau of Weights and Measures in Paris. Since it is not feasible to have access to the International Prototype Kilogram, copies have been made and are held in various standards laboratories throughout the world. Boxes of masses containing multiples and subdivisions of the kilogram are readily available.

Mass may be considered as gravitational mass or inertial mass. The gravitational mass of an object compares the gravitational pull on the object with the gravitational pull on the International Prototype kilogram at the same location. This is carried out with a beam balance where the object is placed on one pan and masses to accurately balance it are placed on the other pan. The gravitational mass is the same for an object regardless of the value of gravity at that location since the same gravity applies to the object and the masses. Gravitational mass is a comparison of the gravitational forces.

Inertial mass compares the acceleration of an object with the acceleration of the International Prototype Kilogram when subjected to the same force. Suppose you were in a space station in a gravity-free situation. You could put a 50 kg mass on one side of a beam balance and a 1 kg mass on the other side and they would balance perfectly since the gravitational pull on both masses is zero. So how could you determine the mass of an object in outer space? You would apply a force to a known mass and measure its acceleration. You would then apply the same force to the unknown mass and measure its acceleration. By applying F = m1a1 = m2a2 it is easy to determine the unknown mass.

Regardless of whether we measure gravitational mass or inertial mass, the mass of an object is the same at all locations, regardless of the value of gravity.

Weight, on the other hand, is a force and should be measured in newtons. When you say your weight is 60 kg you really mean that your mass is 60 kg and your weight is 60 kgF i.e. 60 x 9.8 newtons = 588N. On the Moon where gravity is about a sixth that on Earth you would only weigh 98 newtons.

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

Coasting with no pressure on the accelerator: When there is no pressure on the accelerator there are no forward forces. If there were no backward forces then the car would keep moving forever at the same speed. However there is rolling friction, air resistance and friction between the moving parts of the car acting to oppose the motion of the car. Consequently the car will gradually slow down and stop.

Pressing on the accelerator: Pressing on the accelerator increases the rate at which fuel is fed to the cylinders of the car. This in turn increases the rate of rotation of the wheels and increases static friction. The reaction force to this increases the forward force on the car allowing it to overcome the forces that retard the motion. If the accelerator is pressed gently the car will cruise at constant speed. If it is pressed a little harder the car will increase its speed.

Pressing on the brakes: The hydraulic system transfers the pressure to the pads that grip the discs or drums. The pads exert a frictional force on the discs or drums that slows the rate at which the wheels are rotating. This causes a forward force between the wheels and the road and the reaction force to this forward force causes the car to slow down and stop.

If too much pressure is applied to the brakes they completely stop the rotation of the wheels. This causes the wheels to slide (skid) over the road and since sliding friction is less than static friction the car travels a much greater distance before stopping.

For a more detailed explanation of how brakes work, try one of the following websites:

http://www.e-z.net/~ts/hybrakes.htm

http://www.babcox.com/editorial/cm/cm20119.htm

http://auto.howstuffworks.com/brake.htm

Passing over an icy patch on the road: Ice reduces friction to a very low value so pressing on the accelerator would speed up the rate of rotation of the driving wheels but they would not be able to grip the road. The wheels would spin on the ice but the car would have no traction and would not increase its speed.

The loss of friction would be disastrous for braking and steering. The brakes would stop the wheels from rotating but according to Newton’s first law, a car in motion continues in uniform motion unless acted on by a force. Hence the car would skid with little change in speed until it was clear of the icy patch or until it met an opposing force such as the back of the car in front. Also because of the lack of friction, turning the wheels would not be able to produce a centripetal force. Consequently the driver would not be able to steer and the car would continue in a straight line even if the road didn’t.

Climbing and descending hills: All objects on Earth have a gravitational force on them pulling them towards the centre of the Earth. This is true whether the object is on level ground or on a slope such as a hill. A car on a hill has a gravitational force acting vertically down. This applies whether the car is ascending, descending or parked on the hill. The gravitational force mg, can be divided into components perpendicular to the plane and parallel to the plane. The parallel component applies a force of mg sinq downhill.  The parallel component applies a force mg sinq downhill.

When a car is ascending a hill the driver has to press harder on the accelerator to overcome the gravitational force downhill.

When the car is descending the hill the driver does not have to press the accelerator as hard since some of the downhill force is supplied by gravity. On steep hills the engine may have to supply no force at all and the car will coast down the hill. In these cases the car has to be slowed by the brakes. The problem is that constant hard braking can cause the brakes to overheat and become less effective (in extreme cases complete failure can result). An alternative to constant braking is to choose a lower gear to descend the hill. (Have you seen the signs; “Steep Descent – Trucks Use Low Gear”?)

Using a lower gear ensures that the wheels turn more slowly and are subject to less static friction.

Following a curve in the road: A body will travel with uniform velocity unless acted on by a force. (Newton’s first law) Consequently a force is required to change the direction of a car. This force is called the “centripetal force” and acts towards the centre of the motion i.e. towards the centre of the road’s curve. The force is supplied by the friction between the tyres and the road and has a value of; Fc = mv2/r

Where Fc = centripetal force, m = mass, v = speed, r = radius of curve The centripetal force is always towards the centre.

Interpret Newton’s Second Law of Motion and relate it to the equation SF = ma

Newton’s second law states that the net force on an object is equal to the product of its mass and its acceleration and that the acceleration is in the direction of the force.

What this means is that for the same force, an object with a large mass will experience a bigger acceleration than an object of smaller mass.

At the school athletics carnival, athletes prefer to use a light shot put rather than a heavy one because they are able to give the light shot put a larger acceleration. This means that it will attain a larger velocity as it leaves the hand and so travel further before it hits the ground.

In cars, the acceleration is largely governed by the power to weight ratio. If the same (or equivalent) engine is installed in a heavy and a light car, the light car will be able to accelerate much more rapidly.

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

Note the words “identify” and “explain”

Consider the forces acting in planes and trains and automobiles.

First consider planes. There are two components of the net force that we have to consider. First there are the horizontal components of the motion that provide the planes forward motion and secondly there are the vertical forces acting that cause the plane to lift from the ground.

The vertical motion of the plane is caused by dynamic lift (as opposed to static lift such as a balloon where the object rises in a more dense fluid). The dynamic lift of an aircraft is caused by two factors, both of which rely on the plane moving forward through the air. The first is caused by the wing sloping upward towards the direction of forward motion so that the air is deflected down. The second is due to the shape of the wing. The wing is curved so that air travelling over the top of the wing travels a greater distance than air travelling underneath the bottom of the wing. This results in it occupying a greater volume and so being less dense. While the difference in density would create very little lift for a stationary object, there is substantial lift for a moving object. This is due to the Bernoulli principle that is beyond the scope of the HSC physics course.

The forward thrust of a jet plane is caused by the reaction force to the exhaust gases (Newton’s third law). A bigger engine or more engines gives a greater forward force. There is also the force of air resistance opposing the motion of the aircraft. The net force is forward and the acceleration is given by a = Fnet /m

Now consider trains. There are several different types (steam, diesel, electric) but all rely on the principle of a force causing the wheels to turn and the static friction between the wheels and the rail providing the forward (reaction) force to cause the train to move. There is friction between the moving parts of the train as well as air resistance but the overall result is a net force forward which results in the forward acceleration of the train. The train will attain a constant velocity when the forward force is equal to the sum of the backward forces.

Automobiles experience much the same forces as trains except that there is friction between the tyres and the road rather than between the wheels and rail. Because tyres have a tread pattern the friction between the tyres and the road is usually greater than the corresponding friction between train wheels and the rails. There is friction between the moving parts of the car as well as air resistance but the overall result is a net force forward which results in the forward acceleration of the car. The car will attain a constant velocity when the forward force is equal to the sum of the backward forces.

Analyse the effects of external forces acting on vehicles

Note the word “analyse”; Identify components and the relationship among them; draw out and relate implications.

In each of the following situations the first task is to identify and calculate the net force and then apply the equation Fnet = ma.

Q.1.     A car accelerating at 2.0 ms-1 is pulling a trailer of mass 500kg. What is the tension in the coupling between the car and trailer?

A.1.     The tension in the coupling is equal to the force pulling the trailer and this is given by F = ma.

F = 500 x 2 = 1000N in direction of acceleration.

Q.2.     A girl is riding a bicycle into a strong headwind. The mass of the girl and her bicycle is 150 kg and the road is exerting a forward force of 1100N by means of static friction between the bicycle and road. Wind resistance is 300N.

(i) If the girl is riding at constant velocity, what is the value of other frictional forces that apply?

(ii) What would her acceleration be if the headwind dropped and wind resistance was reduced to 50N?

A.2.     (i) Since the girl is travelling at constant velocity the sum of the forces acting on her is zero.

Forward force = wind resistance + backward frictional forces

1150N = 300N + backward frictional forces

Backward frictional forces = 1100N – 300N = 800N against motion.

(iii)             If wind resistance is reduced to 50N then the forces become

Fnet = forward force – backward forces

Fnet = forward force – (wind resistance + backward frictional forces)

ma = 1000N – (50N + 800N) = 1000 – 850 = 150N              since m = 150 kg

150a = 150      a = 1 ms-2 in direction of travel

Q.3.     A man owns an outboard motor that can exert a force of 500N. He can put in on a dingy that has a mass of 200 kg or a launch that has a mass of 1200 kg. If the frictional forces opposing motion are 90% of the forward forces applied, calculate the acceleration the motor can produce for each boat.

A.3.     Forward force applied in each case is 500N

Friction opposing forward force = 90% of 500N = 450N

Net forward force = 500 – 450 = 50N in each case

F = ma             a = F/m

adingy = 50/200 = 0.25 ms-2

alaunch = 50/1200 = 0.042 ms-2

Gather first-hand information about different situations where acceleration is positive or negative

Note that negative acceleration does not necessarily mean slowing down.

Suppose north is defined as positive.

A car travelling north and speeding up has positive acceleration.

A car travelling north and slowing down has negative acceleration.

A car travelling south and speeding up has negative acceleration.

A car travelling south and slowing down has positive acceleration.

Consider the motion of a lift as it travels from the ground floor to the top floor and then back down again. In most cases it is up to you to choose which direction is positive but having made your choice it is up to you to state it and to be consistent throughout. In this example I will choose up as positive.

As the lift begins to move upwards it has a positive acceleration. The acceleration drops to zero as the lift ascends at constant speed. The acceleration becomes negative as the lift slows down to stop at the top floor. The acceleration is also negative as the lift begins to descend and speeds up (remember that I defined up as positive). The acceleration becomes zero as the lift descends at constant speed and again becomes positive as the lift slows down to stop at the ground floor.

To gain first-hand information about whether acceleration is positive or negative take a set of bathroom scales into a lift and stand on them. Define up as positive. As the lift moves, if your weight is recorded as being heavier than normal you have a positive acceleration and if your weight is less than normal you have a negative acceleration.

Negative acceleration is not to be confused with deceleration which means to become slower.

Plan, choose equipment or resources for and perform a first-hand investigation to demonstrate vector addition and subtraction

The following experiment doesn’t require any fancy equipment and is carried out fairly quickly.

Take three spring balances graduated in newtons. Hook them together as shown in the diagram below. Pull them apart gently and hold them in position. Record the forces F1, F2 and F3. Since the system is at rest the forces are in equilibrium and if the law of vector addition is true then their sum should be zero. Marjorie set up three spring balances as shown in the diagram above. She gently pulled them apart and noted the readings F1, F2 and F3. She recorded the readings as shown below.

F1 = 20N ß

F2 = 25N ß

F3 = 40N à

She then calculated the vector sum of the forces as follows:

Fnet = F1 + F2 + F3 = 20 + 25 – 40 = 5N ß

What should Marjorie conclude from her experiment?

The sum of the vector forces was very close to zero which tends to verify the statement that the vector sum of a number of forces in equilibrium is zero. The deviation was probably due to friction in the spring balances, exacerbated by the fact that the balances were lying flat on the bench.

Solve problems using vector diagrams to determine resultant velocity, acceleration and force

Q.1. A plane is flying at 300 km/h north when it experiences a cross wind of 50 km/h from the east. What is the resultant velocity of the plane relative to someone on the ground?

A.1. Draw arrows joined head to tail to represent the velocities. Join the tail of the first arrow to the head of the second arrow to find the resultant. R2 = 3002 + 502           = 90 000 + 2500 = 92 500      R = 304 km/h

Sin q = 50/300             q = 9o35’

Resultant velocity = 304 km/h 9o35’ west of north.

Plan, choose equipment or resources and perform first-hand investigations to gather data and use available evidence to show the relationship between force, mass and acceleration using suitable apparatus.

Student at a particular school were issued with the following apparatus and told to plan and carry out an experiment to demonstrate Newton’s second law.

Apparatus: dynamics trolley, ticker-tape timer, ticker tape, power pack, electrical leads, sticky tape, 1 kg mass carrier, 1 kg masses, string, pulley, bricks, 5 kg spring balance, plastic bag.

Sarah and Jack carried out the following experiment.

They put a 1 kg mass on the 1 kg mass carrier and connected it to the dynamics trolley by means of a string suspended over a pulley as shown in the diagram below. They set up the ticker-tape timer by connecting it to the A.C. terminals of the power pack set at the recommended 6V.They connected ticker tape to the trolley by means of sticky tape and threaded the ticker tape through the ticker-tape timer. Jack then lifted the masses and moved the trolley close to the ticker-tape timer, ensuring that the string was taut. Sarah turned on the ticker-tape timer and Jack allowed the masses to drop to the floor. Sarah turned off the ticker-tape timer and changed the tape. She labelled the used tape “trolley only” and put it aside to analyse later.

Jack then placed a brick on the trolley and they repeated the experiment. They repeated the experiment again with 2, 3 and 4 bricks with appropriate labels for the tape each time.

They placed the dynamics trolley in the plastic bag and suspended it on the spring balance to determine its mass. They then determined the mass of each of the bricks, one at a time, by placing them in the plastic bag and weighing them. They recorded the masses in the table below.

 Object Mass (kg) Dynamics cart 0.7 Brick 1 1.6 Brick 2 1.6 Brick 3 1.6 Brick 4 1.6

They then analysed the tapes to determine the acceleration in each case. Sarah and Jack determined the acceleration by the following method.

They ignored the first few dots then numbered the others. They measured the distance between dots 1 & 3 as 1.7 cm (0.017m)

Since the ticker-tape timer is connected to a 50 Hz A.C. electricity supply it is oscillating at 50 Hz and the time interval between dots is 1/50 second (0.02s)

They determined the time interval between dots 1 & 3 as 0.04s and the average speed between dots 1 & 3 as 0.017/0.04 = 0.425 ms-1

They measured the distance between dots 9 & 11 as 5.1 cm (0.051m)

They determined the time interval between dots 9 & 11 as 0.04s and the average speed as 0.051/0.04 = 1.275 ms-1

They calculated that the time interval between dots 2 & 10 was 0.16s

Acceleration = = = 5.3 ms-2

Exercise: Print out a copy of the tapes 2 to 5 (above) and verify that the accelerations are; tape 2 = 2.8 ms-2, tape 3 = 1.8 ms-2, tape 4 = 1.4 ms-2, tape 5 = 1.2 ms-2.

Sarah and Jack then constructed a table of mass and acceleration, keeping in mind that the total mass to be accelerated includes the 2 kg mass suspended over the pulley.

 Object Mass of object (kg) Total mass (kg) Acceleration (ms-2) Trolley only 0.7 2.7 5.3 Trolley + 1 brick 2.3 4.3 2.8 Trolley + 2 bricks 3.9 5.9 1.8 Trolley + 3 bricks 5.5 7.5 1.4 Trolley + 4 bricks 7.1 9.1 1.2 Conclusion: The graph of acceleration against mass is a hyperbola and shows that a is inversely proportional to m.

Solve problems and analyse information using SF = ma for a range of situations involving modes of transport

Q.1.     A car accelerating at 2.0 ms-1 is pulling a trailer of mass 500kg. What is the tension in the coupling between the car and trailer?

A.1.     The tension in the coupling is equal to the force pulling the trailer and this is given by F = ma.

F = 500 x 2 = 1000N in direction of acceleration.

Q.2.     A girl is riding a bicycle into a strong headwind. The mass of the girl and her bicycle is 150 kg and the road is exerting a forward force of 1100N by means of static friction between the bicycle and road. Wind resistance is 300N.

(i) If the girl is riding at constant velocity, what is the value of other frictional forces that apply?

(ii) What would her acceleration be if the headwind dropped and wind resistance was reduced to 50N?

A.2.     (i) Since the girl is travelling at constant velocity the sum of the forces acting on her is zero.

Forward force = wind resistance + backward frictional forces

1150N = 300N + backward frictional forces

Backward frictional forces = 1100N – 300N = 800N against motion.

(iv)             If wind resistance is reduced to 50N then the forces become

Fnet = forward force – backward forces

Fnet = forward force – (wind resistance + backward frictional forces)

ma = 1000N – (50N + 800N) = 1000 – 850 = 150N              since m = 150 kg

150a = 150      a = 1 ms-2 in direction of travel

Q.3.     A man owns an outboard motor that can exert a force of 500N. He can put in on a dingy that has a mass of 200 kg or a launch that has a mass of 1200 kg. If the frictional forces opposing motion are 90% of the forward forces applied, calculate the acceleration the motor can produce for each boat.

A.3.     Forward force applied in each case is 500N

Friction opposing forward force = 90% of 500N = 450N

Net forward force = 500 – 450 = 50N in each case

F = ma             a = F/m

adingy = 50/200 = 0.25 ms-2

alaunch = 50/1200 = 0.042 ms-2

Solve problems and analyse information involving F = mv2/r for vehicles travelling around curves

Q.1.     A car of mass 1 tonne is travelling at 20 ms-1 around a curve of radius 400m. What is the centripetal force between the tyres and the road?

A.1.     Fcentripetal = mv2 /r          = 1000 x 202 / 400       = 1000N towards the centre

Q.2.     The maximum frictional force between the tyres and the road for a particular car is equal to half the weight of the car.

What is the maximum speed that a car can travel around a curve of radius 200 metres?

A.2.     Fcentripetal = mv2 /r = ½ mg         v2/r = ½ g        v2 = rg/2          = (200 x 9.8) /2

v2 = 980           v = 31.3 ms-1

“3. Moving vehicles have kinetic energy and energy transformations are an important aspect in understanding motion” (syllabus)

Identify that a moving object possesses kinetic energy and that work done on that object can increase that energy” (syllabus)

Kinetic energy is energy of motion and is given by Ek = ½ mv2. The kinetic energy of a body is increased by making it go faster. Since a body will remain in its state of rest or uniform motion unless acted on by an external force then a force has to be applied.

As the force changes the displacement of the object, work is done on it according to the equation W = Fs. The amount of work done is equal to the change in kinetic energy. Both work and energy are measured in joules.

Describe the energy transformations that occur in collisions” (syllabus)

Energy is “lost” in collisions because it is transformed to other forms. If we can hear the collision then some of the energy is converted to sound energy. Objects also heat up in collisions and if the object changes its shape there is also energy of deformation.

Define the law of conservation of energy” (syllabus)

Energy can not be created nor destroyed but can be transferred from one body to another or transformed from one type to another. If a student is running down the hallway and collides with a teacher then energy is transferred from the student to the teacher. If you rub your hands together quickly you can feel them heat up. Friction transforms mechanical energy into heat energy.

Einstein took the conservation of energy a step further when he outlined his mass-energy equivalence. He proposed that mass can be converted to energy and vice-versa according to the equation E = mc2.

Solve problems and analyse information to determine the kinetic energy of a vehicle and the work done using the formulae: Ek = ½ mv2 and W = Fs” (syllabus)

Q.1. A car of mass 1.2 tonne is pushed with a force of 200N for 50 metres.

(i) How much work is done?

(ii) What is the change in kinetic energy of the car?

A.1.     (i) W = Fs = 200 x 50 = 10 000J or 10kJ

(ii) Work done = change in kinetic energy = 10kJ

Q.2. A car of mass 1.5 tonne doubles its speed from 5 ms-1 to 10 ms-1.

(i) By what factor does its kinetic energy change?

(ii) How much work is done?

A.2.     (i)         Let the original kinetic energy be E so that E = ½ mv2

As the speed doubles the new kinetic energy becomes ½ m(2v)2

= 2mv2 = 4(½ mv2) = 4E

Energy increases by a factor of 4.

(ii)        Work done = change in kinetic energy

DEk = Ek f – Ek I = ½ x 1500 x 102 – ½ x 1500 x 52

= 75000- 18750 = 56250J = 56.25 kJ

Analyse information to trace the energy transfers and transformation in collisions leading to irreversible distortions” (syllabus)

In a collision the most obvious energy transformation is from kinetic energy to sound energy. Heat is also produced. The panels of the car have a force acting on them through a displacement to push them out of shape. The kinetic energy supplies the work for distortion i.e. energy of distortion.

Energy can also be transferred from on vehicle to another in a collision.

In all cases the sum of all types of energy before the collision is equal to the sum of all types of energy after the collision.

“4. Change of momentum relates to the forces acting on the vehicle or the driver” (syllabus)

Define momentum as p = mv” (syllabus)

The momentum of a body is defined as the product of its mass and velocity. p = mv

Momentum is a vector and its direction must be specified.

The units of momentum are kgms-1 or Ns (these two units are identical)

Define impulse as the product of force and time” (syllabus)

The change in speed of an object such as a car is governed by both the magnitude of the force applied and how long it acts. The product of the force and the time for which it acts is known as impulse.

Impulse = force x time            Impulse = Ft

Also Impulse = Ft = mat

But a = So mat = = mv – mu = change in momentum

Explain why momentum is conserved in collisions in terms of Newton’s Third Law of motion” (syllabus)

Newton’s third law states that for every action there is an equal and opposite reaction. Consider a collision between two objects, A and B.

According to Newton’s third law, while the objects are in contact, FA = - FB

Suppose the objects are in contact for time t.

Then FA t = - FBt

i.e. the impulse of A is equal and opposite to the impulse of B

And since impulse is equal to the change of momentum

The change of momentum of A is equal and opposite to the change of momentum of B

So the total momentum remains constant i.e. momentum is conserved.

Solve problems and analyse secondary data using: p = mv and Impulse = Ft” (syllabus)

Q.1.     A car of mass 1200 kg is travelling at 10 ms-1 when a force of 200N acts for 20 seconds in the direction of motion.

(i)         What is the initial momentum of the car?

(ii)        What impulse acts on the car?

(iii)       What is the final momentum of the car?

(iv)       What is the final speed of the car?

A.1.     (i) p = mv = 1200 x 10 = 12000 kgms-1 in direction of motion

(ii) impulse = Ft = 200 x 20 = 4000Ns in direction of force]

(iii) impulse = Dp so final momentum = 12000 + 4000 = 16000kgms-1 in direction of motion

(iv) p = mv      v = p/m            = 16000/1200 = 13.33 ms-1

Perform first-hand investigations to gather data and analyse the change in momentum during collisions” (syllabus)

Purpose: To analyse the change in momentum during collisions.

Equipment: Curved track, retort stand, bosshead and clamp, 2 steel balls, carbon paper, paper, sticky tape, metre rule, cent-o-gram balance, permanent marker.

The curved track can be made from a plastic ruler with a grove down the middle or a length of plastic pipe that has been cut down the centre.

Preamble: If the steel ball leaves the bench in a horizontal direction, its horizontal velocity is uniform until it hits the floor, and the time of fall can be calculated from the formula:          t = where t = time of fall, h = height of bench, g = acceleration due to gravity.

Procedure: Use the permanent marker to label the steel balls “1” and “2” and find the mass of each ball by weighing it on the cent-o-gram balance. Record the masses in your physics practical book.

Set up the curved track on the edge of the bench as shown in the diagram above. Bend the curved track so that it is horizontal at the edge of the bench. This ensures that the balls leave the track in a horizontal direction. Place a mark near the top of the curved track. Roll ball “1” from this mark and note approximately where the ball hits the floor. Place a sheet of white paper at this point and secure it to the floor with sticky tape. Place a sheet of carbon paper, carbon side down, over the white paper and secure it with sticky tape. Roll ball “1” down the track 3 times, ensuring that it begins from the same mark each time. Remove the carbon paper. Carefully measure the distance from the bottom of the bench, directly underneath the edge of the curved track, to the carbon marks on the white paper. Record these distances in your physics practical book and determine the mean as the horizontal displacement. Carefully measure the vertical distance, h, from the edge of the track to the floor. Calculate the time of fall from t = Calculate the horizontal velocity of the ball from vx = s/t where vx is the horizontal velocity, s is the horizontal displacement and t is the time of fall.

Calculate the horizontal momentum of the ball from px = mvx Now carefully place the steel ball “2” at the bottom of the curved track, as close as possible to the edge of the bench. Roll ball “1” from the mark and note where both balls hit the ground. Place white paper at each of these positions and secure it in position with sticky tape. Place carbon paper, carbon side down, over the white paper and secure it with sticky tape. Again place ball 2 at the edge of the track and roll ball 1 from the mark. Remove the carbon paper and measure the forward displacement of each of the balls. Replace the carbon paper and repeat the procedure twice.

Sample Results: Mass of ball “1” = 57.4g, mass of ball “2” = 51.8g

Height of bench = 92.7 cm

Time to fall t = = = 0.435s

Forward displacements of ball 1 = 59.1cm, 58.4cm, 58.6cm. Average = 58.7 cm

Horizontal velocity of ball 1 on leaving bench = s/t = 0.587/0.435 = 1.35 ms-1

Momentum of ball 1 on leaving bench = mv = 0.0574 x 1.35 = 0.0775 Ns forward

Displacements of balls after collisions

 Displacement of Ball 1 (cm) Displacement of Ball 2 (cm) Collision 1 16.7 39.6 Collision 2 17.2 40.2 Collision 3 17.8 40.1

Momentum after collision:

Collision 1: p = m1v1 + m2v      = m1s1/t + m2s2/t

= (0.0574 x 0.167)/0.435 + (0.0518 x 0.396)/0.435

= 0.0220  + 0.0472 = 0.0692 Ns forward

Collision 2: p = m1v1 + m2v      = m1s1/t + m2s2/t

= (0.0574 x 0.172)/0.435 + (0.0518 x 0.402)/0.435

= 0.0227  + 0.0479 = 0.0706 Ns forward

Collision 3: p = m1v1 + m2v      = m1s1/t + m2s2/t

= (0.0574 x 0.178)/0.435 + (0.0518 x 0.401)/0.435

= 0.0235  + 0.0478 = 0.0713 Ns forward

Average momentum after collision = (0.0692 + 0.0706 + 0.0713)/3

= 0.0704 Ns forward

Average loss of momentum during collisions =0.0775 -0.0713 = 0.0062 Ns

% loss of momentum during collisions = (0.0062/0.0775) x 100 = 8%

Conclusion: The experiment verifies the principle of conservation of momentum i.e. the total momentum before the collision is equal to the total momentum after the collision. Errors could include friction and air resistance, the rotational motion of ball 2 after collision, limit of reading of the measuring instruments and energy and momentum absorbed by the apparatus during the experiment.

Solve problems that apply the principle of conservation of momentum to qualitatively and quantitatively describe the collision of a moving vehicle with:

-          a stationary vehicle

-          an immoveable object

-          another vehicle moving in the opposite direction

-          another vehicle moving in the same direction” (syllabus)

Stationary vehicles: When a car crashes into a stationary vehicle several things happen to reduce the severity of the impact on the occupants. Firstly the crumple zone of the moving vehicle collapses, increasing the time it takes to come to rest. Secondly the panels of the stationary vehicle distort and this further increases the distance and so the time over which the moving vehicle comes to rest. Thirdly the stationary vehicle gains some of the momentum of the moving vehicle and moves. This allows the moving vehicle to keep moving but at a slower rate. This further increases the time for the moving vehicle to come to rest and reduces the force.

Q. A car of mass 1200 kg travelling at 10 ms-1 crashes into the back of a stationary car of mass 1500 kg. The stationary car lurches forward at a speed of 5 ms-1. Find the speed of the moving vehicle immediately after the collision.

A.        m1u1 + m2u2 = m1v1+ m2v2

1200 x 10 + 0 = 1200 v1 + 1500 x 5

12000 = 1200v1 + 7500

4500 = 1200v1

v1 = 4500/1200 = 3.75

The moving vehicle will move forward at 3.75 ms-1.

Immoveable object: When a car crashes into an immovable object there is still the crumple zone of the moving vehicle. However the panels of the immoveable object don’t deform and the immoveable object does not move forward. This means that a collision with an immoveable object such as a telegraph pole is much more dangerous than a collision with a stationary vehicle.

Also in the case of a telegraph pole the force is spread over a smaller area so the pressure is greater and the force is greater over the area of the impact.

Q. A 1 tonne car travelling at 10 ms-1 hits a telegraph pole and comes to rest in 0.05 second. What force does the telegraph pole exert on the car?

A. F = Dp/t      = (1000 x 10) /0 .05 = 2 x 105 N against motion.

Head-on Collision: It is sometimes said that a head-on collision between two cars travelling at 60 km/h is like hitting a brick wall at 120 km/h. If a car hits a brick wall it comes to rest over the length of its crumple zone. If two cars collide head-on they come to rest over the length of two crumple zones. Assuming they come to rest at the mid point of the two crumple zones then each car comes to rest over the length of one crumple zone. This means that in a head-on collision between two cars each travelling at 60 km/h is like each of them hitting a brick wall at 60 km/h, not 120 km/h.

Q. Two cars, each of mass 1200 kg travelling at 15 ms-1 (54 km/h) collide head on and come to rest in 0.04s. A newspaper reporter said that it was like each car hitting a brick wall at 108 km/h.

(i) What was the change of momentum of each car?

(ii) What was the change of kinetic energy of each car?

(iii)What average force did each car exert on the other one?

(iv) By what factor did the reporter overstate the change in momentum?

(v)   By what factor did the reporter overstate the change in kinetic energy?

A.        (i) Dp = mDv   = 1200 x 15 = 18000 Ns against motion

(ii) Ek = ½ mv2 = ½ x 1200 x 152 = 135 000 J

(iii) F = Dp/t    = 18000/0.04   = 450 000N against motion.

(iv) x2. Doubling the speed doubles the momentum.

(v) x4. Doubling the speed means we have (2v)2 i.e. 4v2 and 4 x Ek

Same direction: When two cars travelling in the same direction collide the speed of collision can be taken as the difference between the two speeds i.e. the speed of one car relative to the other. Such collisions are not normally very dangerous provided both cars are travelling at approximately the same speed.

Q. A utility of mass 1500 kg travelling at 20 ms-1 runs into the back of a car of mass 1200 kg travelling at 15 ms-1. If the two vehicles coalesce (join together) what is the speed of the two vehicles after the collision.

A.        m1u1 + m2u2 = m1v1+ m2v2

1500 x 20 + 1200 x 15 = (1200 + 1500)v

30000 + 18000 = 2700v

v = 48000/2700 = 17.78 ms-1

“5. Safety devices are utilised to reduce the effects of changing momentum” (syllabus)

Define the inertia of a vehicle as its tendency to remain in uniform motion or at rest” (syllabus)

Newton’s first law is called the law of inertia. The inertia of a vehicle refers to its tendency to remain at rest or in uniform motion unless acted on by an external force.

Discuss reasons why Newton’s First Law of Motion is not apparent in many real world situations” (syllabus)

While the first part of Newton’s first law seems apparent (we don’t expect something at rest to suddenly start moving of its own accord) the second part is not as obvious. The reason objects do not remain in uniform motion is that it is impossible to completely eliminate friction and friction always acts to oppose the motion. Consequently any object in motion will be slowed down by friction between itself and the surface over which it travels and by the resistance of the air that it has to push out of its way.

Assess the reasons 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” (syllabus)

Low speed zones mean that vehicles travelling at low speed have less momentum than faster vehicles. Consequently in a collision their change in momentum is less and the occupants are subjected to less force. Also there is a big reduction in kinetic energy when vehicles travel more slowly. A car travelling at 40 km/h has less than half of the kinetic energy that it would have if travelling at 60 km/h

The key to road safety is to remember that force is equal to the rate of change of momentum and if we can increase the time over which something comes to rest we reduce the force. Whether a car comes to rest suddenly (a collision) or gradually (gentle application of the brakes) the impulse or change of momentum is the same. Crumple zones increase the time it takes for a car in a collision to come to rest and so the change of momentum occurs over a longer time and reduces the force.

Air bags spread the force over a bigger area of the body and so reduce the pressure on points of impact. Since pressure is equal to force per unit area it means that the maximum force to any part of the body is reduced.

Evaluate the effectiveness of some safety features of motor vehicles” (syllabus)

The number of deaths in Australia due to road accidents is about 60% of what it was 30 years ago. This is despite a huge increase in the number of motor vehicles on the road. The reduction in road deaths can be attributed to three factors

(i)                 improvements in car safety features

(ii)               improvement in road design and construction

(iii)             safety legislation and its enforcement

By far the most effective improvement in car safety features was the inclusion of seat belts in all new cars and legislation to make wearing them compulsory. This occurred in the 1970s. In the event of an accident seat belts restrain the body and stretch a little as they bring the body to rest. Without a seat belt the occupant would be thrown forward and would be brought to rest by colliding with the windscreen or other parts of the cars interior. In some cases the windscreen would break and the occupant would be thrown from the vehicle.

Another safety feature in car design is the inclusion of crumple zones. This means that in an accident the front of the car collapses, the engine is pushed down, underneath the car and the steering column collapses and tilts forward so that it will not impale the driver. While older model cars are more rigid and suffer less damage in a comparable collision, their rigidity brings them to rest much more quickly in a collision and because of this they are subject to larger forces. Air bags are designed to inflate almost immediately in a front end collision. Air bags spread the area over which the force is spread, reducing the pressure. They also increase the time it takes to bring persons to rest and reduce the force.

Headrests prevent the head being thrown back in a collision from behind. When a car is struck from behind the car and its seats are suddenly thrown forward. For a seat without headrests (such as the rear seat of most vehicles) the seat will push the shoulders forward but inertia will cause the head to lag behind. The head swing back suddenly causing neck injury called whiplash.

ABS brakes (Antilock Braking Systems) stop the brakes from locking and subsequent skidding of the wheels when the brakes are applied hard. ABS brakes have sensors which detect if a wheel has started to lock and reduces the pressure to that wheel until it rotates again.

Road construction. The amount of dual carriageway has increased in recent years and this overcomes the risk of head-on collisions. Also work has proceeded in eliminating dangerous bends and traffic lights have been introduced at many dangerous intersections.

Legislation such as random breath testing and reduction of speed limits has als helped reduce road deaths.

Seat Belts

http://auto.howstuffworks.com/seatbelt.htm

http://www.lemurzone.com/airbag/belts.htm

http://www.clearview.nrma.com/pub/nrma/motor/car

research/safe_seating/index.shtml

http://www.pbs.org/wgbh/nova/escape/timecar.html#seatbelt

http://www.nhtsa.dot.gov/kids/research/seatbelt/index.html

Gather and process first-hand data and/or secondary information to analyse the potential danger presented by loose objects in a vehicle” (syllabus)

According to Newton’s first law a body in motion will remain in uniform motion unless acted on by an external force. This means that unrestrained objects in a vehicle will be thrown forward if the vehicle stops suddenly. For instance if a car is involved in a minor rear end collision, objects such as an umbrella or groceries that are stored on the shelf behind the back seat can become airborne missiles and can cause serious injury or even death.

Station wagons or panel vans are a particular problem. While wire mesh screens are available that will protect the front seat occupants from being struck by the cargo in the case of an accident, very few wagons and vans have them fitted. Since these vehicles are often used to carry rather heavy items such as building materials, computer monitors and sound systems, a sudden stop in an accident can result in the front seat occupants being struck with heavy lethal objects.

Purpose: To demonstrate the dangers associated with carrying unrestrained objects in motor vehicles.

Equipment: Dynamics cart, Plastic mesh such as used for bags of oranges, egg, rubber bands, 1 kg mass. Procedure: Place the egg on the dynamics cart and cover it with plastic mesh, allowing a generous overlap all around. Use rubber bands to securely tie the plastic mesh to the cart. Place the dynamics cart on the floor and place a 1 kg mass on the cart, about 10 cm behind the egg. Push the cart into a brick wall and note what happens to the unrestrained mass.

Q.1.     How does this situation represent the danger to the occupants of a vehicle containing unrestrained objects?

Q.2.     Would you feel comfortable carrying a small fridge in the back of a station wagon? Justify your answer.

Identify data sources, gather, process, analyse present secondary information and use the available evidence to assess benefits of technologies for avoiding or reducing the effect of a collision” (syllabus)

Seat Belts

http://auto.howstuffworks.com/seatbelt.htm

http://www.lemurzone.com/airbag/belts.htm

http://www.clearview.nrma.com/pub/nrma/motor/car

research/safe_seating/index.shtml

http://www.pbs.org/wgbh/nova/escape/timecar.html#seatbelt

http://www.nhtsa.dot.gov/kids/research/seatbelt/index.html