Physics G481


Work and conservation of energy

Candidates should be able to: (a) define work done by a force;

Work is the force multipied by the distance moved in the direction of the force.

(b) define the joule;

1 Joule is the work done when a force of 1 newton moves its point of application 1 metre in the direction of the force in (1 joule = 1 newton metre)

(c) calculate the work done by a force using W = Fx and W = Fx cos θ;

work = f x 

work = fx cos θ

where f is the force  x is the distance moved i the direction of the force and theta is the angle between the force and the direction of travel

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Work and conservation of energy

(d) state the principle of conservation of energy; 

Energy is the stored ability to do work                                                                                  The Principle of conservation of energy : in any closed system, energy may be converted from one form to another but cannot be created of destroyed.

Conservation of energy is never 100% efficient and sankey diagrams illustrate this perfectly and they show that no energy is lost.

(e) describe examples of energy in different forms, its conversion and conservation, and apply the principle of energy conservation to simple examples .                                                                                                                Basic energy is kinetic :wheere moment is taking place or Potential energy:regions where electric,magnetic,gravitational and nuclear charges exist. Then : chemical energy (energy can be released by altering the arrangemnt of atoms)      Electrical P.D:(electrical energy), EM energy,GPE,Internal energy (heat),Nuclear energy and sound energy

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Kinetic and potential energies

Candidates should be able to: (a) select and apply the equation for kinetic energy e=0.5c mv^2

Kinetic energy is the work an object can do by virtue of its speed and it is calculated by E=0.5 x mv^2

(b) apply the definition of work done to derive the equation for the change in gravitational potential energy;

GPE is the energy stored in an object by virtue of its position in a gravitational field so for a falling person work = weight x distance (height) because w=mg we end up with GPE = mgh

(e) apply the principle of conservation of energy to determine the speed of an object falling in the Earth’s gravitational field. 

Because of the conservation principle when an object loses GPE they gain KE so mgh=0.5 x mv^2 ----> v^2 = 2gh

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Candidates should be able to: (a) define power as the rate of work done;

Power is the rate of doing work (work done /time taken) in joules or watts .                       1kW=1000watts

(b) define the watt;

one watt is equal to one joule per second.

(c) state that the efficiency of a device is always less than 100% because of heat losses;

efficiency = (useful output power/total input power)x100

it is never 100% because energy is lost as heat 

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Behaviour of springs and materials

 (a) describe how deformation is caused by a force in one dimension and can be tensile or compressive Tensile forces - cause tension in the object and are forces that stretch objects such as wires ,spings and rubber bands.Two equal and opposite forces required Comprehensive forces - Forces that reduce the length by squeezing the object   assuming that the object is not accelerating ,two equal and opposite forces are required

(b) describe the behaviour of springs and wires in terms of force, extension, elastic limit, Hooke’s law and the force constant (ie force per unit extension or compression); Hooke's law - The extension of an elastic body is proportional to the force that causes it as long as the elastic limit is not exceeded.

(c) select and apply the equation F = kx, where k is the force constant of the spring or the wire;  F=kx where x is the extenstion and k is the force/spring constant in newtons per metre k is how much force is required per unit extension

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Behaviour of springs and materials

(d) determine the area under a force against extension (or compression) graph to find the work done by the force;

Work done = 0.5x fx  or 0.5 x kx^2                                                                                    This is the area under the force/extension graph.

(f) define and use the terms stress, strain, Young modulus and ultimate tensile strength (breaking stress);

Stress is the force per cross sectional area  it is measured in newtons per square metre or pascals                                                                                                          Strain is the extension per unit length, does not have units because its length/length        Young Modulus - the ratio between stress and strain in a material,in pascals. This is because stress on a material causes strain                                                                        Ultimate Tensile strength - Maximum stress that can be applied to amaetrial before it breaks

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Behaviour of springs and materials

(g) describe an experiment to determine the Young modulus of a metal in the form of a wire;

Experiment to stretch wire: copper wire held firmly in a clamp stand at one end and the other end has a hanger weight after passing over a pulley. The hanger must be heavy enough to keep the wire taut. A marker is added to the wire and a ruler in a fixed position to mark the position of the mark as the wire is stretched.The gradient of the straight line section of the force /extension graph is needed .The length of the wire section for which the extension is being measured is needed. The diameter of the wire is needed to calculate the cross sectional area.                                                                          Young modulus = (f/x x I/A)

(h) define the terms elastic deformation and plastic deformation of a material; When a distorting force is applied to an object the object exhibits

-          Elastic behaviour if it regains its original shape                                             -   -   -          Plastic behaviour if it is permanently distorted

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Behaviour of springs and materials

(i) describe the shapes of the stress against strain graphs for typical ductile, brittle and polymeric materials.

Ductile materials: copper                                                                                                 Can be drawn out into a wire and has a large plastic region. As you pull it increases in length straightens and decreases in the cross sectional area and breaks. Where force is constant the decrease in the area under the graph shows increased stress. The graph has an elastic region and a large plastic region and it will continue to stretch in that region even if stress is reduced the maximum stress that can be applied to it before it breaks is called the ultimate tensile strength.

Brittle materials : concrete                                                                                           Distort very little and snap when subjected to a lot of stress so the graph ceases abruptly .Area under graph is small so it has little elastic potential energy

Polymetric materials :plastic                                                                                           May reach strain of 300 % without snapping but wont stretch .Molecules arrange in squashed long chains when stretching you are straightening out the hains so large strain .Flexible . Used for car tyres .Mixed with impurities such as sulphur to bind the chains of rubber molecules together making them harder and stronger and less easy to stretch in vulcanisation

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Behaviour of springs and materials


Uncertainty in Young modulus experiment

-          Measure extension more accurately by using a travelling microscope .Accurate to the hundredth of a millimetre

-          Use the s.w.g, standard wire gauge to find the cross section area so we do not have to calculate and squaring a value doubles the uncertainty

-          Measure wire form marker to clamp

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Candidates should be able to: (a) explain that some physical quantities consist of a numerical magnitude and a unit; 

Physical quantities have a unit too eg length , area , volume

(b) use correctly the named units listed in this specification as appropriate; 

 Mass - kg , length - m ,time - s, temperature - K , current - A, amount of a substane-mol

(c) use correctly the following prefixes and their symbols to indicate decimal sub-multiples or multiples of units: pico (p), nano (n), micro (µ), milli (m), centi (c), kilo (k), mega (M), giga (G), tera (T); 

10^-12 = pico(p) , 10^-9=nano(n) , 10^-6=micro (µ), 10^-3=milli(m) , 10^-2=centi (c), 10^3=kilo(k) ,10^6=mega(M), 10^9=giga(G), 10^12=tera(T)


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(a) define scalar and vector quantities and give examples

Scalar quantities : ones that have a magnitude but not direction. such as density ,temperature, pressure, potential difference and wavelength                                                                                     Vector quantities: have a magnitude and a direction such as displacement, velocity, acceleration, force, impulse, momentum and currect

(b) draw and use a vector triangle to determine the resultant of two coplanar vectors such as displacement, velocity and force;

You can use vector teiangles to determine a resultant using simple arithmetic like A+B=C

(c) calculate the resultant of two perpendicular vectors such as displacement, velocity and force;

When two vectors are perpendicular we can use pythagoras to find the resultant

(d) resolve a vector such as displacement, velocity and force into two perpendicular components

Have to generate two equivalent vectors from one vector, a horizontal and vertical one. To find the horizontal use Vsin x and vertical use Vcosx but use common sense and apply trigonometry appropriately

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(a) define displacement, instantaneous speed, average speed, velocity and acceleration; 

Displacement : the distance moved in a stated direction                                                                      Instantaneous speed: the speed at a given instant of time Average speed : the distance travelled per unit time Velocity : the displacement per unit time Acceleration : the rate of change of velocity (b) select and use the relationships average speed = distance /time ,acceleration = change in velocity / time to solve problems; Speed = distance/time and acceleration =  change in velocity/time c) apply graphical methods to represent displacement, speed, velocity and acceleration;  Displacement time graph –a straight line is constant velocity is the gradient ,Horizontal line = no displacement so stationary if graph comes back down moving backwards  Since gradient is velocity if graph curves it is instantaneous velocity  The upward curving represents acceleration and getting less steep is decelerationDisplacement time graph –a straight line is constant velocity is the gradient  Horizontal line = no displacement so stationary   If graph comes back down moving backwards  Since gradient is velocity if graph curves it is instantaneous velocity .The upward curving represents acceleration and getting less steep is deceleration (see picture in notes)

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Velocity-time Graph shows the change in velocity gradient shows acceleration area under graph is displacement

(d) determine velocity from the gradient of a displacement against time graph; 

(e) determine displacement from the area under a velocity against time graph; 

(f) determine acceleration from the gradient of a 

velocity against time graph. 

Gradient of a displacement time graph = velocity Area under velocity time graph = displacement Gradient of velocity time graph = acceleration

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(a) derive the equations of motion for constant acceleration in a straight line from a velocity against time graph; 

Velocity time graph – to find displacement area under graph

Area of triangle  =1/2v-u x t                                                                                                     Because v-u=at ,area = 1/2at2                                                                                                                                     Area of rectangle =ut                                                                                                                      so  S = ut + 1/2at2                                                                                                                                                                               - Distance = speed x time    S = (u+v)/2 x t

(b) Select and use the equations of motion for constant acceleration in a straight line: 

v = u + at without s

s=1/2(u+v)t without a

s=ut + 1/2at2 without v

s= vt-1/2at2 without u

v2=u2+2as without t

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c) apply the equations for constant acceleration in a straight line, including the motion of bodies falling in the Earth’s uniform gravitational field without air resistance;

Assuming that gravity is constant and air resistance has a negligible effect                          Acceleration of free fall is vertically down towards the centre of the earth                                           An object thrown upwards still has the constant acceleration g downwards

(d) explain how experiments carried out by Galileo overturned Aristotle’s ideas of motion;

Aristostle thought that heavier objects fall faster than lighter .Galileo dropped balls of equal surface area and different mass from the tower they hit the ground at the same time showing that acceleration is constant no matter what mass

(e) describe an experiment to determine the acceleration of free fall g using a falling body

The trap door and electromagenetic method Electromagnet supports a steel ball .When the current through the magnet is stopped the ball starts to fall and an electronic clock is triggered.the balls falls onto a trap door and when it breaks it open the clock is stopped .The distance the ball drops is measured and the time is taken from the clock . Repeat and get average. Theory = s=ut + ½ at2 ,U=0 ,so s = 1/2gt2 so g = 2s/t2 ,The graph of s and t2 will have g/2 gradient

Notes** If magnet’s current is too large it will delay releasing the ball after current is switched off and clock is triggered –adjust current so it just supports the ball                                                      If distance is too large or ball is too small air resistance ,might have an effect                      Accurately measure the distance from the bottom of the ball to the top of the trap door.

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(f) apply the equations of constant acceleration to describe and explain the motion of an object due to a uniform velocity in one direction and a constant acceleration in a perpendicular direction

For projectiles separate the vertical and horizontal components

Horizontal velocity is constant

Then apply the appropriate equations

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Forces in action

(a) Solve problems using the relationship: net force = mass × acceleration (F = ma) appreciating that acceleration and the net force are always in the same direction;

Force is a push or pull such as weight, drag, tension and thrust                                             Gravitational force: between two objects with a mass                                                                       Magnetic force: between two magnetic objects                                                                                 Electrical force:betweencharged object                                                                                           When an object has a zero resultant force it does not accelerate, force causes acceleration, acceleration does not cause force

So acceleration is proportional to force if the mass is constant and acceleration is inversely proportional to the mass if the force is constant                                                                                  so F=kma where k is a constant

 (b) define the newton;

One newton : the force that causes a mass on one kilogramto have an acceleration of one metre per second every second

1N=k x 1kg x 1ms^-2 = F = ma

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Forces in action

(c) apply the equations for constant acceleration and F = ma to analyse the motion of objects;     (d) recall that according to the special theory of relativity, F = ma cannot be used for a particle travelling at very high speeds because its mass increases.

Einsteins special theory of releativity introduced the idea that at speeds approachign the speed of light the mass of an object increases so F=ma is invalid at high speeds because the mass changes.

Non linear motion

(a) explain that an object travelling in a fluid experiences a resistive or a frictional force known as drag; 

If the force causing acceleration increseas then the acceleration increseas as well. The resistive forces increase as the velocity increases so the resultant force decreseas

(b) state the factors that affect the magnitude of the drag force

Drag depends on : velocity, roughness of surface, cross sectional area and the shape of the oject

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Forces in action

(c) determine the acceleration of an object in the presence of drag;

(g) use and explain the term terminal velocity

When an object falls from a great height through the air drag increases on it as it accelerates, there will reach a point when the drag becomes equal to the weightso the resultant forceon the obect is zero.

At terminal velocity the object travels at a constant velocity.

(d) state that the weight of an object is the gravitational force acting on the object; 

Weight : the gravitational force acting on an object in newtons (e) select and use the relationship: weight = mass × acceleration of free fall because F=ma weight = gravity x mass so w=mg where g is 9.81

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Forces in action

(f) describe the motion of bodies falling in a uniform gravitational field with drag;

As a skydiver jumps out the only force acting on him is weight at that particular momement and he accelerates 

As he continues to fall the drag increases ad he will reach a point where the drag equals the weight and he is no longer accelerating


(a) draw and use a triangle of forces to represent the equilibrium of three forces acting at a point in an object; 

Use triangles similar to vector triangles to add up forces  (b) state that the centre of gravity of an object is a point where the entire weight of an object appears to act; The centre of gravity of an object is the point where the entire weight of the object can be considered to act as a single force

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Forces in action

(c) describe a simple experiment to determine the centre of gravity of an object;

- support the object on a wire passed through a small hole at the edge of the oblect. Hang a plumb plumb line from the wire to show the vertical and mark the line of the string as the centre of gravity must be on it. Hang the plumb line from the wire from another small hole on the edge and mark this line, where the two lines intersect in the centre of gravity

(d) explain that a couple is a pair of forces that tends to produce rotation only; 

A couple is a pair of equal and parallel but oppossite forces which tend to produce rotation only.

 (e) define and apply the torque of a couple;

Torque of a couple describe the turning effect of a couple. one of the forces multipied by the perpendicular distance between the forces in newton metres

(f) define and apply the moment of force;

The moment of a force is the force multiplied by the perpendicular  distance from a stated point in newton metres.

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Forces in action

(g) explain that both the net force and net moment on an extended object in equilibrium is zero; 

the net force and net moments must be zero to be in equilibrium.

(h) apply the principle of moments to solve problems, including the human forearm;

The principle of moments states that for a body in rotational equilibrium the sum of the clockwise momemnts equals the sum of the anticlockwise moments.

moment =fd

(i) select and use the equation for density: 

density = mass/volume  p=m/Vnin kgm^-3

(j) select and use the equation for pressure where F is the force normal to the area A

Pressure = force/area in pascals

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Forces in action

Candidates should be able to: (a) define thinking distance, braking distance and stopping distance

(b) analyse and solve problems using the terms thinking distance, braking distance and stopping distance

Thinking distance is the distance travelled after seeing the need to stop and applying the brakes          Braking distance is the distance travelled after applying the brakes and the car comng to a stop         Stopping distance is the thinking and braking distance combined

(c) describe the factors that affect thinking distance and braking distance;

factors that affect thinking distances : Time of day, how quickly you notice the hazard, your mental capabilities (are you drunk?) and concentration level

Baking distance : the surface of the road, the mass of the car and the efficiency of the brakes

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Forces in action

d) describe and explain how air bags, seat belts and crumple zones in cars reduce impact forces in accidents;

kinetic energy = braking force x braking distance

Crumple zones : a part of the car designed to collapse during a collision .They increase the distance over which the force is acting so the average force is less.The front of the car crumples and stops but the passengers move an extra half a metre

Seat belts : increase the distance over which the forces is felt as they stretch.without a seatbelt you would continue to move after a collision because of your mass and would be stopped by the forces exerted by the windscreen or a rigid part of the car.

Airbags: should be fully inflated before you hit them, without a seaatbelt they wouldnt be fully inflated before you hit them.You can harm yourself as you would be travelling so fast towards an airbag tha is travelling fast torwards you.

(e) describe how air bags work, including the triggering mechanism; 

a flexible nylon bag is folded into the steering wheel or dashboard. An accelerometer , when the front end of the spring is suddenly stopped in a collison the mass on the end of the spring continues to mive forward and makes contact witha switch that starts a chemical reaction between sodium nitride and potassium nitrate to produce nitrogen that quickly fills the airbag. This only occurs when the acceleration is 10g (during accidents)

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Forces in action

(f) describe how the trilateration technique is used in GPS (global positioning system) for cars.

GPS - allows us to know where we are on the earth within a distance of baout 10m

Trilateration -3 satelites used

satelite a sends out a signaland it arrives after a known time at the GPS receiver then we can find the distance of the receiver from the satelite given thespeed of the electromagnetic radation.this can be repeated over two more satellites to fix the position of the receiver and the in car computer can plot this position and guide the car.

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