year 11 science

Year 11 Science

Science 1.1Demonstrate an understanding of aspects

of mechanics in one dimension.

SLO’s

You will be able to give reasons why phenomena, concepts or principles relate to given situations in the context of;

• Distance, speed, interpretation of distance and speed time graphs, average acceleration and deceleration in the context of everyday experiences such as journeys, sport, getting going, etc. Using v =Δd/Δt and a= Δv/ Δt

• Mass, weight and the acceleration due to gravity, balanced and unbalanced forces, in the context of everyday experiences such as being stationary, moving at constant speed, accelerating, etc. The relationship F net = ma.

• Force and pressure in the context of everyday experiences. The relationship P =F/A

• Work and power, gravitational potential energy, kinetic energy, and the conservation of mechanical energy in free fall situations in the context of everyday experiences such as sports performance, dropping things, tossing balls, etc. The relationships Δ EP = mgΔh, EK = ½ mv2, W=Fd and P= W/t.

Units you will need to knowName Description Symbol Unit Unit symbol

Distance How far it moves

d Metre M

Speed How fast v Metres per second

ms-1

Acceleration How speed changes

a Metres per second per second

ms-2

Time How long it takes

t Seconds S

Mass How much matter

m Kilogram Kg

Force How much push/pull

F Newton N

Kinetic Energy

Energy of moving objects

Ek Joule J

Potential energy

Energy stored in objects

Ep Joule J

Power How fast it works P Watt W

Gravitational acceleration

The force which pulls objects to the centre of Earth

g 10N kg-1

Height h

Height Distance off the ground of an object

h meters M

Work What is done when a force moves an object

W Joule J

Area Space an object takes up

A Centimetres squared.

cm2

Pressure The amount of force applied to an area

p Newton’s per metre squared

Nm-2

Weight Mass with the force of gravity acting upon it

w Newton’s N

Key formula’s.

Distance and speed.• Velocity = change in distance v= Δd change in time Δt

• Acceleration= change in speed a= Δv change in time Δt

Mass and weight• Net force = mass x acceleration Fnet = ma

Force and pressure• Pressure = force P= F area A Work and power• Potential energy = mass x gravity x height ΔEP = mgΔh

• Kinetic energy= ½ mass x speed squared Ek = ½ mv2

• Work (energy)= force x distance W=Fd Ew = Fd• Power = work time P = W/t

Using a formula.

• How do we know which formula to choose?• Read the question; look for the data given,

find the equation which fits the data and the question.

• Sometimes you will need to use 2 or 3 formulae.

• What if we need to rearrange the formula?• Put the equation into a ∆ and look for the data

point you want to calculate.

Writing it out.

• When you answer any exam question;1. Write out the formula.2. Substitute the data.3. Write out the answer with units.

W=mgW= 50 x 10W= 500N

Review

% %X

The car starts at rest and travels a distance of 6m in 3s, what is its average speed?Formula; _____________________Answer; ____________________________________________________________ The cars mass is 400g. The car then travels a further 28m at a constant speed of 4ms-1 for 7s, how much kinetic energy does the car have?Formula; _____________________Answer; ____________________________________________________________ An athlete runs for 8s at a speed of 14ms-1, calculate their acceleration.Formula; _____________________Answer; ____________________________________________________________ 

The athlete then does some weight training; he lifts a 100kg dumbbell, what is the weight force of the dumbbell?Formula; _____________________Answer; ____________________________________________________________ The athlete lifts the dumbbell above his head 1.5m, what is the work being done?Formula; _____________________Answer; ____________________________________________________________ While the dumbbell is in mid-air it has gravitational potential energy, calculate how much energy is created.Formula; _____________________Answer; ____________________________________________________________

 While he is holding the dumbbell, he is standing on both of his feet. His feet have an area of 459cm2 then combined weight of the athlete and the dumbbells is 160kg, what pressure is being exerted?Formula; _____________________Answer; ____________________________________________________________ The athlete finishes off his workout by cycling; he accelerates at 1.5ms-2 over a distance of 25m, he and the bike have a mass of 70kg. Calculate the net force acting upon the bike when it is accelerating.Formula; _____________________Answer; ____________________________________________________________ 

If it takes 5s to cover the distance of 25m, calculate the power output of the bike during this time.Formula; _____________________Answer; ____________________________________________________________ Formula; _____________________Answer; ____________________________________________________________ Formula; _____________________Answer; ____________________________________________________________

Distance & Time• Distance is a measurement of how far apart two points are.• The unit for distance is the metre (m), or kilometre (km).• Displacement is a measurement of how far an object has moved

from its starting point.• The equation is:

displacement = final – initial distance distance

• Time is a measurement of the duration of an event.• The unit for time is the second (s), or

hour (h).

d/tMotion Time (s)

Walk 22.15

Skip 7.73

Run 6.00

D/t graph.

Time (s)

Dis

tanc

e (m

)

Motion of the mini.Distance (cm) Time (s) Speed (cms-1)

20 1.6

30 1.8

40 2.1

50 2.2

60 2.5

70 2.6

80 2.8

90 3.5

100 4.0

d/t carsDistance (cm)

Time 1 (s) Time 2 (s) Time 3 (s) Av time Speed (ms-1)

10

15

20

25

30

35

40

45

50

55

60

Speed & Velocity• An object has speed when it travels a distance in a time interval.• At any moment in time, a moving object has instantaneous speed.

Since this is difficult to calculate, we usually use average speed.• Since velocity is speed in a given direction, the equation for average

velocity is: final distance – initial distance

velocity = final time – initial time ∆d

vav = ∆t

• The unit for speed or velocity is metres per second (ms-1) or kilometres per hour (kmh-1).

• Pg 8

Distance/time Graphs

• Time always goes along the horizontal (x) axis• Distance always goes up the vertical (y) axis• A flat line (slope = 0) means

the object is stationary• A slope means the object is

moving. The slope gives thespeed of the objectA straight line indicatesa constant speed.

• A curved line means theobject is accelerating(speeding up or slowingdown)

• Pg11

d

td

t

d

t

What are these objects doing?

d

t

d

t

d

t

d

t

Calculating distance.Time (s) 1 2 3 4

Distance (m) 2.5 5 7.5 10 25m

Time (s) 1 2 3 4 5

Distance (m)

5 10 15 20 25 75m

Speed/time Graphs

• Time always goes along the horizontal (x) axis• Speed or velocity always goes up the vertical (y) axis• A flat line (slope = 0) means

the object is travelling at a constant speed(no acceleration)

• A slope means the object isaccelerating. The slope givesthe acceleration of the object

• A curved line means therate of acceleration is increasing or decreasing

• Pg14

v

tv

tv

t

2cm 3cm 4cmEach is 0.1s2/0.1 = 20ms-1

3/0.1 = 30ms-1

4/0.1 = 40ms-1

s/tTime (min) Speed (ms-2)

0 0

1 5

2 10

3 15

4 20

5 20

6 20

7 20

8 20

9 5

10 0

1 2 3 4 5 6 7 8 9 10 110

5

10

15

20

25

Series1

Area under graphs• The area under a velocity/time graph can be used to calculate the

total displacement of an object• For a simple straight line, simple calculate the area of the rectangle

or triangle under the line• For more complex shapes, split them into rectangles and triangles

and add together the areas for each shape to calculate the total displacementeg. total displacement = ½ base x height

v

t

Area under the graph- 1 shape

½ base x height

V

t

Area under the graph- 2 shapes

Base x height

½ base x height

Then add them together

V

t

Area under the graph- more than 2 shapes.

V

t

Acceleration• An object changing its speed is said to be accelerating. If the

acceleration is:– positive (eg. 2ms-2) = object speeding up– negative (eg. -2ms-2) = object slowing down

• The equation is:final speed – initial speed

acceleration = time taken ∆v

a = ∆t

• The unit for acceleration is metres per second squared (or ms-2).• Pg 9

Deceleration• So deceleration is simple an object slowing down.• The same formula is used, but the result will be a negative number.

• Pg 13, 14

Forces• A force is a push, pull, or twist.• Forces can change the:

– speed– direction, and/or the– shape of an object.

• The unit for force is the Newton (N)• When a force acts on an object (the action force), an opposing force

appears as a reaction force (eg. gravity or friction). eg.

driving force friction force(from engine) (from road/air)

Pg 16

Thrust.

• The force that moves objects in the direction of the force.

• The greater the thrust, the faster the movement.

Thrust

Friction

• Friction is a force that opposes motion.• Friction always works in the opposite direction

to thrust.• Friction between an object and air or water is

called drag.

Pg 17

Friction

Friction• Friction is a force that opposes motion.• It is created when objects rub against each other, releasing energy

as heat.• Friction between an object and air or water is called drag.• Friction can be:

– useful – eg. brakes, tyres– undesirable – eg. engine wear

• Friction can be reduced by using:– lubricants– bearings

Pg 17

Support

• The force that pushes upwards, so objects do not fall to the centre of the Earth.

Support

Gravity

• The force that pulls objects toward the centre of the Earth.

• It is a constant force at 10ms-2.Gravity

Balanced Forces

• When the action and reaction forces are equal in size and opposite in direction, they are balanced.

• If the forces on an object are balanced it will either remain stationary, or continue moving at a constant speed.

•eg.

450N 450NNet force = 0N

Pg 18

Unbalanced forces.• When the action and reaction forces are different in size or

direction, they are unbalanced.• If the forces on an object are unbalanced it will accelerate in the

direction of the resultant or net force.

500N 450N

Net force = 50N

Force, Mass & Acceleration

• When an unbalanced force acts on an object, it accelerates in the direction of the net force.

• The equation is:force = mass x acceleration Fnet = ma

• The same equation also applies to the effect of gravity on mass (ie. weight).weight force = mass x acceleration due

to gravity Fw = mg

= m x10N (if mass is in kg)• The unit for weight force is the Newton (N).

Pg 19

F= maMass (kg) Acceleration (cms-1) Stopping distance

(cm)Force (N)

1 0.6 670

1.2 5.2 700

1.4 16.3 690

1.6 21.3 700

1.8 23.4 723

2 26.2 726

Gravity, Mass & Weight

• Since weight is the result of the force of gravity acting on mass, weight can change depending on the force of gravity

• On Earth, a person with mass 60kg would weigh 600N

• On the moon, a person with the same mass would only weigh 100N

• F=mg

600N

100N

Force and lifting.• Force is needed to lift an object.• This force equals the weight of the object.• The formula is; Force = mass x gravitational acceleration F = mg

• E.g. A mass of 5kg is lifted 2m off the groundF = 5 x 10F = 50 N• Work is also done to lift the objectW = F x dW = 50 x 2W = 100J

5kg

Forces and the shot-puts.

Mass (kg) Acceleration (ms-2)

Force (N) Gravity (ms-2) Distance (m)

1.5

3

6.25

Experiment 1; lifting the shot-put (F=mg)Experiment 2; work and the shot-put (W=Fd)Experiment 3; dropping the shot-put (F=ma)

We also need to measure the area of the impact craters.

Force and pressure• Pressure is created when a force is applied to an area.• The formula is; Pressure = force / area P = F A• The greater the force and the smaller the area, the greater the

pressure. 10N

P = 10/8 P=1.25Nm-2 P= 10/2 P= 5Nm-2

8cm 2 2cm2

Pg 20

High heels.

• A 50kg woman stands in a pair of high heels.• The area of each heel is 2cm x 2cm and she

stands with the front of her foot off the ground.

• How much pressure is going into the ground?

Steps to solve;8cm2 is the area of the shoes.Convert to m2; 0.0008 m2

Calculate her force; F=mg50 x 10 = 500NP = F/AP = 500/ 0.0008P = 625,000Nm-2

Work• Work (W) is done when a force moves an object.• Work is measured in Joules (J).• The equation is; work done = force x distance W= F x d

• Work done on an object equals the energy gained by the object.

• E.g. Pushing a pedal car 3m using a force of 200N• 200N x 3m = 600J• The car gains 600J of kinetic energy.

W

F d

Pg 26

Power

• Power is the rate at which work is done or energy is transferred.

• Power (P) is measured in watts (W).• The equation is; Power = Energy / time• P = E/t My pet has power

E

P t

Energy

• Energy- the capacity to do work; the property of a system that diminishes when the system does work on any other system, by an amount equal to the work so done; potential energy. Symbol: E

• Sorry what now?

Pg 23

Energy

Potential • Gravitational objects that

are lifted can fall.• Chemical- energy stored in

chemical bonds.• Elastic- energy stored in

stretched objects.• Nuclear- energy stored in

the nucleus of an atom

Active• Electrical- currents of

electrons• Light- waves of photons• Heat/Thermal- movement

of particles • Sound- vibrations• Kinetic- movement energy

Vinegar bottle.

• We will now complete the vinegar bottle rocket experiment and observe the energies & forces involved.

Gravitational potential energy• Gravitational potential energy is stored in objects placed above

ground level, energy is released when objects fall to the ground.• The formula is; potential energy = mass x gravity x change in height ΔEp

= mgΔh

Pg 26

Kinetic energy• Kinetic energy is found in moving objects.• The formula is;• Kinetic energy = ½ mass x velocity squared. Ek

= ½ mv2

• Use mass in kg and velocity in ms-1.

Calculating energy.

• Complete the experiment to calculate the energy of a toy car.

Ramp height (cm)

Observations Ep (J) V (ms-1) Ek (J)

10

20

30

40

50

• If the car weighed 0.0288kg, gravity is 10ms-2

and it was set off from 10cm the equation would be;

• Ep = 0.0288 x 10 x 10 = 2.88J• V2 = 2.88 / 0.0144 = 200 • V = √200 = 14.1 ms-1

• Ek = 0.5 x 0.0288 x 14.12 = 2.9J (rounded to 1 d.p.)

What do you notice?

Chemical potential energy.

• Energy stored in chemical bonds.• When you break a bond between 2

atoms, you release the energy.• E.g. digestion of food; breaks the

bonds between food molecules, releasing energy for your body to use.

Elastic potential energy.

• Energy stored in compressed or stretched objects, the energy is released when the object returns to its original shape.

• E.g. Spray cans, rubber bands.

Nuclear potential energy.

• Energy stored in the nucleus of atoms; released when the atom is destroyed.

• E.g. Nuclear power plant

Electrical energy

• The flow of electrons along a conductor.

Light energy

• Waves of photons moving in one direction.

Heat/Thermal energy

• Movement of particles towards a gas state.

Sound energy• Vibrating particles which in turn vibrate the

bones of the middle ear, which is turn causes cells in the cochlea to move.

Energy can change.

• Energy can be transformed or transferred; but no energy is lost as it does this, this is called THE CONSERVATION OF ENERGY.

Energy transformations.

• When one type of energy is transformed into another, without any loss to the net amount of energy.

100J

80J

20J

Energy transformation.

• Think back to the toy car; why was the amount of gravitational potential energy equal to the amount of kinetic energy?

• The gravitational potential energy was transformed into kinetic energy as the car moved, as the amount was conserved.

Push a toy car along a surface

Turning on a light bulb

Push a slinky down stairs

Whisking a batter

Frying pancakes

Bounce a tennis ball

Snow board down a half pipe

Riding a rollercoaster

Listening to your ipod

Eating an apple

Using a magnet to pick up metal

Law of Energy Conservation

1. Energy must come from somewhere.2. Energy may change form but the total

amount is constant.3. Energy cannot be destroyed, it can only be

transformed.

Conservation of mechanical energy in free fall.

• Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.

• E.g objects in motion stay in motion until acted upon by another force.

• For every action there is an equal and opposite reaction.

Pg 28

Energy transference.

• When energy is shifted without changing its form.

• E.g. heat moving along a piece of metal.

To remember

• Power= pet (my pet has power)• Work= fwd (forward)• Kinetic energy= mev (my mate mev)• Pressure= pfa (pfffaaa the sound of a can

opening)• Force= mfg (my force goes)• Force= fma

Q2

• Ep= mgh v2 = Ek / ½ m

• 83x10x1.2= 996J

• 996/41.5=24 √ = 4.9ms-1

Q1

• EK = ½ m v2 h= Ep

/ mg

• 75 x 625 = 46875

• 46875/1500 = 31.25m

Revision questions.

Exam revision.

Toy CarEugene has a toy car which moves along a concrete path when he winds up the spring and lets the car go. The car accelerates constantly from rest to reach a top speed of 6 m s-1 in 0.5 s. a. Calculate the toy car’s acceleration.

Show all working. b. From its top speed, the car slows

steadily and takes 4 s to stop. Sketcha graph of the toy car’s motion from itsrelease until it stops.

c. Describe one energy change whichoccurs during the toy cars motion.

a= ∆v / ∆t a= 6/0.5 a= 12ms-2

c. Elastic potential kinetic energy

The Water Park SlideBen and Ranui were on a slide at a water park.

20 m 10 m

a. It took Ben 10 s to reach thebottom of the slide. Calculate hisaverage speed.

s=d/t s= 20/10 s= 2ms-1

Diving DownRanui has a mass of 70 kg and Ben has a mass of 90 kg. Both of them stepped off a 10 m diving board at the same time.a. Explain why Ranui and Ben hit the

water at the same time. Mass does not effect the rate of acceleration of an object.b. Ranui and Ben both accelerate at 10 m s-2. Calculate the

force on Ranui as she falls through the air. Show all working. State the unit.

F = ma F= 70 x 10 F = 700Nc. Ben started from rest and hit the water with a speed of 8ms-1. Calculate the time it took him to reach the water.∆ t= ∆v / a 8/10 = 0.8s t= d/s t= 10/8 t= 1.25s

Warbirds over WanakaThe Spitfire LF Mk XVI is a spectacular aircraft that takes part in the biennial “Warbirds over Wanaka” airshow.a. During the show, the spitfire accelerates in a low-level

horizontal pass over the crowd.

(i) Draw a diagram of the spitfire showing the relative size and direction of the FOUR forces acting on the plane during this acceleration.

(ii) If it takes 3 s to accelerate from150 m s-1 to 180 m s-1, calculate the acceleration of the spitfire

Warbirds over Wanaka Gravity

Thrust

Drag/friction Support

(ii) If it takes 3 s to accelerate from150 m s-1 to 180 m s-1, calculate the acceleration of the spitfirea= ∆v / ∆t a= 180-150 / 3 a= 30/3 a= 10ms-2

b. A Hurricane aircraft, mass 4 000 kg, repeats the pass over the crowd with an acceleration of 8 m s-2 and covers a distance of 500 m.

(i) Calculate the size of the unbalanced force acting on the Hurricane during its acceleration using the equation F = ma.F=ma F=4000x8 F=32,000N

(ii) Calculate the work done accelerating the Hurricane using the equation W = Fd.W=Fd W=32000x500 w = 16,000,000 watts

Bungy JumpRanui tried the bungy jump at a local water park.At the top and middle of the jump, the bungy cord was slack. At the bottom of the jump, the bungy cord was fully stretched.a. Complete the table below to show the main forms of energy at each point.

b. Ranui did not bounce back up to the platform. Explain why.Some of the kinetic energy was transformed to elastic energy, so the total amount of kinetic energy was not available for her return journey.

Position Main form of energyTop of jump Gravitational potential energyMiddle of jump Kinetic energyBottom of jump Elastic potential energy

Monica’s MorrieThe starter motor of Monica’s Morrie would not work. To start the car, Monica got three of her friends to push it down a gentle slope. The effect of the slope is to balance the friction forces on the car. Each of her friends pushed the car with a force of 450 N. The car with Monica inside it had a mass of 900 kg.a. i) Using the information above, calculate the car’s

acceleration. Show your working.a= ∆v / ∆t a= 18/14 a=1.28ms-2 a= f/m a= 450/900 a= 0.5ms-2

ii) The table below shows how the car’s speed changed with time. Draw a speed-time graph of the motion.

Time (s) 0 2 4 6 8 10 12 14Speed (m s-1) 0 3 6 6 6 10 14 18

0 2 4 6 8 10 12 14 Time (s)

Monica’s Morrie Moves Onb. While recovering from pushing the car, her three friends watched a dog chase after a stick. The distance-time graph for the dog is shown below. d (m) 72

36 t (s) 12 18 36

(i) Calculate the dog’s speed for the first 12 s.36/12=3ms-1

(ii) Find the dog’s speed during the time 12 s to 18 s.0

(iii) Describe the dog’s motion from the time 18 s to 36s. He was accelerating.

Just CruisingThe diagram showsthe four main forcesacting on a car as ittravels to the left ata steady speed.a. Use labels to describe the forces

acting on the car.b. The force to the right is

measured as 550 N. What is thesize of the force to the left?Explain your answer. 550N

Gravity

Drag/friction

Support

Thrust

The Sand YachtThe wind pushes a sand yacht along on dry sand. When the unbalanced force on the yacht is 280 N its acceleration is 2 m s-2.a. Calculate the mass of the yacht. Give the unit with your answer. 140kgb. The yacht is blown onto wet sand where the friction force has the same size as the wind force. Describe what happens to the yacht. It will either stop or move at a constant speed.c. When the yacht moves onto soft sand the friction force is larger than the wind force. Explain what happens to the speed of the yacht. It will slow down.

What force is required to move a 15kg mass at 3ms-2?_______________________________________________________________

A force if 20N causes an object to move at 7ms-2, what is its mass?_______________________________________________________________

If a 10N unbalanced force is applied to a 10kg object; what will be its acceleration?_______________________________________________________________

Calculate the force acting upon an 18kg object here on Earth with a gravity force if 10ms-2._______________________________________________________________

Calculate the size of the force acting upon a 50kg object on the moon._____________________________________________________________________

If a 80kg man stands on a platform which is 50cm2, how much pressure is he applying? _____________________________________________________________________

If an object exerts 10pascals of pressure over 1m2, what force is being created?_____________________________________________________________________

If 50N of force and 6 pascals of pressure are being exerted, what area are the acting upon?_____________________________________________________________________