Distance against time graphs

distance

time

Constant speed

distance

time

The gradient of this graph gives the speed

Getting faster (accelerating)

distance

time

A car accelerating from stop and then hitting a wall

distance

time

Speed against time graphs

speed

time

No movement

speed

time

Constant speed

speed

time

Getting faster? (accelerating)

speed

time

Constant acceleration

Getting faster? (accelerating)

speed

time

a = v – u

t

(v= final speed, u = initial speed)

v

u

The gradient of this graph gives the acceleration

Getting faster? (accelerating)

speed

time

The area under the graph gives the distance travelled

A dog falling from a tall building (no air resistance)

speed

time

Area = height of building

Acceleration/time graphs

acceleration

time

Constant/uniform acceleration?

acceleration

time

Note!

The area under an acceleration/time

graph gives the change in velocity

acceleration

time

Displacement

• Displacement the distance moved in a stated direction (the distance and direction from the starting point). A VECTOR

Displacement/time graphs

• Usually in 1 dimension (+ = forward and - = backwards)

Displacement/m

Time/s

Velocity?

• Velocity is the rate of change of displacement. Also a VECTOR

Velocity/time graphs

• Usually in 1 dimension (+ = forward and - = backwards)

velocity/m.s-1

Time/s

Ball being thrown into the air, gradient = constant = -9.81 m.s-2

Acceleration?

• Acceleration is the rate of change of velocity. Also a VECTOR

Acceleration/time graphs

• Usually in 1 dimension (+ = up and - = down)

accel/m.s-2

Time/s

Acceleration = constant = -9.81 m.s-2

Average speed/velocity?

• Average speed/velocity is change in distance/displacement divided by time taken over a period of time.

Instantaneous speed/velocity?

• Instantaneous speed/velocity is the change in distance/displacement divided by time at one particular time.

The equations of motion

• The equations of motion can be used when an object is accelerating at a steady rate

• There are four equations relating five quantities

u initial velocity, v final velocity,

s displacement, a acceleration, t time

SUVAT equations

The four equations1 This is a re-arrangement of

2 This says displacement = average velocity x time

3 With zero acceleration, this becomes displacement = velocity

x time

4 Useful when you don’t know the time

t)u+v(2

1=s

at+u=v

2at2

1+ut=s

as2+u=v 22

t

u-v=a

NOT in data book

Beware!

• All quantities are vectors (except time!). These equations are normally done in one dimension, so a negative result means displacement/velocity/acceleration in the opposite direction.

Example 1

Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest?

Example 1

Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest?

What does the question tell us. Write it out.

Example 1

Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest?

u = 10 ms-1

v = 0 ms-1

a = -2 ms-2

t = ? s

Example 1

Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest?

u = 10 ms-1

v = 0 ms-1

a = -2 ms-2

t = ? s

Choose the equation that has these quantities in

v = u + at

Example 1

Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest?

u = 10 ms-1 v = 0 ms-1 a = -2 ms-2 t = ? s

v = u + at

0 = 10 + -2t

2t = 10

t = 5 seconds

Example 2

Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan?

Example 2

Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan?

What does the question tell us. Write it out.

Example 2

Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan?

u = 10 ms-1

v = 0 ms-1

a = -2 ms-2

t = 5 ss = ? m

Example 2

Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan?

u = 10 ms-1 v = 0 ms-1 a = -2 ms-2 t = 5 s s = ? m

Choose the equation that has these quantities in

v2 = u2 + 2as

Example 2

Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan?

u = 10 ms-1 v = 0 ms-1 a = -2 ms-2 t = 5 s s = ? m

v2 = u2 + 2as

02 = 102 + 2x-2s

0 = 100 -4s

4s = 100

s = 25m, the car does not hit Jan.

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• When is the velocity of the ball 12 m.s-1?

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• When is the velocity of the ball 12 m.s-1?

u = 24 m.s-1 a = -9.8 m.s-2 v = 12 m.s-1

t = ?

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• When is the velocity of the ball 12 m.s-1?

u = 24 m.s-1 a = -9.8 m.s-2 v = 12 m.s-1

t = ?

v = u + at

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-

1. • When is the velocity of the ball 12 m.s-1?

u = 24 m.s-1 a = -9.8 m.s-2 v = 12 m.s-1

v = u + at

12 = 24 + -9.8t

-12 = -9.8t

t = 12/9.8 = 1.2 seconds

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• When is the velocity of the ball -12 m.s-1?

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• When is the velocity of the ball -12 m.s-1?

u = 24 m.s-1 a = -9.8 m.s-2 v = -12 m.s-1

t = ?

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• When is the velocity of the ball -12 m.s-1?

u = 24 m.s-1 a = -9.8 m.s-2 v = -12 m.s-1

v = u + at

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-

1. • When is the velocity of the ball -12 m.s-1?

u = 24 m.s-1 a = -9.8 m.s-2 v = -12 m.s-1

v = u + at

-12 = 24 + -9.8t

-36 = -9.8t

t = 36/9.8 = 3.7 seconds

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the displacement of the ball at those times? (t = 1.2, 3.7)

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the displacement of the ball at those times? (t = 1.2, 3.7)

t = 1.2, v = 12, a = -9.8, u = 24 s = ?

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the displacement of the ball at those times? (t = 1.2, 3.7)

t = 1.2, v = 12, a = -9.8, u = 24 s = ?

s = ut + ½at2 = 24x1.2 + ½x-9.8x1.22

s = 28.8 – 7.056 = 21.7 m

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the displacement of the ball at those times? (t = 1.2, 3.7)

t = 3.7, v = 12, a = -9.8, u = 24 s = ?

s = ut + ½at2 = 24x3.7 + ½x-9.8x3.72

s = 88.8 – 67.081 = 21.7 m (the same?!)

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the velocity of the ball 1.50 s after launch?

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the velocity of the ball 1.50 s after launch?

u = 24, t = 1.50, a = -9.8, v = ?

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the velocity of the ball 1.50 s after launch?

u = 24, t = 1.50, a = -9.8, v = ?

v = u + at

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the velocity of the ball 1.50 s after launch?

u = 24, t = 1.50, a = -9.8, v = ?

v = u + at

v = 24 + -9.8x1.50 = 9.3 m.s-1

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the maximum height reached by the ball?

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the maximum height reached by the ball?

u = 24, a = -9.8, v = 0, s = ?

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the maximum height reached by the ball?

u = 24, a = -9.8, v = 0, s = ?

v2 = u2 + 2as

0 = 242 + 2x-9.8xs

0 = 242 -19.6s

Example 3

• A ball is thrown upwards with a velocity of 24 m.s-1.

• What is the maximum height reached by the ball?

u = 24, a = -9.8, v = 0, s = ?

0 = 242 -19.6s

19.6s = 242

s = 242/19.6 = 12.3 m

Imagine a dog being thrown out of an aeroplane.

Woof! (help!)

Force of gravity means the dog accelerates

gravityTo start, the dog is falling slowly (it has not had time to speed up).

There is really only one force acting on the dog, the force of gravity.

The dog falls faster (accelerates) due to this force.

Gravity is still bigger than air resistance

gravity

As the dog falls faster, another force becomes bigger – air resistance.

The force of gravity on the dog of course stays the same

The force of gravity is still bigger than the air resistance, so the dog continues to accelerate (get faster)

Air resistance

Gravity = air resistanceTerminal velocity

gravity

As the dog falls faster and air resistance increases, eventually the air resistance becomes as big as (equal to) the force of gravity.

The dog stops getting faster (accelerating) and falls at constant velocity.

This velocity is called the terminal velocity.

Air resistance

Falling without air resistance

gravity

This time there is only one force acting in the ball - gravity

Falling without air resistance

gravity

The ball falls faster….

Falling without air resistance

gravity

The ball falls faster and faster…….

Falling without air resistance

gravity

The ball falls faster and faster and faster…….

It gets faster by 9.81 m/s every second (9.81 m/s2)

This number is called “g”, the acceleration due to gravity.

Falling without air resistance?

distance

time

Falling without air resistance?

speed

time

Gradient = acceleration = 9.8 m.s-2

Velocity/time graphs

Taking upwards are the positive direction

velocity/m.s-1

Time/s

Ball being thrown into the air, gradient = constant = -9.81 m.s-2

Falling with air resistance?

distance

time

Falling with air resistance?

velocity

time

Terminal velocity

Gravity

What is gravity?

Gravity

Gravity is a force between ALL objects!

Gravity

Gravity

The size of the force depends on the mass of the objects. The bigger they are, the bigger the force!

Small attractive force

Bigger attractive force

Gravity

The size of the force also depends on the distance between the objects.

Gravity

We only really notice the gravitational attraction to big objects!

Hola! ¿Como estas?

Gravity

The force of gravity on something is called its weight. Because it is a force it is measured in Newtons.

Weight

Gravity

On the earth, Mr George’s weight is around 800 N.

800 N

I love physics!

Gravity

On the moon, his weight is around 130 N.

Why?

130 N

Gravity

In deep space, far away from any planets or stars his weight is almost zero. (He is weightless). Why?

Cool!

Mass

Mass is a measure of the amount of material an object is made of and also its resistance to motion (inertia). It is measured in kilograms.

Mass

Mr George has a mass of around 77 kg. This means he is made of 77 kg of blood, bones, hair and poo!

77kg

Mass

On the moon, Mr George hasn’t changed (he’s still Mr George!). That means he still is made of 77 kg of blood, bones, hair and poo!

77kg

Gravity

In deep space, Mr George still hasn’t changed (he’s still Mr George!). That means he still is made of 77 kg of blood, bones, hair and poo!

77kgI feel sick!

Calculating weight

To calculate the weight of an object you multiply the object’s mass by the gravitational field strength wherever you are.

Weight (N) = mass (kg) x gravitational field strength (N/kg)

Newton’s 1st Law

An object continues in uniform motion in a straight line or at rest unless a resultant external force acts

Newton’s first law

Galileo imagined a marble rolling in a very smooth (i.e. no friction) bowl.

Newton’s first lawIf you let go of the ball, it always rolls up the opposite side until it reaches its original height (this actually comes from the conservation of energy).

Newton’s first lawNo matter how long the bowl, this always happens.

constant velocity

Newton’s first lawGalileo imagined an infinitely long bowl where the ball never reaches the other side!

Newton’s first lawThe ball travels with constant velocity until its reaches the other side (which it never does!).

Galileo realised that this was the natural state of objects when no (resultant ) forces act.

constant velocity

Another example

Imagine Mr George cycling at constant velocity.

Newton’s 1st law

He is providing a pushing force.

Constant velocity

Newton’s 1st law

There is an equal and opposite friction force.

Constant velocity

Pushing force

friction

Newton’s second law

Newton’s second law concerns examples where there is a resultant force.

Let’s go back to Mr George on his bike.

Remember when the forces are balanced (no resultant force) he travels at constant velocity.

Constant velocity

Pushing force

friction

Newton’s 2nd law

Now lets imagine what happens if he pedals faster.

Pushing force

friction

Newton’s 2nd law

His velocity changes (goes faster). He accelerates!

Pushing force

friction

acceleration

Remember from last year that acceleration is rate of change of velocity. In other words

acceleration = (change in velocity)/time

Newton’s 2nd law

Now imagine what happens if he stops pedalling.

friction

Newton’s 2nd law

So when there is a resultant force, an object accelerates (changes velocity)

Pushing force

friction

Mr George’s Porche

Newton’s 2nd lawThere is a mathematical relationship between the resultant force and acceleration.

Resultant force (N) = mass (kg) x acceleration (ms-2)

FR = maIt’s physics, there’s

always a mathematical relationship!

An example

Resultant force = 100 – 60 = 40 N

FR = ma

40 = 100a

a = 0.4 m/s2

Pushing force (100 N) Friction (60

N)

Mass of Mr George and bike = 100 kg

Newton’s 3rd lawIf a body A exerts a force on body B, body B will exert an equal but opposite force on body A.

Hand (body A) exerts force on table (body B)

Table (body B) exerts force on hand (body A)

Free-body diagrams

Shows the magnitude and direction of all forces acting on a single body

The diagram shows the body only and the forces acting on it.

Examples

• Mass hanging on a rope

W (weight)

T (tension in rope)

Examples

• Inclined slope

W (weight)

R (normal reaction force)F

(friction)

If a body touches another body there is a force of reaction or contact force. The force is perpendicular to the body exerting the force

Examples

• String over a pulley

T (tension in rope)

T (tension in rope)

W

1

W

1

Momentum

• Momentum is a useful quantity to consider when thinking about "unstoppability". It is also useful when considering collisions and explosions. It is defined as

Momentum (kg.m.s-1) = Mass (kg) x Velocity (m.s-1)

p = mv

Law of conservation of momentum

• The law of conservation of linear momentum says that

“in an isolated system, momentum remains constant”.

We can use this to calculate what happens after a collision (and in fact during an “explosion”).

Law of conservation of momentum

• In a collision between two objects, momentum is conserved (total momentum stays the same). i.e.

Total momentum before the collision = Total momentum after

Momentum is not energy!

A harder example!

• A car of mass 1000 kg travelling at 5 m.s-1 hits a stationary truck of mass 2000 kg. After the collision they stick together. What is their joint velocity after the collision?

A harder example!

5 m.s-1

1000kg

2000kgBefore

AfterV m.s-1

Combined mass = 3000 kg

Momentum before = 1000x5 + 2000x0 = 5000 kg.m.s-1

Momentum after = 3000v

A harder example

The law of conservation of momentum tells us that momentum before equals momentum after, so

Momentum before = momentum after

5000 = 3000v

V = 5000/3000 = 1.67 m.s-1

Momentum is a vector

• Momentum is a vector, so if velocities are in opposite directions we must take this into account in our calculations

An even harder example!

Snoopy (mass 10kg) running at 4.5 m.s-

1 jumps onto a skateboard of mass 4 kg travelling in the opposite direction at 7 m.s-1. What is the velocity of Snoopy and skateboard after Snoopy has jumped on?

I love physics

An even harder example!

10kg

4kg-4.5 m.s-1

7 m.s-1

Because they are in opposite directions, we make one velocity negative

14kg

v m.s-1

Momentum before = 10 x -4.5 + 4 x 7 = -45 + 28 = -17

Momentum after = 14v

An even harder example!

Momentum before = Momentum after

-17 = 14v

V = -17/14 = -1.21 m.s-1

The negative sign tells us that the velocity is from left to right (we choose this as our “negative direction”)

“Explosions” - recoil

ImpulseFt = mv – mu

The quantity Ft is called the impulse, and of course mv – mu is the change in momentum (v = final

velocity and u = initial velocity)

Impulse = Change in momentum

ImpulseFt = mv – mu

F = Δp/Δt

Units

Impulse is measured in N.s (Ft)

or kg.m.s-1 (mv – mu)

ImpulseNote; For a ball bouncing off a wall, don’t forget

the initial and final velocity are in different directions, so you will have to make one of them

negative.

In this case mv – mu = -3m -5m = -8m

5 m/s

-3 m/s

Example

• Szymon punches Eerik in the face. If Eerik’s head (mass 10 kg) was initially at rest and moves away from Szymon’s fist at 3 m/s, what impulse was delivered to Eerik’s head? If the fist was in contact with the face for 0.2 seconds, what was the force of the punch?

• m = 10kg, t = 0.2, u = 0, v = 3

• Impulse = Ft = mv – mu = 10x3 – 10x0 = 30 Ns

• Impulse = Ft = 30

Fx0.2 = 30

F = 30/0.2 = 150 N

Another example

• A tennis ball (0.3 kg) hits a racquet at 3 m/s and rebounds in the opposite direction at 6 m/s. What impulse is given to the ball?

• Impulse = mv – mu == 0.3x-6 – 0.3x3

= -2.7kg.m.s-1

3 m/s

-6 m/s

Area under a force-time graph = impulse

Area = impulse

Work

In physics, work has a special meaning, different to “normal” English.

Work

In physics, work is the amount of energy transformed (changed) when a force moves (in the direction of the force)

Calculating work

The amount of work done (measured in Joules) is equal to the force used (Newtons) multiplied by the distance the force has moved (metres). Force (N)

Distance travelled (m)

Work (J)= Force(N) x distance(m)

W = Fscosθ

Important

The force has to be in the direction of movement. Carrying the shopping home is not work in physics!

Work = Fscosθ

s

F

θ

What if the force is at an angle to the distance moved?

Lifting objects

Our lifting force is equal to the weight of the object.

Lifting force

weight

Work done (J) = Force (N) x distance (m)

A woman pushes a car with a force of 400 N at an angle of 10° to the horizontal for a distance of 15m. How much work has she done?

W = Fscosθ = 400x15x0.985

W = 5900 J

Work done (J) = Force (N) x distance (m)

A man lifts a mass of 120 kg to a height of 2.5m. How much work did he do?

Force = weight = 1200N

Work = F x d = 1200 x 2.5

Work = 3000 J

Power!

Power is the amount of energy transformed (changed) per second. It is measured in Watts (1 Watt = 1 J/s)

Power = Energy transformed

time

Work done in stretching a spring

F/N

x/m

Work done in strectching spring = area under graph

Chemical kinetic gravitational

Gain in GPE = work done = m x g x Δh

ΔEp = mgΔh

Joules kgN/kg or m/s2

m

Example

A dog of mass 12 kg falls from an aeroplane at a height of 3.4 km. How much gravitational energy does the dog lose as it falls to the ground

Woof! (help!)

ExampleOn earth g = 10 m/s2

Height = 3.4 km = 3400 m

Mass of dog = 12 kg

ExampleOn earth g = 10 m/s2

Height = 3.4 km = 3400 m

Mass of dog = 12 kg

GPE lost by dog = mgh = 12 x 10 x 3400 = 408 000 J

Example

GPE lost by dog = mgh = 12 x 10 x 3400 = 408 000 J

Just before the dog hits the ground, what has this

GPE turned into?

Kinetic energy

Kinetic energy of an object can be found using the following formula

Ek = mv2

2where m = mass (in kg) and v = speed (in m/s)

Example

A bullet of mass 150 g is travelling at 400 m/s. How much kinetic energy does it have?

Ek = mv2/2 = (0.15 x (400)2)/2 = 12 000 J

Energy changes

Energy transfer (change)

A lamp turns electrical energy into heat and light energy

Sankey Diagram

A Sankey diagram helps to show how much light and heat energy is produced

Sankey Diagram

The thickness of each arrow is drawn to scale to show the amount of energy

Sankey Diagram

Notice that the total amount of energy before is equal to the total amount of energy after (conservation of energy)

Efficiency

Although the total energy out is the same, not all of it is useful.

Efficiency

Efficiency is defined as

Efficiency = useful energy output

total energy input

Example

Efficiency = 75 = 0.15=15%

500

Energy efficient light bulb

Efficiency = 75 = 0.75

100 That’s much better!

Elastic collisions

• No loss of kinetic energy (only collisions between subatomic particles)

Inelastic collisions

• Kinetic energy lost (but momentum stays the same!)

Satellites

How far could you kick a dog?

From a table, medium kick.

How far can you kick a dog?

Gravity

Harder kick

Gravity

Small cannon

Gravity

Woof! (help)

Bigger cannon

GravityGravity

Even bigger cannon

GravityGravity

Gravity

VERY big cannon

Gravity

Humungous cannon?

Dog in orbit!

The dog is now in orbit! (assuming no air resistance of course)

Dog in orbit!

The dog is falling towards the earth, but never gets there!

Dogs in orbit!

The force that keeps an object moving in a circle is called the centripetal force (here provided by gravity)

Gravity

Other examples

Earth’s gravitational attraction on moon

Uniform Circular Motion

• This describes an object going around a circle at constant speed

Direction of centripetal acceleration/force

VA

VB

VAVB

VA + change in velocity = VB

Change in velocity

The change in velocity (and thus the acceleration) is directed towards the centre of the circle.

Uniform circular motion

The centripetal acceleration/force is always directed towards the centre of the circle

Centripetal force/acceleration

velocity

Not uniform velocity

• It is important to remember that though the speed is constant, the direction is changing all the time, so the velocity is changing.

Uniform speed ≠ uniform velocity

How big is the centripetal acceleration?

a = v2 = 4π2r

r T2

where a is the centripetal acceleration (m.s-2), r is the radius of the circle (m), and v is the constant speed (m.s-1).

How big is the centripetal force?

F = mv2

r

from F = ma (Newton’s 2nd law)

Centripetal Force - The Real Force

Work done?

• None! Because the force is always perpendicular to the motion, no work is done by the centripetal force.

 

That’s it!