Marr College - Physics Unit Dynamics and · PDF fileMarr College Physics Department National 4/5 Summary Notes Dynamics and Space Unit 2 Key Area – Velocity and Displacement –

Marr College Physics Department National 4/5 Summary Notes Dynamics and Space Unit

1

Marr College - Physics

National 4/5 Physics

Unit – Dynamics and Space

Key Areas

• Velocity and Displacement – Vectors and Scalars

• Acceleration

• Velocity-time graphs

• Newton’s Laws

• Conservation of Energy

• Projectile Motion

• Satellites

• Space Exploration

• Cosmology


2

Key Area – Velocity and Displacement – Vectors and Scalars

Red Amber Green

1 Describe how to measure an average speed.

2 Carry out calculations involving distance, time and average speed.

3 Describe how to measure instantaneous speed.

4 Identify situations where average speed and instantaneous speed are different.

5 Describe what is meant by vector and scalar quantities.

6 State the difference between distance and displacement.

7 State the difference between speed and velocity.

8 Carry out calculations involving average velocity, displacement and time.


3

Key Area – Velocity and Displacement – Vectors and Scalars

Average Speed

Average speed is the distance travelled per unit time.

�̅� =𝑑

𝑡

SI unit of average speed is m s-1 (metres per second) since the SI unit for distance is metres (m) and

SI unit for time is seconds (s).

Examples of alternative units for average speed are kilometres per hour (km h-1) or miles per hour

(mph).

Measurement of average speed

To measure an average speed, you must:

• measure the distance travelled with a measuring tape, metre stick or trundle wheel

• measure the time taken with a stop clock

• calculate the average speed by dividing the distance by the time

average speed (ms-1)

distance (m)

time (s)


4

Calculations involving distance, time and average speed

Note: care must be taken to use the correct units for time and distance.

Example

Calculate the average speed in metres per second of a runner who runs 1500 m in 5 minutes.

Solution

s = 1500 m

t = 5 minutes = 5 60 seconds = 300 s

v = ?

�̅� =𝑑

𝑡

=1500

300

= 5 𝑚 𝑠−1

Instantaneous speed

The instantaneous speed of a vehicle at a given point can be measured by finding the average speed

during a very short time interval as the vehicle passes that point.

Average speed and instantaneous speed are often very different e.g. the average speed for a car

over an entire journey from Glasgow to Edinburgh could be 40 mph, but at any point in the journey,

the instantaneous speed of the car could be 30 mph, 70 mph, 60 mph or 0 mph if the car is

stationary at traffic lights.


5

Measuring Instantaneous Speed

To measure instantaneous speeds, it is necessary to measure very short time intervals.

With an ordinary stop clock, human reaction time introduces large uncertainties. These can be avoided by using electronic timers. The most usual is a light gate.

A light gate consists of a light source aimed at a photocell. The photocell is connected to an electronic timer or computer.

The timer measures how long an object takes to pass through the light beam.

The distance travelled is the length of the object which passes through the beam.

Often a card is attached so that the card passes through the beam.

Procedure to measure the instantaneous speed of a trolley

• Set up apparatus as shown

• Measure length of card

• Push trolley through light beam

• Timer measures time taken for card to pass through light gate

• Calculate instantaneous speed by

𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑠𝑝𝑒𝑒𝑑 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑎𝑟𝑑

𝑡𝑖𝑚𝑒 𝑡𝑜 𝑝𝑎𝑠𝑠 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑙𝑖𝑔ℎ𝑡 𝑏𝑒𝑎𝑚

TIMER To power supply

Light source Photocell

trolley

TIMER card

light source

photocell


6

Example

In the above experiment, the following measurements are made:

Length of card: 10 cm

Time on timer: 0.25 s

Calculate the instantaneous speed of the vehicle

Solution

Length of card = 10 cm = 0.1 m

Time on timer = 0.25 s

𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑠𝑝𝑒𝑒𝑑 =𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑎𝑟𝑑

𝑡𝑖𝑚𝑒 𝑡𝑜 𝑝𝑎𝑠𝑠 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑙𝑖𝑔ℎ𝑡 𝑏𝑒𝑎𝑚

=0.1

0.25

= 0.4 𝑚 𝑠−1


7

Vector and Scalar Quantities

A scalar quantity is fully described by its magnitude (size) and unit, e.g.

time = 220 s

A vector quantity is fully described by its magnitude, unit, and direction, e.g.

Force = 800 N upwards

Distance and Displacement

Distance is a scalar quantity. It measures the total distance travelled, no matter in which direction.

Displacement is a vector quantity. It is the length measured from the starting point to the finishing

point in a straight line. Its direction must be stated.

Example

A girl walks 3 km due north then turns and walks 4 km due east as shown in the diagram.

quantity

magnitude

unit

quantity

unit

direction

magnitude

START

FINISH

3 km

4 km

000˚

090˚

180˚

270˚


8

Calculate

(a) The total distance travelled

(b) The girl’s final displacement relative to her starting position.

Solution

(a) Total distance = 3 km + 4 km

= 7 km

(b) (to calculate displacement, we need to draw a vector diagram)

(this solution involves using Phythagoras and trig functions (SOH-CAH=TOA), but you can

also solve these types of problems using a scale diagram).

Speed and Velocity

Speed is a scalar quantity. As discussed above, speed is the distance travelled per unit time.

�̅� =𝑑

𝑡

(𝑎𝑣𝑒𝑟𝑎𝑔𝑒)𝑠𝑝𝑒𝑒𝑑 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

Velocity is a vector quantity (the vector equivalent of speed). Velocity is defined as the displacement

per unit time.

�̅� =𝑠

𝑡

(𝑎𝑣𝑒𝑟𝑎𝑔𝑒)𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑡𝑖𝑚𝑒

θ

START

FINISH

3 km

4 km

Magnitude of displacement = ξ32 + 42

= 5 km

𝑡𝑎𝑛𝜃 =4

3

Θ= 53˚

Resultant displacement = 5 km at 053˚

Resultant

displacement


9

Since velocity is a vector, you must state its direction.

The direction of velocity will be the same as the displacement.

Example

The girl’s walk in the previous example took 2.5 hours.

Calculate

(a) The average speed

(b) The average velocity for the walk, both in km h-1

Solution

(a) Distance = 7 km 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 =𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

Time = 2.5 hours

=7

2.5

= 2.8 𝑘𝑚 ℎ−1

(b) Displacement = 5 km (053˚) 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑡𝑖𝑚𝑒

Time = 2.5 hours

= 5

2.5

= 2 𝑘𝑚 ℎ−1 at 053˚

(When performing calculations on speed and velocity like the above, it is best to write the equations in

words as shown, to avoid confusion with symbols. This communicates your understanding in the clearest

way).


10

Key Area – Acceleration

Red Amber Green

1 State that acceleration is the change in velocity per unit time.

2 Carry out calculations involving the relationship between initial velocity, final velocity, time and uniform acceleration.

Acceleration is defined as the change in velocity per unit time.

𝑎 =∆𝑣

𝑡

The change in velocity can be given by the moving object’s final velocity (v) – initial velocity (u)

∆𝑣 = 𝑣 − 𝑢

Therefore we can write our equation for acceleration as

𝑎 =𝑣 − 𝑢

𝑡

Acceleration is a vector quantity. If the final speed is less than the initial speed, the acceleration is in

the opposite direction to motion and will be negative. This indicates a deceleration.

The equation for acceleration can be rearranged in terms of the final velocity, v

𝑣 = 𝑢 + 𝑎𝑡

acceleration (m s-2)

metres per second per second

final velocity (m s-1) initial velocity (m s-1)

time (s)


11

Example

A car is travelling with an initial velocity of 15 m s-1. The driver presses the accelerator pedal and the

car accelerates at a rate of 2 m s-1 for 4 s. Calculate the final velocity of the car.

v = ? v = u + at

u = 15 m s-1 = 15 + (2x4)

a = 2 m s-2 = 23 m s-1

t = 4 s

Scalar Vector

distance displacement

speed velocity

mass acceleration

time force

energy weight


12

Measuring Acceleration

Two light gates are required as an initial and final velocity have to be measured.

• Set up the apparatus as shown

• Measure the width of the card with a ruler

• Set the timers t1 and t2 to measure velocity – input the width of the card

• Let the trolley run down the slope and measure the time taken between the two light gates (t) using timer t3

• Record the initial velocity (u) from timer t1

• Record the final velocity (v) from timer t2

• Calculate the acceleration using 𝑎 =𝑣−𝑢

𝑡


13

Key Area – Velocity-time Graphs Red Amber Green

1 Draw velocity–time graphs involving more-than-one constant acceleration.

2 Describe the motions represented by a velocity–time graph, including a change in direction.

3 Calculate acceleration and displacement from velocity–time graphs involving more-than-one constant acceleration and including a change in direction.


14

Key Area – Velocity-time Graphs

Velocity-time graphs provide a useful way of displaying the motion of an object.

They can be used to:

1) Describe the motion of the object in detail.

2) Calculate values for accelerations and decelerations using gradients.

3) Calculate distances travelled and resultant displacements using area under graph.

4) Calculate the average velocity for a journey.

constant velocity constant acceleration constant deceleration

constant

deceleration to

rest, then constant

acceleration in the

opposite direction

(Note: Speed-time graphs are also used, but a speed-time graph would not show any change in

direction).

velo

city

(m

s-1

)

time (s)

velo

city

(m

s-1

)

time (s)

velo

city

(m

s-1

)

time (s) 0 0

0 0

0 0

0 0

velo

city

(m

s-1

)

time (s)


15

Example

The graph below shows the motion of a car over 35 seconds.

(a) Describe the motion of the car during the 35 seconds

(b) Calculate the acceleration between 0 and 10 seconds

(c) Calculate the acceleration between 30 and 35 seconds

(d) Calculate the final displacement of the car from the starting position

(e) Calculate the average velocity during the 35 seconds.

Solution

(a) 0→10 seconds: constant acceleration from rest to 16 m s-1

10→30 seconds: constant velocity of 16 m s-1

30→35 seconds constant deceleration from 16 m s-1 to rest.

(b) a = ? 𝑎 =𝑣−𝑢

𝑡

v = 16 m s-1 =16−0

10

u= 0 = 1.6 m s-2

t = 10 s

(c) a = ? 𝑎 =𝑣−𝑢

𝑡

v = 0 m s-1 =0−16

5

u= 16 m s-1 = - 3.2 m s-2

t = (35-30)=5 s

0

2

4

6

8

10

12

14

16

18

0 5 10 15 20 25 30 35 40

velo

city

(m

s-1

)

time (s)


16

(d) displacement = area under v-t graph

= (½ x 10 x 16) + (20 x 16) + (½ x 5 x 16)

= 80 + 320 + 40

= 440 m

(e) average velocity = ? 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑡𝑖𝑚𝑒

displacement = 440 m = 440

35

time = 35 s = 12.6 m s-1


17

Key Area – Newton’s Laws Red Amber Green

1 Describe the effects of forces in terms of their ability to change the shape, speed and direction of travel of an object.

2 Describe the use of a newton balance to measure force.

3 State that weight is a force and is the Earth’s pull on an object.

4 Distinguish between mass and weight.

5 State that weight per unit mass is called the gravitational field strength.

6 Carry out calculations involving the relationship between weight, mass and gravitational field strength including situations where g is not equal to 9.8 Nkg-1.

7 State that the force of friction can oppose the motion of an object.

8 Describe and explain situations in which attempts are made to increase or decrease the force of friction.

9 State that force is a vector quantity.

10 State that forces which are equal in size but act in opposite directions on an object are called balanced forces and are equivalent to no force at all.

11 Explain the movement of objects in terms of Newton’s First Law.

12 Describe the qualitative effects of change of mass or of force on the acceleration of an object.

13 Define the newton.

14 Use free body diagrams to analyse the forces on an object.

15 State what is meant by the resultant of a number of forces.

16 Carry out calculations using Newton’s Second Law i.e. the relationship between acceleration, resultant force and mass - involving more than one force but in one dimension only.

17 Use a scale diagram, or otherwise, to find the magnitude and direction of the resultant of two forces acting at right angles to each other.

18 Explain the equivalence of acceleration due to gravity and gravitational field strength.


18

19 Explain free-fall and terminal velocity in terms of Newton’s Laws.

20 Carry out calculations involving vertical motion in gravitational fields.

21 State Newton’s Third Law.

22 Use Newton’s Third Law to explain motion resulting from a ‘reaction’ force.

Key Area – Newton’s Laws

Effects of Forces

We can’t see forces themselves, but we can observe the effects that they have.

Forces can change:

• the shape of an object

• the speed of an object

• the direction of travel of an object

Measuring Force

The unit of force is the Newton.

The size of a force can be measured using a device called a Newton Balance.

A force is applied to the hook and the spring is stretched.

The greater the force applied, the greater the extension of the

spring.

The size of the force is read from the scale.

scale

spring

force to be

measured is applied

to the hook here


19

Mass and Weight

Mass and Weight mean different things!

Mass is a measure of the amount of matter that an object is made of. The units of mass are

kilograms (kg).

Weight is the size of the force that gravity exerts on an object. The units of weight are Newtons (N)

since weight is a force.

Gravitational field strength

Gravitational field strength is the weight per unit mass, i.e. gravitational field strength is the amount

of force that gravity exerts on every kilogram of an object.

𝑔 =𝑊

𝑚

Therefore the units of gravitational field strength are Newtons per kilogram (Nkg-1).

This can be rearranged to give:

𝑊 = 𝑚 𝑔

Example

The gravitational field strength on the surface of Earth is 9.8 Nkg-1.

The gravitational field strength on the surface of the Moon is 1.6 Nkg-1.

Calculate the weight of an astronaut of mass 60 kg

(a) on the surface of the Earth

(b) on the surface of the Moon

Solution

(a) W=? W = m g

m=60 kg = 60 x 9.8

g=9.8 Nkg-1 = 588 N

(b) W=? W = m g

m=60 kg = 60 x 1.6

g=1.6 Nkg-1 = 96 N

weight (N)

mass (kg)

gravitational field

strength (Nkg-1)


20

Friction

Friction is a resistive force, which opposes the motion of an object. This means that it acts in the opposite direction to motion. Friction acts between any two surfaces in contact. When one surface moves over another, the force of friction acts between the surfaces and the size of the force depends on the surfaces, e.g. a rough surface will give a lot of friction.

Friction caused by collisions with particles of air is usually called air resistance. It depends mainly on two factors:

• the shape and size of the object • the speed of the moving object.

Air resistance increases as the speed of movement increases.

Increasing and Decreasing Friction

Where friction is making movement difficult, friction should be reduced.

This can be achieved by:

• lubricating the surfaces with oil or grease • separating the surfaces with air, e.g. a hovercraft • making the surfaces roll instead of slide, e.g. use ball bearings • streamlining to reduce air friction.

Where friction is used to slow an object down, it should be increased.

This can be achieved by:

• choosing surfaces which cause high friction e.g. sections of road before traffic lights have higher friction than normal roads

• increasing surface area and choosing shape to increase air friction, e.g. parachute.


21

Force is a vector

Force is a vector quantity. The direction that a force acts in will dictate the effect that it has.

Balanced Forces

Two forces acting in opposite directions are said to be balanced. The effect is equivalent to no force

acting on the object at all.

Newton’s First Law

“An object will remain at rest or move at a constant velocity in a straight line unless acted on by an unbalanced force”.

This means that if the forces acting on an object are balanced, then the object will remain stationary if it was already stationary. For a moving object, if the forces acting on it are balanced, then it will continue to move in a straight line at a constant velocity.

balanced forces ↔ Object at rest OR object moving at a constant velocity in a straight line

Newton’s Second Law

If an overall resultant (or unbalanced) force is applied to an object, then the object will accelerate in the direction of the unbalanced force.

• The greater the resultant force, the greater the acceleration.

• The smaller the mass, the greater the acceleration.

These two relationships are combined to give

𝐹𝑢𝑛 = 𝑚 𝑎

This equation shows that a force of 1 N is the force needed to accelerate a mass of 1 kg at a rate of 1 ms-2

Example

A trolley of mass 2 kg accelerates at a rate of 2.5 ms-2.

Calculate the resultant force acting on the trolley.

Solution

Fun=? Fun = m a

m=2kg = 2 x 2.5

a=2.5 ms-2 = 5 N

(resultant or

unbalanced) force (N)

mass (kg)

acceleration (ms-2)


22

Free Body Diagrams and Newton’s Second Law

“Free body diagrams” are used to analyse the forces acting on an object. The names of the forces are labelled on the diagram. (It is also useful to mark the sizes of the forces on the diagram).

This enables us to calculate the resultant force, and then apply Fun=ma.

Example

A space rocket of mass 2.5x106 kg lifts off from the earth with a thrust of 3.5x107 N.

(a) Draw a free body diagram of the rocket at lift-off (b) Calculate the resultant force acting on the rocket (c) Calculate the initial acceleration of the rocket.

Solution

(a)

(Initially, no air resistance will act on the rocket at the instant it lifts off. Air resistance will only begin to act once the rocket starts moving).

(b) Fun = thrust – weight W = mg

= 3.5x107 – 2.45x107 = 2.5x106 x 9.8

= 1.05 x 107 N =2.45x107 N

(c) Fun = 1 x107 N 𝐹𝑢𝑛 = 𝑚 𝑎

m = 2.5x106 kg

𝑎 = 𝐹𝑢𝑛

𝑚

a = ? =1.05×107

2.5×106

= 4.2 ms-2

2.5

x10

6 k

g

thrust

weight


23

Forces at Right Angles

The resultant force arising from two forces acting at right angles can be found by

(a) Drawing the force vectors “nose-to-tail” (b) Calculating the magnitude using Pythagoras (c) Calculate the direction using trigonometry (SOH-CAH-TOA)

(Alternatively, you could draw the vectors “nose-to-tail” as a scale diagram. Then find the magnitude using a ruler and your chosen scale, and the direction using a protractor).

Example

Two forces act on an object at right angles to each other as shown.

Calculate

(a) The magnitude of the resultant force (b) The direction of the resultant force, relative to the 40 N force.

Solution

(arrange vectors “nose-to-tail” and draw resultant from tail of first to nose of last)

20 N

40 N

θ

(a) Magnitude of resultant force =

ξ402 + 202

= 44.7 N

(𝑏)𝑡𝑎𝑛𝜃 =20

40

Θ= 26.6˚

Resultant force = 44.7 N at 26.6˚ to 40 N

force

Resultant Force

40 N

20 N

(question asks for

direction relative to 40 N

force)


24

Acceleration Due to Gravity and Gravitational Field Strength

Recall from pages 12-13 that weight is the force which causes an object to accelerate downwards and has the value mg, where g is the gravitational field strength.

The value of the acceleration caused by weight can be calculated from Newton’s second law, using Fun = ma where F is now the weight W, and W = mg.

𝑎 =𝐹𝑢𝑛

𝑚

𝑎 =𝑊

𝑚

𝑎 =𝑚𝑔

𝑚

a = g

This shows that the acceleration due to gravity and the gravitational field strength are numerically equivalent.

Therefore the acceleration due to gravity on Earth is 9.8 m s-2since the gravitational field strength is 9.8 N kg-1.

The acceleration due to gravity on the moon is 1.6 ms-2 since the gravitational field strength is 1.6 N kg-1.

Example

An astronaut drops a hammer on the Moon where the gravitational field strength is 1.6 ms-2.

The hammer lands on the surface of the Moon 2.2 s after being dropped.

Calculate the velocity of the hammer at the instant it strikes the surface of the Moon.

Solution

(a) u=0 v = u + at v= ? = 0 + (1.6x2.2) a=1.6 ms-2 = 3.52 ms-1 t=2.2 s


25

Free-fall and Terminal Velocity

An object falling from a height on Earth will accelerate due to its weight. This is called free-fall.

As the velocity increases, the air resistance acting on the object will increase.

This means that the unbalanced force on the object will decrease, producing a smaller acceleration.

Eventually, the air resistance will balance the object’s weight, meaning that the object will fall with a constant velocity.

This final velocity is called terminal velocity.

The velocity-time graph shown below is for a parachutist undergoing free-fall before opening her parachute.

Point A – parachutist jumps from plane and undergoes acceleration due to gravity (free-fall)

Point B – air resistance has increased to balance weight – constant velocity: terminal velocity 1

Point C – parachutist opens parachute – increased air resistance causes deceleration

Point D – weight and air resistance balanced again so new slower constant velocity reached: terminal velocity 2

Constant

velocity

Constant

velocity

A

B C

D


26

Newton’s Third Law

“For every action, there is an equal and opposite reaction”.

Newton noticed that forces occur in pairs. He called one force the action and the other the reaction.

These two forces are always equal in size, but opposite in direction. They do not both act on the

same object.

If an object A exerts a force (the action) on object B, then object B will exert an equal, but opposite

force (the reaction) on object A.

Examples

These action and reaction forces are also known as Newton Pairs.

Rocket forces fuel

downwards (action force)

Fuel forces rocket upwards

(reaction force)

Skateboarder pushes wall to

the left (action force)

Wall pushes skateboarder

to right (reaction force)

Water pushes swimmer to

left (reaction force)

Swimmer pushes water to

right (action force)

Cannon fires cannonball to

the right (action force)

Cannonball forces cannon

back to the left (reaction

force)


27

Key Area - Conservation of energy

Red Amber Green

1. State the principle of Conservation of Energy.

2. Carry out calculations involving work done, force and distance.

𝐸𝑤 = 𝐹 𝑑

3. Carry out calculations involving work done, power and time.

𝑃 = 𝐸𝑤

𝑡

4. Carry out calculations involving the relationship between change in potential energy (Ep), mass, gravitational field strength and height,

𝐸𝑝 = 𝑚𝑔ℎ

5. Carry out calculations involving the relationship between change in kinetic energy (Ek), mass and velocity,

𝐸𝑘 =1

2𝑚𝑣2, and 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝐸𝑘 = 𝐸𝑘 𝑏𝑒𝑓𝑜𝑟𝑒 − 𝐸𝑘 𝑎𝑓𝑡𝑒𝑟

6. Carry out calculations involving energy transformations, using the principle of Conservation of Energy.

7. Account for the difference between input energy and output energy during energy transformations, e.g. some energy is wasted as heat due to work done by friction.


28

Key Area – Conservation of energy

Conservation of energy

Energy cannot be created or destroyed; it can only be changed from one form to another.

i.e., the total amount of energy in a system is constant.

Work done

“Work done” is a measure of energy transferred

Work is done when a force (F) moves an object over a distance (d).

The ‘mechanical’ work done is given the symbol Ew .

• A greater force will mean a greater amount of work done and

• if the distance is greater it will also mean a greater amount of work done.

The relationship between work done (Ew), Force (F) and distance (d) is,

Ew = Fd

Example

A dog pulls a sledge for a distance of 15 m using a force of 30 N. Calculate the work done by the dog. F = 30 N Ew = F d d = 15 m = 30 x 15 Ew = ? = 450 J

J N

m


29

Power Power is the rate of work done, i.e. work done per unit time. The units of power are Watts (W). 1 Watt means 1 Joule of energy is transferred every second.

𝑃 =𝐸

𝑡

Example A cyclist applies a force of 60 N and travels 2 km.

(a) Calculate the work done by the cyclist. F = 60 N Ew = F d d = 2 km = 2000 m = 60 x 2000 Ew = ? = 1.2x105 J

(b) The journey takes 8 minutes. Calculate the power developed by the cyclist.

P= ? 𝑃 =𝐸

𝑡

Ew= 1.2x105 J

t = 8 min =1.2×105

480

= 8x60 = 480s = 250 W

Power (W)

Energy (J)

time (s)


30

Gravitational Potential Energy (Ep)

If an object is moved upwards by a force (F) through a distance (d) at a constant speed the work

done will be given by,

Ew = Fd

When an object is moved vertically up or down we

use the term ‘height’ (h) rather than distance.

The formula then becomes,

Ew = Fh

The force (F) required to move the

object vertically, up or down, will be equal to

the weight W = mg.

The formula then becomes,

Ew = mgh

Therefore the change in gravitational potential energy Ep is given by

Ep = m g h

Example

A brick of mass 1 kg is carried up the Eiffel Tower to the 275m platform. How much gravitational

potential energy does the brick gain?

h = 275m Ep = m g h

m = 1 kg = 1 x 9.8 x 275

g = 9.8 N kg-1 = 2695 J

Ep = ?

gravitational field strength (N kg-1)

change in height (m) change in gravitational

potential energy (J) mass (kg)

http://www.google.co.uk/imgres?q=gravitational+potential+energy&hl=en&gbv=2&sig=107598370982865288844&biw=1024&bih=546&tbm=isch&tbnid=q3EfGVd8Uviw7M:&imgrefurl=http://resources.yesican-science.ca/energy_flow/energy_forms_activity2_key.html&docid=XvekO84MogqXyM&imgurl=http://resources.yesican-science.ca/energy_flow/images/potential_energy2.png&w=300&h=287&ei=XGm7T__pN8en0QXL983TBw&zoom=1&iact=hc&vpx=747&vpy=142&dur=828&hovh=220&hovw=230&tx=119&ty=148&page=1&tbnh=161&tbnw=167&start=0&ndsp=8&ved=1t:429,r:3,s:0,i:7


31

Kinetic Energy (Ek)

All moving objects have kinetic energy.

The kinetic energy (Ek) of an object has depends on its mass (m) and its speed (v). In fact, the kinetic

energy is directly proportional to both the mass and the (speed)2. The relationship between kinetic

energy (Ek), mass (m) and speed (v) is,

𝐸𝑘 = 1

2𝑚𝑣2

Example

A car has a mass of 1000 kg and is travelling at a speed of 20 m s-1 Calculate the amount of kinetic energy that the car has at this speed.

m = 1000 kg Ek = ½ m v2

v = 20 m s-1 = ½ x 1000 x 202

= 200 000 J

Conservation of energy

• Energy cannot be created or destroyed, it can only be changed from one form to another.

• This means that the total energy in any given situation is the same.

• The main form of “wasted” energy is heat caused by the work done due to the force of

friction

mass (kg) kinetic energy (J)

speed (m s-1)


32

Conservation of energy – example

A trolley rolls towards a ramp with a speed of 20 m s-1. The trolley has a mass of 0·3 kg.

(a) Calculate the kinetic energy of the trolley before it travels up the ramp.

(b) There are no energy losses due to friction. How much potential energy does the trolley gain as it

travels up the ramp and stops?

(c) Calculate the height that the trolley reaches on the ramp.

Solution

(a) 𝐸𝑘 =1

2𝑚𝑣2

=1

2× 0 ∙ 3 × 22

= 1 ∙ 2 𝐽

(b) 1 ∙ 2 𝐽

(c) 𝐸𝑝 = 𝑚 𝑔 ℎ

ℎ =𝐸𝑝

𝑚𝑔

=1 ∙ 2

0.3 × 9.8

= 0 ∙ 4 𝑚

Input and output energies

• No system is truly 100 % efficient

• In a given energy transformation, the useful energy output is less than the total energy input

• The main form of wasted energy is heat (due to work done by friction)

o Energy can also be wasted as unwanted sound


33

Key Area – Projectile Motion

Red Amber Green

1 Explain the curved path of a projectile in terms of independent horizontal and vertical motions.

2 Draw velocity-time graphs for horizontal and vertical projectile motion.

3 Solve numerical problems for an object projected horizontally.


34

Key Area – Projectile Motion

A projectile is an object which has been given a forward motion through the air, but which is also

pulled downward by the force of gravity. This results in the path of the projectile being curved.

A projectile has two separate motions at right angles to each other.

Each motion is independent of the other.

The horizontal motion is at a constant velocity since there are no forces acting horizontally (air

resistance can be ignored).

So 𝑑ℎ = 𝑣ℎ𝑡

The vertical motion is one of constant acceleration, equal to g (9.8 ms-2 on earth).

For projectiles which are projected horizontally, the initial vertical velocity is zero.

For vertical calculations, use

vv = uv + at where uv = 0 and a = g

horizontal

distance (m)

horizontal

velocity (ms-1)

time (s)

final vertical

velocity (ms-1)

initial vertical

velocity (ms-1) acceleration

due to gravity

(ms-2)

time (s)


35

Velocity time graphs for horizontal and vertical motion

Horizontal motion Vertical motion

constant velocity constant acceleration from rest

horizontal distance can be found using vertical height can be found from area

area under graph (equivalent to dh=vht) under graph (equivalent to h=½ vmaxt)

Example

An aircraft flying horizontally at 200 ms-1, drops a food parcel which lands on the ground 12 seconds

later.

Calculate:

a) the horizontal distance travelled by the food parcel after being dropped

b) the vertical velocity of the food parcel just before it strikes the ground

c) the height that the food parcel was dropped from.

Solution

(a useful layout is to present the horizontal and vertical data in a table)

(then select the data you need for the question, and clearly indicate whether you are using horizontal

or vertical data).

Horizontal Vertical

dh = ? uv = 0

vh = 200 ms-1 vv = ?

t = 12 s a = 9.8 ms-2

t = 12 s

velo

city

(m

s-1)

time (s) 0 0

velo

city

(m

s-1)

0 0

time (s)

initial vertical velocity always

0 for a projectile launched

horizontally

acceleration due

to gravity

time of flight the same for

both horizontal and vertical


36

(a) Horizontal:

dh = ? dh = vh t

vh = 200 ms-1 = 200 x 12

t = 12 s = 2400 m

(b) Vertical:

uv = 0 vv = uv + at

vv = ? = 0 + 9.8 x 12

a = 9.8 ms-2 = 118 ms-1

t = 12 s

(c) (Draw a velocity time graph of vertical motion)

height = area under v-t graph

= ½ x 12 x 118

= 708 m

0 0

velo

city

(m

s-1)

time (s)

118

12


37

Key Area – Satellites

Red Amber Green

1 Explain satellite orbits in terms of projectile motion.

2 Explain apparent weightlessness for astronauts in orbit around the Earth

3 Describe the dependence of period of orbit on height.

4 Describe the range of heights and functions of satellites in orbit around the earth, including geostationary and natural satellites.

5 Describe the use of parabolic reflectors to send and receive signals.

6 Describe the range of applications of satellites including telecommunications; weather monitoring; the use of satellites in environmental monitoring.

7 Describe the use of satellites in developing our understanding of the global impact of mankind’s actions.

8 State that a geostationary satellite has a period of 24 hours and an orbital altitude of 36 000 km.


38

Satellites and Projectile Motion

Newton’s Thought Experiment

Apparent Weightlessness

Astronauts in orbit around the Earth are in a constant state of free-fall. The spaceship, the

astronauts and everything inside are all accelerating towards the Earth due to gravity.

This is known as apparent weightlessness. The effects are because of gravity, and not because they

have escaped from the gravitational field of the Earth.

Period of a Satellite

• The period of a satellite is the time it takes to complete one orbit.

• The higher the satellite orbit the longer it takes to orbit.

Orbits, heights, periods and functions of satellites in orbit around the Earth

Name Orbit Height Orbital Period

Function

LANDSAT 7 Polar 725 km 103 min Mapping

TIROS Polar 7000 km 256 min Weather

Echostar 3 Around Equator

36,000 km

24 hours Communication

Newton’s thought experiment allowed us to

understand satellite orbits.

If a projectile is launched with sufficient horizontal

velocity, it will travel so far that the curvature of the

Earth must be taken into account.

Now imagine a projectile launched with such a great

horizontal velocity that it never reaches the ground! It

will continue to circle the Earth until its horizontal

velocity decreases.

The satellite orbits around the Earth because it is in

constant free-fall due to the Earth’s gravity.


39

The Moon

• The Moon is a natural satellite of the Earth

• Orbital height is around 385 000 km

• Orbital period is around 27 days

Geostationary Satellite

• A satellite that stays above the same point on the Earth at all times.

• Its period is 24 hours.

• It has an orbital altitude of 36 000 km.

Intercontinental Communication

• A signal is sent from A to the satellite.

• The satellite amplifies the signal and sends the signal to station B.

• Curved Reflectors are used at each stage

A B


40

Satellite Communication – Parabolic (Curved) Reflectors

Receiving the signal Transmitting the signal

curved

reflector curved

reflector

receiving

aerial

transmitting

aerial

Curved reflectors are used to increase the strength of a received signal from a satellite or other source. The curved shape of the reflector collects the signal over a large area and brings it to a focus. The receiving aerial is placed at the focus so that it receives a strong signal.

Curved reflectors are also used on certain transmitters to transmit a strong, parallel signal beam. In a dish transmitter the transmitting aerial is placed at the focus and the curved shape of the reflector produces a parallel signal beam.


41

Key Area – Space Exploration

Red Amber Green

1 State that our current understanding of the universe is a result of our continual observation and exploration of space.

2 State that the accepted model of the development of the universe is known as the Big Bang theory.

3 Describe the main ideas behind the Big Bang theory in simple terms of origin and expansion.

3 State that the Big Bang Theory describes the universe as having evolved over a long period of time.

4 Provide evidence that supports the Big Bang Theory (e.g. Doppler red-shift, cosmic microwave background).

5 Describe how space exploration has improved our understanding of planet Earth – including the use of satellites for environmental monitoring.

6 Describe some technologies arising from space exploration.

7 Describe the impact that technologies arising from space exploration have on everyday life.

8 Describe the benefits associated with space exploration.

9 Describe the risks associated with space exploration.

10 Awareness of the challenges of space travel.

11 Explain the challenges of re-entry to a planet’s atmosphere in terms of frictional heating.

12 Describe the design of a heat shield to protect spacecraft on re-entry, in terms of a material’s specific heat capacity and latent heat of vapourisation

13 Carry out calculations involving heating on re-entry involving change in kinetic energy (Ek= ½mv2) and work done by friction (Ew = Fd)

Key Area – Space Exploration

Current understanding of the universe

What we know about the Universe is a result of humankind’s continual curiosity about our existence.

This has led to a remarkable understanding of our Universe which is continually developing. Much of

this can be attributed to our continual observation of space along with our exploration of the Solar

System and beyond.


42

Big Bang Theory (http://big-bang-theory.com)

• The Big Bang theory is an effort to explain what happened at the very beginning of our universe.

• Discoveries in astronomy and physics have shown beyond a reasonable doubt that our universe did in fact have a beginning. Prior to that moment there was nothing; during and after that moment there was something: our universe.

• Our universe sprang into existence as a "singularity" around 13.7 billion years ago.

• Singularities are zones which defy our current understanding of physics. They are thought to exist at the core of Black Holes.

• Black holes are areas of intense gravitational pressure. The pressure is thought to be so intense that finite matter is compressed into infinite density.

• Our universe is thought to have begun as an infinitesimally small, infinitely hot, infinitely dense singularity.

• After its initial appearance, the universe inflated, expanded and cooled, going from very, very small and very, very hot, to the size and temperature of our current universe.

• It continues to expand and cool to this day and we are inside of it!

• There is currently debate as to whether the universe will continue expanding, or whether it will start to contract.

Big Bang Theory - Evidence for the Theory

• Galaxies appear to be moving away from us at speeds proportional to their distance. This is called "Hubble's Law," named after Edwin Hubble (1889-1953) who discovered this phenomenon in 1929. This observation supports the expansion of the universe and suggests that the universe was once compacted.

• “Doppler red-shift” also provides evidence that galaxies are moving away from us. The light from galaxies appears to be more red than it should be, and the decreased frequency of the light tells us that the galaxies are moving away from us.

• If the universe was initially very, very hot as the Big Bang suggests, we should be able to find some remnant of this heat. In 1965, Arno Penzias and Robert Wilson discovered a 2.725 degree Kelvin (-270.425 degree Celsius) Cosmic Microwave Background radiation (CMB) which pervades the observable universe. This is thought to be the remnant which scientists were looking for. Penzias and Wilson shared in the 1978 Nobel Prize for Physics for their discovery.

• The abundance of the "light elements" Hydrogen and Helium found in the observable universe are thought to support the Big Bang model of origins.

Space exploration and our understanding of the Earth

Satellites have allowed us to make observations of the Earth.

Important observations of the environment have been made using monitoring satellites in orbit around the Earth, including:

• reduction of rainforest

• melting of polar icecaps

http://big-bang-theory.com/


43

Technologies arising from space exploration

Some examples include

• Weather forecasting

• GPS and Sat Nav

• Global communication

• Satellite TV

• Protective paints

• Scientific discovery and space exploration (for example Hubble telescope and the ISS).

NASA has a website called “Spin-off” which shows how technologies developed in their Space Programme have benefits in everyday life –

http://spinoff.nasa.gov/index.html

http://spinoff.nasa.gov/Spinoff2008/tech_benefits.html

http://www.nasa.gov/externalflash/nasacity/index2.htm

For any technology that you provide, you must be able to describe the impact that it has on our everyday life.

http://spinoff.nasa.gov/index.html

http://spinoff.nasa.gov/Spinoff2008/tech_benefits.html

http://www.nasa.gov/externalflash/nasacity/index2.htm


44

Re-entry into the atmosphere

• When a spacecraft re-enters the Earth’s atmosphere, it is travelling at a velocity of around

11 000 ms-1.

• The force of air resistance from the Earth’s atmosphere is huge at these velocities.

• Air resistance (force of friction) does work on the spacecraft which changes the kinetic

energy to heat (Ek→Eh).

o Recall from S3 – Work done, Ew = F d

o Ek = ½ mv2

• The heat absorbed can cause a temperature increase of around 1300 °C

Heat Shield Design

Case Study – the Shuttle

The Shuttle was the first (and only) reusable spacecraft. The first Shuttle mission was launched in

1981 and the final mission was in July 2011.

The part of the Shuttle that returns to Earth is called the Orbiter and its shape resembles an aircraft.

(For more information, see http://www.nasa.gov/mission_pages/shuttle/main/index.html).

• The Shuttle Orbiter is made from aluminium alloy covered in special tiles to protect it from

the intense heat generated during re-entry.

• The Shuttle needs around 34 000 thermal protection tiles (all of different shapes and sizes).

• The tiles are made of a material called silica, which has a high specific heat capacity and a

high melting point (c=1040 Jkg-1C-1, melting point = 1610 C). (See Electricity and Energy

Unit)

• The tiles are painted black so that heat is lost to the surroundings. The air around the

shuttle heats up. The temperature increase of the shuttle is therefore not as great.

http://www.nasa.gov/mission_pages/shuttle/main/index.html


45

Protection by vapourisation

Heat shields are also designed with coatings that vapourise. The temperature of the craft is

stabilised since the coating absorbs heat at a constant temperature while it melts and boils (Eh=ml).

Example

A heat shield on a spacecraft has a mass of 50 kg.

The spacecraft is travelling at 1000 ms-1. On re-entry into the Earth’s atmosphere, the velocity of the

spacecraft is reduced to 200 ms-1.

(a) Calculate the change in kinetic energy of the heat shield.

(b) The reduction in velocity occurs over a distance of 50 km.

Calculate the average frictional force acting on the heat shield

Solution

(a)

m = 50 kg Change in Ek = initial Ek – final Ek

u = 1000 ms-1 = (½mu2) – (½mv2)

v = 200 ms-1 = (½x50x10002) – (½x50x2002)

= 2.5x107 – 1.0x106

= 2.4x107 J

(b)

Ew = ΔEk = 2.4x107J EW = F d

F = ? 2.4x107 = F x 50000

d = 50 000 m 𝐹 =2.4×107

50000

= 4800 N


46

The Challenges of Space Travel

To travel the vast distances required in space ion drive can be used. This produces a small

unbalanced force that acts over a long time to attain a high velocity. Although the ions are very small

and do not produce a large force there is no friction in space and over a large time very high speeds

can be reached.

Gravitational Catapults can be used to gain kinetic energy. Interplanetary space probes often make

use of the "gravitational slingshot" effect to propel them to high velocities. For example, Voyager 2

performed a close flyby of Saturn on the 27th of August in 1981, which had the effect of slinging it

toward its flyby of Uranus on the 30th of January in 1986.


47

Solar Cells have to be used to produce energy to maintain life support systems in a spacecraft. This

can require very large area solar panels as the distance away from the Sun increases.

The Risks Associated With Manned Space Exploration

• The fuel used in rockets is extremely explosive. At take-off this can easily lead to disaster. In

1986 a leak of pressurised burning gas from O-rings in the Challenger Space Shuttle resulted

in the shuttle disintegrating over the Atlantic Ocean. All seven crew members died on the

mission.

• In space beyond the Earth’s atmosphere there is no protection from the radiation emitted

by the Sun. This can put astronauts at increased risk. Astronauts are exposed to

approximately 50-2,000 mSv while on six-month-duration missions to the International

Space Station (ISS), the moon and beyond. The risk of cancer caused by ionizing radiation is

well documented at radiation doses beginning at 50 mSv and above.[1][3][4]

• Space is a vacuum. This means that the air inside the space-vehicle is constantly exerting a

pressure outwards against the wall of the vehicle. This pressure differential can produce

damage to the craft or create leaks that will leave the astronauts with no air to breath. In

1970, the Soviet, Soyuz 11 spacecraft developed an air leak on re-entry. All three

cosmonauts died of suffocation due to an eleven minute exposure to a vacuum.

• As mentioned earlier in the notes, on re-entry through the atmosphere, the kinetic energy

of the spacecraft is converted to heat due to the force of friction. This heat energy can

produce high temperatures that will melt the spacecraft. In 2003 the space shuttle Columbia

burned up on re-entry killing all seven astronauts.

https://en.wikipedia.org/wiki/Spaceflight_radiation_carcinogenesis#cite_note-Cucinotta_and_Durante_2006-1

https://en.wikipedia.org/wiki/Spaceflight_radiation_carcinogenesis#cite_note-Cucinotta_and_Durante_2006-1

https://en.wikipedia.org/wiki/Spaceflight_radiation_carcinogenesis#cite_note-BIER_2006-4


48

Key Area – Cosmology

Red Amber Green

1 State that ‘light year’ is a measure of distance.

2 State that one light year is the distance that light travels in one year.

3 Calculate the number of metres in one light year.

4 Carry out calculations involving distance, speed and time for light on the light year scale.

5 State that current thinking dates the universe as being approximately 14 billion years old.

6 Describe features of the observable universe correctly in context: planet, dwarf planet, moon, sun, asteroid, solar system, star, exoplanet, galaxy, black hole, universe.

7 Classify as members of the electromagnetic spectrum the following radiations: gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, TV and radio.

8 List the above radiations in order of wavelength (and frequency).

9 Give an example of a detector for each of the above radiations .

10 Explain why different kinds of telescope are used to detect signals from space.

11 State that the line spectrum produced by a source provides information about the atoms within the source.

12 Identify continuous and line spectra.


49

Key Area – Cosmology

The Light Year

• The “light year” is a measurement of distance.

• 1 light year is the distance that light travels in 1 year.

Example

(a) Calculate the number of metres in 1 light year.

(b) Proxima Centauri, the next closest star after the Sun, is 4.3 light years from the Earth.

Calculate this distance in metres.

Solution

(a) d = ? d = vt

v = 3x108 ms-1 = 3x108x(365.25x24x60x60)

t = 1 year = 9.47x1015 m

= 365.25x24x60x60 s

(b) 1 light year = 9.47x1015 m

4.3 light years = 4.3x9.47x1015 m

= 4.07x1016 m

Proxima Centauri is 4.07x1016 m from Earth

Age of the universe

Current thinking dates the age of the universe as approximately 14 billion years.


50

Observable universe

• The universe consists of many galaxies separated by empty space.

• A galaxy is a large cluster of stars (e.g. theMilky Way).

• A star is a massive ball of matter that is undergoing nuclear fusion and emitting light and

heat. The Sun is a star.

• The sun and many other stars have a solar system. A solar system consists of a central star

orbited by planets.

• A planet is a large ball of matter that orbits a star (e.g. Earth or Jupiter). Planets do not emit

light themselves.

• An exoplanet is a planet orbiting around another star (but not our Sun)

• Many planets have moons. A moon is a lump of matter that orbits a planet (e.g. the Moon

orbits the Earth. Deimos and Phobos orbit Mars.

• A dwarf planet is a large ball of matter that orbits the Sun but is too small to be regarded as

a planet. There are a hundred or so objects in the solar system that are believed to be dwarf

planets.

• An asteroids is a large, irregularly shaped object in space that orbits our Sun. Around a

million asteroids are located between Mars and Jupiter in an area called the “asteroid belt.”

They are too small to be regarded as planets.

These are features of the observable universe. The universe also contains matter that cannot be

observed directly, e.g. dark matter.

Exoplanets and Life Beyond Our Solar System

• An exoplanet is a planet orbiting around another star (but not our Sun).

• To support life-as-we-know-it, an exoplanet must

• Not be too hot or too cold

• have liquid water

• have air with oxygen in it

• not be a gas giant (must have rocks)

Electromagnetic Spectrum

Gamma rays X-rays Ultraviolet Visible light Infrared Microwaves Radio + TV

~1020 ~1018 ~1016 ~1015 ~1012 ~108 ~104

increasing wavelength (m)

increasing frequency (Hz)


51

Detectors of electromagnetic radiation

Radiation Detector

Gamma rays Geiger-Müller tube

X-rays Photographic film

Ultraviolet Fluorescent paint

Visible light Photographic film

Infrared Charge-coupled diode (CCD)

Microwaves Diode

Radio + TV Waves Aerial

Spectra

• White light is made up of a range of colours.

• These colours can be separated by splitting white light with a prism to obtain a spectrum.

• A spectrum can also be produced using a diffraction grating.

• A continuous spectrum is a complete spectrum with all visible colours (frequencies)

represented

• A line absorption spectrum consists of a complete (continuous) spectrum with certain

colours missing which appear as black lines in the spectrum.


52

• A line emission spectrum consists of lines of light of distinct colours rather than a

continuous spectrum.

• Every element produces a unique line spectrum. Studying line spectra allows the elements

present in a light source (e.g. a star)to be identified

• This can help to identify the type, distance, age or speed of a star.

Documents

Marr College - Physics Unit Dynamics and · PDF fileMarr College Physics Department National 4/5 Summary Notes Dynamics and Space Unit 2 Key Area – Velocity and Displacement –