Download Unit 1 Section 2 - Belfast Royal Academy

In order to get the G.C.S.E. grade you are capable of, you must make
your own revision notes using your Physics notebook.
When summarising notes, use different colours and draw
diagrams/pictures. If you do, you will find them easier to remember.
Once you have made revision notes for a topic, re-visit these regularly
(on the day of your examination you will not remember something you
revised 4 weeks previously). Each time you re-visit a note tick the top
of the page/card. This will allow you to identify any notes you have
neglected.
WARNING: DO NOT RELY SOLELY ON THE REVISION
POWERPOINTS!
Scalar and Vector Quantities
A scalar quantity is one which has magnitude (size) only, e.g.
distance, mass, speed, density, volume etc.
A vector quantity is one which has magnitude and direction, e.g.
displacement, velocity, acceleration, force and weight.
Distance
Example
Distance is the length of path taken by any
moving object. It is a quantity which only has
magnitude, and is therefore a scalar quantity.
A car travels from town A to town
B, town B to town C and town C
to town D.
5 km
A
B
2 km
2 km
D
C
5 km
Displacement
Displacement is the distance measured in a
straight line between two points. It is a
quantity which has both magnitude and
direction, and is therefore a vector
quantity.
What distance does the car
travel? 12 km
What is the displacement of the
car? 2 km
Speed
The distance travelled in unit time. Speed is a scalar quantity.
Velocity
This is the distance travelled in unit time in a certain direction. Velocity is a
vector quantity.
Average Speed
average speed = distance /time
Unit: m/s
distance
distance = average speed x time
time = distance / average speed
A speed of 5 m/s means the object moves 5 m every second
Speed x time
Average Velocity
average velocity = displacement / time
Unit: m/s
displacement = average velocity x time
time = displacement / average velocity
displacement
Velocity x time
When velocity changes we say there is an acceleration.
Acceleration is the rate of change of velocity (i.e. change in velocity per sec.).
Acceleration (Rate of Change of Speed)
Acceleration = change in vel /time taken for change
= (final vel – initial vel) / time
unit: m/s2
An acceleration of 5 m/s2 means that the velocity increases by 5m/s every second.
A constant or uniform acceleration means that the velocity increases by equal
amounts every second.
If the object or body is slowing down it has a negative acceleration or deceleration.
Paul Biedermann holds the world record for the 200 m freestyle (4 lengths of
an olympic sized swimming pool). The record is 1 minute 42 seconds.
(a) Calculate his average speed
Average speed = distance / time
Average speed = 200 / 102
Average speed = 1.96 m/s
(b) Calculate his average velocity
Average velocity = displacement / time
Since Biedermann finishes at the same end as he started, his total
displacement from his starting position = 0 m
 Average velocity = 0 / 102 = 0 m/s
Definitions and Formulae
Equations of Motion for a body moving with uniform acceleration
The body has:
acceleration = a m/s2
initial velocity = u m/s
final velocity = v m/s
travels for a time = t seconds
and its displacement = s metres
v = u  at
s = ut  ½ at2
v2 = u2  2as
s = ½ (u  v)t
When using these equations remember the following:
•
•
•
•
If a body is slowing down (decelerating) the value for “a” is negative.
All quantities must have the correct units.
Write down any values that you are given.
Choose a direction as positive and ensure all quantities acting in the
opposite direction are negative
• Choose the appropriate equation i.e. the one that will only have one
unknown.
• Check that the signs are correct.
Example 1
A bus, starting from rest, has a constant acceleration of 2m/s2.
Calculate (a) How far it will travel in the first 5 seconds (b) What its
speed is after 5 seconds (c) The average speed during the first 5
seconds (d) How far it will have travelled by the time it has reached
20m/s.
(a) u = 0, a = 2, t = 5, s =?
Positive Direction
a=2
m/s2
Use: s = ut + ½ at2
s = 0 + ½ (2 x 52)
s = 25 m
(b) u = 0, a = 2, t = 5, v =?
Use: v = u + at
u = 0 m/s
v = 0 + (2 x 5)
v = 10 m/s
(c) av. Speed = distance/time
(d) u = 0, a = 2, v = 20, s = ?
Av. Speed = 25/5
Use: v2 = u2 + 2as
Av. Speed = 5 m/s
400 = 0 + 2(2s)
400 = 4s  s = 100 m
Click for
solutions
Example 2
An object is dropped from a height of 45m, calculate
(a) The time taken to reach the ground, and (b) its maximum velocity.
u = 0 m/s
a = 10 m/s2
Click for
solutions
Positive Direction
(a) u = 0, a = 10 , s = 45, t = ?
(b) u = 0, a = 10 , s = 45, v = ?
Use: s = ut + ½ at2
Use: v2 = u2 + 2as
45 = 0 + ½ (10 x t2)
v2 = 0 + 2(10 x 45)
45 = 5t2
v2 = 900
t2 = 9
v = 30 m/s
t=3s
Graphs
1. Gradient of displacement-time graph =
velocity
2. Gradient of velocity-time graph
=
Acceleration
3. Area between graph
and time axis of velocity-time graph
=
Displacement
4. Gradient of distance-time graph
=
Speed
5. Area between graph and time axis
of speed-time graph
=
distance
Graphs:
Distance / time graphs
distance
distance
distance
time
time
Object stationary
Object moving with
constant speed
Gradient of graph
= speed
time
Object’s speed
increasing
Graphs:
Displacement / time graphs
displacement
displacement
displacement
time
Object stationary
time
Object moving with
constant velocity
Gradient of graph
= velocity
time
Object accelerating
(gradient increasing
 velocity increasing)
Graphs:
Velocity / time graphs
Velocity
Velocity
Area under
graph
Area under graph
= displacement
= displacement
time
Object moving with
constant velocity
(gradient is zero)
time
Object is accelerating
Gradient = acceleration
(1) Using the displacement-time graph, describe the motion of the particle.
What is the particle’s velocity during the first 2 seconds? What is the particle’s
velocity during the last 4 seconds?
(a)
Displacement (m)
0 – 2 s  constant positive velocity
6
5
4
3
2
1
2 – 6 s  at rest
6 – 10 s  constant negative velocity
(b)
Velocity = gradient
0
Time (s)
1 2 3 4 5 6 7 8 9 10
Click for
solutions
Velocity = 4/2 = 2 m/s
(c)
Velocity = gradient
Velocity = -4/4 = -1 m/s
(1 m/s in the negative direction)
(2) Using the velocity-time graph, describe the motion of the particle. What is
the particle’s acceleration during the first 2 seconds? What is the particle’s
acceleration during the last 4 seconds? What is the particle’s displacement
after 10s?
(a)
0 – 2 s  constant positive acceleration
Velocity (m/s)
Click for
solutions
6
5
4
3
2
1
6 – 10 s  constant deceleration
(b)
Acceleration = gradient
A
0
(d)
2 – 6 s  constant velocity
B
Acceleration = 4/2 = 2 m/s2
C
Time (s)
1 2 3 4 5 6 7 8 9 1
0
(c)
Acceleration = gradient
Velocity = -4/4 = -1 m/s2
Displacement = area between graph and time axis = Area A + Area B + Area C
Displacement = ½ (4 x 2) + (4 x 4) + ½ (4 x 4) = 28 m
Newton’s Laws of Motion
1. An object will remain at rest or move with a constant velocity
unless acted upon by a resultant force.
2. The acceleration, a, of a body is directly proportional to the
resultant force, F, acting on the body and inversely proportional to
the mass, m, of the body.
One newton is defined as:
the force which gives a 1kg mass an acceleration of 1m/s2
∴
F = ma
3. If object A exerts a force on object B, object B exerts an
equal but opposite force on object A
Click to see
what happens
the following
objects.
Acceleration due to Gravity
Ignoring air resistance, all bodies falling under the force of
gravity do so with a constant acceleration called the
acceleration due to gravity, which is denoted by the letter g.
(It is sometimes called the gravitational field strength).
The value of g is taken as 10m/s2.
The weight of an object is given by the equation:
Weight = mass x acceleration of free fall
W = mg
Note: in reality air resistance will act on any falling object.
Therefore, the acceleration will not be as large as 10 m/s2.
The air resistance increases with speed. DON’T WORRY
THOUGH AS AIR RESISTANCE IS IGNORED IN
CALCULATIONS.
Circular Motion
When a mass moves in a circle it is always changing direction, and
therefore the velocity is always changing. This means there must be
an acceleration and a force (F=ma)
As the speed is not changing:
The (accelerating) force is always towards the centre of the circle and is
called the centripetal force.
m
m = mass
v = velocity
r = radius of the path
r
v
Examples of circular motion:
Example of circular motion
Centripetal force provided by
A car going round a corner
Friction between the tyres and
the road
Earth orbiting the sun
Gravity
Moon orbiting the earth
Gravity
Satellite orbiting earth
Gravity
A mass held in a circle by a
string
Tension in the string
The centripetal force increases:
1. As the mass increases
2. As the speed increases
3. As the radius of the circle decreases
Note: if the centripetal
force is removed all
objects will travel in a
straight line at a tangent
to the circle at the point
where the force has been
removed
Click to see what happens
when the centripetal
force is removed.
Momentum
The momentum of a body depends on its mass and its velocity.
It is the product of the body’s mass and velocity.
momentum = mass (kg) x velocity (m/s)
Unit: kgm/s
Momentum is a vector quantity, i.e. it has magnitude and direction.
Conservation of momentum
The Law of Conservation of Momentum states:
Momentum is conserved when two bodies collide
or act on one another.
This means that the total momentum before the collision or action is
equal to the total momentum after the collision or action.
Two bodies are moving towards each other. One has a mass of 10 kg
and a velocity of 4 m/s, the other has a velocity of –3 m/s and a
momentum of –15 kgm/s. Calculate (a) the momentum of the 4kg
mass, (b) the mass of the other object, and (c) the total momentum
v = 4m/s
v = 3m/s
10 kg
?
M=?
M = -15 kgm/s
(a) Mom. = mass x vel.
(b) Mom. = mass x vel.
Mom. = 10 x 4
-15 = mass x -3
Mom. = 40 kgm/s
mass = 5 kg
(c) Total mom. = mom. of car A + mom. of car B
Total mom. = 40 – 15 = 25 kgm/s to the right
Click for
solutions
If the cars in the previous example collide and stick together what
will their final velocity be?
Momentum before = momentum after
25 = mass x velocity
25 = 15 x velocity
Velocity = 1.67 m/s
Click for
solution
Change in Momentum
In order for the momentum of an object to change a resultant
force must act on it:
change in momentum = force x time
Examples
1.
Calculate the change in momentum when Robin Van Persie
strikes a football with a force of 50N for 0.5s.
change in momentum = force x time
= 50 x 0.5
= 25 kgm/s
Click for
solution
2. A boxer caused a change in momentum of 30kgm/s when he
struck a punch bag for 0.01s. What force did he hit the
bag with?
force = change in momentum / time
= 30 / 0.01
= 3000 N
Click for
solutions
3. A swimmer pushes off the edge of the pool by applying a
force of 100N. What was her time of contact if her change
in momentum was 40kgm/s?
time = change in momentum / force
= 40 / 100
= 0.4 s
Safety Features on the Road
Crumple zones, airbags,
seatbelts and roadside
barriers all increase the
time it takes a car and its
passengers to stop.
Re-arranging the previous
equation we get:
Force = momentum change
time
Therefore, the bigger the
time the smaller the
force and the greater the
chance of survival.
Time = Total distance / Average speed
Time = 25 / 0.8
Time = 31.25 s
Click for
solutions
Constant speed  Resultant force = 0
Therefore, frictional force = 50 N
0.1
Click for
solutions
F = ma
0.3 – 0.1 = 0.04 x a
0.2 = 0.04a
a = 0.2 / 0.04
a = 5 m/s2
Distance is the length of path followed. It is a scalar quantity.
Displacement is straight line distance from start to finish. It is a
vector quantity
Click for
solutions
20 m
Gerry  25 m; Robert  10 m
Separation = 15 m
See previous slide
They meet when their displacement
is the same – 4 s
Click for
solutions
He runs 30 m during the first 6
seconds. He then runs another 30
m between 8 s and 14 s
Therefore he runs a total of 60 m
Total Displacement / Time
Displacement = 0 at start
Displacement = 0 after 14 s
Therefore, Robert is 0 m from his starting position
Average Velocity = Total Displacement / Time
Average Velocity = 0 / 14
Average Velocity = o m/s
Click for solutions
Average speed = total distance / time
Average speed = 3000 / 300
Average speed = 10 m/s
Click for
solutions
A
Acceleration = gradient
Acceleration = 14 / 20
Acceleration = 0.7 m/s2
B
C
F = ma
3000 – Friction = 1500 x 0.7
3000 – Friction = 1050
Friction = 3000 – 1050
Friction = 1950 N
Displacement = area below the graph
Displacement = area A + area B + area C
Displacement = ½(20 x 20) + (80 x 20) + ½(20 x 20)
Displacement = 200 + 1600 + 200
Displacement = 2000 m
Click for
solutions
Constant velocity
Constant deceleration
At rest
Click for
solutions
Momentum = mass x velocity
Momentum = 1500 x 14
Momentum = 21,000 kgm/s
Momentum before = Momentum after
(1200 x 5) + 2000 = mass x velocity
8000 = 2000 x velocity
Velocity = 4 m/s
Conservation of momentum
Click for
solutions
Increases