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1 of 42 © Boardworks Ltd 2009
© Boardworks Ltd 2009
2 of 42
3 of 42 © Boardworks Ltd 2009
What is a scalar?
Scalar quantities are measured with numbers and units.
length
(e.g. 102°C)
time
(e.g. 16cm)
temperature
(e.g. 7s)
4 of 42 © Boardworks Ltd 2009
What is a vector?
Vector quantities are measured with numbers and units, but
also have a specific direction.
acceleration
(e.g. 30m/s2
upwards)
displacement
(e.g. 200miles
northwest)
force
(e.g. 2N
downwards)
5 of 42 © Boardworks Ltd 2009
Comparing scalar and vector quantities
6 of 42 © Boardworks Ltd 2009
Speed or velocity?
Distance is a scalar and displacement is a vector.
Similarly, speed is a scalar and velocity is a vector.
Speed is the rate of change of distance in
the direction of travel. Speedometers in cars
measure speed.
Velocity is a rate of change of displacement and has both
magnitude and direction.
average
speed
average
velocity
Averages of both can be useful:
distance
time
displacement
time
= =
7 of 42 © Boardworks Ltd 2009
Vector or scalar?
8 of 42 © Boardworks Ltd 2009
Adding vectors
Displacement is a quantity that is independent of the
route taken between start and end points.
Any two vectors of the same type can be added in this way
to find a resultant.
Two or more displacement
vectors can be added
‘nose to tail’ to calculate a
resultant vector.
A
B C
resultant vector
If a car moves from A to B and then to C, its total
displacement will be the same as if it had just moved in a
straight line from A to C.
9 of 42 © Boardworks Ltd 2009
Simplifying vectors
Because of the way vectors are added, it is always possible to
simplify a vector by splitting it into components.
A
B
C
Imagine that instead of travelling via B, the car travels via D:
D
The car has a final displacement of x miles east and y miles
north. This can be represented by (x,y).
x
y
Its displacement is the
same, but it is now much
easier to describe.
x component
y
component
How would you describe
the car’s displacement in
component terms?
10 of 42 © Boardworks Ltd 2009
Understanding vector calculations
© Boardworks Ltd 2009
11 of 42
12 of 42 © Boardworks Ltd 2009
Motion under constant acceleration
Calculating vector quantities such as velocity or displacement
can be complicated, but when acceleration is constant, four
equations always apply.
v u + at
=
v2 u2 + 2as
=
s ½(u + v)t
=
s ut + ½at2
=
These are sometimes known as the constant acceleration
equations, or the ‘uvast’ or ‘suvat’ equations.
What do the symbols u, v, a, s and t represent?
13 of 42 © Boardworks Ltd 2009
Representing motion
The symbol a represents acceleration.
The symbol t represents time.
Velocity at time t is represented by v, and u represents the
value of v when t = 0. This is the initial velocity.
 a = acceleration
 t = time
 s = displacement
 v = velocity
 u = initial velocity
Displacement, s is always measured relative to a starting
position, so it is always true that when t = 0, s = 0.
14 of 42 © Boardworks Ltd 2009
Two velocity equations
Two of the constant acceleration equations are velocity
equations.
The first of these can be used to find the velocity at a
particular time t. The second can be used to find the velocity
at a particular displacement s.
When deciding which equation to
use, it is good practice to write
down what you know about the
values of u, v, a, s and t before
you start any calculations.
v2 u2 + 2as
=
v u + at
=
u = ?
v = ?
a = ?
s = ?
t = ?
15 of 42 © Boardworks Ltd 2009
Calculating final velocity
A cyclist accelerates towards the end of the race in order to
win. If he is moving at 6m/s then accelerates by 1.5m/s2 for
the final five seconds of the race, calculate his speed as he
crosses the line.
u = 6
v = ?
a = 1.5m/s2
s =
t = 5s
v = 6 + (1.5 × 5)
v = u + at
v = 13.5m/s
First write down what you know about u, v, a, s and t:
The question gives a value
for t. What is the relevant
equation?
16 of 42 © Boardworks Ltd 2009
Rearranging a velocity equation
17 of 42 © Boardworks Ltd 2009
Rearranging an equation – calculations
A coin is dropped from a window. If it hits the
ground at 10m/s, work out the height of the window.
u = 0m/s
v = 10m/s
a = 10m/s2
s = ?
t =
s = 5m
The question involves the
variables u, v, a and s, so the
relevant equation is v2 = u2 + 2as.
s =
s =
s =
2a
v2 – u2
2 x 10
102 – 02
20
100
First write down what you know about u, v, a, s and t:
Start by rearranging the
equation to find a formula for s:
© Boardworks Ltd 2009
18 of 42
19 of 42 © Boardworks Ltd 2009
Two displacement equations
Two of the constant acceleration equations are displacement
equations.
The first of these gives the displacement as a function of the
initial velocity and velocity at time t. The second gives it as a
function of initial velocity, acceleration and time.
As with the velocity equations, it is
good practice to write down what
you know about the values of u, v,
a, s and t before you attempt any
calculations.
u = ?
v = ?
a = ?
s = ?
t = ?
s ½(u + v)t
= s ut + ½at2
=
20 of 42 © Boardworks Ltd 2009
Calculating displacement
A car travelling at 20m/s takes five seconds to stop.
What is the stopping distance of the car?
u = 20m/s
v = 0m/s
a =
s = ?
t = 5s
s = 50m
s = ½ × 20 × 5
s = ½ × (20 + 0) × 5
First write down what you know about u, v, a, s and t:
The question gives u, v and t and asks for a value for s, so the
relevant equation is s = ½(u + v)t:
21 of 42 © Boardworks Ltd 2009
Rearranging a displacement equation
22 of 42 © Boardworks Ltd 2009
Rearranging an equation – calculations
Calculate the acceleration of an ambulance if it starts at rest
and takes six seconds to travel 50m.
u = 0m/s
v =
a = ?
s = 50m
t = 6s
a =
a = 2.8m/s2
50
a =
18
a =
½ × 6 × 6
50 – (0 × 6)
½t2
s – ut
First write down what you know about u, v, a, s and t:
The question involves the
variables u, a, s and t, so the
relevant equation is s = ut + ½at2.
Start by rearranging the
equation to find a formula for a:
23 of 42 © Boardworks Ltd 2009
Using the equations of motion
© Boardworks Ltd 2009
24 of 42
25 of 42 © Boardworks Ltd 2009
Describing parabolic motion
 A trajectory is the path of any moving object.
 A projectile is an object that is given an initial force,
then allowed to move freely through space.
 A parabola is the name given to the shape of the
curve a projectile follows when gravity is the only
force acting on it.
What are some of the terms that are used to describe
projectile motion?
26 of 42 © Boardworks Ltd 2009
Forces on a projectile
What are the two forces that act on a freely moving projectile?
Air resistance and gravity.
Depending on the shape and density of
an object, it is often possible to ignore the
effects of air resistance.
This reduces the forces on the object to one
constant force in a constant direction, giving:
Horizontal motion is therefore very simple, and vertical motion
can be solved with the constant acceleration equations.
 constant horizontal velocity
 constant vertical acceleration.
27 of 42 © Boardworks Ltd 2009
When is a trajectory a parabola?
28 of 42 © Boardworks Ltd 2009
Understanding projectile motion
29 of 42 © Boardworks Ltd 2009
Forces on a projectile
30 of 42 © Boardworks Ltd 2009
Motion under gravity
31 of 42 © Boardworks Ltd 2009
Solving projectile motion
How should you go about solving a problem involving
projectile motion? Split the problem into two parts:
Horizontal motion at constant velocity:
 Use the equation, speed = distance / time (equation 1).
Vertical motion under constant acceleration:
 As always, start by writing down what you already know
about u, v, a, s and t.
 Choose a constant acceleration equation (equation 2).
The order you use these two equations in will depend on the
nature of the problem. Start by writing down everything you
know, and the rest should follow!
32 of 42 © Boardworks Ltd 2009
Projectile calculations
© Boardworks Ltd 2009
33 of 42
34 of 42 © Boardworks Ltd 2009
Momentum
Momentum is a vector quantity given by the following formula:
To have momentum, an object needs to have mass and to
be in motion.
momentum = mass × velocity
Which of these two
vehicles do you
think has a higher
momentum?
35 of 42 © Boardworks Ltd 2009
Conservation of momentum
initial
momentum
final
momentum
=
Momentum is always conserved in any event or interaction:
When a snooker ball collides with another, what happens to
the momentum of each ball?
What happens when a car brakes and comes to a halt, or
when a rock falls to the ground and stops?
Momentum is transferred to the Earth.
Momentum is not created or lost, but transferred from one
object to the other.
36 of 42 © Boardworks Ltd 2009
Explosions and recoil
Remember: momentum is always conserved.
What happens to the momentum in the following situations?
37 of 42 © Boardworks Ltd 2009
Calculating momentum
Momentum is a vector quantity. The van has a momentum of
1500 × 10 = 15000 kgm/s to the right. The car’s momentum is
15000kgm/s to the left, or –15000kgm/s to the right:
10m/s 10m/s
Two vehicles, each weighing 1.5tonnes, are driving towards
each other at 30mph. If they collide, what will happen?
Both vehicles come to a halt. What has happened to their
momentum?
initial momentum to the right = 15000 + –15000 = 0
final momentum to the right = 0
38 of 42 © Boardworks Ltd 2009
Momentum calculations
39 of 42 © Boardworks Ltd 2009
© Boardworks Ltd 2009
40 of 42 © Boardworks Ltd 2009
Glossary
41 of 42 © Boardworks Ltd 2009
Anagrams
42 of 42 © Boardworks Ltd 2009
Multiple-choice quiz

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Forces and Motion.ppt

  • 1. 1 of 42 © Boardworks Ltd 2009
  • 2. © Boardworks Ltd 2009 2 of 42
  • 3. 3 of 42 © Boardworks Ltd 2009 What is a scalar? Scalar quantities are measured with numbers and units. length (e.g. 102°C) time (e.g. 16cm) temperature (e.g. 7s)
  • 4. 4 of 42 © Boardworks Ltd 2009 What is a vector? Vector quantities are measured with numbers and units, but also have a specific direction. acceleration (e.g. 30m/s2 upwards) displacement (e.g. 200miles northwest) force (e.g. 2N downwards)
  • 5. 5 of 42 © Boardworks Ltd 2009 Comparing scalar and vector quantities
  • 6. 6 of 42 © Boardworks Ltd 2009 Speed or velocity? Distance is a scalar and displacement is a vector. Similarly, speed is a scalar and velocity is a vector. Speed is the rate of change of distance in the direction of travel. Speedometers in cars measure speed. Velocity is a rate of change of displacement and has both magnitude and direction. average speed average velocity Averages of both can be useful: distance time displacement time = =
  • 7. 7 of 42 © Boardworks Ltd 2009 Vector or scalar?
  • 8. 8 of 42 © Boardworks Ltd 2009 Adding vectors Displacement is a quantity that is independent of the route taken between start and end points. Any two vectors of the same type can be added in this way to find a resultant. Two or more displacement vectors can be added ‘nose to tail’ to calculate a resultant vector. A B C resultant vector If a car moves from A to B and then to C, its total displacement will be the same as if it had just moved in a straight line from A to C.
  • 9. 9 of 42 © Boardworks Ltd 2009 Simplifying vectors Because of the way vectors are added, it is always possible to simplify a vector by splitting it into components. A B C Imagine that instead of travelling via B, the car travels via D: D The car has a final displacement of x miles east and y miles north. This can be represented by (x,y). x y Its displacement is the same, but it is now much easier to describe. x component y component How would you describe the car’s displacement in component terms?
  • 10. 10 of 42 © Boardworks Ltd 2009 Understanding vector calculations
  • 11. © Boardworks Ltd 2009 11 of 42
  • 12. 12 of 42 © Boardworks Ltd 2009 Motion under constant acceleration Calculating vector quantities such as velocity or displacement can be complicated, but when acceleration is constant, four equations always apply. v u + at = v2 u2 + 2as = s ½(u + v)t = s ut + ½at2 = These are sometimes known as the constant acceleration equations, or the ‘uvast’ or ‘suvat’ equations. What do the symbols u, v, a, s and t represent?
  • 13. 13 of 42 © Boardworks Ltd 2009 Representing motion The symbol a represents acceleration. The symbol t represents time. Velocity at time t is represented by v, and u represents the value of v when t = 0. This is the initial velocity.  a = acceleration  t = time  s = displacement  v = velocity  u = initial velocity Displacement, s is always measured relative to a starting position, so it is always true that when t = 0, s = 0.
  • 14. 14 of 42 © Boardworks Ltd 2009 Two velocity equations Two of the constant acceleration equations are velocity equations. The first of these can be used to find the velocity at a particular time t. The second can be used to find the velocity at a particular displacement s. When deciding which equation to use, it is good practice to write down what you know about the values of u, v, a, s and t before you start any calculations. v2 u2 + 2as = v u + at = u = ? v = ? a = ? s = ? t = ?
  • 15. 15 of 42 © Boardworks Ltd 2009 Calculating final velocity A cyclist accelerates towards the end of the race in order to win. If he is moving at 6m/s then accelerates by 1.5m/s2 for the final five seconds of the race, calculate his speed as he crosses the line. u = 6 v = ? a = 1.5m/s2 s = t = 5s v = 6 + (1.5 × 5) v = u + at v = 13.5m/s First write down what you know about u, v, a, s and t: The question gives a value for t. What is the relevant equation?
  • 16. 16 of 42 © Boardworks Ltd 2009 Rearranging a velocity equation
  • 17. 17 of 42 © Boardworks Ltd 2009 Rearranging an equation – calculations A coin is dropped from a window. If it hits the ground at 10m/s, work out the height of the window. u = 0m/s v = 10m/s a = 10m/s2 s = ? t = s = 5m The question involves the variables u, v, a and s, so the relevant equation is v2 = u2 + 2as. s = s = s = 2a v2 – u2 2 x 10 102 – 02 20 100 First write down what you know about u, v, a, s and t: Start by rearranging the equation to find a formula for s:
  • 18. © Boardworks Ltd 2009 18 of 42
  • 19. 19 of 42 © Boardworks Ltd 2009 Two displacement equations Two of the constant acceleration equations are displacement equations. The first of these gives the displacement as a function of the initial velocity and velocity at time t. The second gives it as a function of initial velocity, acceleration and time. As with the velocity equations, it is good practice to write down what you know about the values of u, v, a, s and t before you attempt any calculations. u = ? v = ? a = ? s = ? t = ? s ½(u + v)t = s ut + ½at2 =
  • 20. 20 of 42 © Boardworks Ltd 2009 Calculating displacement A car travelling at 20m/s takes five seconds to stop. What is the stopping distance of the car? u = 20m/s v = 0m/s a = s = ? t = 5s s = 50m s = ½ × 20 × 5 s = ½ × (20 + 0) × 5 First write down what you know about u, v, a, s and t: The question gives u, v and t and asks for a value for s, so the relevant equation is s = ½(u + v)t:
  • 21. 21 of 42 © Boardworks Ltd 2009 Rearranging a displacement equation
  • 22. 22 of 42 © Boardworks Ltd 2009 Rearranging an equation – calculations Calculate the acceleration of an ambulance if it starts at rest and takes six seconds to travel 50m. u = 0m/s v = a = ? s = 50m t = 6s a = a = 2.8m/s2 50 a = 18 a = ½ × 6 × 6 50 – (0 × 6) ½t2 s – ut First write down what you know about u, v, a, s and t: The question involves the variables u, a, s and t, so the relevant equation is s = ut + ½at2. Start by rearranging the equation to find a formula for a:
  • 23. 23 of 42 © Boardworks Ltd 2009 Using the equations of motion
  • 24. © Boardworks Ltd 2009 24 of 42
  • 25. 25 of 42 © Boardworks Ltd 2009 Describing parabolic motion  A trajectory is the path of any moving object.  A projectile is an object that is given an initial force, then allowed to move freely through space.  A parabola is the name given to the shape of the curve a projectile follows when gravity is the only force acting on it. What are some of the terms that are used to describe projectile motion?
  • 26. 26 of 42 © Boardworks Ltd 2009 Forces on a projectile What are the two forces that act on a freely moving projectile? Air resistance and gravity. Depending on the shape and density of an object, it is often possible to ignore the effects of air resistance. This reduces the forces on the object to one constant force in a constant direction, giving: Horizontal motion is therefore very simple, and vertical motion can be solved with the constant acceleration equations.  constant horizontal velocity  constant vertical acceleration.
  • 27. 27 of 42 © Boardworks Ltd 2009 When is a trajectory a parabola?
  • 28. 28 of 42 © Boardworks Ltd 2009 Understanding projectile motion
  • 29. 29 of 42 © Boardworks Ltd 2009 Forces on a projectile
  • 30. 30 of 42 © Boardworks Ltd 2009 Motion under gravity
  • 31. 31 of 42 © Boardworks Ltd 2009 Solving projectile motion How should you go about solving a problem involving projectile motion? Split the problem into two parts: Horizontal motion at constant velocity:  Use the equation, speed = distance / time (equation 1). Vertical motion under constant acceleration:  As always, start by writing down what you already know about u, v, a, s and t.  Choose a constant acceleration equation (equation 2). The order you use these two equations in will depend on the nature of the problem. Start by writing down everything you know, and the rest should follow!
  • 32. 32 of 42 © Boardworks Ltd 2009 Projectile calculations
  • 33. © Boardworks Ltd 2009 33 of 42
  • 34. 34 of 42 © Boardworks Ltd 2009 Momentum Momentum is a vector quantity given by the following formula: To have momentum, an object needs to have mass and to be in motion. momentum = mass × velocity Which of these two vehicles do you think has a higher momentum?
  • 35. 35 of 42 © Boardworks Ltd 2009 Conservation of momentum initial momentum final momentum = Momentum is always conserved in any event or interaction: When a snooker ball collides with another, what happens to the momentum of each ball? What happens when a car brakes and comes to a halt, or when a rock falls to the ground and stops? Momentum is transferred to the Earth. Momentum is not created or lost, but transferred from one object to the other.
  • 36. 36 of 42 © Boardworks Ltd 2009 Explosions and recoil Remember: momentum is always conserved. What happens to the momentum in the following situations?
  • 37. 37 of 42 © Boardworks Ltd 2009 Calculating momentum Momentum is a vector quantity. The van has a momentum of 1500 × 10 = 15000 kgm/s to the right. The car’s momentum is 15000kgm/s to the left, or –15000kgm/s to the right: 10m/s 10m/s Two vehicles, each weighing 1.5tonnes, are driving towards each other at 30mph. If they collide, what will happen? Both vehicles come to a halt. What has happened to their momentum? initial momentum to the right = 15000 + –15000 = 0 final momentum to the right = 0
  • 38. 38 of 42 © Boardworks Ltd 2009 Momentum calculations
  • 39. 39 of 42 © Boardworks Ltd 2009 © Boardworks Ltd 2009
  • 40. 40 of 42 © Boardworks Ltd 2009 Glossary
  • 41. 41 of 42 © Boardworks Ltd 2009 Anagrams
  • 42. 42 of 42 © Boardworks Ltd 2009 Multiple-choice quiz