Name…………..……………..
Class……….
science-spark.co.uk
G481
MECHANICS MODULE 1:
MOTION
NOTES AND QUESTIONS
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Lesson 1 notes - Physical Quantities and units
Objectives
(a)explain that some physical quantities consist of a numerical magnitude and
a unit;
(b)use correctly the named units listed in this specification as appropriate;
(c)use correctly the following prefixes and their symbols to indicate decimal
sub-multiples or multiples of units: pico (p), nano (n), micro(µ), milli (m), centi
(c), kilo (k), mega (M), giga (G), tera (T);
(d) Make suitable estimates of physical quantities included within this
specification.
Units
Quantity
Mass
Length
Volume
Time
Electric Current
Temperature
Intensity of light
amount of a substance
Unit
kilograms
meters
litres
seconds
ampere
Kelvin, Celsius
candela
mole
Abbreviation
kg
m
L
s
A
K,C
cd
mol
Numerical prefixes
(Shaded ones to learn)
Factor Name Symbol
1024 yotta
Y
21
10
zetta
Z
18
10
exa
E
15
10
peta
P
12
10
tera
T
109 giga
G
6
10
mega
M
3
10
kilo
k
2
10
hecto
h
1
10
deka
da
10-1 deci
d
-2
10
centi
c
-3
10
milli
m
-6
10
micro
µ
-9
10
nano
n
10-12 pico
p
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10-15 femto
10-18 atto
10-21 zepto
10-24 yocto
f
a
z
y
Lesson 2 notes – Scalars and Vectors
Objectives
(a) define
scalar and vector quantities and give examples
Definitions
Vectors are quantities with magnitude and direction.
Scalar quantities have magnitude only.
Examples
Scalars
Vectors
Mass
Force
Temperature
Displacement
Speed
Velocity
Distance
Momentum
Energy
Acceleration
Adding Vectors and Scalars
Vectors differ because they have direction as well as magnitude. Imagine a
person walking a distance Disth from a from a point and then changing
direction and walking a distance Distv; the vector diagram would look
something like that below:
Disth
O
Distv
DistR
P
Adding Scalars: The total distance (scalar) travelled would be Disth + Distv
Adding vectors: The displacement (vector), which is the Resultant distance
moved from O to P, would be DistR.
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Lesson 2 questions – scalars and vectors
1
The table shows vector and scalar quantities.
Speed, acceleration
Energy, power
Force, pressure
Velocity, displacement
In the blank spaces provided, label the pair of quantities that are both vectors with a V
and the pair that are both scalars with an S.
(2)
2
Explain the difference between a scalar and a vector including one example in
your answer.
Scalar……………………………………………………………………………………
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
Vector…………………………………………………………………………………...
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (4)
3
Put the following quantities into a list of vectors and a list of scalars.
MASS, FORCE, SPEED, VELOCITY, WORK, DISPLACEMENT
Vector
Scalar
(3)
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Lesson 3 notes – vector addition
Objectives
(b) draw
and use a vector triangle to determine the resultant of two coplanar
vectors such as displacement, velocity and force;
(c) calculate the resultant of two perpendicular vectors such as displacement,
velocity and force;
Adding vectors
In one dimension:
a)
+
5N
b)
=
1N
+
5N
6N
=
1N
4N
i.e.
a) 5N + 1N = 6N
b) 5N + (-1N) = 5N – 1N = 4N
Resultant Vectors
Finding the resultant velocity P from an object given a horizontal velocity
(right) Vh and a vertical velocity (down) Vv. Firstly a vector diagram is drawn.
O
Vh
Vv
VR
P
Finding the magnitude of P
Using Pythagoras’ theorem that the square of the 2 sides of a triangle equal
the square of the hypotenuse, (a2 + b2 = c2)
Therefore:
Vh2 + Vv2 = P2
So, P = √(Vh2 + Vv2)
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Lesson 3 questions – Vector addition
1 a) Explain the difference between a scalar and a vector quantity.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
b)
Fig1 shows the path of a car as it travels around a right angles bend.
25ms-1
A
B
25ms-1
The car travels from point A to point B in 7.6 seconds at a constant speed of 25ms-1.
i)
Calculate the distance the car travels in 7.6s.
Distance=………………m (2)
ii)
Draw a line on fig1 to show the displacement of the car having travelled from
A to B.
iii)
Explain why the velocity of the car changes as it travels from A to B although
the speed remains constant.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
………………………………………………………………………………………(2)
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iv)
Using a labelled vector triangle, calculate the magnitude of the change in
velocity of the car (velocity at B – velocity at A).
Magnitude of velocity change = …………………….ms-1. (4)
v)
State and explain whether the car is accelerating as it travels around the bend
from A to B.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
Total (13)
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Lesson 4 notes – Vector components
Objectives
(d) resolve a vector such as displacement, velocity and force into two
perpendicular components.
Resolving Vectors
Resolving a vector is the process of taking the resultant vector and working
out its magnitude in two perpendicular directions.
The easiest way to understand this is with an example:
A ball is kicked at 10 ms-1 at an angle of 30° to the horizontal. What are the
horizontal and vertical components of its velocity?
10 ms-1
Vv ?
30°
Vh ?
The vector triangle would look like this:
10 ms-1
Vv ?
30°
Vh ?
From our knowledge of Trigonometry we can say that:
sin30°= Vv / 10 ms-1
And
cos30°= Vh / 10 ms-1
(Don’t forget soh, cah (and toa))
So we could rearrange these to get:
Vv = 10 sin30°
And
Vh = 10 cos30°
So in this case Vv = 5 ms-1 and Vh = 8.7ms-1
Remember that the hypotenuse is 10 and this is the longest side, so if either
of your components is bigger than 10 then you know you’ve done something
wrong.
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Perpendicular Vectors
Imagine trying to move a toy car by pushing it.
If you wanted to move it to the right you would push it from the left:
Force
If you wanted to move it up you would pull it from the top:
Force
You can see that no matter how hard you pushed it from the left, the car
would never go upwards. And the same is true that no matter how hard you
pulled the car up, it would never go left or right.
These vectors (perpendicular ones) are independent of each other and can be
treated individually when doing calculations. This becomes very important
when looking at projectile motion in lesson 23.
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Lesson 4 questions – Vector components
1a)
Fig 1.1 shows a skier being pulled up a slope at constant speed.
The tension in the wire pulls the skier with a force of 400N that acts at 40° to the
slope.
4N
40°
Fig1.1
i)
Explain with reference to the forces acting on the skier why he travels at
constant speed.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
ii)
1.
Calculate the component of the tension in the wire.
Parallel to the slope
Component = ………………….
2.
Perpendicular to the slope
Component = ………………….
(3)
b)
Describe two possible effects on the skier if the tension in the wire was
suddenly increased.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
Total (7)
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2
Fig 2.1 shows a force of 6.0 N acting at 30° to the horizontal.
6.0N
30°
horizontal
Calculate the component of the force that acts
i)
horizontally,
horizontal component…………………….N
ii)
vertically,
vertical component…………………….N
(2)
Total (2)
3
Fig 3.1 shows a ball kicked from the top of a cliff with a horizontal
velocity of 5.6ms-1. Air resistance can be neglected.
5.6ms-1
After 0.90 seconds the vertical component of the velocity is 8.8ms-1.
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i)
Use a vector triangle to determine the resultant velocity of the ball after 0.90s.
Resultant velocity: magnitude = …………………..ms-1.
Angle to the horizontal……………………° (4)
ii)
Calculate
1. the vertical distance the ball falls in 0.90 s,
2. the horizontal distance the ball travels in this time.
the vertical distance = …………………………
the horizontal distance = …………………… (3)
Total (7)
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Lesson 7 notes – motion graphs: Displacement
Objectives
(a)define displacement, instantaneous speed, average speed, velocity and acceleration;
(b)select and use the relationships average speed = distance /time
acceleration = change in velocity /time to solve problems;
Definitions
Distance is how far an object moves.
Displacement is a vector and so it is how far an object moves in a particular
direction.
Speed is the rate of change of distance. Speed is a scalar.
Velocity is the rate of change of displacement. Velocity is a vector.
Acceleration is the rate of change of velocity. Acceleration is a vector. An
object can accelerate if its speed or direction changes. So an object can
accelerate even though going at the same speed, just by changing its
direction.
Graphical representation: displacement(s/m) – time(t/s) graphs.
The gradient of a
displacement-time graph
gives the velocity of the
object.
From the gradients we can
work out the instantaneous
velocity and then draw a
graph of how this changes
with time – this is considered
in lesson 8.
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Lesson 7 questions – motion graphs: Displacement
1a)i) Define speed.
…………………………………………………………………………………………
…………………………………………………………………………………………
ii)
Define velocity.
…………………………………………………………………………………………
…………………………………………………………………………………………
iii)
State the differences between these quantities.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (4)
b)
Fig 1.1 shows a fairground big wheel. The wheel is rotating in a vertical plane
and carriages travel round a circle of diameter 40m at a constant speed. The carriages
complete one revolution in 3.5 minutes.
i)
A carriage moves half a revolution from X to Y. Calculate
1.
the speed of the carriage
speed = ……………ms-1
2.
the magnitude of the average velocity of the carriage
magnitude of the average velocity= ……………ms-1
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ii)
The carriage in (bi) returns to point X. Calculate, for the complete revolution,
1.
the speed of the carriage
speed = ……………ms-1
2,
the average velocity of the carriage.
average velocity= ……………ms-1
Comment on your answer.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
c)
Describe how the instantaneous velocity of the carriage at Y differs from the
average velocity of the carriage after traveling from X to Y.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
Total [14]
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Lesson 7 notes – motion graphs: velocity and
acceleration
Objectives
(c) apply graphical methods to represent displacement, speed, velocity and acceleration;
(d) determine velocity from the gradient of a displacement against time graph;
(e) determine displacement from the area under a velocity against time graph;
(f) determine acceleration from the gradient of a velocity against time graph.
Velocity-time graphs
The following are examples of graphs where velocity is varying with time.
Remember that velocity is a vector and so if the line is above zero on the yaxis it means there is a positive velocity and if it is below zero it is a negative
velocity.
A slope means that the velocity is changing – acceleration. This can be
getting quicker or getting slower (deceleration is also called retardation)
A positive flat line shows that there is a constant (uniform) positive velocity.
A rising positive line means that there is a constant positive acceleration.
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Displacement
To find the displacement of an object using a velocity-time (v-t) graph we find
the area underneath the graph.
Acceleration
To find the acceleration of an object using a v-t graph we take the gradient. If
we have a sloping line we can find the instantaneous acceleration be taking a
tangent at a point.
These ideas will be looked at in more detail in lesson 9.
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Distance, time and speed calculations
Try these calculations
Graph paper is needed for some. These questions are all based on the connection between
speed, distance and time. Answer in the spaces provided:
Hints
In these questions it is useful to remember that:
– if an object is accelerating steadily from rest its average speed is half the maximum speed
– and that distance travelled = average speed time
1.
You are watching a batsman hit a cricket ball. If 0.375 s passes between the time you
see him strike the ball and the time you hear the sound of this, how far from the
–1
batsman are you sitting? The speed of sound in air is 340 m s . (The speed of light is
nearly a million times bigger than this, so you see the bat hit the ball more or less at
the instant it occurs.)
2.
A girl diving from a 15 m platform wishes to know how fast she enters the water. She
is in the air for 1.75 s and dives from rest (with an initial speed of zero). What can you
tell her about her entry speed?
3.
An experiment performed on the Moon finds that a feather falls 20.75 m from rest in 5
s. What is its speed as it hits the Moon's surface?
4.
The sketch graph shown represents the variation in vertical height with time for a ball
thrown upwards and returning to the thrower.
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time
From this graph sketch a velocity–time graph.
–1
5.
In a Tour de France time trial a cyclist is able to reach a top speed of 100 km h by
starting from rest and pedalling flat out for a distance of 3 km. If the rate at which the
cyclist's speed changes is uniform, how long will this take?
6.
You are travelling in a car moving at 50 km h (just over the 30 mph speed limit).
–1
What is this speed in m s ? You have to brake so the car comes to rest uniformly in
1.4 s, how far will you travel? A cat runs out in front of your car and your reaction time
is 0.6 s. What is the total distance the car will travel before stopping?
7.
A steam traction engine speeds up uniformly from rest to 4 m s in 20 s. It then
travels at a steady speed for 440 m and finally comes to rest uniformly in 10 s having
travelled 500 m in total. Draw a speed–time graph for its motion showing key values
–1
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–1
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of speed and time. What is the total time for the journey? What is the average speed
for the whole journey?
8.
A tennis ball is dropped from a height of 2 m above a hard level floor, and falls to the
floor in 0.63 s. It rebounds to a height of 1.5 m, rising to a maximum height 1.18 s
after it was released. Draw a speed–time graph indicating speed and time at key
points of the motion.
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Lesson 8 notes – reviewing motion graphs
Objectives
(a) define displacement, instantaneous speed, average speed, velocity and acceleration;
(b) select and use the relationships average speed = distance / time
acceleration = change in velocity / time to solve problems;
(c) apply graphical methods to represent displacement, speed, velocity and acceleration;
(d) determine velocity from the gradient of a displacement against time graph;
(e) determine displacement from the area under a velocity against time graph;
(f) determine acceleration from the gradient of a velocity against time graph.
Non-Uniform Acceleration
Imagine a rollercoaster ride with a track shaped as below:
A
B
C
D
E
At different points along the track, the rollercoaster would have different
velocities and therefore its acceleration would be changing – it would not be
constant (or uniform – in fact it would be non-uniform.
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Graphical methods
Look at the graphs below:
620 m
A
B
To find the velocity from a displacement-time graph we have seen that we
take the gradient of the line. But if the acceleration is non-uniform it is, by
definition, changing. So we can only work out the acceleration at specific
points, or instants. We call this taking the instantaneous acceleration.
Graph A shows the V-T graph of an object with non-uniform acceleration. To
work out the instantaneous acceleration we must work out the gradient of the
line at that point. In order to do this we must draw a tangent (a line parallel
with the direction of the graph at that point) and take the gradient of that line in
the normal way. (change in y-coordinates/change in x-coordinates) Δv/Δt.
We can also find the displacement from a V-T graph from the area under the
graph (s=vt) but for non-uniform acceleration this is slightly harder as we have
to count the squares under the graph since the velocity is changing with
changing acceleration.
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Deriving the equations of motion
Look at the graph of motion with uniform acceleration. By finding the area under this graph we
can derive another useful equation for uniformly accelerated motion – because the area under
a velocity–time graph is equal to the displacement:
area of rectangle
=
initial velocity × time interval
area of rectangle
=
uΔt
area of triangle
=
½ × base × height
area of triangle
=
½ Δt × (v - u)
=
a Δt
so area of triangle
=
½ a Δt2
and displacement
=
area of rectangle + area of triangle
=
uΔt + ½ a Δt2
You know that:
(v – u)
Usually you will see this equation written using just t (not Δt) to represent the overall time
taken and s to represent the overall displacement:
s
=
ut + ½ at2
Use the equations that have you already met to show that: (Hint: substitute for t)
v2
=
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u2 + 2as
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Lesson 8 questions– acceleration
1
Fig 1.1 shows a trolley on a bench surface, connected to a mass M by a string.
The mass is released and the trolley moves along the surface. The graph shows the
variation of velocity v of the trolley with time t for the motion from A to B.
a)i)
Calculate the acceleration of the trolley between A and B.
acceleration = ………………… ms-2 (2)
ii)
Show that the distance from A to B is 0.72m.
(2)
b)
When the trolley reaches B the mass M has just reached the floor.
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i)
Ignoring any resistive forces, calculate the time it takes the trolley to
travel from B to C.
time for B to C = …………………. s (3)
ii)
On the graph complete a line showing the trolley moving from B and
coming to rest at the pulley at C.
(3)
Total [10]
2
a)
Define acceleration
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
b)
Fig 2.1 shows a graph of velocity against time for a train that stops at a station.
i)
For the time interval t=40s to t=100s, calculate
1.
The acceleration of the train,
acceleration = ………………..ms-2
2.
The distance travelled by the train.
distance ……………. m
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(5)
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ii)
Calculate the distance travelled by the train during its acceleration from rest to
-1
25ms .
distance………………..m
(2)
iii)
Calculate the journey time that would be saved if the train did not stop at the
station but continued at a constant speed of 25ms-1.
time saved …………………….s
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Lesson 9 notes– uniform acceleration
Objectives
(a) derive the equations of motion for constant acceleration in a straight line from a
velocity against time graph;
(b) Select and use the equations of motion for constant acceleration in a straight line:
1 v = u + at , s = 0.5(u + v)t ,
2 s = ut + 1/2at2 and
v2 = u2 + 2as
SUVAT Equations (equations of motion)
v
=
u+at
s
=
((v + u)/2) x t
(Displacement = average velocity x time)
s
=
ut + ½ a t2
v2
=
u2 + 2as
These equations describe an object that has constant acceleration.
g
An object that has a constant force will experience a uniform (constant)
acceleration like that of g. g is the acceleration due to gravity and is 9.81 ms -2.
As long as air resistance is neglected, all objects will fall with this acceleration.
s  1 gt 2
2
A method to find g uses the equation
(since the initial velocity, u, is
zero) and includes dropping a mass from a height, s, and recording the time
to fall, t.
rearranging: g = 2s/t2. As long as the height is large enough, errors in
recording t will be small and a good approximation for g can be made.
Terminal velocity
If air resistance is not neglected, it will eventually balance the weight of the
object and a terminal (maximum) velocity will be reached.
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Lesson 9 questions – uniform acceleration
1
In this question, two marks are available for the quality of written
communication.
An object falls vertically through a large distance from rest in air. Describe and
explain the motion of the object as it descends in terms of the forces that act and its
resulting acceleration.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………… (6)
b)
Explain how a free-fall diver can increase the rate at which she descends
through the air.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………… (2)
Total [10]
a)i)
Define speed…………………………………………………………(1)
ii)
Distinguish between speed and velocity………………………………..
………………………………………………………………………………………….
……………………………………………………………………………………… (2)
2
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b)
Use the equations given below, which represent uniformly accelerated motion
in a straight line, to obtain an expression for v in terms of u, a and s only.
v = u + at
s = (u + v)t/2
(2)
Fig 2.1 shows a ball kicked from the top of a cliff with a horizontal velocity of 5.6ms1
. Air resistance can be neglected.
5.6ms-1
i)
ii)
Show that after 0.90s the vertical component of the velocity is 8.8ms-1.
(2)
Use a vector triangle to determine the resultant velocity of the ball after 0.90s.
Resultant velocity: magnitude = …………………..ms-1.
Angle to the horizontal……………………° (4)
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iii)
Calculate
3. the vertical distance the ball falls in 0.90 s,
4. the horizontal distance the ball travels in this time.
the vertical distance = …………………………
the horizontal distance = …………………… (3)
Total (14)
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Lesson 10 notes – motion under gravity
Objectives
(c) apply the equations for constant acceleration in a straight line, including
the motion of bodies falling in the Earth’s uniform gravitational field without air
resistance;
(d) explain how experiments carried out by Galileo overturned Aristotle’s
ideas of motion;
(e) describe an experiment to determine the acceleration of free fall g using a
falling body;
Constant Acceleration
Falling objects are accelerated by gravity. If we discount air
resistance, they have the same acceleration since force is
proportional to mass. Beacause of this, we can state g as
9.81 ms-2.
Since falling objects have a constant acceleration we can use
the SUVAT equations to describe their motion.
SUVAT Equations (equations of motion)
v
=
u+at
s
=
((v + u)/2) x t
(Displacement = average velocity x time)
s
=
ut + ½ a t2
v2
=
u2 + 2as
These equations describe an object that has constant acceleration.
g
An object that has a constant force will experience a uniform (constant)
acceleration like that of g. g is the acceleration due to gravity and is 9.81 ms -2.
As long as air resistance is neglected, all objects will fall with this acceleration.
s  1 gt 2
2
A method to find g uses the equation
(since the initial velocity, u, is
zero) and includes dropping a mass from a height, s, and recording the time
to fall, t.
rearranging: g = 2s/t2. As long as the height is large enough, errors in
recording t will be small and a good approximation for g can be made.
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Aristotle
Aristotle thought that all objects had a normal tendency to
fall towards the centre of the universe and as the Earth was
the centre of the Universe all objects fell towards the centre
of the Earth. Heavier objects would obviously fall faster that
lighter ones.
The planets and stars were in orbit around the Earth on
crystal spheres that kept them from falling towards the
Earth.
Galileo
It wasn’t until Galileo tried experiments that this intuitive idea was overturned.
Galileo proposed that all things fell towards Earth with the same acceleration.
The force that pulled them towards the Earth was the same force that held the
planets in their orbits. Galileo said that the gravitational force was proportional
to the mass.
Because of this (and F=ma) the objects would have same acceleration. Then
their final speeds (terminal velocity) would only be different because of their
drag.
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Of course, gravity can be completely neutralised
by attaching a cat (always lands feet-down) to
toast (always lands butter-side-down).
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Lesson 11 notes – Projectiles
Objectives
Know what a projectile is.
Know that the path of a projectile is a parabola.
Know that horizontal and vertical motion can be obtained separately.
What is a projectile?
A projectile is an object which is initially projected by a force, but which then
continues to move freely under the influence of gravity;
A rocket which is firing its motors is not a projectile.
The Parabolic Path
A parabola is shown below:
parabola
A projectile follows a parabolic path because of its definition. It moves under
the influence of gravity and therefore accelerates because of this in the
vertical direction whilst its horizontal movement stays constant.
As it travels up and down (vertically) it is decelerated and accelerated by
gravity. It decelerates to zero at its maximum height and then accelerates
back down.
Side to side (horizontally) there are no forces acting on it whilst in flight and so
its speed is constant.
If these two conditions hold then the path traced out will be a parabola.
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Vertical and horizontal motion
Projectile motion can look very complicated at first because of its curved
trajectory. How can we work out where a projectile will be at a particular time
or how fast it is going? Once you can accept that the horizontal and vertical
components of the motion can be split up the whole thing gets a lot easier.
Let’s look at an example:
In the drawing above, a boy throws a ball up at different angles so that the ball
reaches the same height. The time period between each snapshot of where
the ball is the same. What do we notice:
1
The ball reaches the same height.
2
The ball lands further to the right in the second drawing.
3
In the vertical direction the ball slows down on the way up and
speeds up on the way down.
4
This speeding up and slowing down is the same in both pictures –
So this means that the horizontal motion does not affect the
vertical motion! It slows down on the way up because gravity
is pulling it down and it speeds up on the way down because
gravity is pulling it down!
5
It is easier to see in the second drawing, but in both drawings the
horizontal distance between each picture of the ball is the same –
So the horizontal speed is constant! Once the ball leaves the
boys hand there are no forces acting on the ball horizontally,
(we’ll ignore air resistance for now!)
We’ll see in next lesson that you already have the tools to work out
everything there is to know about projectile motion but it links together
ideas of energy transformation and conservation and the suvat equations.
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Monkey and the Hunter
A scenario that shows that the vertical and horizontal motion can be thought
of separately is the monkey and hunter, or in this case, (brought up to date for
the caring 21st century) the monkey and zookeeper.
The zookeeper wants to feed the monkey so he shoots a banana at the
monkey’s mouth, at the instant the keeper fires the monkey drops from the
tree. Because the banana and the monkey are both accelerated towards the
ground at the same rate due to gravity (9.81 ms-2) the banana can be
yummed up as the monkey falls.
If the monkey had not dropped, the banana would have been shot too low.
Lucky little monkey!
Components of a vector
You should go over lesson 2 and 3 notes to make sure you are happy with
finding components of a vector and resultant magnitudes and directions of
vecto
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Lesson 11 questions – projectile motion 1
Practice with components of a vector and the suvat equations in the context of
projectile motion. Take g = 9.81 ms-2
1
A skier moves at 11.0 ms-1 down a 16º slope. What is the skier’s
a)
vertical velocity
………… ms-1 (2)
b)
horizontal velocity
………… ms-1 (2)
Total [4]
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2
A ball is thrown at 20º to the horizontal at 20.0 ms-1 , What is the ball’s
a)
vertical velocity
………… ms-1 (2)
b)
horizontal velocity
………… ms-1 (2)
c)
Describe the shape of the trajectory that the ball follows.
…………………………………………………………………………………………
…………………………………………………………………………………… (1)
Total [5]
3
A ball is thrown vertically upwards at 19.6 ms-1 .
a)
Fill in the table of the velocity of the ball at different times below:
Time t/s
Velocity v/ ms-1
1.0
2.0
3.0
4.0
(8)
c) What is the displacement after
i)
2.0s
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
ii)
4.0s
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
c)
How far does it travel in the first 4.0 seconds?
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
d)
Draw a sketch velocity-time graph for motion of the ball the instant after it has
left the hand until it is caught again. Take the upward direction to be positive and
assume no air resistance.
(4)
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e)
Explain how the graph can be used to find the distance and displacement.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
Total [20]
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Lesson 12 notes – Projectile motion.
Objectives
(f) apply the equations of constant acceleration to describe and explain the
motion of an object due to a uniform velocity in one direction and a constant
acceleration in a perpendicular direction.
Components of a vector
A projectile is an object which is initially projected by a force, but which then
continues to move freely under the influence of gravity. This could be directed
straight up at 90 degrees to the horizontal or at any angle towards the
horizontal. Because of this we can describe a projectile’s motion by looking at
the vertical and horizontal components of velocity separately.
Remember that vertical motion will be affected by acceleration due to gravity
but horizontal velocity will not change, as there are no forces acting on the
projectile horizontally (if we neglect air resistance).
SUVAT
We can use the suvat equations to describe the vertical motion of a projectile
since there is a constant acceleration (gravity) affecting it.
v
=
u + at
s
=
ut + ½ at2
v2
=
u2 + 2as
For horizontal motion we can just use velocity = displacement / time (since
there is no acceleration in the horizontal plane)
Example (taken from lesson 3 notes)
A ball is kicked at 10 ms-1 at an angle of 30° to the horizontal. What are the
horizontal and vertical components of its velocity?
uv ?
10 ms-1
30°
uh ?
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The vector triangle would look like this:
10 ms-1
uv ?
30°
uh ?
From our knowledge of Trigonometry we can say that:
sin30°= uv / 10 ms-1
And
cos30°= uh / 10 ms-1
(Don’t forget soh, cah (and toa))
So we could rearrange these to get:
uv = 10 sin30°
And
uh = 10 cos30°
So in this case uv = 5 ms-1 and uh = 8.7ms-1.
How long will it be before it hits the ground?
We can use the suvat equations.
The vertical displacement for the ball to come back to the ground will be zero,
so using s = ut + ½ at2
0 = uv t + ½ (-9.8) t2
divide through by t, and substituting in value for initial vertical velocity
0 = 5 + ½ (-9.8) t
t = 5/4.9
t = 1.0(2)s
How far will it go (what is its range, how far horizontally)?
Using s=ut (there are no forces acting on the projectile in the horizontal plane
– we neglect air resistance – so there is no acceleration)
s = uh t
s = 8.7 x 1.02
s = 8.87m
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Lesson 13 notes– projectiles 3, energy and drag
Objectives
(f) apply the equations of constant acceleration to describe and explain the
motion of an object due to a uniform velocity in one direction and a constant
acceleration in a perpendicular direction.
Energy transformation
In a pendulum, if we neglect air resistance and friction,
the total energy transforms between gravitational
potential energy and kinetic energy.
It has maximum gravitational potential energy at the top
of the swing but zero kinetic energy as it is stationary.
At the bottom of the swing it is going at its fastest so it
has maximum kinetic energy but minimum gravitational
potential energy as it is at its lowest point.
This is because the total amount of energy stays the
same, (none is created or destroyed – conservation of
energy)
Similarly in models of projectile motion where air resistance (drag) is
neglected, kinetic energy is transformed into gravitational potential energy and
back to kinetic energy with no energy being gained or lost in the process.
A projectile has maximum kinetic energy the instant it leaves the force
projecting it. It loses kinetic energy as it travels upwards as this energy is
being transformed into gravitational energy and so has minimum kinetic
energy at the top of the trajectory (it is not zero since it still has the same
amount of horizontal speed as it started with). It therefore has maximum
gravitational potential energy at the top of the trajectory. The reverse happens
on the way down as the gravitational energy is transformed into kinetic energy
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on the way down back to the maximum as it hits its target, (this is the same
amount as at the beginning of the flight because of the conservation of
energy.)
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Lesson 13 questions - projectiles
1
Here you can work on predicting the positions of objects that are accelerated.
One archery competition requires archers to fire a total of 90 arrows for a maximum
possible score of 900. The targets are 1.22 m in diameter at 60, 50 and 40 metres
distance. You can model the motion of the arrow to find out what problems the archer
faces.
The arrow leaves the bow at 60 m s–1 and travels at almost this speed horizontally for
the whole of its flight.
The arrow, of course, falls because of the acceleration due to gravity. You can find its
position at any moment by working out how far it has moved horizontally and how far
it has fallen vertically.
a)
The archer shoots the arrow horizontally at the 40 m target. How far does it
drop over this range?
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
………………………………………………………………………………………(2)
b)
How would the archer make allowance for this fall? Think carefully before
committing yourself.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (1)
c)
Now try to calculate the fall at 50 m and 60 m.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (4)
d)
50 years ago the release speed of an arrow was about 30 m s–1. What effect
would this have on the vertical distance the arrow fell? Calculate the drop for a range
of 60 m to check your answer.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (3)
e)
We have ignored the effect of air resistance in these calculations. How could
you take account of it?
…………………………………………………………………………………………..
……………………………………………………………………………………… (1)
Total [11]
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2
A body is projected horizontally from ground level with a speed of 24ms-1 at
an angle of 30 degrees above the horizontal. Neglect air resistance and calculate:
a)
The vertical part of the velocity
……………….. (2)
b)
The time to reach its highest point
……………….. (2)
c)
The greatest height reached
d)
The horizontal range of the body
……………….. (2)
……………….. (2)
Total [8]
3
fig 3.1 shows the trajectory of a projectile fired from the ground if air
resistance is neglected.
parabola
a)
Describe the changes in energy of the projectile that occur throughout its
flight.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
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…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (5)
b)
Describe the changes in speed of the projectile that occur throughout its flight.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (5)
c)
Explain qualitatively what would happen if the projectile experienced air
resistance. Use a diagram to help your explanation if necessary.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (3)
Total [13]
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