1. PROJECTILE MOTION
constant acceleration in a straight line; free fall under gravity, projectile motion; Relative motion, change in velocity, velocity vector
components, circular motion (constant speed with one force only providing centripetal force).
1. Recall that in the absence of friction a falling object will have a constant
acceleration of 10 ms-2 and that this value is referred to as “gravitational
acceleration, g”
2. Use F = mg to show that Nkg-1 is an equivalent unit to ms-2.
3. Explain the effect that air friction has on the acceleration of a falling object
(resulting in free fall)
4. Define the term projectile.
5. Describe projectile motion
in terms of its uniform
horizontal motion and it
accelerated vertical motion.
6. Use equations of motion to
calculate time, distance,
velocity and acceleration.
Monday, 24 May 2010
2. FREE FALL IN THE ABSENCE OF FRICTION
Mass & weight
The mass of an object, m is a measure of the amount of matter in that object.
The weight of an object, Fw is a measure of the force due to gravity on that object.
Gravitational constant, g
The force exerted on an object by the earth’s gravitational field is called the
gravitational constant, g.
What
g = 10 Nkg -1
does this
mean?
This means that for every kg of mass there is a force of 10N acting on it because of
gravity
This causes all objects to fall with an acceleration of 10 ms-2
We can also say g = 10 ms-2 which is called gravitational acceleration
Fw = mg
Monday, 24 May 2010
3. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
Velocity - time graph for the object
v (ms-1)
t (s)
Constant negative
slope shows a
constant negative
acceleration
Monday, 24 May 2010
4. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
Velocity - time graph for the object
v (ms-1)
t (s)
Constant negative
slope shows a
constant negative
acceleration
Monday, 24 May 2010
5. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
Velocity - time graph for the object
v (ms-1)
t (s)
Constant negative
slope shows a
constant negative
acceleration
Monday, 24 May 2010
6. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
A symmetrical path
Velocity - time graph for the object
v (ms-1)
t (s)
Constant negative
slope shows a
constant negative
acceleration
Monday, 24 May 2010
7. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
A symmetrical path
Velocity - time graph for the object
v (ms-1) x
t (s)
Constant negative
slope shows a
constant negative
acceleration
Monday, 24 May 2010
8. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
A symmetrical path
Velocity - time graph for the object
v (ms-1) x
x
t (s)
Constant negative
slope shows a
constant negative
acceleration
Monday, 24 May 2010
9. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
A symmetrical path
Velocity - time graph for the object
v (ms-1) x
x
t (s)
Constant negative
x slope shows a
constant negative
acceleration
Monday, 24 May 2010
10. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
A symmetrical path
Velocity - time graph for the object
v (ms-1) x
x
t (s)
Constant negative
x slope shows a
constant negative
acceleration
Monday, 24 May 2010
11. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
A symmetrical path
Velocity - time graph for the object
Decreasing speed in
v (ms-1) x
a positive direction
x
t (s)
Constant negative
x slope shows a
constant negative
acceleration
Monday, 24 May 2010
12. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
A symmetrical path
Velocity - time graph for the object
Decreasing speed in
v (ms-1) x
a positive direction
Stationary for an instant
x
t (s)
Constant negative
x slope shows a
constant negative
acceleration
Monday, 24 May 2010
13. FOR AN OBJECT THROWN VERTICALLY UPWARDS THE ACCELERATION IS CONSTANT
& NEGATIVE
It slows down as it moves in the upwards
+ up
(positive) direction
and
- down
speeds up as it moves in the downwards
(negative) direction.
A symmetrical path
Velocity - time graph for the object
Decreasing speed in
v (ms-1) x
a positive direction
Stationary for an instant
x
t (s)
Constant negative
Increasing speed in a slope shows a
negative direction x
constant negative
acceleration
Monday, 24 May 2010
14. Examples
1. A plane drops a Red Cross package from a height of 1200 m. If the package had no
parachute (and by this you can assume negligible air friction & g = 10 ms-2)
(a) How fast will the package be travelling just before it hits the ground?
(b) How many seconds will the package take to fall?
2. A 1 kg object is dropped from a tower 120 m high.
(a) Calculate the time it will take for the object to fall to the ground.
(b) Calculate the objects final speed on reaching the ground.
(c) How long does it take to reach a speed of 35 ms-1 ?
Monday, 24 May 2010
15. 3. A shell is fired straight up with an initial speed of 96 ms-1.
(a) Calculate the time it will take for the object to fall to the ground.
(b) When will the shell have an upwards speed of 48 ms-1 ?
(c) Calculate the time for the shell to reach its maximum height.
(d) Calculate the maximum height reached by the shell.
(e) What is the shells acceleration at the top of its motion?
Monday, 24 May 2010
16. FREE FALL - IN THE PRESENCE OF FRICTION
If the object is falling fast:
Air friction is too large to ignore. Here we consider the object to have two forces
acting on it. Air friction and gravity.
A. When the object is initially falling slowly, air friction is much smaller than the force
of gravity on the object.
B. As the object speeds up the air friction increases.
C. The object will eventually reach a speed at which the air friction balances the force
of gravity on the object. The object cannot fall any faster than at this speed (unless
it changes shape). This speed is called terminal velocity.
This situation is illustrated by the following example:
B
A
C
Vector arithmetic with notes
Monday, 24 May 2010
17. WHAT IS A PROJECTILE?
Brainstorm
Examples
of
Projectiles
“What do all these examples have in common?”
“Why do you think that a ball dropping vertically (in free fall) is not
an example of projectile motion?”
Monday, 24 May 2010
18. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
19. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
20. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
21. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
22. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
23. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
24. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
25. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vx
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
26. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vx
vx
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
27. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vx
vx
vx
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
28. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vx
vx
vy vx
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
29. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vy
vx
vx
vy vx
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
30. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vy
vx
vx vy = 0
vy vx
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
31. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vy
vx
vx vy = 0
vy vx
vy
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
32. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vy
vx
vx vy = 0
vy v vx
vy
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
33. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vy v
vx
vx vy = 0
vy v vx
vy
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
34. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vy v v
vx
vx vy = 0
vy v vx
vy
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
35. PROJECTILES - HORIZONTAL AND VERTICAL MOTION
• A projectile is an object that has constant horizontal velocity and constant vertical
acceleration due to gravity. In other words it is moving horizontally at the same time it is in
free fall. The projectiles velocity is the sum of these two velocities.
• The path of a projectile is parabolic and symmetrical
• Air resistance is considered to be negligible
• The force of gravity is the only force acting on the object. Once the projectile is launched there
is no thrust force.
vy v v
vx
vx vy = 0
vy v vx
vy v
vx
v = the size of the velocity of the projectile
To determine the velocity: vx vx = the horizontal component of the
vy projectile’s velocity
add the horizontal and
v vy = the vertical component of the
vertical components
projectile's velocity
Monday, 24 May 2010
36. SYMMETRICAL FLIGHT PATH
• The path of a projectile is parabolic and symmetrical
Symmetrical Path
It is often convenient to work with half the flight path of a projectile.
Remember
When the projectile is at the maximum height the vertical velocity is zero.
At t : vy = 0
2
Monday, 24 May 2010
37. SYMMETRICAL FLIGHT PATH
• The path of a projectile is parabolic and symmetrical
Symmetrical Path
It is often convenient to work with half the flight path of a projectile.
d
Remember
When the projectile is at the maximum height the vertical velocity is zero.
At t : vy = 0
2
Monday, 24 May 2010
38. SYMMETRICAL FLIGHT PATH
• The path of a projectile is parabolic and symmetrical
Symmetrical Path
It is often convenient to work with half the flight path of a projectile.
t
2
d
Remember
When the projectile is at the maximum height the vertical velocity is zero.
At t : vy = 0
2
Monday, 24 May 2010
39. SYMMETRICAL FLIGHT PATH
• The path of a projectile is parabolic and symmetrical
Symmetrical Path
It is often convenient to work with half the flight path of a projectile.
t
2
t
d
Remember
When the projectile is at the maximum height the vertical velocity is zero.
At t : vy = 0
2
Monday, 24 May 2010
40. SYMMETRICAL FLIGHT PATH
• The path of a projectile is parabolic and symmetrical
Symmetrical Path
It is often convenient to work with half the flight path of a projectile.
t
2
t
d
d= the total distance travelled by a projectile
t = the time taken to travel that distance
t/2 = the time taken to reach maximum height
Remember
When the projectile is at the maximum height the vertical velocity is zero.
At t : vy = 0
2
Monday, 24 May 2010
41. Forces The force due to gravity is the only force
acting on the projectile.
Not a projectile:
An object in horizontal flight is not a projectile
Fgrav because Fgrav is balanced by the lift force. An
object that has not been launched but maintains
a high horizontal speed will have air friction
acting on it and therefore must have a thrust
force to balance air friction. This will not be a
projectile.
Vertical motion calculations
• Since the force of gravity is constant (10 N per kg of any object), the acceleration is
also constant. (10 ms-2 for any object)
• Kinematic equations must be used for analysing the vertical motion
Horizontal motion calculations
• A projectile progresses horizontally with constant speed.
• Horizontal speed can be calculated in the usual way:
Speed = distance travelled v=d
time taken t Units of v: ms-1
Monday, 24 May 2010
42. PROCESS FOR PROBLEM-SOLVING [PIA]
1. Read the question and underline the relevant information
2. Draw a diagram of the situation
3. List the information that relates to the vertical motion and the horizontal motion.
(Keeping it separate. Use a table with two columns “V” & “H”, remembering that
vertical and horizontal motion are independent of each other)
4. Show up as a positive direction and down as negative. The signs that you use for
the vertical information should reflect this.
5. Select the appropriate kinematic equation when calculating a vertical quantity.
6. Use v = d/t for calculation of a horizontal quantity.
Example 1
Imagine a car is driven off a 250m cliff at 30 ms-1. How far from the base of the cliff
will the car be when it lands?
Diagram
Monday, 24 May 2010
43. Example 2
A hockey ball is flicked with an initial velocity of 14 ms-1, 45o to the horizontal, as
shown. A spectator 1.6 m tall is standing directly in the path of the ball, 20 m away.
Will the ball hit the spectator?
Diagram
Monday, 24 May 2010
44. 12 PHYSICS PROJECTILES ASSIGNMENT Name ______________________
1. A rock is dropped from the top of a bridge. It takes 4 seconds to reach the water
below.
(b) Explain why this is not an example of projectile motion.
________________________________________________________________
________________________________________________________________
(c) What would need to be done to the rock to turn it into a projectile?
________________________________________________________________
(d) How far has the rock fallen?
________________________________________________________________
________________________________________________________________
________________________________________________________________
(e) A second rock is now thrown horizontally at 6 ms-1 from the same bridge.
________________________________________________________________
(f) How long will it take for the rock to reach the water?
________________________________________________________________
________________________________________________________________
________________________________________________________________
(g) Calculate the range of this second rock
________________________________________________________________
________________________________________________________________
________________________________________________________________
Monday, 24 May 2010
45. 2. A 2 kg bowling ball is launched from ground level and follows the path shown
below:
At each of the labelled positions A, B, C and D, state the size and direction of:
(a) the acceleration of the projectile
(b) the net force on the projectile
3. When an aircraft was travelling in level flight at 200 ms-1, a nut fell off part of the
landing gear. Assume air friction is negligible.
(a) Sketch the paths of the aircraft and the nut.
Sketch
The nut covered a total horizontal distance of 3.0 km.
(b) Find the total time it took to fall to the ground.
________________________________________________________________
________________________________________________________________
Monday, 24 May 2010
46. (c) What was the altitude of the plane when the nut fell off?
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
4. A ball is kicked with an initial velocity as shown below. The angle of inclination is
40o.
21 ms-1
40o
(a) Calculate the horizontal component of the initial velocity. __________________
(b) Calculate the vertical component of the initial velocity. ____________________
(c) What is the instantaneous velocity of the ball at the top of its path?
________________________________________________________________
________________________________________________________________
________________________________________________________________
(d) Find the height of the ball at the top of its path.
________________________________________________________________
________________________________________________________________
________________________________________________________________
Monday, 24 May 2010
47. (e) Find the time taken for the ball to reach its maximum vertical displacement.
________________________________________________________________
________________________________________________________________
________________________________________________________________
(f) Calculate the range of the ball.
________________________________________________________________
________________________________________________________________
If the ball was kicked at an angle of 60o to the horizontal, would it have travelled
as far along the field?
________________________________________________________________
________________________________________________________________
________________________________________________________________
5. A potato was launched from the muzzle of a spud gun with an initial velocity of
30 ms-1 at an angle of 60o to the horizontal. Will it clear a 25 m tree that is 60 m
away on the flight path?
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
6. A cricket ball was hit with the following trajectory. Find the initial velocity of the ball.
25 m
100 m
Monday, 24 May 2010
48. 7. Johnny is competing in the javelin event of his school athletics competition. The
javelin behaves like an ideal projectile.
(a) Describe the shape of the path of the javelin.
________________________________________________________________
(b) Ignoring air resistance, draw arrow(s) on the drawing of the javelin below to
show the force(s) acting on it when it is in the position shown. Name the
forces.
Joe now throws the javelin into the air at an angle of 40o above the horizontal at
an initial velocity of 30 ms-1.
Joe now throws the javelin into the air at an angle of 40° above the horizontal at an initial velocity of 30 m s–1
Monday, 24 May 2010
49. (c) Show that the horizontal component of the initial velocity of the javelin is
23 ms-1.
________________________________________________________________
________________________________________________________________
(d) Calculate the range of the javelin under these conditions.
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
[this is a 2004 NCEA exam question]
Monday, 24 May 2010
50. UNIFORM CIRCULAR MOTION
1. Recognize that even though a body in circular motion may have a constant speed,
its velocity is changing and hence it is accelerating.
2. Demonstrate that the acceleration is towards the centre, i.e. centripetal.
3. Deduce that all things in circular motion must have a centripetal force.
4. Show graphically using velocity vectors that the acceleration is towards the centre.
5. Remember the equations for uniform circular motion questions, and use them to
solve problems:
Read
Chapter 12 (p139 to 145)
Monday, 24 May 2010
51. BRAINSTORM
Give me some
everyday examples of
objects in uniform
circular motion
Complete the functional definition below:
An object in uniform circular object is _____________________________________
___________________________________________________________________
Monday, 24 May 2010
52. CENTRIPETAL ACCELERATION
For an object in uniform circular motion:
• The speed is always the same
• but the direction is always changing
Therefore the velocity is always changing and so the object is always
accelerating
“remember that
acceleration is the rate
of change in velocity.”
REMEMBER
Centripetal acceleration:
THIS -->
• Constant in size
• Direction is always changing
• Direction is towards the centre of the circular
path
Monday, 24 May 2010
53. MORE EVIDENCE FOR AN ACCELERATION TOWARDS THE CENTRE
For each 0.1 s time interval, determine ∆v. When you carry out the vector
~
~
subtractions required, do so by leaving vf in the position drawn (below).
~
~
~
∆t = 0.01 s
vi
~
vf
~
.
vi
~
vf
~
∆t = 0.01 s
Comment on the size and direction of the change in velocity vector:
Monday, 24 May 2010
54. CENTRIPETAL FORCE
“Acceleration is caused by an unbalanced force so now we can say something
about the force acting on an object in circular motion”
Centripetal force:
• Constant in size
REMEMBER
• Direction is always changing
THIS -->
• Direction is towards the centre of the circular
path
“Does this sound familiar??”
Note
• Greater speeds will require greater forces.
• Force in uniform circular motion changes direction not speed
Monday, 24 May 2010
55. The carousel WHAT DO WE MEAN BY UNBALANCED FORCE?
The force acting
through the rope
Tension
The force acting
vertically downwards
Gravity
“Tension on its own does not cause circular motion” UNBALANCED
FORCE
“Gravity on its own does not cause circular motion”
• A combination of these two forces is called the
unbalanced force.
• It is the unbalanced force that causes the uniform
circular motion
Monday, 24 May 2010
56. EQUATIONS FOR UNIFORM CIRCULAR
Consider a mass in uniform circular motion with speed v and a radius of circular path, r:
v = velocity (ms-1)
v ~
~
m
F = Force (N)
F ~
~
P a = acceleration (ms-2)
a ~
~
r r = radius of the circular path (m)
(measured from the centre of the path to
the centre of mass of the object.
m = mass of the object (kg)
velocity is at a tangent to
At any point:
the circular path
Speed
The speed is given by the distance travelled around the circular path divided by
the time taken to travel that distance.
v=d
t
Monday, 24 May 2010
57. Note that the distance travelled in a complete rotation is the circumference, C.
C = 2 πr
Period and frequency
• The Period, T is the time the object takes to move through one complete rotation.
(Unit: second, s)
• The frequency, f is the number of revolutions performed per second.
(Unit: Hertz, Hz or s-1))
T=1 f=1
and
f T
Acceleration and force
Experiment shows that:
ac = v2 ac = centripetal acceleration
r
Fc = centripetal force
Fc = mv2
r since F = ma
Monday, 24 May 2010
58. Examples
1. The Apollo 11 space capsule was placed in a parking orbit around the Earth before
moving onwards to the Moon. The radius of the orbit was 6.56 x 104 m and the
mass of the capsule was 4.4 x 104 kg.
to the moon
moon
parking orbit Earth
(a) If the centripetal force on the capsule was 407 kN while it was in the parking
orbit, what was its acceleration?
(b) What was the speed of the capsule in the parking orbit?
(c) How long did it take the capsule to complete one orbit?
Monday, 24 May 2010
59. 2. A game of swing-ball is played with a 100 g ball. The effective radius of the circular
path of the ball is 1.4 m. Find the tension in the string (centripetal force) when the
ball has a velocity of:
(a) 7.5 ms-1
(b) 15 ms-1
3. A string has a breaking strain of 320 N. Find the maximum speed that a mass can
be whirled around in uniform circular motion with a radius of 0.45 m if the mass is
0.2 kg.
4. An object is in uniform circular motion, tracing an angle of 30o every 0.010 s. Find:
(a) the period of this motion.
(b) the frequency of this motion
If the radius of the object’s path is doubled but the period remains the same, what
happens to:
(c) its speed?
(d) its acceleration?
Monday, 24 May 2010
60. 5. In a circular motion experiment, a mass is whirled around a horizontal circle of
radius 0.5 m. A student times four revolutions to take 1.5 s.
sinker
tube
(a) Calculate the speed of the mass around the circle
(b) What is the direction of the velocity of the mass?
(c) Calculate the centripetal acceleration of the mass.
(d) What is the direction of this acceleration?
(e) How does the value of the centripetal acceleration compare to the acceleration of
gravity?
Ex.12A All Q’s
Monday, 24 May 2010
70. 12 PHYSICS CIRCULAR MOTION ASSIGNMENT Name
1. A car is travelling around a bend in the road and for a few seconds is in uniform
circular motion.
(a) The centripetal force is being provided by the road. Name this force.
________________________________________________________________
The car passes over a patch of oil while it is rounding the bend.
(b) Describe the path the car will take after it hits the oil patch and explain why this
happens in terms of the forces acting.
________________________________________________________________
________________________________________________________________
________________________________________________________________
2. An object in uniform circular motion completes 10 revolutions in 0.4 seconds
(a) Find the frequency of this motion.
________________________________________________________________
________________________________________________________________
(b) Find the period of this motion.
________________________________________________________________
_______________________________________________________________
Monday, 24 May 2010
71. 3. A big wheel at a fair spins in a circular path of radius 20 m. Once the wheel has
reached a steady speed, a student times each revolution at 13 seconds.
(a) Calculate the circumference of the big wheel.
______________________________________
(b) Hence calculate the speed of the big wheel.
______________________________________
______________________________________
(c) Calculate the centripetal acceleration of each passenger.
________________________________________________________________
________________________________________________________________
4. In a circular motion experiment, a mass is whirled around a horizontal circle which
has a 0.50 m radius. A student time 4 revolutions to take 2.0 s.
(a) Calculate the speed of the mass around the circle.
________________________________________
________________________________________
(b) What is the direction of the velocity of the mass?
________________________________________
________________________________________
(c) Calculate the centripetal acceleration of the mass.
________________________________________
________________________________________
Monday, 24 May 2010
72. (d) What is the direction of this acceleration?
_______________________________________________________________
(e) How does the value of the centripetal acceleration compare to the acceleration
of gravity? ______________________________________________________
5. Two students go to a fun park for a day where they pay to drive carts around a
circular track. The track has a radius of 31.8 m and once the carts are at a
maximum speed they complete a lap in 16 s.
(a) What is the frequency of the cart’s motion when
travelling at maximum speed?
__________________________________________
(b) When travelling at maximum speed, calculate the
speed of the cart.
__________________________________________
__________________________________________
(c) Calculate the acceleration of the cart when travelling at maximum speed.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
Monday, 24 May 2010
73. The cart has a mass of 150 kg and one of the students, Chris has a mass of 75 kg.
(d) Calculate the size of the force acting on Chris and his cart at maximum speed.
________________________________________________________________
________________________________________________________________
________________________________________________________________
(e) Chris drives over a patch of oil and loses control of his cart whilst travelling at
this maximum speed. On the diagram, draw his path after driving through the
oil.
6. Jon and Ana are two ice-skaters. In a practiced skating move, Jon spins Ana around
in a horizontal circle.
Ana moves in a circle
as shown: Jon
Ana
(a) Draw an arrow on the diagram to show the direction of the tension force that
Jon’s arm exerts on Ana at the instant shown.
(b) If the radius of the circle is 0.95 m and the tension force in Jon’s arm is
5.00 x 102 N, calculate the speed with which Ana (55 kg) is travelling around the
circle. Give your answer to the correct number of significant figures.
________________________________________________________________
________________________________________________________________
________________________________________________________________
Monday, 24 May 2010
74. (d) While Ana is still moving in a circle on the ice, Jon lets her go.
(i) Describe her velocity (speed and direction) after he releases her.
______________________________________________________________
(ii) Explain why Ana travels with this velocity.
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
Monday, 24 May 2010
