PHYSICAL SCIENCES 12
(PHYSICS: P1)
MECHANICS
METRO SOUTH EDUCATION DISTRICT
INTERVENTION HUBS
2019
2
CONTENT
TOPIC
1
2
3
4
5
WORK, ENERGY
& POWER
NO
GENERAL INFORMATION
APPLICATION OF NEWTO'S LAWS
VERTICAL PROJECTILE MOTION
MOMENTUM-IMPULSE
WORK, ENERGY & POWER
PAGE/S
3
DATA FOR PHYSICAL SCIENCES GRADE 12
PAPER 1 (PHYSICS)
GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12
VRAESTEL 1 (FISIKA)
RE
M
me
-e
k
h
c
G
g
SYMBOL/SIMBOOL
6,38 x 106 m
5,98 x 1024 kg
9,11 x 10-31 kg
-1,6 x 10-19 C
9,0 x 109 N·m2·C-2
6,63 x 10-34 J·s
3,0 x 108 m·s-1
6,67 x 10-11 N·m2·kg-2
9,8 m·s-2
VALUE/WAARDE
TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES
NAME/NAAM
Acceleration due to gravity
Swaartekragversnelling
Universal gravitational constant
Universele gravitasiekonstant
Speed of light in a vacuum
Spoed van lig in 'n vakuum
Planck's constant
Planck se konstante
Coulomb's constant
Coulomb se konstante
Charge on electron
Lading op elektron
Electron mass
Elektronmassa
Mass of Earth
Massa van Aarde
Radius of Earth
Radius van Aarde
vi
vi
2
a t
2a x or/of v f
2
vi
2
2a y
Δx
Δx
v i Δt
vf
1
2
2
vi
TABLE 2: FORMULAE/TABEL 2: FORMULES
2
MOTION/BEWEGING
vf
vf
FORCE/KRAG
p mv
ma
g =G
M
d2
w mg
Fnet
m1m 2
r2
fk = μk N
or/of
F =G
fs max = μsN
Fnet Δt = Δp
m1m 2
d2
Δp = mv f - mvi
F =G
or/of
Ek
Ep
1
mv2
2
U mgh
Kf
W
t
Wnet
K
P
K
Ki
f
fL
v vL
fb
v vb
T
1
f
E hf
a t 2 or/of Δy
or/of
or/of
Wnet
EP
c
Ek
E h
v i Δt
vf
1
2
2
vi
M
r2
Ekf
Δt
a t2
Eki
Ek
mgh
g =G
Δt or/of Δy
or/of
or /of
or/of
WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING
W F x cos
1
K
mv2
2
U or/of Wnc
Ek
Wnc
/ Pgemid = Fvgemid
K
Pav = Fvav
v
v vL
fs
v vs
WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG
fL
1
1
2
2
hf0 and/en E k (max) = mvmax
or/of
K
mvmax
max =
2
2
E = Wo + Ek(max) or/of E = Wo + K max where/waar
E hf and/en W0
kQ1Q 2
r2
E
E
F
q
kQ
r2
ELECTROSTATICS/ELEKTROSTATIKA
F
W
q
Q
qe
V
n=
Q
e
or/of
n=
...
...
/
/
I wgk
Vwgk
Imaks
2
Vmaks
W
Δt
q I
P
P = VI
V2
R
t
P = I 2R
P
/
Pgemiddeld Vwgk I wgk
R
2
Vwgk
I 2wgkR
Pave Vrms Irms
Pgemiddeld
Pgemiddeld
2
Vrms
R
/
Pave
/
emk ( ε )= I(R + r)
emf ( ε )= I(R + r)
ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE
R1 R 2
1
1
R1 R 2
V
R
I
Rs
1
Rp
W = Vq
W = VI t
W = I2R t
V 2 Δt
W=
R
I max
2
Vmax
2
2
Pave Irms
R
ALTERNATING CURRENT/WISSELSTROOM
I rms
Vrms
2
GENERAL INFORMATION:
3
Application of Newton's Laws.
Verical Projectile Motion (Free Fall)
Momentum-Impulse, and
Work, Energy & Power
This tutorial unit covers the knowledge strand MECHANICS, which includes the
following topics:
•
•
•
•
Examination guidelines with definitions
short note/concept map,
multiple choice questions, and
Long questions, drawn from past papers, from Nov2014-Nov2018.
Each of these sections is made-up of
•
•
•
•
Enjoy every moment as you refresh and consolidate what you know.
Remember:
ANCIENT PROVERB
“The teacher opens the door, but you must enter by yourself.”
Many blessings…#Matric2019!
QUESTION 1: MULTIPLE-CHOICE QUESTIONS - MECHANICS
C
B
A
constant momentum.
constant kinetic energy.
constant velocity.
constant acceleration.
N
concrete floor
table
A block rests on a table. The table stands on a concrete floor. The normal
force is represented by N, as shown in the diagram below.
D
Force F will cause the box to move with …
Which ONE of the following statements regarding force F is CORRECT?
A constant horizontal force F is applied to a box resting on a horizontal,
frictionless surface.
Various options are provided as possible answers to the following questions. Choose
the answer and write only the letter (A–D) next to the question number (1.1–1.10) in
the ANSWER BOOK, for example 1.11 D.
1.1
1.2
block
C
B
A
Force of the block on the concrete floor
Force of the table surface on the block
Force of the block on the table
Force of the block on the Earth
Which ONE of the following forces will form an action-reaction pair with the
normal force (N)?
D
(2)
(2)
1.3
A small stone is dropped from a height y above the ground. It strikes the
ground after time t, as shown in the diagram below.
●
y
ground
Take upwards as the positive direction and the ground as zero reference.
Ignore the effects of air resistance.
y
0
A
C
(s)
time
time
y
0
y
0
B
D
time
time
(s)
(s)
Which ONE of the following diagrams shows a correct position-time graph for
the motion of the stone?
y
0
(s)
Position
Position
Position
Position
(2)
1.4
1.5
mm
m
m
m
m
m
m
Compressed spring
Learners perform an experiment using identical trolleys, each of mass m.
The trolleys are arranged, as shown in the diagram below. They are initially at
rest on a frictionless surface and are connected with a compressed, massless
spring.
2p
When the spring is released it falls vertically down and the single trolley
moves with momentum p to the left.
The magnitude of the momentum of the two trolleys moving to the right
will be:
A
p
1
p
4
1
p
2
B
C
D
P
horizontal surface
block
A pendulum bob is released from point P above a horizontal surface. At the
lowest point, Q, of its swing, it collides with a stationary block situated on a
frictionless horizontal surface, as shown below. Ignore air friction.
pendulum bob
Q
SPEED AFTER COLLISION
Conservation of linear
momentum
Conservation of mechanical
energy
Conservation of mechanical
energy
Conservation of linear
momentum
Which ONE of the following combinations of conservation laws can be used
to calculate the speed of the bob at Q immediately before and after colliding
with the block?
A
B
C
D
SPEED AT Q
Conservation of mechanical
energy
Conservation of linear
momentum
Conservation of mechanical
energy
Conservation of linear
momentum
(2)
(2)
1.6
1.7
1.8
A constant net force acts on a trolley.
C
B
A
inversely proportional to
directly proportional to
independent of
equal to
According to Newton's Second Law, the acceleration of the trolley is … the
mass of the trolley.
D
The weight of a man on the surface of the Earth is w. Planet X has the same
radius as the Earth, but half the mass of the Earth.
C
B
A
2w
w
½w
¼w
If the same man goes to Planet X, his weight on the surface will be …
D
An object falls freely in a vacuum near the surface of the Earth.
C
B
A
The rate of change of velocity of the object will remain constant.
The rate of change of velocity of the object will increase uniformly.
The velocity of the object will decrease uniformly.
The velocity of the object will remain constant.
Which ONE of the following statements regarding the motion of the object is
CORRECT?
D
(2)
(2)
(2)
1.9
1.10
v
wall
AFTER COLLISION
A ball, moving horizontally, hits a wall with a speed 2v. The ball then bounces
back horizontally with a speed v, as shown in the diagram below.
BEFORE COLLISION
2v
wall
LINEAR MOMENTUM
Conserved
Conserved
Not conserved
Not conserved
TOTAL KINETIC ENERGY
Not conserved
Conserved
Not conserved
Conserved
Which ONE of the following combinations regarding the linear momentum and
the total kinetic energy of the ball for the collision above is CORRECT?
Assume that the ball-wall system is isolated.
A
B
C
D
∆x
surface
A constant horizontal force F displaces a box by ∆x over a rough horizontal
surface. Study the diagram below.
F
C
B
A
perpendicular to the displacement of the box.
equal and opposite to the weight of the box.
perpendicular to the applied force.
equal to the applied force.
The normal force acting on the box does NO work on the box during the
motion, because it is …
D
(2)
(2)
1.11
1.12
1.13
C
B
A
rate of change of kinetic energy
rate of change of momentum
change in kinetic energy
change in momentum
The net (resultant) force acting on an object is equal to the ... of the object in
the direction of the net force.
D
C
B
A
weight.
acceleration.
force.
inertia.
A physical quantity that is described as a measure of the resistance of a body
to a change in motion is called …
D
I•
II
• IV
• III
•
The diagram below shows a section of the path of a stone projected vertically
upwards.
C
B
A
IV
III
II
I
At which ONE of the positions indicated on the diagram will the magnitude of
the momentum of the stone be the GREATEST? Ignore air resistance.
D
(2)
(2)
(2)
1.14
1.15
1.16
1.17
Two cars, P and Q, moving in a straight line, have the same momentum.
The kinetic energy of Q is greater than the kinetic energy of P.
C
B
A
Q is moving at the same speed as P.
Q is moving slower than P.
Q has the same mass as P.
Q has a smaller mass than P.
Which ONE of the following statements regarding the cars is CORRECT?
D
W
B
2W
C
3W
D
4W
The net work done on an object to increase its speed from rest to v is W.
How much net work must be done on the same object to increase its speed
from v to 2v?
A
The acceleration due to gravity on Earth is greater than that on the moon.
C
B
A
The weight of an object on Earth is less than that on the moon.
The mass of an object on Earth is greater than that on the moon.
The mass of an object on Earth is the same as that on the moon.
The weight of an object on Earth is the same as that on the moon.
Which ONE of the following statements is CORRECT?
D
N
w
θ
F
The force diagram below shows the forces acting on a box.
C
B
A
N = w – Fsinθ
N = w – Fcosθ
N = w + Fsinθ
N = w + Fcosθ
Which ONE of the following equations for the magnitude of the normal
force (N) is CORRECT?
D
(2)
(2)
(2)
(2)
1.18
t1
t (s)
A stone is projected vertically upwards from the top of a building at a speed
of v m∙s-1. The position-time graph below represents the motion of the stone.
Ignore the effects of air resistance.
0
MAGNITUDE OF VELOCITY
(m∙s-1)
0
0
v
v
MAGNITUDE OF ACCELERATION
(m∙s-2)
9,8
0
0
9,8
Which ONE of the combinations below regarding the magnitudes of the
stone's velocity and acceleration, at time t 1 , is CORRECT?
A
B
C
D
m
v
-v
v
½v
v
t
t(s)
t(s)
time t?
B
D
v
-½v
2v
v
t
t
t(s)
t(s)
velocity of the trolley before and after
below CORRECTLY represents the
Which ONE of the velocity-time graphs
The ball of clay sticks to the trolley.
A trolley of mass m is moving at constant velocity v to the right on a
frictionless horizontal surface. A ball of clay, also of mass m, dropped
vertically, falls onto the trolley at time t, as shown in the diagram below.
C
A
Ball of clay m
1.19
y (m)
t
v (m∙s-1)
v (m∙s-1)
v (m∙s-1)
v (m∙s-1)
(2)
(2)
1.20
1.21
1.22
1.23
A person lifts a crate vertically upwards at constant velocity through a
distance h. The person does work x on the crate in time t.
B
x
C
maintain its mass.
D
A
continue in a state of non-uniform motion.
2x
B
remain at rest or in the state of uniform motion.
4x
The person now lifts the same crate vertically upwards at constant velocity
through the same distance, but in time 2t.
½x
The work done by the person on the crate will now be …
A
C
maintain its velocity when a non-zero net force is acting on it.
Inertia is the tendency of an object to …
D
B
A
upwards at a increasing speed.
downwards at a constant speed.
upwards at a constant speed.
B
A
is directed upwards, but its acceleration is directed downwards.
and acceleration are both directed downwards.
and acceleration are both directed upwards.
scale
lift
lift cable
A person stands on a bathroom scale that is fixed to the floor of a lift, as
shown in the diagram below.
C
downwards at a increasing speed.
The reading on the scale is largest when the lift moves …
D
An object is projected vertically upwards. Ignore air resistance.
C
is directed downwards, but its acceleration is directed upwards.
As the object rises, its velocity …
D
(2)
(2)
(2)
(2)
1.24
1.25
½v
Q
Ball P and ball Q, of the same mass, are dropped onto a concrete floor.
Both balls hit the concrete floor at the same speed, v. Ball P rebounds with
the same vertical speed, v, but ball Q rebounds with speed ½v.
Refer to the diagram below. Ignore air resistance.
P
v
concrete floor
C
B
A
Momentum is conserved for the collision of ball P, but not for that of
ball Q.
The contact time with the floor is the same for both balls P and Q.
The change in momentum of ball P is greater than that of ball Q.
Kinetic energy is conserved for both balls P and Q.
Which ONE of the following statements regarding the collision of EACH ball
with the concrete floor is CORRECT?
D
C
B
A
has not changed.
has decreased.
has increased.
is zero.
If the net work done on a moving object is POSITIVE, then we can conclude
that the kinetic energy of the object …
D
(2)
(2)
1.26
1.27
constant speed
An object, of mass m, hangs at the end of a string from the ceiling of a lift
cage. The lift is moving upward at CONSTANT SPEED. The acceleration due
to gravity is g.
T
m
Which ONE of the following statements regarding the tension (T) in the string
is CORRECT?
C
B
A
cannot be determined without knowing the speed of the lift cage.
will be greater than mg.
will be less than mg.
will be equal to mg.
The tension T …
D
Two hypothetical planets, X and Y, have the same mass. The diameter of
planet Y is twice that of planet X.
C
B
A
2g
g
2
g
4
g
16
If the acceleration due to gravity on the surface of planet X is g, then the
acceleration due to gravity on the surface of planet Y will be …
D
(2)
(2)
1.28
1.29
1.30
A ball is projected vertically upwards from a height X above the ground. After
some time, the ball falls to the ground and bounces back to the same height
from which it was projected. Ignore friction and assume that there is a
negligible time lapse during the collision of the ball with the ground.
t (s)
t (s)
B
D
X
0
X
0
t (s)
t (s)
Which ONE of the following is the CORRECT position-time graph for the
motion of the ball as described above?
A
X
0
X
0
Position (m)
Position (m)
C
B
A
Kinetic energy is conserved, but momentum not.
Momentum is conserved, but kinetic energy not.
Both momentum and kinetic energy are not conserved.
Both momentum and kinetic energy are conserved.
D
C
B
A
kinetic energy of the object remains unchanged.
kinetic energy of the object is decreasing.
kinetic energy of the object is increasing.
kinetic energy of the object is zero.
When the net work done on an object is positive (greater than zero), the …
D
Which ONE of the following statements is always TRUE for inelastic collisions
in an isolated system?
C
Position (m)
Position (m)
(2)
(2)
(2)
1.31
1.32
1.33
¼a
B
½a
C
2a
D
4a
A net force F which acts on a body of mass m causes an acceleration a. If
the same net force F is applied to a body of mass 2m, the acceleration of the
body will be …
A
m
Two objects of masses 2m and m are arranged as shown in the diagram
below.
2m
d
C
B
A
Halve the distance between the masses.
Double the distance between the masses.
Halve the smaller mass.
Double the larger mass.
Which ONE of the changes below will produce the GREATEST increase in
the gravitational force exerted by the one mass on the other?
D
(ii)
(i)
A steel ball falls through the air in the absence of air friction.
A box slides along a smooth horizontal surface at constant speed.
A feather falls from a certain height inside a vacuum tube.
The statements below describe the motion of objects.
(iii)
C
B
A
(i), (ii) and (iii)
(ii) and (iii) only
(i) and (iii) only
(i) and (ii) only
Which of the following describes UNIFORMLY ACCELERATED motion
CORRECTLY?
D
(2)
(2)
(2)
1.34
1.35
1.36
1.37
Airbags in modern cars provide more safety during an accident.
(ii)
(i)
The impulse increases.
The impact force decreases.
The time of impact increases.
B
(ii) only
C
(ii) and (iii) only
D
(i) and (ii) only
The statements below are made by a learner to explain how airbags can
ensure better safety in a collision.
(iii)
(i) only
Which of the statements above are CORRECT?
A
The work done by a constant force F applied to an object to increase the
object's speed from v to 2v is W.
⅓W
B
½W
C
2W
D
3W
The work done by the same force to increase the speed of the object from
0 to v will be ...
A
A horizontal force F is applied to a crate, causing it to move over a rough,
horizontal surface as shown below.
A
the surface area of the crate in contact with the floor.
the applied force F.
F
B
how fast the crate moves on the surface.
crate
C
the upward force exerted by the surface on the crate.
The kinetic frictional force
between the crate and the
surface on which it is moving
depends on …
D
C
B
A
moves downward at constant speed.
moves upward at constant speed.
accelerates upward.
accelerates downward.
An object is placed on a bathroom scale in a lift which is stationary on
the third floor of a building. The reading on the scale will be greatest when
the lift ...
D
(2)
(2)
(2)
(2)
1.38
1.39
1.40
1.41
C
B
A
greater than the weight of the ball.
Iess than the weight of the ball.
equal to the weight of the ball.
zero.
A ball is thrown vertically upwards into the air. Ignore the effects of friction.
The NET FORCE acting on the ball when the ball is at its highest point is ...
D
A
decreases
is zero
D
C
remains constant
increases
During a collision an inflated air bag in a car decreases the net force that
would have acted on the driver of the car. This is because the time interval
over which the net force acts on the driver … for the same momentum
change.
B
An object moving horizontally at a constant velocity suddenly encounters a
rough horizontal surface. The object continues to move over this rough
surface. Which ONE of the following statements is CORRECT?
zero.
B
positive.
C
negative.
D
constant.
The net work done on the object during the motion over the rough surface
is …
A
The hooter of a car emits sound of constant frequency as the car moves away
from a stationary listener.
C
B
A
Both frequency and loudness
Both wavelength and frequency
Frequency
Velocity
Which ONE of the following properties of the sound heard by the listener will
NOT change?
D
(2)
(2)
(2)
(2)
1.42
1.43
C
B
A
Tension force
Gravitational force
Frictional force
Normal force
Which ONE of the following forces always acts perpendicular to the surface
on which a body is placed?
D
2m
B
Two isolated bodies, A and B, having masses m and 2m respectively, are
placed a distance r apart.
A
m
r
(iii)
(ii)
(i)
The forces will always be attractive.
The force exerted on body A by B is equal but opposite to that exerted
on body B by A.
The force exerted on the bodies is independent of the masses of the
bodies.
The force exerted by B on body A is half that exerted by A on body B.
Consider the following statements regarding the gravitational force exerted by
the bodies on each other.
(iv)
C
B
A
(iv) only
(iii) and (iv) only
(ii), (iii) and (iv) only
(i), (ii) and (iv) only
Which of the statements above is/are TRUE?
D
(2)
(2)
1.45
0
0
A ball is released from a height above the floor. The ball falls vertically and
bounces off the floor a number of times. Ignore the effects of friction and
assume that the collision of the ball with the floor is elastic. Take the point of
release of the ball as the reference point and downward direction as positive.
time
0
B
D
time
time
Which ONE of the following is a CORRECT representation of the positiontime graph for the motion of the ball?
A
C
0
D
C
B
A
Conserved
Not conserved
Conserved
Not conserved
MOMENTUM
Conserved
Not conserved
Not conserved
Conserved
KINETIC ENERGY
Two bodies undergo an INELASTIC collision in the absence of friction. Which
ONE of the following combinations of momentum and kinetic energy of the
system is CORRECT?
time
position
position
1.44
position
position
(2)
(2)
1.46
1.47
1.48
4.
B
6.
C
8.
D
16.
The speed of a bicycle increases from 2 m∙s-1 to 8 m∙s-1. Its kinetic energy
increases by a factor of …
A
The magnitude of an electric field, a distance r from a point charge is E. The
magnitude of an electric field, a distance 2r from the same point charge
will be …
B
2E
1
E
2
1
E
4
C
4E
A
D
A
accelerate towards the west.
accelerate towards the east.
W
N
S
E
Two forces, F1 and F2, are applied on a crate lying on a frictionless, horizontal
surface, as shown in the diagram below.
B
move at a constant speed towards the east.
F2
The magnitude of force F1 is greater than that of force F2.
F1
C
move at a constant speed towards the west.
The crate will …
D
(2)
(2)
(2)
3
W
4
C
W
D
1
g, where
4
A person stands on a bathroom scale that is calibrated in newton, in a
stationary elevator. The reading on the bathroom scale is W.
B
5
W
4
The elevator now moves with a constant upward acceleration of
g is the gravitational acceleration.
1
W
4
What will the reading on the bathroom scale be now?
A
v (m∙s-1)
B
D
v (m∙s-1)
v (m∙s-1)
Which ONE of the graphs below correctly represents the relationship between
the kinetic energy (K) of a free-falling object and its speed (v)?
A
C
Rocket
W
N
S
E
D
C
B
A
The rocket pushes the exhaust gases to the east and the exhaust gases
push the rocket to the west.
The air outside the rocket exerts a greater force on the back of the
rocket than at the front.
The pressure of the atmosphere at the back of the rocket is less than at
the front.
The speed of the exhaust gases is smaller than the speed of the rocket.
Which ONE of the statements below best explains why the rocket
accelerates?
exhaust gases
The simplified diagram below shows a rocket that has been fired horizontally,
accelerating to the west.
v (m∙s-1)
K (J)
K (J)
1.49
1.50
1.51
K (J)
K (J)
(2)
(2)
(2)
1.52
1.53
1.54
1.55
Time (s)
D
C
B
A
potential energy.
B
A
2F to the west.
F to the east.
F to the west.
B
A
change in momentum.
rate of change in energy.
change in energy.
C
B
A
decreases.
increases.
becomes zero.
remains constant.
kinetic energy.
momentum.
power.
(2)
(2)
(2)
(2)
The gradient of the graph represents the …
The graph below represents the relationship between the work done on an
object and the time taken for this work to be done.
C
2F to the east.
C
rate of change in momentum.
D
If air resistance is negligible, the total mechanical energy of a free-falling
body …
D
Net force is a measure of the …
D
If the sphere of charge Q experiences a force F to the east, then the sphere of
charge 2Q will experience a force …
Two charged spheres of magnitudes 2Q and Q respectively are placed a
distance r apart on insulating stands.
Work done (J)
1.56
1.57
1.58
1.59
1.60
A
doubled.
halved.
D
C
four times greater.
three times greater.
If the momentum of an object is doubled, then its kinetic energy is ...
B
B
C
Q.
D
2Q.
Two small identical metal spheres, each carrying equal charges Q, are
brought into contact and then separated.
zero.
The charge on each sphere will now be …
A
Q
2.
A
Energy
Mass
D
C
Acceleration
Velocity
Which ONE of the following physical quantities is a measure of the inertia of
a body?
B
1F
4
B
1F
2
C
2F
D
4F
The magnitude of the gravitational force exerted by one body on another body
is F. When the distance between the centres of the two bodies is doubled, the
magnitude of the gravitational force, in terms of F, will now be …
A
C
B
A
Zero
Maximum
Zero
Zero
VELOCITY
Downwards
Zero
Upwards
Zero
ACCELERATION
An object is thrown vertically upwards. Which ONE of the following regarding
the object's velocity and acceleration at the highest point of its motion is
CORRECT? Ignore the effects of friction.
D
(2)
(2)
(2)
(2)
(2)
1.61
1.62
1.63
B
A
Conserved
Not conserved
Conserved
MOMENTUM
Not conserved
Not conserved
Conserved
Conserved
MECHANICAL ENERGY
m
v
2m
vf = 0
AFTER COLLISION
An object of mass m moving at velocity v collides head-on with an object of
mass 2m moving in the opposite direction at velocity v. Immediately after the
collision the smaller mass moves at velocity v in the opposite direction and
the larger mass is brought to rest. Refer to the diagram below.
2m
v
BEFORE COLLISION
v
m
Ignore the effects of friction.
C
Not conserved
Which ONE of the following is CORRECT?
D
Two balls, P and Q, are dropped simultaneously from the same height.
Ball P has TWICE the mass of ball Q. Ignore the effects of air friction.
B
1x
2
C
x
D
2x
Just before the balls hit the ground, the kinetic energy of ball P is x. The
kinetic energy of ball Q, in terms of x, will be …
A
1x
4
C
B
A
rate of change in momentum of the object.
change in momentum of the object.
acceleration of the object.
mass of the object.
The net force acting on an object is directly proportional to the ...
D
(2)
(2)
(2)
1.64
1.65
1.66
1.67
C
B
A
Velocity
Momentum
Kinetic energy
Acceleration
A ball is thrown vertically upwards. Which ONE of the following physical
quantities has a non-zero value at the instant the ball changes direction?
D
2m
Q
Two trolleys, P and Q, of mass m and 2m respectively are at rest on a
frictionless horizontal surface. The trolleys have a compressed spring
between them.
P
m
C
B
A
The sum of the final momentum of P and Q is zero.
The sum of the final kinetic energies of P and Q is zero.
The speed of P is less than the speed of Q.
P and Q have equal kinetic energies.
The spring is released and the trolleys move apart. Which ONE of the
following statements is TRUE?
D
C
B
A
Newton's Second Law.
Newton's Third Law.
acceleration.
inertia.
The tendency of an object to remain at rest or to continue in its uniform
motion in a straight line is known as …
D
1
4
M
B
1
9
M
C
M
D
2M
The mass of an astronaut on Earth is M. At a height equal to twice the radius
of the Earth, the mass of the astronaut will be …
A
(2)
(2)
(2)
(2)
1.68
1.69
1.70
An object is thrown vertically upwards from the ground.
OBJECT MOVING UPWARDS
Downwards
Upwards
Downwards
Upwards
OBJECT MOVING DOWNWARDS
Upwards
Downwards
Downwards
Upwards
Which ONE of the following is CORRECT regarding the direction of the
acceleration of the object as it moves upwards and then downwards? Ignore
the effects of air resistance.
A
B
C
D
A person drops a glass bottle onto a concrete floor from a certain height and
the bottle breaks. The person then drops a second, identical glass bottle from
the same height onto a thick, woollen carpet, but the bottle does not break.
AVERAGE FORCE ON
SECOND BOTTLE
Larger
Smaller
Larger
Smaller
TIME OF CONTACT WITH
CARPET
Smaller
Smaller
Larger
Larger
Which ONE of the following is CORRECT for the second bottle compared to
the first bottle for the same momentum change?
A
B
C
D
A block of mass m is released from rest from the top of a frictionless inclined
plane QR, as shown below.
The total mechanical energy of the block is E Q at point Q and E R at point R.
The kinetic energy of the block at points Q and R is K Q and K R respectively.
TOTAL MECHANICAL ENERGY E
EQ > ER
KQ < KR
KINETIC ENERGY K
KQ = KR
m
A
EQ = ER
KQ = KR
Q
B
EQ = ER
KQ > KR
Which ONE of the statements
regarding the total mechanical
energy and the kinetic energy of
the block at points Q and R
respectively is CORRECT?
C
EQ < ER
R
D
(2)
(2)
(2)
-
I
1.71
B
A
directly proportional to its mass.
always equal to its mass.
independent of its mass.
According to Newton's Second Law of Motion, the acceleration of an
o bject is ...
C
I J>8<I
inversely proportional to its mass.
I I
D
,-A--.! I
EXAMPLE:
1.80
The diagram below shows three blocks, P, Q and R, suspended from a
ceiling. The blocks are identical, stationary and have the same mass but are
at different heights a bove the ground.
Tp
p
Q
Ta
R
TR
I
ceiling
The connecting strings are massless and inextensible. The tensions in the
strings attached to blocks P, Q and Rare T p, Ta and TR respectively.
ground
Tp
>
Ta and Ta
<
TR
Which ONE of the following statements about the tensions is CORRECT?
D
(2)
(2 )
A
c
v
v
t
B
D
C
B
A
linear momentum of the o bject is doubled.
potential energy of the o bject is doubled.
kinetic energy of the object is doubled.
net work done by the object is doubled.
v
v
When the velocity of a moving object is doubled, the ...
0
8
A
change in momentum of the object.
change in kinetic energy of the object.
inertia of the object.
The net work required to stop a moving object is equal to the ...
C
change in impulse of the object.
·
t
t
Which ONE of the following velocity-time graphs CORRECTLY describes the
motion of the ball?
1. 81 A ball is projected vertically upwards from the ground. It returns to the ground,
makes an elastic collision with the ground and then bounces to a maximum
height. Ignore air resistance.
1.82
1.83
D
(2)
(2)
(2)
PAPER 1: PHYSICS
Newton's Laws and Application of Newton's Laws (Grade 11)
(This section must be read in conjunction with the CAPS, p. 62–66.)
Different kinds of forces: weight, normal force, frictional force, applied force (push,
pull), tension (strings or cables)
•
Define normal force, N.
•
Define frictional force, f.
Force diagrams, free-body diagrams
•
Draw force diagrams.
•
Draw free-body diagrams.
•
Resolve two-dimensional forces (such as the weight of an object with respect to the
inclined plane) into its parallel (x) and perpendicular (y) components.
Determine the resultant or net force of two or more forces.
•
Newton's first, second and third laws
•
State Newton's first law: A body will remain in its state of motion (at rest or moving at
constant velocity) until a net force acts on it.
•
Discuss why it is important to wear seatbelts using Newton's first law.
•
State Newton's second law: When a net force acts on an object, the object will
accelerate in the direction of the force and the acceleration is directly proportional to the
force and inversely proportional to the mass of the object.
•
Draw force diagrams and free-body diagrams for objects that are in equilibrium or
accelerating.
•
Apply Newton's laws to a variety of equilibrium and non-equilibrium problems including:
o A single object:
- Moving on a horizontal plane with or without friction
- Moving on an inclined plane with and without friction
- Moving in the vertical plane (lifts, rockets, etc.)
o Two-body systems (joined by a light inextensible string):
- Both on a flat horizontal plane with and without friction
- One on a horizontal plane with and without friction, and a second hanging
vertically from a string over a frictionless pulley
- Both on an inclined plane with or without friction
- Both hanging vertically from a string over a frictionless pulley
•
State Newton's third law: When one body exerts a force on a second body, the second
body exerts a force of equal magnitude in the opposite direction on the first body.
Identify action-reaction pairs.
List the properties of action-reaction pairs.
•
•
Solve problems using F =
r2
Newton's Law of Universal Gravitation
•
State Newton's Law of Universal Gravitation: Each body in the universe attracts every
other body with a force that is directly proportional to the product of their masses and
inversely proportional to the square of the distance between their centres.
Gm1m 2
.
•
•
•
•
•
•
Describe weight as the gravitational force the Earth exerts on any object on or near its
surface.
Calculate weight using the expression w = mg.
Calculate the weight of an object on other planets with different values of gravitational
acceleration.
Distinguish between mass and weight.
Explain weightlessness.
The static frictional force is a maximum (f ms ax ) just before the object starts to
move across the surface.
This is a diagram that shows the relative magnitudes and directions of forces
acting on a body/particle that has been isolated from its surroundings
A body will remain in its state of rest or motion at constant velocity unless a
non-zero resultant/net force acts on it.
The resistance of a body to a change in its state of uniform motion or te rest.
Mass is a measure of an obejct’s inetia.
When a resultant/net force acts on an object, the object will accelerate in the
direction of the force at an acceleration directly proportional to the force and
inversely proportional to the mass of the object.
In symbols: Fnet = ma
When object A exerts a force on object B, object B SIMULTANEOUSLY
exerts a force equal in magnitude but opposite in direction on object A.
Each body in the universe attracts every other body with a force that is
directly proportional to the product of their masses and inversely proportional
to the square of the distance between their centres.
Gm 1m 2
In symbols: F 
r2
Define, N, as the force or the component of a force which a surface exerts on
an object with which it is in contact, and which is perpendicular to the surface.
The amount of matter in a body measured in kilogram (kg).
The gravitational force, in newton (N), exerted on an object.
The sensation experienced when all contact forces are removed i.e. no
external objects touch one's body.
TERMS AND DEFINITIONS: PHYSICS – PAPER 1
MECHANICS: NEWTON’S LAWS
Static frictional force The force that opposes the tendency of motion of a STATIONARY object
(f )
relative to a surface.
s
Kinetic frictional
The force that opposes the motion of a MOVING object relative to a surface.
force (fk)
Maximum static
frictional force (f ms ax )
Free-body diagrams
Newton's first law of
motion
Inertia
Newton's second law
of motion
Newton’s Third Law
of motion
Newton's Law of
Universal Gravitation
Normal force
Mass
Weight
Weightlessness
Newton’s Laws of Motion
Grade 12 Science Essentials
SCIENCE CLINIC 2019 ©
FORCES
Non-contact force: A force exerted between
Contact force: A force exerted between
objects over a distance without physical contact.
objects that are in contact with each other.
Electrostatic force (FE)
Applied force (FA)
Gravitational force (w/Fg)
Tension (T or FT)
Magnetic force
Friction (Ff or fs/fk)
A force is a push or a pull action exerted on an object by another object. This action can be exerted
while objects are in contact (contact force) or over a distance (non-contact force).
Because forces have magnitude and direction, they are vectors. Force is measured in newton (N). 1 N is
the force required to accelerate a 1 kg object at 1 m·s-2 in the direction of the force. We can therefore
say that 1 N = 1 kg·m·s-2.
Normal force (N/FN)
Normal force (FN)
Friction (Ff or fs/fk)
The perpendicular force exerted by a surface on an object in contact with it.
Frictional force due to a surface is the force that opposes the motion of an object in contact
with it, parallel to the surface.
FN
The normal force is equal to the perpendicular component of
gravity if there are no other forces acting on the object.
FN = Fg
Friction is the parallel component of the contact force on an object by the surface on which
it rests. The friction between the contact surfaces is determined by the properties of both the contact
surfaces of the object and surface. The coefficient of friction (µs/µk) is a description of the roughness of
the surface. The rougher the surface, the greater the coefficient of friction.
Fg
If alternative forces act on the object, the normal force will change depending on the direction and magnitude of the applied force. All vertical forces must be balanced if there is no acceleration in the vertical
plane.
FN
FA
FN
θ
FA
θ
Static friction (fs)
Kinetic friction (fk)
Static friction is the frictional force on a stationary object that opposes the tendency of
motion of the object. The magnitude of the
static friction will increase from 0N as the parallel
component of the applied force is increased, until
maximum static friction is reached. fs(max) is the
magnitude of friction when the object just starts
to move.
Kinetic friction is the frictional force on a moving object that opposes the motion of the
object. The magnitude of the kinetic friction is
constant for the specific system at all velocities
greater than zero, and irrespective of the applied
force.
fk = μk FN
fs(max) = μs FN
Fg
Fg
FN − Fg − FAy = 0
FN = Fg + FA sin θ
Objects suspended from a rope/string/cable have no
normal force, as there is no surface on which the object rests.
The tension is equal to the perpendicular component of
gravity if there are no other forces acting on the object
OR the full magnitude of Fg for vertically suspended
objects that are stationary/moving at constant velocity.
FT + (− Fg ) = 0
fk = kinetic friction (N)
μs = coefficient of friction (no unit)
μk = coefficient of friction (no unit)
FN = normal force (N)
FN = normal force (N)
If the applied force is greater than the maximum static friction, the object will start to move.
fsmax
FT
Fric%on (N)
FN + FAy − Fg = 0
FN + FA sin θ = Fg
fs(max) = maximum static friction (N)
c
a$
St
c
fri
Fg
n
$o
)
(f s
Kine%c fric%on (fk)
Applied force (N)
13
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Newton’s Laws of Motion
Grade 12 Science Essentials
Newton’s First Law of Motion
An object continues in a state of rest or uniform (moving with
constant) velocity unless it is acted upon by an unbalanced (net
or resultant) force.
Newton’s First Law is due to inertia- the resistance of an object to
change its state of rest or constant/uniform motion.
Fnet = 0 N
a = 0 m⋅s
A 3kg object moves up an incline surface at an angle of 15º with a
constant velocity. The coefficient of friction is 0,35. Determine
the magnitude of the applied force.
FA
FN
Newton’s Second Law of Motion
Newton’s Third Law of Motion
When a net force is applied to an object of mass, it accelerates in the direction of the net force. The acceleration is
directly proportional to the net force and inversely proportional to the mass.
When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A.
Newton’s Second Law is dependent on the resultant forceThe vector sum of all forces acting on the same object.
−2
A 20 N force is applied to a 5 kg object. The object accelerates
up a frictionless incline surface at an angle of 15º. Determine
the acceleration of the object.
FA
FN
Fg//
=
FN
FN
FN
∴ FN
=
=
=
=
Force pairs properties:
• Equal in magnitude
• Opposite in direction
• Acts on different objects (and therefore DO NOT CANCEL each
other out)
Fg ⊥
Fnet//
FA + (− Fg // ) + (− fk )
=
=
0
0
FA
=
FA
FA
∴ FA
=
=
=
Fg // + fk
Fg
15°
m g cos θ
(3)(9,8)cos 15∘
28,40 N
28,40 N ⊥ up from slope
Fearth on man
Take upwards as positive:
Fnet//
=
ma
FA + (− Fg // )
=
ma
m g sin θ + μk FN
(3)(9,8)sin 15∘ + (0,35)(28,40)
17,55 N
Importance of wearing safety belts:
According to Newton’s First Law, an object will remain in motion at a
constant velocity unless a non-zero resultant force acts upon it. When a
car is in an accident and comes to a sudden stop, the person inside the
car will continue with a constant forward velocity. Without a safety
belt, the person will make contact with the windscreen of the car, causing severe head trauma. The safety belt acts as an applied force (new
Fnet), preventing the forward motion of the person.
NOTE:
The force pairs shown
here are gravitational
forces.
Fman on earth
T
Take upwards as positive:
Fnet⊥
=
0
FN + (− Fg ⊥ ) =
0
FN
FA on B = − FB on A
T
Fg
15°
NB!
Newton’s Third Law describes action-reaction force pairs. These
are forces on different objects and can not be added or subtracted.
a ≠ 0 m ⋅ s −2
Fnet = m a
Fg//
fk
SCIENCE CLINIC 2019 ©
20 − (5)(9,8)sin 15∘
20 − 12,68
=
=
a
=
∴a
=
Fman on wall
5a
5a
1,46
7,32
5
m ⋅ s−2 // up
the slope
Effect of Newton’s Second Law on overloading:
According to Newton’s Second Law, the acceleration of an object
is directly proportional to the net force and inversely proportional
to the mass of the object. If a vehicle is overloaded, the stopping
distance will increase which can lead to serious accidents. When
brakes are applied, the force (friction) remains the same, but the
increase in mass causes a decrease in negative acceleration, increasing the time (and distance) it takes for the vehicle to stop.
14
Gravity and Normal force
are NOT force pairs.
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Fwall on man
Newton’s Third Law during an accident
According to Newton’s Third Law, the force that two objects exert
on each other is equal in magnitude but opposite in direction. If
two cars are in an accident, they will both exert the same amount
of force on each other irrespective of their masses.
Newton’s Laws of Motion
ALL EXAMPLES:
DIRECTION OF MOTION POSITIVE
Horizontal
Slopes
The vertical resultant = 0 N.
The perpendicular (⟂) resultant = 0 N.
The horizontal resultant determines acceleration.
The parallel (//) resultant determines acceleration.
FN
FN
θ
Ff
Ff
FA
Horizontal:
Vertical:
Fg//
Fnet = m a
FAx + (− Ff ) = m a
Fnet = 0
(− Fg ) + FN + FAy = 0
Fg
θ
Parallel:
FA
Fg ⊥
FN
Ff
FAy
Horizontal:
Vertical:
Fnet = m a
FAx + (− Ff ) = m a
Fnet = 0
Fg + (− FN ) + FAy = 0
FN
Horizontal resultant = 0 N.
Vertical resultant determines acceleration.
REMEMBER: No normal or friction forces.
θ
Fg
Ff
FA
Fg⟂
Perpendicular:
Fnet = m a
(− Fg ∥) + (− Ff ) + FA = m a
Fg ⊥
Fnet = 0
+ (− FN ) = 0
Fg
θ
Fnet = m a
Fg ∥ + (− Ff ) = m a
Perpendicular:
Fg⟂
Fg ⊥
Lift accelerating
Vertical:
Fnet = 0
Fg + (− FT ) = 0
Fnet = m a
Fg + (− FT ) = m a
Fg
Acceleration will be in the direction of the greatest force.
15
Fnet = 0
+ (− FN ) = 0
Lift in freefall (cable snap)
FT
FT
Vertical:
Fg
Parallel:
Ff
Fg//
Fg//
FT
FN
Ff
Lift stationary/constant velocity
Suspended
Fg//
Fg//
fk
FN
Parallel:
T
Fg
FA
FN
No force applied
FAx
Fg
Fg⟂
Fnet = 0
+ (− FN ) = 0
θ
Ff
Ff
Perpendicular:
Fnet = m a
Fg ∥ + FA + (− Ff ) = m a
Pushed at an angle
Force applied up the slope
FN
Ff
Fg//
Fg
FA
FN
FAx
Fg
FN
Fg ⊥ = Fg cos θ
Force applied down the slope
FAy
T
FA
Fg // = Fg sin θ
REMEMBER: Use components of weight.
Pulled at an angle
SCIENCE CLINIC 2019 ©
T
Grade 12 Science Essentials
FT
Vertical:
Fnet = m a
Fg = m a
Fg
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Fg
Fg
Fg
Newton’s Laws of Motion
Grade 12 Science Essentials
Connected objects (e.g Pulley Systems)
ALL EXAMPLES:
DIRECTION OF MOTION POSITIVE
SCIENCE CLINIC 2019 ©
Objects attached by rope/cable
FN
Do separate free body diagrams for each object.
Horizontal:
The velocity and acceleration of all objects are equal in
magnitude and direction.
Fnet = m a
(− Ff ) + FT = m a
Applied forces are applied to only one object at a time.
FN
FT
Ff
Simultaneous equations for acceleration and tension are
sometimes needed.
FT
Ff
FA
Fnet = m a
FA + (− FT ) + (− Ff ) = m a
FA
FT
Fg
REMEMBER:
Horizontal:
Ff
Ff
Fg
Ropes/cables- The tension forces on
the objects are the same in magnitude
but opposite in direction.
FN
FN
Fg
Fg
Objects in contact
FN
FN
Touching objects- Newton’s Third Law
Horizontal:
Same axis
Can be horizontal (multiple objects on a surface) or vertical
(multiple suspended objects).
Fnet = m a
FA + (− FBA) + (− Ff ) = m a
Ff
FA
FBA
FN
Horizontal (objects on a surface) AND vertical (suspended
objects).
The velocity and acceleration of all objects are equal in
magnitude NOT DIRECTION.
Ff
FBA FAB
A
Ff
Fg
The velocity and acceleration of all objects are equal in
magnitude and direction.
Multiple axes
FA
FN
FN
Horizontal:
B
Ff
Fg
Fnet = m a
(− Ff ) + FAB = m a
FAB
Ff
Fg
Fg
Multiple axes
In these examples,
clockwise is positive:
FT1
Fg
Fg
Vector direction on multiple axes
FN
FN
FT1
Right positive
Ff
Fg
Down positive
FT2
FT2
FT1
Ff
FT2
Fg
Fg
Clockwise:
Right and Down positive
Left and Up negative
OR
Anti-clockwise:
Left and Up positive
Right and Down negative
Fg
FN
FT1
Ff
FT2
FT1
FN
FN
FT1
Ff
Fg
Fg
Fg
Fg
FT2
Horizontal:
Vertical:
Fnet = m a
FT1 + (− Ff ) = m a
Fnet = m a
Fnet = m a
Fg + (− FT1) + FT 2 = m a Fg + (− FT 2 ) = m a
Vertical:
16
FT2
Ff
FT1
FT2
Fg
Fg
Horizontal:
Horizontal:
Vertical:
Fnet = m a
FT1 + (− Ff ) = m a
Fnet = m a
FT 2 + (− FT1) + (− Ff ) = m a
Fnet = m a
Fg + (− FT 2 ) = m a
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Newton’s Law of Universal Gravitation
Grade 12 Science Essentials
Every particle in the universe attracts every other particle with
a force which is directly proportional to the product of their
masses and inversely proportional to the square of the distance
between their centres.
Gm1m2
F=
r2
F =
G =
m=
r =
force of attraction between objects (N)
universal gravitational constant (6,67 ×10−11 N·m2·kg−2 )
object mass (kg)
distance between object centers (m)
A uniform sphere of matter attracts a body that is outside the shell as if
all the sphere’s mass was concentrated at its center.
1.
2.
3.
4.
5.
RATIOS
Write out the original formula.
Manipulate unknown as subject.
Substitute changes into formula (Keep symbols!).
Simplify ratio number.
Replace original formula with unknown symbol.
a) One mass is doubled and the distance between the masses is
halved?
Gm1m2
Write out the formula
F=
r2
G(2m1)m2
Substitute changes into formula
=
( 1 r)2
=
2 Gm1m2
1
r2
4
= 8(
Gm1m2
r2
Simplify ratio number
)
∴ Fnew = 8 F
rmoon
rman
NOTE:
The radius of the earth is added
to the distance between the
earth and the moon.
NOTE:
The radius of object
(man) on the earth is
negligibly small.
KNOW THE DIFFERENCE!
Replace original formula
b) Both the two masses as well as the distance are doubled?
Gm1m2
F=
Write out the formula
r2
G(2m1)(2m2 )
Substitute changes into formula
=
(2r)2
4 Gm1m2
Simplify ratio number
=
4
r2
Gm1m2
= 1(
r2 )
∴ Fnew = 1 F
g vs G
Replace original formula
DETERMINING GRAVITATIONAL ACCELERATION (g)
g: Gravitational acceleration (9,8 m·s−2 on earth)
g is the acceleration due to gravity on a specific planet.
G: Universal gravitational constant (6,67×10−11 N·m2·kg−2)
Proportionality constant which applies everywhere in the universe.
Mass vs Weight
Mass (kg)
A scalar quantity of matter which remains constant everywhere in the
universe.
Weight (N) [gravitation force]
Weight is the gravitational force the Earth exerts on any object. Weight
differs from planet to planet. Fg = mg. Weight is a vector quantity.
F = m object g
and
m og =
g=
∴g=
F=
Gm1m2
r2
REMEMBER:
Mass in kg
Radius in m
Radius: centre of mass to centre of mass.
Direction is ALWAYS attractive.
Both objects experience the same force.
(Newton’s Third Law of Motion)
EXAMPLE:
The earth with a radius of 6,38 x 103 km is 149,6 x 106 km
away from the sun with a radius of 696 342 km. If the earth
has a mass of 5,97 x 1024 kg and the sun has a mass of
1,99 x 1030 kg, determine the force between the two bodies.
r = 6,38 × 10 3 km + 149,6 × 106 km + 696 342 km
= 6,38 × 106 m + 149,6 × 109 m + 696 342 × 10 3 m
= 1,5 × 1011 m
F=
Gm1m2
r2
6,67 × 10−11(5,97 × 10 24 )(1,99 × 10 30 )
F=
(1,50 × 1011)2
F = 3,52 × 10 22 N attraction
The force of gravitational attraction is a vector, therefore all vector rules can be applied:
• Direction specific
• Can be added or subtracted
Gm object mPlanet
Gm o mP
r 2Planet
r 2P
Gm o mP
m o r 2P
GmP
r 2P
Therefore the gravitational acceleration of an object only depends
on the mass and radius of the planet. Object mass is irrelevant!
17
CALCULATIONS
The gravitational force can be calculated using F =
EXAMPLE:
Two objects, m1 and m2, are a distance r apart and experience a
force F. How would this force be affected if:
2
Thus, the distance is determined between the centers of the two bodies.
SCIENCE CLINIC 2019 ©
For more information about Science Clinic’s seminars, classes and resources, visit www.scienceclinic.co.za
Take right as positive:
Fnet on satallite =
∴ (
Fm on s +
= −(
Gm m m s
rms 2
)= (
Gm m m s
rms 2
Gm e m s
res 2
)
Fe on s
)+ (
Gm e m s
res 2
)
QUESTION 1
NEWTON’S LAWS
Two blocks of masses 20 kg and 5 kg respectively are connected by a light inextensible string,
P. A second light inextensible string, Q, attached to the 5 kg block, runs over a light frictionless
pulley. A constant horizontal force of 250 N pulls the second string as shown in the diagram
below. The magnitudes of the tensions in P and Q are T1 and T2 respectively. Ignore the
effects of air friction.
1.2
1.1
Calculate the magnitude of the tension T1 in string P.
Draw a labelled free-body diagram indicating ALL the forces acting on the 5 kg
block.
State Newton's second law of motion in words.
(6)
(3)
(2)
250 N

T2 Q
P
5 kg
T1
1.3
20 kg
1.4
State Newton's third law of motion in words.
(2)
(1)
[12]
When the 250 N force is replaced by a sharp pull on the string, one of the two strings
break. Which ONE of the two strings, P or Q, will break?
QUESTION 2
4 kg
A block of mass 1 kg is connected to another block of mass 4 kg by a light inextensible
string. The system is pulled up a rough plane inclined at 30 o to the horizontal, by means of
a constant 40 N force parallel to the plane as shown in the diagram below.
40 N
1 kg
30°
2.1
The magnitude of the kinetic frictional force between the surface and the 4 kg block
is 10 N. The coefficient of kinetic friction between the 1 kg block and the surface
is 0,29.
2.2
(5)
(3)
Draw a labelled free-body diagram showing ALL the forces acting on the
1 kg block as it moves up the incline.
Kinetic frictional force between the 1 kg block and the surface
(6)
[16]
Calculate the magnitude of the:
2.3.1
Tension in the string connecting the two blocks
2.3
2.3.2
QUESTION 3
A 5 kg block, resting on a rough horizontal table, is connected by a light
inextensible string passing over a light frictionless pulley to another block of mass
2 kg. The 2 kg block hangs vertically as shown in the diagram below.
5 kg
A force of 60 N is applied to the 5 kg block at an angle of 10o to the horizontal, causing the
block to accelerate to the left.
60 N
10°
2 kg
The coefficient of kinetic friction between the 5 kg block and the surface of the table
is 0,5. Ignore the effects of air friction.
(2)
(5)
Vertical component of the 60 N force
(2)
Draw a labelled free-body diagram showing ALL the forces acting on the 5 kg block.
3.2.1
Horizontal component of the 60 N force
3.1
3.2.2
(2)
Calculate the magnitude of the:
State Newton's Second Law of Motion in words.
3.2
3.3
3.4
Tension in the string connecting the two blocks
Normal force acting on the 5 kg block
(7)
[20]
(2)
Calculate the magnitude of the:
3.5
QUESTION 4
4.1
Two blocks of mass M kg and 2,5 kg respectively are connected by a light,
inextensible string. The string runs over a light, frictionless pulley, as shown in the
diagram below. The blocks are stationary.
M
kg
(2)
table
State Newton's THIRD law of motion in words.
(3)
2,5 kg
4.1.1
Calculate the tension in the string.
(5)
4.1.2
Calculate the minimum value of M that will prevent the blocks from
moving.
The coefficient of static friction (μs) between the unknown mass M and the surface of
the table is 0,2.
4.1.3
4.2
Calculate the magnitude of the acceleration of the 5 kg block.
The block of unknown mass M is now replaced with a block of mass 5 kg. The 2,5 kg
block now accelerates downwards. The coefficient of kinetic friction (µk) between the
5 kg block and the surface of the table is 0,15.
4.1.4
A small hypothetical planet X has a mass of 6,5 x 1020 kg and a radius of 550 km.
Calculate the gravitational force (weight) that planet X exerts on a 90 kg rock on this
planet's surface.
5 kg
(5)
5.1.1
Calculate the speed of the 20 kg mass as it strikes the ground.
Calculate the acceleration of the 20 kg mass.
(1)
(4)
(5)
(4)
[19]
5.1.2
At what minimum distance from the pulley should the 5 kg mass be
placed initially, so that the 20 kg mass just strikes the ground?
A person of mass 60 kg climbs to the top of a mountain which is 6 000 m above
ground level.
6 000 m
Ground
level
State Newton's Law of Universal Gravitation in words.
Calculate the difference in the weight of the climber at the top of the
mountain and at ground level.
(2)
5.1.3
When the stationary 5 kg mass is released, the two masses begin to move. The
coefficient of kinetic friction, μk, between the 5 kg mass and the horizontal surface is
0,4. Ignore the effects of air friction.
6m
20 kg
A 5 kg mass and a 20 kg mass are connected by a light inextensible string which
passes over a light frictionless pulley. Initially, the 5 kg mass is held stationary on a
horizontal surface, while the 20 kg mass hangs vertically downwards, 6 m above the
ground, as shown in the diagram below. The diagram is not drawn to scale.
QUESTION 5
5.1
5.2
5.2.1
5.2.2
(6)
[18]
QUESTION 6
The diagram below shows a 10 kg block lying on a flat, rough, horizontal surface of
a table. The block is connected by a light, inextensible string to a 2 kg block hanging over the
side of the table. The string runs over a light, frictionless pulley. The blocks are stationary.
(2)
10 kg
State Newton's FIRST law of motion in words.
(1)
table surface
6.1
Write down the magnitude of the NET force acting on the 10 kg block.
2 kg
6.2
6.3
Calculate the coefficient of kinetic friction between the 10 kg block and the
of the table.
Draw a free-body diagram for the 2 kg block when the 15 N force is applied
to it.
(7)
When a 15 N force is applied vertically downwards on the 2 kg block, the 10 kg block
accelerates to the right at 1,2 m∙s-2.
6.4
(1)
State Newton's Second Law of Motion in words.
(3)
(2)
(2)
[16]
(3)
6.5
How does the value, calculated in QUESTION 2.4, compare with the value of the
coefficient of STATIC friction for the 10 kg block and the table? Write down only
LARGER THAN, SMALLER THAN or EQUAL TO.
surface
6.6
If the 10 kg block had a larger surface area in contact with the surface of the table,
how would this affect the coefficient of kinetic friction calculated in QUESTION 2.4?
Assume that the rest of the system remains unchanged. Write down only
INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer.
QUESTION 7
1,5 kg
30°
A learner constructs a push toy using two blocks with masses 1,5 kg and 3 kg respectively.
The blocks are connected by a massless, inextensible cord. The learner then applies a force of
25 N at an angle of 30o to the 1,5 kg block by means of a light rigid rod, causing the toy to
move across a flat, rough, horizontal surface, as shown in the diagram below.
25 N
3 kg
7.1
Calculate the magnitude of the kinetic frictional force acting on the 3 kg block.
(5)
The coefficient of kinetic friction (µk) between the surface and each block is 0,15.
7.2
Draw a labelled free-body diagram showing ALL the forces acting on the 1,5 kg
block.
(3)
7.3
Kinetic frictional force acting on the 1,5 kg block
(5)
[18]
Calculate the magnitude of the:
7.4.1
Tension in the cord connecting the two blocks
7.4
7.4.2
QUESTION 8(Start on a new page.)
7o
2 kg
1,5 m⋅s-1
In the diagram below, a small object of mass 2 kg is sliding at a constant velocity of
1,5 m⋅s-1 down a rough plane inclined at 7o to the horizontal surface.
Horizontal surface
At the bottom of the plane, the object continues sliding onto the rough horizontal
surface and eventually comes to a stop.
Write down the magnitude of the net force acting on the object.
(3)
(1)
Magnitude of the frictional force acting on the object while it is
sliding down the inclined plane
(3)
(3)
(5)
[15]
Coefficient of kinetic friction between the object and the surfaces
Calculate the:
Draw a labelled free-body diagram for the object while it is on the inclined
plane.
The coefficient of kinetic friction between the object and the surface is the same for
both the inclined surface and the horizontal surface.
8.1
8.2
8.3
8.3.1
8.3.2
8.3.3
Distance the object travels on the horizontal surface before it
comes to a stop
2 kg
F
20o
Draw a labelled free-body diagram showing ALL the forces acting
on the crate.
9.1.2
Force F
Normal force acting on the crate
(3)
(4)
(3)
(4)
9.1.3
Acceleration of the crate
State Newton's Law of Universal Gravitation in words.
A massive rock from outer space is moving towards the Earth.
9.2.1
(2)
[18]
(2)
9.1.4
Calculate the magnitude of the:
A constant frictional force of 3 N acts between the surface and the crate.
The coefficient of kinetic friction between the crate and the surface is 0,2.
9.1.1
The force F is applied at an angle of 20o to the horizontal, as shown in the
diagram below.
A crate of mass 2 kg is being pulled to the right across a rough horizontal
surface by a constant force F.
QUESTION 9 (Start on a new page.)
9.1
9.2
9.2.2
How does the magnitude of the gravitational force exerted by the
Earth on the rock change as the distance between the rock and the
Earth becomes smaller?
Choose from INCREASES, DECREASES or REMAINS THE
SAME.
Give a reason for the answer.
P
(2)
F
State Newton's First Law in words.
(4)
8 kg
10.1.1
Draw a labelled free-body diagram for block P.
Calculate the magnitude of force F.
The kinetic frictional force between the block and the surface of the inclined
plane is 20,37 N.
10.1.3
Calculate the magnitude of the acceleration of the block.
Force F is now removed and the block ACCELERATES down the plane.
The kinetic frictional force remains 20,37 N.
10.1.4
State Newton's Law of Universal Gravitation in words.
(2)
(4)
10.2.1
Calculate the mass of the planet if its radius is 700 km.
A 200 kg rock lies on the surface of a planet. The acceleration due to gravity
on the surface of the planet is 6,0 m·s-2.
(5)
10.1.2
30o
Force F is parallel to the inclined plane, as shown in the diagram below.
An 8 kg block, P, is being pulled by constant force F up a rough inclined plane
at an angle of 30o to the horizontal, at CONSTANT SPEED.
QUESTION 10 (Start on a new page.)
10.1
10.2
10.2.2
(4)
[21]
QUESTION 12 (Start on a new page.)
F
2 m∙s-1
Two boxes, P and Q, resting on a rough horizontal surface, are connected by a light
inextensible string. The boxes have masses 5 kg and 2 kg respectively. A constant
force F, acting at an angle of 30o to the horizontal, is applied to the 5 kg box, as shown
below.
P
5 kg
30o
The two boxes now move to the right at a constant speed of 2 m∙s-1.
Q
string
(2)
2 kg
State Newton's First Law of Motion in words.
(4)
Rough surface
12.1
Draw a labelled free-body diagram for box Q.
Box P experiences a constant frictional force of 5 N and box Q a constant
frictional force of 3 N.
Calculate the magnitude of force F.
The string connecting P and Q suddenly breaks after 3 s while force F is still
being applied.
Z
3
X
Y
7
Time (s)
Learners draw the velocity-time graph for the motion of P and Q before and
after the string breaks, as shown below.
2
2
Write down the time at which the string breaks.
Which portion (X, Y or Z) of the graph represents the motion of box Q, after
the string breaks? Use the information in the graph to fully support
the answer.
(4)
[17]
(1)
(6)
12.2
12.3
12.4
12.5
Velocity (m∙s-1)
QUESTION 13 (Start on a new page.)
•
3 kg
3 kg
ground
0,5 m
Block P, of unknown mass, is placed on a rough horizontal surface. It is connected to
a second block of mass 3 kg, by a light inextensible string passing over a light,
frictionless pulley, as shown below.
P
Define the term acceleration in words.
Initially the system of masses is held stationary with the 3 kg block, 0,5 m above the
ground. When the system is released the 3 kg block moves vertically downwards and
strikes the ground after 3 s. Ignore the effects of air resistance.
13.1
Acceleration of the 3 kg block using equations of motion
(3)
(3)
(2)
13.2
Tension in the string
Calculate the magnitude of the:
13.3
13.4
Calculate the mass of block P.
Draw a labelled free-body diagram for block P.
(4)
The magnitude of the kinetic frictional force experienced by block P is 27 N.
13.5
(3)
[15]
QUESTION 14 (Start on a new page.)
8 kg
C
15°
A block, of mass 8 kg, is placed on a rough horizontal surface. The 8 kg block, which is
connected to a 2 kg block by means of a light inextensible string passing over a light
frictionless pulley, starts sliding from point A, as shown below.
B
(2)
A
State Newton's Second Law in words.
(4)
2 kg
14.1
Draw a labelled free-body diagram for the 8 kg block.
14.3.2
14.3.1
Use the 2 kg mass to calculate the tension in the string.
Give a reason why the system is NOT in equilibrium.
(4)
(3)
(1)
14.3.3
As the 8 kg block moves from B to C, the kinetic frictional force between the
8 kg block and the horizontal surface is not constant.
(2)
[17]
(1)
Calculate the kinetic frictional force between the 8 kg block and the
horizontal surface.
When the 8 kg block reaches point B, the angle between the string and the
horizontal is 15° and the acceleration of the system is 1,32 m·s-2.
14.2
14.3
14.4
Give a reason for this statement.
The horizontal surface on which the 8 kg block is moving, is replaced by another
horizontal surface made from a different material.
14.5
Will the kinetic frictional force, calculated in QUESTION 14.3.3 above,
change? Choose from: YES or NO. Give a reason for the answer.
Momentum and Impulse
(This section must be read in conjunction with the CAPS, p. 99–101.)
Momentum
•
Define momentum as the product of an object's mass and its velocity.
•
Describe linear momentum as a vector quantity with the same direction as the velocity
of the object.
•
Calculate the momentum of a moving object using p = mv.
•
Describe the vector nature of momentum and illustrate it with some simple examples.
•
Draw vector diagrams to illustrate the relationship between the initial momentum, the
final momentum and the change in momentum for each of the cases above.
Newton's second law in terms of momentum
•
State Newton's second law of motion in terms of momentum: The net (or resultant)
force acting on an object is equal to the rate of change of momentum of the object in
the direction of the net force.
∆p
•
Express Newton's second law in symbols: F net =
∆t
•
Calculate the change in momentum when a resultant force acts on an object and its
velocity:
o
Increases in the direction of motion, e.g. 2nd stage rocket engine fires
o
Decreases, e.g. brakes are applied
o
Reverses its direction of motion, e.g. a soccer ball kicked back in the direction it
came from
Impulse
•
Define impulse as the product of the net force acting on an object and the time the net
force acts on the object.
•
Deduce the impulse-momentum theorem: F net Δt = mΔv
•
Use the impulse-momentum theorem to calculate the force exerted, the time for which
the force is applied and the change in momentum for a variety of situations involving
the motion of an object in one dimension.
Explain how the concept of impulse applies to safety considerations in everyday life,
e.g. airbags, seatbelts and arrestor beds.
•
Conservation of momentum and elastic and inelastic collisions
•
Explain what is meant by:
o
An isolated system (in Physics): An isolated system is one on which the net
external force acting on the system is zero.
o
Internal and external forces
•
State the principle of conservation of linear momentum: The total linear momentum of
an isolated system remains constant (is conserved).
•
Apply the conservation of momentum to the collision of two objects moving in one
dimension (along a straight line) with the aid of an appropriate sign convention.
Distinguish between elastic collisions and inelastic collisions by calculation.
•
Δp
Δt
A collision during which kinetic energy is conserved.
A collision during which kinetic energy is not conserved.
A system in which the net external force acting on the system is zero.
The product of the resultant/net force acting on an object and the time the
resultant/net force acts on the object.
In symbols: Impulse = FnetΔt
FnetΔt = mΔv = m(vf – vi)
The TOTAL linear momentum in an isolated system remains constant (is
conserved).
In symbols: Fnet =
MECHANICS: MOMENTUM AND IMPULSE
Contact forces
Contact forces arise from the physical contact between two objects (E.g. a
soccer player kicking a ball.)
Non-contact forces
Non-contact forces arise even if two objects do not touch each other (E.g. the
force of attraction of the earth on a parachutist even when the earth is not in
direct contact with the parachutist.)
Linear momentum is the product of an object’s mass and its velocity.
The net (or resultant) force acting on an object is equal to the rate of change
of momentum of the object in the direction of the net force.
Momentum
Newton’s Second
Law of motion in
terms of momentum
Principle of
conservation of
linear momentum
Isolated system
Impulse
Impulse-momentum
theorem
Elastic collision
Inelastic collision
Momentum and Impulse
Grade 12 Science Essentials
MOMENTUM
Momentum can be thought of as quantifying the
motion of an object. The following equation is used
to calculate momentum:
p = momentum (kg ⋅ m ⋅ s−1)
p = mv
NEWTON’S SECOND LAW OF MOTION
VECTOR NATURE OF MOMENTUM
Momentum: the product of the mass of an
object and its velocity.
m = mass (kg)
−1
v = velocity (m ⋅ s )
Momentum is a vector quantity and has both mag- Newton’s second law in terms of momentum: The resultant/net
nitude and direction. It is therefore important to force acting on an object is equal to the rate of change of moalways include direction in all momentum calcu- mentum of the object in the direction of the resultant/net force.
lations.
According to Newton’s Second Law, a resultant force applied to an object
will cause the object to accelerate. When the net force on an object
EXAMPLE:
changes, so does its velocity and hence the momentum.
A golf ball of mass 0,05 kg leaves a golf club
at a velocity of 90 m·s 1 in an easterly direction. Calculate the momentum of the golf
ball.
Fnet = resultant force (N)
Fnet Δt = Δp
p
=
=
mv
(0,05)(90)
Impulse = FΔt
Impulse = Δp
mΔv = Δp
Δp = change in momentum (kg ⋅ m ⋅ s−1)
pf = final momentum (kg ⋅ m ⋅ s−1)
pi = initial momentum (kg ⋅ m ⋅ s )
Due to the vector nature of momentum, it is very important to choose a positive direction.
Choosing east as positive:
Δp
=
pf − pi
Δp
=
pf − pi
Δp
=
mvf − mvi
Δp
=
mvf − mvi
Δp
Δp
=
=
(1 000)(0) − (1 000)(16)
− 16 000
Δp
Δp
=
=
(0,2)(− 50) − (0,2)(40)
− 18
∴ Δp = 16 000 kg ⋅ m ⋅ s−1 west
Δp is measured in kg·m·s−1
EXAMPLE:
EXAMPLE:
A golf ball with a mass of 0,1 kg is driven from the
tee. The golf ball experiences a force of 1000 N while
in contact with the golf club and moves away from the
golf club at 30 m·s 1. For how long was the golf club
in contact with the ball?
Fnet Δt =
m Δv
1000t = (0,1)(30 − 0)
The following graph shows the force exerted on a hockey ball over time. The
hockey ball is initially stationary and has a
mass of 150 g.
t
Choosing east as positive:
Impulse, FΔt , is measured in N·s.
The change in momentum is directly dependent on the magnitude of the resultant force and the duration for which the force is applied. Impulse is a vector, ∴ direction specific.
−1
A cricket ball with a mass of 0,2 kg approaches a
cricket bat at a velocity of 40 m·s 1 east and
leaves the cricket bat at a velocity of 50 m·s 1
west. Calculate the change in the ball’s momentum during its contact with the cricket bat.
Fnet Δt = Δp
By rearranging Newton’s second law in terms of momentum, we find that impulse is equal to the change
in momentum of an object according to the impulse-momentum theorem:
When a net force acts on an object, it results in a change in velocity for the object and therefore a
change in momentum (p) for the object. The change in momentum can be calculated by using:
A 1000 kg car initially moving at a constant
velocity of 16 m·s 1 in an easterly direction
approaches a stop street, starts breaking and
comes to a complete standstill. Calculate the
change in the car’s momentum.
Fnet = m a
Δv
Fnet = m
Δt
mvf − mvi
Fnet =
Δt
Δp
Fnet =
Δt
Impulse: the product of the net force acting on an object and the time the over which the
net force acts on the object.
CHANGE IN MOMENTUM
EXAMPLE:
Derivation from Newton’s
Second Law
IMPULSE
4,5 kg ⋅ m ⋅ s−1 east
EXAMPLE:
Δp = change in momentum (kg ⋅ m ⋅ s−1)
Δt = time (s)
Choosing east as positive:
=
Δp = pf − pi
SCIENCE CLINIC 2019 ©
=
3 × 10−3 s
EXAMPLE:
∴ Δp = 18 kg ⋅ m ⋅ s−1 west
18
Why can airbags be useful during a collision? State
your answer by using the relevant scientific principle.
Calculate the magnitude of the impulse
(change in momentum) of the hockey ball.
The change in momentum remains constant, but the
use of an airbag prolongs the time (t) of impact during the accident. The resultant force experienced is
inversely proportional to the contact time (F ∝ 1/t),
therefore resulting in a smaller resultant force (Fnet)
(Δp is constant).
Fnet Δt
=
impulse
=
For more information about Science Clinic’s seminars, classes and resources, visit www.scienceclinic.co.za
impulse
=
impulse
=
area under graph
1
b
2
⊥ h
1
(0,5)(150)
2
37,5 N ⋅ s
Conservation of Momentum
Grade 12 Science Essentials
CONSERVATION OF MOMENTUM
SCIENCE CLINIC 2019 ©
NEWTON’S THIRD LAW AND MOMENTUM
Conservation of linear momentum: The total linear momentum of an isolated system re- During a collision, the objects involved will exert forces on each other. Therefore, according to Newton’s
mains constant .
third law, if object A exerts a force on object B, object B will exert a force on object A where the two
forces are equal in magnitude, but opposite in direction.
Σpbefore = Σpafter
pA(before) + pB(before) = pA(after) + pB(after)
mA viA + mBviB + . . . = mA vfA + mBvfB + . . .
The magnitude of the force, the contact time and therefore the impulse on both objects are
equal in magnitude.
Forces are applied between objects during:
System: A set number of objects and their interactions with each other.
External forces: Forces outside of the system.
Isolated system: A system on which the net external force is zero.
Collisions: Move off together, collide and deflect, object dropped vertically on moving object.
Explosions: Explosions, springs, firearms
Collisions
Explosions
Move off together
When objects collide and move off
together, their masses can be
added as one object
Explosions
vA
vB = 0
mA
Objects that are stationary (B)
have an initial velocity of zero.
vA+B
mB
Objects that experience the same
explosion will experience the same
force.
mA+B
Collision
The acceleration, velocity and momentum of the object is dependent
on the mass.
Σp before = Σpafter
mA viA + mBviB = (mA + mB)vf
Objects can collide and move off
separately
REMEMBER: The velocity and
momentum are vectors (i.e. direction specific). Velocity substitution
must take direction into account.
Linear momentum= momentum
along one axis.
A dropped object has a horizontal
velocity of zero,
∴viB= 0m·s
1
Explosion
vB
mA
mB
Σp before = Σpafter
(mA + mB)vi = mA vfA + mBvfB
Springs
vA
vB
mA
mB
vA
Collision
vB
mA
The spring will exert the same
force on both objects (Newton’s
Third Law).
mB
The acceleration, velocity and momentum of the object is dependent
on the mass.
Σp before = Σpafter
mA viA + mBviB = mA vfA + mBvfB
Object dropped vertically on a moving object
Example: A stuntman jumps off a
bridge and lands on a truck.
mA+B
Objects that are stationary (A+B)
have an initial velocity of zero.
Collide and rebounds
vA
vA+B = 0
vA
mA
vA+B
MB
vB = 0
Collision
vB = 0
vB = 0
mA
mB
vA
Push
vB
mA
mB
Σp before = Σpafter
(mA + mB)vi = mA vfA + mBvfB
Objects that are stationary (A+B)
have a velocity of zero.
Firearms/ cannons
The gun and bullet will experience
the same force.
The acceleration of the weapon is
significantly less than the bullet
due to mass difference
mA+B
Σp before = Σpafter
mA viA + mBviB = (mA + mB)vf
Recoil can be reduced by increasing the mass of the weapon.
19
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vG+B = 0
vG
vB
mG
mB
Shoot
mG+B
Σp before = Σpafter
(m G + mB)vi = m G vfG + mBvfB
Grade 12 Science Essentials
ELASTIC VS INELASTIC COLLISIONS
Momentum and Energy
SCIENCE CLINIC 2019 ©
PENDULUMS
Elastic collision: a collision in which both momentum and kinetic
energy are conserved.
Inelastic collision: a collision in which only momentum is conserved.
In an isolated system, momentum will always be conserved. To prove
that a collision is elastic, we only have to prove that kinetic energy is
conserved.
Kinetic energy can be calculated using the mass and velocity of an
object:
EM(top) = EM(bottom)
mg h +
1
2
mv 2 = mg h +
1
2
mv 2
EK = kinetic energy (J)
1
mv 2
2
EK =
DOWNWARD SWING:
Conservation of mechanical energy (EM) to determine velocity at
the bottom of the swing:
h
COLLISION:
m = mass (kg)
Conservation of linear momentum to determine the velocity of the block after impact.
v = velocity (m ⋅ s−1)
Σp before = Σpafter
pA(before) + pA(before) = pA(after) + pA(after)
mA viA + mBviB + . . . = mA vfA + mBvfB + . . .
Elastic collision: ΣEk(before) = ΣEk(after)
Inelastic collision: ΣEk(before) ≠ ΣEk(after)
(some energy is lost as sound or heat)
EXAMPLE:
The velocity of a moving trolley of mass 1 kg is 3 m·s 1. A block of
mass 0,5 kg is dropped vertically on to the trolley. Immediately
after the collision the speed of the trolley and block is 2 m·s 1 in
the original direction. Is the collision elastic or inelastic? Prove your
answer with a suitable calculation.
ΣEk(before)
=
=
ΣEk(after)
1
2
=
=
=
=
mt vt2 +
1
(1)(3)2
2
+
UPWARD SWING:
Conservation of mechanical energy (EM) to determine height that
the pendulum will reach:
1
m v2
2 b b
1
(0,5)(0)2
2
EM(bottom) = EM(top)
mg h +
4,5 J
1
2
1
(1
2
mt+ b vt+2 b
COLLISION:
Conservation of linear momentum to determine the velocity of the pendulum after impact.
+ 0,5)(2)2
3J
Σp before = Σpafter
pA(before) + pA(before) = pA(after) + pA(after)
mA viA + mBviB + . . . = mA vfA + mBvfB + . . .
ΣEk(before) ≠ ΣEk(after)
∴ Kinetic energy is not conserved and the collision is inelastic
20
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1
2
mv 2 = mg h +
1
2
mv 2
QUESTION 1
MOMENTUM AND IMPULSE
1.1
Define the term impulse of a force.
Calculate the momentum with which the dancer reaches the ground.
(3)
(2)
(3)
Dancers have to learn many skills, including how to land correctly. A dancer of mass 50 kg
leaps into the air and lands feet first on the ground. She lands on the ground with a velocity of
5 m∙s-1. As she lands, she bends her knees and comes to a complete stop in 0,2 seconds.
1.2
Calculate the magnitude of the net force acting on the dancer as she lands.
Will the force now be GREATER THAN, SMALLER THAN or EQUAL TO the force
calculated in QUESTION 1.3?
2.1
Define the concept momentum in words.
(4)
(2)
(3)
[12]
(1)
1.3
1.4
Give a reason for the answer to QUESTION 1.4.
Assume that the dancer performs the same jump as before but lands without bending her
knees.
1.5
QUESTION 2
2.2
Calculate the change in momentum of Percy and the bike, from the moment the
brakes lock until the bike comes to a stop.
Percy, mass 75 kg, rides at 20 m∙s-1 on a quad bike (motorcycle with four wheels) with a mass
of 100 kg. He suddenly applies the brakes when he approaches a red traffic light on a wet and
slippery road. The wheels of the quad bike lock and the bike slides forward in a straight line.
The force of friction causes the bike to stop in 8 s.
2.3
0,2 m
A
B
Diagram 2
(2)
(4)
[10]
Calculate the average frictional force exerted by the road on the wheels to stop the
bike.
QUESTION 3
B
Two stationary steel balls, A and B, are suspended next to each other by massless, inelastic
strings as shown in Diagram 1 below.
Diagram 1
A
What is meant by an elastic collision?
Ball A of mass 0,2 kg is displaced through a vertical distance of 0,2 m, as shown in Diagram 2
above. When ball A is released, it collides elastically and head-on with
ball B. Ignore the effects of air friction.
3.1
Immediately after the collision, ball A moves horizontally backwards (to the left). Ball B
acquires kinetic energy of 0,12 J and moves horizontally forward (to the right).
3.2
Speed of ball A immediately after the collision
Kinetic energy of ball A just before it collides with ball B (Use energy principles only.)
(5)
[14]
(4)
(3)
Calculate the:
3.3
Magnitude of the impulse on ball A during the collision
(1)
[10]
(5)
(4)
3.4
QUESTION 4
Calculate the speed at which the bullet leaves the rifle.
A bullet of mass 20 g is fired from a stationary rifle of mass 3 kg. Assume that the bullet moves
horizontally. Immediately after firing, the rifle recoils (moves back) with a velocity of 1,4 m∙s -1.
4.1
after
The bullet strikes a stationary 5 kg wooden block fixed to a flat, horizontal table. The bullet is
brought to rest after travelling a distance of 0,4 m into the block. Refer to the diagram below.
before
Calculate the magnitude of the average force exerted by the block on the bullet.
5 kg
4.2
20 g
4.3
How does the magnitude of the force calculated in QUESTION 3.2 compare to the
magnitude of the force exerted by the bullet on the block? Write down only LARGER
THAN, SMALLER THAN or THE SAME.
QUESTION 5
600 g
Q
The diagram below shows two trolleys, P and Q, held together by means of a compressed
spring on a flat, frictionless horizontal track. The masses of P and Q are 400 g and 600 g
respectively.
P
400 g
When the trolleys are released, it takes 0,3 s for the spring to unwind to its natural length.
Trolley Q then moves to the right at 4 m∙s-1.
(4)
(2)
Velocity of trolley P after the trolleys are released
(4)
State the principle of conservation of linear momentum in words.
5.2.1
Magnitude of the average force exerted by the spring on trolley Q
5.1
5.2.2
(1)
[11]
Calculate the:
Is this an elastic collision? Only answer YES or NO.
5.2
5.3
QUESTION 6
Z
The diagram below shows two sections, XY and YZ, of a horizontal, flat surface.
Section XY is smooth, while section YZ is rough.
A 5 kg block, moving with a velocity of 4 m∙s-1 to the right, collides head-on with a stationary 3
kg block. After the collision, the two blocks stick together and move to the right, past point Y.
The combined blocks travel for 0,3 s from point Y before coming to a stop at point Z.
0 m∙s-1
4 m∙s-1
3 kg
Y
X
State the principle of conservation of linear momentum in words.
5 kg
6.1
Velocity of the combined blocks at point Y
Calculate the magnitude of the:
6.2.1
Net force acting on the combined blocks when they move through section
YZ
6.2
6.2.2
QUESTION 7
40 000 ─
30 000 ─
20 000 ─
14 000 ─
10 000─
20,2
W
N
S
20,3
MOMENTUM VERSUS TIME GRAPH FOR CAR A
20,1
Time (s)
What do you understand by the term isolated system as used in physics?
0
20
E
The graph below shows how the momentum of car A changes with time just before and just
after a head-on collision with car B.
Car A has a mass of 1 500 kg, while the mass of car B is 900 kg.
Car B was travelling at a constant velocity of 15 m∙s-1 west before the collision.
Take east as positive and consider the system as isolated.
7.2
(2)
(4)
(4)
[10]
7.2.2
7.2.1
Magnitude of the net average force acting on car A during the collision
Velocity of car B just after the collision
Magnitude of the velocity of car A just before the collision
(5)
(3)
(1)
7.2.3
(4)
[13]
Calculate the:
Use the information in the graph to answer the following questions.
7.1
Momentum (kg∙m·s-1)
gun
mechanical
support
700 m∙s-1
bullets
W
N
(2)
(3)
E
Magnitude of the momentum of each bullet when it leaves the gun
(5)
S
8.2.1
The net average force that each bullet exerts on the gun
(2)
[12]
8.2.2
Calculate the:
The gun fires 220 bullets per minute. The mass of each bullet is 0,03 kg.
ground
Each bullet travels at a speed of 700 m∙s-1 in an easterly direction when it
leaves the gun.
(Take the initial velocity of a bullet, before being fired, as zero.)
The diagram below shows a gun mounted on a mechanical support which is
fixed to the ground. The gun is capable of firing bullets rapidly in a horizontal
direction.
Define the term impulse in words.
QUESTION 8 (Start on a new page.)
8.1
8.2
8.3
Without any further calculation, write down the net average horizontal force
that the mechanical support exerts on the gun.
QUESTION 9 (Start on a new page.)
6,0 kg
B
A teacher demonstrates the principle of conservation of linear momentum using two
trolleys. The teacher first places the trolleys, A and B, some distance apart on a flat
frictionless horizontal surface, as shown in the diagram below. The mass of trolley A is
3,5 kg and that of trolley B is 6,0 kg.
vi
A
3,5 kg
Trolley A moves towards trolley B at constant velocity. The table below shows the
position of trolley A for time intervals of 0,4 s before it collides with trolley B.
9.1
State the principle of conservation of linear momentum in words.
Use the table above to prove that trolley A is moving at constant velocity
before it collides with trolley B.
(2)
(3)
RELATIONSHIP BETWEEN POSITION AND TIME FOR TROLLEY A
Position of trolley A (m)
0
0,2
0,4
0,6
Time (s)
0
0,4
0,8
1,2
9.2
Calculate the magnitude of the average net force exerted on trolley B by
trolley A.
At time t = 1,2 s, trolley A collides with stationary trolley B. The collision time is 0,5 s
after which the two trolleys move off together.
9.3
(6)
[11]
QUESTION 10 (Start on a new page.)
A 2 kg block is at rest on a smooth, frictionless, horizontal table. The length of the
block is x.
AFTER
0,7 m∙s-1
x
W
N
S
Calculate the magnitude of the velocity of the bullet immediately after it
emerges from the block.
E
A bullet of mass 0,015 kg, travelling east at 400 m∙s-1, strikes the block and passes
straight through it with constant acceleration. Refer to the diagram below. Ignore any
loss of mass of the bullet and the block.
BEFORE
2 kg
x
State the principle of conservation of linear momentum in words.
400 m∙s-1
bullet
10.1
10.2
If the bullet takes 0,002 s to travel through the block, calculate the length, x, of
the block.
The block moves eastwards at 0,7 m∙s-1 after the bullet has emerged from it.
10.3
(2)
(4)
(5)
[11]
QUESTION 11 (Start on a new page.)
cat
State the principle of conservation of linear momentum in words.
B
surface
The diagram below shows two skateboards, A and B, initially at rest, with a cat
standing on skateboard A. The skateboards are in a straight line, one in front of the
other and a short distance apart. The surface is flat, frictionless and horizontal.
A
11.1
3 m∙s-1
EACH skateboard has a mass of 3,5 kg. The cat, of mass 2,6 kg, jumps
from skateboard A with a horizontal velocity of 3 m∙s-1 and lands on skateboard B with
the same velocity of 3 m∙s-1.
cat
B
surface
Calculate the magnitude of the impulse on skateboard B as a result of the
cat's landing.
Immediately after the cat has landed, the cat and skateboard B move
horizontally to the right at 1,28 m∙s-1.
Calculate the velocity of skateboard A just after the cat has jumped from it.
Refer to the diagram below.
A
11.2
11.3
(2)
(5)
(3)
[10]
QUESTION 12 (Start on a new page.)
B
2 kg
A trolley of mass 1,5 kg is held stationary at point A at the top of a frictionless track.
When the 1,5 kg trolley is released, it moves down the track. It passes point P at the
bottom of the incline and collides with a stationary 2 kg trolley at point B. Refer to the
diagram below. Ignore air resistance and rotational effects.
A
1,5 kg
P
1,5 kg
Use the principle of conservation of mechanical energy to calculate the speed
of the 1,5 kg trolley at point P.
2,0 m
12.1
When the two trolleys collide, they stick together and continue moving with constant
velocity.
The principle of conservation of linear momentum is given by the incomplete
statement below.
Rewrite the complete statement and fill in the missing words or phrases.
(4)
(2)
(4)
Calculate the speed of the combined trolleys immediately after the collision.
12.2
12.3
Calculate the distance travelled by the combined trolleys in 3 s after the
collision.
In a/an … system, the … linear momentum is conserved.
12.4
(3)
[13]
QUESTION 13 (Start on a new page.)
Define the term momentum in words.
(2)
(2)
4 m∙s-1
Initially a girl on roller skates is at rest on a smooth horizontal pavement. The girl
throws a parcel, of mass 8 kg, horizontally to the right at a speed of 4 ms-1.
Immediately after the parcel has been thrown, the girl-roller-skate combination moves
at a speed of 0,6 ms-1. Ignore the effects of friction and rotation.
13.1
13.2
Will the girl-roller-skate combination move TO THE RIGHT or TO THE LEFT
after the parcel is thrown?
NAME the law in physics that can be used to explain your choice of direction.
Calculate the mass of the girl.
(3)
(5)
Without any further calculation, write down the change in momentum
experienced by the parcel while it is being thrown.
Calculate the magnitude of the impulse that the girl-roller-skate combination is
experiencing while the parcel is being thrown.
The total mass of the roller skates is 2 kg.
13.3
13.4
13.5
(2)
[14]
Vertical Projectile Motion in One Dimension (1D)
(This section must be read in conjunction with the CAPS, p. 102–103.)
•
•
•
•
•
Change in position. Symbol: ∆x (horizontal displacement) or ∆y (vertical
displacement). Unit: meters (m).
The type of motion in which the only significant vertical force acting on the
body is the body's weight.
A force of attraction of one body on another due to their masses.
Where an object is relative to a reference point. Symbol: x (horizontal
position) or y (vertical position). Unit: meters (m).
An object in free fall.
The rate of change of position. Symbol v. Unit: meters per second (m∙s-1).
MECHANICS: VERTICAL PROJECTILE MOTION
One-dimensional motion. Linear motion. Motion in one line.
The rate of change of velocity. Symbol a. Unit: meters per second squared
(m∙s-2).
The acceleration of a body due to the force of attarction of the earth.
Explain what is meant by a projectile, i.e. an object upon which the only force acting is
the force of gravity.
Use equations of motion to determine the position, velocity and displacement of a
projectile at any given time.
Sketch position versus time (x vs. t), velocity versus time (v vs. t) and acceleration
versus time (a vs. t) graphs for:
o
A free-falling object
o
An object thrown vertically upwards
o
An object thrown vertically downwards
o
Bouncing objects (restricted to balls)
For a given x vs. t, v vs. t or a vs. t graph, determine:
o
Position
o
Displacement
o
Velocity or acceleration at any time t
For a given x vs. t, v vs. t or a vs. t graph, describe the motion of the object:
o
Bouncing
o
Thrown vertically upwards
o
Thrown vertically downward
1-D motion
Acceleration
Acceleration due to
gravity (g)
Displacement
Free fall
Gravitational force
Position
Projectile
Velocity
Vertical Projectile Motion
Grade 12 Science Essentials
PARTS OF PROJECTILE PATH
Only vertical movement (up and down) is considered, no horiA projectile is an object that moves freely under the influ- zontal movement is taken into account. The path of projectile
ence of gravity only. It is not controlled by any mechanism motion can be analysed using the 4 sections as shown below.
(pulley or motor). The object is in free fall, but may move The combination of these 4 parts will depend on the actual
upwards (thrown up) or downwards.
path travelled by the projectile. Example: Dropped projectile is
sections C and D only. Object thrown upwards and falls on roof
Forces on a projectile
is sections A to C.
In the absence of friction, the gravitational force of the earth
is the only force acting on a free falling body. This force alTime:
ways acts downwards.
tA = tD
Because the gravitational force is always downward, a projectB = tC
tile that is moving upward, must slow down. When a projec-
B
C
SAME
HEIGHT
Acceleration due to gravity
All free falling bodies near the surface of the earth have the
same acceleration due to gravity. This acceleration is 9,8
m·s−2 downward.
Ignoring air resistance/friction; If a marble and a rock are
released from the same height at the same time, they will
strike the ground simultaneously, and their final velocity will
be the same.
Their momentum (mv) and kinetic energy (½mv2) are
not the same, due to a difference in mass.
If two objects are released from different heights, they have
the same acceleration, but they strike the ground at different
times and have different velocities.
vf = vi + aΔt
SAME
HEIGHT
x=
−b ±
Graph manipulation:
Δy vs Δt
2(4,9)
OR
3,54*
*time for downward
direction
v vs Δt
Change in reference position:
Shift x-axis (Δy-Δt only)
a vs Δt
Gradient
Ne
ga
%
v (m·s−1)
Δy (m)
a = acceleration (m·s−2)
(9,8 m·s−2 downwards)
Posi%ve
Δy
ve
a
Ne Nega%ve
ga Δy
%v
ea
Δt (s)
Nega%ve Δv
Δt (s)
Δt (s)
Δy (m)
v
v (m·s−1)
Posi%ve
a
%ve
Nega
REMEMBER:
1. Draw a sketch diagram
2. Write down given variables
3. Choose positive direction
4. Solve
b − 4a c
2a
(− 19,6) 2 − 4(4,9)(8)
Area
Change in positive direction:
Flip graph along x-axis
Posi%
ve v
vf = final velocity (m·s−1)
− (− 19,6) ±
GRAPHS OF PROJECTILE MOTION
Posi%
ve v
vi = initial velocity (m·s−1)
2a
(Eg. A ball is thrown into the air and is caught at the same height.)
UP POSITIVE
Δt = time (s)
b 2 − 4ac
−b ±
t = 0,46 s
∴ t = 0,46 s
2
v
2
)Δt
Δy = displacement (m)
t =
t =
%ve
Nega
1
Δy = viΔt + aΔt 2
2
Δy = (
0 = vi + (9,8)(2)
vi = − 19,6
∴ vi = 19,6 m ⋅ s −1 u p
viA = vfD
vfA = viD
viB = vfC
VfB = ViC = 0 m·s−1
1
(9,8)t 2
2
0 = 4,9t 2 − 19,6t + 8
Velocity:
D
at 2
− 8 = − 19,6t +
vf = vi + aΔt
Nega%ve
a
+ 2aΔy
vi + vf
ΔyA = ΔyD
ΔyB = ΔyC
DOWN POSITIVE
=
vi2
A
Displacement:
1
2
Δy = viΔt +
total time = 4 s
∴ t up = 2 s
a (m·s−2)
tile is moving downward, it moves in the direction of the
gravitational force, therefore it will speed up.
EXAMPLE:
An object is projected vertically upwards. 4 seconds later, it is caught at the same
height (point of release) on its way downwards. Determine how long it took the
ball to pass a height of 8 m in the upward direction. Choose downward as positive direction.
a
ve
si% Posi%ve
o
P
Δy
Nega%ve
Δy
Δt (s)
8
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Δt (s)
a (m·s−2)
PROJECTILE MOTION
vf2
SCIENCE CLINIC 2019 ©
Posi%ve Δv
Δt (s)
Path of a Projectile
OBJECT DROPPED FROM HEIGHT (C+D)
OBJECT THROWN DOWN FROM HEIGHT (D)
vi = 0 m·s−1
Δy (m)
Δy (m)
Δt (s)
Δt (s)
Δt (s)
vf(down)
Δt (s)
a (m·s−2)
Also applies to objects
dropped from a downward
moving reference.
a (m·s−2)
vf
Δt (s)
Δt (s)
v (m·s−1)
v (m·s−1)
v (m·s−1)
vi(down) = 0 m·s−1
vi(up) ≠ 0m·s−1
Δt (s)
a (m·s−2)
SCIENCE CLINIC 2019 ©
OBJECT THROWN UP FROM HEIGHT (B+C+D)
vf(up) = 0 m·s−1
vi ≠ 0 m·s−1
Δt (s)
vf
ALL EXAMPLES:
UP POSITIVE, POINT OF RELEASE IS REFERENCE
Δy (m)
Grade 12 Science Essentials
Also applies to objects
dropped from an upward
moving reference.
Δt (s)
OBJECT THROWN UP AND CAUGHT (A+B+C+D) OBJECT THROWN UP, LANDS AT HEIGHT (A+B+C) OBJECT THROWN UP FROM HEIGHT, BOUNCES (B+C+D)
Δt (s)
9
D
Δt (s)
E
v (m·s−1)
A
C
B
D
Δt (s)
C
E
Accera&on due to force
by surface during bounce
a (m·s−2)
v (m·s−1)
vi(up) ≠ 0m·s−1
a (m·s−2)
v (m·s−1)
a (m·s−2)
Δt (s)
A
C
Δt (s)
Δt (s)
Vi(up) = Vf(down)
B
Δy (m)
vf(down)
Δy (m)
vf(up) = 0 m·s−1 = vi(down)
Δy (m)
vf(up) = 0 m·s−1 = vi(down)
Δt (s)
Δt (s)
Δt (s)
A
B
D
E
C
If the collision is perfectly elastic, the downward velocity before the bounce
and the upward velocity after the bounce is equal in magnitude.
Treat the 2 projectile paths (before and after bounce) as separate paths.
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Special Projectile Paths
Contact time
Lift moving down
A
Δyli%
EXAMPLE:
A hot air balloon ascends with a constant
velocity of 5 m·s−1. A ball is dropped from the
hot air balloon at a height of 50 m and falls
vertically towards the ground. Determine (a)
the distance between the hot air balloon and
ball after 2 seconds and (b) the velocity of the
ball when it reaches the ground.
(a) Take downwards as positive:
Distance travelled by balloon :
Δy = vi Δt +
1
2
aΔt 2
1
(0)(22 )
2
= (− 5)(2) +
= − 10
∴ Δy = 10 m up
Distance travelled by ball :
Δy = vi Δt +
1
2
aΔt 2
= (− 5)(2) +
1
(9,8)(22 )
2
= − 10 + 19,6
∴ Δy = 9,6 m down
∴ total distance = 10 + 9,6
= 19,6 m apart
(b) Take downwards as positive:
vf2
vf2
=
vi2
+ 2aΔy
2
= (− 5 ) + 2(9,8)(50)
vf =
A lift accelerates upwards at a rate of 1,4 m·s−2. As the lift
starts to move, a lightbulb falls from the ceiling of the lift.
Determine how long it takes the lightbulb to reach the
lift’s floor. The height from the ceiling of the lift to its floor
is 3m.
Take downwards as positive:
movement of lift :
25 + 980
∴ ylif t = − 0,7t
1
aΔt 2
2
1
(− 1,4)t 2
2
2
3 + ylif t = (0)t +
3 − 0,7t 2 = 4,9t 2
Δt (s)
A
B
C
C
D
Apex
EXAMPLE:
The velocity-time graph below represents the
bouncing movement of a 0,1 kg ball. Use the
graph to answer the questions that follow:
Δt (s)
Contact time
Bounce
a) Which direction of movement is positive?
Downwards
Apex
A
B
C
D
Δt (s)
Bounce
gradient = g = –9,8 m.s–2
10
5
Δt (s)
−8
b) How many times did the ball bounce?
3 times
c) What does the gradient of the graph represent?
Acceleration of the ball
movement of bulb :
Δybu lb = vi Δt +
A
gradient = g = +9,8 m.s–2
EXAMPLE:
ylif t = (0)t +
D
Δt (s)
Δyball = lift height + Δylift
Δylif t = vi Δt +
D
B
v (m·s−1)
Δyball
li% height
Δyball
B
Δy (m)
Δyli%
C
Δy (m)
Lift moving up
li% height
When an object is dropped
from a moving reference
(hot air balloon), the initial
velocity will be equal to that
of the reference. The acceleration of the object will be
downwards at
9,8 m·s−2, regardless of the
acceleration of the reference.
BOUNCING BALL – Ball falls from rest and bounces
LIFT
v (m·s−1)
HOT AIR BALLOON
SCIENCE CLINIC 2019 ©
v (m·s−1)
Grade 12 Science Essentials
d) Are the collisions between the ball and ground elastic or inelastic?
After each bounce there is a decrease in magnitude of the velocity of the ball,
and therefore a change in kinetic energy. The collisions are inelastic as kinetic
energy is not conserved.
1
aΔt 2
2
1
(9,8)t 2
2
e) If the ball is in contact with the ground for a duration of 0,08 s, determine the impulse on
the ball
3 = 5,6t 2
∴ t = 0,73s
Impulse = Δp
= m (vf − vi )
Simultaneous equation is needed because there are 2 unknown variables:
•Distance that lift moved
•Time to reach floor
= (0,1)(− 8 − 10)
= − 1,8
∴ Impulse = 1,8 N ⋅ s upwards
f) Predict why the ball stopped moving.
it was most likely caught
vf = 31,70 m ⋅ s−1 downwards
10
For more information about Science Clinic’s seminars, classes and resources, visit www.scienceclinic.co.za
QUESTION 1
VERTICAL PROJECTILE MOTION
A ball, A, is thrown vertically upward from a height, h, with a speed of 15 m∙s-1.
AT THE SAME INSTANT, a second identical ball, B, is dropped from the same height as ball A
as shown in the diagram below. Both balls undergo free fall and eventually hit the ground.
(2)
B
Explain the term free fall.
(4)
A
1.1
Calculate the time it takes for ball A to return to its starting point.
(7)
15 m∙s-1
1.2
Calculate the distance between ball A and ball B when ball A is at its maximum
height.
h
1.3
Ground
1.4
(1)
(4)
[17]
Sketch a velocity-time graph in the ANSWER BOOK for the motion of ball A from the
time it is projected until it hits the ground.
Clearly show the following on your graph:

The initial velocity

The time it takes to reach its maximum height

The time it takes to return to its starting point
QUESTION 2
Ground
An object is released from rest from a point X, above the ground as shown in the diagram
below. It travels the last 30 m (BC) in 1,5 s before hitting the ground. Ignore the effects of air
friction.
●X
●B
●C
Name the type of motion described above.
30 m
2.1
(4)
Calculate the:
Magnitude of the velocity of the object at point B
(5)
2.2
2.2.1
Height of point X above the ground
(3)
[13]
2.2.2
Sketch an acceleration-time graph for the entire motion of the object.
After hitting the ground, the object bounces once and then comes to rest on the ground.
2.3
QUESTION 3
A hot air balloon is rising vertically at a constant velocity. When the hot air balloon reaches
point A a few metres above the ground, a man in the hot air balloon drops a ball which hits the
ground and bounces. Ignore the effects of friction.
A
GROUND
The velocity-time graph below represents the motion of the ball from the instant it is dropped
until after it bounces for the first time. The time interval between bounces is ignored. THE
UPWARD DIRECTION IS TAKEN AS POSITIVE.
0,6
P
3,2
(1)
t (s)
Write down the magnitude of the velocity of the hot air balloon.
(3)
2,6
USE INFORMATION FROM THE GRAPH TO ANSWER THE QUESTIONS THAT FOLLOW.
v (m∙s-1)
5,88
2,94
0
- 2,94
3.1
Calculate the height above the ground from which the ball was dropped.
-19,60
3.2
3.4
3.3
Maximum height the ball reaches after the first bounce
Time at the point P indicated on the graph
(3)
(2)
Calculate the:
3.5
(4)
[13]
Distance between the ball and hot air balloon when the ball is at its maximum height
after the first bounce
QUESTION 4
4.1
Sketch a velocity-time graph for ball A.
Calculate the time taken by ball A to return to the ground.
Ball A is projected vertically upwards at a velocity of 16 m∙s-1 from the ground. Ignore the
effects of air resistance. Use the ground as zero reference.
4.2
Show the following on the graph:
(a) Initial velocity of ball A
(b) Time taken to reach the highest point of the motion
(c) Time taken to return to the ground
30 m
B
9 m∙s-1
16 m∙s-1
A
ground
ONE SECOND after ball A is projected upwards, a second ball, B, is thrown vertically
downwards at a velocity of 9 m∙s-1 from a balcony 30 m above the ground. Refer to the
diagram below.
4.3
Calculate how high above the ground ball A will be at the instant the two balls pass
each other.
QUESTION 5
(4)
(3)
5.1.2
5.1.1
Initial velocity of ball B
Time it takes for ball B to hit the surface of the water
Speed with which ball A hits the surface of the water
(5)
(3)
(3)
(6)
[13]
5.1.3
Calculate the:
A man throws ball A downwards with a speed of 2 m∙s-1 from the edge of a window, 45 m
above a dam of water. One second later he throws a second ball, ball B, downwards and
observes that both balls strike the surface of the water in the dam at the same time. Ignore air
friction.
5.1
5.2
Initial velocities of both balls A and B
The time of release of ball B
The time taken by both balls to hit the surface of the water
On the same set of axes, sketch a velocity versus time graph for the motion of balls
A and B. Clearly indicate the following on your graph:



(5)
[16]
QUESTION 6
Ball A is projected vertically upwards from the ground, near a tall building, with a speed of
30 m∙s-1. Ignore the effects of air friction.
Explain what is meant by a projectile.
(2)
(4)
6.1
The total time that ball A will be in the air
(4)
Calculate:
6.2.1
The distance travelled by ball A during the last second of its fall
B
ground
TWO SECONDS after ball A is projected upwards, ball B is projected vertically
upwards from the roof of the same building. The roof the building is 50 m above the
ground. Both balls A and B reach the ground at the same time. Refer to the diagram
below. Ignore the effects of air friction.
50 m
30 m∙s-1
A
Calculate the speed with which ball B was projected upwards from the roof.
Sketch velocity-time graphs for the motion of both balls A and B on the same set of
axes. Clearly label the graphs for balls A and B respectively.
(4)
[18]
(4)
6.2.2
6.2
6.3
6.4
Indicate the following on the graphs:
(a) Time taken by both balls A and B to reach the ground
(b) Time taken by ball A to reach its maximum height
QUESTION 7
A ball is dropped from the top of a building 20 m high. Ignore the effects of air resistance.
20 m
(4)
(2)
Speed at which the ball hits the ground
(3)
Define the term free fall.
7.2.1
Time it takes the ball to reach the ground
7.1
7.2.2
(2)
[11]
Calculate the:
Sketch a velocity-time graph for the motion of the ball (no values required).
7.2
7.3
QUESTION 7 (Start on a new page.)
1,2 m·s-1
A hot-air balloon moves vertically downwards at a constant velocity of 1,2 m∙s-1.
When it reaches a height of 22 m from the ground, a ball is dropped from the balloon.
Refer to the diagram below.
22 m
ground
7.1
Is the hot-air balloon in free fall? Give a reason for the answer.
Explain the term projectile motion.
(4)
(2)
(2)
Assume that the dropping of the ball has no effect on the speed of the hot-air balloon.
Ignore air friction for the motion of the ball.
7.2
Calculate the time it takes for the ball to hit the ground after it is dropped.
(6)
[14]
7.3
When the ball lands on the ground, it is in contact with the ground for 0,3 s and then it
bounces vertically upwards with a speed of 15 m∙s-1.
7.4
Calculate how high the balloon is from the ground when the ball reaches its
maximum height after the first bounce.
QUESTION 8 (Start on a new page.)
A ball is projected vertically upwards with a speed of 10 m∙s-1 from point A, which is at
the top edge of a building.
The ball hits the ground after 3 s. It is in contact with the ground for 0,2 s and
then bounces vertically upwards, reaching a maximum height of 8 m at point B.
See the diagram below.
Ignore the effects of friction.
10 m∙s-1
A
B
8m
ground
(3)
(2)
Height of the building
(3)
Why is the ball considered to be in free fall during its motion?
8.2.1
Speed with which the ball hits the ground
(3)
8.1
8.2.2
Speed with which the ball leaves the ground
Calculate the:
8.2.3
8.2
8.3
The magnitude of the velocity with which it hits the ground
The magnitude of the velocity with which it leaves the ground
The time taken to reach the ground, as well as the time at which it leaves
the ground
Draw a velocity versus time graph for the complete motion of the ball from A
to B. Show the following on the graph:
•
•
•
(4)
[15]
QUESTION 9 (Start on a new page.)
Stone A is projected vertically upwards at a speed of 12 m∙s-1 from a height h above
Calculate the time taken for stone A to reach its maximum height.
the ground. Ignore the effects of air resistance.
9.1
v
B
At the same instant that stone A is projected upwards, stone B is thrown vertically
downwards from the same height at an unknown speed, v. Refer to the diagram below.
12 m∙s-1
A
h
ground
Calculate the speed, v, with which stone B is thrown downwards.
When stone A reaches its maximum height, the speed of stone B is 3v.
9.2
9.3
Sketch velocity-time graphs for the complete motions of stones A and B on
the same set of axes. Label your graphs for stones A and B clearly.
Calculate the height h.
At the instant stone A passes its initial position on its way down, stone B hits
the ground.
9.4
The time taken for stone A to reach its maximum height
The velocity with which stone B is thrown downwards
Show the following on the graphs:
•
•
(3)
(4)
(3)
(4)
[14]
QUESTION 10 (Start on a new page.)
A ball is thrown vertically downwards from the top of a building and bounces a few
times as it hits the ground. The velocity-time graph below describes the motion of the
ball from the time it is thrown, up to a certain time T.
2,2
4,955
7,71
8
T
time (s)
(1)
2
Take downwards as the positive direction and the ground as zero reference. The graph
is NOT drawn to scale. The effects of air friction are ignored.
29,6
27
10
0
Write down the speed with which the ball is thrown downwards.
-25
X
10.1
(2)
(3)
ALL parts of the graph have the same gradient. Give a reason for this.
Height from which the ball is thrown
(4)
10.2
10.3.1
Time (T) shown on the graph
10.4.1
10.4.2
10.4.3
Time at which the ball reaches its maximum height after the first
bounce
Time that the ball is in contact with the ground at the first bounce
(2)
(1)
(2)
[16]
(1)
Is the collision of the ball with the ground elastic or inelastic? Give a reason
for the answer using information in the graph.
Value of X
Write down the:
10.3.2
Calculate the:
10.5
10.4
10.3
velocity (m∙s-1)
QUESTION 11 (Start on a new page.)
Define the term free fall.
●A
●B
ground
In the diagram below, point A is at the top of a building. Point B is exactly halfway
between the point A and the ground. Ignore air resistance.
11.1
Calculate the height of point A above the ground.
A ball of mass 0,4 kg is dropped from point A. It passes point B after 1 s.
11.2
11.3
Calculate the magnitude of the average net force exerted on the ball while it is
in contact with the ground.
Calculate the magnitude of the velocity of the ball when it strikes the ground.
When the ball strikes the ground it is in contact with the ground for 0,2 s and then
bounces vertically upwards, reaching a maximum height at point B.
11.4
(2)
(3)
(3)
(6)
[14]
QUESTION 12 (Start on a new page.)
In a competition, participants must attempt to throw a ball vertically upwards past
point T, marked on a tall vertical pole. Point T is 3,7 m above the ground. Point T may,
or may not, be the highest point during the motion of the ball.
7,5 ms-1
T
(2)
1,6 m
In which direction is the net force acting on the ball while it moves towards
point T?
Choose from: UPWARDS or DOWNWARDS. Give a reason for the answer.
(3)
ground
Calculate the time taken by the ball to reach its highest point.
(6)
(2)
[13]
Determine, by means of a calculation, whether the ball will pass point T or
not.
3,7 m
One participant throws the ball vertically upwards at a velocity of 7,5 ms-1 from a point
that is 1,6 m above the ground, as shown in the diagram below. Ignore the effects of
air resistance.
12.1
12.2
12.3
12.4
Draw a velocity-time graph for the motion of the ball from the instant it is
thrown upwards until it reaches its highest point.
The initial velocity and final velocity
Time taken to reach the highest point
Indicate the following on the graph:


Work, Energy and Power
Define the work done on an object by a constant force F as F Δx cos θ , where F is the
magnitude of the force, Δx the magnitude of the displacement and θ the angle
between the force and the displacement. (Work is done by a force – the use of the term
'work is done against a force', e.g. work done against friction, must be avoided.)
Draw a force diagram and free-body diagrams.
Calculate the net work done on an object.
Distinguish between positive net work done and negative net work done on the system.
(This section must be read in conjunction with the CAPS, p. 117–120.)
Work
•
•
•
•
Work-energy theorem
•
State the work-energy theorem: The work done on an object by a net force is equal to
the change in the object's kinetic energy:
W net = Δ K = K f - K i
Apply the work-energy theorem to objects on horizontal, vertical and inclined planes
(for both frictionless and rough surfaces).
•
W
∆t
Calculate the power involved when work is done.
Perform calculations using P ave = Fv ave when an object moves at a constant speed along
a rough horizontal surface or a rough inclined plane.
Calculate the power output for a pump lifting a mass (e.g. lifting water through a height
at constant speed).
In symbols: P =
Conservation of energy with non-conservative forces present
•
Define a conservative force as a force for which the work done in moving an object
between two points is independent of the path taken. Examples are gravitational force,
the elastic force in a spring and coulombic force.
•
Define a non-conservative force as a force for which the work done in moving an object
between two points depends on the path taken. Examples are frictional force, air
resistance, tension in a chord, etc.
•
State the principle of conservation of mechanical energy: The total mechanical energy
(sum of gravitational potential energy and kinetic energy) in an isolated system remains
constant. A system is isolated when the net external force (excluding the gravitational
force) acting on the system is zero.)
•
Solve conservation of energy problems using the equation: W nc = ΔE k + ΔE p
•
Use the relationship above to show that in the absence of non-conservative forces,
mechanical energy is conserved.
Power
•
Define power as the rate at which work is done or energy is expended.
•
•
•
Work
Positive work
Negative work
Work-energy
theorem
Principle of
conservation of
mechanical energy
Conservative force
Non-conservative
force
Power
W
t
MECHANICS: WORK, ENERGY AND POWER
Work done on an object by a constant force is the product of the magnitude of
the force, the magnitude of the displacement and the angle between the force
and the displacement.
In symbols: W = F Δx cos θ
The kinetic energy of the object increases.
The kinetic energy of the object decreases.
The net/total work done on an object is equal to the change in the object's
kinetic energy OR the work done on an object by a resultant/net force is equal
to the change in the object's kinetic energy.
In symbols: W net = Δ K = Kf - Ki.
The total mechanical energy (sum of gravitational potential energy and kinetic
energy) in an isolated system remains constant. (A system is isolated when
the resultant/net external force acting on the system is zero.)
A force for which the work done in moving an object between two points is
independent of the path taken.
Examples are gravitational force, the elastic force in a spring and electrostatic
forces (coulomb forces).
A force for which the work done in moving an object between two points
depends on the path taken.
Examples are frictional force, air resistance, tension in a chord, etc.
The rate at which work is done or energy is expended.
In symbols: P =
Energy
Grade 12 Science Essentials
PRINCIPLE OF CONSERVATION OF MECHANICAL ENERGY
ENERGY
The ability to do work
Unit: joules (J)
Scalar quantity
Gravitational Potential Energy (EP)
The energy an object possesses
due to its position relative to a
reference point.
Kinetic Energy (EK)
The energy an object has
as a result of the object’s
motion
Amount of energy transferred when an
object changes position relative to the
earth’s surface.
Amount of energy transferred
to an object as it changes
speed.
EP = mg h
1
EK = mv 2
2
g = 9,8 m·s–2, m is mass in kg,
h is height in m above the ground
m is mass in kg,
v is velocity in m·s–1
Example:
Determine the gravitational potential
energy of a 500 g ball when it is placed
on a table with a height of 3 m.
Example:
Determine the kinetic energy
of a 500 g ball when it travels
with a velocity of 3 m.s–1.
EP
EK
=
=
=
mg h
(0,5)(9,8)(3)
14,7 J
=
=
=
1
2
m v2
1
(0,5)(32 )
2
2,25 J
Mechanical Energy (EM)
The sum of gravitational potential and kinetic energy of an
object at a point
EM
=
EM
=
=
EM
=
mg h +
EM
=
EM
=
EP + EK
mg h +
1
2
(0,5)(9,8)(2,5) +
m v2
1
(0,5)(1,82 )
2
13,06 J
Principle of conservation of mechanical energy: The total mechanical
energy in an isolated system remains constant.
The law of conservation of mechanical energy applies when there is no friction or air resistance acting on the object. In the absence of air resistance, or
other forces, the mechanical energy of an object moving in the earth’s gravitational field in free fall, is conserved.
EMECHA
=
(EP + EK )A
(m g h +
1
2
EMECHB
=
m v 2 )A
(EP + EK )B
=
(m g h +
1
2
m v 2 )B
General law of conservation of energy: Energy cannot be created or destroyed, merely transferred.
In the following instances the gravitational potential energy of an object is converted to kinetic energy (and vice versa), while the
mechanical energy remains constant.
EXAMPLE 1: Object moving vertically
EXAMPLE 2: Object moving on an inclined plane
A 2 kg ball is dropped from rest at A, determine the maximum velocity
of the ball at B just before impact.
A 2 kg ball rolls at 3 m·s−1 on the ground at A, determine the maximum
height the ball will reach at B.
(EP + EK )A
=
(m g h + 1 m v 2 )A
2
1
(2)(9,8)(4) + (2)(0 2 )
2
=
78,4 + 0
=
vB
=
vB
=
(EP + EK )B
(m g h + 1 m v 2 )B
2
1
(2)(9,8)(0) + (2)vB2
2
0 + 1vB2
=
78,4
8,85 m ⋅ s−1 downwards
(EP + EK )A
=
(EP + EK )B
1
(m g h + m v 2 )A
2
(2)(9,8)(0) + 1 (2)(32 )
2
=
0+ 9
=
9
19,6
=
hB
=
0,46 m
hB
1
m v 2 )B
2
(2)(9,8)(h B ) + 1 (2)(0 2 )
2
(m g h +
=
19,6h B + 0
EXAMPLE 3: Pendulum
EXAMPLE 4: Rollercoaster
The 2 kg pendulum swings from A at 5 m·s−1 to B, on the ground,
where its velocity is 8 m·s−1. Determine the height at A.
The 2 kg ball rolls on a toy rollercoaster from A, at 20 m above the
ground, to B where its height is 8 m and velocity is 14 m·s−1. Calculate
its starting velocity at A.
EP + EK
1
mv 2
2
EXAMPLE:
A ball, mass 500 g, is thrown horizontally through the air. The ball travels
at a velocity of 1,8m·s −1 and is 2,5 m from the ground. Determine the
mechanical energy of the ball.
EM
SCIENCE CLINIC 2019 ©
(EP + EK )A
=
=
=
0 + 64
=
hA
vA
=
=
1,99 m
vA
=
64 − 25
19,6
hA
(EP + EK )A
1
(m g h + m v 2 )A
2
1
(2)(9,8)(20) + (2)(vA2 )
2
392 + vA2
=
19,6h A + 25
(EP + EK )B
(m g h + 1 m v 2 )B
2
1
(2)(9,8)(0) + (2)(82 )
2
(m g h + 1 m v 2 )A
2
1
(2)(9,8)(h A ) + (2)(52 )
2
=
=
21
For more information about Science Clinic’s seminars, classes and resources, visit www.scienceclinic.co.za
=
=
(EP + EK )B
1
m v 2 )B
2
1
(2)(9,8)(16) + (2)(142 )
2
(m g h +
313,6 + 196
313,6 + 196 − 392
10,84 m ⋅ s−1 to the right
Work, Energy and Power
Grade 12 Science Essentials
No Work done on an object (moving at a constant velocity) if the force and
WORK
Work done is the transfer of energy. Work done on an displacement are perpendicular to each other.
object by a force is the product of the displacement Consider a man carrying a suitcase with a weight of 20 N on a ‘travelator’
and the component of the force parallel to the dis- moving at a constant velocity.
placement.
W
F
Δx
θ
W = FΔx cos θ
= work (J)
= force applied (N)
= displacement (m)
= Angle between F and Δx
Fx = F cos θ
Direc)on
of mo)on
NET WORK ON AN OBJECT
A number of forces can act on an object at the same time. Each force
can do work on the object to change the energy of the object. The net
work done on the object is the sum of the work done by each force acting on the object.
If Wnet is positive, energy is added to the system.
If Wnet is negative, energy is removed from the system.
Work and Energy are SCALARS, and NOT direction specific.
Fg = 20 N
EXAMPLE:
The joule is the amount of work done when a force of one
newton moves its point of application one meter in the direction of the force.
Δx
FA is perpendicular to the displacement: θ = 90° ; cos 90° = 0.
Work always involves two things:
No force in the plane of the displacement, hence, NO WORK IS DONE by
FA and Fg and no energy is transferred. We can also say that FA / Fg does
not change the potential energy (height) or kinetic energy (vertical velocity) of the object.
1. A force which acts on a certain object. (F)
2. The displacement of that object. (Δx / Δy)
F
A force/force component in the direction of the displacement does positive
work on the object. The force increases the energy of the object.
Δx
When a resultant force is applied to an object, the resultant
force accelerates the block across distance Δx. Work has
been done to increase the kinetic energy of the block.
Positive work means that energy is added to the system.
Direc+on
of mo+on
F
θ
If a resultant force is applied to an
object vertically, the resultant force
lifts the block through distance Δy.
Work has been done to increase the
potential energy of the block.
“Lifting” usually implies at a constant
velocity.
W = Fx Δx cos θ
FA = 20 N
NOTE:
Work is a scalar quantity,
i.e. NO DIRECTION for
F or x!
SCIENCE CLINIC 2019 ©
W = Fx Δx cos θ
Calculate the net work done on a trolley where a force of 30 N is applied to the trolley. The trolley moves 3 m to the left. The force of
friction is 5 N to the right.
Work done by applied force:
Work done by frictional force:
WA = FΔx cos θ
= (30)(3)cos 0
= 90 J gained
Wf = Ff Δx cos θ
= (5)(3)cos 180
= − 15 J "lost"
Work done by gravity
Wg = FgΔx cos θ
Work done by normal force:
WN = FN Δx cos θ
= (FN )(3)cos 90
= 0J
= (Fg )(3)cos 90
= 0J
Wnet = WA + Wf + WN + Wg
Fx = F cos θ
Δx
= 90 − 15 + 0 + 0
= 75 J nett energy gained
Alternative method for determining net work:
F
0° ≤ θ < 90° ; +1 ≥cos θ > 0
Δy
A force/force component in the opposite direction of the displacement does
negative work on the object. The force decreases the energy of the
object.
1. Draw a free body showing only the forces acting on the object.
2. Calculate the resultant (net) force acting on the object.
3. Calculate the net work using Wnet = FnetΔx cos θ
Step 1: Freebody diagram
Negative work means that energy is being removed from the
system.
FN
W = Fx Δx cos θ
F
Fx = F cos θ
θ
F
θ
Direc+on
of mo+on
Δx
Δx
Work is only done in the direction of the displacement.
Work is done by the component of the force that is parallel
to the displacement. The angle between the force and the
displacement is θ. If no displacement takes place due to
the applied force, no work is done.
FA = 30 N
Fg
90° < θ ≤ 180° ; 0 > cos θ ≥ −1
NB: Never use a – for F in the opposite direction. The cos θ
makes provision for that.
22
Ff = -5 N
Take left as positive:
Step 2: Calculate Fnet
Fnet = FA + Ff
= 30 − 5
= 25 N left
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Step 3: Net work
Wnet = Fnet Δx cos θ
= (25)(3)cos 0
= 75 J gained
Work, Energy and Power
Grade 12 Science Essentials
WORK ENERGY THEOREM
NON-CONSERVATIVE FORCES
According to Newton’s Second Law of Motion, when a resultant force
acts on an object, the object accelerates. This means there is a
change in velocity of the object, and therefore a change in kinetic
energy of the object, since Ek = ½ mv2
WORK-ENERGY THEOREM: The work done by a net force on
an object is equal to the change in the kinetic energy of the
object
Wnet = ΔE K
Fnet Δx =
1
2
m(v 2f − v 2i )
POWER
(FT, FA, Ff, Fair resistance)
Power is the rate at which work is done OR the rate at
which energy is transferred.
A force is a non-conservative force if:
1. The work done by the force in moving an object from point A to
point B is dependent of the path taken.
2. The net work done in moving an object in a closed path which starts
and ends at the same point is not zero.
A non-conservative force does not conserve mechanical energy.
A certain amount of energy is converted into other forms such as internal
energy of the particles which the objects is made of. An example of a
non-conservative force is the frictional force.
CONSERVATIVE FORCES
A force is a conservative force if:
1. The work done by the force in moving an object from point A to point B
is independent of the path taken.
EP = EP
P=
P = power (watt)
W = work (J)
Δt = time (s)
FA
Ff
Calculate the power expended by an engine when an object
of mass 100 kg is lifted to a height of 2,2 m in a time of 3 s,
at a constant velocity.
P=
FA
Ff
=
=
1
2
Ff
The work done by FA is more when the longer path is taken. The work
done to overcome the friction will result in the surface of the crate becoming hotter. This energy is dissipated (as sound and/or heat) and is
very difficult to retrieve, i.e. not conserved.
Note: the total energy of the system is conserved in all cases, whether
the forces are conservative or non-conservative.
Wnc = ΔE K + ΔEP
The work done by gravity on each ball is independent of the path
taken. Only the h⟂ is considered.
EXAMPLE:
100
m
=
ΔEP =
=
=
=
1
2
m(v 2f
−
v 2i )
+ mgΔh
ΔE K = E Kf − E Ki
mgΔh
mg(h f − h i )
(800)(9,8)(0 − 100 sin 30∘ )
− 392 000 J
=
=
AVERAGE POWER (CONSTANT VELOCITY)
Paverage = Fvaverage
Paverage = F
− 482 000
− 100
FA = 4 820 N
23
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Δx
Δt
EXAMPLE:
A man lifts a 50 kg bag of cement from ground level up to a
height of 4 m above ground level in such a way that the bag
of cement moves at constant velocity (i.e. no work is done to
change kinetic energy). Determine his average power if he
does this in 10 s.
Paverage = Fvaverage
= F
= − 90 000 J
FA =
Fnet = 0 N
∴ Wnet = 0 J
∴ Pnet = 0 W
We can calculate the average power needed to keep an object
moving at constant speed using the equation:
1
m(v 2f − v 2i )
2
1
(800)(0 2 − 152 )
2
Wnc = ΔE K + ΔE P
FA Δx cos θ = ΔE K + ΔEP
FA (100)cos 180 = − 90 000 − 392 000
30°
NOTE:
W
Δt
FΔx cos θ
Δt
(100)(9,8)(2,2)cos 0
3
= 718,67 W
FA
Conservative force conserve mechanical energy. Example of
conservative forces are gravitational force and spring force.
W
E
=
Δt
Δt
EXAMPLE:
Consider the crate on a rough surface being pushed with a constant
force FA from position 1 to position 2 along two different paths.
2. The net work done in moving an
object in a closed path which starts
and ends at the same point is zero.
An 800 kg car traveling at
15 m·s−1 down a 30° hill needs
to stop within 100 m to avoid
an accident. Using energy calculations only, determine the magnitude of the average force that
must be applied to the brakes
over the 100 m. Assume that
the surface is frictionless.
SCIENCE CLINIC 2019 ©
4m
Δy
Δt
4
= (50)(9,8)( 10
)
= 196 W
This cannot be used for object in freefall, unless the object has
reached terminal velocity.
Work, Energy and Power
Grade 12 Science Essentials
SCIENCE CLINIC 2019 ©
When to use which formulae
CONSERVATION OF MECHANICAL ENERGY
E MA
E MB
=
(E P + EK )A
1. When an object moves up a slope
with a known angle
(E P + EK )B
=
ALTERNATIVE METHODS TO DETERMINE WORK DONE BY Fg ON A SLOPE
WITHOUT RESOLVING Fg INTO Fg⟂ AND Fg//
2. When an object moves down a slope
with a known angle
• When there are no external force (FA/Ff) present.
• Any track/pendulum with a curve that is not a straight line.
4m
WORK-ENERGY PRINCIPLE
Wnet
=
=
ΔEK
1
mv 2f
2
60°
30°
1
− 2 mv 2i
Wg
=
=
=
• If an object is accelerating on a horizontal/incline plane.
4m
120°
60°
30°
Fg = 100 N
Fg Δx cos θ
Wg
(100)(4)cos 120∘
− 200 J
=
=
=
120°
Fg = 100 N
Fg Δx cos θ
(100)(4)cos 60∘
200 J
Any of the following methods may be used:
FN
FA
3. When an object moves up a slope without a
given angle, but with specified height
30°
Fg
3m
A. Fnet → Wnet
A. Wnet = ∑W
1. Determine Fnet separately
Fnet =
FA − F f − Fg //
1. Determine W
WA =
WN =
Wg =
Fnet
=
Fnet
=
FA − Ff − Fg sin θ
FA − Ff − (100 sin 30∘ )
2. Apply Fnet to Wnet
Wnet
1
1
= Fnet Δx cos θ =
m v 2 − m vi2
2 f
2
4. When an object moves down a slope without a
given angle, but with specified height
3m
Fg = 100 N
of each force seperately
FAΔx cos θ
FN Δx cos θ
Fg Δx cos θ
Wg
=
=
=
Wf = Ff Δx cos θ
Fg Δx cos θ
(100)(3)cos 180∘
− 300 J
2. Apply ∑W to Wnet
WA + Wf + Wg + WA = Wnet =
1
1
m v 2 − m vi2
2 f
2
24
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Fg = 100 N
Wg
=
=
=
Fg Δx cos θ
(100)(3)cos 0∘
300 J
A
4m
WORK, ENERGY AND POWER
8m
C
1.1.3
1.1.2
1.1.1
Is mechanical energy conserved as the object slides from A to C? Write only
YES or NO.
State the principle of conservation of mechanical energy in words.
(6)
(1)
(2)
rope
(6)
[18]
(3)
Using ENERGY PRINCIPLES only, calculate the magnitude of the frictional
force exerted on the object as it moves along BC.
An object of mass 10 kg is released from point A which is 4 m above the ground. It
slides down the track and comes to rest at point C.
B
The diagram below shows a track, ABC. The curved section, AB, is frictionless. The
rough horizontal section, BC, is 8 m long.
QUESTION 1
1.1
1.2
motor
300 kg
Calculate the magnitude of the frictional force acting between the crate and
the surface of the inclined plane.
25°
A motor pulls a crate of mass 300 kg with a constant force by means of a light
inextensible rope running over a light frictionless pulley as shown below. The
coefficient of kinetic friction between the crate and the surface of the inclined plane is
0,19.
1.2.1
The crate moves up the incline at a constant speed of 0,5 m∙s-1.
1.2.2
Calculate the average power delivered by the motor while pulling the crate
up the incline.
QUESTION 2
A 5 kg block is released from rest from a height of 5 m and slides down a frictionless incline to
point P as shown in the diagram below. It then moves along a frictionless horizontal portion PQ
and finally moves up a second rough inclined plane. It comes to a stop at point R which is 3 m
above the horizontal.
5 kg
2.3
2.2
2.1
Calculate the angle (θ) of the slope QR.
Explain the term non-conservative force.
Explain why the kinetic energy at point P is the same as that at point Q.
Using ENERGY PRINCIPLES only, calculate the speed of the block at point P.
(7)
[15]
(2)
(2)
(4)
R
5m
3m
θ
P
Q
The frictional force, which is a non-conservative force, between the surface and the block is
18 N.
2.4
QUESTION 3
The diagram below shows a heavy block of mass 100 kg sliding down a rough
25o inclined plane. A constant force F is applied on the block parallel to the inclined plane as
shown in the diagram below, so that the block slides down at a constant velocity.
F
100 kg
25°
The magnitude of the kinetic frictional force (fk) between the block and the surface of the
inclined plane is 266 N.
(2)
(1)
Friction is a non-conservative force. What is meant by the term non- conservative
force?
Is the learner correct? Answer only YES or NO.
(2)
3.1
3.2.1
Explain the answer to QUESTION 3.2.1 using physics principles.
A learner states that the net work done on the block is greater than zero.
3.2.2
(4)
(6)
[15]
Calculate the magnitude of the force F.
3.2
3.3
Calculate the speed of the block at the bottom of the inclined plane.
If the block is released from rest without the force F being applied, it moves 3 m down the
inclined plane.
3.4
QUESTION 4
The track for a motorbike race consists of a straight, horizontal section that is 800 m long.
Calculate the average power developed by the motorbike for this motion.
A participant, such as the one in the picture above, rides at a certain average speed and
completes the 800 m course in 75 s. To maintain this speed, a constant driving force of 240 N
acts on the motorbike.
4.1
450 m
Another person practises on the same motorbike on a track with an incline. Starting from rest,
the person rides a distance of 450 m up the incline which has a vertical height of 5 m, as
shown below.
5m
4.2
State the WORK-ENERGY theorem in words.
Draw a labelled free-body diagram for the motorbike-rider system on the incline.
(2)
(4)
(3)
4.3
The total frictional force acting on the motorbike is 294 N. The combined mass of rider and
motorbike is 300 kg. The average driving force on the motorbike as it moves up the incline is
350 N. Consider the motorbike and rider as a single system.
4.4
F
20o
BEFORE
200 kg
5.3
5.2
5.1
Draw a free-body diagram indicating ALL the forces acting on the block while it is
being pulled.
State the work-energy theorem in words.
Give a reason why the coefficient of kinetic friction has no units.
(4)
(4)
(2)
(1)
(6)
[15]
Use energy principles to calculate the speed of the motorbike at the end of the 450 m
ride.
QUESTION 5
AFTER
3m
A constant force F, applied at an angle of 20o above the horizontal, pulls a 200 kg block, over a
distance of 3 m, on a rough, horizontal floor as shown in the diagram below.
F
20o
200 kg
5.4
Show that the work done by the kinetic frictional force (W fk) on the block can be
written as W fk = (-1 176 + 0,205 F) J.
Rough floor
The coefficient of kinetic friction, μk, between the floor surface and the block is 0,2.
5.5
(4)
[15]
Calculate the magnitude of the force F that has to be applied so that the net work
done by all forces on the block is zero.
QUESTION 6
20 kg A
5m
6.1 State the principle of conservation of
mechanical energy in words.
6.2.2
Speed of the block at point B at the
bottom of the ramp
6.2 The kinetic frictional force between the 20 kg
block and the surface of the ramp is 30 N.
Use energy principles to calculate the:
Work done by the kinetic frictional
(3)
force on the block
30°
A 20 kg block is released from rest from the top of a ramp at point A at a construction site as
shown in the diagram below. The ramp is inclined at an angle of 30o to the horizontal and its
top is at a height of 5 m above the ground.
B
6.2.1
6.3
A 100 kg object is pulled up the SAME RAMP at a constant speed of 2 m∙s -1 by a
small motor. The kinetic frictional force between the 100 kg object and the surface of
the ramp is 25 N. Calculate the average power delivered by the small motor in the
pulling of the object up the incline.
QUESTION 7
Before
2 kg
After
¼h
2 kg
B
A pendulum with a bob of mass 5 kg is held stationary at a height h metres above the ground.
When released, it collides with a block of mass 2 kg which is stationary at point A.
The bob swings past A and comes to rest momentarily at a position ¼ h above the ground.
he diagrams below are NOT drawn to scale.
h
A
(2)
(5)
(3)
(4)
[14]
Kinetic energy of the block immediately after the collision
(4)
Calculate the:
Immediately after the collision the 2 kg block begins to move from A to B at a constant speed
of 4,95 m∙s-1. Ignore frictional effects and assume that no loss of mechanical energy occurs
during the collision.
7.1
7.1.1
Height h
C
(4)
[13]
(2)
0,5 m
7.1.2
4,95 m·s-1
2 kg
B
State the work-energy theorem in words.
Use energy principles to calculate the work done by the frictional force when the 2 kg
The block moves from point B at a velocity
of 4,95 m·s-1 up a rough inclined plane to
point C. The speed of the block at point C
is 2 m·s-1. Point C is 0,5 m above the
horizontal, as shown in the diagram below.
During its motion from B to C a uniform 7.2
frictional force acts on the block.
7.3
block moves from point B to point C.
QUESTION 8
The diagram below shows a bullet of mass 20 g that is travelling horizontally. The bullet strikes
a stationary 7 kg block and becomes embedded in it. The bullet and block together travel on a
rough horizontal surface a distance of 2 m before coming to a stop.
Use the work-energy theorem to calculate the magnitude of the velocity of the bulletblock system immediately after the bullet strikes the block, given that the frictional
force between the block and surface is 10 N.
(2)
(5)
7 kg
8.1
State the principle of conservation of linear momentum in words.
20 g
8.2
Calculate the magnitude of the velocity with which the bullet hits the block.
2m
8.3
cage
electric motor
(4)
[11]
QUESTION . (Start on a new page.)
A lift arrangement comprises an
electric motor, a cage and its
counterweight. The counterweight
moves vertically downwards as
the cage moves upwards. The
cage and counterweight move at
the same constant speed. Refer to
the diagram below.
counterweight
The cage, carrying passengers, moves vertically
upwards at a constant speed, covering 55 m in 3
minutes. The counterweight has a mass of 950 kg.
The total mass of the cage and passengers is 1 200
kg. The electric motor provides the power needed to
operate the lift system. Ignore the effects of friction.
(2)
(3)
Define the term power in words.
Gravitational force on the cage
(2)
9.1
9.2.1
Counterweight on the cage
Calculate the work done by the:
9.2.2
9.2
9.3
Calculate the average power required by the motor to operate the lift
arrangement in 3 minutes. Assume that there are no energy losses due to
heat and sound.
(6)
[13]
QUESTION 10 (Start on a new page.)
10.1 Draw a labelled free-body diagram,
showing ALL the forces acting on the boyskateboard unit while moving down the ramp
from P to Q.
The diagram below shows a boy skateboarding on a ramp which is inclined at 20o to
the horizontal. A constant frictional force of 50 N acts on the skateboard as it moves
from P to Q. Consider the boy and the skateboard as a single unit of mass 60 kg.
Ignore the effects of air friction.
vi
10.2
Use energy principles to calculate the speed v i of the skateboarder at point P.
State the work-energy theorem in words.
(5)
(2)
(3)
10.3
Points P and Q on the ramp are 25 m apart. The
skateboarder passes point P at a speed vi and
passes point Q at a speed of 15 m∙s-1. Ignore
rotational effects due to the wheels of the
skateboard.
10.4
1,6 m
(4)
[14]
Calculate the average power dissipated by the skateboarder to overcome
friction between P and Q.
QUESTION 11 (Start on a new page.)
6 kg
6 kg
In the diagram below, a 4 kg block lying on a rough horizontal surface is connected to
4 kg
(2)
a 6 kg block by a light inextensible string passing over a light frictionless pulley.
State the work-energy theorem in words.
Initially the blocks are HELD AT REST.
11.1
When the blocks are released, the 6 kg block falls through a vertical
distance of 1,6 m.
Draw a labelled free-body diagram for the 6 kg block.
(2)
11.2
Calculate the work done by the gravitational force on the 6 kg
(3)
11.3
block.
The coefficient of kinetic friction between the 4 kg block and the
horizontal surface is 0,4. Ignore the effects of air resistance.
(5) [12]
11.4 Use energy principles to calculate the speed of the 6 kg block when it
falls through 1,6 m while still attached to the 4 kg block.
QUESTION 12 (Start on a new page.)
A slide, PQR, at an amusement park consists of a curved frictionless section, PQ, and
a section, QR, which is rough, straight and inclined at 30o to the horizontal. The
starting point, P, is 3 m above point Q. The straight section, QR, is 5 m long.
P
3m
(2)
Q
A learner, with mass 50 kg, starting from rest at P, slides down section PQ, then
continues down the straight section, QR.
5m
State the law of conservation of mechanical energy in words.
(4)
30o
12.1
Calculate the speed of the learner at Q.
R
12.2
12.3
(3)
(3)
Draw a labelled free-body diagram for the learner while he/she is on
section QR.
Use energy principles to calculate the speed of the learner at point R.
Calculate the magnitude of the kinetic frictional force acting on the learner
when the learner is on section QR.
The coefficient of kinetic friction (µk) between the learner and the surface of
section QR is 0,08.
12.4
12.5
(5)
[17]
QUESTION 13 (Start on a new page.)
75 kg
0,65 m⋅s-2
height of 12 m.
the speed of the load when it is at a
13.5 Use the work-energy theorem to calculate
13.4 State the work-energy theorem in words.
reached a height of 12 m.
the gravitational force when the load has
13.3 Calculate the work done on the load by
the load.
13.2 Name the non-conservative force acting on
load while it moves upward.
13.1 Draw a labelled free-body diagram for the
A load of mass 75 kg is initially at rest on the ground. It is then pulled vertically
upwards at a constant acceleration of 0,65 m⋅s-2 by means of a light inextensible rope.
Refer to the diagram below. Ignore air resistance, rotational effects and the mass of
the rope.
12 m
ground
QUESTION 14 (Start on a new page.)
200 m
B
The diagram below, not drawn to scale, shows a vehicle with a mass of 1 500 kg
starting from rest at point A at the bottom of a rough incline. Point B is 200 m vertically
above the horizontal.
1 500 kg
A
(2)
(1)
(3)
(2)
(2)
(5)
[13]
Define the term non-conservative force.
The total work done by force F that moves the vehicle from point A to point B in 90 s
is 4,80 x 106 J.
14.1
(1)
(2)
(3)
Is force F a conservative force? Choose from: YES or NO.
Calculate the average power generated by force F.
14.2
14.3
State the work-energy theorem in words.
(5)
The speed of the vehicle when it reaches point B is 25 m∙s-1.
14.4
14.5
Use energy principles to calculate the total work done on the vehicle by the
frictional forces.