MOVING ABOUT Introduction Mechanics Motion Motion Graphs
Relative Velocity Force Newton's Laws of Motion Mass & Weight
Usefulness of Vector Diagrams, Vector Addition, Vector Subtraction,
Vector Components Circular Motion Work Energy - Kinetic, Potential,
Transformation of Energy Momentum Conservation of Momentum
Conservation of Momentum During Collisions Inertia Practicals Page
- lots of good pracs Worksheets Page - Forces, Energy, Momentum
& Vector Analysis INTRODUCTION: Increased access to transport
is a feature of todays society. Most people access some form of
transport for travel to and from school or work and for leisure
outings at weekends or on holidays. When describing journeys that
they may have taken in buses or trains, they usually do so in terms
of time or their starting point and their destination. When
describing trips they may have taken in planes or cars, they
normally use the time it takes, distance covered or the speed of
the vehicle as their reference points. While distance, time and
speed are fundamental to the understanding of kinematics and
dynamics, very few people consider a trip in terms of energy, force
or the momentum associated with the vehicle, even at low or
moderate speeds. The faster a vehicle is travelling, the further it
will go before it is able to stop. Major damage can be done to
other vehicles and to the human body in collisions, even at low
speeds. This is because during a collision some or all of the
vehicles kinetic energy is dissipated through the vehicle and the
object with which it collides. Further, the materials from which
vehicles are constructed do not deform or bend as easily as the
human body. Technological advances and systematic study of vehicle
crashes have increased understanding of the interactions involved,
the potential resultant damage and possible ways of reducing the
effects of collisions. There are many safety devices now installed
in or on vehicles, including seat belts and air bags. Modern road
design takes into account ways in which vehicles can be forced to
reduce their speed.The branch of Physics that deals with phenomena
such as those mentioned above is called mechanics.
MECHANICS: The branch of Physics that is concerned with the
motion and equilibrium of bodies in a particular frame of reference
is called mechanics. Mechanics can be divided into three branches:
(i) Statics which deals with bodies at rest relative to some given
frame of reference, with the forces between them and with the
equilibrium of the system; (ii) Kinematics - the description of the
motion of bodies without reference to mass or force; and (iii)
Dynamics which deals with forces that change or produce the motions
of bodies.Some common terms used in the study of mechanics (and
indeed many other branches of Physics) are: scalars, vectors and SI
Units. 1. Scalars A scalar is a physical quantity defined in terms
of magnitude (size) only - eg temperature, mass, volume, density,
distance, time2. Vectors A vector is a physical quantity defined in
terms of both magnitude and direction - eg force, displacement,
velocity, acceleration, electric field strength.Diagramatically we
can represent a vector by a straight line with an arrow on one end.
The length of the line represents the magnitude of the vector
quantity and the direction in which the arrow is pointing
represents the direction of the vector quantity.3. Systeme
International (SI) Units The internationally agreed system of
units. There are seven fundamental units. The three that we will
use in this topic are the metre (length), the kilogram (mass) and
the second (time). Various prefixes are used to help express the
size of quantities eg a nanometre (1 nm) = 10-9 of a metre, a
gigametre (1 Gm) = 109 metres. See a list of some common SI units
and prefixes. Most texts will contain such information.MOTION: The
following terms are commonly used to describe motion.
1.Displacement is the distance of a body from a point in a given
direction. It is a vector quantity. The SI unit of displacement is
the metre (m).
2.Speed The speed of a body is the rate at which it is covering
distance. It is a scalar quantity. The SI units are m/s, which can
also be written as ms-1.
where vav = average speed, d = total distance travelled and t =
total time taken to travel distance d.
3.Velocity The velocity of a body is its speed in a given
direction. In other words, velocity is the rate of change of
displacement with time. It is a vector quantity with the same SI
units as speed.
where vav = average velocity, r = change in displacement and t =
change in time taken to achieve that change in displacement.
Another way to express average velocity is as the average of the
initial and final velocities.
where vav = average velocity, u = initial velocity of the body
and v = final velocity of the body. Note that this equation applies
ONLY when the velocity of the body is increasing or decreasing at a
constant rate.4.Acceleration The acceleration of a body is the rate
of change of the velocity of the body with time. It is a vector
quantity, with units of (metres/second)/second, written as
ms-2.
where aav = average acceleration, v = change in velocity of the
body and t = change in time over which the change in velocity took
place. Alternatively, we may write:
where aav = average acceleration, v = final velocity of the
body, u = initial velocity of the body, and t = time over which the
change in velocity took place.Note that a body accelerates when: a.
It speeds up; b. It slows down; c.It changes direction.MOTION
GRAPHS: Often the most effective way to describe the motion of a
body is to graph it. Note that in this section the variable s will
be used to represent displacement instead of r. You will find in
Physics that these two variables are both in common use to denote
displacement. Displacement-Time Graphs These may be used to gain
information about the displacement of an object at various times or
about the velocity of the object at various times. Clearly, the
gradient (slope) of a displacement-time graph gives the
velocity.Gradient =s/t = velocity Note that a positive gradient
implies a positive velocity and a negative gradient implies a
negative velocity- (travelling in the opposite direction to the
initial motion).For a curved displacement-time graph, the gradient
of the tangent to the curve at a particular point equals the
gradient of the curve at that point, which in turn equals the
velocity of the object at that particular time. Such a velocity,
that is, the velocity at a particular instant in time, is called
the instantaneous velocity. An example of an instrument that
measures instantaneous velocity is the speedometer in a car. In
older cars the speedometer was linked mechanically to the
transmission. These days, however, a device located in the
transmission produces a series of electrical pulses whose frequency
varies in proportion to the vehicle's speed. The electrical pulses
are sent to a calibrated device that translates the pulses into the
speed of the car. This information is sent to a device that
displays the vehicle's speed to the driver in the form of a
deflected speedometer needle or a digital readout. Note that a
straight line displacement-time graph implies that velocity is
constant. A curved line displacement-time graph implies that
velocity is changing with time (ie the object is accelerating).
Velocity-Time GraphsThese may be used to gain information about the
displacement, velocity and acceleration of an object at various
times.
The gradient is clearly the acceleration of the object.
Gradient =v/t = acceleration Note that a positive gradient
implies a positive acceleration and a negative gradient implies a
negative acceleration-(deceleration).Also, the area under the
graph, In the case above, the units are: seconds x metres per
second = metres. Thus, the area under a velocity-time graph is
equal to the displacement travelled by the object in the time
t.Note that a horizontal straight line velocity-time graph implies
that acceleration is zero ie velocity remains constant.A
non-horizontal, straight line velocity-time graph implies that
acceleration is constant and non-zero.A curved line velocity-time
graph implies that acceleration is varying.Acceleration-Time
GraphsThese may be used to gain information about the velocity and
acceleration of an object at various times. The area under an
acceleration-time graph gives the velocity of an object. Check the
units of the area: (ms-2 x s = ms-1).A horizontal straight line
acceleration-time graph implies that velocity is varying at a
constant rate (ie velocity is increasing or decreasing by the same
amount each second). That is, acceleration is constant.RELATIVE
VELOCITY: Often it is necessary to compare the velocity of one
object to that of another. For instance, two racing car drivers, A
and B, may be travelling north at 150 km/h and 160 km/h
respectively. We could say that the velocity of car B relative to
car A is 10 km/h north. In other words, driver A would see driver B
pull away from her with a velocity of 10 km/h north.Likewise, two
jet aircraft, C and D, flying directly at each other in opposite
directions (hopefully as part of an aerobatics display) may have
velocities of 900 km/h north and 1000 km/h south respectively. We
could say that the velocity of D relative to C is 1900 km/h south.
In other words, jet C will observe jet D flying towards it at a
speed of 1900 km/h. Clearly, when the objects are travelling in the
same direction, the velocity of one relative to the other is the
difference between their speeds, taking due care to state the
appropriate direction. When the objects are travelling in opposite
directions the velocity of one relative to the other is the sum of
their speeds, again taking due care to state the correct direction.
There is a vector equation which can be used to calculate the
relative velocities of objects, even when the objects travel at
various angles to one another:Velocity of A relative to B =
Velocity of A Velocity of BFORCE: What is Force? In simple terms a
force can be defined as a push or a pull. We experience examples of
forces every day. If we push a stationary lawn mower (with enough
force) it begins to move that is, it accelerates and its velocity
increases. If we push on the brakes of a moving bicycle, it slows
down that is it decelerates (or undergoes negative acceleration).
If we apply sufficient force to an aluminum can, by squeezing it
with our hand, we can change the shape of the can.So a force can
cause a change in the state of motion of an object or a change in
the shape of an object. In fact, all accelerations are caused by
forces. Does every force cause acceleration?Again, from our
everyday experience, we know the answer to this question is no. If
a person pushes on the brick wall of a house, the house does not
accelerate. Sometimes when we want to push or pull an object from
one place to another we find that no matter how hard we push or
pull, we just cannot move (accelerate) the object. What is the
relationship between force and acceleration? We could perform an
experiment to determine the relationship between the size of a
force applied to an object at rest on a laboratory bench and the
change in velocity experienced by the object over a set period of
time (ie the acceleration). Such an experiment would produce
results as shown below.
The graph above shows that: The change in velocity does not
happen instantaneously. A certain amount of force is required
before the object begins to accelerate. This makes sense, since the
force of friction between the bench and the object must be overcome
before the object can move. So, we can say that a net external
force is required in order to change the velocity of an object. The
acceleration produced is directly proportional to the force
applied. If we repeated the experiment on a frictionless surface
(eg using a dry-ice puck on a very smooth, polished table top) the
straight-line graph would even pass through the origin. What is the
relationship between acceleration and mass? We could measure the
accelerations produced when the same sized force is applied to
different objects. This would produce results like those below. The
previous graph suggests that there is an inverse relationship
between acceleration and mass. A plot of acceleration versus the
reciprocal of mass, using the same data, would produce a graph
similar to that below.
This graph clearly shows that: The acceleration produced by a
given force is inversely proportional to the mass of the object.
From experiments such as those above, we can say that: andNEWTONS
LAWS OF MOTION Newtons First and Second Laws: By combining the
results above and defining the units of force appropriately, we can
write that: .This can be taken as a statement of Newtons Second
Law. The SI Unit of force is the newton (N), defined so that 1N =
1kgms-2.Note that in this equation, F is the vector sum of all the
forces acting on the object, m is the mass of the object and a is
its vector acceleration. To remind us of that fact we will write:
.Note that if the resultant force on the object is zero, there is
no acceleration. Therefore, in the absence of a resultant force, an
objects velocity will remain unchanged. In other words, an object
at rest will remain at rest, and an object in motion will remain in
motion with uniform velocity, unless acted upon by a net external
force. This is a statement of Newtons First Law. Newtons Third
Law:Forces acting on a body originate in other bodies that make up
its environment. Any single force is only one aspect of a mutual
interaction between two bodies. We find by experiment that when one
body exerts a force on a second body, the second body always exerts
a force on the first. Furthermore, we find that these forces are
equal in size but opposite in direction. A single, isolated force
is therefore an impossibility.If one of the two forces involved in
the interaction between two bodies is called an action force, the
other is called the reaction force. Either force may be called the
action and the other the reaction. Cause and effect is not implied
here, but a mutual simultaneous interaction is implied.This
property of forces is stated by Newtons Third Law: To every action
there is always opposed an equal reaction; or, the mutual actions
of two bodies upon each other are always equal, and directed to
contrary parts. In other words, if body A exerts a force on body B,
body B exerts an equal but oppositely directed force on body A; and
furthermore the forces lie along the line joining the bodies.
Notice that the action and reaction forces, which always occur in
pairs, act on different bodies. If they were to act on the same
body, we could never have accelerated motion because the resultant
force on every body would be zero. Consider the following
examples:1. Imagine a boy kicking open a door. The force exerted by
the boy B on the door D accelerates the door (it flies open); at
the same time, the door D exerts an equal but opposite force on the
boy, which decelerates the boy (his foot loses forward velocity).
The force of the boy on the door and the force of the door on the
boy is an action-reaction pair of forces.2.When you walk, you apply
a force backwards on the earth. Likewise, the earth applies a force
to you of equal magnitude but in the opposite direction. So, you
move forwards.
The force of the person on the earth and the force of the earth
on the person is an action-reaction pair of forces.3.Consider a
body at rest on a horizontal table:
Each of the pairs of forces above is an action-reaction pair of
forces. Definitions of Mass and Weight:The mass of an object is a
measure of the amount of matter contained in the object. Mass is a
scalar quantity. The weight of an object is the force due to
gravity acting on the object. Weight is a vector quantity.The
weight, W, of an object is given by Newtons 2nd Law as: where m is
the mass of the object and g is the acceleration due to gravity
(9.8 ms-2 close to the earths surface).USEFULNESS OF VECTOR
DIAGRAMS: Many of the quantities with which we deal in Physics are
vectors. Sometimes we need to add a number of vectors together. For
instance, we may be trying to calculate the total or resultant
force acting on a car when several forces act on the car
simultaneously the wind, friction, gravity and the force supplied
by the engine. Sometimes we need to subtract two vectors. For
instance, we may be trying to calculate the change in velocity of a
car as it goes around a bend in the road. The change in velocity of
the car equals the final velocity of the car minus initial velocity
of the car. When the need arises to add or subtract vector
quantities, this proves to be easy only when the vector quantities
act along the same straight line. If the vectors act at an angle to
each other we really need to draw a vector diagram to assist in
solving the problem. VECTOR ADDITION:The method we will use is
called the Vector Polygon method. To find the sum of a number of
vectors draw each vector in the sum, one at a time, in the
appropriate direction, placing the tail of the second vector so
that it just touches the head of the first. Continue in this
fashion until all of the vectors in the sum have been included in
the diagram. Note that it does not matter which vector you start
with. The vector that closes the vector polygon in the same sense
as the component vectors is called the equilibrant. It is the
vector which when drawn into the diagram gets you back to where you
started. The vector that closes the vector polygon in the opposite
sense to the component vectors is called the resultant. The
resultant is the answer to the sum of all the vectors. Its size can
be calculated mathematically or measured using a ruler if the
vector polygon has been drawn to scale. The direction of the
resultant can be calculated mathematically or can be measured using
a protractor if the vector polygon has been drawn to scale. Either
way, the direction of the resultant must be stated in an
unambiguous way. Sometimes in Physics our vector additions only
involve two vectors at a time. In this case, the polygon formed is
a triangle, making the mathematical calculation of the magnitude
and direction of the resultant quite straight forward. EXAMPLE 1: A
fighter pilot flies her F-14D Tomcat jet with a true airspeed of
400 km/h North. A crosswind from the East blows at 300 km/h
relative to the ground. Calculate the jets resultant velocity
relative to the ground. Note: For aircraft, the true airspeed (TAS)
is the actual speed of the aircraft through the air (the speed of
the aircraft relative to the air). The wind speed is usually
measured relative to the ground. Groundspeed is the speed of the
aircraft relative to the ground. The groundspeed of the aircraft is
the vector sum of the true airspeed and the wind speed. Obviously,
a vector diagram would be very useful in solving Example Problem 1.
See the diagram below. By Pythagoras Theorem, the magnitude of the
resultant velocity of the jet is: and the direction can be found
using basic trigonometry as follows: So, the velocity of the jet
relative to the ground is 500 km/h N36.9oW. Note that if the angle
between the two vectors being added together is other than 90o,
Cosine Rule and Sine Rule can be used to solve the problem
mathematically. Note also the use of the compass in the diagram to
establish direction. EXAMPLE 2: In the previous problem, in which
direction should the pilot head and with what airspeed in order to
actually fly north at 400 km/h relative to the ground? Again a
vector diagram is useful. Our intuition tells us that the pilot
must fly into the wind. So, when we draw a diagram that shows all
of the information that we know to be true, we obtain the diagram
shown below.
Clearly, if the pilot flies N36.9oE with an airspeed of 500
km/h, the wind will bring her back to a heading of due North at a
speed of 400 km/h relative to the ground. Remember also, there is
usually more than one way to give the direction. The direction the
pilot should fly in this example could just as correctly be given
as E53.1oN or as a True Bearing of 36.9o. EXAMPLE 3: Four children
pull on a small tree stump firmly stuck in the ground. Looking down
on the tree stump from above, the forces applied by the children
are as shown below. Determine the resultant force applied to the
tree stump. To solve this problem mathematically we would need to
add two of the vectors together, then add our answer to the third
vector and finally add our answer to that addition to the fourth
vector. It is actually far quicker and easier to solve this problem
graphically. To do this we construct a vector polygon, using the
rules stated above and simply measure the size and direction of the
resultant. See the following diagram. The resultant force, R, is
found by measurement to be 3.1 N at an angle of 39o clockwise from
the direction of the 4.5 N force. Note that when using a graphical
approach, the scale must be clearly stated on the diagram. Always
choose a sensible numerical scale. Also, choose a scale that will
produce a large diagram. The larger the diagram, the more accurate
the answer. For the example problem above, the scale used was 1 N =
1.5 cm (approximately). Note also, there may be a small discrepancy
between the stated scale and the actual scale on the page.VECTOR
SUBTRACTION: If vector A is as shown below: then vector A is a
vector of the same magnitude as A but opposite direction. In order
to find the difference between two vectors, add the negative of the
second vector to the first. EXAMPLE 4: A car is moving due East at
20 ms-1. A short time later it is moving due North at 20 ms-1.
Calculate the change in velocity of the car. To find the change in
size of any quantity, you subtract the initial size of the quantity
from the final size. Obviously, with vector quantities you must do
a vector subtraction not just an arithmetic one, since vectors
possess both size and direction. Change in velocity = final
velocity + ( initial velocity) since that is how we draw the vector
diagram. We literally add the negative of the initial velocity to
the final velocity. Study the vector diagram below to ensure you
understand the process of vector subtraction. Using Pythagoras
Theorem and basic trigonometry as shown in Example 1 above, we find
that the change in velocity of the Maserati is 28.3 ms-1 at an
angle of 45o West of North. Note that even though the car has the
same initial and final speeds, because the direction of the car has
changed, so too has its velocity.VECTOR COMPONENTS:Sometimes we are
only interested in part of a vector rather than all of it. For
instance, if we push a car that has run out of petrol, we apply a
force to the car. However, if we are not careful some of the force
we apply pushes down vertically on the car and the rest of it
pushes horizontally on the car. Obviously, we are trying to
maximise the component of the force that is applied horizontally.
The angle at which we apply force to the car will determine how
much of our force is applied horizontally.Any vector may be broken
into two component vectors at right angles to each other. These
components are called the rectangular components of the vector. The
rectangular components of a vector add up to the original
vector.Consider, the example we used previously. We may push on the
car with a force F at an angle of to the horizontal, as shown
below. The force F may be broken into its vertical and horizontal
components as shown below. The magnitude of each component can then
be calculated using simple trigonometry. The size of the vertical
component of F is Fsin. The size of the horizontal component of F,
the one that must overcome the force of friction if we are to move
the car forward, is Fcos.EXAMPLE 5: A block of mass 20 kg is being
pulled up an inclined plane by a rope inclined at 30o to the planes
surface as shown in the diagram below. The plane is inclined at 45o
to the horizontal. The friction force, F, opposing the blocks
motion is 10 N. Determine the tension, T, in the rope if the net
acceleration, a, of the block up the plane is 4 ms-2.You will
observe that we have resolved two vectors in the diagram into
rectangular components the tension, T, in the rope and the weight,
W, of the block. The rectangular components we have chosen are
those acting parallel to and perpendicular to the inclined plane.
These components are the most useful ones in a situation like this.
Now the total force acting down the plane is the sum of the
friction force, F, and the component of the weight force of the
block acting down the plane (W sin). So, from the diagram we
have:FD = F + mg sin (since W = mg) FD = 10 + 20 x 9.8 x sin 45o FD
= 148.59 NTotal force acting up the plane must be: FU = ma + FD FU
= (20 x 4) + 148.59 FU = 228.59 N Note that the logic we have used
to obtain an expression for FU is as follows. The force up the
plane MUST be sufficient to overcome the 148.59 N force down the
plane and to provide the required force of 80 N to give the block
the correct acceleration up the plane. Now we can calculate the
tension in the rope. The total force up the plane FU is actually
the component of the tension force parallel to the plane. This is
the part of the tension force that is applied parallel to the
plane. Therefore, we have: cos 30o = FU / T or T = FU /cos 30o T =
228.59 / cos 30oT = 263.95 N (Tension in the rope is 264 N).Now try
Vector Analysis Worksheet No.1.You may also like to check out this
applet; http://surendranath.tripod.com/Applets.html which helps you
to understand vector addition & subtraction. When you get to
the site, move your mouse over the "Applet Menu" button at the top
left of the page. The menu items will appear. Select "Some Math",
"Vectors", "Vector Addition". Read the instructions & run the
applet.Circular Motion An object movingin a circular path at
constant speed is said to be executing uniform circular motion.
Obviously, although the speed is constant, the velocity is not,
since the direction of the motion is always changing. It can be
shown that for an object executing uniform circular motion (UCM),
the acceleration keeping the object in its circular path is given
by:ac = v2/r where ac is called the centripetal ("centre-seeking")
acceleration, v = speed of the object and r = radius of the
circular path. As the name implies, centripetal acceleration is
directed towards the centre of the circle.Using the fact that force
can be written as F = ma, the centripetal force, Fc, acting on an
object undergoing UCM is given by: where m = mass of the object.
This force is also directed towards the centre of the
circle.Example: A car of mass 900 kg moves at a constant speed of
25 m/s around a circular curve of radius 50 m. Calculate the
centripetal force acting on the car. (11250 N, towards the centre
of the circle)WORK A force applied to an object is often capable of
moving that object through a certain distance. Whenever this
happens we say that work has been done on that object.
Work is a scalar quantity defined mathematically as: where W =
work done on object, F = force acting on object along the line of
motion and s = net displacement of object caused by the force along
the line of motion. The SI unit of work is the joule (J). 1J = 1 Nm
EXAMPLE: Calculate the work done when an object is moved through a
displacement of 20m north by a force of 10N north. ANSWER: W = F.s
and so W = 10 x 20 = 200J (No direction, work is a scalar)ENERGY
Energy and work are closely related quantities. An object can do
work only if it has energy. Energy, then, is the property of a
system that is a measure of its capacity for doing work. The amount
of energy an object has is equal to the amount of work it can do.
Like work, energy is a scalar quantity with an SI Unit of the joule
(J).Energy has several forms: electric energy, chemical energy,
heat energy, nuclear energy, radiant energy (ie EM radiation such
as light), mechanical energy (eg kinetic energy) and sound energy
(ie the kinetic energy of the vibration of the air). In a closed
system (ie one in which no mass enters or leaves), energy can
neither be created nor destroyed, although it may be transformed
from one form into another. This is called the Law of Conservation
of Energy. KINETIC ENERGYKinetic energy is the name given to the
energy associated with a moving body. The amount of kinetic energy
possessed by a body is given by: where m = mass of the body and v =
velocity the body. (Ek =KE)Moving vehicles have kinetic energy.
Consider a small car of mass 920 kg moving at a constant speed of
60 km/h (16.67 m/s). The kinetic energy of the car can be
calculated as: Ek = 0.5 x 920 x 16.672 Ek = 1.28 x 105 J To change
the velocity of a moving body, or to set a body at rest into
motion, a net force must be applied to it and work must be done on
it. The work done on the body is equivalent to the change in
kinetic energy of the body. POTENTIAL ENERGY Stored energy is
called potential energy (PE), since it has the potential to do work
for us. Examples of potential energy include: the energy stored in
a stretched (or compressed) spring; the chemical energy stored in a
car battery; the energy stored in the water in a damn above a
hydroelectric power station; and the energy stored in the chemical
bonds holding compounds together.Gravitational Potential Energy:
GPE = mgh(m=mass, g=9.8ms-2, h=height)ENERGY TRANSFORMATIONS Energy
transformations (changes) are an important aspect in understanding
motion. In a car battery, chemical potential energy is transformed
into electrical energy. In an internal combustion engine (such as a
car engine) the chemical potential energy stored in petrol is
transformed among other things into mechanical energy in the form
of kinetic energy of motion.When cars collide, various energy
transformations take place. Some of the kinetic energy (KE) of the
cars is transformed into sound we hear the collision. Some KE is
changed into radiant energy (light energy) we see sparks fly as the
wreckage scrapes along the ground. Some of the KE is transformed
into heat the friction produced by metal scraping on metal and
tyres gripping the road under heavy braking produces heat. Some KE
is transformed into potential energy of deformation the car bodies
are compressed, compacted and twisted during the collision and some
of the KE is stored in the deformed wreckage. In a worst-case
scenario, where an explosion takes place, chemical energy stored in
the fuel is converted into kinetic energy (and sound, heat, light)
as parts of the wreckage are flung far and wide. MOMENTUM: Everyday
experience tells us that both the mass and velocity of an object
are important in determining things like (i) how hard it is to stop
the object or (ii) the effect the object has in a collision with
another object. An 85 kg man running at 5 m/s is a lot harder to
stop than a 15 kg six year old child running at the same speed. A
50 gram bullet fired from a rifle with a muzzle velocity of 500 m/s
will do a lot more damage than an identical bullet thrown at the
target by hand.Isaac Newton spoke of the quantity of motion of an
object. Today we define the momentum of an object to be the product
of mass (m) and velocity (v).
Momentum is a vector quantity with units of kgms-1 or Ns, since
1N = 1kgms-2.Newtons 2nd Law can be re-written as: where p = the
change in momentum of the object and t = the time taken for the
change in momentum to occur.This quantity p (the change in
momentum) is given the name impulse. Clearly, from the above
equation, impulse, I, is defined as the product of force and time
and has SI Units of Ns. Impulse is a vector quantity.CONSERVATION
OF MOMENTUM According to Newtons 1st and 2nd Laws of motion, there
is no change in momentum without the action of a net external
force. Thus, if no net external force acts on a particular system,
the total momentum of the system must be constant. This is known as
the Principle of the Conservation of linear momentum. A system on
which the net external force is zero is given a special name. Such
a system is called an isolated system. So, another way to express
the principle of the conservation of linear momentum is to say that
within an isolated system the total momentum is a constant. This
principle is applicable to many important physical
situations.CONSERVATION OF MOMENTUM DURING COLLISIONS One important
physical situation to which the principle of the conservation of
linear momentum is applicable is the case of collisions between
bodies. In such cases, if we assume that no external net force acts
during the collision, we can say that the total momentum of the
system before collision equals the total momentum of the system
after collision. This proves to be an extremely useful starting
point for analysing many collision situations. To see that momentum
is conserved during collisions we can use Newtons 3rd Law. Consider
a collision between two particles, A and B, as shown below.
During the brief collision these particles exert large forces on
one another. At any instant FAB is the force exerted on A by B and
FBA is the force exerted on B by A. By Newtons 3rd Law these forces
at any instant are equal in magnitude but opposite in direction.The
change in momentum or Impulse of A resulting from the collision is:
in which the bar above the FAB indicates that we are taking the
average value of FAB during the time interval of the collision,
t.The change in momentum of B resulting from the collision is the
Impulse or: in which the bar above the FBA indicates that we are
taking the average value of FBA during the time interval of the
collision, t. Note that it is necessary to take the average value
of the collision forces since the magnitudes of both forces will
vary over the duration of the collision.If no other forces act on
the particles, then pA and pB give the total change in momentum for
each particle. But we have seen that at each instant: FAB = - FBASo
that: And therefore that:pA = - pB If we consider the two particles
as an isolated system, the total momentum of the system is: P = pA
+ pB And the total change in momentum of the system as a result of
the collision is zero, that is: P = pA + pB = 0.Thus, using Newtons
3rd Law and our knowledge of impulse we have shown that if there
are no external forces, the total momentum of the system is not
changed by the collision. Therefore, as we said before, if we
assume that no external net force acts during the collision, we can
say that the total momentum of the system before collision equals
the total momentum of the system after collision.How accurate is it
though to assume that no external net force acts on a system during
a collision? When a golf club strikes a golf ball surely there are
external forces that act on the system of club + ball? Indeed there
are: gravity and friction are two obvious forces that act on both
club and ball. So how can we simply ignore these forces? The answer
is that it is safe to neglect these external forces during the
collision and to assume that momentum is conserved provided, as is
almost always the case, that the external forces are negligible
compared to the impulsive forces of collision. If the external
forces are negligible compared to the impulsive forces, then the
change in momentum of a particle during a collision arising from an
external force is negligible compared to the change in momentum of
that particle arising from the impulsive force of collision. In the
case of the golf club striking the golf ball, the collision lasts
only a tiny fraction of a second. Since the observed change in
momentum is large and the time of collision is small, it follows
from the impulse equation: p = F t that the average impulsive force
F is relatively large. Compared to this force, the external forces
of gravity and friction are negligible. During the collision we can
safely ignore these external forces in determining the change in
motion of the ball; the shorter the collision time, the more
accurate this assumption becomes. In practice, we can apply the
principle of momentum conservation during collisions if the time of
collision is small enough. EXERCISE: Try the "Hollywood Physics"
exercise on the Brainteasers page.INERTIAAs we have seen, Newtons
1st Law states that an object in uniform motion will remain in
uniform motion and an object at rest will remain at rest, unless
acted upon by an external, net force. This ability of a body to
resist changes in its state of motion is called the inertia of the
body. The inertia of a car, for instance, is its tendency to remain
in uniform motion or remain at rest. The fact that bodies possess
inertia has important consequences when dealing with moving
vehicles.A moving vehicle is a complex body. It consists of the
vehicle body itself, the driver (and passengers) and other objects
carried in the car. If the driver or passengers or other objects
are not restrained, they will continue to move at whatever speed
the car is travelling, even after the application of the brakes.
Let us consider some questions: What are some of the dangers
presented by loose objects in vehicles? Most people, when they see
an unrestrained object fly off the car seat when they slam on the
brakes, would say that it got pushed off the seat (ie some sort of
force acted on the object to move it off the seat). Is this an
accurate account of the physics of the situation? Explain in terms
of Newtons 1st Law. Why do you think Newtons 1st Law of Motion is
not applied correctly in many real world situations (like the one
mentioned above)? What is the function of an inertia reel safety
belt? Where could we obtain information on how they work and their
effectiveness? Describe a possible experiment we could do to assess
the relative effectiveness of lap, lap-sash and harness seatbelts
in reducing the effects of inertia in a collision. List the
features of a modern car that are designed to reduce or avoid the
effects of a collision. Where would we obtain information to use in
assessing the relative effectiveness of these features? Assess the
reasons for the introduction of low speed zones in built-up areas
and the addition of air bags and crumple zones to vehicles with
reference to the concepts of impulse and momentum.Try this link for
an interesting video showing a cinder block broken on top of a man
lying on a bed of nails:
http://www.youtube.com/watch?v=icXlTShblj0&feature=relatedVECTOR
ANALYSIS WORKSHEET No.11. A ship is heading due west at a steady
speed of 15 km/h. A current of 3 km/h is running due south.
Calculate the velocity of the ship relative to the seabed. (Hint:
The velocity of the ship plus the velocity of the current will add
up to the total velocity of the ship relative to the seabed.) 2.
Two tractors pull on a large boulder in an effort to shift it out
of the way of a new fence line. One tractor pulls with a force of
3000 N west and the other tractor pulls with a force of 2500 N in a
southerly direction because of the terrain. Determine the resultant
force acting on the boulder. If the boulder has a mass of 1000 kg
calculate the acceleration of the boulder due to the resultant
force acting on it. 3. Three coplanar horizontal forces each of
magnitude 10 N act on a body of mass 5 kg as shown below. Determine
the magnitude of the net force acting on the body and the magnitude
of the resultant acceleration.
4. If vector A = 5 N north and vector B = 10 N east, find the
resultant of vector A vector B. 5. A submarine is travelling at 20
km/h due east. A short time later it is travelling due north at 15
km/h. Calculate the change in velocity of the submarine. 6. An
F-117A Nighthawk stealth fighter jet has a true airspeed of 1000
km/h due east. There is a cross wind blowing in a direction E60oS
at 100 km/h. Calculate the velocity of the jet relative to the
ground. (Hint: You will need to use the cosine & sine rules if
you intend to do this question mathematically.) 7. A sled of mass
10 kg sits on a horizontal surface as shown below.
A force is exerted on the sled by means of a rope inclined at
60o to the horizontal. If the tension in the rope is 150 N and the
frictional force between the sled and the horizontal surface is 55
N, will the sled move under these conditions? Explain. If the sled
does move determine the size of the acceleration with which it
moves. 8. A barge of total mass 300 kg is pulled by a single rope
attached to a tugboat of mass 1000 kg. If the drag on the tug and
the barge is one-tenth of their respective weights, and the total
forward force exerted by the tug is 5130 kg force (ie 5130 x 9.8
N), find the magnitude of: a. the total force resisting the forward
motion of the tug and barge; b. the acceleration of the tug/barge
system; c. the unbalanced accelerating force on the barge; d. the
tension in the towing rope. 9. An astronaut of mass MA = 80 kg
stands on the horizontal floor of a spaceship moving vertically
with acceleration a. If the acceleration due to gravity on the
astronaut is g = 9.8ms-2, write a mathematical expression for and
calculate the value of the reaction force R between the astronaut
and the floor of the spaceship when: a. a = 0; b. a = 8 ms-2
upwards; c. a = 8 ms-2 downwards.ANSWERS TO VECTOR ANALYSIS
WORKSHEET No.11. 15.3 km/h W11.3oS or S78.7oW 2. Resultant force =
3905 N in a direction of W39.8oS & acceleration = 3.9 ms-2 in
the same direction as the resultant force. 3. Net force = 20 N
& resultant acceleration = 4 ms-2. 4. 11.2 N in a direction of
N63.4oW 5. 25 km/h in a direction of N53.1oW 6. 1054 km/h E5oS 7.
Horizontal force to the left on sled is 75 N, which is greater than
the friction force to the right. Therefore the sled will move to
the left with an acceleration of 2 ms-2. If you did not get the
right acceleration, ask yourself how much of the 75 N to the left
actually accelerates the sled all of it or just some of it? Some of
it has to overcome friction! 8. (a) 1274 N, (b) 37.7 ms-2, (c) 11
307 N, (d) 11 601 N. 9. (a) R = MA g = 784 N upwards; (b) R = MA g
+ MA a = 1424 N upwards; (c) R = MA g - MA a = 144 N upwards. FORCE
WORKSHEET 11.Describe the typical effects of the following external
forces acting on bodies: a. Friction between surfaces b. Air
resistance 2. In each of the following situations, forces act to
cause a change in the velocity of a vehicle (car). Outline (ie
sketch in general terms; indicate the main features of) the forces
involved:a. Coasting along a level road with no pressure on the
accelerator b. Pressing on the accelerator c. Pressing on the
brakes d. Passing over an icy patch on the road e. Climbing hills
f. Descending hills and g. Following a curve in the road3. Explain
the difference between mass and weight.4. An unbalanced force of
48N west is applied to a 4 kg cart. Calculate the carts
acceleration. (12ms-2 west)5. A 2200 kg car, travelling at 25 m/s
south, comes to a stop in 10 s. Calculate (a) the cars acceleration
and (b) the unbalanced force required to cause that acceleration.
(2.5 ms-2 north & 5500 N north)6. Astronauts are placed
horizontally in their space capsule during launch. Use Newtons
First Law to explain why this is a good idea.7. A particular
pressure on the accelerator of a 4-wheel drive van of mass 2000 kg,
travelling along a smooth, level road supplies sufficient force
from the engine to accelerate the van at 5 ms-2. When this same van
travels through soft sand, the same pressure on the accelerator
results in a constant velocity of the van. Determine the force due
to friction acting on the van in the soft sand. (1 x 104N)8. The
drivers handbook in a particular country states that the minimum
safe distance between vehicles on the road is the distance a
vehicle can travel in 2 s at constant speed. Assume that a 1200 kg
car is travelling south at 72 km/h when the truck ahead crashes
into a northbound truck and comes to a sudden stop.a. If the car is
at the required safe distance behind the truck, what is the
separation distance between the car and the truck in metres?
(40m)b. If the average braking force exerted by the car is 6400 N
north, how long would it take the car to stop? (3.75s)c. What
additional data would you need to obtain to determine whether or
not the car would be involved in a collision. Assume that the car
driver has a reaction time of 0.1s.ENERGY - WORKSHEET 1 1. A car is
travelling at 27 m/s north and has a mass of 1500 kg. Calculate the
kinetic energy of the car. (5.47 x 105 J)2. A 7500 kg truck
travelling east at 5 ms-1 collides and coalesces with a 1500 kg car
travelling south-west at 20 ms-1. Describe the energy
transformations that may occur during such a collision.3. Determine
the amount of work that must be done by the engine of a 500 kg
racing car to change the velocity of the car from 55 ms-1 east to
60 ms-1 east. If this change in velocity was accomplished in 0.3 s,
calculate the acceleration of the car. Find the net force applied
by the engine to cause this acceleration. (143750 J, 16.67 ms-2,
8335 N)
4. Determine the kinetic energy of a 300 kg space probe launched
from the surface of Mars, once it has reached escape velocity of
5.1 km/s. (3.90 x 109 J)5. An army tank of 2.5 x 104 kg mass has a
kinetic energy of 1.25 x 106 J. Calculate the speed of the tank.
(10ms-1) MOMENTUM WORKSHEET 11. A 2200 kg car, travelling at 25 m/s
south, comes to a stop in 10 s. Calculate (a) the initial momentum
of the car; (b) the final momentum of the car; (c) the impulse of
the net force applied by the brakes; and (d) the magnitude of the
net force applied by the brakes. (55000 kgm/s south, 0 kgm/s, 55000
kgm/s north, 5500 N north) 2. A hammer strikes a nail with a force
of 55 N for a period of 0.2 s. Calculate the impulse of the force.
(11 kgm/s) 3. For how long must an explosive force of 3.3 x 104 N
act on a stationary bullet of mass 0.15 kg to give it a velocity of
220 m/s? (0.001 s) 4. A mass of 4 kg experiences a varying force as
given in the diagram below. Determine the change in velocity of the
mass. (3 m/s) MOMENTUM WORKSHEET 2 1. In experimental tests run by
the manufacturer, a car of mass 1500 kg travelling at 20 ms-1 due
east collides with an identical stationary car. Assuming that all
of the kinetic energy of the moving car is transferred to the
stationary car during the collision, describe quantitatively and
qualitatively the expected results of the collision. 2. Laboratory
Trolley Car A has a mass of 0.9 kg and Laboratory Trolley Car B a
mass of 0.5 kg. For each of the situations below, describe
quantitatively and qualitatively the expected results of the
collision for Car A: a. Car A moves east at 0.5 ms-1 and collides
but does not coalesce with Car B moving west at the same speed.
After collision, Car B is moving east with a speed of 0.79 ms-1.
(Car A: v = 0.22 m/s west) b. Car A moves east at 0.5 ms-1 and
collides but does not coalesce with Car B moving east at 0.3 ms-1.
After collision, Car B is moving east with a speed of 0.56 ms-1.
(Car A: v = 0.36 m/s east) c. Car A moves east at 0.4 ms-1 and
collides but does not coalesce with Car B moving west at 0.3 ms-1.
After collision, Car B is moving east with a speed of 0.6 ms-1.
(Car A: v = 0.1 m/s west) d. Car A moving at 0.3 ms-1 east collides
and coalesces with Car B moving at 0.5 ms-1 west. (Car A & B: v
= 0.014 m/s east)
(Hint: Use appropriate signs, +, -, to designate direction.)3. A
car of mass 1300 kg travelling at 25 m/s south, collides with a
solid rock cliff face. Use your knowledge of force and momentum to
describe a possible result of this collision.REVISION WORKSHEET
No.1 1. Kate is chasing Angela with a large bucket of water. She
runs 100m due east, followed by 100m due north and then 150m due
south. Calculate Kates displacement from her start point at the end
of her run. 2. Brads Lamborghini has an initial velocity of 20ms-1
south and an acceleration of 5 ms-2 south for five seconds.
Determine: a. the total change in velocity of the car b. the final
velocity of the car after five seconds.3. A car changes its
velocity from 5 ms-1 south to 12 ms-1 east. Calculate the change in
velocity of the car. 4. A car of mass 1000kg changes its velocity
from 20ms-1 north to 30ms-1 north in 2 seconds. a. Determine the
amount of work done by the engine of the car to achieve this change
in velocity. b. Calculate the acceleration of the car and the net
force applied by the engine to cause this acceleration.5. A truck
of mass 5000kg travelling at 15ms-1 is brought to rest in a
distance of 100m. a. Calculate the average force exerted by the
trucks brakes to achieve this change in velocity. b. Identify the
energy transformations that occur as the trucks kinetic energy is
reduced to zero.6. A bus travels at constant speed of 12ms-1 around
a circular bend of radius 150m. a. Determine the centripetal
acceleration (magnitude & direction) of the bus as it
negotiates the bend. b. Calculate the centripetal force acting on
the bus (magnitude & direction), if the mass of the bus is
2000kg.7. A ship is heading due west at a steady speed of 15 km/h.
A current of 3 km/h is running due south. Calculate the velocity of
the ship relative to the seabed. (Hint: The velocity of the ship
plus the velocity of the current will add up to the total velocity
of the ship relative to the seabed.) 8. Two trucks pull on a large
boulder in an effort to shift it out of the way of a new fence
line. One truck pulls with a force of 3000 N east and the other
truck pulls with a force of 2500 N south-east. a. Determine the
magnitude of the net force acting on the boulder. (Hint: A scaled
vector diagram would be helpful here.) b. If the boulder has a mass
of 500 kg calculate the size of the acceleration of the boulder due
to this net force.9. Two railway cars are being used in a structure
integrity test. Car A of mass 900 kg moves east at 25 ms-1 and
collides but does not coalesce with Car B of mass 700kg moving west
at 20 ms-1. After collision Car B is moving east with a speed of
30.6 ms-1. Determine: a. The total momentum of the system before
collision. b. The total momentum of the system after collision. c.
The speed and direction of Car A after collision. d. The change in
momentum of Car A as a result of the collision. e. The average
force acting on Car A, given that the change in momentum took 1.5
seconds.10. Study the graph below, which shows the velocity of a
vehicle over a period of time.
a. The average speed of the vehicle from t = 2s to t = 4s. b.
The instantaneous speed of the vehicle at t = 3s. c. The
acceleration of the vehicle from t = 2s to t = 4s. d. The total
distance covered from t = 0 to t = 4s.ANSWERS TO REVISION WORKSHEET
No.1:1. 111.8m E26.6oS 2. (a) 25ms-1 South; (b) 45ms-1 South 3.
13ms-1 E22.6oN 4. (a) 2.5 x 105 J; (b) 5ms-2 north and 5000N north.
5. (a) 5625N in opposite direction to original motion; (b) kinetic
energy is transformed into sound & heat energy. 6. (a) 0.96ms-2
towards centre of circular bend; (b) 1920N in same direction as the
centripetal acceleration. 7. 15.3 km/h W11.3oS (or S78.7oW) 8. (a)
5000N; (b) 10ms-2 9. (a) 8500Ns East; (b) 8500Ns, East; (c)
14.35ms-1 West; (d) 35 415Ns, West; (e) 23 610N, West. 10. (a)
2ms-1; (b) 1.8ms-1; (c) 2ms-2 in opposite direction to motion; (d)
12m. 1