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 today’s 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 vehicle’s 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
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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, time
2. 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) = 10 9 metres. See a list of some common SI units and prefixes. Most texts will contain such information.
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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 v av
= 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 v av
= average velocity, Δr = change in displacement and Δt = 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 v av
= 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.
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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 a av
= average acceleration, Δv = change in velocity of the body and Δt
= change in time over which the change in velocity took place. Alternatively, we may write: where a av
= 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.
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Clearly, the gradient (slope) of a displacement-time graph gives the velocity.
Gradient =Δs/Δt = 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.
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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 Graphs
These 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 =Δv/Δt = acceleration
Note that a positive gradient implies a positive acceleration and a negative gradient implies a negative acceleration-(deceleration).
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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 Δt.
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 Graphs
These 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.
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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 B
FORCE: 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.
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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.
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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:
and
NEWTON’S LAWS OF MOTION
Newton’s 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 Newton’s Second Law. The SI Unit of force is the newton (N), defined so that 1N = 1kgms -2 .
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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 object’s 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 Newton’s First Law.
Newton’s 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 Newton’s 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.
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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 Newton’s 2 nd 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 earth’s surface).
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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 jet’s resultant velocity relative to the ground.
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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 N36.9
o W.
Note that if the angle between the two vectors being added together is other than 90 o , 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?
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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 N36.9
o E 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 E53.1
o N or as a True Bearing of 36.9
o .
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.
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The resultant force, R, is found by measurement to be 3.1 N at an angle of 39 o 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.
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EXAMPLE 4: A car is moving due East at 20 ms -1 moving due North at 20 ms -1
. A short time later it is
. 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 45 o 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.
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Consider, the example we used previously. We may push on the car with a force
F at an angle of Θ 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 FsinΘ. 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 FcosΘ.
EXAMPLE 5: A block of mass 20 kg is being pulled up an inclined plane by a rope inclined at 30 o to the plane’s surface as shown in the diagram below.
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The plane is inclined at 45 o to the horizontal. The friction force, F, opposing the block’s 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 sinΘ). So, from the diagram we have:
F
D
= F + mg sinΘ (since W = mg)
F
D
= 10 + 20 x 9.8 x sin 45 o
F
D
= 148.59 N
Total force acting up the plane must be:
F
U
= ma + F
D
F
F
U
U
= (20 x 4) + 148.59
= 228.59 N
Note that the logic we have used to obtain an expression for F
U
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
F
U
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 30 o = F
U
/ T or T = F
T = 228.59 / cos 30 o
U
/cos 30 o
T = 263.95 N (Tension in the rope is 264 N).
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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 moving in 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: a c
= v 2 /r where a c
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, F c
, 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.
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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 ENERGY
Kinetic 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. (E k
=KE)
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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:
E k
= 0.5 x 920 x 16.67
2
E k
= 1.28 x 10 5 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.
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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 .
Newton’s 2 nd Law can be re-written as: where Δp = the change in momentum of the object and Δt = the time taken for the change in momentum to occur.
This quantity Δp (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 Newton’s 1 st and 2 nd 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.
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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 Newton’s
3 rd 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 F
AB
is the force exerted on A by B and F
BA
is the force exerted on B by A. By Newton’s 3 rd 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 F
of F
AB
AB
indicates that we are taking the average value
during the time interval of the collision, Δt.
The change in momentum of B resulting from the collision is the Impulse or: in which the bar above the F
of F
BA
BA
indicates that we are taking the average value
during the time interval of the collision, Δt.
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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 Δp
A
and Δp
B
give the total
change in momentum for each particle. But we have seen that at each instant:
F
AB
= - F
BA
So that:
And therefore that : Δp
A
= - Δp
B
If we consider the two particles as an isolated system, the total momentum of the system is:
P = p
A
+ p
B
And the total change in momentum of the system as a result of the collision is zero, that is:
ΔP = Δp
A
+ Δp
B
= 0.
Thus, using Newton’s 3 rd 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
25

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:
Δp = F Δt 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.
INERTIA
As we have seen, Newton’s 1 st 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 Newton’s 1 st Law.
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 Why do you think Newton’s 1 st 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=relate d
1.
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.
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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 E60 o S 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 60 o 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;
28

d.
the tension in the towing rope.
9.
An astronaut of mass
M
A = 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.1
1.
15.3 km/h W11.3
o S or S78.7
o W
2.
Resultant force = 3905 N in a direction of W39.8
o S & 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 N63.4
o W
5.
25 km/h in a direction of N53.1
o W
6.
1054 km/h E5 o S
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 = M
A g = 784 N upwards; (b) R = M
A g + M
A a = 1424 N upwards; (c) R = M
A g -
M
A a = 144 N upwards.
1.
Describe the typical effects of the following external forces acting on
bodies: a.
Friction between surfaces b.
Air resistance
29

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 road
3.
Explain the difference between “mass” and “weight”.
4.
An unbalanced force of 48N west is applied to a 4 kg cart. Calculate the cart’s 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 car’s 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 Newton’s 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 10 4 N)
8.
The driver’s 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.
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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 10 5 J)
2.
A 7500 kg truck travelling east at 5 ms -1 travelling south-west at 20 ms -1 occur during such a collision.
collides and coalesces with a 1500 kg car
. Describe the energy transformations that may
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
Find the net force applied by the engine to cause this acceleration. (143750 J, 16.67 ms -2 , 8335 N)
east. If this change in velocity was accomplished in 0.3 s, calculate the acceleration of the car.
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 10 9 J)
5.
An army tank of 2.5 x 10 4 kg mass has a kinetic energy of 1.25 x 10 the speed of the tank. (10ms -1 )
6 J. Calculate
1.
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 10 4 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)
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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 with Car B moving east at 0.3 ms east with a speed of 0.56 ms
and collides but does not coalesce
-1
-1 . After collision, Car B is moving
. (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.
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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 Kate’s displacement from her start point
at the end of her run.
2.
Brad’s 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 truck’s brakes to achieve this change in
velocity. b.
Identify the energy transformations that occur as the truck’s 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.)
33

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.
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d.
The total distance covered from t = 0 to t = 4s.
1.
111.8m E26.6
o S
2.
(a) 25ms -1 South; (b) 45ms -1 South
3.
13ms -1 E22.6
o N
4.
(a) 2.5 x 10 5 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 W11.3
o S (or S78.7
o W)
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.
35