Notes1
To view the Notes, click the following links below:
2. NEWTON’S LAWS OF MOTION AND PROJECTILE MOTION
3. CIRCULAR MOTION, SIMPLE HARMONIC MOTION, AND GRAVITATION
HOW TO SUCCEED IN PHYSICS – PART 1
1.
Build a Strong Foundation of Basic Concepts and Terminologies
Understanding
physics begins with a solid foundation of basic concepts and terminologies,
especially when starting with introductory physics. Akin to learning a new
language, mastering the “vocabulary” of physics — the basic physics
concepts — is essential. This includes understanding key principles such as
Newton’s Laws, the law of conservation of energy, and the concept of entropy,
to name a few.
2.
Brush up on your math
Sometimes
the root of your physics problems—as was the case with me—lies in rusty or even
non-existent math skills. And, unfortunately, the further you advance in
physics the more you’ll encounter mathematical processes you thought you’d long
since left behind you. (Yeah, I’m talking about you, algebra, trigonometry, and
calculus.)
3.
Have a List of Basic Physics Equations as a Resource
In
physics, equations are the tools that translate the language of natural
phenomena into quantifiable terms. Having a quick reference sheet of these
equations helps you prepare for any problem that may come up in your physics
class or an exam.
4.
Think conceptually
Whether
you’re searching for a unified field theory or studying the basic dynamics of
force and motion, taking a conceptual approach to learning physics can
help you break down your problem into a series of basic steps. Focus on the
fundamentals, but keep the big picture in the back of your mind. Bringing
together concepts into a unified understanding of the problem at hand makes
finding a solution much easier.
……..
Motion
Motion occurs when a body covers distance with time. The quantity of motion is the product of mass of a body and the speed at which it moves. Momentum represents the quantity of motion and it is the time rate of change of momentum that determines which keeps-the body moving. Through ages man has observed motions of various bodies both in space and on smooth and rough planes. Out of these observations some laws that govern motion in general. It was Isaac Newton who formulated the three laws we now call Newton’s, laws of motion. By applying these laws with certain conditions motion problems can be solved.
4.2 Laws of motion
(a) First law of motion
“A body remains in a state of rest or uniform motion in a straight line unless acted upon by the external force”.
This law is sometimes referred to as the law of inertia. Inertia means reluctance of a body to be set into motion or to stop if already moving. This inertia depends very much on the mass the body possesses? A body with less mass has small inertia and vice versa. On the other hand mass is the measure of inertia of a given body. The greater the mass of the body the less the acceleration when an external force is applied.
Acceleration
a
If a force F is applied on a body of mass M1, and then on another body of mass M2 as shown in figure, the corresponding accelerations a1 and a2 are related by
/ = / OR / = /
Figure 4.1
Mass is an inherent property of a body. It is independent of its surroundings and the method used in measuring it. Under the first law if there is no external force, a Stationary body is supposed to remain in one position forever; likewise a body in motion continues moving along the same direction indefinitely. However in reality when a body is pushed along a horizontal plane, it just moves for a short time and stops. If the plane is polished and the same body pushed over it, it moves for a little longer time before coming to rest.
(b) Second law of motion.
“The rate of change of momentum of a body is directly proportional to the external force applied and takes place in the direction of the force”We can use the above statement to derive an expression for the force that keeps the body accelerating or decelerating. To do this let us consider the momentum and momentum change of the body under the influence of the force.
Momentum
Momentum is the product of mass of a moving body and the velocity at which it moves Momentum = mass x velocity
The unit of momentum is a
Momentum change
For a solid body, what can possibly change during motion is the velocity because the mass remains constant. If initially a body moves at a velocity V and after sometime t the velocity changes to v then its momentum changes from initial value P, to the final value
Pf as illustrated below
Figure 4.2
The change in momentum = Final momentum – Initial momentum
The time rate of change in momentum = Change in momentum/Time interval =
It is this rate of change of momentum which is proportional to the applied force F.
Since and
The unit of force is Newton by definition:
A Newton is a force that makes a mass of 1kg to move with an acceleration of 1 ms-2
Which means that, when m = 1kg and, the force F = 1N
Substituting for m, a and F we have
And therefore
Thus the expression for the force is given by
(c) The third law
For every action there is an equal and opposite reaction
Fig 4.3 gives an illustration on the third law of motion. A body rest on the top reaction of the table exerts the weight W in return the table through the point of contact supports it by exerting a reaction force R equal to that of W only that the two are acting in opposite direction
Figure 4.3
4.3 Implications of Newton’s laws of motion
The laws we have seen above can be used to explain some of the occurrences that we encounter in daily life. Things like friction, impulse, weightlessness, collisions, motion in fluids and so on. We shall consider few of them in general,In common cases and try to use the appropriate laws to explain the situation.
(a) Friction and Frictional force
Friction is the rubbing between two bodies in contact when moving relative to one another or stationery. If the surfaces in contact are rough there is opposition that takes place giving rise to the friction force. To illustrate this consider a block resting on a plane as in fig 4.4. The molecular moles sticking out of each surface interlock when the surfaces come in contact. The extent of interaction between these protrusions depends on the weight a block exerts on the plane. When the action force F is applied to the body with an intention of pulling it, the obstruction by the interlocking protrusions translates into a friction ‘f’ force / as in fig 4.4.
figure 4.4
To maintain the motion of the block, the applied force F must be greater than factional force “ƒ” such that the difference (F – ƒ) produces the acceleration ‘a’
This means the applied force partly goes into overcoming frictional force and partly accelerates the body. The difference (F – ƒ) is sometimes referred to as the effective force that causes acceleration. If the molecular protrusions are crashed and leveled, the surfaces become smoothed and the opposition to motion is drastically reduced to the extent that very little force is required to keep the body moving. In addition if the surface is polished, frictional force is further minimized to the level where it can be regarded as smooth. In solving mechanics problems the terms smooth or frictionless are used to mean that the frictional force is zero. When there is no frictional force the applied force wholly goes into accelerating the body.
(ii) Relationship between friction force and normal reaction
There is relationship between the friction “f ” and the normal reaction R. The more is the reaction force , the greater the force of friction in another words friction force is direct proportion to the reaction. The ratio of the friction force to the normal reaction is called the coefficient of friction.
Coefficient of friction (μ) =
From which,
(iii) Coefficient of static friction
A body resting on the plane does not begin to move until the external force is applied to overcome friction force. The maximum friction force that is to be overcome by applied force is sometimes referred to as limiting force
=R
One of the methods for determining the value of is shown in fig 4.5.
figure 4.5
In this experimental a block of known mass is placed on the horizontal plane whose coefficient of friction is sought. The plane consists of a frictionless pulley at one end. A string with a pan at one end is tied to the block and made to pass over the pulley. Before adding anything on the pan, the only forces on the plane are the weight “W” of the block and the reaction R from the plane. As soon as the mass is added to the pan, the tension “T” created in the string tends to pull the block and at the same time the frictional force takes charge. Initially the weight “W” arising from the mass on the pan may not be strong enough to move the block but as more and more masses are placed on the pan the frictional force grow bigger and bigger. The time comes, when the frictional force can no longer hold the block and therefore block just begins to slide. It is at this moment we say that the tension “T” is just equal to the frictional force “”. This is the maximum frictional force the plane can offer in preventing the block to slide over it, By recording the total mass on the pan, the magnitude of the weight W’ can be found and from the equation
Where T = W’ and W’ = g the numerical value for is evaluated.
Definition
The coefficient of static friction is the ratio of the frictional force when the body is on the average slide.
Once the block has gain momentum, the coefficient of friction is now referred to as the coefficient of kinetic friction and its value is less than that of . Table 4.1 shows the value of and for the surface of different materials.
Table 4.1: coefficients of friction for some materials
| Surface in contact | ||
| Steel on the steel | 0.74 | 0.57 |
| Copper on steel | 0.61 | 0.47 |
| Aluminium on steel | 0.53 | 0.36 |
| Rubber on concrete | 1.00 | 0.80 |
| Wood on wood | 0.25-0.50 | 0.20 |
| Glass on glass | 0.94 | 0.40 |
| Ice on ice | 0.1 | 0.03 |
(b)Motion on the horizontal plane
Here we shall consider a body sliding along rough plane as a general case. Being rough, the plane offers frictional force to the body on sliding it. If the body is just projected with initial velocity u , its velocity keep on decreasing until it come to the halt due to the frictional force that opposes motion persistently as in fig 4.6
figure 4.6
The question arising from this example would be;
How far does the body go?
What is the deceleration?
What is the frictional force?
How far does the body go?
To find how far the body reaches we can start with the third equation of linear motion in which the acceleration
such that; = =
Where R = mg and therefore S = mu2 / µmg or
(b) Motion on a rough inclined plane
Consider a block projected up the inclined plane with an initial velocity as it climbs, there are two forces which opposes the motion, one being frictional force and other is the component of weight parallel to the plane as shown in fig 4.7
figure 4.7
The two forces opposing the motion simultaneously bring the block to a standstill somewhere along the plane after traveling a distance S. To find the magnitude of the distance covered we can use the same method as in the case of the horizontal plane only that the component of the block has to be
considered. The total force opposing the motion is
Where F = ma, and
a = gsin
a =
From the third equation of linear motion,
Note:
Motion in viscous fluids.
1.
Where v = 0 (given), a = (
From which
If the inclined plane is smooth i.e. there is only one opposing force and that is and the distance becomes
(C)Motion of connected bodies
(i) On smooth horizontal plane
Fig 4.8(a) shows two bodies A and B of masses and respectively connected by the light in extensible string that passes over a frictionless pulley such that the body A rest on the smooth surface and B hang freely.
figure 4.8
As soon as the system is let free and given that m2 > m1, body B falls under its own weight W2 pulling body A via a string and both move with the common acceleration as shown in fig 4.8(b). The tension in the string is uniform throughout. Using the free-body diagram it is easy to determine both the tension and the acceleration of the system. The process is as follows:
Body B falls vertically downwards because
But
………………(1)
Body A moves horizontally in direction of T without opposing because the surface is smooth
……………………… (2)
Solving equations (1) and (2) we have
Substituting for in equation (2) we get
(ii)On rough inclined plane
figure 4.9
For a rough plane frictional force play part by opposing the motion of body A as the system is free to move. Equation (i) above remains the same
…………………(3)
Due to the frictional force equation (2) becomes
But
……………………(4)
Add (3) and (4)
(
+
Substitute for in equation (4) and simplify
T
(iii) Connected bodies on inclined plane
Consider a situation where two bodies connected by an extensible string that passes over a frictionless pulley rest on two planes inclined at an angle to the horizontal as shown in fig 4.10(a)
Figure 4.10 (a) Figure 4.10 (b)
When let it free, the bodies may move as in fig 4.10(b) provided and the surface of inclined plane are alike i.e their coefficients of friction are equal. The forces involved are indicated their direction and as a results both tension T and acceleration of the system can be found using similar procedures as in the case of horizontal plane. Since body A climbs the plane it implies that
T i.e
But, and
………(1)
Likewise since body B moves down the plane it means
a
Where
and
Or
————(2)
Add (1) and (2) to obtain
(
Substitute for in (2) and simplify to get
(iv)Connected bodies on pulley
Again consider two unequal bodies of masses connected by a light inextensible string that passes over a frictionless pulley as in fig 4.11
Figure 4.11 (a) Figure 4.11 (b)
In figure 4.1 1(a) the masses are mounted on a pulley with m2 > m1. When released, the weight of the larger mass overcomes that of smaller mass and the motion takes place as in figi4.1 l(b). Using the observation and reasoning we can determine the acceleration”a” of the system ‘and the tension “T” in the string. For the mass moving downwards W2>T and therefore;
For the mass moving vertically upwards T > W1 and hence
Adding (1) and (2) we have
Since and
From (1)
(v)Motion in a lift
If you have been in a lift, you may have felt a difference in your weight as a lift ascends or descends with constant acceleration. When ascending one feel heavier than normal by pressing hard on the lift floor and when descending, one becomes apparently lighter than normal by losing weight.
To get the idea of what happens consider a body of mass m on a pan of a weighing machine on the floor in a lift. Before the motion along the vertical plane begins, the weight of the body is taken to be normal, say W as in fig 4.12.
Figure 4.12(a) (b) (c)
When a lift accelerates uniformly downwards, the scale of a weighing machine shows that there is a drop in weight by indicating a lower value W1 as in fig 4.12(b) and when it accelerates upwards the machine indicates a higher value in weight Was in fig 4.12(c). To explain these observations, let us consider the body on the scale pan in each case separately.
In the body exerts the normal weight W on the pan and in return the pan exerts force of figure 4.12 (a) reaction R as shown in fig 4.13. Since the lift is at rest, meaning that
Figure 4
But and therefore
figure 4.13
In the lift accelerates uniformly vertically downwards, this causes the apparent loss in weight meaning that the weight W1 indicated on the scale equals to the reaction R1 such that R1 as in fig 4.14
Figure 4.14.
The apparent loss in weight as the lift goes down is
The reason for the apparent loss in weight in a body is that the surface on which the body is resting runs away or falls faster in such a way that the contact between; the pan and the body becomes lighter and hence less reaction from the surface. In fact if the lift falls with acceleration equal to that of gravity, the reaction on the body from the surface would be zero i.e. when a = g, .R, = m(g – g) = 0 and the pointer on the scale would indicate zero weight. Under such condition the body would appear weightless.
In figure 4.12(c), the lift accelerates vertically upwards causing the scale to indicate a larger weight W2. The apparent increase in weight as the lift ascends arises from the fact that the pan on which the body sits tends to rise faster the body itself and therefore pressing harder underneath it and hence the reaction R2 as in fig 4.15. The effect of the pan pressing hard on the body is the apparent gain in weight W2 registered on the scale. The apparent gain in weight is given as
Or
Figure. 4 .15
If the lift ascends with acceleration equals to that of the gravity, the reaction R2 would double. That is if , then R1 =m (
4.4 Collision
A collision is the process in which two or more bodies suddenly smash into each other. An impact at the point of collision causes an impulse on each of the colliding bodies’ results into change in momentum. There are mainly two types of collision.
a) Elastic collision.
b) Inelastic collision.
(a)Elastic collision
The collision whereby the colliding bodies take very short time to separate is known as elastic collision. In this kind of collision, both the momentum and kinetic energy are conserved. Fig 4.18 illustrates the collision of two spherical bodies with masses and m2 initially moving with velocities and respectively where. At the point of collision, the rear body exerts a force on the front body and at the same time the front body exerts an equal but opposite force -F2 on the rear body. After collision the rear body slows down to, velocity whereas the front body picks up the motion attaining velocity.
Figure 4 .17
Impulse,
The impulse of a force is the product of force applied and time interval remain in action, that is
The unit of impulse is Newton-second (Ns). From the Newton’s second law of motion we have also seen that
And therefore. This means the impulse is equal to the change in momentum
This means that the impulse is a kilogram-meter per second ()
Principle of conservation of linear momentum:
“In system of colliding particles, the total momentum before collisions is equal to the total momentum after collision so long as there is no interference to the system”
From Newton’s third law of motion, action and reaction are equal but opposite.
For example at the point of impact in fig 4.18
Since the action and reaction are taken at an equal interval of time to remain in action each body experiences the same impulse that is
Where causes change in momentum on and causes change in momentum on such that ) and
Or
Collecting initial terms together and final terms together we have
Equation (4.27) summarizes principle of conservation of momentum.
Conservation of kinetic energy:
“Work is done when the force moves a body through a distance, In motion the work done is translated into a change kinetic energy as it can be shown from the second law of motion and third equation of linear motion”.
and
But from
In the case of collision we talk in terms virtual distance and therefore virtual work done the forces of action and reaction.
The virtual work done on by is given by
Likewise the virtual done on;
by is and hence
–
Collecting the initial quantities together on one side and the final quantities together on the other side we get
Equation (4.29) is the summary of the conservation of kinetic energy in a system of colliding particle provided the collision is perfectly elastic.
(b)Inelastic collision
There are certain instances whereby the colliding bodies delay in separating after collision has taken place and at times they remain stuck together. Delaying to separate or sticking together after collision is due to in elasticity and hence elastic collision. In this case it only the momentum which is conserved but not kinetic energy. The deformation that takes place while the bodies are exerting onto each other in the process of colliding, results into transformation of energy from mechanical into heat and sound the two forms of energy which are recoverable. Once the energy has changed into heat we say that it has degenerated, it is lost to the surroundings. When the two bodies stick together after impact they can only move with a common velocity and if they do not move after collision then the momentum is .said to -have been destroyed.
Figure 4.18
Coefficient of restitution
One of the measures of elasticity of the body is the ratio of the different in velocity after and before the collision. Before colliding, the space between the particles decreases as the rear body overtakes that in front but after collision the space between them widens as the front particle run away from the rear one. The difference in velocity before collision is called velocity of approach and that after collision is called velocity of separation. The ratio of velocity of separation to velocity of approach is known as coefficient of restitution.
Let coefficient of restitution, (velocity of separation, velocity of approach.
For perfectly elastic collision, in this case perfectly inelastic collision. But collision result into explosion, otherwise in normal circumstances, 0 the coefficient of restitution cannot be 1 due to the fact that it does not matter how hard the colliding bodies are they always undergo deformation at the moment of impact and hence take longer to recover to their original shape while separating. We only assume to make calculation simpler.
4.5 Oblique collision
In the previous discussion on collision we dealt with direct impingement of one body onto the other along the line joining their common center. However there are situations in which bodies collide at an angle. This is known as oblique collision. Fig 4.19 illustrates the oblique collision of two bodies of mass and initially moving at velocities u1 and u2 respectively in the x-direction. After collision the body in front moves along the direction making an-angle with the initial direction whereas the rear body goes in the direction making an angle with initial direction. Given the initial conditions of the colliding bodies, the final velocities and directions after collision can be found
Figure 4.19
Applying the principle of conservation of linear momentum in equation (4.27) we can come up with more,equations for solving problems on oblique considering the motion in x – and y – directions
(a)Motion along x-direction
Where
, ,
(b) Motion along y-direction
If initially the bodies are not moving in y-direction then
and
Which is
Applying the definition of coefficient of restitution in equation (4.30), we have
Or
………………………………………(4.33)
4.6 The ballistic balance
Ballistic balances are used in determining velocities of bullets as well as light comparison of masses. To do this a wooden block of mass M is suspended from light wires so that it hangs vertically. A bullet of mass m is fired horizontally towards a stationary block. If the bullet is embedded inside the block, the two swings together as a single mass this is inelastic collision. The block will swing until the wires make an angle θ with the vertical as in figure 4.20
Figure 4 .20 shows the ballistic pendulum.
Since the collision is inelastic, only the momentum is conserved. If and are the mass and initial velocity of the bullet and M, the mass and initial velocity of the block, then by principle of conservation of linear momentum.
From which
After impact the kinetic energy of the system at the beginning of the swing is transformed into gravitational potential energy at the end of the swing and therefore
Substituting for v in equation (4.34) we get
Thus the initial velocity of the bullet is found to be
In fig 4.21, suppose the length of the wire is before the block swings. After swinging, the center of gravity of the block rises by a distance reducing the vertical distance to. By forming the triangle of displacements the values of and can easily be found as shown in fig 4.21
Figure 4.21
The height is
……………………………………………….(4.37)
Or
The angle the wire makes with vertical is
………………………………………(4.38)
4.7 Reaction from a jet engine
The operation of a jet engine depends on the third law of motion where the escaping mass of hot gases exerts force on jet enabling it to move forward. Air is first sucked in through the front side then compressed, the oxygen contained in this air intake is used in burning the fuel producing gases which when expelled at a very high speed through the rear action forces are created and hence forward thrust . Fig 4.22 illustrate the principle of a jet in which the mass of air Ma is taken in at the rate of with relative velocity and passes through the engine at the rate with relative velocity of.The mass of gases produced at the rate of by combustion is ejected at a relative velocity .The total rate of change of momentum of the system is therefore given as
…………(4.39)
Figure : 4 .22 Jet engine
From Newton’s second law of motion, the rate of change of momentum is equal to force. Therefore equation (4.39) represents the forward thrust on the jet aircraft. Some jet aircraft have two identical engines and others have four. The total thrust is the product of number of engines and thrust of one engine.
4.8 Reaction from a rocket
Unlike the jet engine, the rocket carries all of its propellant materials including oxygen with it. Imagine a rocket that is so far away from gravitational influence of the earth, then all of the exhaust hot gases will be available for the propelling and accelerating the rocket. Fig 4.24 is an illustration of a rocket of mass m carrying the fuel of mass such that the total mass at time t is ( ) moving horizontally far away from the earth surface. As the fuel burns and gases formed expelled from the rocket at a velocity of .∨g after sometime , the mass of the rocket becomes m but its velocity increases to () as in fig.4.23(b) whilst the velocity of the ejected gases decreases from v to (
Figure. 4 .23
From the principle of conservation of linear momentum
(
Hence
Since the time rate of change in momentum is equal to thrust or force (F) on the rocket by the escaping mass of the gases the above relation can be written as
If the large thrust is to be obtained, the rocket designer has to make the velocity at which the hot gases are ejected and the rate at which the fuel is burnt high as possible.
4.9 Rocket moving vertically upwards
Let us consider the rocket fired vertically upwards from the surface of the earth as shown in fig 4.24
Figure. 4. 24
The thrust developed during combustion must be greater than the weight of the rocket if at all it is to accelerate vertically upwards which means that
4.10 Reaction from the hose pipe
If a hose pipe connected to the running tap on the smooth horizontal surface, the free end that issues water seems to move backwards as the water flow out. This is yet another example of action and reaction forces. Again if a jet of water from a horizontal hose pipe is directed at a vertical wall, it exerts an equal but opposite force on the water.
Figure. 4.25
Let be the initial velocity of water when leaving the pipe. On striking the wall its final velocity assuming that the water does not rebound. If is the density of water, A is the cross-sectional area of the pipe, then the mass of water hitting the wall per second is given by
Where
The time rate of change in momentum of water is therefore
Where and
i.e.
The negative sign means the force is the reaction of the wall on water. Thus the force exerted by the water on the wall is
4.11 Reaction on a gun
Consider a gun mass with bullet mass in it initially at rest. Before firing the gun, their total momentum is zero as in fig. 4.27(a). At the point of firing there are equal opposite internal forces as in fig 4.27(b). As the bullet leaves the gun the total momentum of the system is still zero as in fig 4.27(c).
Figure. 4. 26
Initially the velocity of the gun and that of the bullet are zero
Therefore the initial momentum
(0)
When the bullet leaves the gun with final velocity, the gun recoil with velocity of – and the final total momentum
From the principle of conservation of linear momentum
Thus
4.12 Equilibrant forces
A body is said to be in static equilibrium if it does not move under the action of external forces. For example in fig 4.3, a block is in equilibrium since it neither moves up nor down under forces R and W. These two forces are action and reaction which cancel each other out such that the net force on the body is zero. The net external force is an algebraic sum of all the forces acting on the body that is
In this case
The forces that keep the body in equilibrium are called equilibrant forces. These are the forces whose resultant is zero. Fig 4.28(a) shows a body in equilibrium
under the action of three forces, hanging vertically. The weight W of the body establishes the tensions and in the sections AB and BC of the string which make angles and respectively with the horizontal point B along the string experiences three forces as shown In fig 4.28(b). If this point is taken as an origin and two perpendicular axes drawn, the tensions and appear to make angles and with the x – axis. The body remains in equilibrium when no motion occurs either horizontally or vertically and for that reason the net force along the horizontal direction is zero
Likewise the net force on the body along the vertical direction is zero
To obtain the net forces we have to find the x- and y-components of tensions and as shown in fig 4.28 (C). For the horizontal components of two tension gives
Where
For the y-direction the net force
Where
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