Newtons laws of motion

NEWTON'S LAWS OF MOTION
4.1 Motion
Motionoccurs when a body covers distance with time. The quantity of motionis 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 withcertainconditions motionproblems canbe solved.
4.2 Laws ofmotion
(a) Firstlaw ofmotion
“A body remains in a state of rest or uniform motion in a straight line unless acted upon
by the externalforce”.
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
accelerationwhenan external forceis 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 correspondingaccelerationsa1 anda2 are relatedby
/ = / 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 commoncases and try to use the appropriate laws to
explainthe situation.
(a) FrictionandFrictional force

Frictionis the rubbing between two bodies incontact whenmoving 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 infig4.4.
figure4.4
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.
Whenthere is no frictionalforce theappliedforcewhollygoesintoaccelerating thebody.
(ii)Relationshipbetweenfrictionforceandnormal 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 offriction (μ) =
From which,
(iii) Coefficient ofstaticfriction
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: coefficientsoffrictionfor some materials
Surface in contact
Steel onthe 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
Woodon wood 0.25-0.50 0.20

Glass on glass 0.94 0.40
Ice onice 0.1 0.03
(b)Motiononthe 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 motionpersistentlyas 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 opposingthe motionis
Where F = ma, and
a = gsin
a =
Fromthethirdequationoflinearmotion,
Note:
Motioninviscousfluids.
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 onthe smoothsurface 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 fig4.8(b).

The tension in the string is uniform throughout. Using the free-body diagram it is easy to
determineboththetension andthe accelerationofthe system. The processisas follows:
BodyB falls verticallydownwards because
But
………………(1)
Body A moves horizontallyin directionof T without opposingbecause the surface is smooth
……………………… (2)
Solving equations (1) and (2) we have
Substituting for in equation(2) we get
(ii)Onrough inclinedplane

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 stringthat passes over a frictionlesspulleyas in fig4.11

Figure4.11 (a) Figure4.11 (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 verticallyupwards T > W1 and hence
Adding (1) and (2) we have

Since and
From (1)
(v)Motioninalift
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 ideaof 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
takento be normal, say W as in fig 4.12.
Figure4.12(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
explainthese observations, let us considerthe body onthe scalepanin eachcase separately.
In the bodyexerts the normal weight W on the pan and in returnthe panexerts force of figure
4.12 (a) reactionR as shownin fig4.13. Sincethe liftis at rest, meaningthat
Figure 4
But and therefore
figure4.13
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 infig4.14
Figure 4.14.

The apparent lossinweight as the liftgoes downis
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 conditionthe 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 hardon 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 accelerationequals 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 collisionof 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 seenthat
And therefore . This means the impulse is equal to the change in momentum
This means that the impulse is a kilogram-meter per second( )
Principleofconservationoflinear momentum:

“In system of colliding particles, the total momentum before collisions is equal to the
total momentum after collisionso longas there is no interference to the system”
From Newton’s third law of motion, actionand reactionare 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
actioneach body experiences the same impulse that is
Where causes change in momentum on and causes change in momentum
on such that ) and
Or
Collectinginitial terms together andfinal terms together we have
Equation (4.27) summarizes principle of conservationof momentum.
Conservationofkinetic 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
secondlaw of motionand 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 providedthe collisionis perfectlyelastic.
(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 whichis 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
momentumis .saidto -have beendestroyed.
Figure 4.18
Coefficientofrestitution
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
velocityofseparationtovelocityofapproachis knownas coefficientofrestitution.
Let coefficient of restitution, ( velocityof 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
calculationsimpler.
4.5 Obliquecollision
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. Fig4.19 illustrates the oblique collisionof
two bodies of mass and initially moving at velocities u1 and u2 respectively in the x-

direction. After collision the bodyin 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 collisioncanbe found
Figure4.19
Applying the principle of conservation of linear momentum in equation
(4.27) we can come up with more,equations for solving problems on
oblique consideringthe motioninx - 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
Whichis
Applying the definitionof coefficient ofrestitutioninequation(4.30), we have
Or
………………………………………(4.33)
4.6 The ballistic balance
Ballistic balances are usedin determining velocities of bullets as well as light comparisonof
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 infigure 4.20

Figure4 .20 shows the ballisticpendulum.
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 potentialenergyat the endof the swingand 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 Reactionfrom ajet 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 . Fig4.22 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 givenas
…………(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 ejectedgases 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 ejectedandthe rate at which the fuel is burnt high as possible.
4.9 Rocket moving verticallyupwards
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 Reactionfrom 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 velocityof water whenleaving the pipe. Onstriking the wall its final velocity
assuming that the water does not rebound. If is the density of water, A is the cross-
sectional areaof thepipe, thenthe massofwater hittingthe wall per secondis givenby
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
exertedby the water on the wall is
4.11 Reactionona 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 velocityof 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 conservationof 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 actingonthe bodythat 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 alongthe horizontal directionis zero

Likewise the net force onthe bodyalong the vertical directionis zero
To obtain the net forces we have to find the x- and y-components of tensions and as
shown in fig4.28 (C). For the horizontal components oftwo tensiongives
Where
For the y-directionthe net force
Where

Figure. 4. 27
Solving for interms of from equations (4.43)and(4.44 we get
Exercise4.0
4.1
(a) State Newton's laws of motion
(b) Give three examples in which Newton's thirdlaw applies
(c) Withthe aidof labeleddiagram explainwhat causes frictional force.
4.2 (a) A bodyof mass m rests onaroughinclinedplanewithangle of inclination
(i) Explain why the body does not slide down the plane,
(ii)Draw the diagram indicating all the forces acting on the body and give each
force itsname, directionandmagnitude.
(b) If the mass of the body in(a) is 5kg and the angle of inclinationis 20° then find
(i) The force that keeps the body in contact with the plane (ii) The
force that prevents the bodyto slide down the plane.

4.3(a) Differentiatebetween
(i) A rough surface and a smoothsurface
(ii) Static frictionand kinetic friction
(iii) Coefficient of static frictionand coefficient of dynamic friction .
(b) A body of mass 20kg is pulled by a horizontal force P. If it accelerates at 1.5
and the coefficient of friction of the plane is 0.25, what is the magnitude of P?
(Take
4.4 (a) From thesecondlawof motionshowthe expressionfortheforce.
, where and are the mass and accelerationrespectively
(b)Using the third law of motion, show that for the two colliding bodies of masses m{ and m2
moving along the common line at velocities u1 and u2 , before collision and at velocities
v1and v2 respectively just after collision, the total momentum of the system is conserved
and representedbyrelation
. Assume elastic collision.
4.5 (a) Fig 4.28 shows two bodies of masses M and m connected by the light string,
if the bodyof mass m rests on a rough plane whose coefficient of friction is μ on releasing the
system free, obtainanexpressionfor
(i) The accelerationof the system.
(ii) The tensioninthe string.

Figure. 4 .28
4.6 (a) (i) What is the difference between the coefficient of restitution and the coefficient
of friction?
(ii) Explain in details the implications of the following about the
coefficient of restitution e
- when e> 1
- when e =0
- when e< 1
- when = l
(b) Two bodies A and B of masses 3kg and 2.5kg are moving towards each other
along a common line with initial velocities 4 and2.5 respectively, after
sometime theyeventuallycollide elastically.
Determinetheirfinal velocities.
4.7 (a) A body is hanging in equilibrium as shown in fig 4.29, findthe tensions
, and .Take

Figure. 4.
29
(b) A body of mass 2kg sits on a horizontal plane, and then the plane is accelerated
vertically upwards at 4ms-2. Determine the magnitude of the reactionon the body by
the plane.
4.8 The engine of a jet aircraft flying at 400 takes in 1000m3 of air per
second at an operating height where the density of air is 0.5 , The air is used
to burn the fuel at the rate of 50 and the exhaust gases (Including incoming
air) are ejectedat 700 relative to the aircraft. Determine thethrust.

Newtons laws of motion

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Newtons laws of motion