MOTION IN A PLANE 87
strokes normal to the river current? How far down the river does he go when he
reaches the other bank ?
4.14 In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat
anchored in the harbour flutters along the N-E direction. If the boat starts moving at a
speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ?
4.15 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that
a ball thrown with a speed of 40 m s can go without hitting the ceiling of the hall ?
-1
4.16 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much
high above the ground can the cricketer throw the same ball ?
4.17 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a
constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and
direction of acceleration of the stone ?
4.18 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900
km/h. Compare its centripetal acceleration with the acceleration due to gravity.
4.19 Read each statement below carefully and state, with reasons, if it is true or false :
(a) The net acceleration of a particle in circular motion is always along the radius of
the circle towards the centre
(b) The velocity vector of a particle at a point is always along the tangent to the path
of the particle at that point
(c) The acceleration vector of a particle in uniform circular motion averaged over one
cycle is a null vector
4.20 The position of a particle is given by
ˆ
ˆ
2 ˆ
k
i
r = 3.0t − 2.0t j + 4.0 m
where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the v and a of the particle? (b) What is the magnitude and direction of
velocity of the particle at t = 2.0 s ?
4.21 A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in
the x-y plane with a constant acceleration of (8.0i + 2.0 ) j m s . (a) At what time is
-2
the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at
that time? (b) What is the speed of the particle at the time ?
4.22 and j are unit vectors along x- and y- axis respectively. What is the magnitude
i
and direction of the vectors i + , and i − ? What are the components of a vector
j
j
A= 2 i + 3 along the directions of i + and i − ? [You may use graphical method]
j
j
j
4.23 For any arbitrary motion in space, which of the following relations are true :
(a) v = (1/2) (v (t ) + v (t ))
average 1 2
(b) v = [r(t ) - r(t ) ] /(t – t )
average 2 1 2 1
(c) v (t) = v (0) + a t
(d) r (t) = r (0) + v (0) t + (1/2) a t 2
(e) a =[ v (t ) - v (t )] /( t – t )
average 2 1 2 1
(The ‘average’ stands for average of the quantity over the time interval t to t )
1 2
4.24 Read each statement below carefully and state, with reasons and examples, if it is
true or false :
A scalar quantity is one that
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes.
4.25 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at
a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the
speed of the aircraft ?
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Additional Exercises
4.26 A vector has magnitude and direction. Does it have a location in space ? Can it vary
with time ? Will two equal vectors a and b at different locations in space necessarily
have identical physical effects ? Give examples in support of your answer.
4.27 A vector has both magnitude and direction. Does it mean that anything that has
magnitude and direction is necessarily a vector ? The rotation of a body can be specified
by the direction of the axis of rotation, and the angle of rotation about the axis. Does
that make any rotation a vector ?
4.28 Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane
area, (c) a sphere ? Explain.
4.29 A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By
adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume
the muzzle speed to be fixed, and neglect air resistance.
4.30 A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes
directly overhead an anti-aircraft gun. At what angle from the vertical should the gun
be fired for the shell with muzzle speed 600 m s to hit the plane ? At what minimum
-1
-2
altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m s ).
4.31 A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the
road of radius 80 m, he applies brakes and reduces his speed at the constant rate of
0.50 m/s every second. What is the magnitude and direction of the net acceleration of
the cyclist on the circular turn ?
4.32 (a) Show that for a projectile the angle between the velocity and the x-axis as a function
of time is given by
v − gt
1 - 0y
θ ( ) = tant
v ox
(b) Shows that the projection angle θ for a projectile launched from the origin is
0
given by
4h m
1 -
θ
0 = tan
R
where the symbols have their usual meaning.
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CHAPTER FIVE
LAWS OF MOTION
5.1 INTRODUCTION
In the preceding Chapter, our concern was to describe the
motion of a particle in space quantitatively. We saw that
5.1 Introduction uniform motion needs the concept of velocity alone whereas
5.2 Aristotle’s fallacy non-uniform motion requires the concept of acceleration in
5.3 The law of inertia addition. So far, we have not asked the question as to what
governs the motion of bodies. In this chapter, we turn to this
5.4 Newton’s first law of motion
basic question.
5.5 Newton’s second law of
Let us first guess the answer based on our common
motion
experience. To move a football at rest, someone must kick it.
5.6 Newton’s third law of motion
To throw a stone upwards, one has to give it an upward
5.7 Conservation of momentum
push. A breeze causes the branches of a tree to swing; a
5.8 Equilibrium of a particle
strong wind can even move heavy objects. A boat moves in a
5.9 Common forces in mechanics flowing river without anyone rowing it. Clearly, some external
5.10 Circular motion agency is needed to provide force to move a body from rest.
5.11 Solving problems in Likewise, an external force is needed also to retard or stop
mechanics motion. You can stop a ball rolling down an inclined plane by
applying a force against the direction of its motion.
Summary
In these examples, the external agency of force (hands,
Points to ponder
wind, stream, etc) is in contact with the object. This is not
Exercises
always necessary. A stone released from the top of a building
Additional exercises
accelerates downward due to the gravitational pull of the
earth. A bar magnet can attract an iron nail from a distance.
This shows that external agencies (e.g. gravitational and
magnetic forces ) can exert force on a body even from a
distance.
In short, a force is required to put a stationary body in
motion or stop a moving body, and some external agency is
needed to provide this force. The external agency may or may
not be in contact with the body.
So far so good. But what if a body is moving uniformly (e.g.
a skater moving straight with constant speed on a horizontal
ice slab) ? Is an external force required to keep a body in
uniform motion?
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5.2 ARISTOTLE’S FALLACY true law of nature for forces and motion, one has
to imagine a world in which uniform motion is
The question posed above appears to be simple.
However, it took ages to answer it. Indeed, the possible with no frictional forces opposing. This
correct answer to this question given by Galileo is what Galileo did.
in the seventeenth century was the foundation 5.3 THE LAW OF INERTIA
of Newtonian mechanics, which signalled the Galileo studied motion of objects on an inclined
birth of modern science. plane. Objects (i) moving down an inclined plane
The Greek thinker, Aristotle (384 B.C– 322
B.C.), held the view that if a body is moving, accelerate, while those (ii) moving up retard.
is an
horizontal
plane
(iii)
a
Motion
on
something external is required to keep it moving. intermediate situation. Galileo concluded that
According to this view, for example, an arrow an object moving on a frictionless horizontal
shot from a bow keeps flying since the air behind plane must neither have acceleration nor
the arrow keeps pushing it. The view was part of retardation, i.e. it should move with constant
an elaborate framework of ideas developed by velocity (Fig. 5.1(a)).
Aristotle on the motion of bodies in the universe.
Most of the Aristotelian ideas on motion are now
known to be wrong and need not concern us.
For our purpose here, the Aristotelian law of
motion may be phrased thus: An external force
is required to keep a body in motion. (i) (ii) (iii)
Aristotelian law of motion is flawed, as we shall
Fig. 5.1(a)
see. However, it is a natural view that anyone Another experiment by Galileo leading to the
would hold from common experience. Even a same conclusion involves a double inclined plane.
small child playing with a simple (non-electric) A ball released from rest on one of the planes rolls
toy-car on a floor knows intuitively that it needs down and climbs up the other. If the planes are
to constantly drag the string attached to the toy- smooth, the final height of the ball is nearly the
car with some force to keep it going. If it releases same as the initial height (a little less but never
the string, it comes to rest. This experience is greater). In the ideal situation, when friction is
common to most terrestrial motion. External absent, the final height of the ball is the same
forces seem to be needed to keep bodies in as its initial height.
motion. Left to themselves, all bodies eventually If the slope of the second plane is decreased
come to rest. and the experiment repeated, the ball will still
What is the flaw in Aristotle’s argument? The reach the same height, but in doing so, it will
answer is: a moving toy car comes to rest because travel a longer distance. In the limiting case, when
the external force of friction on the car by the floor the slope of the second plane is zero (i.e. is a
opposes its motion. To counter this force, the child horizontal) the ball travels an infinite distance.
has to apply an external force on the car in the In other words, its motion never ceases. This is,
direction of motion. When the car is in uniform of course, an idealised situation (Fig. 5.1(b)).
motion, there is no net external force acting on it:
the force by the child cancels the force ( friction)
by the floor. The corollary is: if there were no friction,
the child would not be required to apply any force
to keep the toy car in uniform motion.
The opposing forces such as friction (solids)
and viscous forces (for fluids) are always present
in the natural world. This explains why forces
by external agencies are necessary to overcome
the frictional forces to keep bodies in uniform
motion. Now we understand where Aristotle Fig. 5.1(b) The law of inertia was inferred by Galileo
went wrong. He coded this practical experience from observations of motion of a ball on a
in the form of a basic argument. To get at the double inclined plane.
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LAWS OF MOTION 91
In practice, the ball does come to a stop after accomplished almost single-handedly by Isaac
moving a finite distance on the horizontal plane, Newton, one of the greatest scientists of all times.
because of the opposing force of friction which Newton built on Galileo’s ideas and laid the
can never be totally eliminated. However, if there foundation of mechanics in terms of three laws
were no friction, the ball would continue to move of motion that go by his name. Galileo’s law of
with a constant velocity on the horizontal plane. inertia was his starting point which he
Galileo thus, arrived at a new insight on formulated as the first law of motion:
motion that had eluded Aristotle and those who Every body continues to be in its state
followed him. The state of rest and the state of of rest or of uniform motion in a straight
uniform linear motion (motion with constant line unless compelled by some external
velocity) are equivalent. In both cases, there is force to act otherwise.
Ideas on Motion in Ancient Indian Science
Ancient Indian thinkers had arrived at an elaborate system of ideas on motion. Force, the cause of
motion, was thought to be of different kinds : force due to continuous pressure (nodan), as the force
of wind on a sailing vessel; impact (abhighat), as when a potter’s rod strikes the wheel; persistent
tendency (sanskara) to move in a straight line(vega) or restoration of shape in an elastic body;
transmitted force by a string, rod, etc. The notion of (vega) in the Vaisesika theory of motion perhaps
comes closest to the concept of inertia. Vega, the tendency to move in a straight line, was thought to
be opposed by contact with objects including atmosphere, a parallel to the ideas of friction and air
resistance. It was correctly summarised that the different kinds of motion (translational, rotational
and vibrational) of an extended body arise from only the translational motion of its constituent
particles. A falling leaf in the wind may have downward motion as a whole (patan) and also rotational
and vibrational motion (bhraman, spandan), but each particle of the leaf at an instant only has a
definite (small) displacement. There was considerable focus in Indian thought on measurement of
motion and units of length and time. It was known that the position of a particle in space can be
indicated by distance measured along three axes. Bhaskara (1150 A.D.) had introduced the concept
of ‘instantaneous motion’ (tatkaliki gati), which anticipated the modern notion of instantaneous
velocity using Differential Calculus. The difference between a wave and a current (of water) was clearly
understood; a current is a motion of particles of water under gravity and fluidity while a wave results
from the transmission of vibrations of water particles.
no net force acting on the body. It is incorrect to The state of rest or uniform linear motion both
assume that a net force is needed to keep a body imply zero acceleration. The first law of motion can,
in uniform motion. To maintain a body in therefore, be simply expressed as:
uniform motion, we need to apply an external If the net external force on a body is zero, its
force to ecounter the frictional force, so that acceleration is zero. Acceleration can be non
the two forces sum up to zero net external zero only if there is a net external force on
force. the body.
To summarise, if the net external force is zero,
Two kinds of situations are encountered in the
a body at rest continues to remain at rest and a
body in motion continues to move with a uniform application of this law in practice. In some
velocity. This property of the body is called examples, we know that the net external force
inertia. Inertia means ‘resistance to change’. on the object is zero. In that case we can
A body does not change its state of rest or conclude that the acceleration of the object is
uniform motion, unless an external force zero. For example, a spaceship out in
compels it to change that state. interstellar space, far from all other objects and
with all its rockets turned off, has no net
5.4 NEWTON’S FIRST LAW OF MOTION external force acting on it. Its acceleration,
Galileo’s simple, but revolutionary ideas according to the first law, must be zero. If it is
dethroned Aristotelian mechanics. A new in motion, it must continue to move with a
mechanics had to be developed. This task was uniform velocity.
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Galileo Galilei (1564 - 1642)
Galileo Galilei, born in Pisa, Italy in 1564 was a key figure in the scientific revolution
in Europe about four centuries ago. Galileo proposed the concept of acceleration.
From experiments on motion of bodies on inclined planes or falling freely, he
contradicted the Aristotelian notion that a force was required to keep a body in
motion, and that heavier bodies fall faster than lighter bodies under gravity. He
thus arrived at the law of inertia that was the starting point of the subsequent
epochal work of Isaac Newton.
Galileo’s discoveries in astronomy were equally revolutionary. In 1609, he designed
his own telescope (invented earlier in Holland) and used it to make a number of
startling observations : mountains and depressions on the surface of the moon;
dark spots on the sun; the moons of Jupiter and the phases of Venus. He concluded
that the Milky Way derived its luminosity because of a large number of stars not visible to the naked eye.
In his masterpiece of scientific reasoning : Dialogue on the Two Chief World Systems, Galileo advocated
the heliocentric theory of the solar system proposed by Copernicus, which eventually got universal
acceptance.
With Galileo came a turning point in the very method of scientific inquiry. Science was no longer
merely observations of nature and inferences from them. Science meant devising and doing experiments
to verify or refute theories. Science meant measurement of quantities and a search for mathematical
relations between them. Not undeservedly, many regard Galileo as the father of modern science.
More often, however, we do not know all the normal force R must be equal and opposite to the
forces to begin with. In that case, if we know weight W ”.
that an object is unaccelerated (i.e. it is either
at rest or in uniform linear motion), we can infer
from the first law that the net external force on
the object must be zero. Gravity is everywhere.
For terrestrial phenomena, in particular, every
object experiences gravitational force due to the
earth. Also objects in motion generally experience
friction, viscous drag, etc. If then, on earth, an
object is at rest or in uniform linear motion, it is
not because there are no forces acting on it, but Fig. 5.2 (a) a book at rest on the table, and (b) a car
because the various external forces cancel out moving with uniform velocity. The net force
i.e. add up to zero net external force. is zero in each case.
Consider a book at rest on a horizontal surface Consider the motion of a car starting from
Fig. (5.2(a)). It is subject to two external forces : rest, picking up speed and then moving on a
the force due to gravity (i.e. its weight W) acting smooth straight road with uniform speed (Fig.
downward and the upward force on the book by (5.2(b)). When the car is stationary, there is no
the table, the normal force R . R is a self-adjusting
force. This is an example of the kind of situation net force acting on it. During pick-up, it
mentioned above. The forces are not quite known accelerates. This must happen due to a net
fully but the state of motion is known. We observe external force. Note, it has to be an external force.
the book to be at rest. Therefore, we conclude The acceleration of the car cannot be accounted
from the first law that the magnitude of R equals for by any internal force. This might sound
that of W. A statement often encountered is : surprising, but it is true. The only conceivable
“Since W = R, forces cancel and, therefore, the book external force along the road is the force of
is at rest”. This is incorrect reasoning. The correct friction. It is the frictional force that accelerates
statement is : “Since the book is observed to be at the car as a whole. (You will learn about friction
rest, the net external force on it must be zero, in section 5.9). When the car moves with
according to the first law. This implies that the constant velocity, there is no net external force.
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LAWS OF MOTION 93
The property of inertia contained in the First It relates the net external force to the
law is evident in many situations. Suppose we acceleration of the body.
are standing in a stationary bus and the driver Momentum
starts the bus suddenly. We get thrown Momentum of a body is defined to be the product
backward with a jerk. Why ? Our feet are in touch of its mass m and velocity v, and is denoted
with the floor. If there were no friction, we would by p:
remain where we were, while the floor of the bus p = m v (5.1)
would simply slip forward under our feet and the
Momentum is clearly a vector quantity. The
back of the bus would hit us. However,
following common experiences indicate the
fortunately, there is some friction between the
importance of this quantity for considering the
feet and the floor. If the start is not too sudden,
i.e. if the acceleration is moderate, the frictional effect of force on motion.
force would be enough to accelerate our feet • Suppose a light-weight vehicle (say a small
along with the bus. But our body is not strictly car) and a heavy weight vehicle (say a loaded
a rigid body. It is deformable, i.e. it allows some truck) are parked on a horizontal road. We all
relative displacement between different parts. know that a much greater force is needed to
What this means is that while our feet go with push the truck than the car to bring them to
the same speed in same time. Similarly, a
the bus, the rest of the body remains where it is
greater opposing force is needed to stop a
due to inertia. Relative to the bus, therefore, we
heavy body than a light body in the same time,
are thrown backward. As soon as that happens,
if they are moving with the same speed.
however, the muscular forces on the rest of the
body (by the feet) come into play to move the body • If two stones, one light and the other heavy,
along with the bus. A similar thing happens are dropped from the top of a building, a
when the bus suddenly stops. Our feet stop due person on the ground will find it easier to catch
to the friction which does not allow relative the light stone than the heavy stone. The
motion between the feet and the floor of the bus. mass of a body is thus an important
But the rest of the body continues to move parameter that determines the effect of force
forward due to inertia. We are thrown forward. on its motion.
The restoring muscular forces again come into • Speed is another important parameter to
play and bring the body to rest. consider. A bullet fired by a gun can easily
pierce human tissue before it stops, resulting
Example 5.1 An astronaut accidentally in casualty. The same bullet fired with
t
gets separated out of his small spaceship moderate speed will not cause much damage.
accelerating in inter stellar space at a Thus for a given mass, the greater the speed,
constant rate of 100 m s . What is the the greater is the opposing force needed to stop
–2
acceleration of the astronaut the instant after the body in a certain time. Taken together,
he is outside the spaceship ? (Assume that the product of mass and velocity, that is
there are no nearby stars to exert momentum, is evidently a relevant variable
gravitational force on him.) of motion. The greater the change in the
momentum in a given time, the greater is the
Answer Since there are no nearby stars to exert force that needs to be applied.
gravitational force on him and the small • A seasoned cricketer catches a cricket ball
spaceship exerts negligible gravitational coming in with great speed far more easily
attraction on him, the net force acting on the than a novice, who can hurt his hands in the
astronaut, once he is out of the spaceship, is act. One reason is that the cricketer allows a
zero. By the first law of motion the acceleration longer time for his hands to stop the ball. As
of the astronaut is zero. t
you may have noticed, he draws in the hands
5.5 NEWTON’S SECOND LAW OF MOTION backward in the act of catching the ball (Fig.
The first law refers to the simple case when the 5.3). The novice, on the other hand, keeps
net external force on a body is zero. The second his hands fixed and tries to catch the ball
law of motion refers to the general situation when almost instantly. He needs to provide a much
there is a net external force acting on the body. greater force to stop the ball instantly, and
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94 PHYSICS
this hurts. The conclusion is clear: force not This force is provided by our hand through
only depends on the change in momentum, the string. Experience suggests that our hand
but also on how fast the change is brought needs to exert a greater force if the stone is
about. The same change in momentum rotated at greater speed or in a circle of
brought about in a shorter time needs a smaller radius, or both. This corresponds to
greater applied force. In short, the greater the greater acceleration or equivalently a greater
rate of change of momentum, the greater is rate of change in momentum vector. This
the force. suggests that the greater the rate of change
in momentum vector the greater is the force
applied.
Fig. 5.3 Force not only depends on the change in
momentum but also on how fast the change
is brought about. A seasoned cricketer draws Fig. 5.4 Force is necessary for changing the direction
in his hands during a catch, allowing greater of momentum, even if its magnitude is
time for the ball to stop and hence requires a constant. We can feel this while rotating a
smaller force. stone in a horizontal circle with uniform speed
by means of a string.
• Observations confirm that the product of
mass and velocity (i.e. momentum) is basic to These qualitative observations lead to the
the effect of force on motion. Suppose a fixed second law of motion expressed by Newton as
force is applied for a certain interval of time follows :
on two bodies of different masses, initially at The rate of change of momentum of a body is
rest, the lighter body picks up a greater speed directly proportional to the applied force and
than the heavier body. However, at the end of takes place in the direction in which the force
the time interval, observations show that each acts.
body acquires the same momentum. Thus
Thus, if under the action of a force F for time
the same force for the same time causes
interval ∆t, the velocity of a body of mass m
the same change in momentum for
changes from v to v + ∆v i.e. its initial momentum
different bodies. This is a crucial clue to the p = m v changes by ∆ =p m∆v . According to the
second law of motion.
Second Law,
• In the preceding observations, the vector
character of momentum has not been evident. F ∝ ∆p or =F ∆p
k
In the examples so far, momentum and change t ∆ t ∆
in momentum both have the same direction.
where k is a constant of proportionality. Taking
But this is not always the case. Suppose a
stone is rotated with uniform speed in a ∆p
the limit ∆t → 0, the term becomes the
horizontal plane by means of a string, the t ∆
magnitude of momentum is fixed, but its derivative or differential co-efficient of p with
direction changes (Fig. 5.4). A force is needed dp
respect to t, denoted by . Thus
to cause this change in momentum vector. dt
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LAWS OF MOTION 95
dp on the particle and a stands for acceleration
F = k (5.2) of the particle. It turns out, however, that the
dt
For a body of fixed mass m, law in the same form applies to a rigid body or,
even more generally, to a system of particles.
dp d dv In that case, F refers to the total external force
= (m v ) = m = m a (5.3) on the system and a refers to the acceleration
dt dt dt
of the system as a whole. More precisely, a is
i.e the Second Law can also be written as the acceleration of the centre of mass of the
F = k m a (5.4)
system about which we shall study in detail in
which shows that force is proportional to the chapter 7. Any internal forces in the system
product of mass m and acceleration a. are not to be included in F.
The unit of force has not been defined so far.
In fact, we use Eq. (5.4) to define the unit of force.
We, therefore, have the liberty to choose any
constant value for k. For simplicity, we choose
k = 1. The second law then is
d p
F = m a (5.5)
=
t d
In SI unit force is one that causes an acceleration
-2
of 1 m s to a mass of 1 kg. This unit is known as
newton : 1 N = 1 kg m s .
-2
Fig. 5.5 Acceleration at an instant is determined by
Let us note at this stage some important points
the force at that instant. The moment after a
about the second law : stone is dropped out of an accelerated train,
1. In the second law, F = 0 implies a = 0. The it has no horizontal acceleration or force, if
second law is obviously consistent with the air resistance is neglected. The stone carries
no memory of its acceleration with the train
first law.
a moment ago.
2. The second law of motion is a vector law. It is
equivalent to three equations, one for each 4. The second law of motion is a local relation
component of the vectors : which means that force F at a point in space
d p x (location of the particle) at a certain instant
F = = ma of time is related to a at that point at that
x x
t d
instant. Acceleration here and now is
d p y
F = = ma y determined by the force here and now, not by
y
t d any history of the motion of the particle
d p z (See Fig. 5.5).
F = = m a
z z (5.6)
t d
This means that if a force is not parallel to t Example 5.2 A bullet of mass 0.04 kg
–1
the velocity of the body, but makes some angle moving with a speed of 90 m s enters a
with it, it changes only the component of heavy wooden block and is stopped after a
velocity along the direction of force. The distance of 60 cm. What is the average
component of velocity normal to the force resistive force exerted by the block on the
remains unchanged. For example, in the bullet?
motion of a projectile under the vertical
gravitational force, the horizontal component Answer The retardation ‘a’ of the bullet
of velocity remains unchanged (Fig. 5.5). (assumed constant) is given by
3. The second law of motion given by Eq. (5.5) is 2
applicable to a single point particle. The force a = – u = – 90 × 90 m s − 2 = – 6750 m s − 2
F in the law stands for the net external force 2s 2 × 0.6
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The retarding force, by the second law of ordinary forces. Newtonian mechanics has no
motion, is such distinction. Impulsive force is like any other
force – except that it is large and acts for a short
= 0.04 kg × 6750 m s = 270 N
-2
time.
The actual resistive force, and therefore,
retardation of the bullet may not be uniform. The t Example 5.4 A batsman hits back a ball
answer therefore, only indicates the average straight in the direction of the bowler without
resistive force. t changing its initial speed of 12 m s .
–1
Example 5.3 The motion of a particle of If the mass of the ball is 0.15 kg, determine
t
the impulse imparted to the ball. (Assume
1 2
mass m is described by y = ut + gt . Find linear motion of the ball)
2
the force acting on the particle. Answer Change in momentum
Answer We know = 0.15 × 12–(–0.15×12)
1 = 3.6 N s,
y ut + gt 2
=
2 Impulse = 3.6 N s,
Now, in the direction from the batsman to the bowler.
dy This is an example where the force on the ball
v = = u + gt by the batsman and the time of contact of the
dt
ball and the bat are difficult to know, but the
dv impulse is readily calculated. t
acceleration, a = = g
dt
5.6 NEWTON’S THIRD LAW OF MOTION
Then the force is given by Eq. (5.5)
F = ma = mg The second law relates the external force on a
Thus the given equation describes the motion body to its acceleration. What is the origin of the
of a particle under acceleration due to gravity external force on the body ? What agency
and y is the position coordinate in the direction provides the external force ? The simple answer
of g. t in Newtonian mechanics is that the external
force on a body always arises due to some other
Impulse body. Consider a pair of bodies A and B. B gives
We sometimes encounter examples where a large rise to an external force on A. A natural question
force acts for a very short duration producing a is: Does A in turn give rise to an external force
finite change in momentum of the body. For on B ? In some examples, the answer seems
example, when a ball hits a wall and bounces
clear. If you press a coiled spring, the spring is
back, the force on the ball by the wall acts for a compressed by the force of your hand. The
very short time when the two are in contact, yet
compressed spring in turn exerts a force on your
the force is large enough to reverse the momentum hand and you can feel it. But what if the bodies
of the ball. Often, in these situations, the force are not in contact ? The earth pulls a stone
and the time duration are difficult to ascertain
downwards due to gravity. Does the stone exert
separately. However, the product of force and time,
which is the change in momentum of the body a force on the earth ? The answer is not obvious
since we hardly see the effect of the stone on the
remains a measurable quantity. This product is
called impulse: earth. The answer according to Newton is: Yes,
the stone does exert an equal and opposite force
Impulse = Force × time duration on the earth. We do not notice it since the earth
= Change in momentum (5.7) is very massive and the effect of a small force on
A large force acting for a short time to produce a its motion is negligible.
finite change in momentum is called an impulsive Thus, according to Newtonian mechanics,
force. In the history of science, impulsive forces force never occurs singly in nature. Force is the
were put in a conceptually different category from mutual interaction between two bodies. Forces
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LAWS OF MOTION 97
always occur in pairs. Further, the mutual forces comes before reaction i.e action is the cause
between two bodies are always equal and and reaction the effect. There is no cause-
opposite. This idea was expressed by Newton in effect relation implied in the third law. The
the form of the third law of motion. force on A by B and the force on B by A act
at the same instant. By the same reasoning,
To every action, there is always an equal and
any one of them may be called action and the
opposite reaction.
other reaction.
Newton’s wording of the third law is so crisp and
3. Action and reaction forces act on different
beautiful that it has become a part of common
bodies, not on the same body. Consider a pair
language. For the same reason perhaps,
of bodies A and B. According to the third law,
misconceptions about the third law abound. Let
us note some important points about the third F = – F BA (5.8)
AB
law, particularly in regard to the usage of the (force on A by B) = – (force on B by A)
terms : action and reaction.
1. The terms action and reaction in the third law Thus if we are considering the motion of any
mean nothing else but ‘force’. Using different one body (A or B), only one of the two forces is
terms for the same physical concept relevant. It is an error to add up the two forces
can sometimes be confusing. A simple and claim that the net force is zero.
and clear way of stating the third law is as However, if you are considering the system
follows : of two bodies as a whole, F AB and F BA are
internal forces of the system (A + B). They add
Forces always occur in pairs. Force on a up to give a null force. Internal forces in a
body A by B is equal and opposite to the
body or a system of particles thus cancel away
force on the body B by A.
in pairs. This is an important fact that
2. The terms action and reaction in the third law enables the second law to be applicable to a
may give a wrong impression that action body or a system of particles (See Chapter 7).
Isaac Newton (1642 – 1727)
Isaac Newton was born in Woolsthorpe, England in 1642, the year Galileo died.
His extraordinary mathematical ability and mechanical aptitude remained hidden
from others in his school life. In 1662, he went to Cambridge for undergraduate
studies. A plague epidemic in 1665 forced the university town to close and Newton
had to return to his mother’s farm. There in two years of solitude, his dormant
creativity blossomed in a deluge of fundamental discoveries in mathematics and
physics : binomial theorem for negative and fractional exponents, the beginning of
calculus, the inverse square law of gravitation, the spectrum of white light, and so
on. Returning to Cambridge, he pursued his investigations in optics and devised a
reflecting telescope.
In 1684, encouraged by his friend Edmund Halley, Newton embarked on writing what was to be one of
the greatest scientific works ever published : The Principia Mathematica. In it, he enunciated the three
laws of motion and the universal law of gravitation, which explained all the three Kepler’s laws of
planetary motion. The book was packed with a host of path-breaking achievements : basic principles of
fluid mechanics, mathematics of wave motion, calculation of masses of the earth, the sun and other
planets, explanation of the precession of equinoxes, theory of tides, etc. In 1704, Newton brought out
another masterpiece Opticks that summarized his work on light and colour.
The scientific revolution triggered by Copernicus and steered vigorously ahead by Kepler and Galileo
was brought to a grand completion by Newton. Newtonian mechanics unified terrestrial and celestial
phenomena. The same mathematical equation governed the fall of an apple to the ground and the
motion of the moon around the earth. The age of reason had dawned.
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98 PHYSICS
magnitude of force cannot be ascertained since
Example 5.5 Two identical billiard balls
t
strike a rigid wall with the same speed but the small time taken for the collision has not
at different angles, and get reflected without been specified in the problem.
any change in speed, as shown in Fig. 5.6. Case (b)
What is (i) the direction of the force on the p = − m u sin 30
p
wall due to each ball? (ii) the ratio of the ( )initial = m u cos 30 , ( ) initialy
x
magnitudes of impulses imparted to the
balls by the wall ?
p
p
( ) final = – m u cos 30 , ( ) final = − m u sin 30
y
x
Note, while p changes sign after collision, p
x y
does not. Therefore,
x-component of impulse = –2 m u cos 30°
y-component of impulse = 0
The direction of impulse (and force) is the same
as in (a) and is normal to the wall along the
negative x direction. As before, using Newton’s
third law, the force on the wall due to the ball is
normal to the wall along the positive x direction.
The ratio of the magnitudes of the impulses
Fig. 5.6
imparted to the balls in (a) and (b) is
Answer An instinctive answer to (i) might be 2
that the force on the wall in case (a) is normal to 2m u / (2m u cos30 ) = 3 ≈ 1.2 t
the wall, while that in case (b) is inclined at 30°
to the normal. This answer is wrong. The force 5.7 CONSERVATION OF MOMENTUM
on the wall is normal to the wall in both cases. The second and third laws of motion lead to
How to find the force on the wall? The trick is
to consider the force (or impulse) on the ball an important consequence: the law of
conservation of momentum.
Take a familiar
due to the wall using the second law, and then example. A bullet is fired from a gun. If the force
use the third law to answer (i). Let u be the speed on the bullet by the gun is F, the force on the gun
of each ball before and after collision with the by the bullet is – F, according to the third law.
wall, and m the mass of each ball. Choose the x The two forces act for a common interval of time
and y axes as shown in the figure, and consider ∆t. According to the second law, F ∆t is the
the change in momentum of the ball in each change in momentum of the bullet and – F ∆t is
case :
the change in momentum of the gun. Since
Case (a) initially, both are at rest, the change in
momentum equals the final momentum for each.
p
( ) initial = mu p = 0
Thus if p is the momentum of the bullet after
( ) initialy
x
b
firing and p is the recoil momentum of the gun,
( ) = − mu p = 0 g
p
x
( ) finaly
p
final
g = – p b i.e. p b + p g = 0. That is, the total
momentum of the (bullet + gun) system is
Impulse is the change in momentum vector.
conserved.
Therefore,
Thus in an isolated system (i.e. a system with
x-component of impulse = – 2 m u no external force), mutual forces between pairs
y-component of impulse = 0 of particles in the system can cause momentum
Impulse and force are in the same direction. change in individual particles, but since the
Clearly, from above, the force on the ball due to mutual forces for each pair are equal and
the wall is normal to the wall, along the negative opposite, the momentum changes cancel in pairs
x-direction. Using Newton’s third law of motion, and the total momentum remains unchanged.
the force on the wall due to the ball is normal to This fact is known as the law of conservation
the wall along the positive x-direction. The of momentum :
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LAWS OF MOTION 99
The total momentum of an isolated system
of interacting particles is conserved.
An important example of the application of the
law of conservation of momentum is the collision
of two bodies. Consider two bodies A and B, with
initial momenta p and p . The bodies collide,
A B
get apart, with final momenta p′ and p′
A B
respectively. By the Second Law
Fig. 5.7 Equilibrium under concurrent forces.
F t ∆ = p′ − p and In other words, the resultant of any two forces
AB A A
say F and F , obtained by the parallelogram
1
2
F t ∆ = p′ − p
BA B B law of forces must be equal and opposite to the
(where we have taken a common interval of time third force, F . As seen in Fig. 5.7, the three
3
forces in equilibrium can be represented by the
for both forces i.e. the time for which the two
sides of a triangle with the vector arrows taken
bodies are in contact.)
in the same sense. The result can be
Since F = − F by the third law, generalised to any number of forces. A particle
AB BA
is in equilibrium under the action of forces F ,
1
p ′ − p A = ( − p ′ − p B ) F ,... F if they can be represented by the sides
A
B
n
2
of a closed n-sided polygon with arrows directed
i.e. p ′ + ′ =p B p A + p B (5.9) in the same sense.
A
which shows that the total final momentum of Equation (5.11) implies that
the isolated system equals its initial momentum.
F + F + F = 0
Notice that this is true whether the collision is 1x 2x 3x
F + F + F = 0
elastic or inelastic. In elastic collisions, there is 1y 2y 3y
a second condition that the total initial kinetic F + F + F = 0 (5.12)
2z
3z
1z
energy of the system equals the total final kinetic where F , F and F are the components of F
1x 1y 1z 1
energy (See Chapter 6). along x, y and z directions respectively.
Example 5.6 See Fig. 5.8. A mass of 6 kg
t
5.8 EQUILIBRIUM OF A PARTICLE is suspended by a rope of length 2 m
from the ceiling. A force of 50 N in the
Equilibrium of a particle in mechanics refers to horizontal direction is applied at the mid-
the situation when the net external force on the point P of the rope, as shown. What is the
particle is zero.* According to the first law, this angle the rope makes with the vertical in
-2
means that, the particle is either at rest or in equilibrium ? (Take g = 10 m s ). Neglect
uniform motion. the mass of the rope.
If two forces F and F , act on a particle,
1 2
equilibrium requires
F = − F (5.10)
1 2
i.e. the two forces on the particle must be equal
and opposite. Equilibrium under three
concurrent forces F , F and F requires that
1 2 3
the vector sum of the three forces is zero.
F + F + F = 0 (5.11)
1 2 3 (a) (b) (c)
Fig. 5.8
* Equilibrium of a body requires not only translational equilibrium (zero net external force) but also rotational
equilibrium (zero net external torque), as we shall see in Chapter 7.
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100 PHYSICS
Answer Figures 5.8(b) and 5.8(c) are known as other types of supports), there are mutual
free-body diagrams. Figure 5.8(b) is the free-body contact forces (for each pair of bodies) satisfying
diagram of W and Fig. 5.8(c) is the free-body the third law. The component of contact force
diagram of point P. normal to the surfaces in contact is called
Consider the equilibrium of the weight W. normal reaction. The component parallel to the
Clearly,T = 6 × 10 = 60 N. surfaces in contact is called friction. Contact
2
forces arise also when solids are in contact with
Consider the equilibrium of the point P under
the action of three forces - the tensions T and fluids. For example, for a solid immersed in a
1 fluid, there is an upward bouyant force equal to
T , and the horizontal force 50 N. The horizontal
2
and vertical components of the resultant force the weight of the fluid displaced. The viscous
must vanish separately : force, air resistance, etc are also examples of
contact forces (Fig. 5.9).
T cos θ = T = 60 N
1 2 Two other common forces are tension in a
T sin θ = 50 N string and the force due to spring. When a spring
1
which gives that is compressed or extended by an external force,
a restoring force is generated. This force is
usually proportional to the compression or
Note the answer does not depend on the length elongation (for small displacements). The spring
of the rope (assumed massless) nor on the point force F is written as F = – k x where x is the
at which the horizontal force is applied. t displacement and k is the force constant. The
negative sign denotes that the force is opposite
5.9 COMMON FORCES IN MECHANICS
to the displacement from the unstretched state.
In mechanics, we encounter several kinds of For an inextensible string, the force constant is
forces. The gravitational force is, of course, very high. The restoring force in a string is called
pervasive. Every object on the earth experiences tension. It is customary to use a constant tension
the force of gravity due to the earth. Gravity also T throughout the string. This assumption is true
governs the motion of celestial bodies. The for a string of negligible mass.
gravitational force can act at a distance without In Chapter 1, we learnt that there are four
the need of any intervening medium. fundamental forces in nature. Of these, the weak
All the other forces common in mechanics are and strong forces appear in domains that do not
contact forces.* As the name suggests, a contact concern us here. Only the gravitational and
force on an object arises due to contact with some electrical forces are relevant in the context of
other object: solid or fluid. When bodies are in mechanics. The different contact forces of
contact (e.g. a book resting on a table, a system mechanics mentioned above fundamentally arise
of rigid bodies connected by rods, hinges and from electrical forces. This may seem surprising
Fig. 5.9 Some examples of contact forces in mechanics.
* We are not considering, for simplicity, charged and magnetic bodies. For these, besides gravity, there are
electrical and magnetic non-contact forces.
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LAWS OF MOTION 101
since we are talking of uncharged and non- exist by itself. When there is no applied force,
magnetic bodies in mechanics. At the microscopic there is no static friction. It comes into play the
level, all bodies are made of charged constituents moment there is an applied force. As the applied
(nuclei and electrons) and the various contact force F increases, f also increases, remaining
s
forces arising due to elasticity of bodies, molecular equal and opposite to the applied force (up to a
collisions and impacts, etc. can ultimately be certain limit), keeping the body at rest. Hence, it
is called static friction. Static friction opposes
traced to the electrical forces between the charged
impending motion. The term impending motion
constituents of different bodies. The detailed
means motion that would take place (but does
microscopic origin of these forces is, however, not actually take place) under the applied force,
complex and not useful for handling problems in
if friction were absent.
mechanics at the macroscopic scale. This is why We know from experience that as the applied
they are treated as different types of forces with force exceeds a certain limit, the body begins to
their characteristic properties determined move. It is found experimentally that the limiting
empirically.
value of static friction ( ) is independent of
f
s
max
5.9.1 Friction
the area of contact and varies with the normal
Let us return to the example of a body of mass m force(N) approximately as :
at rest on a horizontal table. The force of gravity ( ) = µ s N (5.13)
f
s
max
(mg) is cancelled by the normal reaction force where µ is a constant of proportionality
(N) of the table. Now suppose a force F is applied s
depending only on the nature of the surfaces in
horizontally to the body. We know from contact. The constant µ is called the coefficient
experience that a small applied force may not s
of static friction. The law of static friction may
be enough to move the body. But if the applied
thus be written as
force F were the only external force on the body,
f ≤ µ N (5.14)
it must move with acceleration F/m, however s s
f
small. Clearly, the body remains at rest because If the applied force F exceeds ( ) max the body
s
some other force comes into play in the begins to slide on the surface. It is found
horizontal direction and opposes the applied experimentally that when relative motion has
force F, resulting in zero net force on the body. started, the frictional force decreases from the
This force f parallel to the surface of the body in
f
s static maximum value ( ) . Frictional force
s
contact with the table is known as frictional max
force, or simply friction (Fig. 5.10(a)). The that opposes relative motion between surfaces
in contact is called kinetic or sliding friction and
subscript stands for static friction to distinguish
it from kinetic friction f that we consider later is denoted by f . Kinetic friction, like static
k
k friction, is found to be independent of the area
(Fig. 5.10(b)). Note that static friction does not
of contact. Further, it is nearly independent of
the velocity. It satisfies a law similar to that for
static friction:
f k = µ k N (5.15)
where µ the coefficient of kinetic friction,
k′
depends only on the surfaces in contact. As
mentioned above, experiments show that µ is
k
Fig. 5.10 Static and sliding friction: (a) Impending less than µ . When relative motion has begun,
s
motion of the body is opposed by static the acceleration of the body according to the
friction. When external force exceeds the second law is ( F – f )/m. For a body moving with
k
maximum limit of static friction, the body constant velocity, F = f . If the applied force on
k
begins to move. (b) Once the body is in the body is removed, its acceleration is – f /m
k
motion, it is subject to sliding or kinetic friction and it eventually comes to a stop.
which opposes relative motion between the The laws of friction given above do not have
two surfaces in contact. Kinetic friction is the status of fundamental laws like those for
usually less than the maximum value of static
gravitational, electric and magnetic forces. They
friction.
are empirical relations that are only
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102 PHYSICS
approximately true. Yet they are very useful in Answer The forces acting on a block of mass m
practical calculations in mechanics. at rest on an inclined plane are (i) the weight
Thus, when two bodies are in contact, each mg acting vertically downwards (ii) the normal
experiences a contact force by the other. Friction, force N of the plane on the block, and (iii) the
by definition, is the component of the contact force static frictional force f opposing the impending
s
parallel to the surfaces in contact, which opposes motion. In equilibrium, the resultant of these
impending or actual relative motion between the forces must be zero. Resolving the weight mg
two surfaces. Note that it is not motion, but along the two directions shown, we have
relative motion that the frictional force opposes. m g sin θ = f , m g cos θ = N
s
Consider a box lying in the compartment of a train As θ increases, the self-adjusting frictional force
that is accelerating. If the box is stationary
relative to the train, it is in fact accelerating along f s increases until at θ = θ max , f s achieves its
with the train. What forces cause the acceleration maximum value, ( ) = µ N.
f
s
of the box? Clearly, the only conceivable force in max s
the horizontal direction is the force of friction. If Therefore,
there were no friction, the floor of the train would tan θ = µ or θ = tan µ
–1
slip by and the box would remain at its initial max s max s
position due to inertia (and hit the back side of When θ becomes just a little more than θ ,
max
the train). This impending relative motion is there is a small net force on the block and it
opposed by the static friction f . Static friction begins to slide. Note that θ depends only on
s max
provides the same acceleration to the box as that µ and is independent of the mass of the block.
s
of the train, keeping it stationary relative to the
train. For θ max = 15°,
µ = tan 15°
Example 5.7 Determine the maximum s = 0.27 t
t
acceleration of the train in which a box
lying on its floor will remain stationary, t Example 5.9 What is the acceleration of
given that the co-efficient of static friction the block and trolley system shown in a
between the box and the train’s floor is Fig. 5.12(a), if the coefficient of kinetic friction
0.15. between the trolley and the surface is 0.04?
What is the tension in the string? (Take g =
Answer Since the acceleration of the box is due 10 m s ). Neglect the mass of the string.
-2
to the static friction,
ma = f ≤ µ N = µ m g
s s s
i.e. a ≤ µ g
s
∴ a = µ g = 0.15 x 10 m s –2
max s
= 1.5 m s –2 t
Example 5.8 See Fig. 5.11. A mass of 4 kg
t
rests on a horizontal plane. The plane is
gradually inclined until at an angle θ = 15°
with the horizontal, the mass just begins to
slide. What is the coefficient of static friction
between the block and the surface ?
(a)
(b) (c)
Fig. 5.11 Fig. 5.12
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LAWS OF MOTION 103
Answer As the string is inextensible, and the is the reason why discovery of the wheel has
pully is smooth, the 3 kg block and the 20 kg been a major milestone in human history.
trolley both have same magnitude of Rolling friction again has a complex origin,
acceleration. Applying second law to motion of though somewhat different from that of static
the block (Fig. 5.12(b)), and sliding friction. During rolling, the surfaces
30 – T = 3a in contact get momentarily deformed a little, and
this results in a finite area (not a point) of the
Apply the second law to motion of the trolley (Fig.
body being in contact with the surface. The net
5.12(c)),
effect is that the component of the contact force
T – f = 20 a.
k parallel to the surface opposes motion.
Now f = µ N,
k k We often regard friction as something
Here µ = 0.04,
k undesirable. In many situations, like in a
N = 20 x 10 machine with different moving parts, friction
= 200 N.
does have a negative role. It opposes relative
Thus the equation for the motion of the trolley is motion and thereby dissipates power in the form
T – 0.04 x 200 = 20 a Or T – 8 = 20a. of heat, etc. Lubricants are a way of reducing
kinetic friction in a machine. Another way is to
22 –2 -2
These equations give a = m s = 0.96 m s use ball bearings between two moving parts of a
23
and T = 27.1 N. t machine [Fig. 5.13(a)]. Since the rolling friction
between ball bearings and the surfaces in
Rolling friction contact is very small, power dissipation is
A body like a ring or a sphere rolling without reduced. A thin cushion of air maintained
slipping over a horizontal plane will suffer no between solid surfaces in relative motion is
another effective way of reducing friction (Fig.
friction, in principle. At every instant, there is 5.13(a)).
just one point of contact between the body and
In many practical situations, however, friction
the plane and this point has no motion relative
is critically needed. Kinetic friction that
to the plane. In this ideal situation, kinetic or
dissipates power is nevertheless important for
static friction is zero and the body should
quickly stopping relative motion. It is made use
continue to roll with constant velocity. We know, of by brakes in machines and automobiles.
in practice, this will not happen and some Similarly, static friction is important in daily
resistance to motion (rolling friction) does occur, life. We are able to walk because of friction. It
i.e. to keep the body rolling, some applied force is impossible for a car to move on a very slippery
is needed. For the same weight, rolling friction road. On an ordinary road, the friction between
is much smaller (even by 2 or 3 orders of the tyres and the road provides the necessary
magnitude) than static or sliding friction. This external force to accelerate the car.
Fig. 5.13 Some ways of reducing friction. (a) Ball bearings placed between moving parts of a machine.
(b) Compressed cushion of air between surfaces in relative motion.
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104 PHYSICS
5.10 CIRCULAR MOTION is the static friction that provides the centripetal
acceleration. Static friction opposes the
We have seen in Chapter 4 that acceleration of
impending motion of the car moving away from
a body moving in a circle of radius R with uniform the circle. Using equation (5.14) & (5.16) we get
2
speed v is v /R directed towards the centre.
the result
According to the second law, the force f providing
c 2
this acceleration is : mv
f = ≤ µ s N
mv 2 R
f = (5.16)
c
R µ RN
2
v ≤ s = µ s Rg [∵N = mg]
where m is the mass of the body. This force m
directed forwards the centre is called the which is independent of the mass of the car.
centripetal force. For a stone rotated in a circle This shows that for a given value of µ and R,
s
by a string, the centripetal force is provided by there is a maximum speed of circular motion of
the tension in the string. The centripetal force the car possible, namely
for motion of a planet around the sun is the
v = µ Rg (5.18)
max s
(a) (b)
Fig. 5.14 Circular motion of a car on (a) a level road, (b) a banked road.
gravitational force on the planet due to the sun. Motion of a car on a banked road
For a car taking a circular turn on a horizontal
We can reduce the contribution of friction to the
road, the centripetal force is the force of friction.
circular motion of the car if the road is banked
The circular motion of a car on a flat and
(Fig. 5.14(b)). Since there is no acceleration along
banked road give interesting application of the
the vertical direction, the net force along this
laws of motion.
direction must be zero. Hence,
Motion of a car on a level road
N cos θ = mg + f sin θ (5.19a)
Three forces act on the car (Fig. 5.14(a):
(i) The weight of the car, mg The centripetal force is provided by the horizontal
(ii) Normal reaction, N components of N and f.
(iii) Frictional force, f 2
mv
As there is no acceleration in the vertical N sin θ + f cos θ = (5.19b)
R
direction
N – mg = 0 But f µ≤ s N
N = mg (5.17)
Thus to obtain v we put
The centripetal force required for circular motion max
is along the surface of the road, and is provided f = µ s N .
by the component of the contact force between
Then Eqs. (5.19a) and (5.19b) become
road and the car tyres along the surface. This
by definition is the frictional force. Note that it N cos θ = mg + µ s N sin θ (5.20a)
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LAWS OF MOTION 105
2
N sin θ + µ N cos θ = mv /R (5.20b)
s t Example 5.11 A circular racetrack of
From Eq. (5.20a), we obtain radius 300 m is banked at an angle of 15°.
mg If the coefficient of friction between the
N = wheels of a race-car and the road is 0.2,
cosθ – µ s sinθ
what is the (a) optimum speed of the race-
Substituting value of N in Eq. (5.20b), we get car to avoid wear and tear on its tyres, and
mg (sin +θ µ s cosθ ) mv 2 max (b) maximum permissible speed to avoid
= slipping ?
cosθ – µ s sinθ R
Answer On a banked road, the horizontal
1
µ + tanθ 2 component of the normal force and the frictional
or v max = Rg s (5.21) force contribute to provide centripetal force to
1 – µ s tanθ
keep the car moving on a circular turn without
Comparing this with Eq. (5.18) we see that slipping. At the optimum speed, the normal
maximum possible speed of a car on a banked reaction’s component is enough to provide the
road is greater than that on a flat road. needed centripetal force, and the frictional force
is not needed. The optimum speed v is given by
For µ = 0 in Eq. (5.21 ), Eq. (5.22): o
s
v = ( R g tan θ ) ½ (5.22) v = (R g tan θ) 1/2
o O
-2
At this speed, frictional force is not needed at all Here R = 300 m, θ = 15°, g = 9.8 m s ; we
to provide the necessary centripetal force. have
-1
Driving at this speed on a banked road will cause v = 28.1 m s .
O
little wear and tear of the tyres. The same The maximum permissible speed v is given by
equation also tells you that for v < v , frictional Eq. (5.21): max
o
force will be up the slope and that a car can be
parked only if tan θ ≤ µ .
s
t
Example 5.10 A cyclist speeding at
t
18 km/h on a level road takes a sharp
circular turn of radius 3 m without reducing
the speed. The co-efficient of static friction 5.11 SOLVING PROBLEMS IN MECHANICS
between the tyres and the road is 0.1. Will The three laws of motion that you have learnt in
the cyclist slip while taking the turn? this chapter are the foundation of mechanics.
You should now be able to handle a large variety
Answer On an unbanked road, frictional force of problems in mechanics. A typical problem in
alone can provide the centripetal force needed mechanics usually does not merely involve a
to keep the cyclist moving on a circular turn single body under the action of given forces.
without slipping. If the speed is too large, or if More often, we will need to consider an assembly
the turn is too sharp (i.e. of too small a radius) of different bodies exerting forces on each other.
or both, the frictional force is not sufficient to Besides, each body in the assembly experiences
provide the necessary centripetal force, and the the force of gravity. When trying to solve a
cyclist slips. The condition for the cyclist not to problem of this type, it is useful to remember
slip is given by Eq. (5.18) : the fact that we can choose any part of the
assembly and apply the laws of motion to that
2
v ≤ µ R g
s part provided we include all forces on the chosen
Now, R = 3 m, g = 9.8 m s , µ = 0.1. That is, part due to the remaining parts of the assembly.
-2
s
-2
2
-1
µ R g = 2.94 m s . v = 18 km/h = 5 m s ; i.e., We may call the chosen part of the assembly as
s
v 2 = 25 m 2 s . The condition is not obeyed. the system and the remaining part of the
-2
The cyclist will slip while taking the circular assembly (plus any other agencies of forces) as
turn. t the environment. We have followed the same
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106 PHYSICS
method in solved examples. To handle a typical the net force on the block must be zero i.e.,
problem in mechanics systematically, one R = 20 N. Using third law the action of the
should use the following steps : block (i.e. the force exerted on the floor by
(i) Draw a diagram showing schematically the the block) is equal to 20 N and directed
various parts of the assembly of bodies, the vertically downwards.
links, supports, etc. (b) The system (block + cylinder) accelerates
-2
(ii) Choose a convenient part of the assembly downwards with 0.1 m s . The free-body
as one system. diagram of the system shows two forces on
(iii) Draw a separate diagram which shows this the system : the force of gravity due to the
system and all the forces on the system by earth (270 N); and the normal force R′ by the
the remaining part of the assembly. Include floor. Note, the free-body diagram of the
also the forces on the system by other system does not show the internal forces
agencies. Do not include the forces on the between the block and the cylinder. Applying
environment by the system. A diagram of the second law to the system,
this type is known as ‘a free-body diagram’.
270 – R′ = 27 × 0.1N
(Note this does not imply that the system
ie. R′ = 267.3 N
under consideration is without a net force).
(iv) In a free-body diagram, include information
about forces (their magnitudes and
directions) that are either given or you are
sure of (e.g., the direction of tension in a
string along its length). The rest should be
treated as unknowns to be determined using
laws of motion.
(v) If necessary, follow the same procedure for
another choice of the system. In doing so,
employ Newton’s third law. That is, if in the
free-body diagram of A, the force on A due to
B is shown as F, then in the free-body
diagram of B, the force on B due to A should
be shown as –F.
The following example illustrates the above
procedure :
t Example 5.12 See Fig. 5.15. A wooden Fig. 5.15
block of mass 2 kg rests on a soft horizontal
floor. When an iron cylinder of mass 25 kg By the third law, the action of the system on
is placed on top of the block, the floor yields the floor is equal to 267.3 N vertically downward.
steadily and the block and the cylinder Action-reaction pairs
together go down with an acceleration of For (a): (i) the force of gravity (20 N) on the block
–2
0.1 m s . What is the action of the block by the earth (say, action); the force of
on the floor (a) before and (b) after the floor gravity on the earth by the block
yields ? Take g = 10 m s . Identify the (reaction) equal to 20 N directed
–2
action-reaction pairs in the problem.
upwards (not shown in the figure).
(ii) the force on the floor by the block
Answer (action); the force on the block by the
(a) The block is at rest on the floor. Its free-body floor (reaction).
diagram shows two forces on the block, the For (b): (i) the force of gravity (270 N) on the
force of gravitational attraction by the earth system by the earth (say, action); the
equal to 2 × 10 = 20 N; and the normal force force of gravity on the earth by the
R of the floor on the block. By the First Law, system (reaction), equal to 270 N,
2018-19
LAWS OF MOTION 107
directed upwards (not shown in the gravity on the mass in (a) or (b) and the normal
figure). force on the mass by the floor are not action-
(ii) the force on the floor by the system reaction pairs. These forces happen to be equal
(action); the force on the system by the and opposite for (a) since the mass is at rest.
floor (reaction). In addition, for (b), the They are not so for case (b), as seen already.
force on the block by the cylinder and The weight of the system is 270 N, while the
the force on the cylinder by the block normal force R′ is 267.3 N. t
also constitute an action-reaction pair. The practice of drawing free-body diagrams is
The important thing to remember is that an of great help in solving problems in mechanics.
action-reaction pair consists of mutual forces It allows you to clearly define your system and
which are always equal and opposite between consider all forces on the system due to objects
two bodies. Two forces on the same body which that are not part of the system itself. A number
happen to be equal and opposite can never of exercises in this and subsequent chapters will
constitute an action-reaction pair. The force of help you cultivate this practice.
SUMMARY
1. Aristotle’s view that a force is necessary to keep a body in uniform motion is wrong. A
force is necessary in practice to counter the opposing force of friction.
2. Galileo extrapolated simple observations on motion of bodies on inclined planes, and
arrived at the law of inertia. Newton’s first law of motion is the same law rephrased
thus: “Everybody continues to be in its state of rest or of uniform motion in a straight line,
unless compelled by some external force to act otherwise”. In simple terms, the First Law
is “If external force on a body is zero, its acceleration is zero”.
3. Momentum (p ) of a body is the product of its mass (m) and velocity (v) :
p = m v
4. Newton’s second law of motion :
The rate of change of momentum of a body is proportional to the applied force and takes
place in the direction in which the force acts. Thus
dp
F = k = k m a
dt
where F is the net external force on the body and a its acceleration. We set the constant
of proportionality k = 1 in SI units. Then
dp
F = = ma
dt
-2
The SI unit of force is newton : 1 N = 1 kg m s .
(a) The second law is consistent with the First Law (F = 0 implies a = 0)
(b) It is a vector equation
(c) It is applicable to a particle, and also to a body or a system of particles, provided F
is the total external force on the system and a is the acceleration of the system as
a whole.
(d) F at a point at a certain instant determines a at the same point at that instant.
That is the Second Law is a local law; a at an instant does not depend on the
history of motion.
5. Impulse is the product of force and time which equals change in momentum.
The notion of impulse is useful when a large force acts for a short time to produce a
measurable change in momentum. Since the time of action of the force is very short,
one can assume that there is no appreciable change in the position of the body during
the action of the impulsive force.
6. Newton’s third law of motion:
To every action, there is always an equal and opposite reaction
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108 PHYSICS
In simple terms, the law can be stated thus :
Forces in nature always occur between pairs of bodies. Force on a body A by body
B is equal and opposite to the force on the body B by A.
Action and reaction forces are simultaneous forces. There is no cause-effect
relation between action and reaction. Any of the two mutual forces can be
called action and the other reaction. Action and reaction act on different
bodies and so they cannot be cancelled out. The internal action and reaction
forces between different parts of a body do, however, sum to zero.
7. Law of Conservation of Momentum
The total momentum of an isolated system of particles is conserved. The law
follows from the second and third law of motion.
8. Friction
Frictional force opposes (impending or actual) relative motion between two
surfaces in contact. It is the component of the contact force along the common
tangent to the surface in contact. Static friction f opposes impending relative
s
motion; kinetic friction f opposes actual relative motion. They are independent
k
of the area of contact and satisfy the following approximate laws :
f s ≤ ( ) = µ s R
f
s
max
f = µ R
k k
µ (co-efficient of static friction) and µ (co-efficient of kinetic friction) are
s k
constants characteristic of the pair of surfaces in contact. It is found
experimentally that µ is less than µ .
k s
POINTS TO PONDER
1. Force is not always in the direction of motion. Depending on the situation, F
may be along v, opposite to v, normal to v or may make some other angle with
v. In every case, it is parallel to acceleration.
2. If v = 0 at an instant, i.e. if a body is momentarily at rest, it does not mean that
force or acceleration are necessarily zero at that instant. For example, when a
ball thrown upward reaches its maximum height, v = 0 but the force continues
to be its weight mg and the acceleration is not zero but g.
3. Force on a body at a given time is determined by the situation at the location of
the body at that time. Force is not ‘carried’ by the body from its earlier history of
motion. The moment after a stone is released out of an accelerated train, there is
no horizontal force (or acceleration) on the stone, if the effects of the surrounding
air are neglected. The stone then has only the vertical force of gravity.
4. In the second law of motion F = m a, F stands for the net force due to all
material agencies external to the body. a is the effect of the force. ma should
not be regarded as yet another force, besides F.
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LAWS OF MOTION 109
5. The centripetal force should not be regarded as yet another kind of force. It is
simply a name given to the force that provides inward radial acceleration to a
body in circular motion. We should always look for some material force like
tension, gravitational force, electrical force, friction, etc as the centripetal force
in any circular motion.
6. Static friction is a self-adjusting force up to its limit µ N (f ≤ µ N). Do not put
s s s
f = µ N without being sure that the maximum value of static friction is coming
s s
into play.
7. The familiar equation mg = R for a body on a table is true only if the body is in
equilibrium. The two forces mg and R can be different (e.g. a body in an
accelerated lift). The equality of mg and R has no connection with the third
law.
8. The terms ‘action’ and ‘reaction’ in the third Law of Motion simply stand for
simultaneous mutual forces between a pair of bodies. Unlike their meaning in
ordinary language, action does not precede or cause reaction. Action and reaction
act on different bodies.
9. The different terms like ‘friction’, ‘normal reaction’ ‘tension’, ‘air resistance’,
‘viscous drag’, ‘thrust’, ‘buoyancy’, ‘weight’, ‘centripetal force’ all stand for ‘force’
in different contexts. For clarity, every force and its equivalent terms
encountered in mechanics should be reduced to the phrase ‘force on A by B’.
10. For applying the second law of motion, there is no conceptual distinction between
inanimate and animate objects. An animate object such as a human also
requires an external force to accelerate. For example, without the external
force of friction, we cannot walk on the ground.
11. The objective concept of force in physics should not be confused with the
subjective concept of the ‘feeling of force’. On a merry-go-around, all parts of
our body are subject to an inward force, but we have a feeling of being pushed
outward – the direction of impending motion.
EXERCISES
-2
(For simplicity in numerical calculations, take g = 10 m s )
5.1 Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, and free of
electric and magnetic fields.
5.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction
and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers
change if the pebble was thrown at an angle of 45° with the horizontal
direction?
Ignore air resistance.
5.3 Give the magnitude and direction of the net force acting on a stone of mass
0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant
velocity of 36 km/h,
-2
(c ) just after it is dropped from the window of a train accelerating with 1 m s ,
-2
(d) lying on the floor of a train which is accelerating with 1 m s , the stone
being at rest relative to the train.
Neglect air resistance throughout.
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110 PHYSICS
5.4 One end of a string of length l is connected to a particle of mass m and the
other to a small peg on a smooth horizontal table. If the particle moves in a
circle with speed v the net force on the particle (directed towards the centre)
is :
2 2
mv mv
(i) T, (ii) T − , (iii) T + , (iv) 0
l l
T is the tension in the string. [Choose the correct alternative].
5.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving
initially with a speed of 15 m s . How long does the body take to stop ?
-1
5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s -1
to 3.5 m s -1 in 25 s. The direction of the motion of the body remains
unchanged. What is the magnitude and direction of the force ?
5.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N.
Give the magnitude and direction of the acceleration of the body.
5.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child
standing in the middle of the road and brings his vehicle to rest in 4.0 s just
in time to save the child. What is the average retarding force on the vehicle ?
The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
5.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial
-2
acceleration of 5.0 m s . Calculate the initial thrust (force) of the blast.
-1
5.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s to
the north is subject to a constant force of 8.0 N directed towards the south
for 30 s. Take the instant the force is applied to be t = 0, the position of the
body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
5.11 A truck starts from rest and accelerates uniformly at 2.0 m s . At t = 10 s, a
-2
stone is dropped by a person standing on the top of the truck (6 m high from
the ground). What are the (a) velocity, and (b) acceleration of the stone at t =
11s ? (Neglect air resistance.)
5.12 A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is
-1
set into oscillation. The speed of the bob at its mean position is 1 m s .
What is the trajectory of the bob if the string is cut when the bob is (a) at one
of its extreme positions, (b) at its mean position.
5.13 A man of mass 70 kg stands on a weighing scale in a lift which is moving
-1
(a) upwards with a uniform speed of 10 m s ,
-2
(b) downwards with a uniform acceleration of 5 m s ,
(c) upwards with a uniform acceleration of 5 m s .
-2
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled
down freely under gravity ?
5.14 Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is
the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and
t = 4 s ? (Consider one-dimensional motion only).
Fig. 5.16
5.15 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal
surface are tied to the ends of a light string. A horizontal force F = 600 N is
applied to (i) A, (ii) B along the direction of string. What is the tension in the
string in each case?
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LAWS OF MOTION 111
5.16 Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible
string that goes over a frictionless pulley. Find the acceleration of the masses, and
the tension in the string when the masses are released.
5.17 A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates
into two smaller nuclei the products must move in opposite directions.
5.18 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s -1
collide and rebound with the same speed. What is the impulse imparted to each ball due
to the other ?
5.19 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the
-1
shell is 80 m s , what is the recoil speed of the gun ?
5.20 A batsman deflects a ball by an angle of 45° without changing its initial speed which is
equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg.)
5.21 A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius
1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the
string ? What is the maximum speed with which the stone can be whirled around if
the string can withstand a maximum tension of 200 N ?
5.22 If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible
value, and the string breaks suddenly, which of the following correctly describes the
trajectory of the stone after the string breaks :
(a) the stone moves radially outwards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the
speed of the particle ?
5.23 Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops
suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.
Additional Exercises
5.24 Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a
suitable physical context for this motion. What is the time between two consecutive
impulses received by the body ? What is the magnitude of each impulse ?
Fig. 5.17
5.25 Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor
belt that is accelerating with 1 m s . What is the net force on the man? If the
-2
coefficient of static friction between the man’s shoes and the belt is 0.2, up to what
acceleration of the belt can the man continue to be stationary relative to the belt ?
(Mass of the man = 65 kg.)
Fig. 5.18
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112 PHYSICS
5.26 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R.
The net forces at the lowest and highest points of the circle directed vertically
downwards are : [Choose the correct alternative]
Lowest Point Highest Point
(a) mg – T mg + T
1 2
(b) mg + T mg – T
1 2
2
(c) mg + T – (m v ) / R mg – T + (m v ) / R
2
1 1 2 1
2
2
(d) mg – T – (m v ) / R mg + T + (m v ) / R
1 1 2 1
T and v denote the tension and speed at the lowest point. T and v denote
1 1 2 2
corresponding values at the highest point.
-2
5.27 A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s . The crew
and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force on the helicopter due to the surrounding air.
5.28 A stream of water flowing horizontally with a speed of 15 m s gushes out of a tube of
-1
-2
2
cross-sectional area 10 m , and hits a vertical wall nearby. What is the force exerted
on the wall by the impact of water, assuming it does not rebound ?
5.29 Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m.
Give the magnitude and direction of
th
(a) the force on the 7 coin (counted from the bottom) due to all the coins on its top,
th
(b) the force on the 7 coin by the eighth coin,
th
th
(c) the reaction of the 6 coin on the 7 coin.
5.30 An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked
at 15°. What is the radius of the loop ?
5.31 A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h.
6
The mass of the train is 10 kg. What provides the centripetal force required for this
purpose — The engine or the rails ? What is the angle of banking required to prevent
wearing out of the rail ?
5.32 A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in
Fig. 5.19. What is the action on the floor by the man in the two cases ? If the floor
yields to a normal force of 700 N, which mode should the man adopt to lift the block
without the floor yielding ?
Fig. 5.19
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LAWS OF MOTION 113
5.33 A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which
can stand a maximum tension of 600 N. In which of the
following cases will the rope break: the monkey
(a) climbs up with an acceleration of 6 m s -2
(b) climbs down with an acceleration of 4 m s -2
(c) climbs up with a uniform speed of 5 m s -1
(d) falls down the rope nearly freely under gravity?
(Ignore the mass of the rope). Fig. 5.20
5.34 Two bodies A and B of masses 5 kg and 10 kg in contact with
each other rest on a table against a rigid wall (Fig. 5.21). The
coefficient of friction between the bodies and the table is
0.15. A force of 200 N is applied horizontally to A. What are
(a) the reaction of the partition (b) the action-reaction forces
between A and B ? What happens when the wall is removed?
Does the answer to (b) change, when the bodies are in motion?
Ignore the difference between µ and µ . Fig. 5.21
s k
5.35 A block of mass 15 kg is placed on a long trolley. The coefficient of static friction
between the block and the trolley is 0.18. The trolley accelerates from rest with
-2
0.5 m s for 20 s and then moves with uniform velocity. Discuss the motion of the
block as viewed by (a) a stationary observer on the ground, (b) an observer moving with
the trolley.
5.36 The rear side of a truck is open and a box of 40 kg
mass is placed 5 m away from the open end as shown
in Fig. 5.22. The coefficient of friction between the
box and the surface below it is 0.15. On a straight
road, the truck starts from rest and accelerates with
2 m s . At what distance from the starting point Fig. 5.22
-2
does the box fall off the truck? (Ignore the size of
the box).
1
5.37 A disc revolves with a speed of 33 rev/min, and has a radius of 15 cm. Two coins are
3
placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction
between the coins and the record is 0.15, which of the coins will revolve with the record ?
5.38 You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-
well’ (a hollow spherical chamber with holes, so the spectators can watch from outside).
Explain clearly why the motorcyclist does not drop down when he is at the uppermost
point, with no support from below. What is the minimum speed required at the
uppermost position to perform a vertical loop if the radius of the chamber is 25 m ?
5.39 A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of
radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of
friction between the wall and his clothing is 0.15. What is the minimum rotational
speed of the cylinder to enable the man to remain stuck to the wall (without falling)
when the floor is suddenly removed ?
5.40 A thin circular loop of radius R rotates about its vertical diameter with an angular
frequency ω. Show that a small bead on the wire loop remains at its lowermost point
for ω ≤ g / R . What is the angle made by the radius vector joining the centre to
the bead with the vertical downward direction for ω = 2g / R ? Neglect friction.
2018-19
CHAPTER SIX
WORK, ENERGY AND POWER
6.1 INTRODUCTION
The terms ‘work’, ‘energy’ and ‘power’ are frequently used
in everyday language. A farmer ploughing the field, a
construction worker carrying bricks, a student studying for
6.1 Introduction
a competitive examination, an artist painting a beautiful
6.2 Notions of work and kinetic
landscape, all are said to be working. In physics, however,
energy : The work-energy
theorem the word ‘Work’ covers a definite and precise meaning.
Somebody who has the capacity to work for 14-16 hours a
6.3 Work
day is said to have a large stamina or energy. We admire a
6.4 Kinetic energy
long distance runner for her stamina or energy. Energy is
6.5 Work done by a variable
thus our capacity to do work. In Physics too, the term ‘energy’
force
is related to work in this sense, but as said above the term
6.6 The work-energy theorem for
‘work’ itself is defined much more precisely. The word ‘power’
a variable force
is used in everyday life with different shades of meaning. In
6.7 The concept of potential
karate or boxing we talk of ‘powerful’ punches. These are
energy
delivered at a great speed. This shade of meaning is close to
6.8 The conservation of the meaning of the word ‘power’ used in physics. We shall
mechanical energy
find that there is at best a loose correlation between the
6.9 The potential energy of a physical definitions and the physiological pictures these
spring
terms generate in our minds. The aim of this chapter is to
6.10 Various forms of energy : the develop an understanding of these three physical quantities.
law of conservation of energy
Before we proceed to this task, we need to develop a
6.11 Power mathematical prerequisite, namely the scalar product of two
6.12 Collisions vectors.
Summary 6.1.1 The Scalar Product
Points to ponder We have learnt about vectors and their use in Chapter 4.
Exercises Physical quantities like displacement, velocity, acceleration,
Additional exercises
force etc. are vectors. We have also learnt how vectors are
Appendix 6.1
added or subtracted. We now need to know how vectors are
multiplied. There are two ways of multiplying vectors which
we shall come across : one way known as the scalar product
gives a scalar from two vectors and the other known as the
vector product produces a new vector from two vectors. We
shall look at the vector product in Chapter 7. Here we take
up the scalar product of two vectors. The scalar product or
dot product of any two vectors A and B, denoted as A . B (read
2018-19
WORK, ENERGY AND POWER 115
A dot B) is defined as
A = A i + A j + A k
.
A B = A B cos θ (6.1a) x y z
B = B i + B j + B k
where θ is the angle between the two vectors as x y z
shown in Fig. 6.1(a). Since A, B and cos θ are their scalar product is
scalars, the dot product of A and B is a scalar A B = ( A i ˆ + A j ˆ + A k ) ( B i ˆ + B j + B k )
ˆ .
ˆ
.
ˆ
quantity. Each vector, A and B, has a direction x y z x y z
but their scalar product does not have a = A B x + A B y + A B z (6.1b)
y
z
x
direction.
From the definition of scalar product and
From Eq. (6.1a), we have (Eq. 6.1b) we have :
. .. .
.
A B = A (B cos θ ) ( i ) A A = A A + A A + A A z
y
x
y
x
z
2
2
2
= B (A cos θ ) Or, A = A + A + A 2 (6.1c)
x y z
Geometrically, B cos θ is the projection of B onto since A . A = |A ||A| cos 0 = A .
2
A in Fig.6.1 (b) and A cos θ is the projection of A (ii) A . B = 0, if A and B are perpendicular.
.
onto B in Fig. 6.1 (c). So, A B is the product of
the magnitude of A and the component of B along u Example 6.1 Find the angle between force
A. Alternatively, it is the product of the F = (3 + 4 -5 )i ˆ j k
ˆ
ˆ unit and displacement
magnitude of B and the component of A along B.
ˆ
ˆ unit. Also find the
d = (5 + 4 +3 )i ˆ j k
Equation (6.1a) shows that the scalar product
follows the commutative law : projection of F on d.
.
.
.
A B = B A Answer F d = F d + F d + F d z
z
y
y
x
x
= 3 (5) + 4 (4) + (– 5) (3)
Scalar product obeys the distributive
law: = 16 unit
Hence F . d = F d cosθ = 16 unit
A . (B + C) = A . B + A . C
.
+
2
2
Further, A . (λ B) = λ (A . B) Now F F = F = F x 2 F + F z 2
y
where λ is a real number. = 9 + 16 + 25
= 50 unit
The proofs of the above equations are left to
.
+
2
2
you as an exercise. and d d = d = d 2 x d + d z 2
y
For unit vectors i, j,k we have = 25 + 16 + 9
= 50 unit
⋅
j j =
⋅
⋅
i i = k k = 1 16 16
∴ cos θ = = = 0.32 ,
i j = k i = 0 50 50 50
j k =
⋅
⋅
⋅
–1
Given two vectors θ = cos 0.32
.
Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A B = A B cos θ. (b) B cos θ is the projection
of B onto A. (c) A cos θ is the projection of A onto B.
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116 PHYSICS
6.2 NOTIONS OF WORK AND KINETIC
ENERGY: THE WORK-ENERGY THEOREM to be proportional to the speed of the drop
but is otherwise undetermined. Consider
The following relation for rectilinear motion under a drop of mass 1.00 g falling from a height
constant acceleration a has been encountered 1.00 km. It hits the ground with a speed of
in Chapter 3, 50.0 m s . (a) What is the work done by the
-1
2
2
v − u = 2 as (6.2) gravitational force ? What is the work done
where u and v are the initial and final speeds by the unknown resistive force?
and s the distance traversed. Multiplying both Answer (a) The change in kinetic energy of the
sides by m/2, we have
drop is
1 1 1
2
mv − mu = mas = Fs (6.2a) ∆ K = m v − 0
2
2
2 2 2
where the last step follows from Newton’s Second = 1 × -3 × ×
Law. We can generalise Eq. (6.2) to three 2 10 50 50
dimensions by employing vectors = 1.25 J
v − u = 2 a.d where we have assumed that the drop is initially
2
2
at rest.
Here a and d are acceleration and displacement
Assuming that g is a constant with a value
vectors of the object respectively. 2
10 m/s , the work done by the gravitational force
Once again multiplying both sides by m/2 , we obtain
is,
1 1
mv − mu = m a.d = F.d (6.2b) W = mgh
2
2
g
2 2 = 10 ×10 ×10 3
-3
The above equation provides a motivation for = 10.0 J
the definitions of work and kinetic energy. The (b) From the work-energy theorem
left side of the equation is the difference in the
=
quantity ‘half the mass times the square of the ∆ K W + W r
g
speed’ from its initial value to its final value. We where W is the work done by the resistive force
r
call each of these quantities the ‘kinetic energy’, on the raindrop. Thus
denoted by K. The right side is a product of the W = ∆K − W
r g
displacement and the component of the force = 1.25 −10
along the displacement. This quantity is called = − 8.75 J
‘work’ and is denoted by W. Eq. (6.2b) is then
is negative. t
K − K = W (6.3)
f i 6.3 WORK
where K and K are respectively the initial and As seen earlier, work is related to force and the
i f
final kinetic energies of the object. Work refers displacement over which it acts. Consider a
to the force and the displacement over which it constant force F acting on an object of mass m.
acts. Work is done by a force on the body over The object undergoes a displacement d in the
a certain displacement. positive x-direction as shown in Fig. 6.2.
Equation (6.2) is also a special case of the
work-energy (WE) theorem : The change in
kinetic energy of a particle is equal to the
work done on it by the net force. We shall
generalise the above derivation to a varying force
in a later section.
t Example 6.2 It is well known that a
raindrop falls under the influence of the Fig. 6.2 An object undergoes a displacement d
downward gravitational force and the under the influence of the force F.
opposing resistive force. The latter is known
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WORK, ENERGY AND POWER 117
The work done by the force is defined to be Table 6.1 Alternative Units of Work/Energy in J
the product of component of the force in the
direction of the displacement and the
magnitude of this displacement. Thus
W = (F cos θ )d = F.d (6.4)
We see that if there is no displacement, there
is no work done even if the force is large. Thus,
when you push hard against a rigid brick wall,
the force you exert on the wall does no work. Yet t Example 6.3 A cyclist comes to a skidding
your muscles are alternatively contracting and stop in 10 m. During this process, the force
relaxing and internal energy is being used up on the cycle due to the road is 200 N and
and you do get tired. Thus, the meaning of work is directly opposed to the motion. (a) How
in physics is different from its usage in everyday much work does the road do on the cycle ?
language. (b) How much work does the cycle do on
the road ?
No work is done if :
(i) the displacement is zero as seen in the
example above. A weightlifter holding a 150 Answer Work done on the cycle by the road is
kg mass steadily on his shoulder for 30 s the work done by the stopping (frictional) force
does no work on the load during this time. on the cycle due to the road.
(ii) the force is zero. A block moving on a smooth (a) The stopping force and the displacement make
horizontal table is not acted upon by a an angle of 180 o (π rad) with each other.
horizontal force (since there is no friction), but Thus, work done by the road,
may undergo a large displacement. W = Fd cosθ
r
(iii) the force and displacement are mutually = 200 × 10 × cos π
perpendicular. This is so since, for θ = π/2 rad
(= 90 ), cos (π/2) = 0. For the block moving on = – 2000 J
o
a smooth horizontal table, the gravitational It is this negative work that brings the cycle
to a halt in accordance with WE theorem.
force mg does no work since it acts at right
(b) From Newton’s Third Law an equal and
angles to the displacement. If we assume that
the moon’s orbits around the earth is opposite force acts on the road due to the
perfectly circular then the earth’s cycle. Its magnitude is 200 N. However, the
road undergoes no displacement. Thus,
gravitational force does no work. The moon’s
instantaneous displacement is tangential work done by cycle on the road is zero. t
while the earth’s force is radially inwards and The lesson of Example 6.3 is that though the
θ = π/2. force on a body A exerted by the body B is always
Work can be both positive and negative. If θ is equal and opposite to that on B by A (Newton’s
between 0 and 90 , cos θ in Eq. (6.4) is positive. Third Law); the work done on A by B is not
o
o
o
o
If θ is between 90 and 180 , cos θ is negative. necessarily equal and opposite to the work done
In many examples the frictional force opposes on B by A.
o
displacement and θ = 180 . Then the work done
6.4 KINETIC ENERGY
o
by friction is negative (cos 180 = –1).
As noted earlier, if an object of mass m has
From Eq. (6.4) it is clear that work and energy
velocity v, its kinetic energy K is
–2
2
have the same dimensions, [ML T ]. The SI unit
of these is joule (J), named after the famous British 1 1
2
physicist James Prescott Joule (1811-1869). Since K = m v v = mv (6.5)
.
work and energy are so widely used as physical 2 2
concepts, alternative units abound and some of Kinetic energy is a scalar quantity. The kinetic
these are listed in Table 6.1. energy of an object is a measure of the work an
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118 PHYSICS
Table 6.2 Typical kinetic energies (K)
object can do by the virtue of its motion. This This is illustrated in Fig. 6.3(a). Adding
notion has been intuitively known for a long time. successive rectangular areas in Fig. 6.3(a) we
The kinetic energy of a fast flowing stream get the total work done as
has been used to grind corn. Sailing x
f
ships employ the kinetic energy of the wind. Table W ≅ ∑ F ( )∆x (6.6)
x
6.2 lists the kinetic energies for various i x
objects. where the summation is from the initial position
x to the final position x .
Example 6.4 In a ballistics demonstration i f
t
a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach
-1
with speed 200 m s (see Table 6.2) on soft zero, then the number of terms in the sum
plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches
emerges with only 10% of its initial kinetic a definite value equal to the area under the curve
energy. What is the emergent speed of the in Fig. 6.3(b). Then the work done is
bullet ?
lim x f
W = lim ∑ F ( )∆x
x
Answer The initial kinetic energy of the bullet ∆ x → 0
is mv /2 = 1000 J. It has a final kinetic energy i x
2
of 0.1×1000 = 100 J. If v is the emergent speed x f
f
=
of the bullet, ∫ F x (6.7)
( ) xd
1 2 x i
mv f = 100 J
2 where ‘lim’ stands for the limit of the sum when
∆x tends to zero. Thus, for a varying force
2× 100 J the work done can be expressed as a definite
v f = . 0 05 kg integral of force over displacement (see also
Appendix 3.1).
= 63.2 m s –1
The speed is reduced by approximately 68%
(not 90%). t
6.5 WORK DONE BY A VARIABLE FORCE
A constant force is rare. It is the variable force,
which is more commonly encountered. Fig. 6.3
is a plot of a varying force in one dimension.
If the displacement ∆x is small, we can take
the force F (x) as approximately constant and
the work done is then
∆W =F (x) ∆x Fig. 6.3(a)
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WORK, ENERGY AND POWER 119
The work done by the frictional force is
W → area of the rectangle AGHI
f
W = (−50) × 20
f
= − 1000 J
The area on the negative side of the force axis
has a negative sign. t
6.6 THE WORK-ENERGY THEOREM FOR A
VARIABLE FORCE
We are now familiar with the concepts of work
Fig. 6.3 (a) The shaded rectangle represents the
work done by the varying force F(x), over and kinetic energy to prove the work-energy
the small displacement ∆x, ∆W = F(x) ∆x. theorem for a variable force. We confine
(b) adding the areas of all the rectangles we ourselves to one dimension. The time rate of
find that for ∆x → 0, the area under the curve change of kinetic energy is
is exactly equal to the work done by F(x).
dK d 1
= m v 2
Example 6.5 A woman pushes a trunk on dt dt 2
t
a railway platform which has a rough dv
surface. She applies a force of 100 N over a = m v
distance of 10 m. Thereafter, she gets dt
progressively tired and her applied force F= v (from Newton’s Second Law)
reduces linearly with distance to 50 N. The dx
total distance through which the trunk has = F dt
been moved is 20 m. Plot the force applied
Thus
by the woman and the frictional force, which
dK = Fdx
is 50 N versus displacement. Calculate the
Integrating from the initial position (x ) to final
work done by the two forces over 20 m. i
position ( x ), we have
f
Answer K f x f
∫ dK = ∫ F dx
K i x i
where, K and K are the initial and final kinetic
i f
energies corresponding to x and x .
i f
x f
K − K = ∫ F x
d
or f i (6.8a)
Fig. 6.4 Plot of the force F applied by the woman and
x i
the opposing frictional force f versus
From Eq. (6.7), it follows that
displacement.
K − K = W (6.8b)
The plot of the applied force is shown in Fig. f i
6.4. At x = 20 m, F = 50 N (≠ 0). We are given Thus, the WE theorem is proved for a variable
that the frictional force f is |f|= 50 N. It opposes force.
motion and acts in a direction opposite to F. It While the WE theorem is useful in a variety of
is therefore, shown on the negative side of the problems, it does not, in general, incorporate the
force axis. complete dynamical information of Newton’s
The work done by the woman is second law. It is an integral form of Newton’s
W → area of the rectangle ABCD + area of second law. Newton’s second law is a relation
F
the trapezium CEID between acceleration and force at any instant of
1 time. Work-energy theorem involves an integral
+
W = 100 ×10 + (100 50 ) ×10 over an interval of time. In this sense, the temporal
F
2
= 1000 + 750 (time) information contained in the statement of
= 1750 J Newton’s second law is ‘integrated over’ and is
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120 PHYSICS
not available explicitly. Another observation is that are like ‘compressed springs’. They possess a
Newton’s second law for two or three dimensions large amount of potential energy. An earthquake
is in vector form whereas the work-energy results when these fault lines readjust. Thus,
theorem is in scalar form. In the scalar form, potential energy is the ‘stored energy’ by virtue
information with respect to directions contained of the position or configuration of a body. The
in Newton’s second law is not present. body left to itself releases this stored energy in
the form of kinetic energy. Let us make our notion
Example 6.6 A block of mass m = 1 kg, of potential energy more concrete.
t
moving on a horizontal surface with speed The gravitational force on a ball of mass m is
–1
v = 2 m s enters a rough patch ranging mg . g may be treated as a constant near the earth
i
from x = 0.10 m to x = 2.01 m. The retarding surface. By ‘near’ we imply that the height h of
force F on the block in this range is inversely the ball above the earth’s surface is very small
r
proportional to x over this range, compared to the earth’s radius R (h <<R ) so that
E E
− k we can ignore the variation of g near the earth’s
F =
r for 0.1 < x < 2.01 m surface*. In what follows we have taken the
x
= 0 for x < 0.1m and x > 2.01 m upward direction to be positive. Let us raise the
where k = 0.5 J. What is the final kinetic ball up to a height h. The work done by the external
energy and speed v of the block as it agency against the gravitational force is mgh. This
f work gets stored as potential energy.
crosses this patch ?
Gravitational potential energy of an object, as a
function of the height h, is denoted by V(h) and it
Answer From Eq. (6.8a) is the negative of work done by the gravitational
2.01 − ( k) force in raising the object to that height.
K = K + ∫ d x
f i V (h) = mgh
0.1 x If h is taken as a variable, it is easily seen that
the gravitational force F equals the negative of
1
( )
2
= mv − k ln x 2.01 the derivative of V(h) with respect to h. Thus,
i
0.1
2
d
F = − V(h) = − m g
1 dh
2
= mv − k ln (2.01/0.1 )
2 i The negative sign indicates that the
gravitational force is downward. When released,
= 2 − 0.5 ln (20.1)
the ball comes down with an increasing speed.
= 2 − 1.5 = 0.5 J
Just before it hits the ground, its speed is given
v f = 2K f /m = 1 m s − 1 by the kinematic relation,
v = 2gh
2
This equation can be written as
Here, note that ln is a symbol for the natural
logarithm to the base e and not the logarithm to 1 2
the base 10 [ln X = log X = 2.303 log X]. t m v = m g h
e 10 2
which shows that the gravitational potential
6.7 THE CONCEPT OF POTENTIAL ENERGY
energy of the object at height h, when the object
The word potential suggests possibility or is released, manifests itself as kinetic energy of
capacity for action. The term potential energy the object on reaching the ground.
brings to one’s mind ‘stored’ energy. A stretched Physically, the notion of potential energy is
bow-string possesses potential energy. When it applicable only to the class of forces where work
is released, the arrow flies off at a great speed. done against the force gets ‘stored up’ as energy.
The earth’s crust is not uniform, but has When external constraints are removed, it
discontinuities and dislocations that are called manifests itself as kinetic energy. Mathematically,
fault lines. These fault lines in the earth’s crust (for simplicity, in one dimension) the potential
* The variation of g with height is discussed in Chapter 8 on Gravitation.
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WORK, ENERGY AND POWER 121
energy V(x) is defined if the force F(x) can be which means that K + V, the sum of the kinetic
written as and potential energies of the body is a constant.
dV Over the whole path, x to x , this means that
i
f
( ) = −
F x K + V(x ) = K + V(x ) (6.11)
dx i i f f
This implies that The quantity K +V(x), is called the total
mechanical energy of the system. Individually
x f V f the kinetic energy K and the potential energy
∫ F(x) x = − ∫ d V = V − V f V(x) may vary from point to point, but the sum
d
i
x i V i
is a constant. The aptness of the term
The work done by a conservative force such as ‘conservative force’ is now clear.
gravity depends on the initial and final positions Let us consider some of the definitions of a
only. In the previous chapter we have worked conservative force.
on examples dealing with inclined planes. If an l A force F(x) is conservative if it can be derived
object of mass m is released from rest, from the
from a scalar quantity V(x) by the relation
top of a smooth (frictionless) inclined plane of given by Eq. (6.9). The three-dimensional
height h, its speed at the bottom generalisation requires the use of a vector
is 2 gh irrespective of the angle of inclination. derivative, which is outside the scope of this
book.
Thus, at the bottom of the inclined plane it
acquires a kinetic energy, mgh. If the work done l The work done by the conservative force
depends only on the end points. This can be
or the kinetic energy did depend on other factors
seen from the relation,
such as the velocity or the particular path taken
W = K – K = V (x ) – V(x )
by the object, the force would be called non- f i i f
which depends on the end points.
conservative. l A third definition states that the work done
The dimensions of potential energy are
[ML T ] and the unit is joule (J), the same as by this force in a closed path is zero. This is
–2
2
once again apparent from Eq. (6.11) since
kinetic energy or work. To reiterate, the change x = x
in potential energy, for a conservative force, i f .
∆V is equal to the negative of the work done by Thus, the principle of conservation of total
the force mechanical energy can be stated as
∆V = − F(x) ∆x (6.9) The total mechanical energy of a system is
In the example of the falling ball considered in conserved if the forces, doing work on it, are
this section we saw how potential energy was conservative.
converted to kinetic energy. This hints at an The above discussion can be made more
important principle of conservation in mechanics, concrete by considering the example of the
gravitational force once again and that of the
which we now proceed to examine.
spring force in the next section. Fig. 6.5 depicts
6.8 THE CONSERVATION OF MECHANICAL a ball of mass m being dropped from a cliff of
ENERGY height H.
For simplicity we demonstrate this important
principle for one-dimensional motion. Suppose
that a body undergoes displacement ∆x under
the action of a conservative force F. Then from
the WE theorem we have,
∆K = F(x) ∆x
If the force is conservative, the potential energy
function V(x) can be defined such that
− ∆V = F(x) ∆x
The above equations imply that Fig. 6.5 The conversion of potential energy to kinetic
∆K + ∆V = 0 energy for a ball of mass m dropped from a
∆(K + V ) = 0 (6.10) height H.
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122 PHYSICS
The total mechanical energies E , E , and E Answer (i) There are two external forces on
0 h H
of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the
level), h and H, are string. The latter does no work since the
displacement of the bob is always normal to the
E = mgH (6.11 a)
H string. The potential energy of the bob is thus
1
E = mgh + mv 2 (6.11 b) associated with the gravitational force only. The
h
2 h total mechanical energy E of the system is
E = (1/2) mv 2 (6.11 c)
0 f conserved. We take the potential energy of the
The constant force is a special case of a spatially system to be zero at the lowest point A. Thus,
dependent force F(x). Hence, the mechanical at A :
energy is conserved. Thus
E = E
H 0 1 2
1 2 E = mv 0 (6.12)
or, mgH = mv f 2
2
v = 2gH [Newton’s Second Law]
f
a result that was obtained in section 3.7 for a where T is the tension in the string at A. At the
A
freely falling body. highest point C, the string slackens, as the
Further, tension in the string (T ) becomes zero.
E = E C
H h Thus, at C
which implies,
2 1
2
v h = 2 g(H − h) (6.11 d) E = mv + 2mgL (6.13)
c
and is a familiar result from kinematics. 2
At the height H, the energy is purely potential. mv 2
It is partially converted to kinetic at height h and mg= L c [Newton’s Second Law] (6.14)
is fully kinetic at ground level. This illustrates
the conservation of mechanical energy. where v is the speed at C. From Eqs. (6.13) and
C
(6.14)
Example 6.7 A bob of mass m is suspended
5
t
by a light string of length L . It is imparted a E = mgL
horizontal velocity v at the lowest point A 2
o
such that it completes a semi-circular Equating this to the energy at A
trajectory in the vertical plane with the string
5 m
becoming slack only on reaching the topmost mgL = v 2 0
point, C. This is shown in Fig. 6.6. Obtain an 2 2
expression for (i) v ; (ii) the speeds at points or, v = 5gL
o 0
B and C; (iii) the ratio of the kinetic energies
(K /K ) at B and C. Comment on the nature (ii) It is clear from Eq. (6.14)
B C
of the trajectory of the bob after it reaches v C = gL
the point C.
At B, the energy is
1
E = mv + mgL
2
2 B
Equating this to the energy at A and employing
the result from (i), namely v = 5 gL ,
2
0
1 2 1 2
mv + mgL = mv
2 B 2 0
5
= m g L
Fig. 6.6 2
2018-19
WORK, ENERGY AND POWER 123
∴ v = 3 gL 2
B W = + x k m (6.16)
2
(iii) The ratio of the kinetic energies at B and C
is :
1 2
K B = 2 mv B = 3
K 1 2 1
C mv
2 C
At point C, the string becomes slack and the
velocity of the bob is horizontal and to the left. If
the connecting string is cut at this instant, the
bob will execute a projectile motion with
horizontal projection akin to a rock kicked
horizontally from the edge of a cliff. Otherwise
the bob will continue on its circular path and
complete the revolution. t
6.9 THE POTENTIAL ENERGY OF A SPRING
The spring force is an example of a variable force
which is conservative. Fig. 6.7 shows a block
attached to a spring and resting on a smooth
horizontal surface. The other end of the spring
is attached to a rigid wall. The spring is light
and may be treated as massless. In an ideal
spring, the spring force F is proportional to
s
x where x is the displacement of the block from
the equilibrium position. The displacement could
be either positive [Fig. 6.7(b)] or negative
[Fig. 6.7(c)]. This force law for the spring is called
Hooke’s law and is mathematically stated as
F = − kx
s
The constant k is called the spring constant. Its
-1
unit is N m . The spring is said to be stiff if k is
large and soft if k is small.
Fig. 6.7 Illustration of the spring force with a block
Suppose that we pull the block outwards as in
attached to the free end of the spring.
Fig. 6.7(b). If the extension is x , the work done by (a) The spring force F is zero when the
m s
the spring force is displacement x from the equilibrium position
is zero. (b) For the stretched spring x > 0
x m x m and F < 0 (c) For the compressed spring
s
W = ∫ F x = − ∫ kx x x < 0 and F > 0.(d) The plot of F versus x.
d
d
s
s
s
s
0 0 The area of the shaded triangle represents
the work done by the spring force. Due to the
x k 2 opposing signs of F and x, this work done is
= − m (6.15) s
2 2
negative, W = − kx / 2 .
s m
This expression may also be obtained by
considering the area of the triangle as in The same is true when the spring is
Fig. 6.7(d). Note that the work done by the compressed with a displacement x (< 0). The
external pulling force F is positive since it c
overcomes the spring force. spring force does work W = − kx c 2 / 2 while the
s
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124 PHYSICS
2
external force F does work + kx / 2 . If the block and vice versa, however, the total mechanical
c energy remains constant. This is graphically
is moved from an initial displacement x to a
i depicted in Fig. 6.8.
final displacement x , the work done by the
f
spring force W is
s
x f k x 2 k x 2
d
W = − ∫ k x x = 2 i − 2 f (6.17)
s
x i
Thus the work done by the spring force depends
only on the end points. Specifically, if the block
is pulled from x and allowed to return to x ;
i i
x i k x 2 k x 2
W = − ∫ k x x =d i − i
s
2 2
x i
= 0 (6.18) Fig. 6.8 Parabolic plots of the potential energy V and
The work done by the spring force in a cyclic kinetic energy K of a block attached to a
process is zero. We have explicitly demonstrated spring obeying Hooke’s law. The two plots
that the spring force (i) is position dependent are complementary, one decreasing as the
other increases. The total mechanical
only as first stated by Hooke, (F = − kx); (ii)
s energy E = K + V remains constant.
does work which only depends on the initial and
final positions, e.g. Eq. (6.17). Thus, the spring
force is a conservative force. t Example 6.8 To simulate car accidents, auto
We define the potential energy V(x) of the spring manufacturers study the collisions of moving
to be zero when block and spring system is in the cars with mounted springs of different spring
equilibrium position. For an extension (or constants. Consider a typical simulation with
compression) x the above analysis suggests that a car of mass 1000 kg moving with a speed
18.0 km/h on a smooth road and colliding
kx 2 with a horizontally mounted spring of spring
V(x) = (6.19) constant 6.25 × 10 3 N m . What is the
–1
2
You may easily verify that − dV/dx = − k x, the maximum compression of the spring ?
spring force. If the block of mass m in Fig. 6.7 is
extended to x and released from rest, then its Answer At maximum compression the kinetic
m
total mechanical energy at any arbitrary point x, energy of the car is converted entirely into the
where x lies between – x and + x will be given by potential energy of the spring.
m m,
The kinetic energy of the moving car is
1 2 1 2 1 2
x k m = x k + m v 1
2 2 2 K = mv 2
where we have invoked the conservation of 2
mechanical energy. This suggests that the speed 1 3
and the kinetic energy will be maximum at the = 2 × 10 × 5 × 5
equilibrium position, x = 0, i.e.,
K = 1.25 × 10 J
4
1 2 1 2
–1
–1
m v m = x k m where we have converted 18 km h to 5 m s [It is
2 2 useful to remember that 36 km h = 10 m s ].
–1
–1
where v is the maximum speed. At maximum compression x , the potential
m m
energy V of the spring is equal to the kinetic
k energy K of the moving car from the principle of
or v = x m
m
m conservation of mechanical energy.
Note that k/m has the dimensions of [T ] and
-2
our equation is dimensionally correct. The V = 1 x k 2
kinetic energy gets converted to potential energy 2 m
2018-19
WORK, ENERGY AND POWER 125
4
= 1.25 × 10 J
We obtain
x = 2.00 m
m
We note that we have idealised the situation.
The spring is considered to be massless. The
surface has been considered to possess
negligible friction. t
We conclude this section by making a few Fig. 6.9 The forces acting on the car.
remarks on conservative forces.
(i) Information on time is absent from the above 1 2
discussions. In the example considered ∆K = K − K 0= − 2 m v
i
f
above, we can calculate the compression, but The work done by the net force is
not the time over which the compression 1
−
=
−
2
occurs. A solution of Newton’s Second Law W kx m g x m
µ
m
for this system is required for temporal 2
information. Equating we have
(ii) Not all forces are conservative. Friction, for
1 2 1 2
example, is a non-conservative force. The m v = k x m + m g x m
µ
2 2
principle of conservation of energy will have 3 3
to be modified in this case. This is illustrated Now µmg = 0.5 × 10 × 10 = 5 × 10 N (taking
-2
g =10.0 m s ). After rearranging the above
in Example 6.9.
equation we obtain the following quadratic
(iii) The zero of the potential energy is arbitrary.
equation in the unknown x .
It is set according to convenience. For the m
spring force we took V(x) = 0, at x = 0, i.e. the 2 + − m v =
2
k x m 2 m g x m 0
µ
unstretched spring had zero potential
energy. For the constant gravitational force
mg, we took V = 0 on the earth’s surface. In
a later chapter we shall see that for the force
due to the universal law of gravitation, the where we take the positive square root since
zero is best defined at an infinite distance x is positive. Putting in numerical values we
m
from the gravitational source. However, once obtain
the zero of the potential energy is fixed in a x = 1.35 m
m
given discussion, it must be consistently which, as expected, is less than the result in
adhered to throughout the discussion. You Example 6.8.
cannot change horses in midstream ! If the two forces on the body consist of a
conservative force F and a non-conservative
c
Example 6.9 Consider Example 6.8 taking force F , the conservation of mechanical energy
nc
t
the coefficient of friction, µ, to be 0.5 and formula will have to be modified. By the WE
calculate the maximum compression of the theorem
spring. (F + F ) ∆x = ∆K
nc
c
But F ∆x = − ∆V
c
Answer In presence of friction, both the spring Hence, ∆(K + V) = F ∆x
nc
force and the frictional force act so as to oppose ∆E = F ∆x
nc
the compression of the spring as shown in where E is the total mechanical energy. Over
Fig. 6.9. the path this assumes the form
We invoke the work-energy theorem, rather E −− −− − E = W
f i nc
than the conservation of mechanical energy.
where W nc is the total work done by the
The change in kinetic energy is non-conservative forces over the path. Note that
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126 PHYSICS
unlike the conservative force, W depends on Chemical energy arises from the fact that the
nc
the particular path i to f. t molecules participating in the chemical reaction
have different binding energies. A stable chemical
6.10 VARIOUS FORMS OF ENERGY : THE LAW compound has less energy than the separated parts.
OF CONSERVATION OF ENERGY
A chemical reaction is basically a rearrangement
In the previous section we have discussed of atoms. If the total energy of the reactants is more
mechanical energy. We have seen that it can be than the products of the reaction, heat is released
classified into two distinct categories : one based and the reaction is said to be an exothermic
on motion, namely kinetic energy; the other on reaction. If the reverse is true, heat is absorbed and
configuration (position), namely potential energy. the reaction is endothermic. Coal consists of
Energy comes in many a forms which transform carbon and a kilogram of it when burnt releases
into one another in ways which may not often 7
about 3 × 10 J of energy.
be clear to us.
Chemical energy is associated with the forces
6.10.1 Heat that give rise to the stability of substances. These
forces bind atoms into molecules, molecules into
We have seen that the frictional force is not a
polymeric chains, etc. The chemical energy
conservative force. However, work is associated
arising from the combustion of coal, cooking gas,
with the force of friction, Example 6.5. A block of
mass m sliding on a rough horizontal surface wood and petroleum is indispensable to our daily
existence.
with speed v comes to a halt over a distance x .
0 0
The work done by the force of kinetic friction f
6.10.3 Electrical Energy
over x is –f x . By the work-energy theorem
0 0
The flow of electrical current causes bulbs to
2
m v /2 = f x . If we confine our scope to glow, fans to rotate and bells to ring. There are
0
o
mechanics, we would say that the kinetic energy laws governing the attraction and repulsion of
of the block is ‘lost’ due to the frictional force. charges and currents, which we shall learn
On examination of the block and the table we later. Energy is associated with an electric
would detect a slight increase in their current. An urban Indian household consumes
temperatures. The work done by friction is not about 200 J of energy per second on an average.
‘lost’, but is transferred as heat energy. This
raises the internal energy of the block and the 6.10.4 The Equivalence of Mass and Energy
table. In winter, in order to feel warm, we Till the end of the nineteenth century, physicists
generate heat by vigorously rubbing our palms believed that in every physical and chemical
together. We shall see later that the internal process, the mass of an isolated system is
energy is associated with the ceaseless, often conserved. Matter might change its phase, e.g.
random, motion of molecules. A quantitative idea glacial ice could melt into a gushing stream, but
of the transfer of heat energy is obtained by matter is neither created nor destroyed; Albert
noting that 1 kg of water releases about 42000 J Einstein (1879-1955) however, showed that mass
of energy when it cools by10 °C.
and energy are equivalent and are related by
6.10.2 Chemical Energy the relation
One of the greatest technical achievements of E = m c 2 (6.20)
humankind occurred when we discovered how where c, the speed of light in vacuum is
–1
8
to ignite and control fire. We learnt to rub two approximately 3 ×10 m s . Thus, a staggering
flint stones together (mechanical energy), got amount of energy is associated with a mere
them to heat up and to ignite a heap of dry leaves kilogram of matter
8 2
16
(chemical energy), which then provided E = 1× (3 ×10 ) J = 9 ×10 J.
sustained warmth. A matchstick ignites into a This is equivalent to the annual electrical output
bright flame when struck against a specially of a large (3000 MW) power generating station.
prepared chemical surface. The lighted
6.10.5 Nuclear Energy
matchstick, when applied to a firecracker,
results in a spectacular display of sound and The most destructive weapons made by man, the
light. fission and fusion bombs are manifestations of
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WORK, ENERGY AND POWER 127
Table 6.3 Approximate energy associated with various phenomena
the above equivalence of mass and energy [Eq.
(6.20)]. On the other hand the explanation of the t Example 6.10 Examine Tables 6.1-6.3
and express (a) The energy required to
life-nourishing energy output of the sun is also break one bond in DNA in eV; (b) The
based on the above equation. In this case kinetic energy of an air molecule (10 —21 J)
effectively four light hydrogen nuclei fuse to form in eV; (c) The daily intake of a human adult
a helium nucleus whose mass is less than the in kilocalories.
sum of the masses of the reactants. This mass
difference, called the mass defect ∆m is the
Answer (a) Energy required to break one bond
2
source of energy (∆m)c . In fission, a heavy
of DNA is
nucleus like uranium 235 U , is split by a neutron
92 − 20
10 J
into lighter nuclei. Once again the final mass is 1.6 10 − 19 J/eV ~ 0.06 eV
×
less than the initial mass and the mass difference
translates into energy, which can be tapped to
provide electrical energy as in nuclear power Note 0.1 eV = 100 meV (100 millielectron volt).
plants (controlled nuclear fission) or can be (b) The kinetic energy of an air molecule is
employed in making nuclear weapons 10 − 21 J
(uncontrolled nuclear fission). Strictly, the energy − 19 ~ 0.0062 eV
×
∆E released in a chemical reaction can also be 1.6 10 J/eV
related to the mass defect ∆m = ∆E/c . However, This is the same as 6.2 meV.
2
for a chemical reaction, this mass defect is much (c) The average human consumption in a day is
smaller than for a nuclear reaction. Table 6.3 7
lists the total energies for a variety of events and 10 J ~ 2400 kcal
phenomena. 4.2×10 3 J/kcal
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128 PHYSICS
We point out a common misconception created W
by newspapers and magazines. They mention P av =
food values in calories and urge us to restrict t
diet intake to below 2400 calories. What they The instantaneous power is defined as the
should be saying is kilocalories (kcal) and not limiting value of the average power as time
calories. A person consuming 2400 calories a interval approaches zero,
day will soon starve to death! 1 food calorie is dW
1 kcal. t P = dt (6.21)
The work dW done by a force F for a displacement
6.10.6 The Principle of Conservation of
dr is dW = F.dr. The instantaneous power can
Energy
also be expressed as
We have seen that the total mechanical energy
dr
of the system is conserved if the forces doing work P = F.
dt
on it are conservative. If some of the forces
involved are non-conservative, part of the = F.v (6.22)
mechanical energy may get transformed into
where v is the instantaneous velocity when the
other forms such as heat, light and sound.
force is F.
However, the total energy of an isolated system
Power, like work and energy, is a scalar
does not change, as long as one accounts for all 2 –3
quantity. Its dimensions are [ML T ]. In the SI,
forms of energy. Energy may be transformed from its unit is called a watt (W). The watt is 1 J s .
–1
one form to another but the total energy of an The unit of power is named after James Watt,
isolated system remains constant. Energy can one of the innovators of the steam engine in the
neither be created, nor destroyed. eighteenth century.
Since the universe as a whole may be viewed There is another unit of power, namely the
as an isolated system, the total energy of the horse-power (hp)
universe is constant. If one part of the universe
1 hp = 746 W
loses energy, another part must gain an equal
This unit is still used to describe the output of
amount of energy.
automobiles, motorbikes, etc.
The principle of conservation of energy cannot
We encounter the unit watt when we buy
be proved. However, no violation of this principle electrical goods such as bulbs, heaters and
has been observed. The concept of conservation refrigerators. A 100 watt bulb which is on for 10
and transformation of energy into various forms hours uses 1 kilowatt hour (kWh) of energy.
links together various branches of physics, 100 (watt) × 10 (hour)
chemistry and life sciences. It provides a = 1000 watt hour
unifying, enduring element in our scientific =1 kilowatt hour (kWh)
3
pursuits. From engineering point of view all = 10 (W) × 3600 (s)
6
electronic, communication and mechanical = 3.6 × 10 J
devices rely on some forms of energy Our electricity bills carry the energy
transformation. consumption in units of kWh. Note that kWh is
a unit of energy and not of power.
6.11 POWER
Often it is interesting to know not only the work
u Example 6.11 An elevator can carry a
done on an object, but also the rate at which
maximum load of 1800 kg (elevator +
this work is done. We say a person is physically
passengers) is moving up with a constant
fit if he not only climbs four floors of a building
speed of 2 m s . The frictional force opposing
–1
but climbs them fast. Power is defined as the
the motion is 4000 N. Determine the
time rate at which work is done or energy is
minimum power delivered by the motor to
transferred.
the elevator in watts as well as in horse
The average power of a force is defined as the
power.
ratio of the work, W, to the total time t taken
2018-19
WORK, ENERGY AND POWER 129
Answer The downward force on the elevator is by the second particle. F is likewise the force
21
exerted on the second particle by the first particle.
F = m g + F = (1800 × 10) + 4000 = 22000 N
f Now from Newton’s third law, F = − F . This
The motor must supply enough power to balance implies 12 21
this force. Hence,
∆p + ∆p = 0
P = F. v = 22000 × 2 = 44000 W = 59 hp t 1 2
The above conclusion is true even though the
6.12 COLLISIONS forces vary in a complex fashion during the
In physics we study motion (change in position). collision time ∆t. Since the third law is true at
At the same time, we try to discover physical every instant, the total impulse on the first object
quantities, which do not change in a physical is equal and opposite to that on the second.
process. The laws of momentum and energy On the other hand, the total kinetic energy of
conservation are typical examples. In this the system is not necessarily conserved. The
section we shall apply these laws to a commonly impact and deformation during collision may
encountered phenomena, namely collisions. generate heat and sound. Part of the initial kinetic
Several games such as billiards, marbles or energy is transformed into other forms of energy.
carrom involve collisions.We shall study the A useful way to visualise the deformation during
collision of two masses in an idealised form. collision is in terms of a ‘compressed spring’. If
Consider two masses m and m . The particle the ‘spring’ connecting the two masses regains
1 2 its original shape without loss in energy, then
m is moving with speed v , the subscript ‘i’
1 1i the initial kinetic energy is equal to the final
implying initial. We can cosider m to be at rest.
2 kinetic energy but the kinetic energy during the
No loss of generality is involved in making such
a selection. In this situation the mass m collision time ∆t is not constant. Such a collision
1 is called an elastic collision. On the other hand
collides with the stationary mass m and this
2 the deformation may not be relieved and the two
is depicted in Fig. 6.10.
bodies could move together after the collision. A
collision in which the two particles move together
after the collision is called a completely inelastic
collision. The intermediate case where the
deformation is partly relieved and some of the
initial kinetic energy is lost is more common and
is appropriately called an inelastic collision.
6.12.2 Collisions in One Dimension
Consider first a completely inelastic collision
Fig. 6.10 Collision of mass m , with a stationary mass m .
1 2 in one dimension. Then, in Fig. 6.10,
The masses m and m fly-off in different
1 2 θ = θ = 0
directions. We shall see that there are 1 2
relationships, which connect the masses, the m v = (m +m )v (momentum conservation)
1
2
f
1 1i
velocities and the angles. m
v = 1 v
6.12.1 Elastic and Inelastic Collisions f m + m 1i (6.23)
1 2
In all collisions the total linear momentum is The loss in kinetic energy on collision is
conserved; the initial momentum of the system
is equal to the final momentum of the system. ∆ K = 1 2 − 1 m + 2
)
One can argue this as follows. When two objects 2 m v 2 ( 1 m v f
2
1 1i
collide, the mutual impulsive forces acting over
the collision time ∆t cause a change in their 1 2 1 m 2 2
respective momenta : = m v − 1 v [using Eq. (6.23)]
1i
1 1i
2 2 m + m 2
1
∆p = F ∆t
1 12
∆p = F ∆t 1 m
2 21 = m v 2 1 − 1
where F is the force exerted on the first particle 2 1 1i m 1 + m 2
12
2018-19
130 PHYSICS
An experiment on head-on collision
In performing an experiment on collision on a horizontal surface, we face three difficulties.
One, there will be friction and bodies will not travel with uniform velocities. Two, if two bodies
of different sizes collide on a table, it would be difficult to arrange them for a head-on collision
unless their centres of mass are at the same height above the surface. Three, it will be fairly
difficult to measure velocities of the two bodies just before and just after collision.
By performing this experiment in a vertical direction, all the three difficulties vanish. Take
two balls, one of which is heavier (basketball/football/volleyball) and the other lighter (tennis
ball/rubber ball/table tennis ball). First take only the heavier ball and drop it vertically from
some height, say 1 m. Note to which it rises. This gives the velocities near the floor or ground,
2
just before and just after the bounce (by using v = 2gh ). Hence you
will get the coefficient of restitution.
Now take the big ball and a small ball and hold them in your
hands one over the other, with the heavier ball below the lighter
one, as shown here. Drop them together, taking care that they remain
together while falling, and see what happens. You will find that the
heavier ball rises less than when it was dropped alone, while the
lighter one shoots up to about 3 m. With practice, you will be able to
hold the ball properly so that the lighter ball rises vertically up and
does not fly sideways. This is head-on collision.
You can try to find the best combination of balls which gives you
the best effect. You can measure the masses on a standard balance.
We leave it to you to think how you can determine the initial and
final velocities of the balls.
(m − m )
1 m m v = 1 2 v
= 1 2 v 2 1 f m + m 1i (6.27)
2 m + m 2 1i 1 2
1
2m v
1 1i
=
which is a positive quantity as expected. and v 2 f m + m 2 (6.28)
1
Consider next an elastic collision. Using the Thus, the ‘unknowns’ {v , v } are obtained in
above nomenclature with θ = θ = 0, the 1f 2f
1 2 terms of the ‘knowns’ {m , m , v }. Special cases
1
2
1i
momentum and kinetic energy conservation of our analysis are interesting.
equations are
Case I : If the two masses are equal
m v = m v + m v (6.24)
1 1i 1 1f 2 2f v = 0
1f
m v = m v 2 + m v 2 (6.25) v = v 1i
2
2f
1 1 f
1 1i
2 2 f
The first mass comes to rest and pushes off the
From Eqs. (6.24) and (6.25) it follows that, second mass with its initial speed on collision.
m v (v − v ) = m v (v − v ) Case II : If one mass dominates, e.g. m > > m 1
2
1 1i 2 f 1i 1 1 f 2 f 1 f
v ~ − v v ~ 0
1f 1i 2f
2
or, v (v − v ) = v − v 2 The heavier mass is undisturbed while the
2 f 1i 1 f 1i 1 f
lighter mass reverses its velocity.
= (v − v 1 f )(v + v 1 f )
1i
1i
Example 6.12 Slowing down of neutrons:
t
Hence, ∴ v 2 f = v + v 1 f (6.26) In a nuclear reactor a neutron of high
1i
speed (typically 10 m s ) must be slowed
–1
7
Substituting this in Eq. (6.24), we obtain
2018-19
WORK, ENERGY AND POWER 131
to 10 3 m s –1 so that it can have a high 6.12.3 Collisions in Two Dimensions
probability of interacting with isotope 235 U Fig. 6.10 also depicts the collision of a moving
92 with the stationary mass m . Linear
and causing it to fission. Show that a mass m 1 2
neutron can lose most of its kinetic energy momentum is conserved in such a collision.
in an elastic collision with a light nuclei Since momentum is a vector this implies three
like deuterium or carbon which has a mass equations for the three directions {x, y, z}.
of only a few times the neutron mass. The Consider the plane determined by the final
material making up the light nuclei, usually velocity directions of m and m and choose it to
1
2
heavy water (D O) or graphite, is called a be the x-y plane. The conservation of the
2
moderator. z-component of the linear momentum implies
that the entire collision is in the x-y plane. The
Answer The initial kinetic energy of the neutron x- and y-component equations are
is m v = m v cos θ + m v cos θ (6.29)
1 1i 1 1f 1 2 2f 2
1
K = m v 2 0 = m v sin θ − m v sin θ (6.30)
1i 1 1i 1 1f 1 2 2f 2
2
One knows {m , m , v } in most situations. There
while its final kinetic energy from Eq. (6.27) 1 2 1i
are thus four unknowns {v , v , θ and θ }, and
1f 2f 1 2
1 2 1 m − m 2 2 only two equations. If θ 1 = θ 2 = 0, we regain
K = m v = m 1 2 v Eq. (6.24) for one dimensional collision.
1 f 1 1 f 1 i 1
2 2 m + m
1
2
If, further the collision is elastic,
1 2 1 2 1 2
The fractional kinetic energy lost is m v = m v + m v (6.31)
1 1f
2 2 f
1 1i
2 2 2
K m − m 2
f = 1 f = 1 2 We obtain an additional equation. That still
1
K i 1 m + m leaves us one equation short. At least one of
1
2
while the fractional kinetic energy gained by the the four unknowns, say θ , must be made known
1
moderating nuclei K /K is for the problem to be solvable. For example, θ 1
2f 1i
can be determined by moving a detector in an
f = 1 − f (elastic collision)
2 1 angular fashion from the x to the y axis. Given
4m m {m , m , v , θ } we can determine {v , v , θ }
1f
2f
2
1
1
1i
2
= 1 2 2 from Eqs. (6.29)-(6.31).
(m + m 2 )
1
Example 6.13 Consider the collision
t
One can also verify this result by substituting depicted in Fig. 6.10 to be between two
from Eq. (6.28). billiard balls with equal masses m = m .
2
1
For deuterium m = 2m and we obtain The first ball is called the cue while the
2 1
f = 1/9 while f = 8/9. Almost 90% of the second ball is called the target. The
1 2
neutron’s energy is transferred to deuterium. For billiard player wants to ‘sink’ the target
carbon f = 71.6% and f = 28.4%. In practice, ball in a corner pocket, which is at an
1 2
however, this number is smaller since head-on angle θ = 37°. Assume that the collision
2
collisions are rare. t is elastic and that friction and rotational
If the initial velocities and final velocities of motion are not important. Obtain θ .
1
both the bodies are along the same straight line, Answer From momentum conservation, since
then it is called a one-dimensional collision, or the masses are equal
head-on collision. In the case of small spherical
bodies, this is possible if the direction of travel v = v + v 2f
1i
1f
of body 1 passes through the centre of body 2
or 2 )
⋅ v
which is at rest. In general, the collision is two- v 1i = v + v 2 f ) ( 1 f + v 2 f
( 1f
dimensional, where the initial velocities and the = 2 + 2 +
final velocities lie in a plane. v 1 f v 2 f 2v 1 f .v 2 f
2018-19
132 PHYSICS
= { v 2 + v 2 + 2v v cos θ + 37° ) } (6.32) The matter simplifies greatly if we consider
1 f
2 f
( 1
2 f
1 f
spherical masses with smooth surfaces, and
assume that collision takes place only when the
Since the collision is elastic and m = m it follows
1 2
from conservation of kinetic energy that bodies touch each other. This is what happens
in the games of marbles, carrom and billiards.
v 1i 2 = v 1 f 2 + v 2 f 2 (6.33)
In our everyday world, collisions take place only
when two bodies touch each other. But consider
Comparing Eqs. (6.32) and (6.33), we get
a comet coming from far distances to the sun, or
cos (θ + 37°) = 0
1 alpha particle coming towards a nucleus and
or θ + 37° = 90° going away in some direction. Here we have to
1
deal with forces involving action at a distance.
Thus, θ = 53°
1 Such an event is called scattering. The velocities
This proves the following result : when two equal and directions in which the two particles go away
masses undergo a glancing elastic collision with depend on their initial velocities as well as the
one of them at rest, after the collision, they will type of interaction between them, their masses,
move at right angles to each other. t shapes and sizes.
SUMMARY
1. The work-energy theorem states that the change in kinetic energy of a body is the work
done by the net force on the body.
K - K = W
f i net
2. A force is conservative if (i) work done by it on an object is path independent and
depends only on the end points {x , x }, or (ii) the work done by the force is zero for an
i j
arbitrary closed path taken by the object such that it returns to its initial position.
3. For a conservative force in one dimension, we may define a potential energy function V(x)
such that
d V x ( )
F x ( ) = −
d x
x f
( )
V − V = ∫ F x x
d
or i f
x i
4. The principle of conservation of mechanical energy states that the total mechanical
energy of a body remains constant if the only forces that act on the body are conservative.
5. The gravitational potential energy of a particle of mass m at a height x about the earth’s
surface is
V(x) = m g x
where the variation of g with height is ignored.
6. The elastic potential energy of a spring of force constant k and extension x is
1
( )
=
V x k x 2
2
7. The scalar or dot product of two vectors A and B is written as A.B and is a scalar
quantity given by : A . B = AB cos θ, where θ is the angle between A and B. It can be
positive, negative or zero depending upon the value of θ. The scalar product of two
vectors can be interpreted as the product of magnitude of one vector and component
of the other vector along the first vector. For unit vectors :
ˆ ˆ
ˆ ˆ
ˆ ˆ ˆ ˆ k k = 1 and i j = ˆ ˆ k i = 0
ˆ ˆ
⋅
j k =
i i =
⋅
j j =
⋅
⋅
⋅
⋅
Scalar products obey the commutative and the distributive laws.
2018-19
WORK, ENERGY AND POWER 133
POINTS TO PONDER
1. The phrase ‘calculate the work done’ is incomplete. We should refer (or imply
clearly by context) to the work done by a specific force or a group of forces on a
given body over a certain displacement.
2. Work done is a scalar quantity. It can be positive or negative unlike mass and
kinetic energy which are positive scalar quantities. The work done by the friction
or viscous force on a moving body is negative.
3. For two bodies, the sum of the mutual forces exerted between them is zero from
Newton’s Third Law,
F + F = 0
12 21
But the sum of the work done by the two forces need not always cancel, i.e.
W + W ≠ 0
12 21
However, it may sometimes be true.
4. The work done by a force can be calculated sometimes even if the exact nature of
the force is not known. This is clear from Example 6.2 where the WE theorem is
used in such a situation.
5. The WE theorem is not independent of Newton’s Second Law. The WE theorem
may be viewed as a scalar form of the Second Law. The principle of conservation
of mechanical energy may be viewed as a consequence of the WE theorem for
conservative forces.
6. The WE theorem holds in all inertial frames. It can also be extended to non-
inertial frames provided we include the pseudoforces in the calculation of the
net force acting on the body under consideration.
7. The potential energy of a body subjected to a conservative force is always
undetermined upto a constant. For example, the point where the potential
energy is zero is a matter of choice. For the gravitational potential energy mgh,
the zero of the potential energy is chosen to be the ground. For the spring
2
potential energy kx /2 , the zero of the potential energy is the equilibrium position
of the oscillating mass.
8. Every force encountered in mechanics does not have an associated potential
energy. For example, work done by friction over a closed path is not zero and no
potential energy can be associated with friction.
9. During a collision : (a) the total linear momentum is conserved at each instant of
the collision ; (b) the kinetic energy conservation (even if the collision is elastic)
applies after the collision is over and does not hold at every instant of the collision.
In fact the two colliding objects are deformed and may be momentarily at rest
with respect to each other.
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134 PHYSICS
EXERCISES
6.1 The sign of work done by a force on a body is important to understand. State carefully
if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the
bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body
sliding down an inclined plane,
(d) work done by an applied force on
a body moving on a rough
horizontal plane with uniform
velocity,
(e) work done by the resistive force of
air on a vibrating pendulum in
bringing it to rest.
6.2 A body of mass 2 kg initially at rest
moves under the action of an applied
horizontal force of 7 N on a table with
coefficient of kinetic friction = 0.1.
Compute the
(a) work done by the applied force in
10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the
body in 10 s,
(d) change in kinetic energy of the
body in 10 s,
and interpret your results.
6.3 Given in Fig. 6.11 are examples of some
potential energy functions in one
dimension. The total energy of the
particle is indicated by a cross on the
ordinate axis. In each case, specify the
regions, if any, in which the particle
cannot be found for the given energy.
Also, indicate the minimum total
energy the particle must have in each
case. Think of simple physical contexts
for which these potential energy shapes
are relevant.
Fig. 6.11
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WORK, ENERGY AND POWER 135
6.4 The potential energy function for a
particle executing linear simple
harmonic motion is given by V(x) =
2
kx /2, where k is the force constant
-1
of the oscillator. For k = 0.5 N m ,
the graph of V(x) versus x is shown
in Fig. 6.12. Show that a particle of
total energy 1 J moving under this
Fig. 6.12
potential must ‘turn back’ when it
reaches x = ± 2 m.
6.5 Answer the following :
(a) The casing of a rocket in flight
burns up due to friction. At
whose expense is the heat
energy required for burning
obtained? The rocket or the
atmosphere?
(b) Comets move around the sun
in highly elliptical orbits. The
gravitational force on the
comet due to the sun is not Fig. 6.13
normal to the comet’s velocity
in general. Yet the work done by the gravitational force over every complete orbit
of the comet is zero. Why ?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy
gradually due to dissipation against atmospheric resistance, however small. Why
then does its speed increase progressively as it comes closer and closer to the earth ?
(d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig.
6.13(ii), he walks the same distance pulling the rope behind him. The rope goes
over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work
done greater ?
6.6 Underline the correct alternative :
(a) When a conservative force does positive work on a body, the potential energy of
the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential
energy.
(c) The rate of change of total momentum of a many-particle system is proportional
to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after
the collision are the total kinetic energy/total linear momentum/total energy of
the system of two bodies.
6.7 State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is
conserved.
(b) Total energy of a system is always conserved, no matter what internal and external
forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in
nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial
kinetic energy of the system.
6.8 Answer carefully, with reasons :
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved
during the short time of collision of the balls (i.e. when they are in contact) ?
(b) Is the total linear momentum conserved during the short time of an elastic collision
of two balls ?
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136 PHYSICS
(c) What are the answers to (a) and (b) for an inelastic collision ?
(d) If the potential energy of two billiard balls depends only on the separation distance
between their centres, is the collision elastic or inelastic ? (Note, we are talking
here of potential energy corresponding to the force during collision, not gravitational
potential energy).
6.9 A body is initially at rest. It undergoes one-dimensional motion with constant
acceleration. The power delivered to it at time t is proportional to
(i) t 1/2 (ii) t (iii) t 3/2 (iv) t 2
6.10 A body is moving unidirectionally under the influence of a source of constant power.
Its displacement in time t is proportional to
(i) t 1/2 (ii) t (iii) t 3/2 (iv) t 2
6.11 A body constrained to move along the z-axis of a coordinate system is subject to a
constant force F given by
ˆ
ˆ
F = − ˆ i + 2 j + 3 N
k
where ,i ˆ ˆ k , j ˆ are unit vectors along the x-, y- and z-axis of the system respectively.
What is the work done by this force in moving the body a distance of 4 m along the
z-axis ?
6.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic
energy 10 keV, and the second with 100 keV. Which is faster, the electron or the
-31
proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10 kg, proton mass
= 1.67×10 –27 kg, 1 eV = 1.60 ×10 –19 J).
6.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with
decreasing acceleration (due to viscous resistance of the air) until at half its original
height, it attains its maximum (terminal) speed, and moves with uniform speed
thereafter. What is the work done by the gravitational force on the drop in the first
and second half of its journey ? What is the work done by the resistive force in the
entire journey if its speed on reaching the ground is 10 m s –1 ?
6.14 A molecule in a gas container hits a horizontal wall with speed 200 m s and angle 30°
–1
with the normal, and rebounds with the same speed. Is momentum conserved in the
collision ? Is the collision elastic or inelastic ?
6.15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m 3
in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%,
how much electric power is consumed by the pump ?
6.16 Two identical ball bearings in contact with each other and resting on a frictionless
table are hit head-on by another ball bearing of the same mass moving initially with a
speed V. If the collision is elastic, which of the following (Fig. 6.14) is a possible result
after collision ?
Fig. 6.14
2018-19
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