WORK, ENERGY AND POWER.pptx

WORK, ENERGY AND
POWER
CLASS XI
Physics By:
SHIVAM DAVE

INTRODUCTION
We use the words work, energy and power in our day-to-day life often.
Two conditions are need to be satisfied for
work to be done in science :-
• A force should act on object.
• The object must be displaced.
If either of these conditions does not satisfy,
we say work done is zero.
Shivam Dave/Physics/Vision Academy

WORK
Work is said to be done when a force produces motion.
So we can formally define work done as
“Work done by force acting on an object is equal to the product of
force and the displacement of the object in the direction of the
force.”
Shivam Dave/Physics/Vision
Academy

Consider the figure given
below where a block is placed
on a frictionless horizontal
floor.
This block is acted upon by a
constant force F.
Action of this force is to move the body through a distance d in a
straight line in the direction of force.
Shivam Dave/Physics/Vision
Academy

The work done by the force is defined to be the product of
component of the force in the direction of the
displacement and the magnitude of this displacement
𝑾 = 𝑭 𝐜𝐨𝐬 𝜽 𝒅 = 𝑭. 𝒅
No work is done if :
(i) The displacement is zero as seen in the example above. A weightlifter
holding a 150 kg mass steadily on his shoulder for 30 s does no work on
the load during this time.

(ii) The force and displacement are mutually perpendicular then also
there will no work done.
Positive Work:
If a force displaces the object in its direction, then the work
done is positive
So, W=Fd
The example of this kind of work done is motion of ball falling
towards ground where displacement of ball is in the direction
of force of gravity.

Negative work
If the force and the displacement are in opposite directions,
then the work is said to be negative.
For example if a ball is thrown in upwards direction, its
displacement would be in upwards direction but the force due
to earth’s gravity is in the downward direction.
So here in this case gravity is doing negative work when you throw the
ball upwards. Hence the work done by gravitational force is negative.
Si unit is joule (J)
Dimensional formula= 𝑴𝟏𝑳𝟐𝑻−𝟐

NOTIONS OF WORK AND KINETIC ENERGY: THE
WORK-ENERGY THEOREM
For linear motion where acceleration is constant,
𝒗𝟐 − 𝒖𝟐 = 𝟐𝒂𝒔……(1)
Where 𝑣 and 𝑢 are initial and final velocity.
Multiply above equation by 𝑚 2 both sides.
𝟏
𝟐
𝒎𝒗𝟐 −
𝟏
𝟐
𝒎𝒖𝟐 = 𝒎𝒂𝒔 = 𝑭𝒔 … … (𝟐)

Equation (1) can be written in three dimensions by employing
vectors
𝒗𝟐
− 𝒖𝟐
= 𝟐 𝒂. 𝒅
Where 𝒂 and 𝒅 are acceleration and displacement vectors of
the object respectively.
Now equation (2) can be written as
𝟏
𝟐
𝒎𝒗𝟐 −
𝟏
𝟐
𝒎𝒖𝟐 = 𝒎𝒂. 𝒅 = 𝑭𝒅
The left side of the equation is the difference in the quantity ‘half the
mass times the square of the speed’ from its initial value to its final
value. We call each of these quantities the ‘kinetic energy’, denoted
by K

The right side is a product of the displacement and the
component of the force along the displacement.
This quantity is called ‘work’ and is denoted by W
𝑲𝒇 − 𝑲𝒊 = 𝑾
where 𝑲𝒇 and 𝑲𝒊 are respectively the initial and final kinetic
energies of the object

KINETIC ENERGY
Kinetic energy is the energy possessed by the body by virtue of its
motion
Body moving with greater velocity would posses greater K.E in
comparison of the body moving with slower velocity
If object having mass 𝑚 and velocity 𝑣, its kinetic energy is
𝑲 =
𝟏
𝟐
𝒎𝒗𝟐
Kinetic energy is a scalar quantity.

WORK DONE BY VARIABLE FORCE
A variable force is a force whose magnitude or direction or both
vary during the displacement of a body on which it acts.
Let us suppose that a body moves along the direction of variable force
acting on it.
We then plot a graph of force (F) versus displacement (x), which helps
calculate the varying force with the displacement as shown.

THE WORK-ENERGY THEOREM FOR A VARIABLE
FORCE
The relationship between Work and kinetic energy of the object is
called the Work Energy Theorem
It states that the net work done on the system is equal to the change
in Kinetic energy of the system
The time rate of change of kinetic energy is
𝒅𝑲
𝒅𝒕
=
𝒅
𝒅𝒕
𝟏
𝟐
𝒎𝒗𝟐

𝒅𝑲
𝒅𝒕
= 𝒎
𝒅𝒗
𝒅𝒕
𝒗
= 𝑭𝒗
(From newtons second law of
motion)
= 𝑭
𝒅𝒙
𝒅𝒕
𝒅𝑲 = 𝑭𝒅𝒙
Intergrating from the initial
position 𝑥𝑖 to final position 𝑥𝑓
𝑲𝒊
𝑲𝒇
𝒅𝑲 =
𝒙𝒊
𝒙𝒇
𝑭𝒅𝒙
Where 𝐾𝑓 and 𝐾𝑖 are initial and
final kinetic energies
𝑲𝒇 − 𝑲𝒊 =
𝒙𝒊
𝒙𝒇
𝑭𝒅𝒙
𝑲𝒇 − 𝑲𝒊 = 𝑾
Thus, the WE theorem is proved
for a variable force
It is an integral form of Newton’s
second law. Newton’s second law is
a relation between acceleration and
force at any instant of time.
Work-energy theorem involves an
integral over an interval of time.

Newton’s second law for two or three dimensions is in vector
form whereas the work-energy theorem is in scalar form.
In the scalar form, information with respect to directions
contained in Newton’s second law is not present.

THE CONCEPT OF POTENTIAL ENERGY
The potential energy is the energy stored in a body or a system
by virtue of its position in a field of force or by its
configuration
Few examples of bodies possessing Potential energy are given below
i) Stretched or compressed coiled spring
ii) Water stored up at a height in the Dam possess PE
iii) Any object placed above the height H from the surface of the earth
posses PE

The gravitational force on a ball of mass 𝑚 is 𝑚𝑔 .
𝑔 may be treated as a constant near the earth surface.
By ‘near’ we imply that the height ℎ of the ball above the
earth’s surface is very small compared to the earth’s radius
𝑅𝐸 (ℎ << 𝑅𝐸) so that we can ignore the variation of g near
the earth’s surface.
Let us raise the ball up to a height ℎ.
The work done by the external agency against the gravitational force is
𝑚𝑔ℎ.
This work gets stored as potential energy.

Gravitational potential energy of an object, as a function of
the height ℎ, is denoted by 𝑉(ℎ) and it is the negative of
work done by the gravitational force in raising the object to
that height
𝑽 𝒉 = 𝒎𝒈𝒉
gravitational force F equals the negative of the derivative of 𝑉(ℎ) with
respect to ℎ. Thus
𝑭 = −
𝒅
𝒅𝒉
𝑽 𝒉 = −𝒎𝒈
The negative sign indicates that the gravitational force is downward.

When released, the ball comes down with an increasing
speed.
Just before it hits the ground, its speed is given by the
kinematic relation,
𝒗𝟐 = 𝟐𝒈𝒉
This equation can also be written as
𝟏
𝟐
𝒎𝒗𝟐 = 𝒎𝒈𝒉
Mathematically, (for simplicity, in one dimension) the potential
energy 𝑉(𝑥) is defined if the force 𝐹(𝑥) can be written as

𝑭 𝒙 = −
𝒅𝒗
𝒅𝒙
𝒙𝒊
𝒙𝒇
𝑭 𝒙 𝒅𝒙 = −
𝒗𝒊
𝒗𝒇
𝒅𝒗 = 𝒗𝒊 − 𝒗𝒇
The work done by a conservative force such as gravity depends on the
initial and final positions only.

THE CONSERVATION OF MECHANICAL ENERGY
∆𝑲 = 𝑭(𝒙)∆𝒙
The principal states that if only the conservative force are
doing work on a body then its mechanical energy remains
constant.
Let the body undergoes displacement ∆𝒙 under the action of
conservative force 𝐹 𝑥 , then from work energy theorem
As the force is conservative the change in potential energy is given
by
∆𝑼 = 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒘𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = −𝑭(𝒙)∆𝒙

Combining the two equations, we get
∆𝑲 = −∆𝑼
∆𝑲 + ∆𝑼 = 𝟎 𝒐𝒓 ∆ 𝑲 + 𝑼 = 𝟎
𝑲 + 𝑼 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑲𝒊 + 𝑼𝒊 = 𝑲𝒇 + 𝑼𝒇
Total mechanical energy of the system remains constant under the
conservative force.

Conservation of mechanical energy in case of a free
falling body.
At point A:
The body at rest,
K.E of the body 𝑲𝑨 = 𝟎
𝑃. 𝐸 of the body 𝑼𝑨 = 𝒎𝒈𝒉
Total mechanical energy
𝑬𝑨 = 𝑲𝑨 + 𝑼𝑨 = 𝒎𝒈𝒉

At point B:
Suppose the body falls freely through height 𝑥 and
reaches the point B with velocity 𝑣, then
𝒗𝟐
− 𝟎𝟐
= 𝟐𝒈𝒙
𝒗𝟐 = 𝟐𝒈𝒙
𝑲𝑩 =
𝟏
𝟐
𝒎𝒗𝟐
=
𝟏
𝟐
𝒎 × 𝟐𝒈𝒙 = 𝒎𝒈𝒙
𝑼𝑩= 𝒎𝒈 𝒉 − 𝒙
𝑬𝑩 = 𝑲𝑩 + 𝑼𝑩
= 𝒎𝒈𝒙 + 𝒎𝒈 𝒉 − 𝒙
= 𝒎𝒈𝒉

At point C:
Suppose the body finally reaches a point C on the ground
with velocity 𝑣′, the considering motion from A to C
𝒗′𝟐
− 𝟎 = 𝟐𝒈𝒉
𝒗′𝟐= 𝟐𝒈𝒉
𝑲𝑪=
𝟏
𝟐
𝒎𝒗′𝟐
=
𝟏
𝟐
𝒎 × 𝟐𝒈𝒉 = 𝒎𝒈𝒉
𝑼𝑪= 𝒎𝒈𝒉 × 𝟎 = 𝟎
𝑬𝑪= 𝑲𝑪 + 𝑼𝑪 = 𝒎𝒈𝒉
Total mechanical energy is
conserved during free fall of the
body.

THE POTENTIAL ENERGY OF A SPRING
Consider an elastic spring of negligibly
small mass with its one end attached to a
rigid support.
Other end attached with a block of mass
(m) which can slide over a smooth
horizontal surface.

According to Hooke's law, Spring force 𝐹𝑠 is proportional to the
displacement of the block from the equilibrium position.
𝑭𝒔 ∝ 𝒙
𝑭𝒔 = −𝒌𝒙
𝑘 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑝𝑟𝑜𝑝𝑜𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑡𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
SI unit of proportionality constant is 𝑁𝑚−1.
If spring is stiff then 𝑘 is large and if spring is soft then 𝑘 is small.
The negative sign shows 𝐹𝑠 acts in opposite direction.
The work doe by the spring force for the small expansion 𝑑𝑥 is
𝒅𝑾𝒔 = 𝑭𝒔𝒅𝒙 = −𝒌𝒙𝒅𝒙

If the block is moved from an initial displacement 𝑥𝑖 to the
final displacement 𝑥𝑓, the work done by the spring force is
𝑾𝒔 = 𝒅𝑾𝒔 = − 𝒙𝒊
𝒙𝒇
𝒌𝒙𝒅𝒙 = −𝒌
𝒙𝟐
𝟐 𝒙𝒊
𝒙𝒇
𝑾𝒔 =
𝟏
𝟐
𝒌𝒙𝟐
𝒇 −
𝟏
𝟐
𝒌𝒙𝟐
𝒊
If the block is pulled from 𝑥𝑖 and allowed to return to 𝑥𝑖 then
𝑾𝒔 = − 𝒙𝒊
𝒙𝒇
𝒌𝒙𝒅𝒙 =
𝟏
𝟐
𝒌𝒙𝟐
𝒊 −
𝟏
𝟐
𝒌𝒙𝟐
𝒊 = 𝟎

Conclusion:
1. Spring force is position dependent.
2. Work done by spring force depends on initial and final
position.
3. The work done by the spring force is a cyclic process.
So the spring force is conservative force.
The work done by external force will be equal to the increase in
P.E of the spring and is given by
∆𝑼 = 𝑾 =
𝟏
𝟐
𝒌𝒙𝟐
𝒇 −
𝟏
𝟐
𝒌𝒙𝟐
𝒊

If potential energy is zero then block is in equilibrium
position, the P.E of the spring for an extension 𝑥 will be
𝑼 𝒙 − 𝟎 =
𝟏
𝟐
𝒌𝒙𝟐
− 𝟎
𝑼 𝒙 =
𝟏
𝟐
𝒌𝒙𝟐

P.E OF AN ELASTIC SPRING BY
GRAPHICAL METHOD
Spring force for an extension 𝑥𝑚 is
𝑭𝒔 = −𝒌𝒙𝒎
The work done by the spring force
for an extension 𝑥𝑚 is
𝑾𝒔 = 𝑨𝒓𝒆𝒂 𝒐𝒇 ∆𝑶𝑩𝑨
=
𝟏
𝟐
× 𝑨𝑩 × 𝑶𝑩
=
𝟏
𝟐
𝑭𝒔 × 𝒙𝒎
=
𝟏
𝟐
−𝒌𝒙𝒎 × 𝒙𝒎
= −
𝟏
𝟐
𝒌𝒙𝟐
𝒎
In order to stretch the spring slowly, an
external force 𝐹 equal to and opposite
to 𝐹𝑠 has to be applied.
So work done by external force 𝐹 is
𝑾 = −𝑾𝒔 = +
𝟏
𝟐
𝒌𝒙𝒎
𝟐
This work done is stored as P.E

CONSERVATION OF ENERGY IN AN ELASTIC SPRING
When the potential energy is maximum then kinetic
energy will be zero and vice versa.
At the extreme position
𝒙 = ±𝒙𝒎 𝒂𝒏𝒅 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒗 = 𝟎
𝑲 =
𝟏
𝟐
𝒎𝒗𝟐
= 𝟎
𝑼 =
𝟏
𝟐
𝒌𝒙𝟐
𝒎 = 𝒂 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒗𝒂𝒍𝒖𝒆

At the equilibrium position
𝑥 = 0
𝑈 =
1
2
𝑘(0)2= 0
𝐾 =
1
2
𝑚𝑣2
𝑚 =
1
2
𝑘𝑥2
𝑚
Maximum speed
𝑣𝑚 =
𝑘
𝑚
𝑥𝑚
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝐾. 𝐸 + 𝑃. 𝐸
𝟏
𝟐
𝒌𝒙𝟐
𝒎 =
𝟏
𝟐
𝒎𝒗𝟐 +
𝟏
𝟐
𝒌𝒙𝟐
𝒌 =
𝟏
𝟐
𝒎𝒗𝟐 =
𝟏
𝟐
𝒌(𝒙𝟐
𝒎 − 𝒙𝟐)
Where
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒗 =
𝒌
𝒎
(𝒙𝟐
𝒎 − 𝒙𝟐)
At any intermediate position 𝒙
For 𝒙 between −𝒙𝒎 𝒕𝒐 +𝒙𝒎 the
energy is partially kinetic and
partially potential.

POWER
Power is defined as the rate of doing work.
The average power of a force is defined as the ratio of the work, 𝑾, to
the total time 𝒕 taken
𝑷𝒂𝒗 =
𝑾
𝒕
Power is scalar quantity. Dimension is [𝑀𝐿2𝑇−3]

SI Unit of Power is watt.(W)
The power of an agent is one watt if it does work at the rate
of 1 joule per second.
𝟏 𝑾𝒂𝒕𝒕 =
𝟏 𝒋𝒐𝒖𝒍𝒆
𝟏 𝒔𝒆𝒄𝒐𝒏𝒅
𝒐𝒓 𝟏𝑾 = 𝑱𝑺−𝟏
The bigger unit of power are kilowatt (kW) and horse power (hp)
1 kilowatt=1000 watt
1 horse power= 746 watt

INSTANTANEOUS POWER
The instantaneous power is defined as the limiting value of
the average power as time interval approaches zero,
𝑷 = 𝐥𝐢𝐦
∆𝒕→𝟎
∆𝑾
∆𝒕
=
𝒅𝑾
𝒅𝒕
Power as a dot product:
The work done by force for a small displacement is given by
𝒅𝑾 = 𝑭. 𝒅𝒓
So instantaneous power can be expressed as
𝑷 =
𝒅𝑾
𝒅𝒕
= 𝑭.
𝒅𝒓
𝒅𝒕
𝒅𝒓
𝒅𝒕
=𝒗(𝒊𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒆𝒐𝒖𝒔 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚)
𝑷 = 𝑭. 𝒗

Kilowatt hour(kWh):
It is a commercial unit of electric energy.
1 kilowatt hour is the electric energy consumed by all
appliance of 1000 watt in 1 hour.
Relation between kWh and joule
𝟏 𝒌𝑾𝒉 = 𝟏𝒌𝑾 × 𝟏𝒉 = 𝟏𝟎𝟎𝟎𝑾 × 𝟏𝒉
= 𝟏𝟎𝟎𝟎𝑱𝒔−𝟏 × 𝟑𝟔𝟎𝟎 𝒔 = 𝟑. 𝟔 × 𝟏𝟎𝟔𝑱

COLLISION
Several games such as billiards, marbles or carrom involve collisions
A collision is an event in which two or more bodies exert forces on each
other for a relatively short time.
In all collisions the total linear momentum is conserved.
In the collision of two particles law of conservation of momentum
always holds true but in some collisions Kinetic energy is not always
conserved.

ELASTIC AND INELASTIC COLLISION
ELASTIC COLLISION
Those collisions in which both momentum and kinetic energy of
system are conserved are called elastic collisions.
for example elastic collision occurs between the molecules of a gas.
This type of collision mostly takes place between the atoms,
electrons and protons.

INELASTIC COLLISION
Those collisions in which momentum of system is conserved
but kinetic energy of the system is not conserved are
known as inelastic collision.
Two bodies stick to each other after collision as a bullet hit its
target and remain embedded in the target.
In this case some of the kinetic energy is converted into heat
or is used up in in doing work in deforming bodies for example
when two cars collide their metal parts are bet out of shape.

Characteristics of inelastic collision
(a) Total momentum is conserved.
(b) Total energy is conserved.
(c) Total kinetic energy is not conserved.
(d) A part or whole of whole mechanical energy may be
converted into other forms of energy.
(e) Some or all forces involved during interaction are non-
conservative in nature

COLLISION IN ONE DIMESION
Consider first a completely inelastic collision
in one dimension.
A body of mass 𝑚1 is moving with velocity
𝑢1 collides with another body with 𝑚2 at rest.
After collision two bodies move together with a
common velocity 𝑣
𝑚1
𝑢1
Before collision
𝑚1
𝑚2
𝑣
𝑚2

As the linear momentum is conserved so
𝒎𝟏𝒖𝟏 + 𝒎𝟐 × 𝟎 = 𝒎𝟏 + 𝒎𝟐 𝒗
𝒗 =
𝒎𝟏
𝒎𝟏 + 𝒎𝟐
𝒖
The loss of kinetic energy in collision Is
∆𝑲 = 𝒌𝒊 − 𝒌𝒇 =
𝟏
𝟐
𝒎𝟏𝒖𝟐
𝟏 −
𝟏
𝟐
𝒎𝟏 + 𝒎𝟐 𝒗𝟐
=
𝟏
𝟐
𝒎𝟏𝒖𝟐
𝟏 −
𝟏
𝟐
𝒎𝟏 + 𝒎𝟐
𝒎𝟏
𝒎𝟏+𝒎𝟐
𝒖𝟏
𝟐
=
𝟏
𝟐
𝒎𝟏𝒖𝟐
𝟏 −
𝟏
𝟐
𝒎𝟏
𝒎𝟏+𝒎𝟐
𝒖𝟐
𝟏
=
𝟏
𝟐
𝒎𝟏𝒖𝟐
𝟏 𝟏 −
𝒎𝟏
𝒎𝟏+𝒎𝟐
=
𝟏
𝟐
𝒎𝟏𝒎𝟐
𝒎𝟏+𝒎𝟐
𝒖𝟐
𝟏

Consider next an elastic collision.
Using the above nomenclature with θ 1 = θ 2 = 0, the
momentum and kinetic energy conservation equations are
Consider two bodies 𝐴 and 𝐵 of masses 𝒎𝟏
and 𝒎𝟐 moving along the same straight line
in the same direction with velocities 𝒖𝟏 and
𝒖𝟐 respectively as shown in Fig. Let us
assume that 𝒖𝟏 is greater than 𝒖𝟐 .
The bodies 𝐴 and 𝐵 suffer a head on
collision when they strike and continue to
move along the same straight line with
velocities 𝑣1 and 𝑣2 respectively.

From the law of conservation of linear momentum,
Total momentum before collision = Total momentum after collision
𝒎𝟏𝒖𝟏 + 𝒎𝟐𝒖𝟐 = 𝒎𝟏𝒗𝟏 + 𝒎𝟐𝒗𝟐 … … … … … … . (𝟏)
since the kinetic energy of the bodies is also conserved during the
collision
Total kinetic energy before collision = Total kinetic energy after collision

Equation (5) shows that in an elastic one-dimensional collision,
the relative velocity with which the two bodies approach each
other before collision is equal to the relative velocity with
which they recede from each other after collision.

Special case:
When two bodies of equal mass are collide
Let 𝒎𝟏 = 𝒎𝟐 = 𝒎
From equation (5)
𝑣1=
2𝑚𝑢2
2𝑚
= 𝑢2
= 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑏𝑜𝑑𝑦 𝑜𝑓 𝑚𝑎𝑠𝑠 𝒎𝟐 before collision.
From equation (6)
𝑣2=
2𝑚𝑢1
2𝑚
= 𝑢1
= 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑏𝑜𝑑𝑦 𝑜𝑓 𝑚𝑎𝑠𝑠 𝒎𝟏 before collision
Means when two bodies of equal masses suffer one
dimensional elastic collision, their velocities get exchanged
after the collision.

When a body collides against a stationary body of equal mass.
𝒎𝟏 = 𝒎𝟐 = 𝒎 𝒂𝒏𝒅 𝒖𝟐 = 𝟎
From equation (5), 𝑣1 = 0
From equation (6), 𝑣2 = 𝑢1
Means when an elastic body collides against another elastic
body of equal mass, initially at rest, after the collision the
first body comes to rest while second body moves with initial
velocity of first.

When a light body collides against a massive stationary body
Here 𝑚1 ≪ 𝑚2 𝑎𝑛𝑑 𝑢2 = 0.
Neglecting 𝑚1 in equation (5), we get
𝒗𝟏 = −
𝒎𝟐𝒖𝟏
𝒎𝟐
= −𝒖𝟏
From equation (6)
𝒗𝟐 = 𝟎
Means when a light body collides against a massive body at rest,
the light body rebounds after the collision with an equal and
opposite velocity while the massive body practically remains at
rest.

ELASTIC COLLISION IN TWO DIMENSION
Let mass 𝑚1 moving along X-axis
with initial velocity 𝑢1 collide
with another particle of mass
𝑚2 at rest.
After collision, two particles
moves with velocities 𝑣1 and 𝑣2
making angle 𝜃1 and 𝜃2

After the Collison the rectangular components of the momentum
of 𝑚1 are
𝒎𝟏𝒗𝟏 𝒄𝒐𝒔𝜽𝟏, 𝒂𝒍𝒐𝒏𝒈 + 𝒗𝒆 𝑿 − 𝒂𝒙𝒊𝒔
𝒎𝟏𝒗𝟏 𝐬𝐢𝐧 𝜽𝟏 , 𝒂𝒍𝒐𝒏𝒈 + 𝒗𝒆 − 𝒂𝒙𝒊𝒔
After the Collison the rectangular components of the momentum
of 𝑚2 are
𝒎𝟐𝒗𝟐 𝒄𝒐𝒔𝜽𝟐, 𝒂𝒍𝒐𝒏𝒈 + 𝒗𝒆 𝑿 − 𝒂𝒙𝒊𝒔
𝒎𝟐𝒗𝟐 𝐬𝐢𝐧 𝜽𝟐 , 𝒂𝒍𝒐𝒏𝒈 + 𝒗𝒆 − 𝒂𝒙𝒊𝒔
Applying the principle of conservation of momentum along X-axis
𝒎𝟏𝒖𝟏 = 𝒎𝟏𝒗𝟏𝒄𝒐𝒔𝜽𝟏 + 𝒎𝟐𝒗𝟐𝒄𝒐𝒔𝜽𝟐. . (𝟏)

The initial momentum of 𝑚1 or 𝑚2 along Y-axis is zero.
Applying the principle of conservation of momentum along Y-axis.
𝟎 = 𝒎𝟏𝒗𝟏𝒔𝒊𝒏𝜽𝟏 − 𝒎𝟐𝒗𝟐𝒔𝒊𝒏𝜽𝟐 … … … . . (𝟐)
As K.E conserved in elastic collision, so
𝟏
𝟐
𝒎𝟏𝒖𝟐
𝟏 =
𝟏
𝟐
𝒎𝟏𝒗𝟏
𝟐
+
𝟏
𝟐
𝒎𝟐𝒗𝟐
𝟐
… … … … . (𝟑)
The four unknown quantities 𝑣1, 𝑣2, 𝜃1, 𝜃2 can not be calculated using
three eqautions. By measuring one of the four unknowns, say
𝜃1, experimentally; the values of other three unknowns can be solved

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