LAWS OF MOTION

Force (ii) If a carpet is beaten with a stick, carpet comes

Push or pull that changes the state of rest, uniform in motion and move forward, but the dust

motion, shape or size of the object is known as particles remain at rest due to inertia of rest

force. and fall.

Force can be exerted on the objects with or without (iii) A one-rupee coin placed on a smooth

physical contact. cardboard covering a tumbler falls into the

Forces which are exerted on the objects with tumbler when card board is stricken with

physical contact are known as contact forces. finger. Cardboard comes in motion when it is

Examples: Friction, spring force, normal reaction, stricken whereas coin remains at rest due to

buoyant force etc. its inertia and falls into the tumbler.

Forces which are exerted on the objects without Inertia of motion/Direction

physical contact are known as non-contact forces. It is the inherent property of the body by virtue of

Examples: Electric force, Magnetic force, which it resists any change in its state of motion or

gravitational force etc. direction of motion.

--------------------------------------------------------------- Examples:

Inertia (i) The passengers fall forward when a fast-

Inherent property of all the bodies by virtue of moving bus stops suddenly. This is because

which they cannot change their state of rest or the lower part of the body of the passenger

uniform motion along a straight line by their own is comes into rest along with bus while upper

called inertia. part of the body, due to inertia of motion,

Types of inertia continues to move forward.

➢ Inertia of rest (ii) A person falls forward while getting down

➢ Inertia of motion/Direction from a moving bus or train. This is because as

Inertia of rest his feet touch on the ground, the lower part of

It is the inherent property of the body by virtue of his body comes to rest. On the other hand, the

which it resists any change in its state of rest. upper part remains in motion due to inertia of

Examples: motion and hence he falls forward.

(i) A person standing in a bus, falls backward (iii) The sparks produced during sharpening of a

when the bus suddenly starts moving. This is knife against a grinding wheel leave the rim of

because; the lower portion of the person being wheel tangentially because of inertia of

in contact with the floor of the bus comes in direction.

motion along with bus whereas the upper (iv) A stone is rotated in a horizontal circle with

portion of the body remains at rest due to the help of a string tied to the stone. If

inertia of rest. Hence, they fall backward. suddenly, the string breaks during the motion,

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 LAWS OF MOTION

the stone will fly off tangentially at that point Momentum

of the circle due to inertia of direction. The quantity of motion contained in a body is
-------------------------------------------------------------- known as momentum of the body.
❖ Mass is the measure of inertia, Inertia of rest/ ➢ Momentum of the body is defined as the
motion/direction increases with increase in product of mass and its velocity.
mass. ➢ Momentum is a vector quantity; it is in the
Justification: Consider two stationary objects; one direction of the velocity of the body.
is heavier than the other. If we apply the same 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 = 𝒎𝒂𝒔𝒔 × 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚
force on these objects, then it is observed that it is ⃗ = 𝑚𝒗
𝒑 ⃗
difficult to move the heavier object than that of S.I unit of momentum is 𝑘𝑔 𝑚𝑠 −1
the lighter one. It means, the heavier one resists The dimensional formula for linear momentum
more to change in its state of rest than the lighter is [𝑀𝐿𝑇 −1 ]
object. Thus, more is the mass, more the inertia. ----------------------------------------------------------------
----------------------------------------------------------------- Newton’s Second Law of motion
Newton’s Laws of motion Time rate of change of momentum is directly

Newton’s First Law of Motion/ Law of Inertia proportional to the external unbalanced force

Every body continues in its sate of rest or uniform applied on the body and this change takes place in

motion in a straight line unless compelled by some the direction of the applied force.

external force to change that state. Consider that a force 𝑭 is applied on an object of

First law is the definition of inertia, hence first law mass 𝑚 ,

of motion can be termed as law of inertia. According to second law of motion,

➢ If an object is at rest/ uniform motion, 𝑑𝑝

𝐹𝛼
𝑑𝑡
according to first law of motion, either no
Where 𝑝 is instantaneous momentum of the object.
force is acting on the object or the net force
𝑑𝑝
due to number of forces acting on the object 𝐹=𝑘
𝑑𝑡
is zero.
➢ If a non-zero force is acting on the object, Where 𝑘 is the constant of proportionality.
then the object cannot be in the state of rest 𝑑(𝑚𝑣)
𝐹=𝑘 [∵ 𝑝 = 𝑚𝑣]
or in uniform motion. 𝑑𝑡
➢ No net force is required to keep an object in 𝑑𝑣
𝐹 = 𝑘𝑚
𝑑𝑡
uniform motion.
𝑑𝑣
➢ Definition of the force comes from Newton’s 𝐹 = 𝑘𝑚𝑎 [ ∵ =𝑎]
𝑑𝑡
first law. ⃗𝑭 = 𝑚𝒂
𝑘 = 1, therefore ⃗
-----------------------------------------------------------------
➢ If 𝑚 is constant then

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 LAWS OF MOTION

𝑭𝛼𝒂 Define 1 dyne

i.e magnitude of acceleration increases with We know 𝐹 = 𝑚𝑎,
increase in the magnitude of applied force. if 𝑚 = 1 𝑔 𝑎𝑛𝑑 𝑎 = 1 𝑐𝑚/𝑠 2 , then
𝐹
➢ = 𝑎, if 𝐹 is constant then, 𝐹 = 1 × 1 = 1 𝑑𝑦𝑛𝑒
𝑚
1 i.e: 1dyne is that force which produces an
𝒂 𝛼
𝑚 acceleration of 1cm/s2 in a body of mass 1g.
i.e. for the given force, magnitude of the ❖ dyne and newton are absolute units of
acceleration decreases with increase in force.
mass. ❖ A force is said to be absolute unit of force if
➢ Second law of motion gives measure of force. it produces unit acceleration in a body of
⃗ = 𝑚𝒂
➢ 𝑭 ⃗ holds good only if the mass of the unit mass.
body remains constant. Gravitational unit of force
➢ 𝐼𝑓 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑔𝑒𝑠, 𝑡ℎ𝑒𝑛 ❖ A force is said to be gravitational unit of
𝑑(𝑚𝑣 ) 𝑑𝑣 𝑑𝑚 force if it produces acceleration equal to
𝐹= =𝑚 + 𝑣
𝑑𝑡 𝑑𝑡 𝑑𝑡 the acceleration due to gravity in a body of
➢ If the applied force increases the velocity of
unit mass.
the object, then the force is known as
➢ In S.I. system gravitational unit of force is
accelerating force.
kilogram weight (kg wt) or kilogram force
➢ If the applied force decreases the velocity of
(𝑘𝑔 𝑓 ).
the object, then the force is known as
We know 𝐹 = 𝑚𝑎,
decelerating force/ retarding force.
If 𝑚 = 1 𝑘𝑔 𝑎𝑛𝑑 𝑎 = 9.8 𝑚/𝑠 2 , then
➢ For uniform motion 𝑎 = 0, using second
𝐹 = 1 × 9.8 = 9.8 𝑁 = 1𝑘𝑔 𝑤𝑡
law 𝐹 = 𝑚𝑎 = 𝑚 × 0 = 0, i.e. No force is
i.e: 1kg wt is that force which produces an
required to move a body with uniform
acceleration of 9.8m/s2 in a body of mass 1kg.
velocity.
---------------------------------------------------------------
➢ In c.g.s. system gravitational unit of force is
Unit of force
gram weight (g wt) or gram force (𝑔 𝑓 ).
➢ S.I unit of force is newton – denoted by (𝑁)
We know 𝐹 = 𝑚𝑎,
➢ c.g.s unit is dyne.
If 𝑚 = 1 𝑔 𝑎𝑛𝑑 𝑎 = 980 𝑐𝑚/𝑠 2 , then
Define 1 N
𝐹 = 1 × 980 = 980 𝑑𝑦𝑛𝑒 = 1𝑔 𝑤𝑡
We know 𝐹 = 𝑚𝑎,
i.e: 1g wt is that force which produces an
If 𝑚 = 1 𝑘𝑔 𝑎𝑛𝑑 𝑎 = 1 𝑚/𝑠 2 , then
acceleration of 980 cm/s2 in a body of mass 1g
𝐹 =1×1=1𝑁
placed on earth surface.
i.e: 1N is that force which produces an acceleration
---------------------------------------------------------------
of 1m/s2 in a body of mass 1kg.

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 LAWS OF MOTION

➢ Absolute unit remains same throughout the i.e change in momentum is equal to impulse of the
universe, whereas gravitational unit force.
depends upon ′𝑔′ - acceleration due to -----------------------------------------------------------------
gravity. ➢ Impulse is a vector quantity.
----------------------------------------------------------------- ➢ Direction of impulse is same as that of the
Impulse applied force.
Impulse or impulsive force is the measure of impact ➢ S.I unit of impulse is N-s.
of large force in a small interval of time. ----------------------------------------------------------------
➢ Impulse is defined as the product of the Force- Time graph
average force and the time interval for
which the force acts on the object.
𝐹 𝐹
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑓𝑜𝑟𝑐𝑒 × 𝑡𝑖𝑚𝑒
Consider a force 𝑭 which acts on a body for time 𝑡 𝑡
𝑑𝑡. The impulse is given by
❖ Area under force – time graph is equal to
𝑑𝑱 = ⃗𝑭𝑑𝑡
magnitude of impulse or change in
If the force acts on the body for a time interval from
momentum.
𝑡1 to 𝑡2 , then the impulse is given by
---------------------------------------------------------------
𝑡2
∫ 𝑑𝑱 = ∫ ⃗𝑭𝑑𝑡 Applications of Newton’s second law of motion
𝑡1
(i) A cricket player lowers his hand while catching
𝑡2
⃗ ∫ 𝑑𝑡
𝑱=𝑭 a ball.
𝑡1
By doing so, he increases the time of impact(
𝑡2
𝑱 = ⃗𝑭[𝑡] catch) hence reduces the rate of change of
𝑡1
𝑑𝑝
⃗ [𝑡2 − 𝑡1 ]
𝑱=𝑭 momentum ( ) and force required to catch
𝑑𝑡
𝑑𝑝
the ball. ( 𝐹 = ).
𝑑𝑡
⃗ ∆𝑡
𝑱=𝑭
(ii) A person falling on a cemented floor gets
---------------------------------------------------------------
injured but a person falling on a heap of sand is
Impulse- momentum theorem
not injured.
➢ Impulse of force is equal to the change in
When a person falls on a cemented floor, he
momentum of the body.
comes to rest abruptly as the floor does not
We know that
yield under his weight. So the change in
∆𝑃
𝐹= momentum of the person takes place in a small
∆𝑡
interval of time. Hence, according to the second
∆𝑃 = 𝐹∆𝑡
𝑝2 − 𝑝1 = 𝐹∆𝑡

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 LAWS OF MOTION
𝑑𝑝
law a large force ( 𝐹 = ) is exerted on the
𝑑𝑡

person by the floor. 𝐹2 𝐹1

𝐴 𝐵
On the other hand, when a person falls on a
heap of sand, the sand yields under the weight --------------------------------------------------------------
of the person. Therefore he comes to rest in a ➢ Action and reaction are not balanced forces,
longer period of time. So the change in as they act on different bodies.
momentum of the person takes place in a ➢ As they are acting on different bodies, they
longer period of time. Hence, according to the never cancel each other.

second law a small force ( 𝐹 =

𝑑𝑝
) is exerted ➢ Action and reaction are equal in magnitude
𝑑𝑡
but opposite in direction.
on the person by the floor.
➢ Action and reaction act along same line.
(iii) The vehicles are fitted with shock absorbers.
➢ Action and reaction act simultaneously.
The floor of the vehicle is connected to the
----------------------------------------------------------
lower part of the vehicle by springs. When the
Illustration of Newton’s Third law of motion
vehicle moves over a rough road, the force due
(i) A block placed on a horizontal table remains
to jerks is transmitted to the floor of the
at rest even though it experiences force due
vehicle. Shock absorbers increase the time of
to gravity.
transmission of the jerks to the floor of the
Let 𝑚 be the mass of the block and 𝑔 be the
vehicle. Hence, less average force is
acceleration due to gravity.
experienced by the passengers or goods in the
𝐹𝑏 = 𝑚𝑔 (Reaction)
vehicle.
----------------------------------------------------------
𝐹𝑇 = 𝑚𝑔 (Action)
Newton’s Third law of motion
To every action, there is an equal and opposite
reaction. 𝐹𝑔 = 𝑚𝑔
A body 𝐴 exerts force 𝐹1 (push) on body 𝐵. This
Earth exerts a force 𝐹𝑔 = 𝑚𝑔 on the body in
force is known as action. According to third law of
downward direction. Therefore, body exerts a force
motion, in return body 𝐵 exerts a reaction
𝐹𝑇 = 𝑚𝑔 on the table. This force is action. In return,
𝐹2 (push) on body 𝐴.
according to third law of motion, table exerts a force
𝐹𝑏 = 𝑚𝑔 on the block in upward direction. This
𝐹2 𝐹1
𝐴 𝐵 force is the reaction force of action 𝐹𝑇 .

If body 𝐴 exerts force 𝐹1 (pull) on body 𝐵 , then Here block experiences two equal and opposite
according to third law of motion, in return body 𝐵 force (balanced). Therefore, block is at rest.
exerts a reaction 𝐹2 (pull) on body 𝐴.
Jojesh- Physics Page 5 of 17
 LAWS OF MOTION

(ii) When a ball is dropped from some height, it ∴ 𝑊 = 𝑚𝑔

rebounds after striking against the floor. (ii) Weighing machine is in motion (moving lift)
When a ball strikes against the floor, it exerts (a) Accelerating upward
a force (action) on the floor. According to When lift is accelerating upward with
Newton’s third law of motion, the floor exerts acceleration 𝑎 , the net force on the body is 𝐹𝑁 =
an equal and opposite force (Reaction) on the 𝑚𝑎.
ball. Due to this force, the ball rebounds.
𝑁′
(iii) While swimming, a man pushes the water 𝐹𝑁 = 𝑚𝑎
backward with his hands. The reaction force
offered by the water to the man pushes him 𝐹𝑔 = 𝑚𝑔
forward.
𝐹𝑁 = 𝑚𝑎
(iv) The gases produced in the combustion
𝑁 ′ − 𝑚𝑔 = 𝑚𝑎
chamber of a rocket engine come out of the
∴ 𝑁 ′ = 𝑚𝑔 + 𝑚𝑎
nozzle of the rocket in the downward direction.
∴ 𝑊 ′ = 𝑚(𝑔 + 𝑎)
The reaction of these exhaust gases pushes the
𝑊′ > 𝑊
engine in the upward direction.
i.e: apparent weight > true weight.
---------------------------------------------------------------
(b) Accelerating downward
Apparent weight of an object in a lift.
When lift is accelerating downward with
Weight of an object is the measure of normal
acceleration 𝑎 , the net force on the body is 𝐹𝑁 =
reaction ( 𝑁 ) exerted by the floor on the object.
𝑚𝑎.
(i)Weighing machine is at rest (stationary lift)
𝑁′
When a body of mass 𝑚 rests on the surface of the 𝐹𝑁 = 𝑚𝑎
weighing machine, body exerts a force 𝐹𝑔 = 𝑚𝑔 on
the machine, in return, machine offers a normal 𝐹𝑔 = 𝑚𝑔

reaction 𝑅 = 𝑚𝑔 on the object. 𝐹𝑁 = 𝑚𝑎

𝑁 = 𝑚𝑔 𝑚𝑔 − 𝑁 ′ = 𝑚𝑎
𝑎 = 0 ∴ 𝐹𝑁 = 𝑚𝑎 = 0
∴ 𝑁 ′ = 𝑚𝑔 − 𝑚𝑎
∴ 𝑊 ′ = 𝑚(𝑔 − 𝑎)
𝐹𝑔 = 𝑚𝑔 𝑊 ′ < 𝑊 i.e: apparent weight < true weight.
(c) When lift falls freely ( free fall)
Since body is at rest, according to first law of
In free fall acceleration of the object is equal to
motion, we have
acceleration due to gravity.
𝐹𝑁 = 0
𝑁 − 𝑚𝑔 = 0
∴ 𝑁 = 𝑚𝑔

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 LAWS OF MOTION

𝑁′ = 0 Let 𝑇 be the tension in the string. The light mass

𝐹𝑁 = 𝑚𝑔 𝑚1 moves upward with acceleration 𝑎 and the
heavy mass 𝑚2 moves downward with
𝐹𝑔 = 𝑚𝑔
acceleration 𝑎.
𝐹𝑁 = 𝑚𝑔 Equation of motion of mass 𝒎𝟏
′
𝑚𝑔 − 𝑁 = 𝑚𝑔 Net upward force on 𝑚1 is given by
∴ 𝑁 ′ = 𝑚𝑔 − 𝑚𝑔 𝐹1𝑁 = 𝑇 − 𝑚1 𝑔
∴ 𝑊′ = 0 𝑚1 𝑎 = 𝑇 − 𝑚1 𝑔 − − − (1)
i.e: apparent weight of an object in free fall is Equation of motion of mass 𝒎𝟐
zero. Net downward force on 𝑚2 is given by
(d) Moving with uniform velocity 𝐹2𝑁 = 𝑚2 𝑔 − 𝑇
Since body is with constant velocity, according to 𝑚2 𝑎 = 𝑚 2 𝑔 − 𝑇 − − − (2)
first law of motion, we have Adding (1) and (2), we get
𝐹𝑁 = 0 (𝑚2 + 𝑚1 )𝑎 = (𝑚2 − 𝑚1 )𝑔
𝑁 ′ − 𝑚𝑔 = 0 𝒎𝟐 − 𝒎𝟏
𝒂=( )𝒈
∴ 𝑁 ′ = 𝑚𝑔 𝒎𝟐 + 𝒎𝟏
Using the value of 𝒂 in equation (1), we get
∴ 𝑊 ′ = 𝑚𝑔
𝟐𝒎𝟐 𝒎𝟏
𝑊 ′ = 𝑊 i.e: apparent weight = true weight. 𝑻=( )𝒈
𝒎𝟐 + 𝒎𝟏
----------------------------------------------------------------
------------------------------------------------------
Motion of connected bodies (Masses and pulley)
Free body diagram is a diagram that shows the
given body or object as a point and the various
forces acting on it.
---------------------------------------------------------------
𝑇
➢ Tension is a force exerted on an object by a
𝑎 string which is connected to that object.
𝑇
➢ Direction of tension is always directed
𝑎
𝑚1 𝑔 outward from the object.
➢ A string can only pull the objects, it cannot

𝑚2 𝑔 push the objects.

---------------------------------------------------------------
Consider two masses 𝑚1 and 𝑚2 connected to the Newton’s Second law is the real law of motion
ends of an inextensible string of negligible mass (i) First law is contained in second law
which passes over a smooth pulley as shown in the According to second law
figure. 𝐹 = 𝑚𝑎

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 LAWS OF MOTION

If 𝐹 = 0, i.e no external force acts on the body, 𝑑

⃗⃗⃗⃗ + ⃗⃗⃗⃗
(𝑃 𝑃𝐵 ) = 0
then 𝑑𝑡 𝐴
Therefore equ (3) becomes
𝐹 = 𝑚𝑎 = 0 ⟹ 𝑎=0
This shows that body is at rest or in uniform
motion. 𝐹𝐴𝐵 + 𝐹𝐵𝐴 = 0

i.e. In the absence of external unbalanced force 𝐹𝐴𝐵 = − 𝐹𝐵𝐴

body at rest will remain at rest or a body moving This is Newton’s third law of motion.

with uniform velocity will continue to move As both laws are present in second law, second law

with same velocity. This is first law. of motion is the real law of motion.

(ii) Third law is contained in second law ----------------------------------------------------------------

Consider an isolated system (free from external Law of conservation of linear momentum

forces) consisting of two particles bodies A and B. Total linear momentum of an isolated system
remains constant.
OR
𝐹𝐴𝐵 𝐹𝐵𝐴
𝐴 𝐵
In the absence of external unbalanced force, total
linear momentum of the system is constant.
Let 𝐹𝐴𝐵 be the force on the body 𝐴 by the body 𝐵 .
---------------------------------------------------------------
𝐹𝐵𝐴 be the force on the body 𝐵 by the body 𝐴 .
Justification of (C.L.M) using second law
Rate of change of momentum of the body 𝐴
According to Newton’s second law of motion,
⃗⃗⃗⃗𝐴
𝑑𝑃
= 𝑑𝑝
𝑑𝑡 𝐹=
𝑑𝑡
Rate of change of momentum of the body 𝐵 𝑑𝑝
If 𝐹 = 0, then = 0 or 𝑃⃗ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
⃗⃗⃗⃗𝐵
𝑑𝑃 𝑑𝑡
= ---------------------------------------------------------------
𝑑𝑡
According to Newton’s second law Justification of (C.L.M) using third law
⃗⃗⃗⃗𝐴
𝑑𝑃 Consider two bodies 𝐴 and 𝐵 of masses 𝑚1 and
𝐹𝐴𝐵 = − − − (1)
𝑑𝑡 𝑚2 constituting an isolated system. Let the bodies
⃗⃗⃗⃗𝐵
𝑑𝑃 are moving with velocities 𝑢1 and 𝑢2 respectively
𝐹𝐵𝐴 = − − − (2)
𝑑𝑡 along straight line in the same direction (𝑢1 > 𝑢2 ).
Adding equ (i) and (ii), we get
After collision, let these bodies move with velocities
𝑑𝑃⃗⃗⃗⃗𝐴 𝑑𝑃 ⃗⃗⃗⃗𝐵
𝐹𝐴𝐵 + 𝐹𝐵𝐴 = + 𝑣1 and 𝑣2 respectively in the same direction.
𝑑𝑡 𝑑𝑡
𝑑
𝐹𝐴𝐵 + 𝐹𝐵𝐴 = (𝑃 ⃗⃗⃗⃗ + 𝑃
⃗⃗⃗⃗𝐵 ) − − − −(3)
𝑑𝑡 𝐴
Since no external force acts on the system, therefore
according to Newton’s second law, the rate of
change of momentum of the system is zero
Jojesh- Physics Page 8 of 17
 LAWS OF MOTION

𝐴 𝐵 𝑢 Examples – conservation of linear momentum

𝑢1 2
𝑚1 𝑚2 (i) Recoil of a Gun
𝐴 𝐵 When a bullet is fired from a gun, gun moves
𝐹𝐴𝐵 𝐹𝐵𝐴
𝑚1 𝑚2 backward. This is known as recoil of a gun.
The velocity with which a gun moves
𝐴 𝐵
𝑣1 𝑣2 backward after firing a bullet is known as
𝑚1 𝑚2
recoil velocity.
Total momentum of the system before Before firing, the system is at rest. So the linear
collision = 𝑚1 𝑢1 + 𝑚2 𝑢2 − − − −(1) momentum of the system before firing is zero,
Total momentum of the system after After firing, the bullet leaves the gun and it
collision = 𝑚1 𝑣1 + 𝑚2 𝑣2 − − − − − (2) moves with a large velocity in the forward
Change in momentum of the body 𝐴 after direction. Thus, the bullet has large linear
collision = 𝑚1 𝑣1 − 𝑚1 𝑢1 − − − −(4) momentum. Since, no external force acts on the
Change in momentum of the body 𝐵 after system, so its linear momentum after firing must
collision = 𝑚2 𝑣2 − 𝑚2 𝑢2 − − − −(5) be zero. This can happen if the gun has linear
During collision body 𝐵 exerts force 𝐹𝐴𝐵 on 𝐵 . momentum in the backward direction.
During collision body 𝐴 exerts force 𝐹𝐵𝐴 on 𝐴 . Mass of the bullet = 𝑚𝑏
According to third law of motion Mass of the bullet = 𝑚𝑔
𝐹𝐴𝐵 = −𝐹𝐵𝐴 − − − (6)
Velocity of the gun and bullet before

Impulse acting on body 𝐴 = 𝐹𝐴𝐵 ∆𝑡 − − − −(7) firing = 𝑢𝑏 = 𝑢𝑔 = 0

Impulse acting on body 𝐵 = 𝐹𝐵𝐴 ∆𝑡 − − − −(8) Velocity of the bullet after firing = 𝑣𝑏
We know that Velocity of the gun after firing = 𝑣𝑔
Impulse = change in momentum Since external force is zero, total linear momentum
Therefore using equ (4) and (7) is conserved
𝐹𝐴𝐵 ∆𝑡 = 𝑚1 𝑣1 − 𝑚1 𝑢1 − − − (9) 𝑚𝑏 𝑢𝑏 + 𝑚𝑔 𝑢𝑔 = 𝑚𝑏 𝑣𝑏 + 𝑚𝑔 𝑣𝑔
Therefore using equ (5 and (8) 0 = 𝑚𝑏 𝑣𝑏 + 𝑚𝑔 𝑣𝑔
𝐹𝐵𝐴 ∆𝑡 = 𝑚2 𝑣2 − 𝑚2 𝑢2 − − − (10) 𝑚𝑔 𝑣𝑔 = −𝑚𝑏 𝑣𝑏
Substituting equ (9) and (10) in equ (6), we get 𝒎𝒃 𝒗𝒃
𝒗𝒈 = −
𝑚2 𝑣2 − 𝑚2 𝑢2 = −( 𝑚1 𝑣1 − 𝑚1 𝑢1 ) 𝒎𝒈
𝑚1 𝑢1 + 𝑚2 𝑢2 = 𝑚1 𝑣1 + 𝑚2 𝑣2 Negative sign shows that velocity of gun is in the
It shows that total momentum before collision = opposite direction of that of the velocity of the
total momentum after collision bullet.
-------------------------------------------------------------- --------------------------------------------------------------

Jojesh- Physics Page 9 of 17

 LAWS OF MOTION

(ii) Explosion of a bomb the particle will be in equilibrium under the action
Suppose a bomb is at rest before it explodes. Its of all the three forces.
momentum is zero. When it explodes, it breaks ∴ ⃗⃗⃗
𝐹1 + ⃗⃗⃗
𝐹2 + ⃗⃗⃗
𝐹3 = 0
up into many parts, each part having a 𝐶
particular momentum. A part flying in one
⃗⃗⃗ 𝐶
direction with a certain momentum, there is 𝐹2 ⃗⃗⃗
𝐹1
⃗⃗⃗
𝐹2
another part moving in the opposite direction ⃗⃗⃗
𝐹2 ⃗⃗⃗
𝐹1
⃗⃗⃗3
−𝐹
with the same momentum. If the bomb explodes 𝑂 𝑂
⃗⃗⃗
𝐹1
into two equal parts, they will fly off in exactly 𝑂
opposite directions with the same speed, since ⃗⃗⃗
𝐹3 ⃗⃗⃗
𝐹3
each part has the same mass.
------------------------------------------------------------
𝑣3
Friction
𝑚3
𝑣1 𝑚1 𝑚2 𝑣2 Friction is the contact force which opposes the
Bomb at rest
𝑚4 relative motion or tendency of relative motion
𝑚1 𝑣1 between two bodies in contact and acts tangentially
𝑣4
𝑚4 𝑣4 along the surfaces of contact.
𝑚3 𝑣3
➢ Direction of friction is opposite to the
𝑚2 𝑣2 direction of relative motion between the two
-------------------------------------------------------------- surfaces in contact.
Concurrent forces ➢ Friction acts parallel to the surfaces in
Concurrent forces refer to a set of forces that all act contact.
on a single point on body and whose lines of action ----------------------------------------------------------------
all intersect at a common point. In other words, Cause of friction
these forces have the same point of application (i) Old view
(where they are applied) and their directions might Friction is caused by the irregularities on the
vary, but they all intersect at one common point. two surfaces in contact. Even those surfaces
which appear very smooth have a large

Let three concurrent forces ⃗⃗⃗

𝐹1 , ⃗⃗⃗
𝐹2 𝑎𝑛𝑑 ⃗⃗⃗
𝐹3 are number of minute irregularities on them.

acting at a common point (O) of a particle. The Irregularities on the two surfaces lock into one
another. When we attempt to move any
resultant of ⃗⃗⃗ ⃗⃗⃗2 is given by the diagonal of
𝐹1 and 𝐹
surface, we have to apply a force to overcome
the parallelogram. i.e. by OC as shown in figure. If
interlocking. The magnitude of the force
the third force ⃗⃗⃗⃗
𝐹3 is equal and opposite to the
required to break the irregularities is the
resultant of ⃗⃗⃗ ⃗⃗⃗2 i.e. ⃗⃗⃗⃗
𝐹1 and 𝐹 ⃗⃗⃗1 + 𝐹
𝐹3 = − (𝐹 ⃗⃗⃗2 ) , then
measure of force of friction.
Jojesh- Physics Page 10 of 17
 LAWS OF MOTION

(ii) Modern view Limiting force of friction

When one surface is placed over another Static friction increases with increase in applied
surface, molecules of the surfaces press against force, therefore static friction is known as self
each other and get interlocked. The points at adjusting friction.
which molecules are attracting each other are The maximum value of the static frictional force
known as actual contact points. Pressure at which acts between the surfaces when one surface
these points is high and a type of cold welding is just going to slide over other is called limiting
takes place. When one surface slides over the value force of friction.
other, the bonds between the molecules are -----------------------------------------------------
continuously broken and re-build at other Dynamic friction
points. The force required to break the bonds The force of friction which acts between the
between the molecules is the measure of surfaces when one surface slides/ rolls over the
frictional force. other is called dynamic friction/ kinetic friction.
---------------------------------------------------------------- There are two types of dynamic friction
Types of friction (i) Sliding friction
(i) Static friction (ii) Rolling friction
(ii) Dynamic or kinetic friction -----------------------------------------------------------
Static friction ➢ Kinetic friction < limiting friction
The force of friction which opposes the ➢ Rolling friction < sliding friction
relative motion and applied force when during the ----------------------------------------------------------
stationary state of a body is called static friction. Limiting friction
Force of friction

𝑁 = 𝑚𝑔 Kinetic friction
Body at Rest
Static
𝐹
𝑓𝑠

Applied Force
𝑊 = 𝑚𝑔
--------------------------------------------------------------
𝑓𝑠 = 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
𝐹 = 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑓𝑜𝑟𝑐𝑒
In the above diagram 𝑓𝑠 = 𝐹 and 𝑊 = 𝑁
Therefore net force on the block =0
-------------------------------------------------------------

Jojesh- Physics Page 11 of 17

 LAWS OF MOTION

Laws of friction ➢ Coefficient of friction between two surfaces is

(i) Direction of force of limiting friction is defined as the ratio of limiting friction to the
always opposite to the direction of normal reaction acting between the two
relative motion between the surfaces in surfaces in contact.
contact. ➢ Coefficient of friction has no unit.
(ii) Force of friction acts tangentially along ➢ Coefficient of friction depends up on the nature
the surfaces of contact of two bodies. (roughness) of surfaces and materials of
(iii) Magnitude of limiting friction is always surfaces in contact.
directly proportional to the normal -----------------------------------------------------------------
reaction between the two surfaces. Coefficient of kinetic friction
𝑓𝑙 𝛼 𝑁 We know that kinetic friction is directly
(iv) Force of limiting friction is independent proportional to the normal reaction.
of area of surfaces in contact as long as 𝑓𝑘 𝛼 𝑁
normal reaction is constant. 𝒇𝒌 = 𝝁𝒌 𝑵 − − − −(1)
(v) Force of limiting friction depends on 𝒇𝒌 = 𝝁𝒌 𝒎𝒈
nature of the surfaces in contact. Where 𝜇𝑘 is known as coefficient of kinetic
------------------------------------------------------------ friction.
Coefficient of limiting force of friction From equ (1)

𝑁 = 𝑚𝑔 𝒇𝒌 𝒇𝒌
𝝁𝒌 = =
𝑵 𝒎𝒈
----------------------------------------------------------
𝐹
𝑓𝑙

𝑊 = 𝑚𝑔

We know that limiting friction is directly

proportional to the normal reaction.
𝑓𝑙 𝛼 𝑁
𝒇𝒍 = 𝝁𝒔 𝑵 − − − −(1)
𝒇𝒍 = 𝝁𝒔 𝒎𝒈
Where 𝜇𝑠 is known as coefficient of static friction.
From equ (1)
𝒇𝒍 𝒇𝒍
𝝁𝒔 = =
𝑵 𝒎𝒈
----------------------------------------------------------------

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 LAWS OF MOTION

Angle of friction At the angle of repose

The angle between the normal reaction and the 𝑚𝑔 𝑠𝑖𝑛𝛼 = 𝑓𝑙 − − − (1)
resultant of limiting force of friction and normal And 𝑚𝑔 cos 𝛼 = 𝑁 − − − −(2)
reaction is called angle of friction. 𝑓𝑙
(1) ÷ (2) ⇒ tan 𝛼 =
The resultant of limiting friction (𝑓𝑙 ) and normal 𝑁
𝑓𝑙
reaction (𝑁) is represented by 𝑁 ′ = 𝑂𝐶. The But = 𝜇𝑠
𝑁
angle 𝜃 between 𝑁 and 𝑁 ′ is called as angle of ∴ 𝐭𝐚𝐧 𝜶 = 𝝁𝒔
friction. Coefficient of limiting friction is numerically equal
𝑁 = 𝑚𝑔 to the tangent of the angle of repose.
𝐶
-------------------------------------------------------------
𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛
𝜃
➢ Angle of repose = angle of friction.
𝑂 𝑓𝑙 ➢ 𝐭𝐚𝐧 𝜶 = 𝐭𝐚𝐧 𝜽 = 𝝁𝒔
-----------------------------------------------------
𝑚𝑔
Friction is a friend and foe
𝑓𝑙 Friction as a friend:
Now tan 𝜃 =
𝑁 1. Friction plays a vital role in our daily life.
𝑓𝑙
But = 𝜇𝑠 2. Construction of buildings is possible only
𝑁
∴ 𝐭𝐚𝐧 𝜽 = 𝝁𝒔 − − − (1) because of friction.
Coefficient of limiting friction is numerically equal 3. We are able to write because of friction
to the tangent of the angle of friction. between the paper and the tip of the pen.
--------------------------------------------------------------- 4. Without friction between our shoes and the
Angle of repose ground, we will not be able to walk.
The minimum angle made by the inclined plane 5. A horse cannot pull a cart unless friction
with the horizontal surface such that the body lying furnishes him a secure Foothold.
on the inclined plane is just at the verge of sliding Friction as a foe:
down along the inclined plane is called angle of 1. The main disadvantage of friction is that it
repose. produces heat in the moving parts of machines
Let 𝛼 be the angle of repose. which causes wear and tear of the parts.
2. It opposes the motion.
𝑁
3. A lot of energy is wasted in overcoming friction.
4. It causes wear and tear of the soles of shoes.
𝑓𝑙 5. Due to friction, noise is also produced in machines.
𝑚𝑔 sin 𝛼
𝛼 𝑚𝑔 cos 𝛼 ----------------------------------------------------------------

𝑚𝑔

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 LAWS OF MOTION

Methods of reducing friction Magnitude of centripetal acceleration and force

1. Polishing: If we polish a surface, it becomes Speed of the object = 𝑣
smooth and friction is reduced. Through Radius of the circular path = 𝑟
polishing, unevenness of the surfaces is Mass of the object = 𝑚
reduced. Centripetal acceleration,
2. Lubricating: By applying lubricants (like oil) to 𝒗𝟐
𝒂𝒄 =
surfaces, friction is reduced. When we apply 𝒓
Centripetal force
lubricant to surfaces, a thin layer of lubricant is
formed over there and moving surfaces do not 𝒎𝒗𝟐
𝑭𝒄 = 𝒎𝒂𝒄 =
𝒓
directly rub against each other.
-----------------------------------------------------------------
3. Using ball bearings: This way of reducing
Centrifugal force
friction involves the principle that an object is
Centrifugal force is a pseudo (fictitious) force which
rolled instead of sliding. The use of ball bearing
is acting radially outward on an object in circular
converts sliding friction into rolling friction.
motion.
4. Separation of surfaces by air: Another way of
➢ This force tends to move the object away
reducing friction is to separate the surfaces by
from the axis of rotation.
air. This is how, a hovercraft works. A hovercraft
➢ Magnitude of the centrifugal force is equal to
moves on a layer of air between its hull and the
the magnitude of the centripetal force.
water. The layer of air reduces friction allowing
----------------------------------------------------------------
the hovercraft to move easily.
Motion of a car on level road
5. Giving a streamlined shape: Boats, cars, planes
Consider a car of mass 𝑚 moving with a constant
and rockets are streamlined to reduce friction
speed 𝑣 on a level horizontal circular path of radius
with water or air.
𝑟.
---------------------------------------------------------------
When car moves along the circular path, it
Circular motion
experiences centrifugal force. This force tends the
➢ Motion of a body in a circular path is known
car to slide radially outward. This produces a
as circular motion.
friction between tyres and road directed towards
➢ The force which keeps the object in circular
the centre of the circular path. This friction
motion is called centripetal acceleration.
provides the centripetal force for the circular
➢ The acceleration which is acting toward the
motion.
centre of the circular path is called
The various forces acting on the car are:
centripetal acceleration.
(i) Weight 𝑊 = 𝑚𝑔 of the car in vertically
---------------------------------------------------------------
downward direction.

Jojesh- Physics Page 14 of 17

 LAWS OF MOTION

(ii) The normal reaction 𝑁 = 𝑚𝑔 of the necessary centripetal force and vehicle will
ground on the car in vertically upward skid and go off the road.
direction. -----------------------------------------------------------
(iii) Force of friction 𝑓 between the tyres Banking of roads
and road. When a car goes round a level curve, the force of
Maximum friction between the tyres and road friction between the tyres and the road provides
𝑓 = 𝜇𝑘 𝑁 the necessary centripetal force. If the frictional
Centripetal force required for the circular force, which acts as centripetal force and keeps
motion is the body moving along the circular road is not
𝑚𝑣 2 enough to provide the necessary centripetal
𝐹𝑐 =
𝑟 force, the car will skid.
This force must be given by friction
At the curved path, the outer edge of the road is
2
𝑚𝑣
∴ 𝑓≥ raised above the level of the inner edge. This is
𝑟
known as banking of curved roads or tracks.
𝑚𝑣 2
𝜇𝑘 𝑁 ≥ ➢ By banking the road, vehicles get a component
𝑟
𝑚𝑣 2 of normal reaction of ground towards the centre
𝜇𝑘 𝑚𝑔 ≥
𝑟 of the circular path. This component provides
√𝜇𝑘 𝑟𝑔 ≥ 𝑣 the centripetal force in addition to the
𝒗 ≤ √𝝁𝒌 𝒓𝒈 centripetal force given by the friction.
➢ The angle between the surface of the road and
horizontal is called angle of banking
𝑟 𝑁
----------------------------------------------------------------
𝑚𝑣 2 𝑓 𝑐𝑎𝑟 Motion of a car on banked road
𝑟

𝑣 𝑁 cos 𝜃
𝑚𝑔 𝑁
𝜃
𝑁 sin 𝜃
𝑓 cos 𝜃 𝜃
----------------------------------------------------------------
𝑚𝑔
➢ Maximum speed the vehicle can have on the 𝑓
𝑓 sin 𝜃
level road is 𝒗𝒎𝒂𝒙 = √𝝁𝒌 𝒓𝒈. 𝜃

➢ In the absence of friction i.e. 𝝁𝒌 = 𝟎,

maximum possible speed 𝒗𝒎𝒂𝒙 = 0.
➢ If the speed of the vehicle is 𝒗 > √𝝁𝒌 𝒓𝒈,
then friction will not be able to provide the

Jojesh- Physics Page 15 of 17

 LAWS OF MOTION

Various forces acting on the car are: ➢ Maximum possible safe speed for the given
(i) Weight 𝑊 = 𝑚𝑔 of the car in vertically tan 𝜃+ 𝜇
𝜇 is 𝑣𝑚𝑎𝑥 = √𝑟𝑔 (1−𝜇 𝑡𝑎𝑛𝜃)
downward direction.
(ii) The normal reaction 𝑁 of the ground on ➢ The optimum speed 𝝁 = 𝟎

the car. 𝒗 = √𝒓𝒈 𝒕𝒂𝒏𝜽

(iii) Force of friction 𝑓 between the tyres ➢ There will be minimum wear and tear of the
and road. tyres with this speed. [ friction is minimum]
Resolve 𝑁 into two components: ➢ It is easier to drive at sea level than at
(𝑖) 𝑁 𝑠𝑖𝑛𝜃 and (𝑖𝑖)𝑁 cos 𝜃 mountain. [𝒗 = √𝒈 ]
Similarly, 𝑓 cos 𝜃 and 𝑓 sin 𝜃 are the horizontal ➢ It is easier to negotiate a curve with bigger
and vertical components of the force of friction 𝑓. radius than a curve with smaller radius.
[𝒗 = √𝒓 ]
For the equilibrium of the car, -------------------------------------------------------------
𝑚𝑔 + 𝑓 𝑠𝑖𝑛𝜃 = 𝑁 cos 𝜃 Bending of a cyclist
𝑚𝑔 = 𝑁 cos 𝜃 − 𝑓 𝑠𝑖𝑛𝜃 − − − (1)
𝑁 sin 𝜃 + 𝑓 cos 𝜃 acts towards the centre of the
circular path. This provides the necessary
centripetal force.
𝑚𝑣 2
≤ 𝑁 sin 𝜃 + 𝑓 cos 𝜃 − − − (2)
𝑟
(2) ÷ (1)
𝑚𝑣 2 𝑁 sin 𝜃 + 𝑓 cos 𝜃
≤
𝑟𝑚𝑔 𝑁 cos 𝜃 − 𝑓 𝑠𝑖𝑛𝜃
sin 𝜃 𝑓 cos 𝜃
𝑣2 𝑁 cos 𝜃 (cos 𝜃 + 𝑁 cos 𝜃)
≤ cos 𝜃 𝑓 𝑠𝑖𝑛𝜃
𝑟𝑔 𝑁 cos 𝜃 (cos 𝜃 − 𝑁 cos 𝜃) A cyclist has to bend slightly towards the centre of
𝑓
𝑣2 tan 𝜃 + 𝑁 the circular track in order to take a safe turn
≤ 𝑓
𝑟𝑔 1 − 𝑁 𝑡𝑎𝑛𝜃 without slipping.
Radius of the circular path = 𝑟.
𝑣2 tan 𝜃 + 𝜇 𝑓
≤ [ ∵ =𝜇] Mass of the cyclist = 𝑚.
𝑟𝑔 1 − 𝜇 𝑡𝑎𝑛𝜃 𝑁
tan 𝜃 + 𝜇 Speed of the cyclist = 𝑣.
𝑣 2 ≤ 𝑟𝑔 ( )
1 − 𝜇 𝑡𝑎𝑛𝜃 When the cyclist negotiates the curve, he bends

𝐭𝐚𝐧 𝜽 + 𝝁 inwards from the vertical, by an angle θ.

𝒗 ≤ √𝒓𝒈 ( )
𝟏 − 𝝁 𝒕𝒂𝒏𝜽 Let R be the reaction of the ground on the cyclist.

------------------------------------------------------- The reaction R may be resolved into two

components: (i) the component R sin θ, acting
Jojesh- Physics Page 16 of 17
 LAWS OF MOTION

towards the centre of the curve providing Consider a heavy mass 𝑚 tied to a string and
necessary centripetal force for circular motion and suspended from a rigid support as shown in figure.
(ii) The component R cos θ, balancing the weight of The mass is whirled round in a horizontal circle of
the cyclist along with the bicycle. radius 𝑟 with constant speed 𝑣 such that string
i.e makes and angle 𝜃 with the vertical.
𝑚𝑣 2
= 𝑅 sin 𝜃 − − − (1)
𝑟 Various forces acting on the mass are:
𝑚𝑔 = 𝑅 cos 𝜃 − − − (2)
(i) The weight 𝑤 = 𝑚𝑔 of the mass acting
(2) ÷ (1)
vertically downward.
𝑚𝑣 2 𝑅 sin 𝜃
= (ii) Tension 𝑇 in the string acting along the
𝑟𝑚𝑔 𝑅 cos 𝜃
string towards the point of suspension.
𝑣2
= tan 𝜃 The tension 𝑇 can be resolved into two
𝑟𝑔
components. i.e. 𝑇 sin 𝜃 𝑎𝑛𝑑 𝑇 cos 𝜃.
𝒗 = √𝒓𝒈 𝒕𝒂𝒏𝜽
𝑇 cos 𝜃 balances weight of the mass.
---------------------------------------------------------
𝑇 sin 𝜃 provides the centripetal force to move
Thus for less bending of cyclist (i.e for θ to be
along horizontal circular path.
small), the velocity v should be smaller and radius r
𝑚𝑔 = 𝑇 cos 𝜃 − − − (1)
should be larger.
𝑚𝑣 2
---------------------------------------------------------------- = 𝑇 sin 𝜃
𝑟
Conical pendulum 𝑚𝜔2 𝑟 = 𝑇 sin 𝜃 − − − (2) [ ∵ 𝑣 = 𝑟𝜔 ]
(2) ÷ (1)
𝑚𝜔2 𝑟 𝑇 sin 𝜃
=
𝑚𝑔 𝑇 cos 𝜃
𝜃
𝜔2 𝑟
= tan 𝜃
𝑔
𝑇𝑐𝑜𝑠𝜃
𝑔 tan 𝜃
𝜃 𝜔 =√
𝑟
𝑇𝑠𝑖𝑛𝜃
𝑟 Therefore time period of suspended mass is given
by

𝒓 2𝜋
𝑚𝑔 𝑻 = 𝟐𝝅 √ [∵ 𝑇 = ]
𝒈 𝐭𝐚𝐧 𝜽 𝜔

Conical pendulum is a small heavy mass suspended

by a string from a rigid support and whirled in a ***********************************************
horizontal circle with a constant speed.

Jojesh- Physics Page 17 of 17