Laws of Motion.pdf

NEWTON’S LAWS OF MOTION
Inertia: Resistive property of an object while change in it's state of motion
or rest is called Inertia.
INERTIA OF REST
The property of a
body due to which it
cannot change its
state of rest by it
self.
The property due to
which a body cannot
change its direction of
motion by itself.
The tendency of a
body to remain in
a state of uniform
motion in a
straight Line.
INERTIA OF DIRECTION INERTIA OF MOTION
A body Continues its sate of
rest or motion until unless an
external force is acted on it.
Newton’s 1st law
Newton’s 2nd Law
The rate of change of linear momentum of a body
is directly proportional to the external force applied on
the body in the direction of force.
dp
F ma
dt
= =

 
S.I . Unit of force = Newton (N)
Impulse
avg
I F t P
= ∆ = ∆
  

⇒ I = P F.dt
∆ = ∫ = area under F– t curve
Newton’s 3rd law
To every action there is always an
equal and opposite reaction.
AB BA
F F
= −
 
• Action & Reaction act on different
bodies and not on the same body.
• action – reaction forces are of
same type.
dm
v constant F=v
dt
= ⇒

 
Force(N)
Time (s)
Impulse (Ns)
Consevation of linear momentum:
When net external force on system is zero, the total
linear momentum of an isolated system of interacting
particles is conserved.
ext
initial final
If F 0 p constant
p p
= ⇒ =
∴ =

Conditions
for
action-reaction
pairs
Equal magnitude and nature
Act along the same line
Act in opposite direction on different objects
Occur simultaneously
FORCES
(i) Normal Contact force
(1) always acts along the
common Normal of two surface
in contact.
(2) Always directed towards the
system.
(3) Normal force on
block is N. N = mg
(ii) Tension Force
(1) Acts along the string and
away from the system on
which it acts.
(2) Tension in a massless string
remains constant throughout
the string if no tangential
force acts along the string.
(3) This is force applied by a
string on an object or force
applied by one part of string
on the remaining part of
string.
The force which opposes the relative
motion of two contact surfaces is
friction. That is a frictional force.
(iii) Frictional Force
f = friction
F = contact force
N = normal force
fs < µsN
acts when a body is just at the verge of movement.
acts when a body is at rest on application
of a force.
f l = µsN
acts when a body is actually moving.
fk = µKN
Static friction
Limiting friction
Kinetic friction
Rolling friction
The force of friction which comes into play when one
body Ralls or tends to roll on the surface of
anorther body.
The friction force depends
upon the nature of surfaces
in Contact and independent
of the area of Contact.
For Non – inertial frame
ext Pseudo
F F ma
+ =
  
pseudo frame
F Ma
= −
 
MOTION OF A CAR on LEVEL
ROAD (by friction only):-
f
N
mg
R
Speed = v
a = v² / R
O
max s
v Rg
≤ µ
MOTION OF A CAR ON BANKED ROAD
(i) Optimum speed of a vehicle on a banked road.
(ii) maximum safe speed on a banked frictional road.
(iii) minimum safe speed on a banked frictional road
V rgtan
= θ
max
rg( tan )
V
1 tan
µ + θ
=
−µ θ
min
Rg(tan )
V
(1 tan )
θ − µ
=
+µ θ
N
f
mg
θ
Ο′

1 ACTIVE SITE EDUTECH-9844532971
NEWTON’S LAW OF MOTION
Total Sessions – 09
SESSION – 1
AIM
✓ To study Newton’s three laws of motion
✓ Concept of Momentum
NEWTON’S FIRST LAW OF MOTION:
According to first law, “everybody continues in the state of rest or of uniform motion in
a straight line unless an external force acts on it to change the state.”
We draw the following inferences from Newton’s first law of motion:
i] If a body is at rest and we want to set it in motion, then a force has to be applied.
ii] If the body is moving with a constant speed along a straight line, then in order to
increase or decrease its speed, a force has to be applied in the direction of motion or
opposite to the direction of motion.
iii] If the body is moving with constant speed along a straight line, then in order to
change its direction of motion, a force has to be applied normal to the direction of
motion.
Therefore, a body is reluctant to change its present state. This property of the body is
known as inertia. Hence Newton’s first law of motion is also known as law of inertia. It is also
obvious from Newton’s first law of motion that a force has to be applied on the body to
change its state of rest or state of uniform motion along a straight line. Hence, Newton’s
first law of motion defines force.
THEORY:
Force is an external effort in the form of push and pull which either changes or tend to
change the state of rest or of uniform motion of a body.
Newton’s first law of motion can also be stated in the following manner.
“If the vector sum of all the forces acting on particle is zero then and only then the
particle remains at rest or moves with constant velocity (unaccelerated).” i.e., a = 0 if
and only if F = 0.
LINEAR MOMENTUM
The linear momentum of a body is defined as total quantity of motion contained in the
body, depends upon the mass and velocity of the body, is measured by the product of
mass m and velocity of the body i.e.,
Momentum = mass x velocity
𝐩 = 𝐦𝐯

ACTIVE SITE EDUTECH-9844532971
MOMENTUM IS A VECTOR QUANTITY
Consider a ball of mass m and a car of mass (M > m) are moving with the same velocity v. If p1
and p2be their momentum respectively, then
P1
P2
=
m
M
or
P1
P2
=
m
M
As M > m, hence p2 > p1
Therefore, if a ball and car be moving with the velocity, then the momentum of car will be
greater than the momentum of ball.
The above facts are represented in the following graphs (fig)
NEWTON’S SECOND LAW OF MOTION
According to Newton’s second law of motion, the rate of change of linear momentum of a
body is directly proportional to the external force applied on it and the change of
momentum takes place in the direction of force.
Let p be the linear momentum of a body and an external force F is applied on it. Then
F  (dp/dt) or F = k (dp/dt)
where k is a constant of proportionality
In S.I. and C.G. S. systems the unit of force is selected in such a way that k= 1.
A unit force is defined as that force which when applied on a body of unit mass accelerates
it with unit acceleration.
∴ F =
dp
dt
=
d
dt
(mv)
OrF = m
dv
dt
= ma [if m is constant]
F = ma
In scalar form, F = m a
The second law of motion gives a measure of force.
In S.I. system, the absolute unit of force is newton.
1 newton = 1 kg x 1 m/s
2
= 1 kg m/s
2
In C.G.S. system, the unit of force is dyne.
1 dyne = 1 g x 1 cm/s
2
= 1 g cm/s
2
Further 1 newton = 105 dyne
v=constant
Y
P
O
m
X
m=constant
Y
P
O
v
X
p=constant
Y
v
O
m
X

In S.I. unit, the gravitational unit of force is kilogram weight.
In C.G.S., system, the gravitational unit of force is gram weight
1 kg wt (or kg f) = 9.8 N
and 1 g wt ( or g f ) = 980 dyne
NEWTONS THIRD LAW OF MOTION
According to Newton’s third law of motion, to very action, there is always an equal and
opposite reaction.
To explain the Newton’s third law, let us consider the following two examples:
1. Consider a body A of weight W resting on another body B as shown in fig. The body A
exerts force F1 (called action) which is equal to its weight W on body B. According
to Newton’s third law, the body B also exerts an opposite force F2 (called
reaction) which is equal to −W on body A. Thus F1 = −F2
2. Consider free fall of a body A on the earth as shown in fig. The body is acted upon by
the
force of gravity of the earth (action). The body also exerts equal and opposite force
on the earth (reaction). The force of gravity produces acceleration in the body while
reaction produces an acceleration in the earth (negligibly small).
Following are few examples of action and reaction:
• When a man jumps on the shore from a boat (action), the boat is simultaneously
pushed away from the shore (reaction)
• When a rubber ball is struck against the wall (action) then it is rebounded back by the
wall (reaction).
B
F1 (Action)
F2
(Reaction)
A
Earth
A
F1
F2

• When a rifle is fired, the bullet moves forward (action) while the rifle recoils
backward (reaction).
• When a swimmer pushes the water backwards (action), the water pushes the swimmer
forward (reaction).
• A book lying on the table exerts a force on the table (action). The table also exerts an
equal force on the book (reaction).
CLASS EXERCISE
1] Two forces 𝐹1
⃗⃗⃗ act 𝐹2
⃗⃗⃗ on a 2 kg mass. If 𝐹1 = 10𝑁and 𝐹2 = 5𝑁, find the acceleration.
a) 2.5 √3 𝑚/𝑠2
b)2.5 m/s2 c) 2√3m/s2 d) √3m/s2
2] Two particles with inertial masses 3 kg and 9 kg are acted upon by the same force. If
the acceleration of the 9 kg particle is3 𝑚/𝑠2
, the acceleration of the 3 kg particle
is
a) 1𝑚/𝑠2
b)9 𝑚/𝑠2
c)0.9 𝑚/𝑠2
d)0.1 𝑚/𝑠2
3] A train reduces its speed uniformly from 75 to 15 kmph in 40s in a straight line. To
come to rest, travelling at the same rate it will take a further time of
a) 40s b) 20s c) 15s d) 10s
4] The horizontal and vertical components of a force are 30N and 40N. If the force acts
on a body of mass 5 kg, acceleration produced is
a) 5 𝑚/𝑠2
b) 0.5 𝑚/𝑠2
c)1 𝑚/𝑠2
d)10 𝑚/𝑠2
HOME EXERCISE
1] A 5 gm bullet acquires a speed of 120 m/s in a gun with barrel length 2.0m. The
average force exerted on the bullet is
a) 3.6 N b) 18 N c) 36 N d) 72N
2] A force of 100N stops a body moving with a velocity of 20 𝑚𝑠−1
. The force required to
stop the same body when moving with 30 ms-1 in the same distance is
a) 550 N b) 225N c) 112.5 N d) 65N
3] A body of mass 5 kg is dropped from the top of a tower. The force acting on the body
during motion is
a) 0 b) 9.8N c) 5 kg wt d) none

4] A 1.5 kg hammer moving with velocity 10 ms-1strikes a nail for 0.005, seconds.
Average force exerted on the nail is
a) 1000N b) 1500 N c) 750 N d) 3000N
5] A 0.6 kg ball strikes a wall with a velocity 5ms-1 at an angle of 300 with the wall and
rebounds at the same angle with the same speed. The change in the momentum
of the ball perpendicular to the wall is:
a) 15 kg ms-1 b)10 kg ms-1 c) 5 kg ms-1 d) 3 kg ms-1
6] A ball of mass m strikes a wall with a speed x and retraces its path with the speed y.
If the ball is in contact with the wall for time t, then the magnitude of average
force exerted by the wall on the ball is
a)
𝑚(𝑥−𝑦)
𝑡
b)
𝑚𝑡
(𝑥+𝑦)
c)(
𝑥+𝑦
𝑚
) d)
𝑚(𝑥+𝑦)
𝑡
7] A ball moving with momentum 10 kg ms-1 strikes a wall at an angle 45°and is reflected
at the same angle, with the same magnitude of momentum. The magnitude
of change in momentum.
a) 0 b) 20 kg ms-1 c) 20√2kg ms-1 d) 10√2 kg ms-1

SESSION – 2
AIM - To study about impulse and impulsive forces
Impulse
Sometimes a large force acts for a very short duration and produces a finite change in
momentum of the body. For example, hitting a cricket ball by a bat. Here, the ball is
reflected back and the force on the ball acts for a very short time when the two are in
contact. Practically, it is not possible to measure either the magnitude of force or the time
for which it acts. In such a case, the total effect of force is measured by the impulse of the
force. The impulse of the force is defined as the product of the force and its time of
action.
If a force F acts on a body for a very short time t, then
Impulse= Force x time duration = F xt ...
= Change in momentum
The force acting for a short duration is called as impulsive force.
Impulse = p2 − p1
Thus, impulse of the force is equal to the total change in momentum of the body.
Impulse is a vector quantity and its direction is same as that of force. The impulse of
the force may be positive, negative or zero according as the momentum of the body
increases, decreases or remains unchanged under the effect of force.
When a graph is drawn between force and time, then it is called as force- time graph.
If the force F is constant, the force-time graph will be a straight line parallel to X-
axis as shown in fig. If force F1 acts for time interval (t2
- t1
) and another force F2
acts for time interval (t3
- t2
), then force-time graph will be as shown in fig. If the
force is constantly varying, then the graph will be as shown in fig.
From fig.
Impulse = F ×t = Area ABDC
From fig
Impulse = Area ABCD + Area GHID
A B
B
C D
Y
O X
t1
t2
Time(t)
F
A B
G
Y
O X
t1 t2
Time(t)
F
H
F2
F1
C D
t3
1
B
Y
F
A C
O t1 t2
Time(t)
X

Few applications of the concept of impulse are:
• While catching a cricket ball, the player lowers his hands.
• Vehicles like cars trucks, bogies of trains etc. are provided with a spring system
(known as shockers) to avoid severe jerks.
• China- clay wares and glass wares are wrapped in paper or straw pieces.
• An athlete is advised to come to stop slowly after finishing a fast race.
• person falling on cemented floor is likely to receive more injuries than one falling on a
heap of and.
CLASS EXERCISE
1] A 1kg ball drops vertically into the floor with a speed of 25 ms-1 and rebounces with a
speed of 10ms-1. What is the impulse acting on the wall?
a) 15kg m/s b) 35kg m/s c) 45 m/s d) 60 m/s
2] A constant force acts on a body of mass 50 gm at rest for 2 seconds. If the body
moves through 27m during that time, impulse of the force is
a) 1.35 kg m/s b) 13.5 Ns c) 135 Ns d) 2.7 kg m/s
HOME EXERCISE
1] A ball strikes the bat with a force 400N.if both are in contact for a time 0.01s. the
impulse of the force
a) 400Ns b) 4Ns c) 40000Ns d) 0.004 Ns
2] A player catches a cricket ball of mass150kg moving at 20m/s. if the catching process
is completed in 0.1s. the force exerted by the ball on the hands is
a) 3000N b) 300N c) 30N d)30000N
3] The area of the force time graph gives
a) velocity b) acceleration c) displacement d) change in momentum.
4] A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet
releasing water at a rate of 1 kg/s and at a speed of 5 m/s. The initial acceleration
of the block will be:
a) 2.5 m/s2 b) 5 m/s2 c) 10 m/s2 d) 5 m/s
5] A body of mass 40 kg, moves with a uniform velocity under the action of a force 50 N
on a surface. If a force of 70 N now acts on the same body in the same direction as
that of 50 N, moving on the same surface, the acceleration of the body is
a) 1.75m/s2 b) 1.5 m/s2 c) 1.0 m/s2 d) 0.5 m/s2
6] A body moving with constant velocity is brought to rest in 0.25 sec by applying a
retarding force 100N. The initial momentum of the body

a) 25 kg m/s b) 50 kg m/s c) 100 kg m/s d) 125 kg m/s
7] A 2 kg plate is kept suspended in air by allowing 10 marbles hitting per second with a
speed v from the downward direction. If the mass of each marble is 20 g then
determine v.
a) 20 ms-1 b) 100 ms-1 c) 40ms-1 d)50ms-1
8] Two stones of masses m1 and m2 are let fall from heights 2h and h, their momenta on
reaching the ground are in the ratio.
a) 1 : 1 b) 𝑚1: √2𝑚2 c) 2𝑚1: 𝑚2 d)𝑚1: 𝑚2
9] A body of mass 100 kg is moving with a velocity 1m/s the frictional force offered by
the surface is 5 kg wt. If the body is pushed by a force of 50 N for one minute,
velocity of the body after one minute is
a) 0.16 m/s b)16 m/s c) 1.6m/s d) 3.2 m/s
10] A body of mass 1.5 kg falls vertically downwards with an acceleration of 29.4 m/s2.
The force acting on the body in addition to force of gravity is
a) 9.8 N b) 19.6N c) 29.4 N d) 49N
11] A body of mass 5 kg started from rest with an acceleration of 4 ms-2. Its momentum
after 5s is
a) 20 kg ms-1 b) 100 kg ms-1 c) 4 kg ms-1 d) 25 kg ms-1
12] A 6 kg balls strikes a vertical wall with a velocity 34 ms-1 and rebounds with a
velocity of 26 ms-1. The impulse is
a) 60 Ns b) 180 Ns c) 48Ns d) 360Ns
13] A block of a mass 4 kg is sliding on a smooth inclined plane of inclination 30°. Its
momentum after 2 sec is
a) 0 b) 39.2 kg ms-1 c) 19.6 kg ms-1 d) 9.8 kg ms-1
14] Two bodies of mass 5 kg and 20 kg at rest are acted upon by the same force. The
ratio of the times for which the force must act to produce the same impulse is
a) 1 : 1 b) 1 : 4 c) 4 : 1 d) 1 : 16

SESSION – 3
AIM - To study concept of Normal reaction force and to learn free body diagram.
FREE BODY DIAGRAM
When several bodies are connected by strings, springs, surfaces of contact, then all the
forces acting on the body are considered and sketched on the body under consideration by
just isolating it. Then the diagram so formed is called the Free Body Diagram (FBD) In order
to sketch free body diagram, the following points are kept in mind.
a] The reaction force also called Normal Reaction (N) always acts normal to surface on
which the body is kept.
(A) (B)
(C)
b] When two blocks (M1 and M2 say) are connected by a string, the tension in string for
block M1 is towards M2 and for block M2, it is towards M1
.
c] When a block is connected by a spring and the spring is stretched by a distance x,
then force on block F = kx ( opposite to stretching of spring, where k is the force constant
of the spring)
N
Block
W
Table
Block
N
N
Block
Inclined Plane
M1
M2
T T

d] A pulley changes the direction of force in a convenient direction. If the pulley is light
and frictionless and a string passes over it without any kinks, then tension on either side
of the string is the same.
e] While drawing a FBD always take into account the forces which are acting on the body.
Never take into account the forces which the body is exerting on others.
f] Always remember that friction is a tangential force acting tangentially to the
surfaces in contact.
g] All quantities involved in writing the Cause-Effect equation must have units same as
that of force. Finally we write the Cause-Effect equation which is a consequence of Newton
2nd Law according to which
∑ F
⃗ = ma
⃗
Sometimes you may even have to select a convenient co-ordinate axis for writing
Cause-Effect equations. In that case we may write
∑ Fx = max , ∑ Fy = may, ∑ Fz = maz
As an illustration of FBD consider two bodies M and m in contact placed on a table. If
a force F is applied on M from the left and the force of interaction between bodies is
f, then free body diagram of M and m will be as shown in figure.
For horizontal motion of M
F − f = Ma and for horizontal motion of m
f = ma
(a) (b)
(c)
For the FBD of system shown above in figure
As system is having vertical equilibrium, so
N1 = Mg, N2 = mg
M
m
F
M
F
N1
M
f
Mg
m
N2
f
mg

EQUILIBRIUM
A body is said to be in the equilibrium state when
Note:
a] No net force acts on the body
∑ F
⃗ = 0
⃗
 ∑ Fx = 0 ∑ Fy = 0 ∑ Fz = 0
(CONDITION FOR TRANSLATIONAL EQUILIBRIUM)
b] No net torque acts on the body
∑  = 0
⃗
 ∑ y = 0 ∑ y = 0 ∑ z = 0
(CONDITION FOR ROTATIONAL EQUILIBRIUM)
This statement is none other than law of conservation of moments according to which the
above condition can be restated as
∑ (
𝑡𝑜𝑡𝑎𝑙 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑚𝑜𝑚𝑒𝑛𝑡𝑠
) = ∑ (
𝑡𝑜𝑡𝑎𝑙 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑚𝑜𝑚𝑒𝑛𝑡𝑠
)
THEORY: MOTION OF A BLOCK ON A HORIZONTAL SMOOTH SURFACE
a] When subjected to a horizontal pull
Consider a block of mass m placed on a horizontal smooth surface as shown in fig.
The block is subjected to a horizontal pull by applying a force F. The different forces
on block are also shown is figure.
As there is no motion along vertical direction, hence
R = mg
For horizontal motion,F = ma or a =
F
m
b] When subjected to a pull acting at an angle  to the horizontal
This situation is shown in fig. Resolving F in horizontal and vertical directions, we have
the components Fcosθ and Fsinθ respectively.
R
a
F
m

In this case, we have
R + F sinθ = mg or R = mg − F sinθ ....(1)
Thus R ≠ mg but it has a value less than mg. For horizontal motion,
F cosθ = ma or a =
F cos
m
....(2)
c] When subjected to a push acting at an angle  to the horizontal
The situation is shown in fig. Resolving F in horizontal and vertical directions, we have
the components Fcosθ and F sinθ respectively. In this case, we have
R = mg + Fsinθ
Thus R is greater than mg.
Further Fcosθ = ma,a =
F cos
m
CLASS EXERCISE
1]Two blocks of masses 2 kg and 1 kg are in contact with each other on a frictionless table.
When a horizontal force of 3.0 N is applied to the block of mass 2 kg , the value of the
force of contact between the two blocks is:
a) 4N b) 3 N c) 2 N d) 1 N
2]Two bodies A and B of masses 3 kg and 2 kg are connected by a string as shown in the
figure. If the table and pulley are frictionless, the common acceleration of the system is (g =
10 ms-1)
R
a
Fcos
m  
F
Fsin
mg
R
a
Fcos
m


F
Fsin
mg

a) 4 𝑚𝑠−2
b) 6 𝑚𝑠−2
c)2 𝑚𝑠−2
d) zero
3] In the above question the tension in the string is
a) 8 N b) 12 N c) 10 N d) 4 N
4] In the above question B moves down with acceleration if
a) 𝑚𝐵 < 𝑚𝐴 b)𝑚𝐵 > 𝑚𝐴 c)𝑚𝐵 = 𝑚𝐴 d) all the above
5] Three bodies A(1 kg), B (2kg) and C (3kg) are connected by light inextensible strings
and the system is pulled with a force F. If the tension in the string connecting B
and C is 4.5 N, the force F is
a) 6 N b) 4.5 N c) 18 N d) 9N
6] In the above question the ratio of the accelerations of A, B and C is
a) 1: 2: 3 b) 3: 2:1 c) 1:1:1 d) 1:3:2
HOME EXERCISE
1] A bird of mass 0.1 kg, rising vertically with an acceleration 0.2 𝑚𝑠−2
. The muscular
force exerted by it is (𝑔 = 9.8 𝑚𝑠−2)
a) 1 Newton b) 9.8 kg.wt c) 1/9.8 kg.wt d) 980 newton’s
2] Three identical blocks of masses m = 2 kg are drawn of force F = 3.6 N with an
acceleration of 0.6 𝑚𝑠−2
on a frictionless surface then what is tension in string
between blocks B & C is
a) 1.2 N b) 8N c) 4 N d) 1.8 N
3] A string of negligible mass going over a clamped pulley of mass m supports a block of
mass M as shown in figure. The force on the pulley by the clamp is given by
a) √2 𝑚𝑔 b) √2 𝑚𝑔 c)√[(𝑀 + 𝑚)2 + 𝑀2]𝑔 d) √[(𝑀 + 𝑚)2 + 𝑚2]𝑔
B
A
A B C F
A
B
C F
M
m

SESSION – 4
AIM - Concept of normal reaction force, contact force and tension in case of
connected bodies
Tension in case of connected bodies: Two bodies
Let us consider the case of two bodies of masses m1 and m2 connected by a thread and
placed on a smooth horizontal surface as shown in fig. A force F is applied on the body of
mass m2 in forward direction as shown. Our aim is to consider the acceleration of the system
and the tension T in the thread. The force acting separately on two bodies are also shown in
the fig
From fig T = m1a
F − T = m2a
Adding (1) and (2)
F = (m1 + m2)a or a =
F
m1+m2
The acceleration of the system can be calculated from eq. (3) and tension in the thread by
eq. (1)
Three bodies: In case of three bodies, the situation is shown in fig.
Acceleration a =
F
m1+m2+m3
T1 = m1a =
m1F
m1 + m2 + m3
F − T2 = m3a
T2 = F −
m3F
m1 + m2 + m3
//////////////////////////////////////////
m1 m2
T T F
a
//////////////////////////////////////////
m1 m2
T F
T
N1
m g
1 m g
2
N2
Motion
Motion
///////////////////////////////////////////////////////////////////////
m1 m2
m3
T1 T2
F
a
////////////////////////////////
M1
T1
R1
m g
1
////////////////////////////////
M3
T2
R3
m g
3
F
a

MOTION OF BODIES IN CONTACT: FORCE OF CONTACT
Let us consider the case of two bodies of masses m1
and m2
respectively in contact placed on
a frictionless table as shown in fig. Let a force
F be applied on mass m1
as shown. Our aim is to calculate the acceleration produced and the
force of contact between the bodies.
The acceleration a of both the bodies =
Force
Total mass
=
F
(m1+m2)
The free body diagram of mass m1
is also shown in fig.
We have F − f = m1a
or f = F − m1a = F − F −
m1×F
m1+m2
When the force is applied on the body of mass m2
as shown in fig. then we have
acceleration=
𝐹
𝑚1+𝑚2
The free body diagram of the body of mass m2
is also shown in the fig.
Now we have
F − f = m2a
f = F − m2a
= F −
m2 × F
m1 + m2
MOTION OF CONNECTED BODIES:
Before discussing the motion of connected bodies, let us first consider few fundamentals
regarding strings.
Fundamentals of Strings:
• A string is assumed to be inextensible, i.e., length remain constant.
//////////////////////////////////////////
f f
m1 m2
a
F
(a)
//////////////////////////////////////////
f
m1
a
F
(b)
//////////////////////////////////////////
m1 m2
a
F (a)
//////////////////////////////////////////
F
m2
a
(b)
'
f
'
f
'
f

• The magnitude of accelerations of a number of bodies connected through strings is
always same.
• The tension in every part of the string remains the same. This is equal to the applied
force. In order to produce a tension in the string, two equal and opposite
stretching forces must be applied.
• If the tension in the string is increased continuously, it breaks down. The maximum
tension which a string can bear without breaking is called breaking strength.
• When a string is fixed horizontally by clamping its free ends and loaded at the middle,
as shown in fig. the tension T is given by
T =
W
(2 cosθ)
2 T cosθ = W or T =
W
(2cosθ)
It is clear from this expression, that tension in the string will be lesser, equal or
greater than load depending upon the value of θ.
i) The tension in the string will be minimum when cos θ = max
i.e., θ = 0°as shown in fig.
∴ Tmin = W/2 ∴
This shows that the tension loaded at the middle can never be lesser than half the
suspended load.
ii)The tension in the string will be maximum when cos= minimum,
i.e., = 900 as shown in fig.
In this case, Tmax = 
2 T cos 
T
T T
T
T sin 
T sin 
R R

W
W
T
T
T
0
0
=

W=2 T
T
W
T T T T
90
0

0
90
=

T = horizontal
W = vertical

This situation cannot be realized practically because the string can bear a finite
tension while the tension here is infinite. Therefore, a string loaded at middle can
never remain horizontal.
CLASS EXERCISE
1] A body of mass 4kg is suspended at the middle of a string clamped horizontally such
that now the string makes an angle 60. Find the Tension in the string.
2] A block of mass M is pulled along a horizontal frictionless surface by a rope of mass
m. If a force F is applied at one end of the rope, the which the rope exerts on the
block is:
a) 𝐹/(𝑀 + 𝑚) b) 𝐹 c)𝐹𝑀/(𝑚 + 𝑀) d) Zero
3] Two blocks of masses 2 kg and 3 kg are on a smooth horizontal table, in contact with
each other. A horizontal force 𝐹 = 6𝑁 is applied as shown. Then the force of
contact is
a) 2.4 𝑁 b) 3 𝑁 c)6 𝑁 d) 3.6 𝑁
HOME EXERCISE
1] Two blocks are attached to the two ends of string passing over a smooth pulley as
shown in the figure. The acceleration of the block will be (𝑖𝑛 𝑚/𝑠2
)
a) 0.33 b) 1.33 c)1 d) 0.66
2] A 40 N force pulls a system of three masses on a horizontal frictionless surface as
shown in the fig. Tension 𝑇1 is
a) 40 N b) 20 N c) 10N d) 5N
3] Find the tension 𝑇2 in the system shown in fig.
2 T cos 
T
T T
T
T sin 
T sin 
R R

W
2 3
F
50kg
370
53
0
100kg
6kg 4kg
10kg
40N T1

a) 1𝑔 𝑁 b) 2g N c) 5𝑔 𝑁 d) 6 𝑔 𝑁
4] Three blocks are connected as shown in the fig. on a horizontal frictionless table. If
𝑚1 = 1 kg,
𝑚2 = 8 kg, 𝑚3 = 27 kg and 𝑇3 = 36 N, T2 will be:
a) 18 N b) 9N c) 3.375N d) 1.75N
5] A load is suspended by a heavy rope from a rigid support A. If the tension in the rope
at A, B and C are TA, TB and TC respectively
a) TA = TB– TC b) TA> TB> TC c) TA< TB< TC d) TA < TB> TC
6] Five identical cubes each of mass m are on a straight line with two adjacent faces in
contact on a horizontal surface as shown in the fig.
Suppose the surface is frictionless and a constant force P is applied from left to right
to the end face of A; which of the following statements are correct?
a) The acceleration of the system is
5𝑃
𝑚
b) The resultant force acting on each cube is
𝑃
5
c) The force exerted by C and D is
2𝑃
5
d) The acceleration of the cube D is
𝑝
5
𝑚
1 kg
2 kg
3 kg
T1
T2
T3
m1
m2 m3
T1 T2 T3
A
B
C
M
A B C D E
P

SESSION – 5
AIM - Calculation of tension in case of pulley systems
MOTION OF TWO BODIES CONNECTED BY A STRING
Case 1:Let us consider the case of two bodies of masses m1
and m2
which are connected by
light inextensible string passing over a light smooth pulley as shown in fig. Here it is assumed
that
m1 > m2. Our aim is to find out the acceleration of the system and tension in the string.
The mass m1 has downward acceleration hence m1g − T = m1a
The mass m2 has upward acceleration hence T − m2g = m2a
Solving eq. (1) and eq. (2) we get a = (
m1−m2
m1+m2
) g
T = (
2m1m2
m1 + m2
) g
Case 2:
ATWOOD’S MACHINE:
Masses M1 and M2 are tied to a string, which goes over a frictionless light pulley.
The string is light and inextensible.
a] If M1 > M2and they move with acceleration 𝑎, then
M1g − T = M1a
T − M2g = M2a
Where T is the tension in the string. It gives
m g
1
m g
2
m1
m2
a
a
T
T
T
T
T
M2
M1
T

a = (
M1−M2
M1+M2
) g and T = (
2M1M2
M1+M2
) g
Thrust is given by 2T. So,
Thrust = (
4M1M2
M1+M2
) g
b] If the pulley begins to move with acceleration 𝑎0
⃗⃗⃗⃗ , then
a = (
M1−M2
M1+M2
) (g
⃗ + a
⃗ 0)and
T = (
2M1M2
M1+M2
) (g
⃗ + a
⃗ 0)
So, if pulley accelerates up, then
a = (
M1−M2
M1+M2
) (g + a0) and T = (
2M1M2
M1+M2
) (g + a0)
c] When it accelerates down, then
a = (
M1−M2
M1+M2
) (g − a0)and T = (
2M1M2
M1+M2
) (g − a0)
and thrust in the pulley is
T
⃗
⃗ =
4M1M2
M1+M2
(g
⃗ − a
⃗ 0)
Case 3: Let us consider the case of a body of mass 𝑚1 to which a light and inextensible
string is
attached, rests on a smooth horizontal table. The string passes over a frictionless pulley
fixed at the end of table. Another end of the string carries a mass m2 as shown in fig. Our
aim is to calculate the acceleration of the system and tension in the string. Here we have
(m2g − T) = m2a
And T = m1a
Solvingeqs. (1) and (2) we have
a = (
m1
m1+m2
) g T = (
m1m2
m1+m2
) g
//////////////////////////////////////////
Motion
M
ot
io
n
m1
m1
m2
T
m g
2
m g
1
T
T

Case 4: Let us consider the case of a body of mass 𝑚3 to which two light and inextensible
strings are attached, rests on a smooth horizontal table. The string passes over a
frictionless pulley fixed at the end of table. Another end of the string carries a mass 𝑚1 and
the second string passes over a frictionless pulley fixed at the other end of table. Another
end of the string carries a mass 𝑚2as shown in fig. Our aim is to calculate the acceleration
of the system. Here we have
For body of mass M1, M1g − T1 = M1a
For body of mass M2 T2 − M2g = M2A
For body of mass M3T1 − T2 = M3a
Solving these equations we get, 𝑎 = (
𝑚1−𝑚2
𝑚1+𝑚2+𝑚3
) 𝑔
Case 5: Here we shall consider the above case with a difference that m1 is placed on
smooth inclined plane making an angle  with horizontal as shown in fig.
In this case, T − m1gsinθ = m1a1
And (m2g − T) = m2 a Solving, we get
𝑎 =
(𝑚2−𝑚1 sin 𝜃)𝑔
(𝑚1+𝑚2)
and T = m2g − m2a
or T = mg [1 −
(m2 − m1 sin θ)
(m1 + m2)
]
M
ot
ion
m2
T
m g
2

m1
T
Motion
.
R T
Horizontal

m
g
1
m g
1
 m g
1
cos 
sin 
m2
m g
2
T
m1

Case 6: Here we shall consider the above case with a difference that m1 is placed on
smooth inclined plane making an angle  with horizontal and m2 is placed on the other side on
the smooth inclined plane as shown in fig.
In this case, for body of mass m1, m1gsinα − T = m1a
For body of mass m2 we get, T– m2gsinθ = m1a
On solving, we get
𝑎 =
(𝑚1sinα−𝑚2 sin 𝛽)𝑔
(𝑚1+𝑚2)
Tension in the string will be: T = [
(m1m2 (sin α+sinβ)g)
(m1+m2)
]
CLASS EXERCISE
1] Three equal weights A, B and C each of mass 2 kg are hanging on a string passing over
a fixed frictionless pulley as shown in figure. The tension in string connecting B and C is
a) zero b) 13 N c) 3.3 N d) 19.6 N
2] A light string passes over a frictionless pulley. To one of its ends a mass of 6 kg is
attached. To its other end a mass of 10 kg is attached. The tension in the thread
will be (g=9.8m/s2)
a)24.5 N b)2.45 N c)79 N d)73.5 N

3] The elevator is going up with an acceleration of g/10, the pulley and the string are
light and the pulley is smooth. Find the reading of the spring balance.
4] Two masses of 4 kg and 5 kg are connected by a string passing through a frictionless
pulley and are kept on a frictionless table as shown in the figure. The
acceleration of 5 kg mass is (g=9.8m/s2)
a) 4.9 𝑚/𝑠2
b) 5.44𝑚/𝑠2
c)12.2 𝑚/𝑠2
d) 9.8𝑚/𝑠2
5] Three masses of 1 kg, 6 kg and 3 kg are connected to each other with threads and are
placed on a table as shown in figure. What is the acceleration with which the
system is moving? Take
𝑔 = 10 𝑚 𝑠−2
a) Zero b) 𝑙 𝑚𝑠−2
c) 2 𝑚 𝑠−2
d) 3 𝑚 𝑠−2
6] To calculate the acceleration of the system and tension in the string when mass 𝑚1 =
1𝑘𝑔 and mass 𝑚2 = 4𝑘𝑔 angle of incline is 30o
.

7] Consider the situation shown in figure. All the surfaces are frictionless and the string
and the pulley are light. Find the magnitude of the acceleration of the two blocks
and tension in the string.
HOME EXERCISE
1] In the figure, the masses of the blocks A and B are same and each equal to ‘m’. The
tensions in the strings OA and OB are 𝑇1 and 𝑇1 respectively. The system is in
equilibrium with a constant horizontal force mg on B. The tension 𝑇1 is
a) mg b)√2 𝑚𝑔 c)√3 𝑚𝑔 d)√5 𝑚𝑔
2] Two masses of 8 kg and 4 kg are connected by an inextensible light thread passing
over a smooth fixed pulley. The force felt by the axle of the pulley is
a) 8 g b) 12 g c) less than 12g d) greater than 12 g
3] Three blocks of masses 𝑚1, 𝑚2 and 𝑚3 are connected with weightless string and are
placed on a frictionless table. If the mass 𝑚3 is dragged with a force T, the
tension in the string between 𝑚2and 𝑚3 is:
a)
𝑇𝑚3
𝑚1+𝑚2+𝑚3
b)
𝑇𝑚3
𝑚1+𝑚2+𝑚3
c)
(𝑚1+𝑚2)
𝑚1+𝑚2+𝑚3
𝑇 d) none of these
O
2

T2
A
1

T1
B
mg
8kg
T T
4kg
XT

4] For the system shown in the figure, the pulleys are light and frictionless. The tension
in the string will be:
a)
2
3
𝑚𝑔 𝑠𝑖𝑛𝜃 b)
3
2
𝑚𝑔 𝑠𝑖𝑛𝜃 c)
1
2
𝑚𝑔 𝑠𝑖𝑛𝜃 d) 2 𝑚𝑔 𝑠𝑖𝑛𝜃
5] A string of length L and mass M is lying on a horizontal table. A force F is applied at
one of its ends. Tension in the string at a distance x from the end at which force is
applied is:
a) Zero b) F c) 𝐹(𝐿 − 𝑥)/𝐿 d) 𝐹(𝐿 − 𝑥)/𝑀
6] A chain of mass M and length L is held vertical by fixing its upper end to a rigid
support. The tension in the chain at a distance y from the rigid support is:
a) mg b) 𝑀𝑔 (𝐿 − 𝑦)/𝐿 c) 𝑀𝑔 𝐿/(𝐿 − 𝑦) d) 𝑀𝑔 𝑦/𝐿
7] A load 5kg is suspended from a rigid support B using a uniform rope of length 2m and
4 kg mass. The tension in the rope at the point B which is 0.5m below A is (see the figure in
the above question)
(𝑔 = 10 𝑚𝑠−2)
a) 60 N b) 80N c) 8N d) 4N
8] Two blocks of masses 5kg and 2kg are connected by a uniform rope of mass 3kg. If an
upward force F is applied on the system, the acceleration of the system is found
to be 2ms-2 upward, then F is equal to
a) 82.6N b) 148N c) 118N d) 78N
9] Two bodies A (400 gm) and B are connected by a light inextensible string which passes
over a frictionless pulley as shown. If B comes down with acceleration 4 𝑚𝑠−2
, its mass is
a) 600 gm b) 1200 gm c) 300gm d) 800gm
10] Two bodies A and B each of mass m are connected by a light inextensible string which
passes over a frictionless pulley as shown. If the angle of inclination is 37° and
the inclined plane is frictionless, then (𝑠𝑖𝑛 37 = 3/5)
m
 m
B
A
5kg
A
B
30
0

a) The bodies will be in equilibrium and the tension in the string is zero
b) B moves down with acceleration g/5 and the tension in the string is 4 mg/5
c) A moves down with acceleration g/5 and the tension in the string is 4 mg/5
d) None of the above is true
11] In the fig. given below two masses m and mare tied with a thread passing over a pulley
m is on a frictionless horizontal surface. If acceleration due to gravity is 𝑔 , the
acceleration of 𝑚 in this arrangement will be:
a) 𝑔 b) mg/(m + m) c) mg/ m d) mg/ (m - m)
12] Two bodies of masses 5 kg and 4 kg are arranged in two positions as shown in fig. (A)
and (B), if the pulleys and the table are perfectly smooth, the accelerations of
the 5 kg body in case (A) and (B) are:
a) g and (5/9) g b) (4/9) g and (1/9) g
c) g/5 and g/5 d) (5/9) g and (1/9) g
13] Three equal weights of mass 2.5 kg each are hanging on a string passing over a fixed
pulley as shown in fig. What is the tension in the string connecting weights B and C?
a) 3.3N b) 13N c) 16.3N d) 19.6N
14] Two blocks each of mass M are resting on a frictionless inclined plane as shown in fig.
a) The block A moves down the plane b) The block B moves down the plane
c) Both the blocks remain at rest d)Both the blocks move down the plane
A
B
m
m’
5 kg
4 kg
(A)
4 kg 5 kg
(B)
.
A
B
C
60
0
300
A
M
M
B

SESSION – 6
AIM - To study pulley constraints.
Pulley Constraints
The problems in which few bodies are connected with one or more strings and strings are
passed through pulleys, some of which are fixed and some are movable. A motion under
some bounded condition is called constraint motion.
LET US DISCUSS WITH FEW EXAMPLES:
Example 1: First start with simple cases of pulleys. Consider the situation shown in fig. Two
bodies are connected with a string which passes over a pulley at the corner of a table.
Here if string is inextensible, we can directly state that the displacement of A in
downward direction is equal to the displacement of B in horizontal direction on table, and
if displacements of A and B are equal in equal time, their speeds and acceleration
magnitude must also be equal.
(1)
L = l1 + l2 = l1 + x + l2 − y
X = y i.e a1 = a2 = a
FBD m1g − T = m1a
T = m2a
a = (
m1
m1 + m2
) g
T = (
m1m2
m1 + m2
) 𝑔
Example 2:
Similarly consider the situation shown in fig.(2) Two masses are hanging over a pulley with a
string. Here if mass A is heavy, it goes down and B goes up by same distance. Thus here also
the displacement, speed and acceleration magnitude of the two are equal. Hence when pulley
is fixed two blocks have same speed, dispend and acceleration.
If M1 > M2and they move with acceleration 𝑎, then
M1g − T = M1a
T − M2g = M2a
Solving eq. (1) and eq. (2) we get a = (
m1−m2
m1+m2
) g
A
B

T = (
2m1m2
m1 + m2
) g
Example 3:
Now consider the case shown in fig. (3) Two masses A and B are tied to strings and arranged
in the situation shown. Here mass B is connected to a movable pulley Y supported by a string
which passes over a fixed pulley X and to which mass A is connected.
(3) (4)
To analyze the motions of A and B, you should look carefully at analysis shown in fig(4) If
mass B goes up by a distance x, we can observe that the string lengths ab and cd are slack,
due to the weight of block A, this length (ab + cd = 2x) will go on this side and block A will
descend by a distance 2x. As in equal time duration. A has travelled a distance twice that of
B, thus its speed and acceleration must also be twice that of B.
M1g − T = M12a
2T − M2 g = M2a
a = (
2m1−m2
4m1+m2
) g
Example 4:
L = L1 + L2 + L3
L = (L1 + X) + (L2 − Y) + (L3 − Y)
X = 2Y ⇒ A1 = 2A2
Let a1 = a
A2 =
A
2
B
A
B
A
Y
X
B
A
c
X
x
b d
a
2x

For m1 m1g − T = m1a … (1)
For m2 2T = M2
A
2
… (2)
---------------------------
4T = M2A
4M1G − 4T = 4M1A
--------------------------A =
4M1G
M2+4M1
⇒
M1M2G
4M1+M2
Example 5:
In such cases it is not necessary that block B will go up. It may also be possible that B will go
down and A will go up with twice the speed and acceleration, it depends on the masses of the
two objects.
Here we consider few more examples of pulley constraints. Consider the situation shown in
figure (5). In this case we find relation in acceleration of masses A and B. Let we analyze the
motion of A and B as shown in fig(6). In this case we find relation in acceleration of masses
A and B. Let we analyze the motion of a and B as shown in fig(6). If we consider that mass B
is going ups by a distance x, pulley Y which is attached to the same string will go down by the
same distance x. Due to this the string which is connected to mass A will now have free
lengths ab and cd (ab = cd =x) which will go on the side of mass A due to its weight as the
other end is fixed at point P. Thus mass A will go down by 2x hence its speed and
acceleration will be twice that of block B.
(5) (6)
l = l1 + l2
l = (l1 + x − y) + (l2 − y)
x = 2y
a1 = 2a2
Let a1 = a
a2 =
a
2
For mass 𝑚1 𝑚1𝑔 − 𝑇 = 𝑚1𝑎 … (1)
For mass m2 2T − m2g = m2
a
2
… (2)
------------------------
4T − 2m2g = m2a
B
X
Y
A
c
d
B
X
Y
A
X
a
b
x
2x

4m1g − 4T = 4m1a
--------------------------
𝑎 = (
4𝑚1−2𝑚2
4𝑚1+𝑚2
) 𝑔
Example 6:
Now consider a situation shown in fig.(7) Which is an extension of the previous problem. A
plank A is tied to two strings which pass over two pulleys X and Y and another mass B as
shown. Here we develop constraint relation between the motion of bodies A and B. It is
analyzed in situation shown in fig.(8) If mass A will go up by a distance x, points P and Q will
also go up by the same distance x and the pulley Y which is connected to point P will go down
by x and hence the strings lengths ab and cd (𝑎𝑏 = 𝑐𝑑 = 𝑥)which become free plus the length
x due to movement of Q upward will go on the side of mass B, hence it will go down by a
distance 3x. Thus its speed and acceleration are thrice that of mass A.
(7) (8)
l = l1
′
+ l2
′
l = l1
′
+ x + x + l2 + x − y
3x = y
3a1 = a2
Let a1 = a, a2 = 3a
For mass m1 m1g − 3T = m1a …(1)
For mass m2 T − m2g = m2(3a) …(2)
--------------------------
3T − 3m2g = 9m2a
m1g − 3T = m1a
--------------------------
𝑎 =
(𝑚1−3𝑚2)𝑔
(𝑚1+9𝑚2)
Example 7:
l = l1 + l2
l = l1 + x − y1 + l2 + x − y2
2x = y1 + y2
B
X
Y
Q
P
A
P
X
Y
A
x
B
x
Q
3x

2a1 = a2 + a3
a2 = a1 − a
a3 = a1 + a
a2 + a3 = 2a1 …(1)
m1g − 2T = m1a1 … (2)
m2(g + a1) − T = m1a
T − m3(g + a1) = m2a
Example 8:
l = l1 + l2 + l3 + l4
l = (l1 − x1) + (l2 + x3) + (l3 + x3) + (l4 − x2)
x1 + x2 = 2x3
a1 + a2 = 2a3 … (1)
For mass m1 T = m1a1 … (2)
For mass m2 T = m2a3 … (3)
T
m1
= a1 ,
T
m2
= a2
T
m1
+
T
m2
= a1 + a2 = 2a3
T =
2m1m2
m1 + m2
a3
m3g −
4m1m2
m1 + m2
a3 = m3a3
a3 =
(m1m3+m2m3)g
4m1m2+m3m1+m3m2
Example 9:
l = l1 + l2 + l3
l = (l1 − y) + (l2 + x) + (l3 + x)
y = 2x Let a1 = a
a2 = 2a1 a2 = 2a

For m1 m1g − 2T = m1a … (1)
For m2 T − m2gsmθ = m2(2a) … (2)
a = [
m1 − 2m2smθ
m1 + 4m2
] g
Example 10:
l = l1 + l2 + l3 + l4 + l5
l = (l1 + x) + (l2 − y) + (l3 − y) + (l4 − y) + (l5 − y)
𝑥 = 4y Let a1 = a
a1 = 4a2 a2 =
𝑎
4
For mass m1 m1g − T = m1a … (1)
For mass 𝑚2 4𝑇 = 𝑚2
𝑎
4
… (2)
𝑎 =
8𝑚1𝑔
8𝑚1 + 𝑚2
Systems of variable Mass, Rocket Propulsion:
Initial momentum at any time ts Pi = dm u
⃗ + M v
⃗
Final momentum after (t + dt)s Pf = (M + dm)(v
⃗ + d
⃗ v)
Change in momentum dp = Pf − Pi = [(M + dm)(V
⃗
⃗ + dv
⃗⃗⃗⃗ ) − (dm u
⃗ + M v
⃗ )]
𝑑𝑝 = 𝑀𝑑𝑣 + 𝑉𝑑𝑚 − 𝑢
⃗ 𝑑𝑚
Rate of change in momentum
𝑑𝑝
𝑑𝑡
= 𝐹𝑒𝑥𝑡
∑ 𝐹𝑒𝑥𝑡 = 𝑀
𝑑𝑣
𝑑𝑡
− (𝑢
⃗ − 𝑣)
𝑑𝑚
𝑑𝑡
= 𝑀
𝑑𝑉
⃗
𝑑𝑡
− 𝑉𝑟𝑒𝑙
𝑑𝑀
𝑑𝑡
∑ 𝐹𝑒𝑟𝑡 = 𝑚𝑎 − 𝐹𝑡ℎ𝑟𝑢𝑠𝑡
The Velocity of rocket in presence and in the absence of gravity:
1] Presence of gravity:

M
dv
dt
= Fext + Vrel
dM
dt
m
dV
dt
= −mg + Vrel
dM
dt
∫ dV
V
V0
= − ∫ gdt
t
0
+ Vrel ∫
dM
m
M
M0
V(t) = V0 − gt + V
⃗
⃗rel ln (
M
m0
)
2] Absence of gravity
M
DV
DT
= FEXT + VREL
DM
DT
∫ DV
V
V0
= VREL ∫
DM
M
M
M0
V(T) = V0 + VREL ln (
M
M0
)
V(T) = V0 − VREL ln (
M0
M
)
𝐅𝐭𝐡𝐫𝐮𝐬𝐭 = 𝐕𝐫𝐞𝐥 (
𝐝𝐌
𝐝𝐭
)

CLASS EXERCISE
1] Find out the constrain relation between 𝐴1 and 𝐴2 in the figure
2] Consider the situation shown in fig. All the surfaces are smooth. The tension in the
string connected to 2𝑚 is
a)
𝑚𝑔
3
b)
4𝑚𝑔
3
c)
2𝑚𝑔
3
d) 𝑚𝑔
3] In the given figure all the pulleys are smooth and massless, string is inextensible. Find
out acceleration of each block
HOME EXERCISE
1] Two bodies of masses m1 and m2 are connected by a light string passing over a smooth
light fixed pulley. The acceleration of the system is g/7. The ratio of their masses is
a) 7 : 1 b) 7 : 2 c) 4 : 3 d) 4 : 5
2] In the system shown below, friction and mass of the pulley are negligible. Find the
acceleration of 𝑚2 if 𝑚1 = 300 𝑔, 𝑚2 = 500 𝑔and 𝐹 = 1.50 𝑁
a) 16/17 m/s2 b) 15/17m/s2 c) 18/17 m/s2 d) 19/17 m/s2

3] A mass of 15 kg and another of mass 6 kg are attached to a pulley system as shown. A
is a fixed pulley while B is a movable one. Both are considered light and frictionless. Find
the acceleration of 6 kg mass.
a) 20/13m/s2 b) 30/13 m/s2 c) 35/13 m/s2 d) 25/13 m/s2
4] Two particles of masses m and 2m are placed on a smooth horizontal table. A string,
which joins them hangs over the edge supporting a light pulley, which carries a
mass 3m. The two parts of the string on the table are parallel and
perpendicular to the edge of the table. The parts of the string outside the table are
vertical. Find the acceleration of the particle of mass 3m.
5] Find the relation between accelerations of blocks A and B.
6] Using constraint method find the relation between accelerations of 1 and 2.
7] Find the constraint relation between the acceleration of block 1, 2 and 3. Their
acceleration are 𝑎1, 𝑎2 and 𝑎3 respectively
3m
2m m

8] In the shown figure masses of the pulleys and strings as well as friction between the
string and pulley is negligible. Find the acceleration of the masses 𝑚1and 𝑚2.
9] Using constraint equations find the relation between acceleration of 1 and 2.
10] Consider the situation of block pulley arrangement shown in fig. A plank is connected
to three strings and an electric motor M is fitted on to it and a string is wound on it
according to the arrangement shown in figure. Given that the string is winding on
shaft of motor at a speed v. Find the speed with which the plank would be going
up.
11] Shows a system of four pulleys with two masses A and B. Find, at an instant:
a) Speed of block A when the block B is going up at 1 m/s and pulley Y is going up at
2m/s.
b) Acceleration of block an if block B is going up at 3 m/s2 and pulley Y is going down
at 1 m/s2.
A
m1
m2
B
y2
y1
yA
yB Fixed
X
Y
A
Z
M

12] Block C shown in fig. is going down at acceleration 2 m/s2. Find the acceleration of
blocks. A and B
13] Block A shown in fig. move by a distance 3m toward left. Find the distance and
direction in which the point P on string shown in fig. is displaced.
14] Consider the situation shown in fig. All the surfaces are smooth. The tension in the
string connected to 2m is
a)
𝑚𝑔
3
b)
4𝑚𝑔
3
c)
2𝑚𝑔
3
d)mg
15] A block A of mass m is tied to a fixed point C on a horizontal table through a string
passing round a massless smooth pulley B. A force F is applied by pulley B as shown
in fig. (B) Find the accelerations of the pulley and mass A.
a)
𝐹
2𝑚
&
𝐹
𝑚
b)
𝐹
𝑚
&
𝐹
2𝑚
c)
𝐹
𝑚
&
𝐹
𝑚
d)
𝐹
2𝑚
&
𝐹
2𝑚
B
X
Z
A
C
A B
2m
m
m
C B
A
F
C
A B

SESSION – 7
AIM - Concept of inertial and non-inertial frames
INERTIAL FRAME OF REFERENCE
Any system relative to which the motion of an object can be described is called a frame of
reference. The frame of reference may also be defined as the coordinate system relative to
which the motion of an object may be described. A coordinate system in which Newton’s laws
of motion in their simplest form are valid is called as inertial frame of reference.
Let there be no external force acting on a body in such a frame. Then the body continues to
be in its state of rest or of uniform motion along a straight line due to inertia. Hence the
frame is called as inertial frame of reference.
When a frame of reference is moving with linear acceleration relative to an inertial frame or
having uniform rotatory motion with respect to inertial frame is called as non-inertial frame
or an accelerated frame.
Properties of Inertial Frames
Following are the properties of inertial frames:
• When no external force acts on the body, the body moves with uniform velocity in this
frame
• The observation in different frames gives different numerical values but the relations
between these quantities are same for all frames. Thus, all inertial frames are
equivalent.
• They are also called as Galilean frames.
WEIGHT OF A BODY IN A LIFT
Earth attracts everybody towards its center. The force of attraction exerted by the earth
on the body is called gravity force. If m be the mass of the body then the gravity force on it
will be mg. Generally, the weight of a body is equal to the gravity force W = mg. But when
the body is on an accelerated platform, the weight of a body appears changed. The new
weight is called as apparent weight of a man standing in a lift which is in motion. We consider
the following cases:
1. The lift is unaccelerated( i.e. a = 0 or constant) : The situation is shown in fig. In this
case R = ma = 0

Hence apparent weight W= Actual weight = mg
2. When the lift is accelerated upward: In this case, there will be two forces acting on
the man, i.e., weight m g and reaction R = m a both acting in the downward direction as
shown in fig.
Apparent weight
W= mg + R = mg + ma = m (g + a) Or
Apparent weight W> Actual weight W
3. When the lift is accelerated downward:
This situation is shown in fig. (5c). Here the weight mg acts downward while the
reaction R = ma acts upward. We assume that a < g.
Hence apparent weight
W = W - R = mg - ma = m (g - a)
Apparent weight W< Actual weight W.
Now we consider the special case when g = a .
In this case,
Apparent weight W = 0
Thus in a freely falling lift, the man will experience a state of weightlessness.
4. When the lift is accelerated downward such that > g : In this case R = ma is greater
than the weight mg
Apparent weight W = m(g − a) = negative
So the man will be accelerated upward and will stay at the ceiling of the lift.
Let us consider the tension when a mass is suspended from a vertical string in a
carriage. We consider the following cases:
1. When the carriage is at rest, then T0 = mg
a=0
W
a
W=mg
R=ma
(a)
(b)
a
W=mg
(c)
g
F=ma

2. When the carriage is accelerated up, then mass has an upward acceleration a. In this
case,
T1 − mg = ma or T1 = m(g + a) (T1 > T0)
3. When the carriage is accelerated down, then mass has a downward acceleration. In
this case,
mg − T2 = ma
or T2 = m(g − a)(T2 < T0)
If a = g, then tension in string becomes zero.
4. When the carriage is accelerated horizontally. The situation is shown in fig.
The tension in the string can be calculated by considering an inertial frame of
reference (ground) or considering non-inertial frame of reference (carriage).
Considering inertial frame of reference, the free body diagram of mass is shown in
fig.
Here, we have T cosθ = mg
and T sin θ = ma
tanθ=
𝑎
𝑔
or θ = tan- 1(
a
g
)
and T = m√(g2 + a2)(TT0)
Let us consider the same situation in non-inertial frame (carriage). The free body diagram is
shown in fig.
Again, we have T cosθ = mg and Tsin = ma. Hence the same results can be obtained.
a

mg
T
mg
T cos 
=
T sin 
a

mg
ma
T
=
T cos 
T sin 
ma
mg

CLASS EXERCISE:
1] An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of 4.9
m/s2. The tension in the cable is
a) 6000 N b) 6000 gN c) 9000 N d) 9000 gN
2] With what acceleration ‘a’ should the box in the figure descend so that a body of mass
M placed in it exerts a force
𝑀𝑔
4
on the base of the box?
a)
3𝑔
4
b)
𝑔
4
c)
𝑔
2
d)
𝑔
8
3] A block can slide on a smooth inclined plane of inclination 𝜃 kept on the floor of a lift.
When the lift is descending with retardation, a m/s2 the acceleration of the
block relative to the incline will be
a) (g – a)sin 𝜃 b) (g + a)sin 𝜃 c) gsin 𝜃 d) g – a
4] A man of mass 90 kg is standing in an elevator whose cable broke suddenly. If the
elevator falls, apparent weight of the man is ___ .
a) 90 N b) 90 x 9.8 N c) zero d) 90 kg
5] A reference frame attached to the earth with respect to an observer in space
a) is an inertial frame because Newton’s laws of motion are applicable in it.
b) is an inertial frame by definition
c) cannot be an inertial frame because earth is rotating about its axis
d) can be an inertial frame because earth is revolving around the sun
HOME EXERCISE:
1] A 60 kg man stands on an elevator floor. The elevator is going up with constant
acceleration of
1.96 m/s2. Percentage change in the apparent weight of the person is
a) 10 b) 15 c) 20 d) 25
2] A spring balance suspended from the roof of a elevator indicates 90 kg as the weight
of a 120 kg body. Acceleration of the elevator
a) g/4 upwards b) g/4 downwards c) g/2 upwards d) g/2 downwards
3] A lamp hangs vertically from a cord in a descending lift. The lift has a retardation of
5.2 m/s2 before coming to a halt. If the tension in cord is 30 N, mass of the lamp is
a) 2 kg b) 1 kg c) 9.8 kg d) 4.9 kg

4] The apparent weight of a mass in a lift moving up is 80 kg when its weight in the
stationary lift is
60 kg. If the same lift now moves up with sameretardation, the weight of the mass
will be
a) 40 kg b) 60 kg c) 80 kg d) 100 kg
5] A lift of mass 500 kg is descending with an acceleration of 2ms-2. If g = 10 ms-2, the
tension in the cable is
a) 4000 N b) 8000 N c) 2000 N d) 6000 N
6] A boy is sliding down a rope, which can with stand to a maximum tension of 2/3 of the
weight of the boy. The maximum acceleration with which the boy can slide down rope safely
is
a) g/2 b) 3g/4 c) g/5 d) g/3
7] Two bodies A and B masses 3 kg and 5 kg are connected by a string as shown in the
figure. If the pulley is frictionless, (𝑔 = 10 𝑚𝑠−2)
a) A moves down with acceleration of 2.5 𝑚𝑠−2
b) B moves up with acceleration
1
4
𝑚𝑠−2
c) B moves down with acceleration 2.5 𝑚𝑠−2
d) A moves up with acceleration
1
4
𝑚𝑠−2
8] In the above question tension in the string is
a) 17 N b) 37.5N c) 10N d) 27.5 N
9] Three blocks A (1 kg), B (2 kg) and C (3 kg) are hanging on two strings as shown in the
figure. The tension 𝑇1 and 𝑇2 are respectively (𝑔 = 10 𝑚𝑠−2)
a) 50 N and 10 N b)3.3 N and 20 N c) 16.7 N and 10 N d) 16.7N and 6.7 N
10] In the above question the acceleration B is
a) 0.67𝑚𝑠−2
b) 9.8 𝑚𝑠−2
c)4.9 𝑚𝑠−2
d) 6.7 𝑚𝑠−2
A
B
A
T1
C
T2
B

11] A reference frame attached to the earth
a) is an inertial frame by definition
b) Cannot be an inertial frame because the earth is revolving around the sun
c) is an inertial frame because Newton’s laws are applicable in this frame
d) cannot be an inertial frame because the earth is rotating about its axis
a) A, B, C are correct b) b only correct
c) b and d are correct d) All are correct
12] A) In a frame of reference 𝑆1 though the net force is zero, the net acceleration is not
zero.
B) In a frame of reference 𝑆2, though the net force is not zero, the net acceleration
is zero.
C) In a frame of reference 𝑆3, the net acceleration is zero whenever the net force is
zero.
a) 𝑆1and𝑆3 are inertial and 𝑆2 is non - inertial
b) 𝑆1and𝑆2 are non-inertial and 𝑆3 is inertial
c)𝑆1, 𝑆2, 𝑆3 are non- inertial
d) 𝑆1, 𝑆2, 𝑆3 are inertial
13] A body is kept on the floor of a lift at rest. The lift starts descending at acceleration
a :
a) if 𝑎 > 𝑔, the displacement of body in time 𝑡 is
1
2
𝑔𝑡2
b) if 𝑎 < 𝑔, the displacement of body in time 𝑡 is
1
2
𝑔𝑡2
c) if 𝑎 < 𝑔, the displacement of body in time 𝑡 is
1
2
𝑎𝑡2
d) if 𝑎 < 𝑔, the displacement of body in time 𝑡 is
1
2
(𝑎 + 𝑔)𝑡2
14] A 5 kg block is supported by a cord and pulled upward with an acceleration of 2 𝑚/𝑠2
.
What is the tension in the string?(𝑔 = 9.8 𝑚/𝑠2
)
a) 59 N b) 61 N c) 70 N d) 10N

SESSION –8
AIM - Concept of Wedge Constraints
WEDGE CONSTRAINTS
Example 1:
For mass m:
Along x-axis mgsinθ = m(a − Acosθ) …(i)
Along y-axis mgcosθ = mAsinθ …(ii)
For wedge of mass M.
Along x-axis Nsinθ = MA …(iii)
Along y-axis N′
= Ncosθ + Mg … (iv)
From eq (iii) N =
MA
Sinθ
⇒ mgcosθ −
MA
sinθ
= mAsinθ
A = [
mgcosθsinθ
M + msin2θ
]
a =
(M + m)gsinθ
M + msin2θ
Along the incline mgsinθ + mAcosθ = ma …(1)
⊥to the incline N + mAsinθ = mgcosθ …(2)
Example 2:
l = l1 + l2
l = (l1 + x) + (l2 − y)
x = y
Vm = VM
am = aM = a
For body of mass m:
mgsinθ − T = m(a − acosθ) …(i)
mgcosθ − N = masinθ …(ii)
For body of mass M
T + Nsinθ − Tcosθ = Ma … (iii)
Tsinθ + Mg + Ncosθ = N′
…(iv)
T(1 − cosθ) + Nsinθ = Ma
T(1 − cosθ) + (mgcosθ − masinθ)sinθ = Ma
T +
mgcosθsinθ
1 + cosθ
=
(M + msin2
θ)a
1 + cosθ
mgsinθ − T = m(a − acosθ)
a =
mgsinθ
M + 2m(1 − cosθ)

Example 3:
tan𝜃 =
𝑦
𝑥
y = xtanθ
dy
dt
=
dx
dt
tanθ
Vrod = Vwedgetanθ
arod = awedgetanθ
a = Atanθ
For the wedge
Nsinθ = MA …(i)
N′
= Mg + Ncosθ …(ii)
For the rod
mg − NCosθ = ma …(iii)
NSinθ = N′
…(iv)
From Equation (1) N =
MA
Sinθ
mg −
MA
Sinθ
Cosθ = m(Atanθ)
A =
mgtanθ
M + mtan2θ
a = tanθ
a =
mgtan2
θ
M + mtan2θ
Example 4:
tanθ =
y
x
y = xtanθ
arod = Awedgetanθ
a = Atanθ … (i)
For wedge of mass M
Nsinθ = MA … (ii)
Ncosθ + Mg = N′
… (iii)
For rod of mass 𝑚
mg − 2Ncosθ = ma …(iv)
Nsinθ − Nsinθ = 0 …(1)
From equation (ii) N =
MA
Sinθ

Sub in Eq (iv) mg − 2
MA
Sinθ
Cosθ = mAtanθ
A =
mgsinθcosθ
msin2θ+Mcos2θ
Example 5:
For body of mass 𝑚
Along x-axis O = m(a − Acosθ)
a = Acosθ …(i)
Along y-axis mg − N = mAsinθ …(ii)
For body of mass M
Along x-axis Mgsinθ + Nsinθ = MA …(iii)
Along y-axis Ncosθ + mgcosθ = N′
… (iv)
From eq (iii) Mg + N =
MA
Sinθ
Mg − N = mASinθ
----------------------------
(M + m)g = (msinθ +
M
Sinθ
) A
A −
(M + m)gsinθ
M + mSin2θ
a = Acosθ =
(M + m)gsinθcosθ
M + mSin2θ
CLASS EXERCISE:
1] A block of mass 𝑚 = 10𝑘𝑔 is placed on a wedge of mass 𝑀 = 20𝑘𝑔 and released.
Determine the acceleration of the 20 𝑘𝑔block (𝜃 = 60°).
2] All the surfaces are smooth. Rod is moving down vertically with an acceleration 9 𝑚/𝑠2
, mass of wedge is 10 𝑘𝑔 and 𝜃 = 37°. Force exerted on the rod by the wedge will be
a) 120 N b) 200 N c) 135 N d) 225 N
3] A block of mass 𝑚 = 2𝑘𝑔 kept on a wedge of mass 𝑀 = 9𝑘𝑔. A force of 210 𝑁 is
applied on wedge horizontally. Determine the distance moved by the wedge when the block
leaves the wedge 𝜃 = 45°
4] A block of mass m is moving on a wedge with the acceleration 𝑎0. The wedge is
moving with the acceleration 𝑎1 . The observer is situated on wedge. The
magnitude of pseudo force on the block is?
a) 𝑚𝑎0 b) 𝑚𝑎1 c) 𝑚√𝑎0
2
+ 𝑎1
2
d) m (
𝑎
1
+𝑎
2
2
)

HOME EXERCISE
1] A wedge is moving horizontally with uniform acceleration a towards right and a block
of mass 𝑚 stays freely at rest on its smooth inclined surface as shown in fig. What is the
acceleration of the wedge?
a) g cot  b) g cos c) g sin  d) g tan 
2] Consider the shown arrangement. Assume all surfaces to be smooth. If 'N' represents
magnitudes of normal reaction between block and wedge then acceleration of 'M'
along horizontal equals.
a)
𝑁𝑠𝑖𝑛𝜃
𝑀
𝑎𝑙𝑜𝑛𝑔 + 𝑣𝑒 𝑥 − 𝑎𝑥𝑖𝑠 b)
𝑁𝑐𝑜𝑠𝜃
𝑀
𝑎𝑙𝑜𝑛𝑔 − 𝑣𝑒 𝑥 − 𝑎𝑥𝑖𝑠
c)
𝑁𝑠𝑖𝑛𝜃
𝑀
− 𝑣𝑒 𝑥 − 𝑎𝑥𝑖𝑠 d)
𝑛𝑠𝑖𝑛𝜃
𝑚+𝑀
𝑎𝑙𝑜𝑛𝑔 − 𝑣𝑒 𝑥 − 𝑎𝑥𝑖𝑠
3] In the situation shown in fig. if mass M is going down along the incline at an
acceleration of 5 𝑚/𝑠2
and m is moving toward right relative to M horizontally with 3 𝑚/
𝑠2
. Find the net acceleration
of 𝑚.
4] Imagine the situation in which a horizontal force F is applied on the wedge. If F0 is
the force required to keep the body stationary,
Choose the correct statement(s)
A) If F  F0, the block remains stationary with respect to wedge
B) If F < F0, the block slides down the wedge
C) If F > F0, the block slides up the wedge
D) If F = F0, the block is accelerating with respect to ground

M
m

SESSION – 9
AIM - To Discuss Typical Problems
• Find the relation of velocities 𝑣1 and 𝑣2
• If vehicle A moves with velocity 𝑣1, then find velocity of block B when angle between
strings is θ.
• One end of rope ACB, is attached by a block B (Which is moving vertically downward)
and other end is attached by ring A (which is moving on a horizontal rod).
• String BAC is attached by roof and ring B, and this string is passes through ring A also.
If ring A slides on a vertical rod in downward direction with velocity 𝑣1. Then find
velocity of ring B at given moment.
• A chain of mass M and length l is held vertical, such that its lower end just touches
the floor. It is released from rest. Find the force executed by the chain on the
table when upper end is about to hit the floor.
• A block of mass 25 kg is raised by a 50kg man in two different ways as shown. In
which way he will find easy to raise the block of mass and why?
• The pulley arrangements of figure (a) and (b) are identical. In (a) the mass 𝑚 is lifted
up by attaching a mass 2m to the other end of the rope, In (b), 𝑚 is lifted up by
pulling the other end of the rope with a constant downward force F = 2 mg. Is the
acceleration of 𝑚 is the same in both cases.
• A monkey is holding on a rope that goes over a smooth light pulley and supports a
mirror of equal mass at the other end. Initially both are at rest and monkey can see
him image in the mirror. With what acceleration the monkey should climb the rope to
escape seeing his image.
• Find the accelerations of block A and block B in the given figure.
• A man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg.
If the man manages to keep the box at rest, what is the weight shown by the
machine? What force should he exert on the rope to get his correct weight on
the machine?
• In the pulley system shown the movable pulleys A, B and C have mass m each, D and E
are fixed pulleys. The string are vertical, light and inextensible. Find the
acceleration of the pulleys and tensions in the strings.
• At the moment t = 0 force F = kt is applied to a small body of mass m resting on a
smooth horizontal plane at an fixed angle with horizontal. Find the velocity and
the distance travelled by the body upto the moment of its breaking off the plane.

• A chain consisting of five links, each of mass 0.1 kg is lifted vertically by pulling the
5
th
link with a constant acceleration
• a = 2.5 m/s2
. Find the force acting between 3
rd
and 4
th
link and the force exerted
on the 5
th
link.
• A cart carries two blocks of mass 2kg and 1kg which are connected by a string passing
over a pulley. The cart is moving right with an acceleration of a = 1m/s2
. Find
the acceleration of block with respect to ground and tension in the string.

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