CONTENTS
4.1
Point mass
4.2
Inertia
4.3
Linear momentum
4.4
Newton’s first law of motion
4.5
Newton’s second law of motion
4.6
Force
4.7
Equilibrium of concurrent forces
4.8
Newton’s third law of motion
4.9
Frame of reference
4.10
Impulse
4.11
Law of conservation of linear momentum
4.12
Free body diagram
4.13
Apparent weight of a body in a lift
4.14
Motion of block on smooth horizontal surface
4.15
Motion of block on smooth inclined surface
4.16
Motion of blocks in contact
4.17
Motion of blocks connected by massless string
4.18
Motion of connected blocks over a pulley
4.19
Motion of massive string
4.20
Spring balance and physical balance
4.21
Modification of Newton’s laws of motion
Sample Problems
Practice Problems (Basic and Advance Level)
Answer Sheet of Practice Problems
Rocket propulsion is based upon the law
of conservation of linear momentum. The
fuel is burnt in the ignition chamber of the
rocket engine. The gases so produced are
ejected from an orifice in the rear of the
rocket. A thrust acts on the rocket due to
reaction of the movement of gases and thus
the rocket starts moving in the forward
direction.
Chapter
4
216 Newton’s Laws of Motion
216
4.1 Point Mass.
(1) An object can be considered as a point object if during motion in a given time, it covers
distance much greater than its own size.
(2) Object with zero dimension considered as a point mass.
(3) Point mass is a mathematical concept to simplify the problems.
4.2 Inertia.
(1) Inherent property of all the bodies by virtue of which they cannot change their state of
rest or uniform motion along a straight line by their own is called inertia.
(2) Inertia is not a physical quantity, it is only a property of the body which depends on
mass of the body.
(3) Inertia has no units and no dimensions
(4) Two bodies of equal mass, one in motion and another is at rest, possess same inertia
because it is a factor of mass only and does not depend upon the velocity.
4.3 Linear Momentum.
(1) Linear momentum of a body is the quantity of motion contained in the body.
(2) It is measured in terms of the force required to stop the body in unit time.
(3) It is measured as the product of the mass of the body and its velocity i.e., Momentum =
mass × velocity.
If a body of mass m is moving with velocity
v
then its linear momentum
p
is given by
vmp
(4) It is a vector quantity and it’s direction is the same as the direction of velocity of the
body.
(5) Units : kg-m/sec [S.I.], g-cm/sec [C.G.S.]
(6) Dimension :
][ 1
MLT
R cos
R
R sin
m
p = constant
v
Newton’s Laws of Motion 217
217
(7) If two objects of different masses have same momentum, the lighter body possesses
greater velocity.
2211 vmvmp
= constant
1
2
2
1m
m
v
v
i.e.
m
v1
[As p is constant]
(8) For a given body
vp
(9) For different bodies at same velocities
mp
4.4 Newton’s First Law.
A body continue to be in its state of rest or of uniform motion along a straight line, unless it
is acted upon by some external force to change the state.
(1) If no net force acts on a body, then the velocity of the body cannot change i.e. the body
cannot accelerate.
(2) Newton’s first law defines inertia and is rightly called the law of inertia. Inertia are of
three types :
Inertia of rest, Inertia of motion, Inertia of direction
(3) Inertia of rest : It is the inability of a body to change by itself, its state of rest. This
means a body at rest remains at rest and cannot start moving by its own.
Example : (i) A person who is standing freely in bus, thrown backward, when bus starts
suddenly.
When a bus suddenly starts, the force responsible for bringing bus in motion is also
transmitted to lower part of body, so this part of the body comes in motion along with the bus.
While the upper half of body (say above the waist) receives no force to overcome inertia of rest
and so it stays in its original position. Thus there is a relative displacement between the two
parts of the body and it appears as if the upper part of the body has been thrown backward.
Note : If the motion of the bus is slow, the inertia of motion will be transmitted to the
body of the person uniformly and so the entire body of the person will come in
motion with the bus and the person will not experience any jerk.
p
v
m =
constant
p
m
v =
constant
218 Newton’s Laws of Motion
218
(ii) When a horse starts suddenly, the rider tends to fall backward on account of inertia of
rest of upper part of the body as explained above.
(iii) A bullet fired on a window pane makes a clean hole through it while a stone breaks the
whole window because the bullet has a speed much
greater than the stone. So its time of contact with glass
is small. So in case of bullet the motion is transmitted
only to a small portion of the glass in that small time.
Hence a clear hole is created in the glass window, while
in case of ball, the time and the area of contact is large.
During this time the motion is transmitted to the entire
window, thus creating the cracks in the entire window.
(iv) In the arrangement shown in the figure :
(a) If the string B is pulled with a sudden jerk then it will experience tension while due to
inertia of rest of mass M this force will not be transmitted to the string A and
so the string B will break.
(b) If the string B is pulled steadily the force applied to it will be
transmitted from string B to A through the mass M and as tension in A will be
greater than in B by Mg (weight of mass M) the string A will break.
(v) If we place a coin on smooth piece of card board covering a glass and
strike the card board piece suddenly with a finger. The cardboard slips away
and the coin falls into the glass due to inertia of rest.
(vi) The dust particles in a durree falls off when it is beaten with a stick. This is because
the beating sets the durree in motion whereas the dust particles tend to remain at rest and
hence separate.
(4) Inertia of motion : It is the inability of a body to change itself its state of uniform
motion i.e., a body in uniform motion can neither accelerate nor retard by its own.
Example : (i) When a bus or train stops suddenly, a passenger sitting inside tends to fall
forward. This is because the lower part of his body comes to rest with the bus or train but the
upper part tends to continue its motion due to inertia of motion.
(ii) A person jumping out of a moving train may fall forward.
(iii) An athlete runs a certain distance before taking a long jump. This is because velocity
acquired by running is added to velocity of the athlete at the time of jump. Hence he can jump
over a longer distance.
(5) Inertia of direction : It is the inability of a body to change by itself direction of motion.
Example : (i) When a stone tied to one end of a string is whirled and the string breaks
suddenly, the stone flies off along the tangent to the circle. This is because the pull in the string
M
A
B
Cracks by the
ball
Hole by the
bullet
Newton’s Laws of Motion 219
219
was forcing the stone to move in a circle. As soon as the string breaks, the pull vanishes. The
stone in a bid to move along the straight line flies off tangentially.
(ii) The rotating wheel of any vehicle throw out mud, if any, tangentially, due to directional
inertia.
(iii) When a car goes round a curve suddenly, the person sitting inside is thrown outwards.
Sample problem based on Newton’s first law
Problem 1. When a bus suddenly takes a turn, the passengers are thrown outwards because of
[AFMC 1999; CPMT 2000, 2001]
(a) Inertia of motion (b) Acceleration of
motion
(c) Speed of motion (d) Both (b) and (c)
Solution : (a)
Problem 2. A person sitting in an open car moving at constant velocity throws a ball vertically up into
air. The ball fall
[EAMCET (Med.) 1995]
(a) Outside the car (b) In the car ahead of
the person
(c) In the car to the side of the person (d) Exactly in the hand which threw it up
Solution : (d) Because the horizontal component of velocity are same for both car and ball so they cover
equal horizontal distances in given time interval.
4.5 Newton’s Second Law.
(1) The rate of change of linear momentum of a body is directly proportional to the external
force applied on the body and this change takes place always in the direction of the applied
force.
(2) If a body of mass m, moves with velocity
v
then its linear momentum can be given
by
vmp
and if force
F
is applied on a body, then
dt
pd
KF
dt
pd
F
or
dt
pd
F
(K = 1 in C.G.S. and S.I. units)
or
am
dt
vd
mvm
dt
d
F
)(
(As
dt
vd
a
acceleration produced in the body)
amF
Force = mass acceleration
Sample problem based on Newton’s second law
220 Newton’s Laws of Motion
220
Problem 3. A train is moving with velocity 20 m/sec. on this, dust is falling at the rate of 50 kg/min.
The extra force required to move this train with constant velocity will be [RPET 1999]
(a) 16.66 N (b) 1000 N (c) 166.6 N (d) 1200 N
Solution : (a) Force
dt
dm
vF
N66.16
60
50
20
Problem 4. A force of 10 Newton acts on a body of mass 20 kg for 10 seconds. Change in its momentum
is [MP PET 2002]
(a) 5 kg m/s (b) 100 kg m/s (c) 200 kg m/s (d) 1000 kg m/s
Solution : (b) Change in momentum
sec/1001010timeforce mkg
Problem 5. A vehicle of 100 kg is moving with a velocity of 5 m/sec. To stop it in
10
1
sec, the required
force in opposite direction is [MP PET 1995]
(a) 5000 N (b) 500 N (c) 50 N (d) 1000 N
Solution : (a)
kgm100
,/5 smu
0v
t = 0.1 sec
t
uvm
dt
mdv
Force )(
1.0
)50(100
NF 5000
4.6 Force.
(1) Force is an external effect in the form of a push or pulls which
(i) Produces or tries to produce motion in a body at rest.
(ii) Stops or tries to stop a moving body.
(iii) Changes or tries to change the direction of motion of the body.
Body remains at rest. Here force is trying to change the state of
rest.
Body starts moving. Here force changes the state of rest.
In a small interval of time, force increases the magnitude of
speed and direction of motion remains same.
In a small interval of time, force decreases the magnitude of
speed and direction of motion remains same.
F
u = 0
v = 0
F
u
v < u
F
u = 0
v > 0
F
v > u
u 0
Newton’s Laws of Motion 221
221
In uniform circular motion only direction of velocity changes,
speed remains constant. Force is always perpendicular to
velocity.
In non-uniform circular motion, elliptical, parabolic or
hyperbolic motion force acts at an angle to the direction of
motion. In all these motions. Both magnitude and direction of
velocity changes.
(2) Dimension : Force = mass acceleration
][]][[][ 22 MLTLTMF
(3) Units : Absolute units : (i) Newton (S.I.) (ii) Dyne (C.G.S)
Gravitational units : (i) Kilogram-force (M.K.S.) (ii) Gram-force (C.G.S)
Newton : One Newton is that force which produces an acceleration of
2
/1 sm
in a
body of mass 1 Kilogram.
1
Newton
2
/1 smkg
Dyne : One dyne is that force which produces an acceleration of
2
/1 scm
in a
body of mass 1 gram. 1 Dyne
2
sec/1 cmgm
Relation between absolute units of force 1 Newton
5
10
Dyne
Kilogram-force : It is that force which produces an acceleration of
2
/8.9 sm
in a
body of mass 1 kg. 1 kg-f = 9.81 Newton
Gram-force : It is that force which produces an acceleration of
2
/980 scm
in a body of
mass 1gm. 1 gm-f = 980 Dyne
Relation between gravitational units of force : 1 kg-f =
7
10
gm-f
(4)
amF
formula is valid only if force is changing the state of rest or motion and the
mass of the body is constant and finite.
(5) If m is not constant
dt
dm
v
dt
vd
mvm
dt
d
F
)(
(6) If force and acceleration have three component along x, y and z axis, then
kFjFiFF zyx ˆ
ˆˆ
and
kajaiaa zyx ˆ
ˆˆ
From above it is clear that
zzyyxx maFmaFmaF ,,
(7) No force is required to move a body uniformly along a straight line.
v
F
v
F
v
F = mg
222 Newton’s Laws of Motion
222
maF
0 F
(As
0a
)
(8) When force is written without direction then positive force means repulsive while
negative force means attractive.
Example : Positive force – Force between two similar charges
Negative force – Force between two opposite charges
(9) Out of so many natural forces, for distance
15
10
metre, nuclear force is strongest while
gravitational force weakest.
nalgravitationeticelectromagnuclear FFF
(10) Ratio of electric force and gravitational force between two electron
43
10/
ge FF
ge FF
(11) Constant force : If the direction and magnitude of a force is constant. It is said to be a
constant force.
(12) Variable or dependent force :
(i) Time dependent force : In case of impulse or motion of a charged particle in an
alternating electric field force is time dependent.
(ii) Position dependent force : Gravitational force between two bodies
221
r
mGm
or Force between two charged particles
2
0
21
4r
qq
.
(iii) Velocity dependent force : Viscous force
)6( rv
Force on charged particle in a magnetic field
)sin(
qvB
(13) Central force : If a position dependent force is always directed towards or away from a
fixed point it is said to be central otherwise non-central.
Example : Motion of earth around the sun. Motion of electron in an atom. Scattering of
-
particles from a nucleus.
-
particle
F
+
+
Nucleus
F
Electron
Nucleus
–
+
F
Sun
Earth
Newton’s Laws of Motion 223
223
(14) Conservative or non conservative force : If under the action of a force the work done in
a round trip is zero or the work is path independent, the force is said to be conservative
otherwise non conservative.
Example : Conservative force : Gravitational force, electric force, elastic force.
Non conservative force : Frictional force, viscous force.
(15) Common forces in mechanics :
(i) Weight : Weight of an object is the force with which earth attracts it. It is also called the
force of gravity or the gravitational force.
(ii) Reaction or Normal force : When a body is placed on a rigid surface, the body
experiences a force which is perpendicular to the surfaces in contact. Then force is called
‘Normal force’ or ‘Reaction’.
(iii) Tension : The force exerted by the end of taut string, rope or chain against pulling
(applied) force is called the tension. The direction of tension is so as to pull the body.
(iv) Spring force : Every spring resists any attempt to change its length. This resistive force
increases with change in length. Spring force is given by
KxF
; where x is the change in
length and K is the spring constant (unit N/m).
4.7 Equilibrium of Concurrent Force.
(1) If all the forces working on a body are acting on the same point, then they are said to be
concurrent.
T = F
x
F = –
Kx
mg
R
mg cos
R
mg
224 Newton’s Laws of Motion
224
(2) A body, under the action of concurrent forces, is said to be in equilibrium, when there is
no change in the state of rest or of uniform motion along a straight line.
(3) The necessary condition for the equilibrium of a body under the action of concurrent
forces is that the vector sum of all the forces acting on the body must be zero.
(4) Mathematically for equilibrium
0
net
F
or
0
x
F
;
0
y
F
; ,
0
z
F
(5) Three concurrent forces will be in equilibrium, if they can be represented completely by
three sides of a triangle taken in order.
(6) Lami’s Theorem : For concurrent forces
sinsinsin 321 FFF
Sample problem based on force and equilibrium
Problem 6. Three forces starts acting simultaneously on a particle moving with velocity
.v
These forces
are represented in magnitude and direction by the three sides of a triangle ABC (as shown).
The particle will now move with velocity [AIEEE 2003]
(a)
v
remaining unchanged
(b) Less than
v
(c) Greater than
v
(d)
v
in the direction of the largest force BC
Solution : (a) Given three forces are in equilibrium i.e. net force will be zero. It means the particle will
move with same velocity.
Problem 7. Two forces are such that the sum of their magnitudes is 18 N and their resultant is
perpendicular to the smaller force and magnitude of resultant is 12. Then the magnitudes of
the forces are [AIEEE 2002]
A
B
C
2
F
1
F
3
F
2
F
1
F
3
F
C
A
B
Newton’s Laws of Motion 225
225
(a) 12 N, 6 N (b) 13 N, 5N (c) 10 N, 8 N (d) 16 N, 2 N
Solution : (b) Let two forces are
1
F
and
)( 212 FFF
.
According to problem:
18
21 FF
…..(i)
Angle between
1
F
and resultant (R) is 90°
cos
sin
90tan 21
2FF
F
0cos
21
FF
2
1
cos F
F
…..(ii)
and
cos2 21
2
2
2
1
2FFFFR
cos2144 21
2
2
2
1FFFF
…..(iii)
by solving (i), (ii) and (iii) we get
NF 5
1
and
NF 13
2
Problem 8. The resultant of two forces, one double the other in magnitude, is perpendicular to the
smaller of the two forces. The angle between the two forces is [KCET (Engg./Med.) 2002]
(a)
o
60
(b)
o
120
(c)
o
150
(d)
o
90
Solution: (b) Let forces are F and 2F and angle between them is
and resultant makes an angle
with
the force F.
90tan
cos2
sin2
tan
FF
F
0cos2
FF
2/1cos
or
120
Problem 9. A weightless ladder, 20 ft long rests against a frictionless wall at an angle of 60o with the
horizontal. A 150 pound man is 4 ft from the top of the ladder. A horizontal force is needed
to prevent it from slipping. Choose the correct magnitude from the following [CBSE PMT 1998]
(a) 175 lb (b) 100 lb (c) 70 lb (d) 150 lb
Solution: (c) Since the system is in equilibrium therefore
0
x
F
and
0
y
F
2
RF
and
1
RW
Now by taking the moment of forces about point B.
).().( ECWBCF
)(
1ACR
[from the figure EC= 4 cos 60]
)60cos20()60cos4()60sin20.( 1
RWF
1
102310 RWF
WR
1
As
lb
W
F70
310
1508
310
8
F1
F2
R =12
16 ft
4 ft
R1
A
F
E
C
B
Wall
R2
W
60o
D
226 Newton’s Laws of Motion
226
Problem 10. A mass M is suspended by a rope from a rigid support at P as shown in the figure. Another
rope is tied at the end Q, and it is pulled horizontally with a force F. If the rope PQ makes
angle
with the vertical then the tension in the string PQ is
(a)
sinF
(b)
sin/F
(c)
cosF
(d)
cos/F
Solution: (b) From the figure
For horizontal equilibrium
FT
sin
sin
F
T
Problem 11. A spring balance A shows a reading of 2 kg, when an aluminium block is suspended from it.
Another balance B shows a reading of 5 kg, when a beaker full of liquid is placed in its pan.
The two balances are arranged such that the Al – block is completely immersed inside the
liquid as shown in the figure. Then [IIT-JEE 1985]
(a) The reading of the balance A will be more than 2 kg
(b) The reading of the balance B will be less than 5 kg
(c) The reading of the balance A will be less than 2 kg. and that
of B will be more than 5 kg
(d) The reading of balance A will be 2 kg. and that of B will be 5
kg.
Solution: (c) Due to buoyant force on the aluminium block the reading of
spring balance A will be less than 2 kg but it increase the reading of balance B.
Problem 12. In the following diagram, pulley
1
P
is movable and pulley
2
P
is fixed. The value of angle
will be
(a) 60o
(b) 30o
(c) 45o
(d) 15o
Solution: (b) Free body diagram of pulley
1
P
is shown in the figure
For horizontal equilibrium
coscos 21 TT
21 TT
and
WTT 21
For vertical equilibrium
M
F
P
Q
v
W
W
P1
P2
A
m
B
F
T
T
cos
mg
T sin
T1
T2
T1 cos
T2 cos
W
T1 sin
T2 sin
Newton’s Laws of Motion 227
227
WTT
sinsin 21
WWW
sinsin
2
1
sin
or
30
Problem 13. In the following figure, the pulley is massless and frictionless. The relation between
1
T
,
2
T
and
3
T
will be
(a)
321 TTT
(b)
321 TTT
(c)
321 TTT
(d)
321 TTT
Solution : (d) Since through a single string whole system is attached so
1232 TTTW
Problem 14. In the above problem (13), the relation between
1
W
and
2
W
will be
(a)
cos2 1
2W
W
(b)
cos2 1
W
(c)
12 WW
(d)
1
2cos2
W
W
Solution : (a) For vertical equilibrium
121 coscos WTT
221
As WTT
12 cos2 WW
.
cos2 1
2
W
W
Problem 15. In the following figure the masses of the blocks A and B are same and each equal to m. The
tensions in the strings OA and AB are
2
T
and
1
T
respectively. The system is in equilibrium
with a constant horizontal force mg on B. The
1
T
is
(a) mg
(b)
mg2
(c)
mg3
(d)
mg5
Solution : (b) From the free body diagram of block B
mgT
11 cos
……(i)
mgT
11 sin
…..(ii)
by squaring and adding
2
1
2
1
22
12cossin mgT
mgT2
1
W1
W2
P2
T2
T1
T3
2
1
A
mg
T1
m
m
B
O
T2
T1
T2
W1
T1 cos
T2 cos
T1
m
g
T1 cos
1
1
B
m
g
T1 sin
1
228 Newton’s Laws of Motion
228
Problem 16. In the above problem (15), the angle
1
is
(a) 30o (b) 45o (c) 60o (d)
2
1
tan 1
Solution : (b) From the solution (15) by dividing equation(ii) by equation (i)
mg
mg
T
T
11
11 cos
sin
1tan
or
45
1
Problem 17. In the above problem (15) the tension
2
T
will be
(a) mg (b)
mg2
(c)
mg3
(d)
mg5
Solution : (d) From the free body diagram of block A
For vertical equilibrium
1122 coscos
TmgT
45cos2cos 22 mgmgT
mgT2cos 22
…..(i)
For horizontal equilibrium
1122 sinsin
TT
45sin2mg
mgT
22 sin
…..(ii)
by squaring and adding (i) and (ii) equilibrium
22
2)(5 mgT
or
mgT5
2
Problem 18, In the above problem (15) the angle
2
will be
(a) 30o (b) 45o (c) 60o (d)
2
1
tan 1
Solution : (d) From the solution (17) by dividing equation(ii) by equation (i)
mg
mg
2cos
sin
2
2
2
1
tan 2
2
1
tan 1
2
Problem 19. A man of mass m stands on a crate of mass M. He pulls on a light rope passing over a
smooth light pulley. The other end of the rope is attached to the crate. For the system to be
in equilibrium, the force exerted by the men on the rope will be
(a) (M + m)g
(b)
gmM )(
2
1
(c) Mg
(d) mg
Solution : (b) From the free body diagram of man and crate system:
M
m
T2
m
g
T2 cos
2
2
A
T1
T1 sin
1
T2 sin
2
1
T1 cos
1
(M +
T
T
Newton’s Laws of Motion 229
229
For vertical equilibrium
gmMT )(2
2
)( gmM
T
Problem 20. Two forces, with equal magnitude F, act on a body and the magnitude of the resultant force
is
3
F
. The angle between the two forces is
(a)
18
17
cos 1
(b)
3
1
cos 1
(c)
3
2
cos 1
(d)
9
8
cos 1
Solution : (a) Resultant of two vectors A and B, which are working at an angle
, can be given by
cos2
22 ABBAR
[As
FBA
and
3
F
R
]
cos2
3222
2FFF
F
cos22
922
2FF
F
cos2
9
17 22 FF
18
17
cos
or
18
17
cos 1
Problem 21. A cricket ball of mass 150 gm is moving with a velocity of 12 m/s and is hit by a bat so that
the ball is turned back with a velocity of 20 m/s. The force of blow acts for 0.01s on the
ball. The average force exerted by the bat on the ball is
(a) 480 N (b) 600 N (c) 500 N (d) 400 N
Solution : (a)
smv /12
1
and
smv /20
2
[because direction is reversed]
kggmm15.0150
,
sec01.0t
Force exerted by the bat on the ball
01.0
)]12(20[15.0
][ 12
t
vvm
F
= 480 Newton
4.8 Newton’s Third Law.
To every action, there is always an equal (in magnitude) and opposite (in direction)
reaction.
(1) When a body exerts a force on any other body, the second body also exerts an equal and
opposite force on the first.
(2) Forces in nature always occurs in pairs. A single isolated force is not possible.
(3) Any agent, applying a force also experiences a force of equal magnitude but in opposite
direction. The force applied by the agent is called ‘Action’ and the counter force experienced by
it is called ‘Reaction’.
(4) Action and reaction never act on the same body. If it were so the total force on a body
would have always been zero i.e. the body will always remain in equilibrium.
230 Newton’s Laws of Motion
230
(5) If
AB
F
= force exerted on body A by body B (Action) and
BA
F
= force exerted on body B by
body A (Reaction)
Then according to Newton’s third law of motion
BAAB FF
(6) Example : (i) A book lying on a table exerts a force on the table which is equal to the
weight of the book. This is the force of action.
The table supports the book, by exerting an equal force on the book.
This is the force of reaction.
As the system is at rest, net force on it is zero. Therefore force of
action and reaction must be equal and opposite.
(ii) Swimming is possible due to third law of motion.
(iii) When a gun is fired, the bullet moves forward (action). The gun recoils backward
(reaction)
(iv) Rebounding of rubber ball takes place due to third law of motion.
(v) While walking a person presses the ground in the backward
direction (action) by his feet. The ground pushes the person in forward
direction with an equal force (reaction). The component of reaction in
horizontal direction makes the person move forward.
(vi) It is difficult to walk on sand or ice.
(vii) Driving a nail into a wooden block without holding the block is
difficult.
Sample problem based on Newton’s third law
Problem 22. You are on a frictionless horizontal plane. How can you get off if no horizontal force is
exerted by pushing against the surface
(a) By jumping (b) By splitting or sneezing
(c) By rolling your body on the surface (d) By running on the plane
Solution : (b) By doing so we can get push in backward direction in accordance with Newton’s third law
of motion.
4.9 Frame of Reference.
(1) A frame in which an observer is situated and makes his observations is known as his
‘Frame of reference’.
R cos
R
R sin
R
mg
Newton’s Laws of Motion 231
231
(2) The reference frame is associated with a co-ordinate system and a clock to measure the
position and time of events happening in space. We can describe all the physical quantities like
position, velocity, acceleration etc. of an object in this coordinate system.
(3) Frame of reference are of two types : (i) Inertial frame of reference (ii) Non-inertial
frame of reference.
(i) Inertial frame of reference :
(a) A frame of reference which is at rest or which is moving with a uniform velocity along a
straight line is called an inertial frame of reference.
(b) In inertial frame of reference Newton’s laws of motion holds good.
(c) Inertial frame of reference are also called unaccelerated frame of reference or
Newtonian or Galilean frame of reference.
(d) Ideally no inertial frame exist in universe. For practical purpose a frame of reference
may be considered as inertial if it’s acceleration is negligible with respect to the acceleration of
the object to be observed.
(e) To measure the acceleration of a falling apple, earth can be considered as an inertial
frame.
(f) To observe the motion of planets, earth can not be considered as an inertial frame but
for this purpose the sun may be assumed to be an inertial frame.
Example : The lift at rest, lift moving (up or down) with constant velocity, car moving with
constant velocity on a straight road.
(ii) Non inertial frame of reference :
(a) Accelerated frame of references are called non-inertial frame of reference.
(b) Newton’s laws of motion are not applicable in non-inertial frame of reference.
Example : Car moving in uniform circular motion, lift which is moving upward or
downward with some acceleration, plane which is taking off.
4.10 Impulse.
(1) When a large force works on a body for very small time interval, it is called impulsive
force.
An impulsive force does not remain constant, but changes first from zero to maximum and
then from maximum to zero. In such case we measure the total effect of force.
(2) Impulse of a force is a measure of total effect of force.
(3)
2
1
t
tdtFI
.
(4) Impulse is a vector quantity and its direction is same as that of force.
232 Newton’s Laws of Motion
232
(5) Dimension : [
1
MLT
]
(6) Units : Newton-second or Kg-m-
1
s
(S.I.) and Dyne-second or gm-cm-
1
s
(C.G.S.)
(7) Force-time graph : Impulse is equal to the area under F-t curve.
If we plot a graph between force and time, the area under the curve and time axis gives the
value of impulse.
I
Area between curve and time axis
2
1
Base
Height
tF
2
1
(8) If
av
F
is the average magnitude of the force then
tFdtFdtFI av
t
t
av
t
t 2
1
2
1
(9) From Newton’s second law
dt
pd
F
or
2
1
2
1
p
p
t
tpddtF
pppI 12
i.e. The impulse of a force is equal to the change in momentum.
This statement is known as Impulse momentum theorem.
(10) Examples : Hitting, kicking, catching, jumping, diving, collision etc.
In all these cases an impulse acts.
ptFdtFI av.
constant
So if time of contact t is increased, average force is decreased (or diluted) and vice-versa.
(i) In hitting or kicking a ball we decrease the time of contact so that large force acts on the
ball producing greater acceleration.
(ii) In catching a ball a player by drawing his hands backwards
increases the time of contact and so, lesser force acts on his hands and
his hands are saved from getting hurt.
(iii) In jumping on sand (or water) the time of contact is increased
due to yielding of sand or water so force is decreased and we are not
injured. However if we jump on cemented floor the motion stops in a very short interval of time
resulting in a large force due to which we are seriously injured.
(iv) An athlete is advised to come to stop slowly after finishing a fast race. So that time of
stop increases and hence force experienced by him decreases.
t
F
Fav
t1
t2
t
Impuls
e
Force
Tim
e
t
F
Newton’s Laws of Motion 233
233
(v) China wares are wrapped in straw or paper before packing.
Sample problem based on Impulse
Problem 23. A ball of mass 150g moving with an acceleration
2
/20 sm
is hit by a force, which acts on it
for 0.1 sec. The impulsive force is [AFMC 1999]
(a) 0.5 N-s (b) 0.1 N-s (c) 0.3 N-s (d) 1.2 N-s
Solution : (c) Impulsive force
time force
tam
1.02015.0
= 0.3 N-s
Problem 24. A force of 50 dynes is acted on a body of mass 5 g which is at rest for an interval of 3
seconds, then impulse is
[AFMC 1998]
(a)
sN -1015.0 3
(b)
sN -1098.0 3
(c)
sN -105.1 3
(d)
sN -105.2 3
Solution : (c)
timeforcepulseIm
31050 5
s-105.1 3N
Problem 25. The force-time (F – t) curve of a particle executing linear motion is as shown in the figure.
The momentum acquired by the particle in time interval from zero to 8 second will be [CPMT 1989]
(a) – 2 N-s
(b) + 4 N-s
(c) 6 N-s
(d) Zero
Solution : (d) Momentum acquired by the particle is numerically equal to the area enclosed between the
F-t curve and time Axis. For the given diagram area in a upper half is positive and in lower
half is negative (and equal to the upper half). So net area is zero. Hence the momentum
acquired by the particle will be zero.
4.11 Law of Conservation of Linear Momentum.
If no external force acts on a system (called isolated) of constant mass, the total
momentum of the system remains constant with time.
(1) According to this law for a system of particles
dt
pd
F
In the absence of external force
0F
then
p
constant
i.e.,
....
321 pppp
constant.
or
....
332211 vmvmvm
constant
2
4
6
8
+ 2
– 2
Force (N)
Time (s)
234 Newton’s Laws of Motion
234
This equation shows that in absence of external force for a closed system the linear
momentum of individual particles may change but their sum remains unchanged with time.
(2) Law of conservation of linear momentum is independent of frame of reference though
linear momentum depends on frame of reference.
(3) Conservation of linear momentum is equivalent to Newton’s third law of motion.
For a system of two particles in absence of external force by law of conservation of linear
momentum.
21 pp
constant.
2211 vmvm
constant.
Differentiating above with respect to time
0
2
2
1
1 dt
vd
m
dt
vd
m
0
2211 amam
0
21 FF
12 FF
i.e. for every action there is equal and opposite reaction which is Newton’s third law of
motion.
(4) Practical applications of the law of conservation of linear momentum
(i) When a man jumps out of a boat on the shore, the boat is pushed slightly away from the
shore.
(ii) A person left on a frictionless surface can get away from it by blowing air out of his
mouth or by throwing some object in a direction opposite to the direction in which he wants to
move.
(iii) Recoiling of a gun : For bullet and gun system, the force exerted by trigger will be
internal so the momentum of the system remains unaffected.
Let
G
m
mass of gun,
B
m
mass of bullet,
G
v
velocity of gun,
B
v
velocity of bullet
Initial momentum of system = 0
Final momentum of system
BBGG vmvm
By the law of conservation linear momentum
0 BBGG vmvm
So recoil velocity
B
G
B
Gv
m
m
v
(a) Here negative sign indicates that the velocity of recoil
G
v
is opposite to the velocity of
the bullet.
G
v
B
v
Newton’s Laws of Motion 235
235
(b)
G
Gm
v1
i.e. higher the mass of gun, lesser the velocity of recoil of gun.
(c) While firing the gun must be held tightly to the shoulder, this would save hurting the
shoulder because in this condition the body of the shooter and the gun behave as one body.
Total mass become large and recoil velocity becomes too small.
man
1
mm
v
G
G
(iv) Rocket propulsion : The initial momentum of the rocket on its launching pad is zero.
When it is fired from the launching pad, the exhaust gases rush downward at a high speed and
to conserve momentum, the rocket moves upwards.
Let
0
m
initial mass of rocket,
m = mass of rocket at any instant ‘t’ (instantaneous mass)
r
m
residual mass of empty container of the rocket
u = velocity of exhaust gases,
v = velocity of rocket at any instant ‘t’ (instantaneous velocity)
dt
dm
rate of change of mass of rocket = rate of fuel consumption
= rate of ejection of the fuel.
(a) Thrust on the rocket :
mg
dt
dm
uF
Here negative sign indicates that direction of thrust is opposite to the direction of escaping
gases.
dt
dm
uF
(if effect of gravity is neglected)
(b) Acceleration of the rocket :
g
dt
dm
m
u
a
and if effect of gravity is neglected
dt
dm
m
u
a
(c) Instantaneous velocity of the rocket :
gt
m
m
uv e
0
log
and if effect of gravity is neglected
m
m
uv e0
log
m
m
u0
10
log303.2
(d) Burnt out speed of the rocket :
r
eb m
m
uvv 0
max log
The speed attained by the rocket when the complete fuel gets burnt is called burnt out
speed of the rocket. It is the maximum speed acquired by the rocket.
u
u
v
m
236 Newton’s Laws of Motion
236
Sample Problem based on conservation of momentum
Problem 26. A wagon weighing 1000 kg is moving with a velocity 50 km/h on smooth horizontal rails. A
mass of 250 kg is dropped into it. The velocity with which it moves now is [MP PMT 1994]
(a) 12.5 km/hour (b) 20 km/hour (c) 40 km/hour (d) 50 km/hour
Solution : (c) Initially the wagon of mass 1000 kg is moving with velocity of 50 km/h
So its momentum
h
kmkg
501000
When a mass 250kg is dropped into it. New mass of the system
kg12502501000
Let v is the velocity of the system.
By the conservation of linear momentum : Initial momentum = Final momentum
v 1250501000
./40
1250
000,50 hkmv
Problem 27. The kinetic energy of two masses m1 and m2 are equal their ratio of linear momentum will
be [RPET 1988]
(a) m1/m2 (b) m2/m1 (c)
21 /mm
(d)
12 /mm
Solution : (c) Relation between linear momentum (P), man (m) and kinetic energy (E)
mEP2
mP
[as E is constant]
2
1
2
1m
m
P
P
Problem 28. Which of the following has the maximum momentum
(a) A 100 kg vehicle moving at 0.02 ms–1 (b) A 4 g weight moving at 10000 cms–1
(c) A 200 g weight moving with kinetic energy 10–6 J (d) A 20 g weight after
falling 1 kilometre
Solution : (d) Momentum of body for given options are :
(a) P
sec/202.0100 kgmmv
(b) P
sec/4.01001043kgmmv
(c) P
sec/103.6102.022 46 kgmmE
(d) P
sec/82.2101021020233 kgmghm
So for option (d) momentum is maximum.
Problem 29. A rocket with a lift-off mass
4
105.3
kg is blasted upwards with an initial acceleration of
./10 2
sm
Then the initial thrust of the blast is [AIEEE 2003]
(a)
N
5
1075.1
(b)
N
5
105.3
(c)
N
5
100.7
(d)
N
5
100.14
Solution : (c) Initial thrust on the rocket
)( agmF
)1010(105.3 4
N
5
100.7
Newton’s Laws of Motion 237
237
Problem 30. In a rocket of mass 1000 kg fuel is consumed at a rate of 40 kg/s. The velocity of the gases
ejected from the rocket is
sm /1054
. The thrust on the rocket is [MP PMT 1994]
(a)
N
3
102
(b)
N
4
105
(c)
N
6
102
(d)
N
9
102
Solution : (c) Thrust on the rocket
dt
udm
F
)40(1054
N
6
102
Problem 31. If the force on a rocket moving with a velocity of 300 m/s is 210 N, then the rate of
combustion of the fuel is
[CBSE PMT 1999]
(a) 0.7 kg/s (b) 1.4 kg/s (c) 0.07 kg/s (d) 10.7 kg/s
Solution : (a) Force on the rocket
dt
udm
Rate of combustion of fuel
./7.0
300
210 skg
u
F
dt
dm
Problem 32. A rocket has a mass of 100 kg. 90% of this is fuel. It ejects fuel vapours at the rate of 1
kg/sec with a velocity of 500 m/sec relative to the rocket. It is supposed that the rocket is
outside the gravitational field. The initial upthrust on the rocket when it just starts moving
upwards is [NCERT 1978]
(a) Zero (b) 500 N (c) 1000 N (d) 2000 N
Solution : (b) Up thrust force
dt
dm
uF
N5001500
4.12 Free Body Diagram.
In this diagram the object of interest is isolated from its surroundings and the interactions
between the object and the surroundings are represented in terms of forces.
Example :
4.13 Apparent Weight of a Body in a Lift.
When a body of mass m is placed on a weighing machine which is
placed in a lift, then actual weight of the body is mg.
This acts on a weighing machine which offers a reaction R given by the
reading of weighing machine. This reaction exerted by the surface of contact
on the body is the apparent weight of the body.
mg
R
m1
m1a
m1g sin
T
m2a
m2g sin
T
m2
a
m1
m2
T
T
a
Free body
diagram of mass
m1
Free body
diagram of mass
m2
238 Newton’s Laws of Motion
238
Condition
Figure
Velocity
Acceleration
Reaction
Conclusion
Lift is at rest
v = 0
a = 0
R – mg = 0
R = mg
Apparent weight
= Actual weight
Lift moving
upward or
downward
with constant
velocity
v = constant
a = 0
R – mg = 0
R = mg
Apparent weight
= Actual weight
Lift
accelerating
upward at the
rate of 'a’
v = variable
a < g
R – mg = ma
R = m(g + a)
Apparent weight
> Actual weight
Lift
accelerating
upward at the
rate of ‘g’
v = variable
a = g
R – mg = mg
R = 2mg
Apparent weight
= 2 Actual
weight
Lift
accelerating
downward at
the rate of ‘a’
v = variable
a < g
mg – R = ma
R = m(g – a)
Apparent weight
< Actual weight
Lift
accelerating
downward at
the rate of ‘g’
v = variable
a = g
mg – R = mg
R = 0
Apparent weight
= Zero
(weightlessness)
Spring
Balance
R
mg
LIFT
Spring
Balance
R
mg
LIFT
a
Spring
Balance
R
mg
LIFT
g
Spring
Balance
R
mg
LIFT
a
Spring
Balance
R
mg
LIFT
g
Spring
Balance
R
mg
LIFT
Newton’s Laws of Motion 239
239
Lift
accelerating
downward at
the rate of
a(>g)
v = variable
a > g
mg – R = ma
R = mg – ma
R = – ve
Apparent weight
negative means
the body will
rise from the
floor of the lift
and stick to the
ceiling of the lift.
a > g
Spring
Balance
R
mg
LIFT
Newton’s Laws of Motion 237
237
Sample problems based on lift
Problem 33. A man weighs
.80kg
He stands on a weighing scale in a lift which is moving upwards with a
uniform acceleration of
./5 2
sm
What would be the reading on the scale.
)/10(2
smg
[CBSE 2003]
(a) 400 N (b) 800 N (c) 1200 N (d) Zero
Solution : (c) Reading of weighing scale
)( agm
)510(80
N1200
Problem 34. A body of mass 2 kg is hung on a spring balance mounted vertically in a lift. If the lift
descends with an acceleration equal to the acceleration due to gravity ‘g’, the reading on
the spring balance will be
(a) 2 kg (b) (4 g) kg (c) (2 g) kg (d) Zero
Solution : (d)
)( agmR
0)( gg
[because the lift is moving downward with a = g]
Problem 35. In the above problem, if the lift moves up with a constant velocity of 2 m/sec, the reading
on the balance will be
(a) 2 kg (b) 4 kg (c) Zero (d) 1 kg
Solution : (a)
mgR
Newtong2
or
kg2
[because the lift is moving with the zero acceleration]
Problem 36. If the lift in problem, moves up with an acceleration equal to the acceleration due to
gravity, the reading on the spring balance will be
(a) 2 kg (b) (2 g) kg (c) (4 g) kg (d) 4 kg
Solution : (d)
)( agmR
)( ggm
[because the lift is moving upward with a =g]
mg2
NgR 22
Ng4
or
kg4
Problem 37. A man is standing on a weighing machine placed in a lift, when stationary, his weight is
recorded as 40 kg. If the lift is accelerated upwards with an acceleration of
sm /2
, then
the weight recorded in the machine will be
)/10(2
smg
[MP PMT 1994]
(a) 32 kg (b) 40 kg (c) 42 kg (d) 48 kg
Solution : (d)
)( agmR
)210(40
N480
or
kg48
Problem 38. An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of
2
5
ms
.
Taking g to be
2
10
ms
, then the tension in the cable is [Manipal MEE 1995]
(a) 6000 N (b) 9000 N (c) 60000 N (d) 90000 N
Solution : (d)
)( agmT
)510(6000
NT 000,90
Problem 39. The ratio of the weight of a man in a stationary lift and when it is moving downward with
uniform acceleration ‘a’ is 3 : 2. The value of ‘a’ is (g- Acceleration due to gravity on the
earth) [MP PET 1997]
(a)
g
2
3
(b)
3
g
(c)
g
3
2
(d) g
Solution : (b)
2
3
)( liftmoving downward in man aof weight
liftstationaryinmanaof weight
agm
mg
238 Newton’s Laws of Motion
238
2
3
ag
g
agg 332
or
3
g
a
Problem 40. A 60 kg man stands on a spring scale in the lift. At some instant he finds, scale reading has
changed from 60 kg to 50 kg for a while and then comes back to the original mark. What
should we conclude
(a) The lift was in constant motion upwards
(b) The lift was in constant motion downwards
(c) The lift while in constant motion upwards, is stopped suddenly
(d) The lift while in constant motion downwards, is suddenly stopped
Solution : (c) For retarding motion of a lift
)( agmR
for downward motion
)( agmR
for upward motion
Since the weight of the body decrease for a while and then comes back to original value it
means the lift was moving upward and stops suddenly.
Note : Generally we use
)( agmR
for upward motion
)( agmR
for downward motion
here a= acceleration, but for the given problem a= retardation
Problem 41. A bird is sitting in a large closed cage which is placed on a spring balance. It records a
weight placed on a spring balance. It records a weight of 25 N. The bird (mass = 0.5kg)
flies upward in the cage with an acceleration of
2
/2 sm
. The spring balance will now record
a weight of [MP PMT 1999]
(a) 24 N (b) 25 N (c) 26 N (d) 27 N
Solution : (b) Since the cage is closed and we can treat bird cage and air as a closed (Isolated) system. In
this condition the force applied by the bird on the cage is an internal force due to this
reading of spring balance will not change.
Problem 42. A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the
spring balance be
1
W
. If the bird flies about inside the cage, the reading of the spring
balance is
2
W
. Which of the following is true
(a)
21 WW
(b)
21 WW
(c)
21 WW
(d) Nothing definite can be predicted
Solution : (b) In this problem the cage is wire-cage the momentum of the system will not be conserved
and due to this the weight of the system will be lesser when the bird is flying as compared
to the weight of the same system when bird is resting is
12 WW
.
4.14 Acceleration of Block on Horizontal Smooth Surface.
(1) When a pull is horizontal
R = mg
and F = ma
a = F/m
(2) When a pull is acting at an angle (
) to the horizontal (upward)
mg
R
m
F
a
Newton’s Laws of Motion 239
239
R + F sin
= mg
R = mg – F sin
and F cos
= ma
m
F
a
cos
(3) When a push is acting at an angle (
) to the horizontal (downward)
R = mg + F sin
and F cos
= ma
m
F
a
cos
4.15 Acceleration of Block on Smooth Inclined Plane.
(1) When inclined plane is at rest
Normal reaction R = mg cos
Force along a inclined plane
F = mg sin
ma = mg sin
a = g sin
(2) When a inclined plane given a horizontal acceleration ‘b’
Since the body lies in an accelerating frame, an inertial force (mb) acts on it in the opposite
direction.
Normal reaction R = mg cos
+ mb sin
and ma = mg sin
– mb cos
a = g sin
– b cos
Note : The condition for the body to be at rest relative to the inclined plane : a = g sin
– b
cos
= 0
b = g tan
4.16 Motion of Blocks in Contact.
Condition
Free body diagram
Equation
Force and acceleration
amfF 1
21 mm
F
a
mg cos
+mb
sin
R
mg
a
b
mg
R
m
F cos
F
F sin
F sin
R
m
F cos
a
F
F
mg
m1
F
f
m1a
m2
f
m2a
m1
m2
F
A
B
mg cos
R
mg
a
F
240 Newton’s Laws of Motion
240
amf 2
21
2mm
Fm
f
amf 1
21 mm
F
a
amfF 2
21
1mm
Fm
f
amfF 11
321 mmm
F
a
amff 221
321
32
1)(
mmm
Fmm
f
amf 32
321
3
2mmm
Fm
f
amf 11
321 mmm
F
a
amff 212
321
1
1mmm
Fm
f
amfF 32
321
21
2)(
mmm
Fmm
f
Sample problems based on motion of blocks in contact
Problem 43. Two blocks of mass 4 kg and 6 kg are placed in contact with each other on a frictionless
horizontal surface. If we apply a push of 5 N on the heavier mass, the force on the lighter
mass will be
(a) 5 N
m1
f
m1a
m2
f
m2a
F
m1
f1
m1a
F
m2a
f1
m2
f2
m3
f2
m3a
m1
f1
m1a
m2a
f1
m2
f2
m3
f2
m3a
F
m1
m3
F
B
m2
C
m1
m2
F
A
B
m1
m3
F
A
B
m2
C
4
kg
6
kg
5N
Newton’s Laws of Motion 241
241
(b) 4 N
(c) 2 N
(d) None of the above
Solution : (c) Let
kgmkgm4,6 21
and
NF 5
(given)
Force on the lighter mass =
21
2mm
Fm
46
54
N2
Problem 44. In the above problem, if a push of 5 N is applied on the lighter mass, the force exerted by
the lighter mass on the heavier mass will be
(a) 5 N (b) 4 N (c) 2 N (d) None of the above
Solution : (d) Force on the heavier mass
21
1mm
Fm
46
56
N3
Problem 45. In the above problem, the acceleration of the lighter mass will be
(a)
2
5.0
ms
(b)
2
4
5
ms
(c)
2
6
5
ms
(d) None of the above
Solution : (a)
system of the massTotal
system theon force Net
onAccelerati
2
/5.0
10
5sm
Problem 46. Two blocks are in contact on a frictionless table one has a mass m and the other 2 m as
shown in figure. Force F is applied on mass 2m then system moves towards right. Now the
same force F is applied on m. The ratio of force of contact between the two blocks will be in
the two cases respectively.
(a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) 1 : 4
Solution : (b) When the force is applied on mass 2m contact force
32
1g
g
mm
m
f
When the force is applied on mass m contact force
gg
mm
m
f3
2
2
2
2
Ratio of contact forces
2
1
2
1
f
f
4.17 Motion of Blocks Connected by Mass Less String.
Condition
Free body diagram
Equation
Tension and acceleration
amT 1
21 mm
F
a
m1
T
m1a
m2
F
m2a
T
m1
m2
F
T
B
A
F
m
2m
F
242 Newton’s Laws of Motion
242
amTF 2
21
1mm
Fm
T
amTF 1
21 mm
F
a
amT 2
21
2mm
Fm
T
amT 11
321 mmm
F
a
amTT 212
321
1
1mmm
Fm
T
amTF 32
321
21
2)(
mmm
Fmm
T
amTF 11
321 mmm
F
a
amTT 221
321
32
1)(
mmm
Fmm
T
amT 32
321
3
2mmm
Fm
T
Sample problems based on motion of blocks connected by mass less string
Problem 47. A monkey of mass 20 kg is holding a vertical rope. The rope will not break when a mass of
25 kg is suspended from it but will break if the mass exceeds 25 kg. What is the maximum
acceleration with which the monkey can climb up along the rope
)/10(2
smg
[CBSE 2003]
m1
T
m1a
F
m1
T1
m1a
m3
F
m3a
T2
T2
m2a
T1
m2
m1
T1
m1a
F
T2
m2a
T1
m2
m3
m3a
T2
m2
m2a
T
m1
m2
F
T
B
A
m1
m3
F
A
B
m2
C
T1
T2
m3
F
A
B
m2
C
T1
T2
m1
Newton’s Laws of Motion 243
243
(a)
2
/10 sm
(b)
2
/25 sm
(c)
2
/5.2 sm
(d)
2
/5 sm
Solution : (c) Maximum tension that string can bear (Tmax) = 25 g N = 250 N
Tension in rope when the monkey climb up
agmT
For limiting condition
max
TT
250)( agm
2501020 a
2
/5.2 sma
Problem 48. Three blocks of masses 2 kg, 3 kg and 5 kg are connected to each other with light string and
are then placed on a frictionless surface as shown in the figure. The system is pulled by a
force
,10NF
then tension
1
T
[Orissa JEE 2002]
(a) 1N (b) 5 N (c) 8 N (d) 10 N
Solution : (c) By comparing the above problem with general expression.
321
32
1mmm
Fmm
T
532
1053
Newton8
Problem 49. Two blocks are connected by a string as shown in the diagram. The upper block is hung by
another string. A force F applied on the upper string produces an acceleration of
2
/2 sm
in
the upward direction in both the blocks. If T and
T
be the tensions in the two parts of the
string, then [AMU (Engg.) 2000]
(a)
NT 8.70
and
NT 2.47
(b)
NT 8.58
and
NT 2.47
(c)
NT 8.70
and
NT 8.58
(d)
NT 8.70
and
0
T
Solution : (a) From F.B.D. of mass 4 kg
gTa 4'4
.….(i)
From F.B.D. of mass 2 kg
gTTa 2'2
.….(ii)
For total system upward force
agTF 42
N2186
= 70.8 N
by substituting the value of T in equation (i) and (ii)
and solving we get
NT 2.47'
Problem 50. Three masses of 15 kg. 10 kg and 5 kg are suspended vertically as shown in the fig. If the
string attached to the support breaks and the system falls freely, what will be the tension
2kg
5kg
3kg
10N
T1
T2
T
F
2kg
4kg
T
4kg
T
4g
4a
2kg
T
2g
2a
T
15k
g
10k
g
244 Newton’s Laws of Motion
244
in the string between 10 kg and 5 kg masses. Take
2
10
msg
. It is assumed that the string
remains tight during the motion
(a) 300 N (b) 250 N (c) 50 N (d) Zero
Solution : (d) In the condition of free fall, tension becomes zero.
Problem 51. A sphere is accelerated upwards with the help of a cord whose breaking strength is five
times its weight. The maximum acceleration with which the sphere can move up without
cord breaking is
(a) 4g (b) 3g (c) 2g (d) g
Solution : (a) Tension in the cord = m(g + a) and breaking strength = 5 mg
For critical condition
mgagm 5
ga 4
This is the maximum acceleration with which the sphere can move up with cord breaking.
4.18 Motion of Connected Block Over a Pulley.
Condition
Free body diagram
Equation
Tension and acceleration
gmTam 111
g
mm
mm
T
21
21
12
122 Tgmam
g
mm
mm
T
21
21
24
12 2TT
g
mm
mm
a
21
12
m1
m1a
T1
m1g
m2
m2a
T1
m2g
T1
T1
T2
P
T1
T1
T2
m1
m2
A
B
a
a
Newton’s Laws of Motion 245
245
gmTam 111
g
mmm
mmm
T
321
321
1][2
1222 TTgmam
g
mmm
mm
T
321
31
22
233 Tgmam
g
mmm
mmm
T
321
321
3][4
13 2TT
321
132 ])[(
mmm
gmmm
a
Condition
Free body diagram
Equation
Tension and acceleration
When pulley have a
finite mass M and
radius R then tension
in two segments of
string are different
111 Tgmam
2
21
21 M
mm
mm
a
gmTam 222
g
M
mm
M
mm
T
2
2
2
21
21
1
Torque
IRTT )( 21
R
a
IRTT )( 21
R
a
MRRTT 2
21 2
1
)(
2
21 Ma
TT
g
M
mm
M
mm
T
2
2
2
21
12
2
m1
m1a
T1
m1g
m3
m3a
T2
m3g
T1
T1
T3
m1
m1a
T1
m1g
m2
m2a
T2
m2g
T1
T2
R
p
T3
m1
A
a
T1
T1
C
T2
B
m3
m2
M
T2
m2
B
T2
T1
A
m1
R
m2
m2a
T1
m2g +
T2
246 Newton’s Laws of Motion
246
amT 1
g
mm
m
a21
2
Tgmam 22
g
mm
mm
T21
21
sin
11 gmTam
g
mm
mm
a
21
12 sin
Tgmam 22
g
mm
mm
T21
21 )sin1(
amgmT 11 sin
g
mm
mm
a21
12 )sinsin(
Tgmam
sin
22
g
mm
mm
T21
21 )sin(sin
Condition
Free body diagram
Equation
Tension and acceleration
amTgm 11 sin
21
1sin
mm
gm
a
m1
A
T
P
m2
a
T
B
m1
m1a
T
m1
m1a
T
m1g sin
m2
m2g
T
m2a
m2
m2g
T
m2a
m2a
T
m2g
sin
m2
m1
A
T
P
m2
a
T
B
m1
A
P
m2
a
T
B
m1
m1a
T
m1g sin
m2
T
m2a
m1
m1a
T
m1g
sin
a
m
1
m
2
T
T
a
A
B
Newton’s Laws of Motion 247
247
amT 2
g
mm
mm
T21
21
4
2
As
22
2)(
dt
xd
21
2)(
2
1
dt
xd
2
1
2a
a
1
a
acceleration of
block A
2
a
acceleration of
block B
amT 1
21
2
14
2
mm
gm
aa
21
2
24mm
gm
a
21
21
4
2
mm
gmm
T
Tgm
a
m2
222
111 Tgmam
g
Mmm
mm
a][
)(
21
21
gmTam 222
g
Mmm
Mmm
T][
)2(
21
21
1
MaTT 21
g
Mmm
Mmm
T][
)2(
21
22
2
Sample problems based on motion of blocks over pulley
Problem 52. A light string passing over a smooth light pulley connects two blocks of masses
1
m
and
2
m
(vertically). If the acceleration of the system is g/8 then the ratio of the masses is [AIEEE 2002]
m1
T1
m1g
m1a
m2
2T
m2g
m2(a/
2)
m1
T
m1a
m2
T2
m2g
m2a
M
T2
T1
Ma
P
B
T
T
a2
a1
m1
A
m2
M
m1
A
T1
m2
B
T1
a
a
T2
T2
C
248 Newton’s Laws of Motion
248
(a) 8 : 1 (b) 9 : 7 (c) 4 : 3 (d) 5 : 3
Solution : (b)
g
mm
mm
a
21
12
=
8
g
; by solving
7/9
1
2
m
m
Problem 53. A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a
frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of
the system is (given g = 10
)
2
ms
[Kerala (Engg.) 2000]
(a)
2
100
ms
(b)
2
3
ms
(c)
2
10
ms
(d)
2
30
ms
Solution : (b)
g
mm
m
a
21
2
2
/310
37
3sm
Problem 54. Two masses
1
m
and
2
m
are attached to a string which passes over a frictionless smooth
pulley. When
,10
1kgm
,6
2kgm
the acceleration of masses is [Orissa JEE 2002]
(a) 20
2
/sm
(b)
2
/5 sm
(c) 2.5
2
/sm
(d)
2
/10 sm
Solution : (c)
g
mm
mm
a
21
21
10
610
610
2
/5.2 sm
Problem 55. Two weights
1
W
and
2
W
are suspended from the ends of a light string passing over a
smooth fixed pulley. If the pulley is pulled up with an acceleration g, the tension in the
string will be
(a)
21
21
4
WW
WW
(b)
21
21
2
WW
WW
(c)
21
21 WW
WW
(d)
)(2 21
21 WW
WW
Solution : (a) When the system is at rest tension in string
g
mm
mm
T
21
21
2
If the system moves upward with acceleration g then
gg
mm
mm
T
21
21
2
g
mm
mm
21
21
4
or
21
21
4
ww
ww
T
Problem 56. Two masses M1 and M2 are attached to the ends of a string which passes over a pulley
attached to the top of an inclined plane. The angle of inclination of the plane in
. Take g =
10 ms–2.
If M1 = 10 kg, M2 = 5 kg,
= 30o, what is the acceleration of mass M2
(a)
2
10
ms
(b)
2
5
ms
m1
m2
10 kg
6 kg
M1
M2
B
A
Newton’s Laws of Motion 249
249
(c)
2
3
2
ms
(d) Zero
Solution : (d) Acceleration
g
mm
mm
21
12 sin
g
105
30sin.105
g
105
55
= 0
Problem 57. In the above problem, what is the tension in the string
(a) 100 N (b) 50 N (c) 25 N (d) Zero
Solution : (b)
g
mm
mm
T
21
21 sin1
510
10.30sin1510
N50
Problem 58. In the above problem, given that M2 = 2M1 and M2 moves vertically downwards with
acceleration a. If the position of the masses are reversed the acceleration of M2 down the
inclined plane will be
(a) 2 a (b) a (c) a/2 (d) None of the above
Solution : (d) If
12 2mm
, then
2
m
moves vertically downward with acceleration
g
mm
mm
a
21
12 sin
g
mm
mm
11
11 2
30sin2
= g/2
If the position of masses are reversed then
2
m
moves downward with acceleration
g
mm
mm
a
21
12 sin
'
0.
2
30sin2
11
11
g
mm
mm
[As m2 = 2m1]
i.e. the
2
m
will not move.
Problem 59. In the above problem, given that M2 = 2M1 and the tension in the string is T. If the positions
of the masses are reversed, the tension in the string will be
(a) 4 T (b) 1 T (c) T (d) T/2
Solution : (c) Tension in the string
g
mm
mm
T
21
21 sin1
If the position of the masses are reversed then there will be no effect on tension.
Problem 60. In the above problem, given that M1 = M2 and
o
30
. What will be the acceleration of the
system
(a)
2
10
ms
(b)
2
5
ms
(c)
2
5.2
ms
(d) Zero
Solution : (c)
g
mm
mm
a
21
12 sin
g
2
30sin1
2
/5.2
4sm
g
[As m1 = m2]
Problem 61. In the above problem, given that M1 = M2 = 5 kg and
o
30
. What is tension in the string
(a) 37.5 N (b) 25 N (c) 12.5 N (d) Zero
Solution : (a)
g
mm
mm
T
21
21 sin1
10
55
30sin155
N5.37
Problem 62. Two blocks are attached to the two ends of a string passing over a smooth pulley as shown
in the figure. The acceleration of the block will be (in m/s2) (sin 37o = 0.60, sin 53o = 0.80)
(a) 0.33
(b) 0.133
100
kg
50
kg
37o
53o
250 Newton’s Laws of Motion
250
(c) 1
(d) 0.066
Solution : (b)
g
mm
mm
a
21
12 sinsin
g
50100
37sin10053sin50
2
/133.0 sm
Problem 63. The two pulley arrangements shown in the figure are identical. The mass of the rope is
negligible. 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 of 2mg. The ratio of
accelerations in two cases will be
(a) 1 : 1
(b) 1 : 2
(c) 1 : 3
(d) 1 : 4
Solution : (c) For first case
32
2
21
12
1g
mm
mm
g
mm
mm
a
…..(i)
For second case
from free body diagram of m
mgTma
2
mgmgma 2
2
[As T= 2mg]
ga
2
……..(ii)
From (i) and (ii)
3/1
3/
2
1 g
g
a
a
Problem 64. In the adjoining figure m1 = 4m2. The pulleys are smooth and light. At time t = 0, the system
is at rest. If the system is released and if the acceleration of mass m1 is a, then the
acceleration of m2 will be
(a) g
(b) a
(c)
2
a
(d) 2a
Solution : (d) Since the mass
2
m
travels double distance in comparison to mass
1
m
therefore its acceleration
will be double i.e. 2a
Problem 65. In the above problem (64), the value of a will be
(a) g (b)
2
g
(c)
4
g
(d)
8
g
Solution : (c) By drawing the FBD of
1
m
and
2
m
Tgmam 2
11
…..(i)
gmTam 22 2
…..(ii)
m
2m
m
2mg
(a)
(b)
m1
m2
20
cm
m
T
mg
ma2
m2
T
m2g
m2(2a)
m1
2T
m1g
m1a
Newton’s Laws of Motion 251
251
by solving these equation
4/ga
Problem 66. In the above problem, the tension T in the string will be
(a) m2g (b)
2
2gm
(c)
gm 2
3
2
(d)
gm 2
2
3
Solution : (d) From the solution (65) by solving equation
gmT 2
2
3
Problem 67. In the above problem, the time taken by m1 in coming to rest position will be
(a) 0.2 s (b) 0.4 s (c) 0.6 s (d) 0.8 s
Solution : (b) Time taken by mass
2
m
to cover the distance 20 cm
5.2
2.02
4/
2.022
ga
h
t
sec4.0
Problem 68. In the above problem, the distance covered by m2 in 0.4 s will be
(a) 40 cm (b) 20 cm (c) 10 cm (d) 80 cm
Solution : (a) Since the
2
m
mass cover double distance therefore S = 2 20 = 40 cm
Problem 69. In the above problem, the velocity acquired by m2 in 0.4 second will be
(a) 100 cm/s (b) 200 cm/s (c) 300 cm/s (d) 400 cm/s
Solution : (b) Velocity acquired by mass
2
m
in 0.4 sec
From
atuv
[As
2
/5
2
10
2/ smga
]
4.050 v
./200/2 seccmsm
Problem 70. In the above problem, the additional distance traversed by m2 in coming to rest position
will be
(a) 20 cm (b) 40 cm (c) 60 cm (d) 80 cm
Solution : (a) When
2
m
mass acquired velocity 200 cm/sec it will move upward till its velocity becomes
zero.
cm
g
u
H20
1002
200
2
2
2
Problem 71. The acceleration of block B in the figure will be
(a)
)4( 21
2mm
gm
(b)
)4(
2
21
2mm
gm
(c)
)4(
2
21
1mm
gm
(d)
)(
2
21
1mm
gm
Solution : (a) When the block
2
m
moves downward with acceleration a, the acceleration of mass
1
m
will
be
a2
because it covers double distance in the same time in comparison to
2
m
.
m1
m
2
A
B
252 Newton’s Laws of Motion
252
Let T is the tension in the string.
By drawing the free body diagram of A and B
amT 2
1
……..(i)
amTgm 22 2
……..(ii)
by solving (i) and (ii)
21
2
4mm
gm
a
4.19 Motion of Massive String.
Condition
Free body diagram
Equation
Tension and acceleration
1
T
force applied by
the string on the block
amMF )(
MaT
1
mM
F
a
)(
1mM
F
MT
F
mM
mM
T)(2
)2(
2
2
T
Tension at mid
point of the rope
a
m
MT
2
2
m = Mass of string
T = Tension in string at
a distance x from the
end where the force is
applied
maF
mFa /
F
L
xL
T
a
L
xL
mT
L
Mxa
TF
1
M
FF
a21
T1
M
a
F
m
M
T2
m/2
M
F
m
a
F
L
T
x
m [(L –
x)/L]
a
T
F1
F2
L
x
B
A
F1
T
(M/L)x
A
B
a
F2
F1
a
M
2T
m
2
m2g
m2a
m1
T
m1(2a)
Newton’s Laws of Motion 253
253
M = Mass of uniform
rod
L = Length of rod
MaFF 21
L
x
F
L
x
FT 21 1
Mass of segment BC
x
L
M
F
L
xL
T
F
L
xL
T
4.20 Spring Balance and Physical Balance.
(1) Spring balance : When its upper end is fixed with rigid support and body of mass m
hung from its lower end. Spring is stretched and the weight of the body can be
measured by the reading of spring balance
mgWR
The mechanism of weighing machine is same as that of spring balance.
Effect of frame of reference : In inertial frame of reference the reading of
spring balance shows the actual weight of the body but in non-inertial frame of
reference reading of spring balance increases or decreases in accordance with
the direction of acceleration
[for detail refer Article (4.13)]
(2) Physical balance : In physical balance actually we
compare the mass of body in both the pans. Here we does
not calculate the absolute weight of the body.
Here X and Y are the mass of the empty pan.
(i) Perfect physical balance :
Weight of the pan should be equal i.e. X = Y
and the needle must in middle of the beam i.e. a = b.
Effect of frame of reference : If the physical balance is perfect then there will be no effect of
frame of reference (either inertial or non-inertial) on the measurement. It is always errorless.
L
T
T
x
C
B
F
A
T
B
A
m
R
X
Y
A
B
O
a
b
254 Newton’s Laws of Motion
254
(ii) False balance : When the masses of the pan are
not equal then balance shows the error in measurement.
False balance may be of two types
(a) If the beam of physical balance is horizontal
(when the pans are empty) but the arms are not equal
YX
and
ba
For rotational equilibrium about point ‘O’
YbXa
……(i)
In this physical balance if a body of weight W is placed in pan X then to balance it we have
to put a weight
1
W
in pan Y.
For rotational equilibrium about point ‘O’
bWYaWX )()( 1
…..(ii)
Now if the pans are changed then to balance the body we have to put a weight
2
W
in pan X.
For rotational equilibrium about point ‘O’
bWYaWX )()( 2
……(iii)
From (i), (ii) and (iii)
True weight
21 WWW
(b) If the beam of physical balance is not horizontal (when the pans are empty) and the
arms are equal
i.e.
YX
and
ba
In this physical balance if a body of weight W is placed in X Pan then to balance it.
We have to put a weight W1 in Y Pan
For equilibrium
1
WYWX
…..(i)
Now if pans are changed then to balance the body
we have to put a weight
2
W
in X Pan.
For equilibrium
WYWX 2
…..(ii)
From (i) and (ii)
True weight
221 WW
W
4.21 Modification of Newton’s Laws of Motion.
According to Newton, direction and time i.e., time and space are absolute. The velocity of
observer has no effect on it. But, according to special theory of relativity Newton’s laws are
X
Y
A
B
O
a
b
X
Y
A
B
O
a
b
Newton’s Laws of Motion 255
255
true, as long as we are dealing with velocities which are small compare to velocity of light.
Hence the time and space measured by two observers in relative motion are not same. Some
conclusions drawn by the special theory of relativity about mass, time and distance which are
as follows :
(1) Let the length of a rod at rest with respect to an observer is
.
0
L
If the rod moves with
velocity v w.r.t. observer and its length is L, then
22
0/1 cvLL
where, c is the velocity of light.
Now, as v increases L decreases, hence the length will appear shrinking.
(2) Let a clock reads T0 for an observer at rest. If the clock moves with velocity v and clock
reads T with respect to observer, then
2
2
0
1c
v
T
T
Hence, the clock in motion will appear slow.
(3) Let the mass of a body is
0
m
at rest with respect to an observer. Now, the body moves
with velocity v with respect to observer and its mass is m, then
2
2
0
1c
v
m
m
m0 is called the rest mass.
Hence, the mass increases with the increases of velocity.
Note : If
,cv
i.e., velocity of the body is very small w.r.t. velocity of light, then
.
0
mm
i.e., in the practice there will be no change in the mass.
If v is comparable to c, then
,
0
mm
i.e., mass will increase.
If
,cv
then
2
2
0
1v
v
m
m
or
.
00 m
m
Hence, the mass becomes infinite, which is
not possible, thus the speed cannot be equal to the velocity of light.
The velocity of particles can be accelerated up to a certain limit. In cyclotron the
speed of charged particles cannot be increased beyond a certain limit.
Sample problems (Miscellaneous)
Problem 72. A body weighs 8 gm, when placed in one pan and 18gm, when placed in the other pan of a
false balance. If the beam is horizontal (when both the pans are empty), the true weight of
the body is
(a) 13 gm (b) 12 gm (c) 15.5 gm (d) 15 gm
256 Newton’s Laws of Motion
256
Solution : (b) For given condition true weight =
21WW
188
=12 gm.
Problem 73. A plumb line is suspended from a ceiling of a car moving with horizontal acceleration of a.
What will be the angle of inclination with vertical [Orissa JEE 2003]
(a)
)/(tan 1ga
(b)
)/(tan 1ag
(c)
)/(cos 1ga
(d)
)/(cos 1ag
Solution : (a) From the figure
g
a
tan
ga/tan 1
Problem 74. A block of mass
kg5
is moving horizontally at a speed of 1.5 m/s. A perpendicular force of
5 N acts on it for 4 sec. What will be the distance of the block from the point where the
force started acting [Pb PMT 2002]
(a) 10 m (b) 8 m (c) 6 m (d) 2 m
Solution : (a) In the given problem force is working in a direction perpendicular to initial velocity. So the
body will move under the effect of constant velocity in horizontal direction and under the
effect of force in vertical direction.
mtuS xx 645.1
2
2
1attuS yy
2
/
2
1
0tmF
2
45/5
2
1
m8
6436
22 yx SSS
m10100
Problem 75. The velocity of a body of rest mass
0
m
is
c
2
3
(where c is the velocity of light in vacuum).
Then mass of this body is [Orissa JEE 2002]
(a)
0
)2/3( m
(b)
0
)2/1( m
(c)
0
2m
(d)
0
)3/2( m
Solution : (c) From Einstein’s formula
2
2
0
1c
v
m
m
2
2
0
2
3
1C
C
m
0
02
4
3
1
m
m
Problem 76. Three weights W, 2W and 3W are connected to identical springs suspended form rigid
horizontal rod. The assembly of the rod and the weights fall freely. The positions of the
weights from the rod are such that
[Roorkee 1999]
(a) 3 W will be farthest (b) W will be farthest
(c) All will be at the same distance (d) 2 W will be farthest
Solution : (c) For W, 2W, 3W apparent weight will be zero because the system is falling freely. So there
will be no extension in any spring i.e. the distances of the weight from the rod will be
same.
Problem 77. A bird is sitting on stretched telephone wires. If its weight is W then the additional tension
produced by it in the wires will be
a
g
a
Newton’s Laws of Motion 257
257
(a) T = W (b) T > W (c) T < W (d) T = 0
Solution : (b) For equilibrium
WT
sin2
sin2
W
T
lies between 0 to 90° i.e.
1sin
T > W
Problem 78. With what minimum acceleration can a fireman slides down a rope while breaking strength
of the rope is
3
2
his weight [CPMT 1979]
(a)
g
3
2
(b) g (c)
g
3
1
(d) Zero
Solution : (c) When fireman slides down, Tension in the rope
agmT
For critical condition m(g – a) = 2/3 mg
mgmamg 3
2
3
g
a
So, this is the minimum acceleration by which a fireman can slides down on a rope.
Problem 79. A car moving at a speed of 30 kilometres per hour’s is brought to a halt in 8 metres by
applying brakes. If the same car is moving at 60 km. per hour, it can be brought to a halt
with same braking power in
(a) 8 metres (b) 16 metres (c) 24 metres (d) 32 metres
Solution : (d) From
asuv 2
22
asu20 2
a
u
s2
2
2
us
(if a = constant)
4
30
60 2
2
1
2
1
2
u
u
s
s
844 12 ss
32
metres.
T
T
W
2T sin
