newton’s laws of mot

NEWTON’S LAWS OF MOTION–I OP-MI-P-1

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NEWTON’S LAWS OF MOTION – I

LESSON 1

This lesson is the beginning of study of Mechanics.

Mechanics can be defined as the science, which describes

and predicts the condition of rest or motion of bodies under

the action of forces. Mechanics is divided into statics and

dynamics, the former deals with bodies at rest, while the

latter with bodies in motion. We study Dynamics in two

parts kinetics and kinematics. In kinetics we study about

cause of motion and in kinematics we study about geometry

of motion irrespective of its cause.

The present lesson is the study about forces and its

effect on motion, i.e., kinetics. The first section of this lesson

is devoted to the study of three laws of Newton, and the

second section to the application of these laws for solving

questions based on motion of connected bodies.

IITJEE Syllabus: Newton’s laws of motion; inertial frame of reference.

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SECTION I







1 NEWTON’S LAWS OF MOTION

Till the mid of 17th century most of the philosophers thought that some influence was

needed to keep a body moving. They thought that a body was in its „natural state‟ when it was at

rest and some external influence was needed to continuously move a body; otherwise it would

naturally stop moving.

Confusions about these issues were solved in 1687 when Newton presented his three laws

of motion. According to him influence is needed not for all kind of motion it is needed for

accelerated motion only. Before going in details about these three laws, let us summerise these

three laws first.

Law 1: An object/body will remain at rest or continue to move with uniform velocity unless an

external force is applied to it.

Law 2: When an external force is applied on a body of constant mass the force produces an

acceleration, which is directly proportional to the force and inversely proportional to the mass of

the body.

Law 3: When a body A exerts a force on another body B, B exerts an equal and opposite force

on A.

1.1 FRAME OF REFERENCE

Before going in details about Newton‟s laws, let us first define frame of reference. Suppose

you are standing on your school bus with one of your friend who is properly seated in his seat.

There is another friend of yours standing on bus stop, waves his hand to stop the bus. The driver

applies brakes and your friend in bus observes you to move forward but your friend outside the

bus observes bus and you to stop together. So your two friends one in the bus and other outside

the bus observe you differently. The person in bus finds you initially at rest and then starts moving,

while a friend outside the bus observes nothing unusual. Each observer such as your friend in bus

or your friend outside bus defines a reference frame. A reference system requires a co-ordinate

system (made of origin and co-ordinate axes) and a set of clocks, which enable an observer to

measure positions, velocities and accelerations in his or her particular reference frame. Observers

in different frame may measure different displacements, velocities and accelerations.

Newton‟s laws are applicable for a special kind of frame of reference. In the example given

earlier, the friend outside the bus is in a frame which observes you moving with bus and then

comes to rest. But the friend inside the bus finds you to come in motion without any cause. So we

can say that your motion can‟t be analysed using Newton‟s law with respect to your friend in bus.

The first law of Newton is called “law of inertia” and the frame in which this law is applicable is

called as inertial frame. In the said example your friend outside the bus defines an inertial frame.

Any reference frame which is not accelerated (either at rest or moving with uniform

velocity) is called an inertial frame. Newton‟s first law is applicable only in an inertial frame. We

generally apply Newton‟s first law with respect to earth by assuming it an inertial frame. In actual

practice earth experiences an accelerations of 4.4 × 103 m/s2 towards the sun due to its circular

motion around sun. In addition earth rotates about its own axis once every 24 hours, a point on the







equator experiences an addition acceleration of 3.37×102m/s2 towards the center of earth.

However these accelerations are small compared with g and can often be neglected. In most

situations we shall assume that a set of nearby points on earth‟s surface constitutes an inertial

frame. At a later stage we will study about accelerated frame also.







1.2 NEWTON‟S FIRST LAW OF MOTION

If a body is observed from an inertial frame which is at rest or moving with uniform velocity

then it will remain at rest or continue to move with uniform velocity until an external force is applied

on it.

The property due to which a body remains at rest or continue its motion with uniform

velocity is called as Inertia.

Force is a push or pull that disturbs or tends to disturb inertia of rest or inertia of uniform

motion with uniform velocity of a body.

Hence first law of motion defines inertia, force and inertial frame of reference.

Illustration 1

Question: The diagram shows the forces that are

acting on a particle. Find whether the

acceleration of the particle is zero or

non-zero.

4 N 4 N

120° 120°

6 N

Solution: To check whether the particle will have any

acceleration or not, let us see net force is

zero or not. Resolving the forces in

horizontal and vertical directions.

Net force in horizontal direction

= 4 cos 30° - 4 cos 30°

= 0

Net force in vertically downward direction

= 6 – 4 sin 30° - 4 sin 30° = 2 N

As net force is not zero, so the particle

will have acceleration.

4 N 4 s

in 3

0°

4cos 30° 4 cos 30°

6 N

4 s

in 3

0°

4 N

Is it possible to have motion in the absence of the force?

1.3 NEWTON‟S SECOND LAW OF MOTION

Newton‟s first law gives definition of force and inertia. Newton‟s second law of motion

defines magnitude of force. Before stating Newton‟s Second‟s Law, Let us know about Mass.







If we attempt to change the state of rest or motion with uniform velocity, the object resists

this change. Inertia is solely a property of an individual object; it is a measure of response of an

object to an external force. If we take two blocks identical in shape and size; one of wood and the

other of steel, the same force causes more acceleration in the wooden block. Therefore we say

steel block has more inertia than the wooden block.

Mass is measure of inertia of a body. It is an internal property of a body and is independent

of the body‟s surrounding and of the method used to measure it. Its SI unit is kg.

Mass should not be confused with weight. Mass and weight are different quantities. We will

see later, the weight of a body is equal to magnitude of force exerted by the earth on the bodies

and varies with location. For example a body, which weighs 60 N on earth weights 10 N on moon.

But its mass is 6 kg on earth as well as on moon.

If we push a block of ice on a smooth surface by applying a horizontal force F, the block

will move with some acceleration. If we double the force the acceleration doubles, likewise if we

make the force 3F the acceleration triples. From such observations we conclude that the

acceleration of an object is directly proportional to the resultant force acting on it.

Also if we push a block of ice on a smooth surface by applying a force F, the block moves

with an acceleration of a. If we double the mass, the same force causes an acceleration of a/2. If

we triple the mass of block, the acceleration will be a/3.

These observations are summarised, as follows: „the acceleration of an object is directly

proportional to the net force acting on it and is inversely proportional to its mass’. Thus we can

relate mass, force and acceleration through following mathematical relation,

aMFext

. … (1)

It is important to note here that it is a vector relation that is acceleration is in the direction of

net force. Also the relation is valid when mass of the object remains constant. More detailed cases

will be dealt later.

Illustration 2

Question: Two forces 1F

and 2F

act on a 5.0 kg

mass in mutually perpendicular

directions. If F1 = 20.0 N and F2 = 15.0 N,

find the acceleration.

F2

F1

90°

m

Solution: Acceleration will be in the direction of net force and will have the magnitude given by

aMF

F

= 21 FF

N251520|| 22 F







|| a = 50.5

||

F

ms-2

If the resultants force is at angle with 1F

.

tan = 20

15 = 37°

Therefore, acceleration is 5 ms2 at an angle 37° with the direction of 1F .







1.4 NEWTON‟S THIRD LAW OF MOTION

We state this law as, “To every action there is an equal and opposite reaction”.

But what is meaning of action and reaction and which force is action and which force is

reaction?

Every force that acts on a body is due to the other bodies in environment. Suppose that a

body A experiences a force ABF

due to other body B. The body B will also experience a force BAF

due to A. According to Newton‟s third law two forces are equal in magnitude and opposite in

direction. Mathematically we write it as

ABF

= – BAF

…..(2)

Here we can take either ABF

or BAF

as action force and the other will be the reaction force.

Another important thing is these two forces always acts on different bodies.

Illustration 3

Question: A horse pulls a cart with a horizontal

force, causing it to accelerate as shown

in figure. Newton’s third law says that

the cart exerts an equal and opposite

force on the horse. In view of this, how

can the cart accelerate?

Solution: The motion of any object is determined by the external forces that acts on it. If resultant of

external force is non-zero, the object accelerate in the direction of resultant force. In this

situation, the horizontal forces exerted on the cart are forward force exerted by the horse (F)

and the backward contact force (f1) due to roughness of surface. When forward force

exerted on the cart exceeds the backward force, the resultant force on it is in the forward

direction. This resultant force causes the cart to accelerate to the right. The horizontal force

that acts on the horse are the forward contact force (f2) due to roughness of surface and the

backward force of the cart (F). The resultant of these two forces causes the horse to

accelerate.

f1 f2

F F







Which one of the three laws of Newton do you think as most important?







PROFICIENCY TEST - I

The following questions deal with the basic concepts of this section. Answer the

following briefly. Go to the next section only if your score is at least 80%. Do not consult

the Study Material while attempting these questions.

1. Define the following terms:

(i) force,

(ii) inertia,

(iii) mass and weight,

(iv) equilibrium.

2. Is there any relation between the total force acting on a body and the direction in which it

moves?

3. If a small sports car collides head-on with a massive truck, which vehicle experiences the

greater impact force? Which vehicle experiences the greater acceleration?

4. A force, F, applied to an object of mass m1 produces an acceleration of 3.0 ms-2. The same

force applied to another object of mass m2 produces an acceleration of 1.0 ms-2.

What is the value of the ratio 1

2

m

m? If m1 and m2 are combined, find their acceleration under

the action of force F.

5. Two forces 1F

and 2F

of equal magnitudes act as shown

in figure on a 5.0 kg mass. If |||| 21 FF

= 5 N. Find the

magnitude and direction of the acceleration with 1F

.

60°

2F

1F







6. What is wrong with the statement, “Since the car is at rest, there is no force acting on it”?

How would you correct this statement?

7. Identify the action reaction pairs in the following situations

(a) A man takes a step

(b) A snowball hits a girl in the back

(c) A cricketer catches the ball

8. The diagram shows the forces acting on a particle.

Find R and if the particle is moving with uniform velocity.

R

8N

6N

ANSWERS TO PROFICIENCY TEST - I

4. 3

5. 1 N, 600

8. 10 N, 1430







SECTION II







2. MOTION OF CONNECTED BODIES

Before knowing how to apply Newton‟s laws of motion to solve questions based on motion

of connected bodies and to know about the stepwise procedure to solve the same, let us know

about the commonly used forces in such situations.

Also in a particular question some assumptions will be given and one should know how to

use them while analysing the problem. Such assumptions definitely simplify the analysis at the

cost of some physical reality. But in later stages we add some new techniques that permit us to be

more realistic in our analysis.

2.1 COMMONLY USED FORCES

(i) Weight of a body: It is the force with which Earth attracts a body towards its center. If

M is mass of body and g is acceleration due to gravity, weight of the body is Mg. We take its

direction vertically downward.

(ii) Normal Force: Let us consider a book resting on the table.

It is acted upon by its weights in vertically downward direction and is at

rest. It means that there is another force acting on the block in opposite

direction, which balances its weight. This force is provided by the table

and we call it as normal force. Hence, if two bodies are in contact a

contact force arises, if the surface is smooth the direction of force is

normal to the plane of contact. We call this force as Normal force. We

take its direction towards the body under consideration.

Fig. 1

(iii) Tension in string: Let a block is hanging from a string. Weight of the block is acting in vertically downward but it is not moving, hence its weight is balanced by a force due to string. This force is called ‘tension in string’. Tension is a force in a stretched string. Its direction is taken along the string and away from the body under consideration.

2.2 ASSUMPTIONS AND THEIR BENEFITS

Fig. 2

(i) If the bodies are rigid and moving together then their accelerations, velocities and

displacements will be same. As in the figure acceleration of blocks A, B and C will be same

A aA = aB = aC B

C

Fig. 3

(ii) If the surface is smooth the contact force will only be the normal force.

(iii) If the string is inextensible then accelerations, velocities and displacements of two blocks

moving together will be same as in figure







F A B C aA = aB = aC

Fig. 4







(iv) If string is massless, tension throughout the string will be same.

(v) If the pulley is massless and frictionless then tension on the two sides of pulley will be

same.

In later stage we will discuss about flexible string, massive and rough pulley also.

2.3 STEPS TO BE FOLLOWED TO SOLVE QUESTIONS BASED ON MOTION OF

CONNECTED BODIES

In such questions you will be given a system of bodies under the action of forces and you

will need to find out accelerations of different bodies and unknown forces on bodies. The following

steps are needed by you to apply while solving such questions.

Step 1: Identify the unknown accelerations and unknown forces involved in the question.

Step 2: Draw free body diagram of different bodies in the given system.

Free body diagram (FBD). It is a diagram that shows forces acting on the body making it

free from other bodies applying forces on the body under consideration. Hence free body diagram

will include the forces like weight of the body, normal force, tension in string and the applied force.

The important thing while drawing FBD is the shape of the body should be taken under

consideration and force should be shown in a particular way. For example weight should be

applied from center of gravity of body, normal force(s) should be applied on the respective

surface(s), tension should be applied on the side(s) of string(s).

Examples

(i) Free body diagram of a block resting on table

Mg

N

Fig. 5

(ii) Free body diagram of bodies in contact and moving together on smooth surface.







F F

N

N1

M1g

N

N2

M2g

M1 M2

Fig. 6

Note that, normal force is taken normal to the surface of contact and towards the body

under consideration

(iii) Free body diagram of bodies connected with strings and moving under the action of external force, on a smooth surface.

F M2 M1

M2g

N1

T

M1g

N2

F T

Fig. 7

Note that, tension is acting along the string and away from the body under consideration.

Step 3: Identify the direction of acceleration and resolve the forces along this direction and perpendicular to it.

Step 4: Find net force in the direction of acceleration and apply F = Ma to write equation

of motion in that direction. In the direction of equilibrium take net force zero.

Step 5: If needed write relation between accelerations of bodies given in the situation

Step 6: Solve the written equations in steps 4 and 5 to find unknown accelerations and

forces.

Illustration 4

Questions: A body suspended with the help of

strings.

A body of mass 12.5 kg is suspended with

the help of strings as shown in figure.

Find tension in the string connected with

12.5 kg block. Strings are light

[g = 10 ms2]

37° 53° a c

b

d

12.5 kg

Solution: Let the tensions in strings ab, bc and bd are respectively T1, T2, and T3. As the body is

hanging in equilibrium, we can use the condition that net force on block is zero. This will

give the value of T3. To know the values of T1 and T2 we need to draw FBD of knot b also.







T3

12.5 g

(FBD of hanging body)

b

T2cos53°

T3

T1cos37°

T1

T2

T1sin

37

°

T2sin

53

°

(FBD of knot b)

For equilibrium of hanging body. T3 = 12.5 g = 125 N

Illustration 5

Questions: Motion of a block on a frictionless incline

A block of mass M = kg3 is placed on a

frictionless, inclined plane of angle = 300, as

shown in the figure.

Determine the acceleration of the block after it

is released. What is force exerted by the

incline on the block? [take g = 10 m/s2]

M

= 300

Solution: When the block is released, it will move down the incline. Let its acceleration be a. As the

surface is frictionless, so the contact force will be normal to the plane. Let it be N.

Here, for the block we can apply equation for

motion along the plane and equation for

equilibrium perpendicular to the plane.

i.e., Mg sin = Ma

a = g sin = 10 sin 300 = 10 1/2

a = 5 m/s2

Also, Mg cos – N = 0

N = Mg cos = 2

3103

N = 15 N

N

Mg Mg cos

Mg sin

a

(FBD of Block)

Illustration 6







Question: One block pushes other

Two blocks of masses M1 = 2 kg and M2 = 4kg

are placed in contact with each other on a

frictionless horizontal surface as shown in

figure. A constant force F = 24 N is applied on

M1 as shown. Find magnitude of acceleration of

the system. Also calculate the contact force

between the blocks.

F

M1 M2

Solution: Here accelerations of both blocks will be same as they are rigid and in contact. As the

surfaces are frictionless, contact force on any surface will be normal force only. Let the

acceleration of each block is a and contact forces are N1, N2 and N as shown in free body

diagrams of blocks.

F

N1

N

M1g

N

N2

M2g

(FBD of M1) (FBD of M2)

Applying, Newton‟s Second Law for M1

F – N = M1a … (i)

M1g – N1 = 0 … (ii)

Applying, Newton‟s second law for M2

N = M2a …(iii)

M2g – N2 = 0 … (iv)

Solving (1) and (3) a = 21 MM

F

42

24 4 m/s

2

N = 21

2

MM

FM

42

244 16 N

Illustration 7

Question: Bodies connected with strings

A light, inextensible string as shown in

figure connects two blocks of mass

M1 = 2kg and M2 = 8kg. A force F = 20 N

as shown acts upon M1. Find

acceleration of the system and tension

F M1 M2







in string.

Solution: Here as the string is inextensible, acceleration of two blocks will be same. Also, string is

massless so tension throughout the string will be same. Contact force will be normal force

only.

Let acceleration of each block is a, tension in string is T and contact force between M1 and

surface is N1 and contact force between M2 and surface is N2.

a

T

N2

M2g

(FBD of M2)

a

F

N1

M1g

T

(FBD of M1)

Applying Newton‟s second law for the blocks;

For M1 , F – T = M1 a … (i)

M1g – N1 = 0 … (ii)

For M2 T = M2a … (iii)

M2g – N2 = 0 … (iv)

Solving (i) and (iii)

21 MM

Fa

28

20

= 2 m/s

2

and 82

208

21

2

MM

FMT = 16 N







Illustration 8

Question: Bodies connected with a string and the string

passes over a pulley. (Atwood’s Machine)

Two blocks of unequal masses M1 = 3 kg and

M2 = 2kg are suspended vertically over a

frictionless pulley of negligible mass as shown

in figure. Find accelerations of each block and

tension in the string. [take g = 10 m/s2]

M2

M1

Solution: As the string is inextensible, the magnitude of acceleration of two blocks will be same.

Pulley in question is massless and frictionless so tension in strings on two sides of pulley

will be same.

Let acceleration of M1 be „a‟ (downward) then acceleration of M2 will be „a’ (upward).

Let the tension in string be T.

a

T

M1g

(FBD of block M1)

a

T

M2g

(FBD of block M2)

Applying Newton‟s second law for the blocks,

For M1, M1g – T = M1a … (i)

For M2, T – M2g = M2a … (ii)

Solving equation (i) & (ii),

a =

1023

23

21

21

MM

g)M(M

a = 2 m/s2

and gMM

M2MT

21

21

10

23

232

T = 24 N







Illustration 9

Question: Weighing a body in an Elevator

A block of mass M = 2 kg is suspended

with the help of a spring balance. The

spring balance is attached to the ceiling

of an elevator moving with upward

acceleration a0 = 1 m/s2 as shown in

figure. What is reading of spring

balance? [take g = 10 m/s2]

M

a0

Solution: A person outside the elevator will observe

the block moving with the elevator upward

with an acceleration a0. Also spring balance

will give the reading according to tension in

spring. So calculating reading of spring

balance means to find tension in the spring

of spring balance.

Let tension in spring is T.

Applying Newton‟s Second law for the block,

a0

Mg

T

(FBD of block)

T – Mg = Ma0 T = M (g + a0) = 1102 T = 22 N

This will be the reading of spring balance. Note that the reading given by spring balance is

different from the weight of block.

2.4 RELATED MOTION

Till now we had seen the case when accelerations of the different parts of a system are

same. There are situations in which the accelerations of different parts of the system may not be

same. We get such situations in case of moveable pulleys or bodies in contact where each body is

free to move.







B A

P

Q

Fig. 8

B

A

Fig. 9

For example in the figure 8, pulley P is movable which leads to different accelerations of

block B and A.

In the figure 9, triangular wedge A and sphere B will not have same acceleration.

In such cases, a relationship between accelerations can be found by considering physical

properties of system. We call such relations as constrained relation.

Illustration 10

Question: Find the ratio of acceleration of blocks A

and acceleration of block B.

B A

Solution: The physical property that we can use is the

inextensibility of string.

i.e., ab + bc + cd + de + ef = constant … (i)

Let at any moment A and B are at distances

xA and xB from the support as shown in

figure.

Let us take gh = 1 and i k = 2 and express

the length in equation (i) in terms of xA, xB, 1

and 2

we get,

a

b c g

h

B

A

f

xA

d e

i

k xB







xB – 1 + bc + (xB – 1 – 2)+ de + (xA – 2) = constant

Here 1, 2, bc and de are constant

2xB + xA = constant … (ii)

let in time t, xB change to xB + xB and xA changes to xA - xA

[B is assumed to move downward]

then, 2(xB + xB) + (xA – xA) = constant … (iii)

From (ii) and (iii)

2xB - xA = 0

Also,

t

x

t

x AB2= 0

2VB – VA = 0

Also, 2VB – VA = 0

t

VB

2

t

VA

= 0

2aB = aA B

A

a

a2

NOTE: Hence magnitude of acceleration of A is two times magnitude of acceleration of B.

Here we get the relation between the acceleration by using the inextensibility of string but

after some practice such relation can easily be written by observation.

Let us think B moves by a distance x during an interval of time, this will cause movement of

pulley g by x. an extra length of 2x of string will come to the left of pulley k. This must be

coming from right side of pulleys. Hence displacement of A will be 2x. On the basis of this

discussion we can say if the acceleration of block B is a, then the acceleration of A will be

2a.

Illustration 11

Question: In the arrangement shown A is a wedge and B

is a rod. The rod is constrained to move

vertical. The acceleration of wedge A is aA and

that of rod B is aB. Find B

A

a

a.

=450

A

B







Solution: Here the physical property that we can use is the

rigidity of body. Let xA and xB are displacement of

wedge and rod as shown in figure.

In pqr, qr = xA and pr = xB

tan = A

B

x

x

xB = xA tan

VB = VA tan

aB = aA tan

p

r

xA

xB

q

=450

B

A

a

a045tan

1

B

A

a

a1







PROFICIENCY TEST - II

The following questions deal with the basic concepts of this section. Answer the

following briefly. Go to the next section only if your score is at least 80%. Do not consult

the Study Material while attempting these questions.

1. In a tug-of-war between two athletes, each pulls on the rope with a force of 200 N. What is

tension in the rope?

2. Find tension T1 and T3 in the cords shown in the figure [g = 10 ms2]

30° T1

T2

T3

10 kg

3. Define free body diagram.

4. Draw free body diagram of bodies in the given system. All surfaces are frictionless and strings are massless.

M1 M2 M3

F1

F1

Fig. (a) Fig. (b)

M2 M3 M1 F2 F2

5. Write the advantage of following physical properties given in a system.

(a) rigidity of body

(b) inextensibility of string

(c) masslessness of string

(d) frictionless surface







ANSWERS TO PROFICIENCY TEST - II

1. 200 N

2. T1 = 200 N, T3 = 100 N







SOLVED OBJECIVE EXAMPLES

Example 1:

If a force

F = ( k10j4i3 ˆˆˆ ) N produces an acceleration of 1 m/sec2 in a given mass, then

the mass is

(a) 55 kg (b) 10 kg (c) 27 kg (d) 125 kg

Solution:

Magnitude of the force = |

F | = N125)10()4()3( 222

Mass is equal to 125 kg = 55 kg

(a)

Example 2:

Three equal weights A, B & C each of mass 2kg are

hanging on a string passing over a fixed frictionless pulley

as shown in figure. The tension in string connecting B and

C is nearly

(a zero (b) 13 N

(c) 3.3 N (d) 19.6 N

A B

C

Solution:

For mass C, resultant force on C = 2g T2

2g T2 = 2a … (i)

For mass B, 2g + T2 T1 = 2a … (ii)

For mass A, T1 2g = 2a … (iii)

Adding (1), (2) and (3)

a 36

2 gg

A B

C

T1 T1

2 kg

2g T2

2kg

2 g

2 g

2 kg

T2







Tension, T2 = 2g 2a = 2g 3

2g

3

2.39

3

8.94

3

4

g = 13 N

(b)







Example 3:

A rope of length L is pulled by a constant force F. What is the tension in the rope at a

distance x from one end where the force is applied?

(a) x

FL (b)

L

x)(LF (c)

x)(L

FL

(d)

x)(L

Fx

Solution:

Let AB be a rope of length L. Let F be the force onstant force acting on end B

pulling the rope to the right.

Mass per unit length of rope = L

M where M is the total mass.

F

P

A B

Let P be a point at a distance x from B. If T is the tension in the rope at P then for the part AP, the

tension is towards right while for the part PB it is towards left. If a is the acceleration produced in the

rope, then for part PB

F – T = mass of PB a

F – T = aL

Mx

Also for the rope, F = Ma

L

X)F(L T

(b)

Example 4:

A block of mass M is pulled along a horizontal

frictionless surface by a rope of mass m. The force P is

applied at one end of the rope. What is the force, which

the rope exerts on the block?

P M

(a) mM

P

(b)

M)M(m

P

(c)

mM

PM

(d)

mM

PM

Solution:

The situation is shown in Figure. Let a be the common

acceleration of the system.

Here T = Ma for block.

P T = ma for rope.

P T

a

T







P Ma = ma

P = a (M + m) amM

P

T = m)(M

MP

(c)







Example 5:

A body of mass 1 kg is suspended from a spring balance graduated at g = 10 m/s2. The

spring balance is fixed in a lift, which is moving up with an acceleration of 5 m/s2. What is the

reading in the spring balance?

(a) 0.5 kg (b) 1.5 kg (c) 1 kg (d) 3.5 kg

Solution:

The tension in spring balance T is given by

T mg = ma T = m (g + a) newton

= kg

1

g

g

amg

= kg10

51

= 1.5 kg

(b)

Example 6:

With what acceleration ‘a’ should the box in the figure

descend so that a body of mass M placed in it exerts a

force 4

Mg on the base of the box?

(a) 4

3g (b)

4

g

(c) 2

g (d)

8

g

a

Box

Solution:

If the box is accelerated downwards, from the frame outside the

elevator, equation of motion can be written as

Mg N = Ma

Here, 4

MgN

4

3ga

(a)

a

N Mg

FBD of block

Example 7:

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 g N (c) 9000 N (d) 9000 g N

Solution:

For the elevator

T 6000g = 6000 a = 6000 2

g

T = 9000g N

(d)

Example 8:

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

Solution:

a = 721

21 gg

mm

mm

7m1 7m2 = m1 + m2

6m1 = 8m2

or, 2

1

m

m =

3

4

(c)

Example 9:

A force F1 acting on a free mass m at rest produces in it an acceleration of 1 m/s2. Another

force F2 acting on the same mass at rest can produce in it a velocity of 10 m/s after 5 s. The

greatest acceleration of the mass m when both forces F1 and F2 act on it together will be

(a) 2 m/s2 (b) 4 m/s

2 (c) 3 m/s

2 (d) 1 m/s

2

Solution:

F1 produces an acceleration of 1 m/s2.

F2 produces a velocity 10 m/s after 5 s. If acceleration produced by F2 is a then,

v = at, 10 = 5a, a = 2 m/s2

Hence both together can produce a maximum acceleration of 3 m/s2

(c)







Example 10:

A body (mass 0.5 kg) is constrained to move Eastwards. A force of 20 N acts on the body

directed 30° east of north. The acceleration produced in the body due to the force will be

(a) 40 m/s2 (b) 20 m/s

2 (c) 320 m/s

2 (d) zero

Solution:

The body is constrained to move only in the East direction. Hence only the component of 20 N in the

East direction can be effective on the body. The component force is 20 cos 60° = 10 N. This is

acting on mass of 0.5 kg will produce an acceleration of 10

0 5. = 20 m/s

2.

(b)







SOLVED SUBJECIVE EXAMPLES

Example 1:

An aeroplane, which together with its load has a mass M =9600kg, is falling with an

acceleration of a = 5 m/s2. If a part of the load equal to m kg be thrown out, the aeroplane will

begin to rise with an acceleration of a = 5 m/s2. Find the value of m?

Solution:

Given M is the mass of the aeroplane.

Let R be the upthrust acting on it. Since it is falling down with an acceleration a,

Mg R = Ma ... (i)

Let a mass m kg be thrown out. The remaining mass is (M m) kg and now the plane

begins to rise up with an acceleration a m/s2.

Now R (M m) g = (M m)a ... (ii)

Adding equations (i) and (ii),

mg = (2M m)a

or, m (g + a) = 2Ma

m kgga

2Ma

=

105

596002

= 6400 kg

Example 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 g and F = 3.4 N

m1 m2

F

Solution:

When the pulley moves a distance d, m1 will move a distance 2d. Hence m2 will have twice as large

an acceleration as m2 has. Also because the total force on the pulley must be zero,

T1 = (T2/2).







m 1 m 2

F T 2

T 1

T 1

For mass m1, T1 = m1 (2a) … (i)

For mass m2, F – T2 = m2 (a) … (ii)

Putting T1 = 2

2T, (i) gives T2 = 4m1 a

Substituting in equation (ii), F = 4m1a + m2a = (4m1 + m2)a

Hence a = 5.0)3.0(4

4.3

4 21

mm

F = 2 m/s

2

Example 3:

A light inextensible string passing over a smooth fixed pulley attaches two masses of

magnitudes m and xm. Find the product of two possible values of x if the acceleration of the

system is g/4.

Solution:

Two cases will arise according as x < 1 or x > 1

Case 1:

When x < 1, xm < m and the mass m will fall while the mass xm

will rise.

The equations of motion will be

for mass m, mg T = ma … (i)

for mass xm, T xmg = (xm) a … (ii)

xm m

T T

Adding, mg (1 x) = (1 + x) ma

or, g(1 x) = a (1 + x)

It is given a = 4

g. Putting this value,

(1 x) =

4

1 x

or, 5x = 3 Hence x = 5

3







Case 2:

When x > 1, xm > m and the mass xm will fall while mass m will rise. The equations of motion will be

for mass m, T mg = ma … (iii)

for mass xm, xmg T = (xm)a … (iv)

Adding, (x 1) mg = (x + 1) ma

Putting a = 4

g, 4(x 1) = (x + 1)

or, 3x = 5 giving x = 3

5

Thus the two possible values of x for which the acceleration of the system will be 4

g are

5

3 and

3

5.

Therefore their product is 3

5

5

3

= 1

Example 4:

A mass of 2 kg hangs freely at the end of a string, which passes over a smooth pulley fixed

at the edge of a smooth table. The other end of the string is attached to a mass M on the

table. If the mass on the table is doubled the tension in the string increases by one-half. Find

the mass M.

Solution:

The tension in the string is given by

T = gMm

mM

… (i)

In the second case M changes to 2M and T changes

to 2

3 T

T2

3g

Mm

Mm

)2(

)2( … (ii)

Dividing (i) by (ii), we get

2

12

3

2

Mm

Mm

Substituting m = 2 kg, M = 1 kg

m

M

a

T

T

a

mg







Example 5:

A mass of 7 kg and another of mass 2 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 2 kg mass. [take g = 10 m/s2]

A

B

2 kg

7 kg

Solution:

Tension is the same throughout the string. It is clear that M1

will descend downwards while M2 rises up. If the acceleration

of M1 is a downwards, M2 will have an acceleration 2a

upwards.

Now, M1g 2T = M1a

T M2g = M2 2a

or, M1g 2 M2g = a (M1 + 4 M2)

a = gMM

MM

21

21

4

2

10

247

227

a = 2 m/s2

acceleration of 2 kg mass = 2a = 4 m/s2

M 1 M

1 g

a

B

T

M 2 g

M 2

A

T T

2a

Example 6:

Two masses m and 2m are connected by a massless string,

which passes over a pulley as shown in figure. The masses

are held initially with equal lengths of the strings on either

side of the pulley. Find the velocity of masses at the instant

the lighter mass moves up a distance of 15m. The string is

suddenly cut at that instant. Calculate the time taken by

heavier mass to reach the ground. (g = 10 m/s2)

m 2m

30 m Ground

Solution:







The masses A and B of m and 2m respectively are initially along

the horizontal position through the line AB.

When the masses are left free, B comes down, A moves up with

acceleration a.

Now, a = 32

)2( g

mm

gmm

The initial velocities of both of them is zero.

When the lighter mass A moves up through a height 15 m, its

velocity v is given by

v = 153

1022 Sa 10 m/s

C

2m

13.08 m

A B

m

Both the masses A and B have the velocity of same magnitude 10 m/s. At this instant the string

snaps.

Calculation of the time taken by B to reach the ground

u = 10 m/s

a = 10 m/s2

S = (30–15) = 15 m

S = ut + 1

2 at

2 15 = 10 t +

2

1 10 t

2

Solving t = 1 s

Example 7:

A lift is going up. The total mass of the lift and the passengers is 1500 kg. The variation in the

speed of the lift is given in the graph.

3.6

0 2 10 12 t sec

Speed m

/s

What will be the tension in the rope pulling the lift at t equal to (i) 1 s (ii) 6 s and

(iii) 11 s? [take g = 9.8 m/s2]







Solution:

The velocity-time graph of the motion of the lift is given below.

0 2 10 12 t s

B A 3.6

v m/s

From t = 0 to t = 2 s, the lift moves up with uniform acceleration a = 2m/s8.12

6.3

If T be the tension in the rope pulling the lift with the passengers up, then

T Mg = Ma

or T = M (g + a)

(i) At t = 1 s, T = 1500 (9.8 + 1.8) N

= 17400 N

(ii) At t = 6 s, the lift moves with uniform speed of 3.6 m/s and hence a = 0

or T = Mg

= 1500 9.8 = 14700 N







(iii) At t = 11 s, the lift is decelerating.

The deceleration = 3 6

218

2.. m/s

and hence the tension T in the string is

T = M (g + a) = 1500 (9.8 1.8) = 12000 N

Example 8:

Two particles of masses m1 = 9kg and m2 = 18 kg

are placed on a smooth horizontal table. A string,

which joins them, hangs over the edge supporting

a light pulley, which carries a block of mass 16 kg.

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 block of mass 16kg.

[ take g = 10 m/s2]

9kg 18kg

16kg

Solution:

Let T be the tension in the string; a be the acceleration of the mass 18 kg; 2a be the acceleration of

mass 9 kg.

T = m 2a

The mass 16 kg will come down with an acceleration 2

3

2

2 aaa

2

316216

aTg

2

3169416

aag

3

8a m/s

2

the acceleration of 16 kg mass = a2

3 = 4 m/s

2

3 mg

9kg 18kg

a 2a

T T

2T 3 2 a

16kg

Example 9:







A body m1 of mass 9 kg and another body m2 of mass 6 kg are connected by a light

inextensible string. Consider a smooth inclined plane of inclination 30° over which one of

them can be placed while the other hangs vertically and freely. If t1 and t2 are the time taken

in dragging m1 and m2 up the whole length of the plane, find t2/t1.







Solution:

30°

m 2

m 1

(9 kg) m 1

m 2

(6 kg)

Case (i):

Let a1 be the acceleration of the system when 9 kg mass hangs freely and T the tension in the

string.

m1g T = m1a1

T m2g sin 30 = m2a1

g (m1 m2 sin 30) = a1 (m1 + m2)

5

2

15

6

15

2

169

1

ggg

a

Case (ii):

Let a2 be the acceleration of the system when 6 kg mass hangs freely and T the tension in the

string.

m2g T = m2a2

T m1g sin 30 = m1a2

g(m2 m1sin 30) = a2(m2 + m1)

= )96(2

196 2

ag

1030

32

gga

If S is the length of the plane,

In case (i), 211

2

1taS

In case (ii), 222

2

1taS 2

22211 tata

2

1

1

2

a

a

t

t2







Example 10:

Find the acceleration of rod B and the

wedge A in the arrangement shown in

figure, if the ratio of the mass of wedge to

that of rod equals n and there is no friction

between any contact surfaces.

=450

A

B







Solution:

Let m be the mass of rod B and M that of wedge nm

M

If acceleration of rod B and wedge A are aA and aB then

aB = aA tan … (i)

mg

N

FBD of rod

Mg

aA

N

N1

FBD of wedge

aB

=450

Writing equation for motion.

for rod, mg N cos = maB … (ii)

for wedge, N sin = MaA … (iii)

Solving equation (i), (ii) and (iii)

ncotαtanα

gtanαaB

ncotαtanα

gaA







MIND MAP

NEWTON’S LAWS OF MOTION

(Applicable for inertial frame only)

Law 1. (Defines force and Inertia)

Everybody remains at rest or continues to move with uniform velocity unless an external force is applied to it.

Law 2. Gives relation between force and acceleration

i.e., aMF ext

Law 3. When a body A exerts a force on another body B, B exerts an equal and opposite force on A. If one of these two forces is considered as action, then other will be reaction.

COMMONLY USED FORCES Normal force: Normal to the surfaces of contact and towards the body under consideration. Weight of body: Equals to Mg and acts vertically downward. Tension in string: along the string, away from the body under consideration.

Stepwise procedure to solve questions based on motion of connected bodies:

1. Identify the unknown forces and accelerations.

2. Draw FBD of bodies in the system.

3. Resolve forces in the direction of motion and perpendicular to it.

4. Apply F = aM in the direction of motion and F = 0 in the direction of equilibrium.

5. Write constraint relation if required and possible.

6. Solve the equations written in steps 4 and 5 to get the results.

COMMONLY USED ASSUMPTIONS

Rigid body

Inextensible string

Massless string

Massless and frictionless pulley







EXERCISE – I

AIEEE-SINGLE CHOICE CORRECT

1. A block of mass m is placed on a smooth inclined plane of inclination with the horizontal.

The force exerted by the plane on the block has a magnitude

(a) mg (b) mg/cos (c) mg cos (d) mg tan

2. A particle of mass 5 kg is pulled along a smooth horizontal surface by a horizontal string.

The acceleration of the particle is 10 ms-2. The tension in the string is

(a) 2 N (b) 50 N (c) 15 N (d) 10 N

3. The pulley in the diagram is smooth and light.

The masses of A and B are 5 kg and 2 kg. The

acceleration of the system is

(a) g (b) 3

7g

(c) 7

3 g (d)

7

1g

A

B

4. At a certain moment of time velocity of A is

10 m/s upward. The velocity of B at that time will

be

(a) 30 m/s downward (b) 20 m/s downward

(c) 10 m/s down ward (d) 5 m/s down ward

B

10 m/s A

5. The ratio of T1 and T2 is (see figure)

(neglect friction)

(a) 2:3 (b) 3:1

(c) 1 : 5 (d) 5 : 1

3 kg 12kg 15kg

F

T2 T1 300

6. A body is placed on a rough inclined plane of inclination . As the angle is increased from 0° to 90°, the contact force between the block and the plane

(a) remains constant (b) first remains constant then decreases

(c) first decreases then increases (d) first increases then decreases







7. Two weights w1 and w2 are suspended form the end of a light string passing over a smooth

pulley. If the pulley is pulled up at an acceleration g, the tension in the string connecting the

two weights will be

(a) 21

214

ww

ww

(b)

21

212

ww

ww

(c)

21

21

ww

ww

(d)

21

21

2 ww

ww

8. A ship of mass 3 × 107 kg initially at rest is pulled by a force of 5 × 104 N through a distance

of 3 m. Assuming that the resistance due to water is negligible, the speed of the ship is

(a) 1.5 m/s (b) 60 m/s (c) 0.1 m/s (d) 5 m/s

9. A fireman wants to slide down a rope. The breaking load for the rope is 75% of the weight

of the man. With what acceleration should the fireman slide down? (Acceleration due to

gravity is g).

(a) 4

g (b)

2

g (c)

4

3g (d) g

10. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown. The force on pulley by the clamp is given by

M

m

(a) 2 Mg (b) 2 mg (c) 22)( mmM g (d) 22)( MmM g

11. A body of mass 2 kg is acted upon by two force each of magnitude 1 N, making an angle of

60° with each other. The net acceleration of the body (in m/s2) is

(a) 0.5 (b) 1 (c) 2

3 (d)

3

2

12. A body of mass 10 kg moves in positive x direction at a constant speed of 10 m/s. A

constant force then acts for 4 second on the body and gives it a speed of 2 m/s in opposite

direction. The force acting on the body is

(a) –30 N (b) –20 N (c) –10 N (d) –15 N







13. Same spring is attached with 2 kg, 3kg and 1 kg blocks in three different cases as shown in the figure. If x1, x2 and x3 be the extensions in the spring in these three cases respectively then

2kg 2kg 3kg 2kg 1kg 2kg

(1) (2) (3)

(a) 231 ,0 xxx (b) 312 xxx

(c) 213 xxx (d) 321 xxx

14. With what acceleration „a‟ should the box of figure

moving up so that the block of mass M exerts a force

7Mg/4 on the floor of the box?

(a) g/4 (b)g/2 (c) 3g/4 (d) 4g

(c) 3g/4 (d) 4g

a M

15. Two blocks of masses 5 kg and 2 kg are initially at rest

on the floor as shown in figure. A light string, passing

over a light frictionless pulley, connects them. An

upward force F is applied on the pulley and maintained

constant. Find the maximum value of F applied so that

the accelerations of 5 kg is zero

(g = 10 ms-2)

(a) 50 N (b) 100 N.

(c) 200 N. (d) none

2 kg 5 kg

F







16. In the figure, the block A, B and C

of mass m each, have

acceleration a1, a2 and a3

respectively. F1 and F2 are

external forces of magnitudes

2 mg and mg respectively. Then

(a) a1 = a2 = a3

(b) a1 > a3 > a2

(c) a1 = a2, a2 > a3

(d) a1 > a2, a2 = a3

F1 = 2mg

m

A

m

B

2m

B

m

C

m

F2 = mg

17. A bead is free to slide down a smooth wire tightly

stretched between points A and B on a vertical circle. If

the bead starts from rest at A, the highest point on the

circle

(a) its velocity v on arriving at B is proportional to sin

(b) its velocity v on arriving at B is proportional to tan

(c) time to arrive at B is proportional to cos

(d) time to arrive at B is independent of

B

A

18. In the given figure, the acceleration of block A with

respect to ground is (Neglect friction)

(a) 3

g (b) 10

3

g

(c) 3

2g (d) g

m C

m

B

m

A







19. In the arrangement shown in figure if the surface is smooth, the acceleration of the block m2 will be

(a) 21

2

4 mm

gm

(b)

21

2

4

2

mm

gm

(c) 21

2

4

2

mm

gm

(d)

21

12

mm

gm

m1

m2

20. On a smooth plane surface (figure), two blocks A and B are accelerated by applying a force 15 N on A. If mass of B is twice that of A, the force on B is

A

B

15N

(a) 30 N (b) 15 N (c) 10 N (d) 5 N

21. A particle stays at rest as seen in a frame. We can conclude that

(a) the frame is inertial

(b) resultant force on the particle is zero

(c) the frame may be inertial but the resultant force on the particle is zero

(d) the frame may be noninertial but the resultant force on the particle is zero

22. The force exerted by the floor of an elevator on the foot of a person standing there is more

than the weight of the person if the elevator is

(a) going up and slowing down (b) going up and speeding up

(c) at rest (d) going down and speeding up.

23. Action and reaction

(a) act on two different objects (b) have equal magnitude

(c) have opposite directions (d) all are correct







24. A block of mass 10 kg is suspended through two light spring

balances as shown in figure.

(a) Both the scales will read 10 kg

(b) Both the scales will read 5 kg

(c) The upper scale will read 10 kg and the lower zero

(d) The readings may be anything but their sum will be 10 kg

10 kg

25. In the situation shown in figure, the tension in the string connecting the two blocks will be

(string is massless and frictional force is negligible)

4kg 6kg 30N 10N

(a) 20 N (b) 25 N (c) 10 N (d) 18 N







EXERCISE – II

IIT-JEE- SINGLE CHOICE CORRECT

1. Two bodies A and B each of mass M are attached by a

massless spring. At that instant when acceleration of A is

a, force F acts on the mass B as shown in figure, the

acceleration of B at that instant is

A B F

(a) aM

F (b) a (c) -a (d)

M

F

2. From the fixed pulley, masses 2 kg, 1 kg and 3 kg are

suspended as shown is figure. Find the extension in the

spring when acceleration of 3kg and 1kg is same if spring

constant of the spring k = 100 N/m. (g = 10 m/s2)

(a) 10 cm (b) 20 cm

(c) 30 cm (d) 25 cm

2 kg

3 kg

1 kg

3. Three blocks A, B and C are suspended as shown in

figure. Mass of each of blocks A and B is m. If system is in

equilibrium and mass of C is M, then

A C B

(a) M = 2 m (b) M < 2 m (c) M > 2 m (d) M > 3 m

4. The horizontal force F to be applied on the triangular

wedge of mass M so that the block of mass m placed on it

appears stationary w.r.t. wedge, is (neglect friction)

m

M F

(a) tanmg (b) tangmM (c) cosgmM (d) singmM

5. A light string fixed at one end to a clamp on ground passes

over a fixed pulley and hangs at the other side. It makes

an angle of 300 with the ground. A monkey of mass 5 kg

climbs up the rope. The clamp can tolerate a vertical force

of 40 N only. The maximum acceleration in upward

direction with which the monkey can climb safely is

a

300







(neglect friction and take g = 10 m/s2)

(a) 2 m/s2 (b) 4 m/s2 (c) 6 m/s2 (d) 8 m/s2

6. In the diagram shown, what should be the value of force F applied so that the whole system can remain in the state of rest? Both the wedges are of the same mass M and the

angle inclination is . (Neglect friction )

(a) Mg tan (b) Mg/tan

(c) 2Mgtan (d) Mgsin

F F

7. A balloon of mass M is descending at a constant acceleration . When a mass m is

released from the balloon it starts raising with the same acceleration . Assuming that its

volume does not change, what is the value of m?

(a) g

M (b)

g

2M (c)

gM (d)

2

gM

8. A block of mass 3 kg is kept on a smooth plane. A variable force of magnitude (3t) N is applied on the block as shown in the figure at t = 0, where t is the time in second. Horizontal acceleration of the block when it losses the contact with the plane is

300

F=3t N

(a) 320 m/s2 (b) 310 m/s2 (c) 35 m/s2 (d) none of these

9. Two blocks A and B of masses 5 kg and 7 kg are connected by a heavy rope of mass 2 kg as shown in figure. An upward force of 200 N is applied on A. Then the tension at the middle of rope is

(a) 114.3 N (b) 112.2 N

(c) 10.3 N (d) none of these

A 5 kg

7 kg

2 kg

B

200 N







10. A wire ABC supports a body of mass m as shown in the

figure. The wire passes over a fixed pulley at B and is firmly

attached to a vertical wall at A. The line AB makes an angle

with the vertical, and the pulley at B exerts on the wire a

force F, then the value of force F is

(a) 2 mg sin 2

(b) 2 mg cos

2

(c) 3 mg cos 2

(d) none of these

m

F

B

C

A

11. In the figure, the block A, B and C

of mass m each, have

acceleration a1, a2 and a3

respectively. F1 and F2 are

external forces of magnitudes 2

mg and mg respectively. Then

(a) a1 = 2g

(b) a2 = g/3

(c) a3 = g

(d) a1 = a2 = a3 = g

F1 = 2mg

m

A

m

B

2m

B

m

C

m

F2 = mg

12. A particle P is sliding down a frictionless hemispherical

bowl. It passes the point A at t = 0. At this instant of

time, the horizontal component of its velocity is v. A

bead Q of the same mass as P is ejected from A at

t = 0 along the horizontal string AB, with the speed v.

Friction between the bead and the string may be

neglected. Let tP and tQ be the respective times taken

by P and Q to reach the point B. Then

(a) tP < tQ (b) tP = tQ

(c) tP > tQ (d) ABcordoflength

ACBarcoflength

t

t

Q

p

A

Q

B

P C







13. An inclined plane makes an angle 30° with the

horizontal. A groove OA = 5 m cut in the plane makes

an angle 30° with OX. A short smooth cylinder is free to

slide down under the influence of gravity. The time

taken by the cylinder to reach from A to O is

(g = 10 m/s2)

(a) 4 s (b) 2 s

(c) 22 s (d) 1 s

30°

cylinder A

30° O

X

14. In the arrangement shown in the figure neglect the

masses of the pulley and string and also friction. The

accelerations of blocks A and B are

(a) g, g/2 (b) g/2, g

(c) 3g/2, 3g/4 (d) g, g

m1 A

B m2

C

15. A man thinks about 4 arrangements as shown to raise two small bricks each having mass

m. Which of the arrangement would take minimum time?

m

m

F

(a)

F

m

m

(b)







m

m

F

(c)

F

m (d) m

16. A trolley is being pulled up an incline plane by a man

sitting on it (as shown in figure). He applies a force

of 250 N. If the combined mass of the man and

trolley is 100 kg, the acceleration of the trolley will be

[sin 15° = 0.26]

(a) 2.4 m/s2 (b) 9.4 m/s2

(c) 6.9 m/s2 (d) 4.9 m/s2

15°

250 N

17. In the given figure, the acceleration of block A with

respect to ground is (Neglect friction)

(a) 2

g (b)

2

2g

(c) 3

2g (d) 2g

2m C

m

B

m

A







18. Consider the three connected blocks shown in diagram. If the inclined plane is frictionless, and the system is in equilibrium, then force exerted by axle on the pulley is

m m

m

(a) mg (b) 2

mg (c) mg3 (d) 2mg

19. In the given figure, all strings and pulleys are ideal

and acceleration of m1 is 3

gm/s2 upward. Then find

the ratio of m1/m2.

(a) 3

1 (b) 1

(c) 2

1 (d)

4

1

m2 m1

20. For the system shown in figure, calculate force

exerted by the axle on the pulley. Assuming all

surface are frictionless (g = 10 m/s2)

(a) 40 2 N (b) zero

(c) 30 2 N, (d) 40 N

5 kg

20 kg

21. A big boulder of mass M has

fallen into a ditch of width 2d. Two

persons are slowly pulling it out

using a light rope and two fixed

pulleys as shown in Figure.

Assuming the force exerted by

two persons are equal, calculate

the force when the boulder is at a

depth h.

M

h

2d

(a) 22 44

hdh

Mg (b) 22 4

4hd

Mg (c) 22

2hd

h

Mg (d) 22

2hd

Mg







22. Two blocks of masses 5 kg and 2 kg are initially at rest

on the floor as shown in figure. A light string, passing

over a light frictionless pulley, connects them. An upward

force F is applied on the pulley and maintained constant.

Find the maximum value of F applied so that the

accelerations of 5 kg is zero (g = 10 ms-2)

(a) 50 N (b) 400 N

(c) 200 N (d) 100 N

5kg 20kg

F

23. A particle of mass m is resting on a wedge of angle as

shown in figure. The wedge is given an acceleration a.

The value of a so that the mass m just falls freely is

(a) g (b) g cot

(c) g sin (d) g tan

m

a

B

A

C

24. Three blocks of mass m, 2m and 3m are accelerating upward. If the normal force between 2m and 3m is k, then the value of force F applied on 3m is

(a) k (b) 6 k

(c) 3 k (d) 2 k

m 2m 3m

F

g

25. A trolley T of mass 5 kg on a horizontal smooth surface

is pulled by a load of 2 kg through a uniform rope ABC of

length 2 m and mass 1 kg. As the load falls from BC =0

to BC =2m, its acceleration (in m/s2) changes from

C

B A

T

(a) 5

30to

6

20 (b)

8

30to

8

20 (c)

6

30to

5

20 (d) none of these







EXERCISE – III

ONE OR MORE THAN ONE CHOICE CORRECT

1. A block of mass m slides down on a wedge of mass M as

shown in figure. Let 1a

be the acceleration of the wedge

and 2a

the acceleration of block. N1 is the normal reaction

between block and wedge and N2 is the normal reaction between wedge and ground. Friction is absent everywhere. Select the correct alternative(s)

m

M

(a) gmMN 2 (b) sin||cos 11 agmN

(c) ||sin 11 aMN

(d) 12 aMam

2. In the arrangement shown in figure all surfaces are smooth. Select the correct alternative(s).

(a) for any value of acceleration of A and B are equal

(b) contact force between the two blocks is zero if

mA /mB = tan

(c) contact force between the two is zero for any value of mA or mB

(d) normal reactions exerted by the wedge on the blocks are equal

B

A

fixed

3. Five concurrent forces are acting on a body. For the body to remain in equilibrium under these five forces;

(a) F = 10 N (b) F = 5 N

(c) 900 < < 1800 (d) 1800 < < 2700

3N 6N

F

4N

8N

4. A block of mass 1 kg is at rest relative to a smooth wedge moving leftwards left with constant acceleration a = 5 m/s2. Let N be the normal reaction between the block and the wedge. Then (g = 10 m/s2)

1kg

a

(a) N = 55 N (b) N = 15 N (c) 2

1tan (d) 2tan







5. Two blocks A and B of equal mass m are connected through a massless string and arranged as shown in figure. Friction is absent everywhere. When the system is released from rest

A

fixed 30

0 B

(a) tension in string is 2

mg (b) tension in string is

4

mg

(c) acceleration of A is 2

g (d) acceleration of A is g

4

3

6. The two ends of a spring are displaced along the length of the spring. All displacements

have equal magnitudes. In which case or cases the tension or compression in the spring

will have a maximum magnitude?

(a) the right end is displaced towards right and the left end towards left

(b) both ends are displaced towards right

(c) both ends are displaced towards left

(d) the right end is displaced towards left and the left end towards right

7. The acceleration of blocks of masses 5 kg and 10 kg are a1 and a2 respectively, then

(a) a1 = a2 = zero if F = 100 N

(b) a1 = 5 m/s2 and a2 = 0 if F = 300 N

(c) a1 = 15 m/s2 a2 = 2.5 m/s2 if F = 500 N

(d) acceleration of the masses is independent of F

A

B a1 a2

10kg 5kg m

F

8. Choose the correct statement/s from the following. No net force acts on

(a) a drop of rain falling vertically with a constant speed

(b) a wood piece floating on water

(c) a car moving with a constant velocity on a rough road.

(d) a body moving in a circular path at constant speed

9. A monkey of mass 40 kg climbs up/down on a rope, which can withstand a maximum

tension of 600N. The rope will not break if







(a) monkey climbs up with an acceleration of 6 m/s2

(b) monkey climbs down with an acceleration of 4 m/s2

(c) monkey climbs up with a uniform speed of 5 m/s

(d) monkey falls down the rope nearly freely under gravity







10. In which of the following situations a force of 9.8 N, would act on a stone of mass 1 kg?

Neglect air resistance.

(a) Just after it is dropped from the windows of a stationary train

(b) Just after if is dropped from the window of a train running at a constant speed of

36 km/m

(c) Just after it is dropped from the window of a train accelerating at 1 m/s2

(d) When it is lying at rest on the floor of a train which is accelerating at 1 m/s2

11. In the figure shown all the surfaces are smooth. All the strings are either horizontal or vertical. A horizontal force of magnitude F newton is acting at the end of the string. Select the correct alternatives if m = 1 kg

(a) acceleration of the block C is zero

(b) acceleration of the block C is 5

2Fm/s2

(c) net acceleration of the block B is 3

Fm/s2

(d) net acceleration of the block B is 5

Fm/s2

2m

2m C

A

B

m

F P1

P2 P3

P4

12. In figure-1 and figure-2 match A and block B are connected by mass less string and

acceleration of wedge A in both cases is towards right and equal to a with respect to earth

then select the correct alternatives

A

B

A

Figure-1 Figure-2

B

(a) magnitude of acceleration of B with respect to ground in the first figure is 2a

(b) magnitude of acceleration of B with respect to ground in the first figure is a

(c) magnitude of acceleration of B with respect to ground in the second figure is

cos610a







(d) magnitude of acceleration of B with respect to ground in the second figure is

cos106a

13. Two men of unequal masses hold on to the two sections of a light rope passing over a smooth light pulley. Which of the following are possible?

(a) the lighter man is stationary while the heavier man slides with some acceleration

(b) the heavier man is stationary while the lighter man climbs with some acceleration

(c) the two men slide with the same acceleration in the same direction

(d) the two men slide with accelerations of the same magnitude in opposite directions

14. A monkey of mass m kg slides down a light rope attached to a fixed spring balance, with an

acceleration a. The reading of the spring balance is W kg. [g = acceleration due to gravity]

(a) ag

Wgm

(b)

g

aWm 1

(c) the tension in the rope is Wg N (d) none of these

15. The blocks B and C in figure have mass m each. The strings AB and BC are light, having tensions T1 and T2 respectively. The system is in equilibrium with a constant horizontal force mg acting on C.

(a) 2

1tan 1 (b) 1tan 2

(c) mgT 51 (d) mgT 22

1

2

T1

T2

C

B

A

F=mg







EXERCISE – IV

MATCH THE FOLLOWING

Note: Each statement in column – I has one or more than one match in column –II.

1. Some observations are made inside a moving lift. Match the following

Column-I Column-II

I. The force exerted by the floor of the elevator is more than the weight of the person standing inside.

A. elevator is going up and slowing down

II. The force exerted by the floor of the elevator is less than the weight of the person standing inside.

B. elevator is going down and slowing down

III. The tension in the cable supporting the elevator is equal to the weight of the elevator and the person.

C. elevator is going up and speeding up.

IV. Newton‟s first law is violated D. elevator is going up and uniform speed.

E. elevator is moving horizontally

Note: Each statement in column – I has only one match in column –II

2. A heavy block C of mass M kept, on a frictionless surface and being pulled by two ropes A and B of equal

mass m. At ,0t force on the left rope )( 1F

is

withdrawn but the force on the right end )( 2F

continues

to act. Match the following based on the above

information. It is given .|||| 12 FF

1F

A B

m m

2F

C

M

Column-I Column-II

I. Magnitude of net force acting on the rope B at 0t is A.

Mm

mF

22

II. Magnitude of net force acting on the rope A at 0t is B.

mM

FFm

2

)( 12

III. Tension force applied by rope B on block C is (at )0t C.

mM

FFMm

2

))(( 12

IV. Tension force applied by rope A on block C is (at )0t D.

)2(

)( 2

Mm

FmM

E. )2(

)2( 2

Mm

FmM







3. A block of mass m = 5kg is put on a smooth triangular wedge fixed inside an accelerating box.

The acceleration of the box is )ˆ4( bjia

m/s2

where b is a positive constant. The acceleration of the box is such that block is not moving relative to wedge. Take 10g m/s2 and match the following.

Fixed wedge

Box

m

º45

Column-I Column-II

I. Contact force exerted by the wedge on the block is (in newton)

A. 6

II. A person makes free body diagram of block m relative to box. The magnitude of pseudo force applied by the person on the block must be (in Newton)

B. 7

III. The value of b is (in m/s2) C. 220

IV. The possible value of b for which block moves up the inline plane of wedge is (in m/s2).

D. 1310

E. 210

REASONING TYPE

Directions: Read the following questions and choose

(A) If both the statements are true and statement-2 is the correct explanation of

statement-1.

(B) If both the statements are true but statement-2 is not the correct explanation of

statement-1.

(C) If statement-1 is True and statement-2 is False.

(D) If statement-1 is False and statement-2 is True.

1. Statement-1: A body falling freely becomes weightless.

Statement-2: For freely falling body a = g, R = mg – ma R = 0.

(a) (A) (b) (B) (c) (C) (d) (D)

2. Statement-1: If you jump barefooted on a hard surface, your legs will get injured. But they

will not be injured if you jump on a soft surface like sand or pillow.

Statement-1: Change in velocity in less time requires high acceleration, which ultimately

results in high value of force.

(a) (A) (b) (B) (c) (C) (d) (D)







3. Statement-1: The mass of a body in a lift moving with constant acceleration changes

Statement-2: Mass of a body is a constant quantity and it does not change, when velocity of

the body is much less than velocity of light.

(a) (A) (b) (B) (c) (C) (d) (D)

4. Statement-1: Two objects of equal masses rest on the opposite pans of an arm balance.

Scale will remains balanced, when it is accelerated up or down in a lift.

Statement-2: Both masses experience unequal fictitious forces in magnitude as well as in

direction

(a) (A) (b) (B) (c) (C) (d) (D)

5. Statement-1: A car accelerates on a horizontal road due to the force exerted by the engine

of the car.

Statement-2: To accelerate a body force is always needed in the direction of required

acceleration.

(a) (A) (b) (B) (c) (C) (d) (D)

LINKED COMPREHENSION TYPE

A wagon is going horizontally with an acceleration a0 m/s2. A system inside the wagon contains two blocks of mass M and m connected by a thread which passes over a pulley as shown in the figure (all the surfaces are smooth)

a0 m/s2

M

m

1. Find the downward acceleration of mass m with respect to wagon.

(a)

Mm

agm

0 (b)

Mm

agM

0 (c) Mm

Mamg

0 (d) mM

Mamg

0

2. The tension in the string is given by

(a)

Mm

agmM

02 (b)

mM

agmM

02 (c)

mM

agmM

0 (d)

Mm

agmM

2

0

3. Magnitude of normal reaction between the block m and vertical wall when acceleration of

wagon became a0/2

(a) 2

0Ma (b)

3

0ma (c) ma0 (d)

2

0ma







4. What is the total reaction force on the pulley as seen from the ground?

(a)

Mm

agmM

02 (b)

mM

agmM

02 (c)

mM

agmM

2

0 (d)

Mm

agmM

2

0







EXERCISE – V

SUBJECTIVE PROBLEMS

1. Determine the accelerations of bodies A and B

and the tension in the cable due to application of

the 300 N force. Neglect all friction and the

masses of pulleys.

2. A string passing over a light frictionless pulley carries at its ends two variable unequal

masses whose sum is constant. If the breaking tension of string is 15/32 of the weight of

the sum of masses, show that least acceleration is g/4 and least value of greater mass is

5/8th of total mass.

3. In the arrangement shown in the figure the ball of mass m

can slide along the thread with some friction. The mass of rod

attached at other end is M. The mass of pulley and the friction

in its axle is negligible. At the initial moment the ball is located

opposite to the lower end of the rod. When set free, the ball

gets opposite to the upper end of the rod t second after the

beginning of motion. Find the frictional force between the ball

and the thread. The hanging portion of rod is of length l.

m

M l

4. In the arrangement shown in the figure, the mass of ball 1 is

= 1.8 times as great as rod 2. The length of the latter is

l = 100 cm. The mass of the pulley and the threads, as well

as the friction, is negligible. The ball is set on the same level

as the lower end of rod and then released. How soon will the

ball be opposite the upper end of the rod?

1 2 l







5. Two blocks A and B having masses m1 = 1 kg and

m2 = 4 kg are arranged as shown in figure. The pulleys P and

Q are light and frictionless. All blocks are resting on the

horizontal floor and pulleys are held such that strings remain

just taut. At moment t = 0, a force F = 30 t N starts acting on

the pulley P along vertically upward direction as shown in

figure; Determine

(a) the time when blocks A and B lose contact with ground,

(b) the velocity of A when B loses contact with ground,

(c) the height raised by A till this instant.

B A

F = 30 t N

P

Q

6. The monkey B shown in figure is holding on to tail of the

monkey, A which is climbing up a rope. The masses of monkey

A and B are 5 kg and 2 kg respectively. If A can tolerate a

tension of 30 N in its tail, what force should it apply on the rope

in order to carry the monkey B with it?

(g = 10 ms-2)

B

A

7. Figure shows three movable pulleys of masses m1, m2 and m3

connected by a single string. If the pulleys are frictionless,

string is light inextensible and pulleys P1 and P2 are light, find

tension in the string.

m3

m2

m1

P2 P1

8. In an arrangement shown in figure, the pulleys are light and

strings are mass less. Find acceleration of M. Neglect the

dimensions of pulleys and masses.

M

m1 m2

9. Two particles of equal masses m and m are connected by a

light string of length 2l as shown in figure. A constant force F is







applied continuously at the middle of the string, always along

the perpendicular bisector of the line joining the two particles.

Show that when the distance between the particles is 2 x, the

acceleration of approach of particles is 2/122 )( xl

x

m

Fa

F

m m

10. Figure shows a man of mass 60 kg standing on a light

weighting machine kept in a box of mass 30 kg. The box is

hanging from a pulley fixed to the ceiling through a light rope,

the other end of which is held by the man himself. If the man

manages to keep the box at rest, what is weight shown by the

machine? What force should be exert on the rope to get his

correct weight on the machine?







ANSWERS

EXERCISE – I

AIEEE-SINGLE CHOICE CORRECT

1. (c) 2. (b) 3. (c) 4. (a) 5. (d)

6. (b) 7. (a) 8. (c) 9. (a) 10. (d)

11. (c) 12. (a) 13. (b) 14. (c) 15. (b)

16. (b) 17. (d) 18. (b) 19. (a) 20. (c)

21. (c) 22. (b) 23. (d) 24. (a) 25. (d)

EXERCISE – II

IIT-JEE-SINGLE CHOICE CORRECT

1. (a) 2. (b) 3. (b) 4. (b) 5. (c)

6. (a) 7. (b) 8. (b) 9. (a) 10. (a)

11. (b) 12. (a) 13. (b) 14. (d) 15. (a)

16. (d) 17. (b) 18. (c) 19. (c) 20. (a)

21. (c) 22. (b) 23. (b) 24. (d) 25. (b)

EXERCISE – III

ONE OR MORE THAN ONE CHOICE CORRECT







1. (a,b,c) 2. (a,c) 3. (b,c) 4. (a,c) 5. (b,d)

6. (a,d) 7. (a,b,c) 8. (a,b,c) 9. (b,c,d) 10. (a,b,c)

11. (a,c) 12. (b,c) 13. (a,b,d) 14. (a,c) 15.(a,b,c,d)

EXERCISE – IV

MATCH THE FOLLOWING

1. I – B,C; II – A ; III – D ; IV – A,B,C, E

2. I – B ; II – B ; III –D ; IV – A

3. I – C ; II – D ; III – A ; IV – B

REASONING TYPE

1. (a) 2. (a) 3. (d) 4. (c) 5. (d)

LINKED COMPREHENSION TYPE

1. (d) 2. (c) 3. (d) 4. (b)

EXERCISE – V

SUBJECTIVE PROBLEMS

1. aA = 2.34 ms-2 ; aB = 1.558 ms-2; T = 81.8 N

3. Ffriction = 2)(

2

tmM

Mml

4. t = 1.4 sec







5. (a) tA = 1 sec, tB = 2 sec ; (b) vA = 5 ms1 ; (c) m3

5h

6. between 70 N and 105 N

7. T = 313221

321

4

4

mmmmmm

gmmm

8. g

m

M

m

M

36

481

21

upward

10. 15 kg, 1800 N




newton’s laws of mot

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