27
SOME USEFUL MATHEMATICAL FORMULAE ........................... 1-18 UNITS DIMENSIONS ERRORS AND INSTRUMENTS .............. 19-48 1 CHAPTER 0 CHAPTER VECTORS .................................................................................... 49-80 2 CHAPTER INDEX MOTION IN 1D .......................................................................... 81-158 3 CHAPTER 1.1 Fundamental Quantities ............................................................................................................................................... 19 1.2 Derived Quantities ....................................................................................................................................................... 19 1.3 The SI system of units ................................................................................................................................................... 19 1.4 Dimensions of a physical quantities ........................................................................................................................... 20 1.5 Errors in Measurement .................................................................................................................................................. 26 1.6 Vernier Callipers and Screw Gauge ............................................................................................................................. 35 2.1 Scalar quantity or scalar ............................................................................................................................................... 49 2.2 Vector quantity or vector .............................................................................................................................................. 49 2.3 Vectors operations ........................................................................................................................................................ 54 3.1 Concept of a Point Object ......................................................................................................................... 81 3.2 Rest and Motion are Relative Terms ........................................................................................................ 81 3.3 Motion ..................................................................................................................................................... 81 3.4 Motion Parameters ................................................................................................................................... 82 3.5 Uniform motion ........................................................................................................................................ 85 3.6 Equations of Motion ................................................................................................................................. 92 3.7 Motion with variable acceleration ........................................................................................................... 102 3.8 Problems based on maxima and minima ................................................................................................ 109 3.9 Study of motion by graphs ..................................................................................................................... 115 3.10 Relative Velocity : .................................................................................................................................. 136 MOTION IN TWO DIMENSION AND CONSTRAINT RELATIONS ............................................ 159-244 4 CHAPTER 4.1 Introduction ............................................................................................................................................ 159 4.2 Position Vector and Displacement ......................................................................................................... 159 4.3 Average velocity ..................................................................................................................................... 160 4.4 Instantaneous velocity ............................................................................................................................ 160 4.5 Average acceleration ................................................................................................................................ 161 4.6 Instantaneous acceleration ....................................................................................................................... 161 4.7 Motion in a plane with constant acceleration .......................................................................................... 162 4.8 Relative velocity in Two Dimension ...................................................................................................... 162 4.9 Motion along a curved path .................................................................................................................... 175 4.10 Projectile Motion .................................................................................................................................. 182 4.11 Constraint Relations ............................................................................................................................... 221

IIT-JEE PHYSICS SUPER SUCCESS SERIES: MECHANICS 1 for IIT-JEE / D. C. Gupta

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The first book in the new Disha`s IIT-JEE PHYSICS SUPER SUCCESS SERIES, Mechanics 1 has been written in the flavour of the new pattern of IIT-JEE. The present book "Mechanics 1 for IIT-JEE" contains ten chapters. In each chapter, a large number of examples along with numerous method of analysis and short-cuts have been given in a systematic manner to assist the students to understand the concepts of the subject. The book has been enriched with inclusion of a number of latest and new problems in 6 variety of exercises, be it Multiple Choice Questions, More than one correct, Statement Questions, Passage & Matrix type Questions, Subjective Integer Type Questions, Subjective Questions. Further the book provides hints and solutions to each and every problem of the book. All the 6 kinds of exercises have been fully solved - Apenas 10 exemplares foram importados da Índia, garanta o seu agora mesmo !

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Page 1: IIT-JEE PHYSICS SUPER SUCCESS SERIES: MECHANICS 1 for IIT-JEE / D. C. Gupta

SOME USEFUL MATHEMATICAL FORMULAE ........................... 1-18

UNITS DIMENSIONS ERRORS AND INSTRUMENTS.............. 19-48

1CHAPTER

0CHAPTER

VECTORS .................................................................................... 49-802CHAPTER

I N D E X

MOTION IN 1D .......................................................................... 81-1583CHAPTER

1.1 Fundamental Quantities ............................................................................................................................................... 191.2 Derived Quantities ....................................................................................................................................................... 191.3 The SI system of units ................................................................................................................................................... 191.4 Dimensions of a physical quantities ........................................................................................................................... 201.5 Errors in Measurement .................................................................................................................................................. 261.6 Vernier Callipers and Screw Gauge ............................................................................................................................. 35

2.1 Scalar quantity or scalar ............................................................................................................................................... 49

2.2 Vector quantity or vector .............................................................................................................................................. 49

2.3 Vectors operations ........................................................................................................................................................ 54

3.1 Concept of a Point Object ......................................................................................................................... 813.2 Rest and Motion are Relative Terms ........................................................................................................ 813.3 Motion ..................................................................................................................................................... 813.4 Motion Parameters ................................................................................................................................... 823.5 Uniform motion ........................................................................................................................................ 853.6 Equations of Motion ................................................................................................................................. 923.7 Motion with variable acceleration ........................................................................................................... 1023.8 Problems based on maxima and minima ................................................................................................ 1093.9 Study of motion by graphs ..................................................................................................................... 1153.10 Relative Velocity : .................................................................................................................................. 136

MOTION IN TWO DIMENSIONAND CONSTRAINT RELATIONS ............................................ 159-2444

CHAPTER 4.1 Introduction ............................................................................................................................................ 1594.2 Position Vector and Displacement ......................................................................................................... 1594.3 Average velocity ..................................................................................................................................... 1604.4 Instantaneous velocity ............................................................................................................................ 1604.5 Average acceleration ................................................................................................................................ 1614.6 Instantaneous acceleration ....................................................................................................................... 1614.7 Motion in a plane with constant acceleration .......................................................................................... 1624.8 Relative velocity in Two Dimension ...................................................................................................... 1624.9 Motion along a curved path .................................................................................................................... 1754.10 Projectile Motion .................................................................................................................................. 1824.11 Constraint Relations ............................................................................................................................... 221

Page 2: IIT-JEE PHYSICS SUPER SUCCESS SERIES: MECHANICS 1 for IIT-JEE / D. C. Gupta

LAWS OF MOTION AND EQUILIBRIUM................................. 245-3625CHAPTER

CIRCULAR MOTION............................................................... 363-3986CHAPTER

5.1 Mass ............................................................................................................................................................................. 2455.2 Weight .......................................................................................................................................................................... 2465.3 Inertia ........................................................................................................................................................................... 2465.4 Linear Momentum ...................................................................................................................................................... 2465.5 Newton’s Laws of Motion ......................................................................................................................................... 2465.6 Impulse ......................................................................................................................................................................... 2485.7 Forces in Nature .......................................................................................................................................................... 2515.8 D' Alembert’s Principle ............................................................................................................................................. 2755.9 Newton’s Second Law in Non-inertial Frame ......................................................................................................... 2965.10 Friction ........................................................................................................................................................................ 3005.11 Static and kinetic friction ........................................................................................................................................... 3015.12 Conservation of linear momentum ............................................................................................................................. 3215.13 Position vector ........................................................................................................................................................... 3265.14 Displacement Vector ................................................................................................................................................... 3265.15 Moment of force or Torque .......................................................................................................................................... 3265.16 Equilibrium ................................................................................................................................................................. 3285.17 Lami’s theorem ............................................................................................................................................................ 3315.18 Varignon’s theorem..................................................................................................................................................... 331

6.1 Angular Velocity ......................................................................................................................................................... 3636.2 Angular acceleration .................................................................................................................................................. 3646.3 Relationship between Angular and Linear Parameters ........................................................................................... 3656.4 Acceleration in Circular Motion ............................................................................................................................... 3676.5 Centripetal acceleration ............................................................................................................................................. 3676.6 Centripetal force .......................................................................................................................................................... 3696.7 Centrifugal force .......................................................................................................................................................... 3696.8 Analysis of conical pendulum ................................................................................................................................... 3716.9 Coriolis Force ............................................................................................................................................................. 3736.10 Motion of Cyclist on Circular Road .......................................................................................................................... 3776.11 Banking of Road ......................................................................................................................................................... 3776.12 Motion on a plane circular path ................................................................................................................................ 3786.13 Motion in a Vertical Circle ......................................................................................................................................... 381

WORK, ENERGY AND POWER .............................................. 399-4607CHAPTER

7.1 Introduction ................................................................................................................................................................ 399

7.2 Work done by constant force ..................................................................................................................................... 399

7.3 Work done by variable force ...................................................................................................................................... 402

7.4 The principle of invariance : Work depends on frame of reference .......................................................................... 404

7.5 Kinetic energy ............................................................................................................................................................. 409

7.6 Work-Energy Theorem ................................................................................................................................................ 409

7.7 Potential energy ......................................................................................................................................................... 415

7.8 Mechanical energy ...................................................................................................................................................... 416

7.9 Principle of conservation of energy .......................................................................................................................... 420

7.10 Einstein's Mass-Energy Equivalence ....................................................................................................................... 420

7.11 Power ........................................................................................................................................................................... 421

7.12 Lifting machines ......................................................................................................................................................... 423

Page 3: IIT-JEE PHYSICS SUPER SUCCESS SERIES: MECHANICS 1 for IIT-JEE / D. C. Gupta

COLLISION AND CENTRE OF MASS .................................... 461-5208CHAPTER

GRAVITATION.......................................................................... 521-5709CHAPTER

8.1 Introduction ................................................................................................................................................................ 461

8.2 Types of collision ........................................................................................................................................................ 462

8.3 Analysis of 1-D or head-on elastic collision ........................................................................................................... 463

8.4 Perfectly inelastic collision in 1 D ........................................................................................................................... 464

8.5 Elastic oblique collision ........................................................................................................................................... 468

8.6 Centre of mass .............................................................................................................................................................. 487

8.7 Newton’s second law for system of Particles ........................................................................................................... 487

8.8 Centre of mass of n - particle system .......................................................................................................................... 489

8.9 Centre of mass of a rigid body .................................................................................................................................... 489

8.10 Centre of Gravity ......................................................................................................................................................... 492

8.11 Centroid ....................................................................................................................................................................... 493

9.1 Heliocentric Model ..................................................................................................................................................... 521

9.2 Kepler’s Laws ............................................................................................................................................................. 521

9.3 Newton’s Law of gravitation .................................................................................................................................... 523

9.4 Shell Theorem .............................................................................................................................................................. 524

9.5 Derivation of Kepler’s Laws ..................................................................................................................................... 524

9.5 Intensity of gravitational field or gravitational field strength ............................................................................... 529

9.6 Gravitational Potential Energy ................................................................................................................................ 533

9.7 Gravitational potential .............................................................................................................................................. 539

9.8 Escape velocity (Escape Speed) ............................................................................................................................... 543

9.9 Black Hole ................................................................................................................................................................... 545

9.10 Satellite ........................................................................................................................................................................ 546

9.11 Orbital velocity .......................................................................................................................................................... 546

9.12 Time period of revolution ........................................................................................................................................... 547

9.13 Energy of a satellite .................................................................................................................................................... 548

9.14 Binding Energy .......................................................................................................................................................... 548

9.15 Geostationary satellite (GSS) .................................................................................................................................... 549

9.16 Reduction of two-body problem to one-body problem: reduced mass .................................................................. 553

REVISION (FORMULAE AND IMPORTANT POINTS) ........... 571-61610CHAPTER

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5.1 MASSThe mass of a body is the quantity of matter contained in it. Its SI unit is kg. The mass of the body whichdetermines its inertia in translatory motion is called its inertial mass. This is the mass that appears in Newton’ssecond law, which can be written as

F = miai or mi = i

Fa

.

Fig. 5.1The mass of a body which determines the gravitational pull due to the earth is called its gravitational mass.This is the mass that appears in Newton’s law of gravitation, which we can write as

F = 2

GMm

Ror mg =

2FRGM

.

Fig. 5.2

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266

The principle of equivalence

Inertial and gravitational masses need not be equal to each other. Experiment shows that, to a precision ofperhaps 1 part in 1012, these two masses are identical. In Newtonian physics, the experimental fact that mi = mgcould be regarded as nothing but an astonishing coincidence. In Einstein theory of relativity, it enter in a naturalway through the principle of equivalence; if acceleration and gravitation are equivalent, then masses measuredvia acceleration and gravitation must be equal.

5.2 WEIGHT

The weight of a body is the force exerted by earth on the body towards the centre of earth. If g is the gravity atany place, then the weight of body at that place, W = mg. As the value of gravity changes from place to place,so the weight of a body is different at different places. The SI unit of weight is newton (N).

5.3 INERTIA

Inertia is the inherent property of material body by virtue of which it resists in change of state of rest or ofuniform motion. It is not a physical quantity, it is the sensation. Mass of a body is the measure of its inertia. Ifa body has large mass, it has more inertia. Different kinds of inertia are :(i) Inertia of rest : The tendency of a body to remain in its position of rest is called inertia of rest.(ii) Inertia of motion : The tendency of a body to remain in its state of uniform motion along a straight line is

called inertia of motion.(iii) Inertia of direction : The inability of a body to change by itself its direction of motion is called inertia of

direction.

5.4 LINEAR MOMENTUM

Momentum of a body is the amount of motion possessed by the body. Mathematically, it is equal to the productof mass and velocity of the body. Thus∴ momentum = mass × velocity

or P = vuur ur

m .

Momentum is a vector quantity, its direction is along the direction of velocity. Its SI unit is kg-m/s or N-s.

5.5 NEWTON’S LAWS OF MOTION

Sir Issac Newton (1642-1727) made systematic study of motion of objects and presented three laws of motionwhich are called Newton’s Laws of Motion. These are :First law : Every body remain in its state of rest or of uniform motion in a straight line unless it is compelled bysome external force to change that state.

Second law : The rate of change of linear momentum of a body is directly proportional to the applied force andthe change takes place in the direction of the applied force. That is

=P

F

uur uurddt

.

Third law : To every action, there is always an equal and opposite reaction. Action and reaction act on twodifferent bodies. For two bodies A and B, we can write

–=F Fuur uur

AB BA .

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267Laws of Motion & Equilibrium

More about Newton’s first law of motionNewton’s first law defines force : According to Newton’s first law of motion a body maintains its state of restor uniform motion unless an external force acts on it. This shows that force is an agent which changes the stateof rest or uniform motion. But sometime a force applied on a body can not change its state of rest or uniformmotion in a straight line. Hence first law of motion gives a qualitative definition of force.Newton’s first law defines inertia : According to Newton’s first law of motion, body remains in its state of restor uniform motion unless an external force acts on it. This shows that body by itself can not change its state.This inability of body to change its state of rest or of uniform motion in a straight line is called inertia of a body.Thus Newton’s first law defines inertia and hence it is also called the law of inertia.

More about Newton’s second law of motion

Newton’s second law gives the measurement of force. Consider a body of mass m moving with a velocity v ,its momentum

=P vuur ur

m .

Differentiating above equation w.r.t. time, we get

( )=P v

uur urd d mdt dt

=

vur

dm

dt = a

urm .

According to Newton’s second law

=P

F

uur uurddt

∴ =F auur ur

m .

Note :

(i) The Newton’s second law =F auur ur

m is strictly applicable to a single particle. The force Fuur

in the law

stands for the net external force. Any internal force in the system is not to be included in F .

(ii) The second law is a local relation. It means the force F at a point in space at a certain instant of time is

related to a at the same point at the same instant, Where a is the acceleration produced in the body..

Absolute unit of force : In SI, the absolute unit of force is newton (N).Thus1 N = 1 kg × 1 m/s2 or 1 N = 1 kg-m/s2

Gravitational unit of force : In SI , the gravitational unit of force is kilogram weight or kilogram force (kgf). Thus1 kg -wt = 1 kg-f = 1 kg × 9.8 m/s2 = 9.8 kg-m/s2,

or 1 kg-wt = 9.8 N

Note :The absolute unit of force remains the same every where. But the gravitational unit of force varies from placeto place because it depends on the value of g.

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268

Newton’s second law in component form : In terms of components, the momentum can be written as

ˆ ˆ ˆ= + +P i j kuur

x y zP P P .

Thus Newton’s second law in component form, can be written as

ˆ ˆ ˆ= + +F i j kr

x y zF F F

=pr

ddt

ˆ ˆ ˆ( )= + +i j kx y zd

P P Pdt

Equating the three components along the three coordinate axes, we get

= =xx x

dPF ma

dt,

= =yy y

dPF ma

dt,

= =zz z

dPF ma

dt.

5.6 IMPULSEA large force acting for a short time to produce a finite change in momentum is called an impulsive force. Whensuch a force acts for a short time, the product of the force and the time for which it acts is called impulse of force.Thus

Impulse = Force × time duration.

Suppose a force F acts for small time dt, the impulse of the force is given by,,

=J Fur uur

d dt .

For a finite interval of time, we can write

2

1

= ∫J Fur uurt

t

dt .

For constant force 2 1( )= −J Fur uur

t t

or = ∆J Fur uur

t .

SI unit of impulse is kgm/s or N-s.

Impulse - momentum theorem : According to Newton’s second law of motion, we have

=P

F

uuruur ddt

or =F Puur uur

dt d

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269Laws of Motion & Equilibrium

or 2 2

1 1

=∫ ∫Fuurt t

t t

dt dP

Here 2

1

=∫ F Juur urt

t

dt , is the impulse of force and

2

1

2 1= −∫ P Puur uurt

t

dP , is the change of momentum in time from t1 to t2. Thus

12 PPJ −= .

The avove relation is known as impulse momentum theorem.Calculation of impulse by graphical method(a) When a constant force acts on a body : Suppose a constant force F acts on a body from time t1 to t2. The

magnitude of impulseJ = F (t2 – t1) = Area under the force time curve between t1 and t2.

(a) (b)Fig. 5.3

(b) When a variable force acts on a body: Suppose a varying force acts for time t2 – t1 = t. The magnitude of

impulse of force J = 2

1

∫t

t

Fdt = Area under the force time curve between t1 and t2.

More about Newton’s third law of motionAccording to Newton’s third law of motion, every action is associated with an equal and opposite reaction.Thus in nature forces always occur between pairs of bodies. Thus two forces act simultaneously. Any one ofthem may be called the action and the other reaction. It is clear from third law that a single force can never exist.Following are few examples which are based on Newton's third law :1. Let us consider a block is suspended by the string. Earth exerts gravitational force on the block (let

action); the equal amount of force the block exerts on the earth (reaction). Similarly block does action onthe support through string, the equal amount of reaction is exerted by support on the block (reaction).Inthis case action and reaction are equal to mg.

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270

Fig. 5.4

2. While walking, we press the ground (action) with our feet slanted in the backward direction. The groundexerts an equal and opposite force on us. The vertical component of reaction force balances our weightand the horizontal component provides us forward motion.

Fig. 5.53. The horse and the cart : Let us consider horse starts pulling the cart from rest and gaining speed with time.

The FBD of whole system is shown in the fig. 5.6

Fig. 5.6Ground exerts force on the horse, whose horizontal component is FH (It is provided by friction). The forceexerted by the horse on the cart in forward direction is T. Cart exerts a force T on the horse in backwarddirection. In addition to this there is frictional force on the wheels of cart in backward direction. Thus

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271Laws of Motion & Equilibrium

(a) For motion of horse, we have

acceleration, aH = −H

H

F TM

…(i)

MH → mass of the horse.(b) For motion of cart, we have

aC = −

C

T fM

…(ii)

MC → mass of the cart.As both move together,∴ aH = aC

or − −= H

H C

F T T fM M

.

(c) For a system as a whole, we have

=+

H

H C

F fa

M M. [aH = aC = a]

4. Tug-of-war: In tug-of-war each team pulls the opposite team with equal force. But winning team exertsgreater force on the ground and hence ground provides equal reaction force. This is very clear from FBD.

Here wf f> l .

Fig. 5.75.7 FORCES IN NATUREAt the present stage of our understanding, we know about four fundamental forces in nature. These are :1. Gravitational force : It is the force of mutual attraction between the objects by virtue of their masses. It is

a universal force. The gravitational force acts over long distances and does not need any interveningmedium. Compared to other fundamental forces, gravitational force is the weakest force of nature.

2. Electromagnetic force : Electromagnetic force is the force between the objects due to charges on them. Itmay be attractive or repulsive. It acts over long distances and does not need any intervening medium. Itis very much stronger than gravitational force. The forces spring, tension or compression, friction areelectromagnetic in nature.

3. The strong nuclear force : The strong nuclear force binds protons and neutrons in nucleus. It does notdepend on charge and acts equally between a proton and a proton, a neutron and a neutron, and a protonand a neutron. Electron does not experience this force. It acts for very short distance, order of 10–15 m.

Note : Recent discovery indicated that the strong nuclear force between nucleons is not a fundamental force ofnature.

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272

4. The weak nuclear force : The weak nuclear force appears only in certain nuclear process such as the β -

decay of a nucleus. The weak nuclear force is not as weak as gravitational force, but much weaker thanstrong nuclear force. The range of weak nuclear force is very small, of the order of 10–15 m.Each fundamental force is thought to arise from the exchange of its characteristic particles.(i) The gravitational force is thought to be caused due to exchange of an undetected particles called

gravitons.(ii) Electromagnetic force arises due to the exchange of photons between the charged particles.

(i) Gravitational force (ii) Electromagnetic force

(iii) Strong nuclear force (iv) Weak nuclear forceFig. 5.8

(iii) The strong nuclear force arises from the exchange of mesons ( )−π .

(iv) Weak nuclear force arises from the exchange of bosons (W–).

EXAMPLE 1 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed of 6 m/s

collide and rebound with the same speed. What is the impulse imparted to each ball by theother ?

SOLUTION Figure shows the motion of balls A and B.For ball A :

Puur

i = 0.05 × 6 i Ns = 0.3 i Ns

and Puur

f = 0.05 × (– 6) i Ns = – 0.3 i Ns

Impulse imparted to ball A due to ball B

= −J P Pur uur uur

AB f i = – 0.3 i – 0.3 i = – 0.6 i Ns. Fig. 5.9

For ball B :

iP = 0.05 × (– 6) i = – 0.3 i Ns

and Puur

f = 0.05 × (6) i = 0.3 i Ns

Impulse imparted to ball B due to ball A

Jur

BA = Puur

f – Puur

i = 0.3 i – (– 0.3 i ) = 0.6 i Ns. Ans.

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273Laws of Motion & Equilibrium

EXAMPLE 2 A rubber ball of mass 50 g falls from a height of 1 m and rebounds to a height of 0.5 m. Find

the impulse and the average force between the ball and the ground if the time for which theyare in contact was 0.1 s.

SOLUTION Velocity of ball just before collision; v2 = 0 + 2gh

v = vi = 2gh = 2 9.8 1× × = 4.43 m/s

∴ Momentum of ball before collision,

=P vuur r

ii m

= 0.050 × (– 4.43) j Ns

= – 0.22 j Ns

Velocity of ball just after collision, vf = 2gh Fig. 5.10

= 2 9.8 0.5× × = 3.13 m/s

∴ Momentum of ball just after collision,

fPuur

= m fvur

= 0.050 × (3.13 j ) = 0.16 j Ns

Now impulse imparted by ground on the ball

J = P∆ = if −P Puur ur

= 0.16 j – (– 0.22 j ) = 0.38 j Ns

The force between them F = Pt

∆∆

= 0.380.1

= 3.8 N. Ans.

EXAMPLE 3 A machine gun fires a bullet of mass m with a speed of v m/s. The person holding the gun can

exert a maximum force F on it. What is the number of bullets that can be fired from the gunper second ?

SOLUTION : The change in momentum of each bullet∆P = m(v – u).

As u = 0,∴ ∆P = mv.If n is the number of bullets fired per second, then rate of change of momentum of the gun

Pt

∆∆

= n mv.

Thus by Newton′s second law

F =Pt

∆∆

= n mv

∴ n = .F

mvAns.

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274

EXAMPLE 4 Fig. 5.11 shows the position time graph of a particle of mass 4 kg. What is the (i) force

acting on the particle for t < 0, t > 4, 0 < t < 4 s ? (ii) impulse at t = 0 and t = 4 s ? Assume thatthe motion is one dimensional.

Fig. 5.11SOLUTION (i) For t < 0 and t > 4 s, the position of particle is not changing i.e. the particle is at rest. So

the force during these intervals is zero.For 0 < t < 4s, the position of particle is changing. But its velocity is constant as clearfrom x - t graph. Thus acceleration of particle is zero. Hence no force acts on the particleduring this interval also.

(ii) Just before t = 0, the particle is at rest, u = 0.Just after t = 0, the particle has constant velocity.

v = 34

m/s

∴ Impulse J = Change in momentum

= m (v – u) = 43

04

= 3 Ns

Just before t = 4s, the particle has constant velocity

u = 34

m/s

Just after t = 4s, the particle is at rest, so v = 0

∴ Impulse J = m (v – u) = 4 3

04

= – 3 Ns Ans.

EXAMPLE 5 Fig. 5.12 below shows the position-time graph of a particle of mass 0.04 kg. Suggest asuitable physical context for this motion. What is the time between two consecutive impulsesreceived by the particle? What is the magnitude of each impulse ?

Fig. 5.12

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275Laws of Motion & Equilibrium

SOLUTION Just before t = 2 s, the velocity of particle,

u = 2 02 0

−−

= 1 cm/s = 0.01 m/s.

Just after t = 2 s, the velocity of particle,

v = 0 24 2

−−

= – 1 cm/s = – 0.01 m/s.

The magnitude of impulse J = | m (v – u) | = | 0.04 (– 0.01 – 0.01) | = 8 × 10–4 Ns.

The given x - t graph may represent the repeated rebounding of a particle between two elasticwalls at x = 0 and x = 2 cm. The particle will get an impulse of 8 × 10–4 Ns after every 2 s.

EXAMPLE 6 Two identical billiard balls strike a rigid wall with the same speed but at different angles,

and get reflected without any loss of speed, as shown in fig. 5.13. What is (i) the direction ofthe force on the wall due to each ball ? and (ii) the ratio of the magnitudes of the impulseimparted on the two balls by the wall ?

Fig. 5.13SOLUTION (i) Let ball strikes the wall with a speed of u. If its mass is m, then its momentum

iPuur

= mu i

and ˆf mu= −P i

r

∴ Impulse x f i= −J P Pur uur uur

= – mu i – mu i

= – 2mu i .As there is no motion along y-axis, the impulse along this direction will be zero;

Jy = 0. Thus resulting impulse is J = x y+J Jur ur

= – 2mu i .

The direction of force is along the direction of impulse. So the force exerted by wall onball is along negative x-axis. By Newton’s third law the direction of force exerted by ballon wall is along positive x-axis.

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276

(ii) Resolve the velocity of the ball along x and y-axis, we have

Fig. 5.14

xuur

= u cos 30° i ; yur

= – u sin 30° j

and xvur

= – u cos 30° i ; yvur

= – u sin 30° j

The impulse exerted by the wall on the ball;

xJur

= m ( – )x xv uur ur

= m (– u cos30° i – u cos30° i )

= – 2mu cos 30° i ˆ3 mu= i

and yJur

= m ( )y y−v uur ur

= m [– u sin30° j – (– u sin30° j )] = 0.∴ Resultant impulse

J = x y+J Jur ur

= – 3 mu i .

The force exerted by wall on ball is along negative x-axis. By Newton’s third law thedirection of force on wall by ball is along positive x-axis.The ratio of impulse imparted by the wall on balls

1

2

JJ

= 23mumu

= 23

. Ans.

Problem - solving procedure by using Newton’s second law of motion :

1. In using Newton’s second law our usual unknown is acceleration. After knowing acceleration, we can getother unknowns of the problem.

2. System may have only a body and it is acted by a number of forces. Find the net force ( netFuur

) acting on thebody. Then use Newton’s second law for the body. If net force is along x-axis, then we have [Fx]net = max

3. (a) If there are many bodies in the system and they are placed or connected in such a way that all bodiesin the system have same magnitude of acceleration, then use Newton’s second law for the system asa whole Fnet = mtotala.

(b) After getting acceleration (a), draw FBD for each body in the system. Then use Newton’s secondlaw for each of them (or any of them) to get unknowns.

4. If system have number of bodies and they may have different accelerations, then first find constraintrelations (relationship between their accelerations). Draw FBD for each of the body in the system. After ituses Newton’s second law for each of them to get the unknowns.

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277Laws of Motion & Equilibrium

Normal reactionWhen a body is pressed against a surface, the body experiences a force that is perpendicular to the surface at thepoint of contact between the body and the surface. This force is called normal reaction and can be denoted by N.(i) In case when a block is placed on a table, the normal reaction on the block by the table is, N = mg.(ii) In case, when block is placed on an inclined plane the normal reaction force, N = mg cos θ .

(i)

(ii)

Fig. 5.15

TensionWhen a rope (string, cord etc.) is connected to a body and pulled taut, the rope is said to be under tension. Itpulls the body with a force T, whose direction is away from the body and along the length of the rope. A rope isusually regarded to be massless and unstretchable. The rope exists only as a connection between two bodies.It pulls on the body at each end with the same magnitude T. Here rope pulls the block and support each with aforce T as shown in fig. 5.16.

Fig. 5.16.

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278

Free body diagram (FBD) :To ensure correct use of the equations of statics or equations of dynamics, we isolate the body in a simplediagram and show all the forces from the surroundings that act on the body. Such a diagram is called freebody diagram. Following are few examples of FBD:

1.

Fig. 5.172.

Fig. 5.18

3.

Fig. 5.19

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279Laws of Motion & Equilibrium

Free bodies involving interior section : If any body is in equilibrium, then each part of thebody must be in equilibrium.Figure shows the FBD of the whole body and its one part.

(a) Whole body (b) One part of the body with interior sectionFig. 5.20

EXAMPLE 7 Two monkeys of masses 10 kg and 8 kg are moving along a

vertical rope as shown in fig. 5.21. The former climbing up withan acceleration of 2 m/s2, while the later coming down with auniform velocity of 2 m/s. Find the tension in the rope at thefixed support.

SOLUTION Let tension in the rope is T. By Newton’s second law, we haveT – (10g + 8g) = 10 × 2 + 8 × 0

or T = 18 g + 20 = 196.4 N. Ans. Fig. 5.21

EXAMPLE 8 Two disks of masses m1 and m2 are connected by a spring of

force constant k. The lower disk of mass m2 lies on a table andthe upper disk is vertically above it (see fig 5. 22). What verticalforce F should be applied to the upper disk so that when theforce is withdrawn, the lower disk is lifted off the table ?

Fig. 5.22

SOLUTION

Fig. 5.23

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280

The situation is shown in fig. 5.23. Let applied force F compresses the spring by x, so F = kx.After removal of force, the spring recovers its compressed length and further extends by x. Atthis instant the force exerted by spring is along upward direction. Thus to lifted of the lowerdisk, normal reaction N becomes zero. Consider the system of two disks together, we have

N = (m1 + m2)g – kx = (m1 + m2) g – F.

As we have, N = 0∴ Fmin = (m1 + m2)g.So to lift off the lower disk,

1 2( )F m m g≥ + .

Motion in a lift /elevatorConsider a man of mass m standing on a weighing machine placedin the lift. The actual weight of the man is mg (weight of man at rest).The reading of weighing machine indicates the force experiencedby it which is equal to the reaction on the man standing on it.Machine at rest with a man standing on it gives N = mg.

Fig. 5.24When a man is in an accelerated lift, his weight appears to change. This changed weight is known as apparentweight.1. (a) When the lift moves upward with acceleration a, we have

N ′ – mg = ma

or N ′ = m (g + a)

∴ Apparent weight, N ′ = m(g + a). (b) When the lift moves upward with retardation a, we have

N ′ – mg = m(–a)

or N′= m (g – a).Fig. 5.25

2. (a) When lift moves downward with constant acceleration a,

we have mg – N′ = ma

or N ′= m(g –a). (b) When lift moves downward with constant retardation a

we have, mg – N ′= m(–a)

or N ′ = m(g + a)3. If the supporting cable of the lift breaks, the lift falls freely

with an acceleration a = g. Thus we have, N = m(g – g) = 0. Fig. 5.26

Hence apparent weight of man in a freely falling lift becomes zero.

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281Laws of Motion & Equilibrium

EXAMPLE 9 Two identical point masses, each of mass m are connected to one another by a massless

string of length L. A constant force F is applied at the mid-point of the string. If l be theinstantaneous distance between the two masses, what will be the acceleration of each mass?

SOLUTION The situation is shown in fig. 5.27.Let T be the tension in the string. For point O, we have

2T sin θ = F

or 2sin

FT =θ

.

Now acceleration of each mass;

/x xa F m=

or cos

xT

am

θ=

( /2sin )cosF

mθ θ

= Fig. 5.27

cot2

Fm

θ=

2 2

/ 22 ( /2) ( /2)

Fm L

=−

ll

2 22Fm L

=

ll . Ans.

EXAMPLE 10 A particle of mass m, originally at rest, is subjected to a force whose direction is constant

but whose magnitude varies with the time according to the relation

-–

=

2

0 1t T

F FT

where F0 and T are constants. The force acts only for the time interval 2T.(a) Make a rough graph of F versus t.

(b) Prove that the speed v of the particle after a time 2T has elapsed is equal to 043F Tm

.

SOLUTION Given that

(a)2

0 1t T

F FT

− = −

At, t = 0, 2

00

1 0T

F FT

− = − =

t = T, 2

0 01T T

F F FT

− = − =

Fig. 5.28

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282

t = 2T, 2

02

1 0T T

F FT

− = − =

The graph of F versus t is shown in fig. 5.28.(b) We have acceleration of the particle

20 1

FF t Ta

m m T

− = = −

or

20 1

Fdv t Tdt m T

− = −

or 2 2

0

0

1T F t T

v dtm T

− = − ∫

23

0

03

Tt T

TF T

tm

− = −

3 3

0

2 0

2 03 3

T T TT T

F T TT

m

− − = − − −

043F Tm

= . Proved

EXAMPLE 11 At the moment t = 0 the force F = at is applied to a small body of mass m resting on a smooth

horizontal plane (a is constant). The permanent direction of this force form an angle α withhorizontal. Find :(a) The velocity of the body at the moment of its breaking off the plane.(b) The distance travelled by the body upto this amount.

SOLUTION (a) The normal reaction on the body N = mg – F sin α = mg – (at)sin α

At the instant of breaking off the plane, N = 0∴ 0 = mg – (at) sin α

or sinmg

ta

= α

.

The acceleration of the block

cos ( )cosdv F atdt m m

α α= =

or( )cosat

dv dtm

α= …(i) Fig. 5.29

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283Laws of Motion & Equilibrium

Integrating equation (i), we have

0

0 0

cosv t

adv tdt

mα=∫ ∫

or 02

00

cos2

tv a t

vm

α=

or v0 2cos2

at

= …(ii)

2cos

2 sina mg

m aα = α

2

2cos

2 sin

mg

a

α=

α. Ans.

(b) We also have

cosdv F

vds m

α=

cosatm

α= …(iii)

From equation (ii), we have

2cos2

a tv

=

∴1 / 22

cosmv

ta

= α .

Thus from (iii) 1 / 22

coscos

dv a mvv

ds m a = α α

or 1 / 2

1 / 2

2 cosm

ds v dva

= α

After integrating, we get

1 / 2

3 / 223 2 cos

ms v

a = α

Here, v = 0 for s = 0. At the moment of break off the plane

2

0 2cos

2 sin

mgv

a

α=

α.

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284

Hence, the distance travelled by the body upto break off

3 / 21 / 2 2

0 22 cos3 2 cos 2 sin

m mgs

a a

α = α α

2 3

2 3cos

6 sin

m g

a

α=

α . Ans.

EXAMPLE 12 A bar of mass m resting on a smooth horizontal plane starts moving due to the force

F = mg/3 of constant magnitude. In the process of its rectilinear motion the angle α betweenthe direction of this force and the horizontal varies as α = as, where a is a constant, and s isthe distance traversed by the bar from its initial position. Find the velocity of the bar as afunction of the angle α.

SOLUTION Let at any distance s, the α is the inclination of force. From Newton’s second law, we have

cosdv

F mdt

α =

or cos3

mg dvas m v

ds

=

or cos( )3g

vdv as ds= …(i)

Integrating equation (i), we get

0 0

cos( )3

v sg

vdv as ds=∫ ∫

or2

00

sin2 3

vsv g

asa

= Fig. 5.30

or ( )2

0 sin sin02 3

v gas

a

− = −

1 / 22

sin3

gv as

a =

1 / 22

sin3

ga

= α . Ans.

EXAMPLE 13 A block of mass 25kg is raised by a 50 kg man in two different ways as shown in fig. 5.31.

What is the action on the floor by the man in the two cases ? If the floor yields to a normalforce of 700 N, which mode should the man adopt to lift the block without the floor yielding.

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285Laws of Motion & Equilibrium

Fig. 5.31SOLUTION The FBD for the two cases are shown in figure.

In I case; let the force exerted by the man on the floor is N1. Consider the forces inside thedotted box, we have

N1 = T + 50 g.Block is to be raised without acceleration, so T = 25 g.∴ N1 = 25 g + 50 g

= 75g = 75 × 9.8 = 735 NIn II case; let the force exerted by the man on the floor is N2. Consider the forces inside thedotted box, we have

N2 = 50g – Tand T = 25 g∴ N2 = 50g – 25g

= 25 g = 25 × 9.8 = 245 N.As the floor yields to a downward force of 700 N, so the man should adopt mode II.

EXAMPLE 14 Fig. 5.32 shows a man of mass 60 kg standing on a lift weighing machine kept in a box ofmass 30 kg. The box is hanging from a pulley fixed to the ceiling through a light rope, theother end of which is held by the man himself. If the man manages to keep the box at rest,what is the weight shown by the machine ? What force should he exert on the rope to get hiscorrect weight on the machine?

Fig. 5.32

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286

SOLUTION The FBD is drawn in figure (b). From the FBD :For man, N + T = 60 g …(i)For box, T = N + 30 g …(ii)

Solving equations (i) and (ii), we get N = 15 g = 150 N

Thus, weight of man as shown by machine is 15 kg. Ans.To get his correct weight let the man exerts force T ′ on the rope. He together with the boxmove up with an acceleration a. Here we have N = 60g = 600 N.For man, (N + T ′) – 60 g = 60 aor 600 + T ′ – 60 × 10 = 60 a … (iii)and for box T ′ – (N + 30g) = 30 aor T ′ – (600 + 30 × 10) = 30 a … (iv)Solving equations (iii) and (iv), we get

a = 3 m/s2

and T ′ = 1800 N. Ans.

Spring : Spring is the mechanical device which is used to store mechanical energy (elastic potential energy).

It stores energy both in tension and compression. The energy stored in the spring is given by 212

kx , where x

is the deformation in the spring. The force in a spring is given by Hooke’s law; –k=F xuur ur

, where k is called forceconstant.

EXAMPLE 15 (i) A 10 kg block is supported by a cord that runs to a spring scale, which is supported byanother cord from the ceiling figure (a). What is the reading on the scale ?

(ii) In figure (b) the block is supported by a cord that runs around a pulley and to a scale.The opposite end of the scale is attached by a cord to a wall. What is the reading of thescale ?

(iii) In figure (c) the wall has been replaced with a second 10 kg block on the lift, and theassembly is stationary. What is the reading on the scale now ?

Fig. 5.33SOLUTION In all the three cases the spring balance reads 10 kg. To understand this

let us cut a section inside the spring as shown;

Fig. 5.34As each part of the spring is at rest, so F = T . As the block is stationary, so T = 10g = 100 N.

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287Laws of Motion & Equilibrium

EXAMPLE 16 What is the reading of the spring balance in the following device ?

Fig. 5.35SOLUTION Let T be the reading of the spring balance, then

for 20 kg block ; 20 g – T = 20 a …(i)for 10 kg block; T – 10 g = 10 a …(ii)

Solving equations (i) & (ii), we get 2m / s3g

a = and 40N

3g

T =

So the spring balance reading is 40

kg3

.

Group I Problem : When each body in the system has same acceleration.

EXAMPLE 17 Two blocks of masses 1 kg and 2 kg are placed in contact on a smooth horizontal surface asshown in figure. A horizontal force of 3N is applied (i) on 1 kg block (ii) on 2kg block. Findforce of interaction between the blocks.

SOLUTION Since both the blocks are in contact therefore they will move together with an acceleration

32 1

netx

total

Fa

M= =

+ = 1 m/s2.

(i) (ii)

Let force of interaction between them is F1 Let force of interaction between them is F2

 

By Newton’s second law for 2kg block, By Newton’s second law for 1 kg block,we have, F = 2a we have, F2 = 1a

= 2 × 1 = 2N = 1 × 1 = 1N.The same result can also be obtainedfrom 1 kg block

3 – F1 = 1a = 1 × 1 ⇒ F = 2 N.

Fig. 5.36

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288

EXAMPLE 18 Find the force of interaction between the bodies as shown in figure.(i) Blocks are in contact

(a)(ii) Blocks are connected by strings

(b)Fig. 5.37

SOLUTION All the bodies move together along x-axis with an acceleration

[ ]x netx

total

Fa

m=

1 2 3

cosFm m m

θ=

+ +

(i) (ii)

By newtons second law for block m1 By Newton’s second law for block m1F1 = m1a F1 = m1a

11 2 3

cosFm

m m m θ

= + + 11 2 3

cosFm

m m m θ

= + +

and for block m2, we have and for block m2, we haveF2 – F1 = m2a F2 – F1 = m2a

or F2 = F1 + m2a or F2 = F1 + m2a

1 21 2 3

cosFF m

m m m θ

= + + + . 1 21 2 3

cosFF m

m m m θ

= + + + .

Here force of interaction is of compressive Here force of interaction is of tensile nature.nature.

Fig. 5.38

EXAMPLE 19 A homogeneous rod of length L and mass M is placed on smooth horizontal surface. It is

acted by a force F at its one end. Find the stretching force in the cross section of a rod at adistance x from the end where the force is applied.

SOLUTION

Fig. 5.39