Quick Revision Mechanics XI 2014

7/26/2019 Quick Revision Mechanics XI 2014

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2013-2015

Notes for School Exams

Physics XI

Quick Revision

Units & dimensions

Vectors

KinematicsLaws of Motion

Work, Energy & Power

Centre of Mass & Collision

P. K. Bharti, B. Tech., IIT Kharagpur

2007 P. K. Bharti

All rights reserved.

www.vidyadrishti.org


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Quick Revision Author: P. K. Bharti (B. Tech., IIT Kharagpur) www.vidyadrishti.org

P a g e | 2 CONCEPT: JB-20, 3rd

Floor, Near Jitendra Cinema, Sec 4, Bokaro Mb: 7488044834

Important questions for 1stterminal Examinations

1. Different type of errors and derivations on productand exponents

Units and dimensions

2. Least count3. Significant digits4. Finding dimensions of physical quantity5. Numerical on checking correctness of an equation

using dimensional analysis6. Numerical on derivation of formula using

dimensional analysis

7. Triangle law of vector addition and parallelogramlaw of vector addition with proof.

Vectors

8. Numerical on component of vectors, vector addition,subtraction, dot product and cross product.

9. To prove triangular inequalities |a+b| < |a| + |b| etc.

10. To provesin sin sin

a b c

A B C= =

11. To prove one of these using calculus method

Kinematics

2

2 2

1

2

2

v u at

s ut at

v u as

= +

= +

= +

Distance covered in nth second

12. Projectile motion derivationsEquation of trajectory (parabolic path)Maximum heightRangeTime of flightSpeed at a particular instantTime of ascent = time of descentCondition for maximum range etc.Particle thrown horizontally from a height

13.Numerical on relative motion and projectile motion14. Simple problem on graphs

15.Newtons 2nd law of motion in terms of linearmomentum

Newtons Laws of Motion

16. To prove Newtons 2ndlaw is the real law of motion17.Numerical on Conservation of Linear Momentum18.Numerical on Free Body Diagram, Newtons 2ndLaw

and pseudo force

19. Graph of friction

Friction

20. Laws of limiting friction21. Friction is necessary evil22. Friction increases even after polishing a surface

23. Cause of friction24. Rolling friction25. Angle of repose and angle of friction26.Numerical

27. Derivation of centripetal acceleration

Circular Motion

2

c

va

r

=

28. Banking of roads without and with friction29.Numerical

30. Derivation of Work-Kinetic Energy theorem

Work, Energy and Power

31. Derivation of spring potential energy32.Numerical on Conservation of Mechanical Energy33.Numerical on ballistic pendulum34. Conservative force35. Potential, kinetic and mechanical energy graph

36. Derivation of centre of mass of semicircular ring andsemicircular disc

Centre of Mass & Collision

37. Perfectly elastic collisions and different cases38. Oblique collision and different cases39. Coefficient of restitution

40. Derivation of moment of inertia of simple bodies(please note that it is not in NCERT and should not

be asked in schools exam)

Rotational Motion

41. Derivation of parallel and perpendicular axis theorem

42.

To proveo t = +

21

2ot t = +

2 2

0 2 = + 43. Geometrical meaning of angular momentum44. Conservation of angular momentum45. Deduction of Keplers 2nd law from conservation o

angular momentum46.Numerical on pure rolling, radius of gyration, torque

equation

47.Newtons law of gravitation

Gravitation

48. Variation in acceleration due to gravity with heightdepth and rotation

49. Gravitational potential energy50. Escape velocity51. Orbital velocity52. Keplers Laws53. Gravitational field and potential54. Geostationary and polar satellites55. Gravitational field and potential (please note that it is

not in NCERT and should not be asked in schoolsexam)


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56. Definitions of all types of stress and strains

Mechanical properties of solids

57. Hookes law58. Stress vs strain graph (imp)59. Elastic potential energy

60. Variation of pressure with depth in hydrostatic

Mechanical properties of fluids

61. Pascal law62. Archimedes Principle63. Bernoullis Theorem (imp)64. Speed of efflux: Torricellis theorem65. Venturimeter66. Stokes law and terminal velocity67. Reynolds number68. Capillary rise69. Drops and bubbles70. Reasoning on surface tension (eg why droplets are

spherical etc)

71. Zeroth law of thermodynamics

Thermodynamics

72.2 3

= =

73. Assumptions of Kinetic theory of gases (imp)

74. 21

3

P = (imp)

75. Mean free path76. Degrees of freedom77. Equipartition of energy78. Work done by a gas for different processes and

graphically (imp)

79. Cp Cv = R80. Why Cp> Cv

81. constantPV = (imp)

82. Why is adiabatic curve steeper than isothermalcurve?

83. Carnot Engine & its efficiency (imp)84. Second law of thermodynamics85.Newtons Law of cooling86. Conduction

87.

SHM

Oscillations (SHM)

88. Energy in SHM89. Spring mass system90. Simple pendulum91. Damped SHM92. Resonance93. Physical Pendulum

94. Oscillations of a liquid column in a U-tube

95. Oscillations of a body dropped in a tunnel along the

diameter of the earth

96. Oscillation of a floating cylinder

97. Oscillation of a ball in the neck of an air chamber

98. Types of waves

Waves

99. Travelling wave & standing wave100.Standing waves on string with both end fixed and one

end fixed101.Standing waves on open and closed organ pipe102.Speed sound (Newton) and Laplaces correction

(imp)103.Beats (imp)104.Doppler effect (imp)

Email:

All the best!

From P. K. Bharti, B. Tech. (IIT Kharagpur)

HOD Physics @ Concept, Bokaro CentreJB-20, Near Jitendra Cinema, Sec4, BokaroPh: 7488044834

[email protected]
mailto:[email protected]:[email protected]:[email protected]:[email protected]


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Units

Measurement of any physical quantity involves comparisonwith a certain basic, arbitrarily chosen, internationallyaccepted reference standard called unit.

Fundamental or base units

The units for the fundamental or base quantities are calledfundamental or base units.

Derived units

The units of all other physical quantities can be expressed ascombinations of the base units. Such units obtained for thederived quantities are called derived units.

System of units

A complete set of these units, both the base units and derivedunits, is known as the system of units.

The International system of units

CGS (centimetre, gram and second)

The base units for length, mass and time in CGS system werecentimetre, gram and second respectively.

FPS (foot, pound and second)

The base units for length, mass and time in FPS system werefoot, pound and second respectively.

MKS (metre, kilogram and second)

The base units for length, mass and time in MKS system were

metre, kilogram and second

S. I. Units (Systme Internationale d Unites)

The system of units which is at present internationallyaccepted for measurement is the Systme Internationale dUnites (French for International System of Units), abbreviatedas SI. We shall follow the SI units in our syllabus.

Base SI Units

Sl. No. Quantity SI unit Symbol

1. Length metre m2. Mass kilogram kg

3.

Time second s4. Electric Current ampere A5. Temperature kelvin K6. Amount of substance mole mol7. Luminous Intensity candela cd

Two more units

Sl.

No.

Quantity SI unit Symbol

1. Plane angle radian rad2. Solid angle steradian sr

Dimensions

The dimensions of a physical quantity are the powers (or

exponents) to which the base quantities are raised to

represent that quantity.

Sl.

No.

Bases Quantity SI unit Dimension

1.

Length m [L]2. Mass kg [M[3. Time s [T]4. Electric Current A [A]5. Temperature K [K]6. Amount of substance mol [mol]7. Luminous Intensity cd [cd]

Application of dimensional analysis

Following are the three main uses of dimensiona

analysis:

(a) To check the correctness of a given physical relation

It is also known as principle of homogeneity ofdimensions.

(b) To convert a physical quantity from one system o

units to another.

(c) To derive a relationship between different physica

quantities.

a) Principle of homogeneity of dimensions:According to

this principle, a physical equation will be dimensionally

correct if the dimensions of all the terms occurring on

both sides of the equation are the same.

To check the dimensional correctness of a physica

equationwe make use of the principle of homogeneity of

dimensions. If the dimensions of all the terms on the two

sides of the equation are same, then the equation is

dimensionally correct.

Example.Let us check the dimensional accuracy of the

equation of motion,

21

2s ut at = +

Dimensional of different terms are

[ ] [ ]

[ ] [ ] [ ]

[ ]

1

2 2 2

1

2

s L

ut LT T L

at LT T L

=

= =

= =

As all the terms on both sides of the equations have the

same dimensions, so the given equation is dimensionally

correct.


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The a calculated above may be positive in certain cases andnegative in some other cases. But absolute error |a| willalways be positive.

b) Mean or final absolute error: The arithmetic mean of the

positive magnitudes of all the absolute errors is called mean

absolute error. It is given by

1 2

1

... 1 nni

i

a a aa a

n n =

+ + + = =

Thus the final result of the measure of a physical quantity can

be expressed as .a a a=

c) Relative error: The ratio of the mean absolute error to the

true value of the measured quantity is called relative error.

Relative error,a

aa

=

d) Percentage error: The relative error expressed in percent is

called percentage error.

Percentage error 100%a

a

=

COMBINATION OF ERRORS

(a) Error in the sum or difference of two quantities: Let

A and B be the absolute errors in the two quantities A

andBrespectively. Then

Measured value ofA=A A

Measured value ofB=BB

Consider the sumZ=A+Bor differenceZ=AB,

The error ZinZis then given by

or Z= A+ B

Hence the rule:The maximum possible error in the sum

or difference of two quantities is equal to the sum of the

absolute errors in the individual quantities.b) Error in the product or quotient of two quantities: Consider

the product,

Z =AB or Z = A/B

The maximum fractional error inZis

Z A B

Z A B

= +

Hence the rule: The maximum fractional error in the product

or quotient of two quantities is equal to the sum of the

fractional errors in the individual quantities.

c) Errors in the power of a quantity:

If ,p q

r

A BZ

C= then maximum fractional error inZis given by

Z A B Cp q r

Z A B C

= + +

The percentage error inZis given by

100 100 100 100Z A B C

p q rZ A B C

= + +

Derivation of general rule using differentiation.

We have

p q

r

A BZ

C=

Taking logarithms, we get

logZ=plogA + qlogB rlog C

On differentiating both sides, we get

dZ dA dB dC p q r

Z A B C= +

Writing the above equation in terms of fractional errors.

Z A B CqZ A B C =

The maximum permissible error inZis given by

.Z A B C

p q rZ A B C

= + +


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VECTORS

AB

Representation of a vector (Arrow over letters or bold letters)

=AB=P

=P

AB

Magnitude of a vector (with modulus or without bold)

= |AB|= P

=|P| = AB = P

Note: Magnitude is the length of the vector.

, , ,i j k n

Unit VectorA vector whose magnitude is unity (1 unit) is called a unitvector. Generally it is represented by cap over letter. e.g.,

etc.

Unit Vector in the direction of a given vector A

i

Unit Vectors along coordinate axes:

Unit vector along X-axis =

Unit vector along Y- axis = j

Unit vector along Z- axis = k

Clearly,

1i j k= = =

Suppose the coordinates of tip of position vector

Representation of a position vector in unit vector form (3D)

r

is (x, y, z),

then that vector is represented in unit vector form as

r xi y j zk = + +

Here,x=x-component of vector r

y=y-component of vector r

z=z-component of vector r

The magnitude (or length) of this vector is

2 2 2r r x y z= = + +

Component of a vector

Component of a vector

A

along a

direction making an angle with it is

A cos .

A

Rectangular component of a vector

Let a vector makes an angle of with positive direction ofx- axis. Then,

X-componentof A

= A cos

Y-componentof A

=Asin

We can write A

in the unit vector notation ascos sinA A i A j = +

Let a vector in unit vector form as

Magnitude and direction of vector (when given in unit vector

form in 2D)

A xi y j= +

Here,x=x-component of vector A

y=y-component of vector A

The magnitude (or length) of this vector is

2 2A A x y= = +

Angle with positive direction of x-axis (Use trigonometry)

1tan y

x =

Vector Addition

1 1 1 1

2 2 2 2

1 2

1 2 1 2 1 2

...........

Then,

...

( ... ) ( ... ) ( ... )

n n n n

n

n n n

A a i b j c k

A a i b j c k

A a i b j c k

R A A A

R a a a i b b b j c c c k

= + +

= + +

= + +

= + + +

= + + + + + + + + + + +

Method 1 : Analytically

Here, R

= resultant (i.e., vector sum of all vectors)

B

P

Note: A is tail and B is

tip (or head) ofAB=PA

AAA

=

A

A

x

z

i

j

k

x

y

z

r

x, , z


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Vector Addition

AB BC AC+ =

Method 2: Triangle law of vector addition(when tail of one

vector is on the head of another vector)

Statement: It states that if two vectors are represented in

magnitude and direction by the two sides of a triangle taken in

the same order, then the resultant is represented in magnitude

and direction by the third side of the triangle taken in opposite

order.

Clearly, (See graphical method for

explanation)

R P Q= +

Magnitude ofR

:

2 2 2 cosR R P Q PQ = = + +

Note is the angle between vectors &P Q (Explained in class)

Angle whichR

makes withP

:

sintan

cos

Q

P Q

=

+

Proof of triangle law of vector addition

Draw a perpendicular CM on AB. Clearly

BC = Q cos and CM = Q sin .

Now using Pythagoras theorem in right angled triangle

ACM, we have

( )

( ) ( )

( ) ( )

( )

( )

22 2 2 2

2 22

2 2 2 2 2 2

2 2 2 2 2

2 2 2

2 2

cos sin

cos 2 cos sin

cos sin 2 cos

2 cos

cos sin 1

AC AM CM AB BM CM

R P Q Q

R P Q PQ Q

R P Q PQ

R P Q PQ

= + = + +

= + +

= + + +

= + + +

= + +

+ =

Again, in right angled triangle ACM

tan

sintan

cos

perpendicular CM CM

base AM AB BM

Q

P Q

= = =+

=+

Vector Addition

R P Q= +

Method 3: Parallelogram law of vector addition(when twovectors are placed tail to tail i.e., co-initial vector)

Statement:It states that if two vectors acting simultaneously

are represented in magnitude and direction by the two adjacen

sides of a parallelogram taken in the same order, then the

resultant is represented in magnitude and direction by the

diagonal of the parallelogram passing through that point.

Magnitude of R

2 2 2 cosR R P Q PQ = = + +

Note is the angle between vectors &P Q

Angle whichR

makes withP

: sintan cos

Q

P Q

= +

Angle whichR

makes with Q

:sin

tancos

P

Q P

=

+

Clearly, + =

Proof is same as that of triangle law of vector addition.

NOTE: In general, we should use parallelogram /triangle lawfor adding two vectors only. For addition of more than twovectors we should use analytical method.

Special cases of vector addition

2 2 2 cosR R P Q PQ = = + +

Case 1: Maximum magnitude of resultant

Clearly R is maximum when cos is maximum. Maximum

value of cos is 1 when = 0o.

( )22 2

max 2R P Q PQ P Q P Q= + + = + = +

maxR P Q = + when two vectors lies in same direction.

2 2 2 cosR R P Q PQ = = + +

Case 2: Minimum magnitude of resultant

Clearly R is minimum when cos is minimum. Minimum value

of cos is 1 when = 180o.

( )22 2

min 2R P Q PQ P Q P Q= + = =

minR P Q = when two vectors are in opposite direction.

P

R

Q

BA

C

P

R

Q

M

P

Q

P

Q


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Therefore,

Case 3: When two vectors have equal magnitude

If P = Q, then parallelogram will be a rhombus. We know that

diagonal of a rhombus is equally inclined to each sides.

2

= =

( )P Q P Q = +

Vector subtraction

i.e. vector subtraction is obtained by taking negative of

another vector and adding it to first vector.

For example: To subtract Q from P, we take negative of

vector Q by reversing its direction. Then add P and Q by

parallelogram law or triangle law.

Shortcut:

Angle betweenP and Q=

Therefore, angle betweenP and Q=

( )

( )2 2

2 2

2 cos

2 cos

P Q P Q

P Q PQ

P Q PQ

= +

= + +

= +

2 2 2 cosP Q P Q PQ = +

Angle whichR

makes withP

:sin

tancos

Q

P Q

=

| || | coA B A B =

Dot Product or Scalar Product

(Dot product)

where is the angle betweenAandB.

Dot product of two vectors gives a scalar quantity.

Using definition of dot product we can find the angle

between two vectors.

cos| || |

A B

A B

=

(Angle between two vectors)

1 1 1

1 2 1 2 1 2

2 2 2

A a i b j c k

A B a a b b c cB a i b j c k

= + + = + +

= + +

Dot product of two vectors (when given in unit vector form):

| || | sinA B A B n =

Cross Product or Vector Product

Cross product gives a vector

quantity.

Here, n is a unit vector

perpendicular to both & .A B

Clearly, direction ofA B

is

alongn , i.e., perpendicular to both & .A B

Direction ofA B

is obtained by using Right hand thumb

rule.

( ) ( ) ( )

1 1 1

2 2 2

1 1 1

2 2 2

1 2 2 1 1 2 2 1 1 2 2 1

&

A a i b j c k

B a i b j c k

i j k

A B a b c

a b c

A B i b c b c j a c a c k a b a b

= + +

= + +

=

= +

Cross product of two vectors (in unit vector form):

Important points

Let us consider two vectors

1 1 1 2 2 2 &A a i b j c k B a i b j c k= + + = + +

1. They are parallel if

1 1 1

2 2 2

a b c

a b c

= =

2. They are perpendicular, if 0A B =

1 2 1 2 1 2 0a a b b c c+ + =

3. A unit vector perpendicular to both &A B

is .| |

A B

A B

1. Component of A

alongA B

BB

=

Vector form: Component of A

alongB=

A B B

B B

2. Physical meaning of A B

A B

= area of parallelogram

P

Q

Q

R

P

R

Q


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KINEMATICS

Distance

Actual length of the path

Denoted by ors rScalar

S.I. unit: m

Displacement

Shortest distance between two points

Denoted by ors r

.

Vector

S.I. unit: m

Average Speed (between two instants of time)

Distance travelled per unit time

Denoted by v

distanceavg speed =

time

sv

t

=

We can use this formula for constant speed also.

Scalar

S.I. unit: m/s

Instantaneous Speed (at a particular instant of time)

Denoted by v

0lim

t

s dsv v

t dt

= =

NOTE: Speed means instantaneous speed. This is main

formula for speed. We can use this formula for any case of

speed.

Average Velocity (between two instants of time)

Displacement per unit time

Denoted by v

displacementavg velocity =

time

rv

t

=

We can use this formula for constant velocity also.

Vector

S.I. unit: m/s

Instantaneous velocity (at a particular instant of time)

Denoted by v

0lim

t

r d rv v

t dt

= =

NOTE:

Velocity means instantaneous velocity. This is main

formula for velocity. We can use this formula for any

case of velocity.

Velocity is tangential to path

Magnitude of instantaneous velocity = instantaneous

speed v v=

Average Acceleration (between two instants of time)

Change in velocity per unit time

Denoted by a

change in velocityavg acceleration=

time

va

t

=

We can use this formula for constant acceleration also.

Vector

S.I. unit: m/s2

Instantaneous Acceleration (at a particular instant of time)

Denoted by a

0lim

t

v d va a

t dt

= =

We can also use

dv dva v

dt dr = =

NOTE: Acceleration means instantaneous acceleration. These

are main formulae for acceleration. We can use these

formulae for any case of acceleration.

Graphs

Case 1: Average value (From slope of chord)

Form: x

yt

=

Supposex

yt

=

and you want y between two instants

1 2andt t graphically. Draw a chord AB between 1 2andt t

Suppose this chord (sometimes after extension of chord

makes an angle with positive taxis. Then slope of thischord gives average of y i.e. y between 1 2andt t .

tany =

Using this method we can find

average speed from distance time graph

average velocity from displacement time graph

average acceleration from velocity time graph

t

x

A

B

t1 t2


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Case 2: Instantaneous value (From slope of tangent)

Form:dx

ydt

=

Supposedx

ydt

= and you want y at an instant tgraphically.

Draw a tangent at t. Suppose this tangent (sometimes after

extension of chord) makes an angle with positive taxis.

Then slope of this tangent gives instantaneous value of y at

t.

tany =


instantaneous speed from distance time graph

instantaneous velocity from displacement time graph

instantaneous acceleration from velocity time graph

Case 3: Change in value (From area)

Form:2 2 2

1 1 1

2 1

y t t

y t t

dy xdt y y xdt = =

Suppose2 2 2

1 1 1

2 1

y t t

y t t

dy xdt y y xdt = = and you want 2 1y y

between two instants 1 2andt t graphically. Draw two linesABand CD parallel to x axis at 1 2andt t . Then area of the

region between curve,AB, CDand taxis gives instantaneous

2 1y y between 1 2andt t .

2

1

2 1

area of the region between curve, , and axis

t

t

y y xdt

AB CD t

=

=


change in velocity from acceleration time graph

change in position (i.e. displacement) from velocity time

graph

distance from speed time graph

Uniformly accelerated motion: constanta =

1. v u at = +

[ ] [ ]0

0 0

( 0)

...(i)

v t tv t

u

u

dva

dt

dv adt a dt v a t

v u a t

v u at

=

= = =

=

= +

2.21

2s ut at = +

From (i), we have

( ) [ ] [ ]

2

0 00 0 0

2

2

1 ...(ii)

2

ts ts t

v u at

ds dsu at v

dt dt

tds u at dt s u t a

s ut at

= +

= + =

= + = +

= +

3.2 2 2v u as= +

We know that

[ ]2 2 2

00

2 2

2 2 2

2 ...(iii)

vs vs

u u

dv dva v

dt ds

dva v

ds

v v uads vdv a s as

v u as

= =

=

= = =

= +

4. Distance travelled innth second

( )

[ ] [ ]

( ) ( )

( )

0 1

2

0 11

22

From (i)

2

11 1

2

12 1 ...(ii)

2

s n

n

n

s n

n

n

v u at

dsu at ds u at dt

dt

t

s u t a

s u n n a n n

s u a n

= +

= + = +

= +

= +

= +

t

x

t

t

x

t1 t2

A

C

B D


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Sign convention

You can take your axis along any side at your will. All the

physical quantities which are along that axis are positive and

quantities opposite to axis direction are negative.

Motion in a plane (when acceleration is constant)

Motion in a plane (two-dimensions) can be treated as two

separate simultaneous one-dimensional motions with constantacceleration along two perpendicular directions.

Along x axis

2

2 2

1

2

2

x x x

x x

x x x

v u a t

x u t a t

v u a x

= +

= +

= +

Along y-axis

2

2 2

12

2

y y y

y y

y y y

v u a t

y u t a t

v u a y

= +

= +

= +

Note we have assumed that acceleration is constant.

Motion in a plane (when acceleration is not constant)

Along x axis

x

x x

x x

dxv

dt

dv dv

a vdt dx

=

= =

Along y axis

y

y y

y y

dyv

dt

dv dva v

dt dy

=

= =

Projectile motion

When a particle is thrown obliquely near the earths surface,

it moves along a curved path. Such a particle is called a

projectile & its motion is called projectile motion. Its motion

is an example of 2D motion. We shall break equation of

motion along x and y axes.

Some points to note about projectile motion

Acceleration in y direction, ay = g (Negative g,

because g is opposite to our +ve y axis.)

Acceleration in x direction ax = 0 (because there is noacceleration in x direction).

Velocity in x direction, vx = u cos is always same

(because ax = 0); i.e., vx does not change with time.

At maximum height H, velocity is parallel to x-axis

(because velocity is tangential to path). Therefore, y

component of velocity at maximum height, vy = 0

To prove:

Equation of trajectory:2

2 2tan

2 cos

gxy x

u

=

Maximum height:2 2sin

2

uH

g

=

Horizontal Range:2 sin2u

Rg

=

Time of Flight:2 sinu

Tg

=

Data to be used:

Initial velocity:

uat an angle with horizontal (x-axis)x component of initial velocity: cos

xu u =

y component of initial velocity: sinyu u =

Acceleration:

( ) acceleration due to gravity along negative -axisa g y=

x component of acceleration:

( )0 no acceleration component along -axisxa x=

y component of acceleration: ya g=

Position at time t : ( ),x y

y component of velocity at maximum height: vy = 0


2 2

1tan

2 cos

gxy x

u

=

Alongx axis

( )

212

cos cos and 0

...(1)cos

x x

x x

x u t a t

x u t u u a

xt

u

= +

= = =

=

Alongy axis

( )

2

2

2

2

2 2

1

2

1sin sin and

2

1sin

cos 2 cos

from (i)cos

1tan

2 cos

y y

y y

y u t a t

y u t gt u u a g

x xy u g

u u

xt

u

gxy x

u

= +

= = =

=

=

=

...(A)

x

y

u

R

H


13/29




Maximum height:2 2sin

2

uH

g

=

At maximum height H, velocity is parallel to x-axis (because

velocity is tangential to path). Therefore, y component of

velocity at maximum height, vy = 0.Alongy axis

( )( )

2 2

22

2 2

2

0 sin 2sin , for max height and

sin

2

y y y

x y

v u a y

u gH

u u y H a g

uH

g

= +

= = = =

=

Time of Flight:2 sinu

Tg

=

For time of flight we should use y = 0 and t =T, becausedisplacement along y axis for complete motion is zero.

( )

2

2

1

2

10 sin 0, , sin and2

1sin 0

2

2 sin0 or,

y y

y y

y u t a t

u T gT y t T u u a g

u gT T

uT T

g

= +

= = = = =

=

= =

2 sinNeglecting 0, we get,

uT T

g

= =

Horizontal Range:2 sin2u

R

g

=

Putting x = R, 0xa = , and t =

2 sinuT

g

= along x-axis we

get,

( )

( )

2

2

2

1

2

cos

2 sin cos2 sincos =

sin2= sin 2 2sin cos

x xx u t a t

R u T

uuR u

g g

uR

g

= +

=

=

=

Condition for maximum range

We know that2 sin2u

Rg

=

Clearly range will be maximum when0 0sin 2 1 2 90 45 = = =

Therefore,

2

max

uR

g=

To prove there are two angles of projection for same range

We know that2 sin2u

Rg

=

Also from Trigonometry,

( )

( )

0

0

sin 2 sin 180 2

sin 2 sin 2 90

=

=

Therefore there are two angles and 900 for same range

To prove time of ascent = time of descent

For upward motion upto maximum height:

Let the required time be t1. We have,

sin , , 0y y y

u u a g v= = =

Now we have,

1

1

0 sin

sin ...(i)

y y yv u a t

u gt

utg

= +

=

=

For downward motion from maximum height to ground:

Let the required time be t2. We have,2 2sin

0, ,2

y y

uu a g y H

g

= = = =

Now we have,

2

2 22

2

2

1

2

sin 1

2 2

sin= ...(ii)

y yy u t a t

ugt

g

utg

= +

=

Clearly from (i) and (ii), we have t1= t2.

Hence, time of ascent = time of descent

Relative Motion:

Terminology

12

12

12

1=object, 2 = observer

= relative position of particle 1 wrt particle 2

= relative velocity of particle 1 wrt particle 2

= relative acceleration of particle 1 wrt particle 2

x

v

a

General equation

12 10 02

12 10 20

m m m

m m m

= +

=

i.e., Relative motion of 1 wrt 2

= (Relative motion of 1 wrt 0)

(Relative motion of 2 wrt 0)

where O is a third object wrt which we know the motion of 1

and 2.


14/29




Projectile given horizontal projection

Data to be used:

Initial velocity:

uin horizontal direction

x component of initial velocity:x

u u=

y component of initial velocity: 0yu =

Projection height above the ground: H

Acceleration:( ) -axis is downwarda g y=

x component of acceleration:

( )0 no acceleration component along -axisxa x=

y component of acceleration: ya g=

Position at time t : ( ),x y


2

2

gxy

u=

Alongx axis

( )

212

and 0

...(1)

x x

x x

x u t a t

x ut u u a

xt

u

= +

= = =

=

Alongy axis

( )

2

2

2

2

2

1

2

1 0 and

2

1 from (i)2

...(A)2

y y

y y

y u t a t

y gt u a g

x xy g tu u

gxy

u

= +

= = =

= =

=

Time of Flight:2H

Tg

=

For time of flight we should use y = H and t =T, because

displacement along y axis for complete motion isH.

( )

2

2

1

2

1 , , 0 and

2

2

y y

y y

y u t a t

H gT y H t T u a g

HT

g

= +

= = = = =

=

Horizontal Range:2H

R ug

=

Puttingx=R, x

u u= , 0x

a = , and t=2H

Tg

= along x-axis

we get,

212

2

x xx u t a t

R u T

HR u

g

= +

=

=

Speed at timet:2H

R ug

=

Alongx axis

...(i)

x x x

x

v u a t

v u

= +

=

Alongy axis

...(ii)

y y y

y

v u a t

v gt= +

=

Speed at time t

y

x

u

H

R


15/29




Laws of Motion

Newtons 1st

law of motion

If the vector sum of all the forces acting on a particle is

zero then and only then the particle remains

unaccelerated.

a= 0if and only ifF= 0.

Newtons 1stlaw gives definition of force.

Inertial frame of reference

A frame of reference in which Newtons 1st

Law is valid

is called an inertial frame of reference. It is non-

accelerating frame. i.e., a=0.

In simple language, an inertial frame of reference is such

a frame which has no acceleration.

Linear MomentumMomentum, of a particle is defined to be the product ofits mass m and velocity v, and is denoted byp:p= m v

Denoted by p

orpVectorS.I. unit: =kg-m/s

Newtons 2nd

law of motion

The rate of change of momentum of a body is directly

proportional to the applied force and takes place in the

direction in which the force acts.

According to the Second Law

d pF

dt

d pF k dt

=

where k is a constant of proportionality.In S. I. system, we choose k = 1. Hence,

d pF

dt=

For a particle of fixed mass m,

d p dmv dvm ma

dt dt dt = = =

Thus, Newtons second law can be written as

d pF ma

dt

= =

Newtons 2ndlaw measures force.

Conditions for Newtons 2ndLaw to hold1. Inertial frame for reference2. Particle or particle like object3. Speed of particles well below speed of light c4. The second law of motion is a local relation which

means that force F at a point in space (location of theparticle) at a certain instant of time is related to a atthat point at that instant. Acceleration here and nowis determined by the force here and now.

Impulse

From Newtons 2ndlaw, we know that

2 1

d pF Fdt d p

dt

Fdt p p

= =

=

Here, Fdt

is known as impulse which is equal to change

in linear momentum. We can write for a constant force

2 1

2 1

Fdt F dt F t p p

F t p p

= = =

=

A large force acting for a short time to produce a finite

change in momentum is called an impulsive force.

Conservation of linear Momentum

The total momentum of an isolated system of interactingparticles is conserved.

Isolated system means, net external force acting on thesystem is zero.

Newtons 3rdlaw of motion

There is an equal and opposite reaction to every action.

Conditions for Newtons 3rdLaw to hold1. Forces always occur in pairs. Force on a body A by B is

equal and opposite to the force on the body B by A.2. There is no cause effect relation implied in the third law.

The force onAbyB and the force onBbyA act at thesame instant.

3. Action and reaction forces act on different bodies, not onthe same body. Consider a pair of bodiesA andB.According to the third law,

AB BAF F=

(force onAbyB) = (force onBbyA)

In scalar form AB BAF F=

NOTE:

1.If you are considering the system of two bodies as a

whole, andAB BAF F

are internal forces of the system (A+ B). They add up to give a null force.Internal forces ina body or a system of particles thus cancel away inpairs. This is an important fact that enables the secondlaw to be applicable to a body or a system of particles

To prove Newtons 2nd

law is the real law of motion

To prove that Newtons 1st law is contained in

Newtons 2nd

law

Newtons 2ndLaw states that net external force F

exerted

on a particle is equal to mass m times acceleration a

o

the particle, i.e., F ma=

If 0F=

then 0 0ma a= =

, which is Newtons 1stlaw

i.e., if the net external force acting on a particle is zero

then and only then the particle remains unaccelerated

0a=

if and only if 0F=

.


16/29




To prove that Newtons 3rd

law is contained in

Newtons 2nd

law

Let us consider two isolated bodies A and B interacting

with each other only. Now from Newtons 2nd law we

have,

Force on B by A

= rate of change in linear momentum of B, i.e.,

BAB

d pF

dt =

Force on A by B

= rate of change in linear momentum of A, i.e.,

ABA

d pF

dt =

Sum of these forces,

( )B A

AB BA A B

d p d p d

F F p pdt dt dt + = + = +

As no external forces acting on our system, rate of

change of total momentum must be zero, i.e,

( ) 0A Bd

p pdt

+ =

.

Hence,

0AB BAF F+ =

AB BAF F =

, which is Newtons 3rdlaw.

Important forces

1. Weight

Direction of weight is always vertically downwards,however the orientation of body may be. It originatesfrom centre of mass of a body.

2. Normal force

When a body presses against a surface, the surfacedeforms and pushes on the body with a force N that is

perpendicular to the surface . This force N is calledNormal force.

Normal force comes into existence when there is somecontactbetween two bodies.

Number of normal forces acting on a body = Number

of contacts. In general Normal reaction is perpendicular to the surface

of contact.

3. Tension

When a cord (or a rope, cable, or other such object) isattached to a body and pulled taut, the cord pulls on the

body with a force T directed away from the body andalong the cord.

This force Tis called as tension force. Tension can never push a body. Tension is always pullingin nature. It means direction of tension is always along the rope

such that it pulls the body (does not push).

Tension in all part of the string will be same, only when, thestring is

Massless, and Inextensible.

Massless means mass of string is negligible compared to massof block.

For a massless or frictionless pulley, tension in two partsof the string will have same magnitude T = T

1

=T2

, even if

the bodies are accelerating. Cord should be masslessand unstretchable.


17/29




Free Body Diagram (F.B.D.)

F.B.D. is nothing, but a diagram showing all the forcesacting on a body.

Lamis Theorem

When three coplanar forces A, Band Cact on a particlesuch that particle remains in equilibrium, then, LamisTheorem states that:

sin sin sin

A B C

= =

where A, B and C are the magnitude of forces A, Band Crespectively.= angle opposite to force A= angle opposite to force B= angle opposite to force C

Pseudo force

To write Newtons law correctly in non-inertial frame weapply a force ma opposite to the acceleration of frame.This ma is called as pseudo force. Here m is the mass ofthe particle (not the mass of frame) whose FBD we aredrawing in non-inertial frame.

Pseudo force

= (mass of body of interest)

x (acceleration of non-inertial frame)

SSttaattiiccFFrriiccttiioonn

The magnitude of the static Friction fs is given by

fs sN

s = a constant known as coefficient of static

friction between the body and the surface o

contact. It depends upon roughness of the surface

More rough the surface is more will be s .

N= Normal force on the object from the surface.

Magnitude of static friction = magnitude o

applied force.

Limiting Friction = Maximum value of Static Friction

Limiting Friction = fs, max = sN.

KKiinneettiiccFFrriiccttiioonn

The magnitude of the kinetic friction fkis given by

fk=

kN

k = a constant known as coefficient of kinetic

friction between the body and the surface of contact

It depends upon roughness of the surface. More

rough the surface is more will be k .

N= Normal force on the object from the surface.

GGrraapphhooffffrriiccttiioonnvvss..aapppplliieeddffoorrccee

From O to A particle remains stationary. Thus static

friction acts in part OA.

Here, static friction = applied force, therefore angle =45 o

Length AB represents maximum friction, i.e., Limitingfriction,fs, max = sN .

After A, point start moving thus in this part kinetic

friction acts.

Length CD represents kinetic friction which is almost

constant; fk = kN in this part.


18/29




Important points

The coefficients of friction depend on the nature of thesurface.

The frictional force is nearly independent of the contactareabetween the objects.

The kinetic friction force is usually less than themaximum static friction force (Limiting friction).

Static friction opposes impending motion. The termimpending motion means motion that would take place(but does not actually take place) under the applied force,

if friction were absent. Note that it is not motion, but relative motion that the

frictional force opposes.

Contact Force (R) and Angle of Friction ():

Resultant of friction and

normal force is called

contact force R.

The angle between the

resultant contact force and

the normal is called angle of

friction .

ANGLE OF REPOSE

The angle of repose is defined as the angle of the inclined

plane at which a body placed on it just begins to slide.

Consider an inclined plane, whose inclination with horizontal

is gradually increased till the body placed on its surface just

begins to slide down. If is the inclination at which the body

just begins to slide down, then is called the angle of repose

The body is under the action of the following forces:

1. The weight M gof the body acting vertically downwards.

2. The limiting friction f in upward direction along the

inclined plane, which in magnitude is equal to the

component of the weight M g acting along the inclinedplane i.e.

f= M gsin (i)

3. The normal reactionNacting at right angle to the inclined

plane in upward direction, which is equal to the

component of weight acting perpendicular to the inclined

plane i.e.

N= Mg cos (ii)

Dividing equation (i) by (ii), we have

sintan

cos

f Mg

N Mg

= =

Since

f N

f

N

=

=

Thus, tan =

Therefore, coefficient of limiting friction is equal to thetangent of the angle of repose.

CIRCULAR MOTION

When a particle moves along acircular path, its motion is saidto be circular motion.

At any instant of time velocity vis always tangent to the circularpath. (Why tangent? Because

velocity is tangent to its pathat each instant of time.)Therefore, velocity is alwaysperpendicular to radius duringcircular motion.

(Because, from Mathematics, we know that a tangent of a

circle is perpendicular () to radius.)

vR

Acceleration in circular motion

In a circular motion, when the particles velocity isv,there is an accelerationa.

This acceleration a has twocomponents:

Tangential acceleration(at)

Centripetal acceleration

or Radial acceleration

or Normal acceleration

(ac or an or ar )

Tangential acceleration at:Its direction is along tangent.

t

dva

dt= (magnitude of tangential accln)

Centripetal acceleration or Radial acceleration orNormal acceleration(ac or an or ar )Its direction along tangent towards the center.

2

c

va

R= (magnitude of centripetal accln)

Mgcos

N

Mg

Mgsin


19/29




Therefore, net accelerationa,2 2

t ca a a= + (magnitude of net accln)

tan t

n

a

a = (direction ofawith an )

NOTE: Speed vmay or may not be uniform.

Uniform Circular Motion

When a particle moves in a circle with constant speed v,

its motion is said to be uniform circular motion. It means magnitude of velocity vector v is constant, but

the directionof vchanges continuously.

(NOTE: Here speed remains same but velocity changes.)

It means velocity is changing with time (because velocity

is a vector, which changes when either direction or

magnitude changes). It means there is acceleration.

Thus a particle moving in a circle, undergoes an

acceleration.

For uniform circular motion speed, v= constant. Therefore, tangential acceleration becomes

0tdv

a dt= =

Therefore, there is only centripetal acceleration, which is2

c

va

R=

Thus, net acceleration a is radial acceleration here. Since, net acceleration direction is towards the centre

here, it has got a special name centripetal acceleration. Thus, during uniform circular motion, net acceleration =centripetal (radial) acceleration

2

c

va a

R= =

Derivation of centripetal acceleration

Change in velocity v when velocity changes from v1 to v2 is

given by v = v2 v1.

This change in velocity is along the centre of the circle.

For small angle we can uses

r

=

Alsov

v

=

From (i) and (ii), we have,

vs sv v

r v r

= =

Dividing both sides by time

v v s

t r t

=

Nowc

va

t

=

ands

vt

=

. Hence,

2

.c

c

va v

r

va

r

=

=

Angu lar var iables

Angular velocity and angular acceleration

For the time being, it is sufficient to know that angular velocityrepresents the change in angle with time and angulaacceleration is the change in angular velocity per unit time.Physically, if angular velocity is higher means the body rotates

faster.Relation between linear and angular variables Linear displacement (l) and angular displacement ()

l

R=

Linear velocity (v) and angular velocity ()

v R=

Linear acceleration (l) and angular acceleration ()

ta R=

SI units of angular variables

SI unit of angular displacement : rad (radian)

SI unit of angular velocity : rad/s

Another unit of is rpm (revolution per minute)

21 rpm rad/s

60

=

SI unit of angular acceleration : rad/s2

Centripetal acceleration in terms of

angular velocity

Since, v R= Therefore, centripetal acceleration,

22

c

va R

R= =

v1

v2

r

r

r

v1

v2v

v2 v

v1


20/29




Centripetal Force

We have just seen that a particle in a uniform circular motionhas centripetal acceleration directed towards its centre.Thus, from Newtons 2ndlaw, a force of magnitude F = mac willact on the particle which will be directed towards the centre.This force Fis known as centripetal force.

Centripetal Force2

2

c

mvF ma m R

R= = =

NOTE: Centripetal force direction is always towards thecentre of the circular path.

Vehicle moving on a circular track with uniform speed v

Suppose a vehicle is taking a circular level turn of radiusRwith uniform speedv. We are assuming vehicle to be aparticle like object.

Therefore, it will have a centripetal acceleration2

v

R

directed towards the centre of the turning.

This centripetal acceleration on the vehicle must beprovided by some external force.

Let us see the forces acting here. Forces are :N (upward)mg (downward)

But both of N and mg are in vertical direction. No forcein radial direction.

But from Newtons 2nd law there must be some externalforce in the radial inward direction to provide centripetalacceleration. Which force is this??

This external force is static friction f from the road onthe vehicle. Thus, friction force provides the necessarycentripetal acceleration.

Therefore, 2 ...(i)

c

mvf ma

R= =

Condition for safe turn

Static friction force provides the necessary centripetalacceleration here. Therefore,

f sN.

Also, along verticle direction,N= mg

Therefore,f sN fsmg (ii) Therefore, from (i) & (ii)

2

s

mvmg

R

2

(Condition for safe turn)sv

Rg

Thus for safe turn, coefficient of static frictions must be

greater than or equal to2

v

Rg

.

In day to day life, when speed of a vehicle taking a turnincrease sufficiently, this condition may be violated.

Therefore, friction force is not very reliable for turningon a circular road.

This also shows that for a given value of s and R, there

is a maximum speed of circular motion of the carpossible, namely

max sv Rg=

Banking of roads

The angle through which the outer edge of the road israised above the inner edge is called angle of banking.

Banking of road analysis without friction

Let the banking angle of the road be Forces are:

mg (downward)N (perpendicular to road surface)

Now , we have to resolve N into radial and tangentiacomponents.

Along radial direction:

Necessary centripetal accln is provided by N sinTherefore,

2

sin ...(i)mv

NR

=

Thus, sin

component of normal force provides thenecessary centripetal accln to vehicles in this case. This

force is quite greater than friction force. So, banked roadsare safer for vehicles than plane roads during turning.

Along vertical:N cos= mg (ii)

Dividing (1) by (2), we get:2

tan v

Rg=

N

f

mg

N

mg

N sin

N cos

ca ca

ca


21/29




21tan (angle of banking)

v

Rg =

Thus, angle of banking depends upon velocity of thevehicle and radius of the road.

Banking of road is designed in such a way, that most ofvehicles may go with their average speed safely.

Banking of road analysis with friction

Let us first call the speed of the vehicle as videalwhen therewas no fiction from road.

Here, we can have two cases:CASE 1: If the speed is well below a videal, vehicle willtend to skid downward the road. Therefore, friction f willact in upward the inclination of the road.CASE 2: If the speed is well above videal, vehicle will tendto skid upward. Therefore, friction will f act in downwardthe inclination of road.

Case 1: To find minimum speed

In this case friction will act upward the incline as thevehicle will have a tendency to move downward the

incline. Let us draw FBD.

Now we have to resolve f and N into radial and verticalcomponents.

Along vertical:

Ncos+fsin mg=0 N cos + Nsin mg = 0 (1)

(because, limiting frictionf =N)

Along radial direction:2

sin cosmv

N fR

=

2

sin cos (2)mv

N NR

=

Solving (1) and (2) simultaneously, we get,1/2

tan

1 tan

v Rg

=

+

This is the minimum speed with which a vehicle can takecircular turn on a banked road safely.

Case 2: To find maximum speed

In this case friction will act upward the incline as thevehicle will have a tendency to move downward theincline.

Let us draw FBD.

Now we have to resolve f and N into radial and verticacomponents.

Along vertical:

Ncos+fsin mg=0 N cos + Nsin mg = 0 (1)

(because, limiting frictionf =N)

Along radial direction:2

sin cosmv

N fR

+ =

2

sin cos (2)mv

N NR

+ =

Solving (1) and (2) simultaneously, we get,1/2

tan1 tan

v Rg

+=

Hence, maximum speed for safe turn on a banked road isgiven by the expression

1/2

max

tan

1 tanv Rg

+=

Bending of a cyclist

A cyclist provides himself the necessary centripetal force by

leaning inward on a horizontal track, while going round acurve. Consider a cyclist of weight M g taking a turn of radius

rwith velocity v. In order to provide the necessary centripetal

force, the cyclist leans through angle inwards as shown in

Fig. The cyclist is under the action of the following forces:

(a) The weight M gacting vertically downward at the centre

of gravity of cycle and the cyclist.

(b) The reaction R of the ground on cyclist. It will act along a

line making angle with the vertical.

The vertical component R cos of the normal reaction R

will balance the weight of the cyclist, while the horizonta

component R sin will provide the necessary centripetaforce to the cyclist.

( )

( )2

cos ... 1

and sin ... 2

R Mg

MvR

r

=

=

Dividing equation (2) by (1), we have

N

mg

N sin

N cos

ca

f cos

fsinf

N

mg

N sin

N cos

ca

f cos

fsin

f


22/29




( )

2

2

sin /

cos

tan ... 3

R Mv r

R Mg

v

rg

=

=

Therefore, the cyclist should bend through an angle2

1tan v

rg

=

So as to have the necessary centripetal force while going

round a curved path. It follows that the angle through

which cyclist should bend will be greater, if

(a) the radius of the curve is small i.e. the curve is

sharper and

(b) the velocity of the cyclist is large.

It may be pointed out that for the same reasons, an

ice skater or an aeroplane has to bend inwards, while

taking a turn.

WORK ENERGY & POWER

Work:When a force Fis applied to a particle, and because ofthe application of this force if there is a displacement orcomponent of net displacement in the direction of appliedforce, then we say that force Fdo work on that particle.Work doneWby a constant forceF in a displacementr at

an angle:

. cosW F r Fr = =

Two important conditions about this formula:i. Force should be constant.

ii. Valid for particles.

We can apply this for those bodies, which are particlelike. That is when a force is applied to a particle likeobject, each part of it moves with a constant velocity. Itmeans that the body is in pure translation motion. Norotation is there in any part of the body

Units of work

Work done is a scalarquantity. S. I. Unit : joule = J.

1 J = 1 Nm = 1 kgm2/s2. CGS Unit: erg

1 erg = 1 dyne cm = 1 gcm2/s2

1 J = 107erg.

Work done by a variable force

Work done by a variable force F

in displacing a particle from

position vector 1r

to 2r

is given by

2

1

.r

r

W F d r =

Work done by spring force

Let us find the work done by spring of spring constant k

If the spring is displaced from 1 2.tox x x x= =

2 2

2 1

2 21 2

1 = ( )

2

x x

x x

W Fdx kx dx k x x= =

Special case:When spring is initially at normal position

Putting 1 2=0 tox x x= we get

( )2 2 21 21 1

2 2W k x x kx= =

Spring potential energy: U = - W = 21

2kx

Energy: The ability to do work is called energy.S. I. Unit ofenergy: J

Kinetic Energy(KEorK or T) =The energy due to motion ofa particle or body.

21

2K mv=

where v= velocity magnitude (speed).Kinetic energy depends on frame of reference.

Work Kinetic Energy Theorem

r

Mg

R sin

RR cos

VERTICAL


23/29




It states that work done by all the forces acting on a body is

equal to the change produced in the kinetic energy of the

body.

Suppose that a body in initially at rest and net force F

is

applied on the body to displace along its own direction. Then,

work done

( ). . 0W F ds F ds = = =

According to Newtons second law of motion,

F = m a,

where ais acceleration produced (in the direction of force) onapplying the force. Therefore,

( ) W Fds mads F ma

dv dv dvW mv ds mvdv a v

ds dt ds

= = =

= = = =

Therefore, work done by the force in order to increase its

velocity from u(initial velocity) to (final velocity) is given

by

2

2 2

2 1

2

1 12 2

vv v

u u u

vW mvdv m vdv m

W mv mu

W K K

= = =

=

=

Hence, work done on a body by all forces is equal to the

change in its kinetic energy.

Conservative Field

A force is said to be conservative force if work done by it is

path independent and depends only on net change of position

and not on the particular path followed in reaching the new

position.

or

A force is said to be conservative force if work done by it in a

closed path is zero. Examples of conservative forces are

. Gravitational force

. Spring force

. Electrostatic force

Potential Energy

Energy associated with position or configuration is known as

potential energy. It is denoted by U. Note that the potential

energy is a property of system of two or more particles rather

than of either particle alone.

Gravitational potential energy

gU mgh=

We have to take reference level in problems where

gravitational potential energy is zero. Gravitational potential

energy above reference level is taken as positive and it is

negative below reference level.

Spring potential energy

21

2sU kx=

Mechanical Energy: Sum of kinetic energy and potential

energy is known as mechanical energy.

ME = K + U

Conservative Field

A field is said to be conservative if:

1. A system where all the forces acting are conservative innature

OR

2. A system where some forces are conservative and some

are non-conservative such that work done by all non-

conservative forces = 0.

Relation between potential energy and work done in a

conservative field:

When a particle is displaced from position 1 to 2 in a

conservative field by the application of a conservative

force, work done by conservative force is given by

W= U= (U2 U

1)

As work done by non-conservative forces in a

conservative field is zero, therefore, work done by total

forces in a conservative field is given by

WT= U= (U

2 U

1)

Conservation of Mechanical Energy in a conservative

field:

Suppose a particle is displaced from point 1 to 2 in a

conservative field. Then,

Change in Potential Energy + Change in K.E. = 0 U+ K = 0

U1+ K

1= U

2+ K

2

Power

Average power:Average power is defined as work

done per unit time.

Denoted by P

work doneAverage Power =

time

WP

t

=

We can use this formula for constant power also.

Instantaneous Power (Power)

Denoted by P

dWP

dt=

Power is a scalar quantity

S.I. unit: Watt

Another unit: horsepower (hp)

1 hp = 746 W

Collision


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1. Elastic collision. Those collisions, in which both

momentum and kinetic energy of the system are

conserved, are called elastic collisions.

The collision between atomic and subatomic particles are

elastic in nature. In daily life, the collisions between two

glass or preferably ivory balls may be taken as elastic

collisions.

Characteristics of elastic collisions:

(a) The momentum is conserved.

(b) Total energy is conserved.

(c) The kinetic energy is conserved.(d) The mechanical energy is not converted into any

other form (sound, heat, light) of energy.

(e) Forces involved during the interaction are of

conservative nature.

2. Inelastic collision. Those collisions, in which the

momentum of the system is conserved but the kinetic

energy is not conserved, are called inelastic collisions.

Most of collisions in every-day life are inelastic

collisions.

3. Perfectly inelastic collision. Those collisions, in which

the colliding particles stick together after the collisionand then move with a common velocity, are called

perfectly inelastic collision.

Mud thrown on the wall and sticking to it, a bullet fired

into wooden block and remaining embedded in it, are the

examples of perfectly inelastic collision.

Characteristics of inelastic collisions:

(a) The momentum is conserved.

(b) The total energy is conserved.

(c) Loss in the kinetic energy is maximum.

(d) A part of whole of the mechanical energy may be

converted into other forms (heat, light, sound) of

energy.(e) Some or all of the forces involved are non-

conservative in nature.

It may be pointed out that inall types of collisions,

(a) momentum is conserved;

(b) total energy is conserved and

(c) it is the kinetic energy which may or not be

conserved.

ELASTIC COLLISION IN ONE DIMENSION

The collision between two bodies is said to be head-on or in

one dimension, if the colliding bodies continue to move alongthe same straight line after the collision.

Consider two perfectly elastic bodies A and B of masses M1and M2moving along the same straight line with velocities u1

and u2respectively. The two bodies will collide, only if u1>

u2. The two bodies undergo a head-on collision and continue

moving along the same straight line with velocities 1and 2

along the same direction. The two bodies will separate after

the collision, only if 2> 1.

As in elastic collision momentum is conserved, we have

M1u1+ M2u2= M11+ M22

M1(u11) = M2(2 u2) (i)

Since, kinetic energy is also conserved in an elastic collision

we have

( ) ( ) ( )

2 2 2 2

1 1 2 2 1 1 2 2

2 2 2 21 1 1 2 2 2

1 1 1 12 2 2 2

...

M u M u M M

M u M u ii

+ = +

=

Dividing equation (ii) by equation (i), we have

( )

2 2 2 21 1 2 2

1 1 2 2

1 1 2 2

1 2 2 2

...

u u

u u

u u

u u iii

=

+ = +

=

From equation (iii), it follows that in one dimensional elastic

collision, the relative velocity of approach (u1 u2) before

collision is equal to the relative velocity of separation (2 1)after collision.

Let us first and velocity of body A after collision. From

equation (iii), we have

2= u1 u2+ 1

Substituting for 2in equation (i), we get

( )

( ) ( )

( )( )

1 1 2 2 1 1 2 1 2 1

1 1 2 2 1 1 2 1 2 2 2 1

1 2 1 2 2 1 2 1

1 2 1 2 2

1

1 2

2

2 ...

M u M u M M u u

M u M u M M u M u M

M M u M u M M

M M u M uiv

M M

+ = + +

+ = + +

+ =

+ =

+

Again from equation (iii), have

1= 2 u1+ u2

Substituting for 1in equation (1.22), we have

( )( )2 1 2 1 12

1 2

2 ...

M M u M uv

M M

+=

+

Let us calculate the final velocities of the two bodies after

collision in the following special cases:

Au1

M1 M2u2

B

BEFORE COLLISION

A 1

M1 M2

2B

AFTER COLLISION


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1. When the two bodies are of equal masses: Let us

consider that

M1= M2= M (say)

From equation (iv), we have

( ) 1 2 21

1 1

2 0 2

2

M M u Mu Mu

M M M

u

+ += =

+ =

Also, from equation (v), we have

( ) 2 1 12

2 1

2 0 2

2

M M u Mu Mu

M M M

+ += =

+ =

Thus, if two bodies of equal masses suffer elastic collision

in one dimension, then after the collision, the bodies will

exchange their velocities.

2. When the target body is at rest: In this case, the body B

is at rest i.e. u2= 0. Then, substituting u2= 0 in equations

(iv) and (v), we have

( )( )

( )

1 2

1 1

1 2

12 1

1 2

...

2 ...

M Mu vi

M M

M

u viiM M

=

+

= +

When the target body B is at rest, let us find the final

velocities of the two bodies in the following subcases:

(a) When the two bodies are of equal masses: Setting M1=

M2= M in equations (vi) and (vii), we get

1= 0 and 2= u1

Therefore, when body A collides against body B of equal

mass at rest, the body A comes to rest and the body B

moves on with the velocity of the body A. This is

sometimes observed, when one of the two boys playing

with glass balls, shoots a stationary glass ball with a ball

with the help of his fingers; his own glass ball comes torest, while the stationary ball of the other boy starts

moving with the same velocity.

(b) When the mass of body B is negligible as compared to

that of A: When M2 < < M1, then in equations (vi) and

(vii), M2can be neglected as compared to M1i.e. M1 M2

M1and M1+ M2 M1. Therefore, we have

1 11 1 1 2 1 1

1 1

2and 2

M Mu u u u

M M = = = =

Therefore, when a heavy body A collides against a light

body B at rest, the body A should keep on moving with

same velocity and the body B will come in motion withvelocity double that of A. Thus, in principle, if a moving

truck (heavy body) collides against a stationary drum,

then the truck would keep on moving with the same

velocity, while the drum would come in motion with a

velocity double the velocity of the truck.

(c) When the mass of body B is very large as compared to

that of A: When M2 > > M1, then in equations (vi) and

(vii), M1can be neglected in comparison to M2 i.e. M1

M2 M2and M1+ M2M2. Therefore, we have

( )2 11 1 1 2 1 2 12 2

2and 0

M Mu u u M M

M M

= = = >>

Therefore, when a light body A collides against a heavy

body B at rest, the body A should start moving with equa

velocity in opposite direction, while the body B should

practically remain at rest. This result is in accordancewith the observation that when a rubber ball hits a

stationary wall, the wall remains at rest, while the bal

bounces back with the same speed.

COEFFICIENT OF RESTITUTION

The coefficient of restitution is defined as the ratio of the

velocity of separation to the velocity of approach of the

colliding particles. It is denoted by e.

According to definition, the coefficient of restitution is given

by

velocity of separationvelocity of approach

e=

The coefficient of restitution can be used to distinguish

between the three types of collision as below:

1. For elastic collision, e = 1

i.e. in an elastic collision, velocity of separation is equal

to velocity of approach.

2. For inelastic collision, 0 < e< 1

i.e. in an inelastic collision, the two particles possess non

zero velocity of separation, which is always less than the

velocity of approach.

3.

For perfectly inelastic collision, e= 0i.e. in a perfectly inelastic collision, the two bodies do no

get separated and move with a common velocity.


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ELASTIC COLLISION IN TWO DIMENSIONS

Consider two perfectly elastic bodies A and B of masses M1

and M2moving along the same straight line (say X-axis) with

velocities u1and u2. If the body A is moving with a velocity

greater than that of B i.e. if u1 > u2, then two bodies will

collide. After the collision, the two bodies A and B travel with

velocities making angles 1and 2with the incident direction

(X-axis) as shown in Fig.

Since the collision is perfectly elastic, the kinetic energy must

be conserved.

Therefore,

( )2 2 2 21 1 2 2 1 1 2 21 1 1 1

...2 2 2 2

M u M u M M i + = +

Momentum is a vector quantity. As the two bodies move along

different directions after the collision, the momentum of thetwo bodies is separately conserved along X-axis and Y-axis.

The component of momentum of body A after collision along

X-axis = M11cos 1

The component of momentum of body B after collision along

X-axis = M22cos 2

Applying the law of conservation of momentum along X-axis,

we have

M1u1+ M2u2= M11cos 1+ M2 2cos 2 (ii)The component of momentum of body A after collision along

Y-axis = M1 1 sin 1 (along OY)

The component of momentum of body B after collision along

Y-axis = M22sin 2 (along OY)

= M22sin 2(along OY)

Before collision, the component of momentum of body A or of

body B along Y-axis is zero.

Therefore, applying the law of conservation of momentum

along Y-axis, we have

0 + 0 = M11sin 1+ (M22sin 2)

or M11sin 1= M22sin 2 (iii)

Special cases:

(a) Glancing collision. For such collisions, 1 0 =

and

2 90 . =

From equations (1) and (2), we get

u1= v1and v2= 0

K.E. of the target particle 22 21

02

m v= =

Hence in a glancing collision, the incident particle does

not lose any kinetic energy and is scattered almos

undeflected.

(b) Head-on collision. In such a collision, the target particle

moves in the direction of the incident particle, i.e., 2= 0o

Then equations (1) and (2) take forms:

m1u1= m1v1cos 1+ m2v2and 0 = m1v1sin 1

Equation (3) for the kinetic energy remains unchanged.

(c) Elastic collision of two identical particles. As the two

particles are identical, so m1= m2= m(say). By

conservation of K.E. for elastic collision,

2 2 2 2 2 21 1 2 1 1 2

1 1 1or

2 2 2mu mv mv u v v= + = +

By conversation of linear momentum,

( ) ( )

2 2 21 1 2 1 1 2

1 1 1 2 1 2

1 1 1 2 2 1 2 2

2 2 21 1 2 1 2

2 2 2 2 21 1 1 2 1 2 1

or

. .

. . . .

or 2 .

or 2 .

o

mu mv mv u v v

u u v v v v

v v v v v v v v

u v v v v

u u v v v v u

= + = +

= + +

= + + +

= + +

= + + =

1 2r . 0.v v =

This shows that the angle between 1v

and 2v

is 90o

Hencetwo identical particles move at right angles to each

other after elastic collision in two dimensions.

u1M1

M2

u1

B

A

1X

A B

2

M1

M2

1

2

Y

Y

X


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Physics Classes by Pranjal Sir

(Admission Notice for XI & XII - 2014-15)

Batches for Std XIIth

Batch 1 (Board + JEE Main + Advanced): (Rs. 16000)Batch 2(Board + JEE Main): (Rs. 13000)Batch 3(Board): (Rs. 10000)Batch 4(Doubt Clearing batch): Rs. 8000

About P. K. Bharti Sir (Pranjal Sir)

B. Tech., IIT Kharagpur (2009 Batch) H.O.D. Physics, Concept Bokaro Centre Visiting faculty at D. P. S. Bokaro Produced AIR 113, AIR 475, AIR 1013 in JEE -

Advanced Produced AIR 07 in AIEEE (JEE Main)

Address:Concept, JB 20, Near Jitendra Cinema, Sec 4Bokaro Steel CityPh: 9798007577, 7488044834Email:[email protected]

Website:www.vidyadrishti.org

Physics Class Schedule for Std XIIth (Session 2014-15) by Pranjal Sir

Sl. No. Main Chapter Topics Board level JEE Main Level JEE Adv Level

Basics from XIth Vectors, FBD, Work, Energy, Rotation,SHM

3r Mar to 4t Apr 14

1. Electric Charges andFields

Coulombs Law 5th& 6thApr 5th& 6thApr 5th& 6thAprElectric Field 10t & 12t Apr 10t & 12t Apr 10t & 12t AprGausss Law 13th& 15thApr 13th& 15thApr 13th& 15thAprCompetition Level NA 17th& 19thApr 17th& 19thApr

2. Electrostatic Potentialand Capacitance

Electric Potential 20t & 22n Apr 20t & 22n Apr 20t & 22n AprCapacitors 24t & 26t Apr 24t & 26t Apr 24t & 26t AprCompetition Level NA 27th& 29thApr 27th& 29thApr, 1st, 3r

& 4thMayPART TEST 1 Unit 1 & 2 4

thMay NA NA

NA 11t

May 11t

May3. Current Electricity Basic Concepts, Drift speed, Ohms

Law, Cells, Kirchhoffs Laws,Wheatstone bridge, Ammeter,Voltmeter, Meter Bridge, Potentiometeretc.

6th, 8th, 10th, 13thMay



Competition Level NA 15t & 16t May 15t , 16t , 17t , 18t &19thMay

PART TEST 2 Unit 3 18th

May NA NANA 20thMay 20thMay

SUMMER BREAK 21stMay 2013 to 30

thMay 2013

4. Moving charges andMagnetism

Force on a charged particle (Lorentzforce), Force on a current carrying wire,Cyclotron, Torque on a current carryingloop in magnetic field, magneticmoment

31stMay, 1st&3rdJun

31stMay, 1st&3rdJun

31stMay, 1st& 3rdJun

Biot Savart Law, Magnetic field due to acircular wire, Ampere circuital law,Solenoid, Toroid

5th, 7th& 8thJun 5th, 7th& 8thJun 5th, 7th& 8thJun

Competition Level NA 10th& 12thJun 10th, 12th, 14th& 15thJun

PART TEST 3 Unit 4 15th

Jun NA NANA 22

ndJun 22

ndJun
mailto:[email protected]:[email protected]:[email protected]://www.vidyadrishti.org/http://www.vidyadrishti.org/http://www.vidyadrishti.org/http://www.vidyadrishti.org/mailto:[email protected]


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5. Magnetism and Matter 17t , 19t & 21stJun

17t , 19t & 21stJun

Not in JEE AdvancedSyllabus

6. ElectromagneticInduction

Faradays Laws, Lenzs Laws, A.C.Generator, Motional Emf, Induced Emf,Eddy Currents, Self Induction, MutualInduction

24t , 26t & 28t Jun

24t , 26t & 28t Jun

24t , 26t & 28t Jun

Competition Level NA 29t Jun & 1stJul 29t Jun, 1st, 3r & 5t

JulPART TEST 4 Unit 5 & 6 6

thJul NA NA

NA 13th

Jul 13th

Jul

7.

Alternating current AC, AC circuit, Phasor, transformer,resonance, 8th

, 10th

& 12th

Jul 8th

, 10th

& 12th

Jul 8th

, 10th

& 12th

Jul

Competition Level NA 15thJuly 15th& 17thJuly8. Electromagnetic Waves 19t & 20t July 19t & 20t July Not in JEE Advanced

SyllabusPART TEST 5 Unit 7 & 8 27

thJul 27thJul 27thJul

Revision Week Upto unit 8 31stJul & 2ndAug

31stJul & 2ndAug

31stJul & 2ndAug

Grand Test 1 Upto Unit 8 3rd

Aug 3rd

Aug 3rdAug9.

Ray Optics

Reflection 5th& 7thAug 5th& 7thAug 5th& 7thAugRefraction 9t & 12t Aug 9t & 12t Aug 9t & 12t AugPrism 14t Aug 14t Aug 14t AugOptical Instruments 16thAug 16thAug Not in JEE Adv

SyllabusCompetition Level NA 19t & 21stAug 19t , 21st, 23r , 24t Au

10.

Wave Optics

Huygens Principle 26t Aug 26t Aug 26t AugInterference 28th & 30th Aug 28th & 30th Aug 28th & 30th AugDiffraction 31stAug 31stAug 31stAugPolarization 2n Sep 2n Sep 2n SepCompetition Level NA 4th& 6thSep 4th, 6th, 7th, 9th, 11thSe

PART TEST 6 Unit 9 & 10 14th

Sep 14thSep 14thSepREVISION ROUND 1 (For JEE Main & JEE Advanced Level): 13

thSep to 27

thSep

Grand Test 2 Upto Unit 10 28th

Sep 28th

Sep 28thSep

DUSSEHRA & d-ul-Zuha Holidays: 29th

Sep to 8th

Oct

11. Dual Nature ofRadiation and Matter

Photoelectric effect etc 9th& 11thOct 9th& 11thOct 9th& 11thOct

Grand Test 3 Upto Unit 10 12th

Oct 12th

Oct 12thOct

12. Atoms 14t & 16t Oct 14t & 16t Oct 14t & 16t Oct

13. Nuclei 18th& 19thOct 18th& 19thOct 18th& 19thOctX-Rays NA 21stOct 21st& 25thOct

PART TEST 7 Unit 11, 12 & 13 26t Oct NA NA14. Semiconductors Basic Concepts and Diodes, transistors,

logic gates26th, 28th, 30thOct & 1stNov

26th, 28th, 30thOct & 1stNov

Not in JEE AdvSyllabus

15. Communication System 2n & 4t Nov 2n & 4t Nov Not in JEE AdvSyllabus

PART TEST 8 Unit 14 & 15 9t Nov 9t Nov NA

Unit 11, 12 & 13 Competition Level NA 8th

, 9th

& 11th

Nov 8th

, 9th

, 11th

, 13th

& 15NovPART TEST 9 Unit 11, 12, 13, X-Rays NA 16

thNov 16

thNov

Revision Round 2

(Board Level)

Mind Maps & Back up classes for late

registered students18

thNov to

Board Exams

18th

Nov to

Board Exams

18th

Nov to Board

Exams

Revision Round 3

(XIth portion for JEE)

18th

Nov to JEE 18th

Nov to JEE 18th

Nov to JEE

30 Full Test Series Complete Syllabus Date will be published after Oct 2014


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Documents

Quick Revision Mechanics XI 2014