WorkPower Energy Notes1

8/20/2019 WorkPower Energy Notes1

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Work, Energy Power, and Collision 1

6.1 Introduction.

The terms 'work', 'energy' and 'power' are frequently used in everyday language. A

farmer learing weeds in his !eld is said to "e working hard. A woman arrying water

from a well to her house is said to "e working. #n a drought a$eted region she may "e

required to arry it over large distanes. #f she an do so, she is said to have a large

stamina or energy. Energy is thus the apaity to do work. The term power is usually

assoiated with speed. #n karate, a powerful punh is one delivered at great speed. #n

physis we shall de!ne these terms very preisely. We shall !nd that there is at "est a

loose orrelation "etween the physial de!nitions and the physiologial pitures theseterms generate in our minds.

Work is said to "e done when a fore applied on the "ody displaes the "ody

through a ertain distane in the diretion of fore.

6.2 Work Done by a Constant Force.

%et a onstant fore F "e applied on the "ody suh that it makes an angle θ with

the hori&ontal and "ody is displaed through a distane s

y resolving fore F into two omponents (

)i* F osθ in the diretion of displaement of the "ody.

)ii* F sinθ in the perpendiular diretion of displaement of the

"ody.

+ine "ody is "eing displaed in the diretion of θ osF ,

therefore work done "y the fore in displaing the "ody through a distane s is given "y

θ θ os*os) FssF W ==

or sF W .=

Thus work done "y a fore is equal to the salar or dot produt of the fore and the

displaement of the "ody.

#f a num"er of fore nF F F F ......,, - are ating on a "ody and it shifts from

position vetor r to position vetor -r then *.)*....) -,- r r F F F F W n −+++=

6.3 Nature of Work Done.

Positive work Negative work

Positive work means that fore )or its omponent*is parallel to displaement

/egative work means that fore )or itsomponent* is opposite to displaement i.e.

Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

F sinθ

F osθ

s

θ

F


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2 Work, Energy, Power and Collision

oo 011


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a magneti !eld as fore *6)5 !" #F ×= is

always perpendiular to motion, work done"y this fore is always &ero.

)-* If t!ere is no dis"#ace$ent &s ' (%

xample$ )i* When a person tries to displae a wall orheavy stone "y applying a fore then it doesnot move, the work done is &ero.

)ii* A weight lifter does work in lifting theweight o$ the ground "ut does not work inholding it up.

)* If t!ere is no force acting on t!e body &F ' (%

xample( 4otion of an isolated "ody in free spae.

Sample Problems based on work done by constant force

P roblem 1. A "ody of mass : %g is plaed at the origin, and an move only on the x ;a2is. Afore of 1 & is ating on it in a diretion making an angle of o


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6.0 Work Done by a ariab#e Force.

When the magnitude and diretion of a fore varies with position, the work done "ysuh a fore for an in!nitesimal displaement is given "y sd F dW .=

The total work done in going from + to ! as shown in the !gure is

∫ ∫ == !

+

!

+dsF sd F W *os). θ

#n terms of retangular omponent % F )F iF F , y x ??? ++=

% d, )dy idx sd ??? ++=

*???.)*???) % d, )dy idx % F )F iF W !

+ , y x ++++=∴ ∫

or ∫ ∫ ∫ ++= !

+

!

+

!

+

,

, ,

x

x

y

y y x d, F dy F dx F W

Sample Problems based on work done by variable force

P roblem 6. A position dependent fore & x x F *,->) -+−= ats on a small a"Det of mass -%g to displae it from 1= x to m x := . The work done in Doule is

&C* P)+ 1,,0%

)a* >1 ' )"* ->1 ' )* : ' )d*

(olution ( )d* Work done ' x x x dx x x dx F x

x ,:-:-:,:6>5*,->) :1

,-:

1

--

=+−=+−=+−== ∫ ∫ P roblem 4. A partile moves under the e$et of a fore F Cx from x 1 to x x . The work

done in the proess is

&CP)+ 1,52%

)a* -Cx )"* --

Cx )* Cx )d*

(olution ( )"* Work done -1

-

1 -

-

-

x C x

Cdx Cx dx F

x x x

x =

=== ∫ ∫

P roblem 5. The vessels + and ! of equal volume and weight are immersed in water to a depthh. The vessel + has an opening at the "ottom through whih water an enter. #f the

work done in immersing + and ! are +W and !W respetively, then

)a* ! + W W = )"* ! + W W < )* ! + W W > )d*

! + W W

(olution ( )"* When the vessels are immersed in water, work has to "e done against up;thrustfore "ut due to opening at the "ottom in vessel +, up;thrust fore goes ondereasing. +o work done will "e less in this ase.

P roblem ,. Work done in time t on a "ody of mass m whih is aelerated from rest to a speed" in time t as a funtion of time t is given "y

)a* -

-

t

t

" m )"*

-

t t

" m )* -

-

-

t

t

m"

)d*

-

-

-

-

t

t

" m

(olution ( )d* Work done F .s

--

. t ama --

-

t am= -

-

-

t

t

" m

=

= given*)onaeleratiAs

t

" a


+

!

ds

F θ


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Work, Energy Power, and Collision -

6.- Di$ension and nits of Work .

Di$ension 7 As work Bore × displaement

∴ 5W 6 5Bore6 × 59isplaement6

656565

--- −−

=×= - M..M.-

nits 7 The units of work are of two types

8bso#ute units 9ravitationa# units

'oule 5+.#.6( Work done is said to "e one 'oule, when &ewton fore displaes the "ody through meter inits own diretion.

Brom W / F.s

'oule &ewton × metre

%g;m 5+.#.6( Kg0m of work is done when afore of %g0wt. displaes the "ody throughm in its own diretion.

Brom W F s

%g0m %g0wt × metre

0.3 & × metre 0.3

'oule

rg 5C..+.6 ( Work done is said to "e one erg when dyne fore displaes the "ody through cm inits own diretion.

Brom W F s

cm1ynerg ×=

2elation between 'oule and erg

'oule & × m 1: dyne × 1- cm

1

>

dyne ×

cm 1

>

rg

gm;cm 5C..+.6 ( gm0cm of work is donewhen a fore of gm0wt displaes the "odythrough cm in its own diretion.

Brom W F s

gm0cm gm0wt × cm. 03

dyne × cm

03 erg

6.6 Work Done Ca#cu#ation by Force Dis"#ace$ent 9ra"!.

%et a "ody, whose initial position is i x , is ated upon "y a varia"le fore )whose

magnitude is hanging ontinuously* and onsequently the "ody aquires its !nal

position f x .

%et F "e the average value of varia"le fore within the interval dx from position x to ) x 3 dx * i.e. for small displaement dx . The work done will "e the area of the shaded

strip of width dx. The work done on the "ody in displaing it from position i x to f x will

"e equal to the sum of areas of all the suh strips

dx F dW =

∫ ∫ ==∴ f

i

f

i

x

x

x

x dx F dW W

∫ =∴ f

i

x

x dx W *widthof stripof Area)

f i x x W and'etweenurveunderArea=∴ i.e. Area under fore displaement urve with proper alge"rai sign represents work

done "y the fore.

Sample problems based on force displacement graphP roblem 1(. A 1 %g mass moves along x ;a2is. #ts aeleration as a funtion of its position is

shown in the !gure. What is the total work done on the mass "y the fore as the

mass moves from 1= x to 3= x cm &8) :)ed.; 2(((%)a* '-13 −× )"* '-1< −×

)* '=1= −× )d* '1


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(olution ( )a* Work done on the mass mass × overed area "etween the

graph and displaement a2is on a0t graph.

-- 1-1*13)-

1 −− ×××× -13 −× '.

P roblem 11. The relationship "etween fore and position is shown in the !gure given )in one

dimensional ase*. The work done "y the fore in displaing a "ody from = x cmto := x cm is &CP)+ 1,46%

)a* -1 ergs

)"* 1 ergs

)d* >11 ergs

(olution ( )a* Work done Covered area on fore;displaement graph × 1 8 × -1 @ × -1

8 × 1 -1 erg.

P roblem 12. The graph "etween the resistive fore F ating on a "ody and the distaneovered "y the "ody is shown in the !gure. The mass of the "ody is -: %g and

initial veloity is - m*s. When the distane overed "y the "ody is m: , its kinetienergy would "e

)a* :1 ' )"* =1 ')* -1 ' )d* 1 '

(olution ( )d* #nitial kineti energy of the "ody :1*-)-:-

-

-- =××== mu '

Binal kineti energy #nitial energy @ work done against resistive fore )Area "etween

graph and displaement a2is* 1=1:1-1=-

:1 =−=××−= '.

6.4 Work Done in Conservative and Non


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)-* #n onservative !eld work done "y the fore )line integral of the fore i.e. ∫ ld F . *over a losed pathFloop is &ero.

1=+ →→ +!! + W W

or ∫ = 1. ld F

Conservative force 7 The fores of these type of !elds are known as onservativefores.

xample $ Eletrostati fores, gravitational fores, elasti fores, magneti foresetc and all the entral fores are onservative in nature.

#f a "ody of man m lifted to height h from the ground level "y di$erent path asshown in the !gure

Work done through di$erent paths

mghmgsF W 5 =×== .

mgh

mglmgsF W 55 =×=×==θ

θ θ sin

sinsin.

=- 111 mghmghmghmghW 555 ++++++= mghhhhmg =+++= *) =- mgsd F W 56 == ∫ .

#t is lear that mgW W W W 56 555555 ==== .Burther if the "ody is "rought "ak to its initial position +, similar amount of work

)energy* is released from the system it means mgW +! = and mgW !+ −= .Gene the net work done against gravity over a round strip is &ero.

!+ +!&et W W W += 1*) =−+= mghmgh

i.e. the gravitational fore is onservative in nature.

Non


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P roblem 13. #f -,W W and W represent the work done in moving a partile from + to !along three di$erent paths , - and respetively )as shown* in the gravitational

!eld of a point mass m, !nd the orret relation

)a* ,- W W W >> )"* ,- W W W ==

)* ,- W W W >

(olution ( )"* As gravitational !eld is onservative in nature. +o work done in moving a partile

from + to ! does not depends upon the path followed "y the "ody. #t always

remains same.

P roblem 10. A partile of mass 1.1 %g travels along a urve with veloity given "y

% i ?


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Work, Energy Power, and Collision ,

#f %gamum ->1. −×== then 'ouleMe6 11:.0, −×== .

#f %gm = then 'oule eat e#ectrica#

Coalurning 4irophone Thermo;ouple


e@ 8 e8γ 7 Photon*

& ( @ 8CathodeAnode

Got

G

Fe

Cu

Cold


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1( Work, Energy, Power and Collision

>eat )ec!anica# *#ectrica# )ec!anica# *#ectrica# >eat

Engine 4otor Geater

*#ectrica# ound *#ectrica# C!e$ica# *#ectrica# =ig!t

+peaker Holtameter ul"

Sample problems based on energy

P roblem 1-. A partile of mass LmK and harge L#K is aelerated through a potentialdi$erene of L6 K "olt . #ts energy is

&P*8+ 2((1%

)a* #6 )"* m#6 )* 6 m

#

)d*

(olution ( )a* Energy of harged partile harge × potential di$erene #6

P roblem 16. An ie ream has a marked value of >11 %cal. Gow many kilowatt hour of

energy will it deliver to the "ody as it is digested&8) :)ed.; 2(((%

)a* 1.3 %Wh )"* 1.01 %Wh )* . %Wh )d* 1.>

(olution ( )a* >11 % cal -.=1>11 , ×× ' %Wh3.1111<

,

=×

××= 5As

%W ' 1


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⇒ as" -1- += a

" s

-

-

=∴

+ine the displaement of the "ody is in the diretion of the applied fore, then workdone "y the fore is

sF W ×= a

" ma -

-

×=

⇒ --

m" W =

This work done appears as the kineti energy of the "ody --

m" W K ==

)-* Ca#cu#us $et!od 7 %et a "ody is initially at rest and fore F is applied on the"ody to displae it through sd along its own diretion then small work done

dsF sd F dW == .

⇒ dsamdW = 5As F ma6

⇒ ds

dt

d" mdW =

=dt

d" a +s

⇒ dt

dsmd" dW .=

⇒ d" " mdW = MM.)i*

= "

dt

ds +s

Therefore work done on the "ody in order to inrease its veloity from &ero to " isgiven "y

∫ ∫

===

" " "

" md" " md" m" W

1 11

-

- -

-

m" =

This work done appears as the kineti energy of the "ody -

-

m" K = .

#n vetor form *.)-

" " mK =

As m and " " . are always positive, kineti energy is always positive salar i.e. kinetienergy an never "e negative.

)* ?inetic energy de"ends on fra$e of reference 7 The kineti energy of aperson of mass m, sitting in a train moving with speed ", is &ero in the frame of train "ut

-

-

m" in the frame of the earth.

)=* ?inetic energy according to re#ativity 7 As we know-

-

m" = .

ut this formula is valid only for )" 99 c* #f " is ompara"le to c )speed of light infree spae smF1, 3× * then aording to Einstein theory of relativity

-

--

-

*F)mc

c"

mc −

−=

):* Work


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This is work energy theorem, it states that work done "y a fore ating on a "ody isequal to the hange produed in the kineti energy of the "ody.

This theorem is valid for a system in presene of all types of fores )e2ternal orinternal, onservative or non;onservative*.

#f kineti energy of the "ody inreases, work is positive i.e. "ody moves in the

diretion of the fore )or !eld* and if kineti energy dereases work will "e negative ando"Det will move opposite to the fore )or !eld*.

xamples $ )i* #n ase of vertial motion of "ody under gravity when the "ody isproDeted up, fore of gravity is opposite to motion and so kineti energy of the "odydereases and when it falls down, fore of gravity is in the diretion of motion so kinetienergy inreases.

)ii* When a "ody moves on a rough hori&ontal surfae, as fore of frition atsopposite to motion, kineti energy will derease and the derease in kineti energy isequal to the work done against frition.

)


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-. Iineti energy of a system of partiles is &ero

Then &8I*** 2((3%

)a* implies - and - implies )"* does not

imply - and - does not imply

)* implies - "ut - does not imply )d* does not imply - "ut - implies (olution ( )d* 4omentum is a vetor quantity whereas kineti energy is a salar quantity. #f the

kineti energy of a system is &ero then linear momentum de!nitely will "e &ero "ut

if the momentum of a system is &ero then kineti energy may or may not "e &ero.

P roblem 1,. A running man has half the kineti energy of that of a "oy of half of his mass. The man speeds up "y m*s so as to have same K.. as that of "oy. The original

speed of the man will "e &Pb. P)+ 2((1%

)a* smF- )"* smF*-) − )* smF*-)

−)d*

(olution ( )* %et m mass of the "oy, M mass of the man, " veloity of the "oy and 6

veloity of the man

#nitial kineti energy of man

== ---

-

-

" mM6

= ---

-

"

M

= given-

As M

m

⇒

=

-- " 6 =

-

" 6 =⇒ .....)i*

When the man speeds up "y m*s ,---

--

-

*)

-

"

M" m6 M

==+ ⇒

-*)

-- " 6 =+

⇒ -

"

6 =+ .....)ii*

Brom )i* and )ii* we get speed of the man sm6 F-

−= .

P roblem 2(. A "ody of mass 1 %g at rest is ated upon simultaneously "y two fores =& and &at right angles to eah other. The kineti energy of the "ody at the end of 1 se is

&?era#a :*ngg.; 2((1%

)a* 11 ' )"* 11 ' )* :1 ' )d*

(olution ( )d* As the fores are working at right angle to eah other therefore net fore on the

"ody &F :,= -- =+=

Iineti energy of the "ody work done F × s --

-

-

t

m

F F t aF

×=×=

-:*1)1

:

-

: - =

×= '.

P roblem 21. #f the momentum of a "ody inreases "y 1.1N, its kineti energy will inrease"y &)P P*+ 2((1%

)a* 1.1N )"* 1.1- N )* 1.1= N )d*



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(olution ( )"* Iineti energym

P

-

-

= ∴ -P ∝

Perentage inrease in kineti energy -)N inrease in momentum* 5#f

hange is very small6

-)1.1N* 1.1-N.

P roblem 22. #f the momentum of a "ody is inreased "y 11 N, then the perentageinrease in the kineti energy is &NC*/+ 1,,( > 1,,, Pb. P)+ 1,,, CP)+1,,,A 2((( C* P)+ 2((1%

)a* :1 N )"* -11 N )* --: N )d*

(olution ( )d*m

P

-

-

= ⇒ =-

--

-

- =

=

=

P

P

P

P

- = = 11 +=+= N of .

P roblem 23. A "ody of mass : %g is moving with a momentum of 1 %g0m*s. A fore of 1.- &ats on it in the diretion of motion of the "ody for 1 seonds. The inrease in its

kineti energy is &)P P*+ 1,,,%

)a* -.3 ' )"* .- ' )* .3 ' )d*

(olution ( )d* Change in momentum t F PP ×=−= - ⇒

sm%gt F PP F;-1-.11- =×+=×+=

#nrease in kineti energy 65-

-

-- PP

m −=

.=.=1

==611==5

:-

6*1)*-5)

-

-- 'm

==−×

=−=

P roblem 20. Two masses of g and 0g are moving with equal kineti energies. The ratio of the magnitudes of their respetive linear momenta is

&C* P)+ 1,,3 CP)+ 1,,-%

)a* ( 0 )"* 0 ( )* ( )d*

(olution ( )* mP -= ∴ mP ∝ if onstant . +o

0

-

-

===m

m

P

P.

P roblem 2-. A "ody of mass - %g is thrown upward with an energy =01 '. The height at whihits kineti energy would "eome half of its initial kineti energy will "e &

-F3.0 smg = %

)a* : m )"* -: m )* -.: m )d*

(olution ( )* #f the kineti energy would "eome half, then Potential energy -

)#nitial kineti

energy*

⇒ 6=015-

=mgh ⇒

6=015-

3.0- =×× h ⇒

mh :.-=

P roblem 26. A 11 g mass has a veloity of *?=?) )i + m*sec at a ertain instant. What is its

kineti energy



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Work, Energy Power, and Collision 1-

)a* .: ' )"* -.= ' )* .>: ' )d*

(olution ( )* *?=?,) )i" += ∴ sm" F:=, -- =+= . +o kineti energy

'm" >:.,*:),.1-

-

-- =××=

6.11 to""ing of e!ic#e by /etarding Force.

#f a vehile moves with some initial veloity and due to some retarding fore it stops

after overing some distane after some time.

)* to""ing distance 7 %et m / 4ass of vehile, " / Heloity, P /

4omentum, / Iineti energy

F / +topping fore, x / +topping distane, t / +topping

time

Then, in this proess stopping fore does work on the vehile and destroy the

motion.

y the work; energy theorem

-

-

m" K W =∆=

⇒ +topping fore )F * × 9istane ) x * Iineti energy )*

⇒ +topping distane ) x **)fore+topping

*)energyIineti

F

=

⇒

F

m" x

-

-

= M..)i*

)-* to""ing ti$e 7 y the impulse;momentum theorem

Pt F Pt F =×⇒∆=×

∴ F

Pt =

orF

m" t = M..)ii*

)* Co$"arison of sto""ing distance and ti$e for two ve!ic#es 7 Two vehiles

of masses m and m- are moving with veloities " and " - respetively. When they are

stopped "y the same retarding fore )F *.

The ratio of their stopping distanes ---

-

-

-

" m

" m

x

x ==

and the ratio of their stopping time--

-

-

" m

" m

P

P

t

t ==

#f vehiles possess same veloities

" " --

-

m

m

x

x

= -

-

m

m

t

t

=

#f vehile possess same kineti momentum


#nitial veloity "

x

Binal veloity 1


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P P-

---

-

-

-

-

-

- m

m

P

m

m

P

x

x =

==

-

-

==P

P

t

t

#f vehile possess same kineti energy

- -

-

==

x x

-

--

-

-

-

-mm

m

mPP

t t ===

&ote ( #f vehile is stopped "y frition then

+topping distane

F

m" x

-

-

=ma

m" -

-

= g"

µ -

-

= 6As5 ga µ =

+topping timeF

m" t =

gm

m"

µ =

g

"

µ =

Sample problems based on stopping of vehicle

P roblem 24. Two arts on hori&ontal straight rails are pushed apart "y an e2plosion of apowder harge : plaed "etween the arts. +uppose the oeOients of frition

"etween the arts and rails are idential. #f the -11 %g art travels a distane of <

metres and stops, the distane overed "y the art weighing 11 %g is &CP)+ 1,5,%

)a* - metres )"* -= metres

)* < metres )d* - metres

(olution ( )* Iineti energy of art will goes against frition. ∴ smgm

P ×== µ

-

-

⇒

-

-

- gm

Ps

µ =

As the two arts pushed apart "y an e2plosion therefore they possess same linearmomentum and oeOient of frition is same for "oth arts )given*. Therefore thedistane overed "y the art "efore oming to rest is given "y

-

ms∝ ∴ metre(

m

m

s

s


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(olution ( )a* Jetarding fore )F * × distane overed ) x * Iineti energy

--

m"

#f " and F are onstants then x ∝ m ∴ :.:.

-

- ===m

m

m

m

x

x ⇒ x - .: s.

P roblem 3(. A vehile is moving on a rough hori&ontal road with veloity " . The stoppingdistane will "e diretly proportional to

)a* " )"* " )* -" )d*

(olution ( )* Asa

" s

-

-

= ∴ -" s∝ .

6.12 Potentia# *nergy.

Potential energy is de!ned only for onservative fores. #n the spae oupied "y

onservative fores every point is assoiated with ertain energy whih is alled the

energy of position or potential energy. Potential energy generally are of three types (

Elasti potential energy, Eletri potential energy and ravitational potential energy etc.)* C!ange in "otentia# energy 7 Change in potential energy "etween any two

points is de!ned in the terms of the work done "y the assoiated onservative fore in

displaing the partile "etween these two points without any hange in kineti energy.

∫ −=−=− -

.-r

r W r d F 88

MM)i*

We an de!ne a unique value of potential energy only "y assigning some ar"itrary

value to a !2ed point alled the referene point. Whenever and wherever possi"le, we

take the referene point at in!nite and assume potential energy to "e &ero there, i.e. if

take ∞=r and r r =- then from equation )i*

∫ ∞ −=−= r

W r d F 8

.

#n ase of onservative fore )!eld* potential energy is equal to negative of work

done in shifting the "ody from referene position to given position.

This is why in shifting a partile in a onservative !eld )say gravitational or eletri*,

if the partile moves opposite to the !eld, work done "y the !eld will "e negative and so

hange in potential energy will "e positive i.e. potential energy will inrease. When the

partile moves in the diretion of !eld, work will "e positive and hange in potential

energy will "e negative i.e. potential energy will derease.

)-* +!ree di$ensiona# for$u#a for "otentia# energy7 Bor only onservative!elds F equals the negative gradient *) ∇− of the potential energy.

+o 8F ∇−= )∇ read as 9el operator or /a"la operator and

% d,

d )

dy

di

dx

d ??? ++=∇ *

⇒

++−= %

d,

d8 )

dy

d8i

dx

d8F ???

where =dx

d8 Partial derivative of 8 w.r.t. x )keeping y and onstant *

=dy

d8 Partial derivative of 8 w.r.t. y )keeping x and onstant*

=d,

d8 Partial derivative of 8 w.r.t. & )keeping x and y onstant*



18/56


)* Potentia# energy curve 7 A graph plotted "etween the potential energy of apartile and its displaement from the entre of fore is alled potential energy urve.

Bigure shows a graph of potential energy funtion 8) x * for one dimensional motion.As we know that negative gradient of the potential energy

gives fore.

∴ F dx d8 =−

)=* Nature of force 7

)i* Attrative fore ( n inreasing x , if 8 inreases

positiv=dx

d8

then F is negative in diretion i.e. fore is attrative in nature. #n graph this is represented inregion !C.

)ii* Jepulsive fore ( n inreasing x , if 8 dereases negativ=dx

d8

then F is positive in diretion i.e. fore is repulsive in nature. #n graph this is represented inregion +!.

)iii* ero fore ( n inreasing x , if 8 does not hanges 1=dx

d8

then F is &ero i.e. no fore works on the partile. Point !, C and 1 represents thepoint of &ero fore or these points an "e termed as position of equili"rium.

):* +y"es of eBui#ibriu$ 7 #f net fore ating on a partile is &ero, it is said to "e inequili"rium.

Bor equili"rium 1=dx

d8, "ut the equili"rium of partile an "e of three types (

tab#e nstab#e Neutra#

When a partile is displaed

slightly from a position, then

a fore ating on it "rings it

"ak to the initial position, it

is said to "e in sta"le

equili"rium position.

When a partile is displaed

slightly from a position, then

a fore ating on it tries to

displae the partile further

away from the equili"rium

position, it is said to "e in

unsta"le equili"rium.

When a partile is slightly

displaed from a position then

it does not e2periene any

fore ating on it and

ontinues to "e in equili"rium

in the displaed position, it is

said to "e in neutral

equili"rium.

Potential energy is minimum. Potential energy is ma2imum. Potential energy is onstant.

1=−=dx

d8F 1=−=

dx

d8F 1=−=

dx

d8F

positiv-

-

=dx

8d

i.e. rate of hange ofdx

d8 is

positive.

negativ-

-

=dx

8d


d8 is

negative.

1-

-

=dx

8d


d8 is

&ero.


8) x *

+

!

C 1

4 x


19/56

Work, Energy Power, and Collision 1,

xample (

A mar"le plaed at the

"ottom of a hemispherial

"owl.

xample (

A mar"le "alaned on top of

a hemispherial "owl.

xample (

A mar"le plaed on hori&ontal

ta"le.

Sample problems based on potential energy

P roblem 31. A partile whih is onstrained to move along the x ;a2is, is su"Deted to a fore inthe same diretion whih varies with the distane x of the partile from the origin as

,*) ax %x x F +−= . Gere % and a are positive onstants. Bor 1≥ x , the funtional

from of the potential energy *) x 8 of the partile is

:creening; 2((2%

)a* )"* )* )d*

(olution ( )d*dx

d8F −= ⇒ dx F d8 .−= ⇒ ∫ +−−=

x

dx ax %x 81

,*) ⇒

=-

=- ax %x 8 −=

∴ We get 1=8 at x 1 anda

% x

-= Also we get =8 negative for

a

% x

->

Brom the given funtion we an see that F 1 at x 1 i.e. slope of 8; x graph is&ero at x 1.

P roblem 32. The potential energy of a "ody is given "y -!x + − )where x is thedisplaement*. The magnitude of fore ating on the partile is

&> 2((2%

)a* Constant )"* Proportional to x

)* Proportional to - x )d* #nversely proportional to x

(olution ( )"* !x !x +dx

d

dx

d8F -*) - =−−=

−= ∴ F ∝ x .

P roblem 33. The potential energy of a system is represented in the !rst !gure. The fore atingon the system will "e represented "y

)a* )"* )* )d*

(olution ( )* As slope of pro"lem graph is positive and onstant upto distane a then it "eomes

&ero. Therefore fromdx

d8F −= we an say that upto distane a fore will "e

onstant )negative* and suddenly it "eomes &ero.

P roblem 30. A partile moves in a potential region given "y =11=3 - +−= x x 8 '. #ts state of equili"rium will "e

)a* m x -:= )"* m x -:.1= )* m x 1-:.1= )d* m x :.-=


8) x *

x

8) x *

x

8) x *

x

8) x *

x

a4 x

8) x *

a

x

F ) x *

a

x

F ) x *

a

x

F ) x *

a x

F ) x *


20/56


(olution ( )"* *=11=3) - +−−=−= x x dx

d

dx

d8F

Bor the equili"rium ondition 1=−=dx

d8F ⇒ 1=< =− x ⇒


21/56


simple harmoni motion a"out its mean position then its potential energy

at any position ) x * an "e given "y

-

-

%x 8 = M.)i*

+o for the e2treme position

-

-

%a8 = 5As x ± a for e2treme6

This is ma2imum potential energy or the total energy of mass.

∴ Total energy-

-

%a = M.)ii*

5eause veloity of mass 1 at e2treme ∴ 1-

- == m" K 6

/ow kineti energy at any position 8K −= --

-

-

x % a% −=

*)-

-- x a% K −= M.)iii*

Brom the a"ove formula we an hek that

-ma2

-

%a8 = 5At e2treme x ± a6 and 1min =8 5At mean x 16

-ma2

-

%aK = 5At mean x 16 and 1min =K 5At e2treme x ± a6

== -

-

%a onstant )at all positions*

#t mean kineti energy hanges para"olially w.r.t. position "ut total energy remain always

onstant irrespetive to position of the mass

Sample problems based on elastic potential energy

P roblem 3-. A long spring is strethed "y - cm, its potential energy is 8. #f the spring isstrethed "y 1 cm, the potential energy stored in it will "e

&CP)+ 1,46A 56A ,6 )P P)+ 2((2 C*

P)+ 2((3%

)a* 8 F -: )"* 8 F : )* : 8 )d*

(olution ( )d* Elasti potential energy of a spring -

-

%x 8 = ∴ - x 8 ∝

+o

-

-

-

= x

x

8

8 ⇒

-

-

-

1

=

cm

cm

8

8 ⇒ 88 -:- =

P roblem 36. A spring of spring onstant m& F1: × is strethed initially "y : cm from theunstrethed position. Then the work required to streth it further "y another : cm is

&8I*** 2((3%

)a*


22/56


P roblem 34. Two springs of spring onstants m& F:11 and m& F111 respetively arestrethed with the same fore. They will have potential energy in the ratio

&)P P*+P)+ 1,,5 Pb. P)+ 2((2%

)a* = ( )"* ( = )* - ( )d*

(olution ( )* Potential energy of spring%

F 8-

-

= ⇒

-

-

%

%

8

8= (-:11

,111== 5#f F

onstant6

P roblem 35. A "ody is attahed to the lower end of a vertial spiral spring and it is graduallylowered to its equili"rium position. This strethes the spring "y a length x . #f the

same "ody attahed to the same spring is allowed to fall suddenly, what would "e

the ma2imum strething in this ase

)a* x )"* - x )* x )d*

(olution ( )"* When spring is gradually lowered to it's equili"rium position

%x mg ∴ %

mg

x = .

When spring is allowed to fall suddenly it osillates a"out it's mean position

%et y is the amplitude of vi"ration then at lower e2treme, "y the onservation of

energy

⇒ mg%y =--

⇒

%

mg y

-= - x .

P roblem 3,. Two equal masses are attahed to the two ends of a spring of spring onstant % . The masses are pulled out symmetrially to streth the spring "y a length x over itsnatural length. The work done "y the spring on eah mass is

)a* --%x

)"* --%x − )*

-=%x

)d*

(olution ( )d* #f the spring is strethed "y length x , then work done "y two equal masses -

-

%x

+o work done "y eah mass on the spring -

=

%x ∴ Work done "y spring on

eah mass -

=

%x − .

6.10 *#ectrica# Potentia# *nergy.

#t is the energy assoiated with state of separation "etween harged partiles that interat

via eletri fore. Bor two point harge # and -# , separated "y distane r .

r

##8 -

1

.=

πε =

While for a point harge # at a point in an eletri !eld where the potential is 6

8 #6

As harge an "e positive or negative, eletri potential energyan "e positive or negative.

Sample problems based on electrical potential energy

P roblem 0(. A proton has a positive harge. #f two protons are "rought near to one another,the potential energy of the system will

)a* #nrease )"* 9erease



23/56


)* Jemain the same )d* Equal to the kineti energy

(olution ( )a* As the fore is repulsive in nature "etween two protons. Therefore potential energy

of the system inreases.

P roblem 01. Two protons are situated at a distane of 11 fermi from eah other. The

potential energy of this system will "e in e6 )a* .== )"* ,1==. × )* -1==. × )d* =1==. ×

(olution ( )d*

e6 e6 'r

##8 =

0

::

:

-00-

1

1==.1 ×=×× − ⇒ ,1:. −×=r m.

P roblem 03. A harged partile + moves diretly towards another harged partile !. Bor the*) ! + + system, the total momentum is P and the total energy is

)a* P and are onserved if "oth + and ! are free to move

)"* )a* is true only if + and ! have similar harges

)* #f ! is !2ed, is onserved "ut not P

)d* #f ! is !2ed, neither nor P is onserved

(olution ( )a, * #f + and ! are free to move, no e2ternal fores are ating and hene P and

"oth are onserved "ut when ! is !2ed )with the help of an e2ternal fore* then is

onserved "ut P is not onserved.

6.1- 9ravitationa# Potentia# *nergy.

#t is the usual form of potential energy and is the energy assoiated with the state of

separation "etween two "odies that interat via gravitational fore.

Bor two partiles of masses m and m- separated "y a distane r

ravitational potential energyr

mmG8 -−=

)* #f a "ody of mass m at height h relative to surfae of earth then

ravitational potential energy

2

h

mgh8

+=

Where 2 radius of earth, g aeleration due to gravity at the surfae of the earth.

)-* #f h QQ 2 then a"ove formula redues to 8 mgh.

)* #f 6 is the gravitational potential at a point, the potential energy of a partile of mass m

at that point will "e

8 m6


F -m

m

F -

r


24/56


)=* Energy height graph ( When a "ody proDeted vertially upward from the ground level

with some initial veloity then it possess kineti energy "ut its

potential energy is &ero.

As the "ody moves upward its potential energy inreases

due to inrease in height "ut kineti energy dereases )due to

derease in veloity*. At ma2imum height its kineti energy

"eomes &ero and potential energy ma2imum "ut through out

the omplete motion total energy remains onstant as shown in

the !gure.

Sample problems based on gravitational potential energy

P roblem 00. The work done in pulling up a "lok of wood weighing -%& for a length of 1 m

on a smooth plane inlined at an angle of o: with the hori&ontal is )sin :o

1.-:0* &8F)C 1,,,%

)a* =.< % ' )"* :.> % ' )* 3.0 % ' )d*

(olution ( )"* Work done mg × h

θ sin1- l××=

%' 'o >.::>


25/56


P roblem 06. #f g is the aeleration due to gravity on the earthKs surfae, the gain in thepotential energy of an a"Det of mass m raised from the surfae of earth to a height

equal to the radius of the earth 2, is &II+


26/56


∴ Work done Bore × displaement 01 × 1 011 '.

P roblem -1. A rod of mass m and length l is lying on a hori&ontal ta"le. The work done inmaking it stand on one end will "e

)a* mgl )"*

-

mgl)*

=

mgl)d*

(olution ( )"* When the rod is lying on a hori&ontal ta"le, its potential energy 1

ut when we make its stand vertial its entre of mass rises upto high-

l. +o it's

potential energy-

mgl=

∴ Work done harge in potential energy-

1-

mgllmg =−= .

P roblem -2. A metre stik, of mass =11 g, is pivoted at one end displaed through

an angle o


27/56


i.e. dW mgy )@ dy *

+o the work done in pulling the hanging portion on the ta"le.

1

F

-1

F -n.

n.

y mgdy mgy W

=−= ∫ -

-

-n

.mg=

∴ --n

Mg.W = 5As m MF6

+lternati"e method $

#f point mass m is pulled through a height h then work done W mgh

+imilarly for a hain we an onsider its entre of mass at the middle point of the hanging

part i.e. at a height of F)-n* from the lower end and mass of the hanging part of hainn

M=

+o work done to raise the entre of mass of the hain on the ta"le is given "y

n

.g

n

MW

-××= 5As W mgh6

or--n

Mg.W =

6.14 e#ocity of C!ain W!i#e =eaving t!e +ab#e.

Taking surfae of ta"le as a referene level )&ero potential energy*

Potential energy of hain when Fnth length hanging from the edge--n

Mg.−=

Potential energy of hain when it leaves the ta"le-

Mg.−=

Iineti energy of hain loss in potential energy

⇒ --

---

nMg.Mg.M" −=

⇒

−=

-

- --

n

Mg.M"

∴ Heloity of hain

−=

-

ng."

Sample problem based on chainP roblem -0. A uniform hain of length and mass M is lying on a smooth ta"le and one third

of its length is hanging vertially down over the edge of the ta"le. #f g is

aeleration due to gravity, the work required to pull the hanging part on to theta"le is &II+


28/56


)a* Mg. )"*,

Mg.)*

0

Mg.)d*

(olution ( )d* As F part of the hain is hanging from the edge of the ta"le. +o "y su"stituting n in standard e2pression

--n

Mg.W =

3*,)- -

Mg.Mg.== .

P roblem --. A hain is plaed on a fritionless ta"le with one fourth of it hanging over theedge. #f the length of the hain is -m and its mass is =%g, the energy need to "espent to pull it "ak to the ta"le is)a* - ' )"*


29/56


P roblem -4. Two stones eah of mass :%g fall on a wheel from a height of 1m. The wheelstirs -%g water. The rise in temperature of water would "e

&/P*+ 1,,4%

)a* -..< m*s )"* 0.3 m*s )* -.1

:.===

µ

i.e. "lok ome to rest at the mid point "etween P and :.

P roblem 6(. #f we throw a "ody upwards with veloity of = −ms at what height its kinetienergy redues to half of the initial value V Take -F1 smg = )a* =m )"* - m )* m )d*these

(olution ( )d* We know kineti energy -

-

m" K = ∴ K " ∝

When kineti energy of the "ody redues to half its veloity "eomes "

smu

F---

=

-==

Brom the equation ghu" --- −= ⇒ h1-*=)*--) -- ×−= ∴ mh =.1-1

3<=

−= .

P roblem 61. A -%g "lok is dropped from a height of 1.= m on a spring of fore onstant0


30/56


P roblem 62. A "lok of mass -%g is released from + on the trak that is one quadrant of airle of radius m. #t slides down the trak and reahes ! with a speed of = −ms

and !nally stops at C at a distane of m from !. The work done against the foreof frition is

)a* 1 ' )"* -1 ')* - ')d* =


31/56


i.e. the "ody whih perform the given work in lesser time possess more power and vie;

versa.

)=* As power workFtime, any unit of power multiplied "y a unit of time gives unit of work

)or energy* and not power, i.e. Iilowatt;hour or watt;day are units of work or energy.

'oulesecsec 'KWh


32/56


Sample problems based on power

P roblem 60. A ar of mass LmK is driven with aeleration LaK along a straight level roadagainst a onstant e2ternal resistive fore L2K. When the veloity of the ar is L" K,

the rate at whih the engine of the ar is doing work will "e5)P P)+P*+ 1,,5 I)P*/ 2(((6

)a* 2" )"* ma" )* " ma2 *) + )d* " 2ma *) −

(olution ( )* The engine has to do work against resistive fore 2 as well as ar is moving with

aeleration a.

Power Bore × veloity )28ma*" .

P roblem 6-. A wind;powered generator onverts wind energy into eletrial energy. Assumethat the generator onverts a !2ed fration of the wind energy interepted "y its

"lades into eletrial energy. Bor wind speed " , the eletrial power output will "e

proportional to &II+


33/56


P roblem 6,. A partile moves with a veloity ?


34/56


Collision is an isolated event in whih a strong fore ats"etween two or more "odies for a short time as a result of whihthe energy and momentum of the interating partile hange.

#n ollision partiles may or may not ome in real touh e.g. in ollision "etween two "illiard"alls or a "all and "at there is physial ontat while in ollision of alpha partile "y a nuleus ) i.e.

Jutherford sattering e2periment* there is no physial ontat.)* tages of co##ision 7 There are three distint identi!a"le stages in ollision, namely,

"efore, during and after. #n the "efore and after stage the

interation fores are &ero. etween these two stages, the

interation fores are very large and often the dominating

fores governing the motion of "odies. The magnitude of the

interating fore is often unknown, therefore, /ewtonKs

seond law annot "e used, the law of onservation of

momentum is useful in relating the initial and !nal veloities.

)-* )o$entu$ and energy conservation in co##ision 7

)i* 4omentum onservation ( #n a ollision the e$et of e2ternal fores suh as gravity or

frition are not taken into aount as due to small duration of ollision )∆t * average impulsive

fore responsi"le for ollision is muh larger than e2ternal fore ating on the system and sinethis impulsive fore is '#nternal' therefore the total momentum of system always remainsonserved.

)ii* Energy onservation ( #n a ollision 'total energy' is also always onserved. Gere total

energy inludes all forms of energy suh as mehanial energy, internal energy, e2itation

energy, radiant energy or even mass energy.

These laws are the fundamental laws of physis and applia"le

for any type of ollision "ut this is not true for onservation of

kineti energy.

)* +y"es of co##ision 7 )i* n the "asis of onservation of kineti energy.

Perfect#y e#astic co##ision Ine#astic co##ision Perfect#y ine#astic

co##ision

#f in a ollision, kineti energyafter ollision is equal to kinetienergy "efore ollision, theollision is said to "e perfetlyelasti.

#f in a ollision kineti energyafter ollision is not equal tokineti energy "efore ollision,the ollision is said to inelasti.

#f in a ollision two "odiesstik together or move withsame veloity after theollision, the ollision is saidto "e perfetly inelasti.

CoeOient of restitution e CoeOient of restitution 1 Q e Q

CoeOient of restitution e 1

)IE*!nal )IE*initial

Gere kineti energy appears inother forms. #n some ases)IE*!nal Q )IE*initial suh as wheninitial IE is onverted intointernal energy of the produt)as heat, elasti or e2itation*while in other ases )IE*!nal Y)IE*initial suh as when internalenergy stored in the ollidingpartiles is released

The term 'perfetly inelasti'does not neessarily meanthat all the initial kinetienergy is lost, it implies thatthe loss in kineti energy isas large as it an "e.)Consistent with momentumonservation*.

xamples ( )* Collision"etween atomi partiles

)-* ouning of "all with sameveloity after the ollision withearth.

xamples ( )* Collision"etween two "illiard "alls.

)-* Collision "etween twoautomo"ile on a road.

#n fat all maDority of ollision"elong to this ategory.

xample ( Collision "etweena "ullet and a "lok of wood

into whih it is !red. Whenthe "ullet remains em"ededin the "lok.


m

-

m

u-

u

m

m

-

m

-

m

" -

"

m

m

-

F

eforeollision

After ollision9uringollision

t F

e2t

∆t


35/56


)ii* n the "asis of the diretion of olliding "odies

>ead on or one di$ensiona# co##ision Eb#iBue co##ision

#n a ollision if the motion of olliding partiles

"efore and after the ollision is along the sameline the ollision is said to "e head on or onedimensional.

#f two partile ollision is LglaningK i.e. suh that

their diretions of motion after ollision are notalong the initial line of motion, the ollision isalled o"lique.

#f in o"lique ollision the partiles "efore andafter ollision are in same plane, the ollision isalled -;dimensional otherwise ;dimensional.

#mpat parameter b is &ero for this type of

ollision.

#mpat parameter b l ies "etween 1 and

*) - r r + i.e .1 Q b Q *) - r r + where r and -r are radii of olliding "odies.

xample ( ollision of two gliders on an air trak. xample ( Collision of "illiard "alls.

6.22 Perfect#y *#astic >ead on Co##ision.

%et two "odies of masses m and -m moving with initial veloities u and -u in the same

diretion and they ollide suh that after ollision their !nal veloities are " and -"

respetively.

Aording to law of onservation of momentum

---- " m" mumum +=+ MM)i*

⇒ *)*) --- u" m" um −=− MM)ii*

Aording to law of onservation of kineti energy

---

-

---

-

-

-

-

-

" m" mumum +=+ M..)iii*

⇒ *)*) -----

-

- u" m" um −=− ..M)iv*

9ividing equation )iv* "y equation )ii*

-- u" u" +=+ M..)v*

⇒ -- " " uu −=− M..)vi*

Jelative veloity of approah Jelative veloity of separation


eforeollision

After ollisionm-

m

u

θ

"

" -

u-φ

b

m-

m

eforeollision

After ollision

m

u

u-

m-

m

"

" -

m-

eforeollision

After ollision

m

u

u-

m-

m

"

" -

m-


36/56


/ote ( The ratio of relative veloity of separation and relative veloity of approah is

de!ned as oeOient of restitution.-

-

uu

" " e

−−

= or

*) -- uue" " −=−

Bor perfetly elasti ollision e ∴ -- uu" " −=− )As shownin eq. )vi*

Bor perfetly inelasti ollision e 1 ∴ 1- =−" " or - " " =

#t means that two "ody stik together and move with same veloity.

Bor inelasti ollision 1 Q e Q ∴ *) -- uue" " −=−

#n short we an say that e is the degree of elastiity of ollision and it is dimensionless quantity.

Burther from equation )v* we get -- uu" " −+=

+u"stituting this value of -" in equation )i* and rearranging we get

-

--

-

-

-

mm

umu

mm

mm"

++

+−

= MM)vii*

+imilarly we get-.

..

-

-.

.-

-

-

mm

umu

mm

mm"

++

+−

= MM)viii*

)* "ecia# cases of !ead on e#astic co##ision

)i* If "roecti#e and target are of sa$e $ass i.e. m1 ' m2

+ine --

-

-

-

-u

mm

mu

mm

mm"

++

+−

= and-.

..

-

-.

.-

-

-

mm

umu

mm

mm"

++

+−

=

+u"stituting - mm = we get

- u" = and - u" =

#t means when two "odies of equal masses undergo head on elasti ollision, theirveloities get interhanged.

xample ( Collision of two "illiard "alls

)ii* If $assive "roecti#e co##ides wit! a #ig!t target i.e. m1 GG m2

+ine-

--

-

-

-

mm

umu

mm

mm"

++

+−

= and-.

..

-

-.

.-

-

-

mm

umu

mm

mm"

++

+−

=

+u"stituting 1- =m , we get

u" = and -- - uu" −=

xample ( Collision of a truk with a ylist


+u" ase ( 1-

=u i.e. target is atrest

1 =" and - u" =

+u" ase ( 1- =u i.e. target is

at rest

u

:1mFs

1%g

eforeollision

u-

-1mFs

1%g

After ollision

" -1

mFs

1%g

" - :1 mFs

1%g

" -1 %m*hr

" - -1

%m*hr

m 1

%g

m-


37/56


efore ollision After ollision

)iii* If #ig!t "roecti#e co##ides wit! a very !eavy target i.e. m1 HH m2

+ine-

--

-

-

-

mm

umu

mm

mm"

++

+−

= and-.

..

-

-.

.-

-

-

mm

umu

mm

mm"

++

+−

=

+u"stituting 1 =m , we get

- -uu" +−= and -- u" =

xample ( Collision of a "all with a massive wall.

)-* ?inetic energy transfer during !ead on e#astic co##ision

Iineti energy of proDetile "efore ollision-

-

umK i =

Iineti energy of proDetile after ollision-

-

" mK f =

Iineti energy transferred from proDetile to target ∆K derease in kineti energy in

proDetile

-

-

-

-

" mumK −=∆ *)

-

-

- " um −=

Brational derease in kineti energy-

-

-

-

*)-

um

" um

K

K −

=∆ -

−=

u

" M..)i*

We an su"stitute the value of " from the equation-

--

-

-

-

mm

umu

mm

mm" ++ +−=

#f the target is at rest i.e. u- 1 then -

- u

mm

mm"

+−

=

Brom equation )i*

-

-

-

+−

−=∆mm

mm

K

K M..)ii*

or --

-

*)

=

mm

mm

K

K

+=

∆M..)iii*


ub case 7 1- =u i.e. target is at

rest" @ u and " - 1

i.e. the "all re"ounds with samespeed in opposite diretion when itollide with stationar and ver

m :1gm

u 1 mFs

efore ollision

" - - mFsu

- - mFs

m- 11

%g

" @ -<mFs

After ollision


38/56


or-

--

-

=*)

=

mmmm

mm

K

K

+−=

∆M..)iv*

/ote ( reater the di$erene in masses less will "e transfer of kineti energy and

vie versa Transfer of kineti energy will "e ma2imum when the di$erene in masses is

minimum

i.e. 1- =−mm or - mm = then N11==∆K

K

+o the transfer of kineti energy in head on elasti ollision )when target is at rest*

is ma2imum when the masses of partiles are equal i.e. mass ratio is and the

transfer of kineti energy is 11N.

#f - mnm = then from equation )iii* we get -*)

=

n

n

K

K

+

=∆

Iineti energy retained "y the proDetile −=

∆

JetainedK

K kineti energy

transferred "y proDetile

⇒ =

∆

JetainedK

K

+−

−−-

-

-mm

mm

-

-

-

+

−=

mm

mm

)* e#ocityA $o$entu$ and kinetic energy of stationary target after !ead

on e#astic co##ision

)i* Heloity of target ( We know-.

..

-

-.

.-

-

-

mm

umu

mm

mm"

++

+−

=

⇒ -

-

-

mm

um"

+=

-

F

-

mm

u

+= 5As 1- =u and %et nm

m=

-6

∴

n

u"

+

=

- -

)ii* 4omentum of target ( --- " mP =n

unm

+=

-

+==

n

u" nmm

- andAs --

∴*F)

- -

n

umP

+=

)iii* Iineti energy of target (----

-

" mK =

-

-

-

+

=n

umn

-

-

*)

-

n

num

+=

nn

nK

=*)

*)=-

+−

=

= --

As umK

)iv* Jelation "etween masses for ma2imum veloity, momentum and kineti energy


eforeollision

Afterollision

m

u

u-

m-

m

"

" -

m-


39/56


Heloity

n

u"

+=

- -

Bor -" to "e ma2imum n must "e

minimum

i.e. 1

- →=m

mn ∴ - mm >

Target should "e

massive.

Iineti

energy

n

nK K

*)

=-

-

+−=

Bor -K to "e ma2imum-*) n−

must "e minimum.

i.e.

-1m

mnn ==⇒=− ∴

- mm =

Target and proDetile

should "e of equal

mass.

Sample problem based on head on elastic collision

P roblem 4-. n small "alls eah of mass m impinge elastially eah seond on a surfae withveloity u. The fore e2periened "y the surfae will "e

&)P P)+P*+ 1,,5 /P*+ 2((1 > 2((1 )P P)+ 2((3%

)a* mnu )"* - mnu )* = mnu )d* mnu-

(olution ( )"* As the "all re"ounds with same veloity therefore hange in veloity -u and themass olliding with the surfae per seond nm

Bore e2periened "y the surfaedt

d" mF = ∴ F - mnu.

P roblem 46. A partile of mass m moving with hori&ontal speed < mFsec. #f mQQM then forone dimensional elasti ollision, the speed of lighter partile after ollision will "e

&)P P)+ 2((3%

)a* - mFsec in original diretion )"* - mFsec opposite to the original

diretion

)* = mFsec opposite to the original diretion )d* = mFsec inoriginal diretion

(olution ( )a*-.

--

.

-.

-.

.

-

mm

umu

mm

mm"

++

+−

=

+u"stituting m 1, - -uu" +−=

⇒ *=)-


40/56


)a*-

of its initial kineti energy )"*

0

of its initial kineti energy

)*0

3 of its initial kineti energy )d*

=

of its initial kineti energy

(olution ( )* %oss of kineti energy of the olliding "ody

---

-

-

-

-

−=

+−

−=

+−

−=∆

mm

mm

mm

mm

K

K

K K K 0

3

0

=

−=∆ ∴ %oss of kineti energy is

0

3 of its initial kineti energy.

P roblem 45. A "all of mass m moving with veloity 6 , makes a head on elasti ollision witha "all of the same mass moving with veloity -6 towards it. Taking diretion of 6 as

positive veloities of the two "alls after ollision are

&)P P)+ 2((2%

)a* @ 6 and -6 )"* -6 and @ 6 )* 6 and < -6 )d* @ -6 and 6

(olution ( )d* #nitial veloities of "alls are 86 and @ -6 respetively and we know that for given

ondition veloities get interhanged after ollision. +o the veloities of two "alls

after ollision are @ -6 and 6 respetively.

P roblem 4,. Consider the following statements

8ssertion ) +* ( #n an elasti ollision of two "illiard "alls, the total kineti energy is

onserved during the short time of ollision of the "alls )i.e., when they are in

ontat*

/eason )2* ( Energy spent against frition does not follow the law of onservationof energy of these statements

&8II) 2((2%

)a* oth + and 2 are true and the 2 is a orret e2planation of +

)"* oth + and 2 are true "ut the 2 is not a orret e2planation of the +

)* + is true "ut the 2 is false

)d* oth + and 2 are false

(olution ( )d* )i* When they are in ontat some part of kineti energy may onvert in potential

energy so it is not onserved during the short time of ollision. )ii* %aw of

onservation of energy is always true.

P roblem 5(. A "ig "all of mass M, moving with veloity u strikes a small "all of mass m,whih is at rest. Binally small "all attains veloity u and "ig "all " . Then what is the

value of " &/P*+ 2((1%

)a* umM

mM

+−

)"* umM

m

+)* u

mM

m

+-

)d* umM

M

+

(olution ( )a* Brom the standard equation umM

mMu

mm

mm"

+−

=

+−

= -

- .

P roblem 51. A ar of mass %g=11 and travelling at >- %mph rashes into a truk of mass

%g=111 and travelling at 0 %mph, in the same diretion. The ar "ounes "ak at aspeed of 3 %mph. The speed of the truk after the impat is

&*8)C*+ :*ngg.; 1,,4%

)a* 0 %mph )"* 3 %mph )* -> %mph )d* < %mph



41/56


(olution ( )"* y the law of onservation of linear momentum ---- " m" mumum +=+

⇒ -=111*3)=110=111>-=11 " ×+−×=×+× ⇒ h%m" F3- = .

P roblem 52. A smooth sphere of mass M moving with veloity u diretly ollides elastiallywith another sphere of mass m at rest. After ollision their !nal veloities are 6 and

" respetively. The value of " is &)P P*+ 1,,-%

)a*m

uM-)"*

M

um-)*

M

m

u

+

-

)d*

m

M

u

+

-

(olution ( )* Binal veloity of the target-.

..

-

-.

.-

-

-

mm

umu

mm

mm"

++

+−

=

As initially target is at rest so "y su"stituting 1- =u we get

M

m

u

mM

Mu"

+=

+=

--- .

P roblem 53. A sphere of mass 1. %g is attahed to a ord of m length. +tarting from theheight of its point of suspension this sphere hits a "lok of same mass at rest on a

fritionless ta"le, #f the impat is elasti, then the kineti energy of the "lok after

the ollision is &/P*+ 1,,1%

)a* ' )"* 1 '

)* 1. ' )d* 1.: '

(olution ( )a* As two "loks are of same mass and the ollision is perfetly elasti therefore their

veloities gets interhanged i.e. the "lok + omes into rest and omplete kineti

energy transferred to "lok !./ow kineti energy of "lok ! after ollision Iineti energy of "lok + "efore

ollision

Potential energy of "lok + at the original

height

mgh 1. × 1 × '.

P roblem 50. A "all moving hori&ontally with speed " strikes the "o" of a simple pendulum atrest. The mass of the "o" is equal to that of the "all. #f the ollision is elasti the

"o" will rise to a height

)a*g

"

-

)"*g

" -

-

)*g

" =

-

)d*g

" 3

-

(olution ( )"* Total kineti energy of the "all will transfer to the "o" of simple pendulum. %et it

rises to height LhK "y the law of onservation of energy.

mgm" =--

∴ g

" h

-

-

=


+

m

m m!

h" mm

m


42/56


P roblem 5-. A moving "ody with a mass m strikes a stationary "ody of mass m-. The

masses m and -m should "e in the ratio-

m

m so as to derease the veloity of

the !rst "ody .: times assuming a perfetly elasti impat. Then the ratio-

m

m

is

)a* F -: )"* F: )* : )d* -:

(olution ( )*-

--

-

-

-

mm

umu

mm

mm"

++

+−

= -

- umm

mm

+−

= 5As u- 1 and

=

:.

u" given6

⇒ -

-

:. u

mm

mmu

+−

= ⇒ *):. -- mmmm −=+ ⇒ :-

=m

m.

P roblem 56. +i2 idential "alls are lined in a straight groove made on a hori&ontal fritionless

surfae as shown. Two similar "alls eah moving with a veloity " ollide with therow of < "alls from left. What will happen

)a* ne "all from the right rolls out with a speed -" and the remaining "alls will

remain at rest

)"* Two "alls from the right roll out with speed " eah and the remaining "alls will

remain stationary

)* All the si2 "alls in the row will roll out with speed " F< eah and the two olliding

"alls will ome to rest

)d* The olliding "alls will ome to rest and no "all rolls out from right

(olution ( )"* nly this ondition satis!es the law of onservation of linear momentum.

P roblem 54. A moving mass of 3 %g ollides elastially with a stationary mass of - %g. #f "ethe initial kineti energy of the mass, the kineti energy left with it after ollision

will "e

)a* 1.31 )"* 1.


43/56


4ass of neutron )m* and 4ass of atom )m-* + ∴ -

+−

=∆

+

+

K

K or

-

+−

+

+.

P roblem 5,. A neutron with 1. Me6 )* > Me6 )d* ero

(olution ( )"* Iineti energy transferred to stationary target )ar"on nuleus*

+−

−=∆

-

-

-mm

mm

K

K

+−

−=∆ -

-

-

K

K

−= and m- at rest, after ollision the "all of mass m- moves with fourtimes the veloity of m

)* When - mm = and -m at rest, there will "e ma2imum transfer of kinetienergy

)d* When ollision is o"lique and -m at rest with ,- mm = after ollision the "allsmove in opposite diretions

(olution ( )", d* We know that transfer of momentum will "e ma2imum when target is massive

and transfer of kineti energy will "e ma2imum when target and proDetile are

having same mass. #t means statement )a* and )* are orret, "ut statement )"*

and )d* are inorret "eause when target is very light, then after ollision it will

move with dou"le the veloity of proDetile and when ollision is o"lique and m- at

rest with ,- mm = after ollision the "all move perpendiular to eah other.

6.23 Perfect#y *#astic Eb#iBue Co##ision.

%et two "odies moving as shown in !gure.

y law of onservation of momentumAlong x ;a2is, φ θ osos ---- " m" mumum +=+ .....)i*

Along y ;a2is, φ θ sinsin1 -- " m" m −= .....)ii*Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

eforeollision

Afterollision

m-

m

u

θ

"

" -

u-φ

m-

m


44/56


y law of onservation of kineti energy

---

-

---

-

-

-

-

-

" m" mumum +=+ .....)iii*

#n ase of o"lique ollision it "eomes diOult to solve pro"lem when some e2perimental

data are provided as in these situations more unknown varia"les are involved than equationsformed.

"ecia# condition 7 #f - mm = and 1- =u su"stituting these values in equation )i*, )ii* and)iii* we get

φ θ osos - " " u += .....)iv*

φ θ sinsin1 - " " −= .....)v*

and ---

- " " u += .....)vi*

+quaring )iv* and )v* and adding we get

*os)- ---

-

- φ θ +++= u" " " u .....)vii*

7sing )vi* and )vii* we get 1*os) =+ φ θ

∴ -Fπ φ θ =+

i.e. after perfetly elasti o"lique ollision of two "odies of equal masses )if the seond "ody

is at rest*, the sattering angle φ θ + would "e o01 .

Sample problems based on oblique elastic collision

P roblem ,2. A "all moving with veloity of smF0 ollides with another similar stationary

"all. After the ollision "oth the "alls move in diretions making an angle of o,1with the initial diretion. After the ollision their speed will "e

)a* smF


45/56


P roblem ,0. Ieeping the priniple of onservation of momentum in mind whih of the followingollision diagram is not orret

)a* )"* )* )d*

(olution ( )d* #n this ondition the !nal resultant momentum makes some angle with x ;a2is.

Whih is not possi"le "eause initial momentum is along the x ;a2is and aording

to law of onservation of momentum initial and !nal momentum should "e equal in

magnitude and diretion "oth.

P roblem ,-. Three partiles +, ! and C of equal mass are moving with the same veloity " along the medians of an equilateral triangle. These partile ollide at the entre G

of triangle. After ollision + "eomes stationary, ! retraes its path with veloity "

then the magnitude and diretion of veloity of C will "e

)a* " and opposite to !

)"* " and in the diretion of +

)* " and in the diretion of C

)d* " and in the diretion of !

(olution ( )d* Brom the !gure )#* it is lear that "efore ollision initial momentum of the system

1

After the ollision, + "eomes stationary, ! retraes its path with veloity " . %et C

moves with veloity 6 making an angle θ from the hori&ontal. As the initial

momentum of the system is &ero, therefore hori&ontal and vertial momentum after

the ollision should also "e equal to &ero.

Brom !gure )##* Gori&ontal momentum1,1osos =+

o

" " θ M..)i*

Hertial momentum 1,1sinsin =− o" " θ M..)ii*y solving )i* and )ii* we get o,1−=θ and 6 " i.e. the C will move with veloity " in the diretion of !.

P roblem ,6. A "all ! of mass M moving northwards with veloity " ollides elastially with

another "all -! of same mass "ut moving eastwards with the same veloity " .

Whih of the following statements will "e true

)a* ! omes to rest "ut -! moves with veloity " -

)"* ! moves with veloity " - "ut -! omes to rest

)* oth move with veloity -F" in north east diretion

)d* ! moves eastwards and -! moves north wards

(olution ( )d* Gori&ontal momentum and vertial momentum "oth should remain onserve "eforeand after ollision. This is possi"le only for the )d* option.


M

α β

M-

M-

M

M-

α

β

M

M-M

M01o

β

M-

M-

M

M01o

M-

M-

M

"

"

"

!

C

+

m"

m"

m"

C

!

-1o

-1o

-1o

+6

!

C

"

θ +

1o


46/56


6.20 >ead on Ine#astic Co##ision.

)* e#ocity after co##ision 7 %et two "odies + and ! ollide inelastially and oeOient of restitution is e.

Where

approaofveloityJelative

separatioofveloityJelative

-

- =

−

−=

uu

" " e

⇒ *) -- uue" " −=−

∴ *) -- uue" " −+= MM)i*

Brom the law of onservation of linear momentum

---- " m" mumum +=+ MM)ii*

y solving )i* and )ii* we get

-

-

-

-

-

*)u

mm

meu

mm

emm"

+

++

+−

=

+imilarly --

-

-

-

*)u

mm

memu

mm

me"

+−+

+

+=

y su"stituting e , we get the value of " and -u for perfetly elasti head on ollision.

)-* /atio of ve#ocities after ine#astic co##ision 7 A sphere of mass m moving with veloity

u hits inelastially with another stationary sphere of same mass.

∴ 1

-

-

-

−−

=−−

=u

" "

uu

" " e

⇒ eu" " =− - MM)i*

y onservation of momentum (

4omentum "efore ollision 4omentum after ollision

- m" m" mu +=

⇒ u" " =+ - MM)ii*

+olving equation )i* and )ii* we get *)-

eu

" −= and *)-

- eu

" +=

∴ e

e

"

"

+−

=

-

)* =oss in kinetic energy%oss )∆K * Total initial kineti energy @ Total !nal kineti energy

+−

+ ---

-

---

-

-

-

-

-

" m" mumum

+u"stituting the value of " and -" from the a"ove e2pression

%oss )∆K * -

--

-

- *)*)-

uue

mm

mm−−

+

y su"stituting e we get ∆K 1 i.e. for perfetly elasti ollision loss of kineti energy

will "e &ero or kineti energy remains onstant "efore and after the ollision.

Sample problems based on inelastic collision


efore

ollision

After

ollision

mu

u

u-

1m

"

" -

m m


47/56


P roblem ,4. A "ody of mass %g=1 having veloity smF= ollides with another "ody of

mass %g


48/56


+imilarly we an alulate the ratio of seond "all "efore and after ollision,

<

:

-

-

-

- ===u

"

u

"

u

" .

P roblem 1((. Two idential "illiard "alls are in ontat on a ta"le. A third idential "all strikesthem symmetrially and ome to rest after impat. The oeOient of restitution is

)a*

-)"*

)*

<

)d*

-

,

(olution ( )a*-

-sin ==

r

r θ ⇒ θ 1o

Brom onservation of linear momentum om" mu ,1os-= or,

u" =

/owapproahofveloityJelative

separatioofveloityJelative=e in ommon normal diretion.

Gene,,

-

-F,

,F

,1os===

u

u

u

" e

o

P roblem 1(1. A "ody of mass %g , moving with a speed of = −ms , ollides head on with a

stationary "ody of mass %g- . Their relative veloity of separation after the

ollision is - −ms . Then

)a* The oeOient of restitution is 1.: )"* The impulse of the ollision is >.-

&0s

)* The loss of kineti energy due to ollision is .


49/56


(olution ( )a* #nitial hori&ontal momentum of the system .-

os-

os θ θ

m" m" +=

#f after the ollision ars move with ommon veloity 6 then !nal hori&ontalmomentum of the system -m6 .

y the law of onservation of momentum, -m6 .-

os-

os-

os θ θ θ

" 6 m" m" =⇒+=

6.2- /ebounding of a## 8fter Co##ision Wit! 9round.

#f a "all is dropped from a height h on a hori&ontal Xoor, then it strikes with the Xoor with a

speed.

11 -gh" = 5Brom 6--- ghu" +=

and it re"ounds from the Xoor with a speed

1 " e" = 1-ghe= = ollision"eforeveloityollisionafterveloity Ase

)* First !eig!t of rebound 7 1-

-

-

heg

" h ==

∴ h e-h1

)-* >eig!t of t!e ba## after nt! rebound 7 "viously, the veloity of "all after nth re"ound

will "e

1" e" n

n =

Therefore the height after nth re"ound will "e 1-

-

-

he

g

" h nnn ==

∴ 1- heh nn =

)* +ota# distance trave##ed by t!e ba## before it sto"s bouncing

......--- -1 ++++= hhhh; .....--- 1<

1=

1-

1 ++++= heheheh

....*)-5


50/56

-( Work, Energy, Power and Collision

∴g

h

e

e- 1

-

−+

=

Sample

problems based on rebound of ball after collision with ground

P roblem 1(3. The hange of momentum in eah "all of mass gm,.1 )* sm%g F;=3.1 )d*

sm%g F;--.1

(olution ( )* 4omentum "efore ollision m" , 4omentum after ollision @ m"

∴ Change in momentum

sm%gsm%gm" F;=3.1F;1=31=1


51/56

Work, Energy Power, and Collision -1

%oss in kineti energy-

----

- *)

-

-

-

comb" mmumumK +−

+=∆

--

-

- *)-

uu

mm

mmK −

+

=∆ 5y su"stituting the value of " comb6

)-* W!en t!e co##iding bodies are $oving in t!e o""osite direction

y the law of onservation of momentum

om"--- *)*) " mmumum +=−+ )Taking left to right as positive*

∴ -

--om"

mm

umum"

+−

=

when -- umum > then 1om" >" )positive*

i.e. the om"ined "ody will move along the diretion of motion of mass m .

when -- umum < then 1om"


52/56

-2 Work, Energy, Power and Collision

Heloity of omposite mass:-1

1:1-1

-

--

+×+×

=++

=mm

umum6 smF3=

∴ Iineti energy of omposite mass .3113*:-1)-

*)

-

--- '6 mm =×+=+=

P roblem 1(,. A neutron having mass of %g->1. −× and moving at smF13 ollides with

a deutron at rest and stiks to it. #f the mass of the deutron is %g->1=. −× Z the

speed of the om"ination is &C* P)+ 2(((%

)a* smF1:->->

3->

-

-- smmm

umum6

×=×+×

+××

=+

+

= −−

−

P roblem 11(. A partile of mass m moving eastward with a speed " ollides with anotherpartile of the same mass moving northward with the same speed " . The twopartiles oalese on ollision. The new partile of mass m- will move in thenorth;easterly diretion with a veloity &NC*/+ 1,5( CP)+ 1,,1 )P P*+ 1,,,DP)+ 1,,,%

)a* -F" )"* " - )* -F" )d* " (olution ( )* #nitially "oth the partiles are moving perpendiular to eah other with momentum

m" .

+o the net initial momentumm" m" m" -*)*) -- =+= .

After the inelasti ollision "oth the partiles )system*

moves with veloity 6 , so linear momentum -m6

y the law of onservation of momentum

m6 m" -- =

∴ .-F" 6 =

P roblem 111. A partile of mass ''m moving with veloity ''" ollides inelastially with astationary partile of mass '-' m . The speed of the system after ollision will "e

&8II) 1,,,%

)a*-

" )"* " - )*

" )d* "

(olution ( )* y the onservation of momentum m6 mm" ,1- =×+ ∴ .,

" 6 =

P roblem 112. A "all moving with speed " hits another idential "all at rest. The two "alls stiktogether after ollision. #f spei! heat of the material of the "alls is (, thetemperature rise resulting from the ollision is &/oorkee 1,,,%

)a*(

"

3

-

)"*(

"

=

-

)*(

"

-

-

)d*(

" -

Collection by Pradeep Kshetrapal

WorkPower Energy Notes1

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