workpower energy notes1
TRANSCRIPT
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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$eted region she may "e
required to arry it over large distanes. #f she an do so, she is said to have a large
stamina or energy. Energy is thus the apaity to do work. The term power is usually
assoiated with speed. #n karate, a powerful punh is one delivered at great speed. #n
physis we shall de!ne these terms very preisely. We shall !nd that there is at "est a
loose orrelation "etween the physial de!nitions and the physiologial pitures theseterms generate in our minds.
Work is said to "e done when a fore applied on the "ody displaes the "ody
through a ertain distane in the diretion of fore.
6.2 Work Done by a Constant Force.
%et a onstant fore F "e applied on the "ody suh that it makes an angle θ with
the hori&ontal and "ody is displaed through a distane s
y resolving fore F into two omponents (
)i* F osθ in the diretion of displaement of the "ody.
)ii* F sinθ in the perpendiular diretion of displaement of the
"ody.
+ine "ody is "eing displaed in the diretion of θ osF ,
therefore work done "y the fore in displaing the "ody through a distane s is given "y
θ θ os*os) FssF W ==
or sF W .=
Thus work done "y a fore is equal to the salar or dot produt of the fore and the
displaement of the "ody.
#f a num"er of fore nF F F F ......,, - are ating on a "ody and it shifts from
position vetor r to position vetor -r then *.)*....) -,- r r F F F F W n −+++=
6.3 Nature of Work Done.
Positive work Negative work
Positive work means that fore )or its omponent*is parallel to displaement
/egative work means that fore )or itsomponent* is opposite to displaement i.e.
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F sinθ
F osθ
s
θ
F
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2 Work, Energy, Power and Collision
oo 011
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Work, Energy Power, and Collision 3
a magneti !eld as fore *6)5 !" #F ×= is
always perpendiular to motion, work done"y this fore is always &ero.
)-* If t!ere is no dis"#ace$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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0 Work, Energy, Power and Collision
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$et of a fore F Cx from x 1 to x x . The work
done in the proess is
&CP)+ 1,52%
)a* -Cx )"* --
Cx )* Cx )d*
(olution ( )"* Work done -1
-
1 -
-
-
x C x
Cdx Cx dx F
x x x
x =
=== ∫ ∫
P roblem 5. The vessels + and ! of equal volume and weight are immersed in water to a depthh. The vessel + has an opening at the "ottom through whih water an enter. #f the
work done in immersing + and ! are +W and !W respetively, then
)a* ! + W W = )"* ! + W W < )* ! + W W > )d*
! + W W
(olution ( )"* When the vessels are immersed in water, work has to "e done against up;thrustfore "ut due to opening at the "ottom in vessel +, up;thrust fore goes ondereasing. +o work done will "e less in this ase.
P roblem ,. Work done in time t on a "ody of mass m whih is aelerated from rest to a speed" in time t as a funtion of time t is given "y
)a* -
-
t
t
" m )"*
-
t t
" m )* -
-
-
t
t
m"
)d*
-
-
-
-
t
t
" m
(olution ( )d* Work done F .s
--
. t ama --
-
t am= -
-
-
t
t
" m
=
= given*)onaeleratiAs
t
" a
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+
!
ds
F θ
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Work, Energy Power, and Collision -
6.- Di$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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6 Work, Energy, Power and Collision
(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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Work, Energy Power, and Collision 4
)-* #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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5 Work, Energy, Power and Collision
P roblem 13. #f -,W W and W represent the work done in moving a partile from + to !along three di$erent paths , - and respetively )as shown* in the gravitational
!eld of a point mass m, !nd the orret relation
)a* ,- W W W >> )"* ,- W W W ==
)* ,- W W W >
(olution ( )"* As gravitational !eld is onservative in nature. +o work done in moving a partile
from + to ! does not depends upon the path followed "y the "ody. #t always
remains same.
P roblem 10. A partile of mass 1.1 %g travels along a urve with veloity given "y
% i ?
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Work, Energy Power, and Collision ,
#f %gamum ->1. −×== then 'ouleMe6 11:.0, −×== .
#f %gm = then 'oule eat e#ectrica#
Coalurning 4irophone Thermo;ouple
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e@ 8 e8γ 7 Photon*
& ( @ 8CathodeAnode
Got
G
Fe
Cu
Cold
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1( Work, Energy, Power and Collision
>eat )ec!anica# *#ectrica# )ec!anica# *#ectrica# >eat
Engine 4otor Geater
*#ectrica# ound *#ectrica# C!e$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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Work, Energy Power, and Collision 11
⇒ 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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12 Work, Energy, Power and Collision
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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Work, Energy Power, and Collision 13
-. 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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10 Work, Energy, Power and Collision
(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
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#nitial veloity "
x
Binal veloity 1
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16 Work, Energy, Power and Collision
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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Work, Energy Power, and Collision 14
(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$u#a for "otentia# energy7 Bor only onservative!elds F equals the negative gradient *) ∇− of the potential energy.
+o 8F ∇−= )∇ read as 9el operator or /a"la operator and
% d,
d )
dy
di
dx
d ??? ++=∇ *
⇒
++−= %
d,
d8 )
dy
d8i
dx
d8F ???
where =dx
d8 Partial derivative of 8 w.r.t. x )keeping y and onstant *
=dy
d8 Partial derivative of 8 w.r.t. y )keeping x and onstant*
=d,
d8 Partial derivative of 8 w.r.t. & )keeping x and y onstant*
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15 Work, Energy, Power and Collision
)* Potentia# energy curve 7 A graph plotted "etween the potential energy of apartile and its displaement from the entre of fore is alled potential energy urve.
Bigure shows a graph of potential energy funtion 8) x * for one dimensional motion.As we know that negative gradient of the potential energy
gives fore.
∴ F dx d8 =−
)=* Nature of force 7
)i* Attrative fore ( n inreasing x , if 8 inreases
positiv=dx
d8
then F is negative in diretion i.e. fore is attrative in nature. #n graph this is represented inregion !C.
)ii* Jepulsive fore ( n inreasing x , if 8 dereases negativ=dx
d8
then F is positive in diretion i.e. fore is repulsive in nature. #n graph this is represented inregion +!.
)iii* ero fore ( n inreasing x , if 8 does not hanges 1=dx
d8
then F is &ero i.e. no fore works on the partile. Point !, C and 1 represents thepoint of &ero fore or these points an "e termed as position of equili"rium.
):* +y"es of eBui#ibriu$ 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
i.e. rate of hange ofdx
d8 is
negative.
1-
-
=dx
8d
i.e. rate of hange ofdx
d8 is
&ero.
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
8) x *
+
!
C 1
4 x
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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 :.-=
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
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 *
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2( Work, Energy, Power and Collision
(olution ( )"* *=11=3) - +−−=−= x x dx
d
dx
d8F
Bor the equili"rium ondition 1=−=dx
d8F ⇒ 1=< =− x ⇒
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Work, Energy Power, and Collision 21
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*
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22 Work, Energy, Power and Collision
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
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Work, Energy Power, and Collision 23
)* 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
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F -m
m
F -
r
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20 Work, Energy, Power and Collision
)=* 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 >.::>
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Work, Energy Power, and Collision 2-
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+
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26 Work, Energy, Power and Collision
∴ 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
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Work, Energy Power, and Collision 24
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+
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25 Work, Energy, Power and Collision
)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* - ' )"*
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Work, Energy Power, and Collision 2,
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
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3( Work, Energy, Power and Collision
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* =
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Work, Energy Power, and Collision 31
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
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32 Work, Energy, Power and Collision
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+
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Work, Energy Power, and Collision 33
P roblem 6,. A partile moves with a veloity ?
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30 Work, Energy, Power and Collision
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.
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
m
-
m
u-
u
m
m
-
m
-
m
" -
"
m
m
-
F
eforeollision
After ollision9uringollision
t F
e2t
∆t
-
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Work, Energy Power, and Collision 3-
)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
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
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-
-
8/20/2019 WorkPower Energy Notes1
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36 Work, Energy, Power and Collision
/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
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
+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-
-
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Work, Energy Power, and Collision 34
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*
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
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
-
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35 Work, Energy, Power and Collision
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
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
eforeollision
Afterollision
m
u
u-
m-
m
"
" -
m-
-
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Work, Energy Power, and Collision 3,
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" +−=
⇒ *=)-
-
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0( Work, Energy, Power and Collision
)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
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
-
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Work, Energy Power, and Collision 01
(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
-
-
=
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
+
m
m m!
h" mm
m
-
8/20/2019 WorkPower Energy Notes1
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02 Work, Energy, Power and Collision
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.
-
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Work, Energy Power, and Collision 03
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
-
8/20/2019 WorkPower Energy Notes1
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00 Work, Energy, Power and Collision
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
-
8/20/2019 WorkPower Energy Notes1
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Work, Energy Power, and Collision 0-
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.
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
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
-
8/20/2019 WorkPower Energy Notes1
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06 Work, Energy, Power and Collision
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
Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda
efore
ollision
After
ollision
mu
u
u-
1m
"
" -
m m
-
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Work, Energy Power, and Collision 04
P roblem ,4. A "ody of mass %g=1 having veloity smF= ollides with another "ody of
mass %g
-
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05 Work, Energy, Power and Collision
+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 .
-
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Work, Energy Power, and Collision 0,
(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
-
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-( 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
-
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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"
-
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-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