Physics Disha

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I. HEAD-ON ELASTIC COLLISIONIf colliding bodies before and after collision remain in a line, the collision is said to behead-on collision. This will happen when bodies move along the line joining theirgeometric centres.Consider two bodies of masses m1 and m2 moving with velocities 1 2 1 2 and ( )>u u u ur r r r

along the same straight line. Let after collision their velocities become 1 2 and v vr r in thesame initial direction. Then

according to conservation of linear momentum, we have

1 1 2 2m m+r ru u = 1 1 2 2+ .m mr rv v

Since all the colliding bodies before and after collision remain in the same line, so we candrop the vector signs from them. Thus we can write

m1u1 + m2u2 = m1v1 + m2v2 … (i)or m1(u1 – v1) = m2(v2 – u2) …(ii)

As kinetic energy before collision = kinetic energy after collision

\ 2 21 1 2 2

1 12 2

m u m u+ = 2 21 1 2 2

1 12 2

m v m v+ …(iii)

or m1(u12 – v1

2) = m2(v22 – u2

2)or m1(u1 + v1)(u1 – v1) = m2(v2

+ u2) (v2 – u2) (iv)Dividing equation (iv) by (ii), we get

u1 + v1 = v2 + u2

or 1 2u u- = 2 1.v v- …(v)

Thus velocity of m1 w.r.t. m2 before collision = velocity of m2 w.r.t. m1 after collision.or velocity of approach = velocity of separation

Also we have,

1v =1 2 2

1 21 2 1 2

2m m mu u

m m m mæ ö æ ö-

+ç ÷ ç ÷+ +è ø è ø …(vi)

and 2v =2 1 1

2 11 2 1 2

2m m mu u

m m m mæ ö æ ö-

+ç ÷ ç ÷+ +è ø è ø ... (vii)

Special cases :(i) When colliding bodies are of equal masses, let m1 = m2 = m. From equation (vi) and

(vii), we getv1 = u2 and v2 = u1

Hence when two bodies of equal masses collide elastically, their velocities getexchanged.

PHYSICS

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(ii) If m1 = m2 = m and u2 = 0, thenv1 = 0 and v2 = u1.

(iii) When a light body collides with a massive stationary body. Here m1 << m2 andu2 = 0\ v1 = – u1 and v2 ; 0.Hence when a light body collides with a massive stationary body, the light bodyrebounds after the collision with an equal speed while the massive body remains atrest.

(iv) When a massive body collides with a light body at rest. Here m1 >> m2 and u2 = 0\ v1 = u1 and v2 = 2u.

Transfer of kinetic energy during collision: Kinetic energy transferred from projectileto the target

DK = decrease in K.E. of projectile

= 2 21 1 1 1

1 12 2

m u m v- .

Fractional decrease in K.E.

KK

D=

2 21 11 1 1 12 2

211 12

m u m v

m u

-

orK

KD

=2

1

11

vu

æ ö- ç ÷è ø

. …(viii)

Perfectly inelastic collision in 1-DConsider two bodies of masses m1 and m2 moving with velocities u1 and u2 along astraight line. They make perfectly inelastic collision. Let after collision, their commonvelocity becomes v, then by conservation of momentum, we have

m1u1 + m2u2 = (m1 + m2)v

\ v =1 1 2 2

1 2

m u m um m

é ù+ê ú+ë û

.

The loss of K.E. in collision

KD =2 2 2

1 1 2 2 1 21 1 1

( )2 2 2

m u m u m m væ ö+ - +ç ÷è ø

=2 2

1 1 2 2 1 21 1 1

( )2 2 2

m u m u m mæ ö+ - +ç ÷è ø

21 1 2 2

1 2

m u m um m

æ ö+ç ÷+è ø

=21 2

1 21 2

1 ( )2

m mu u

m mæ ö

-ç ÷+è ø .

The loss of K.E.. will appear as heat and sound.

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General analysis of 1-D collisionNewton's experimental law : Coefficient of restitutionIt is defined as;

e =velocity of separationvelocity of approach

=2 1

1 2

v vu u

--

or e =2 1 1 2

2 1 1 2.

v v v vu u u u

é ù é ù- -- = -ê ú ê ú- -ë û ë û

The value of e depends on materials of colliding bodies. The value of e can be 1.e £(i) For perfectly elastic collision, e = 1.(ii) For perfectly inelastic collision, e = 0.

Note:

The coefficient of restitution is a 1–D concept. Thus in problem involving oblique collision,'e' is defined only along the line of collision. In the absence of tangential forces thecollision in the perpendicular direction is taken as elastic.

Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 along a line.Let the coefficient of restitution between the bodies is e. After collision their velocitiesbecome v1 and v2 respectively. Then we have,

m1u1 + m2u2 = m1v1 + m2v2 … (i)

and e = 1 2

1 2

v vu u

--

-. … (ii)

Solving equations (i) and (ii), we get

1v = 1 2 21 2

1 2 1 2

(1 )m em e mu u

m m m mæ ö æ ö- +

+ç ÷ ç ÷+ +è ø è ø … (iii)

and 2v =2 1 1

2 11 2 1 2

(1 )m em e mu u

m m m mæ ö æ ö- +

+ç ÷ ç ÷+ +è ø è ø …(iv)

Special caseIf m1 = m2 = m and u1 = u, u2 = 0, then

mu = mv1 + mv2

and e = 1 20

v vu

--

-After solving above equations, we get

1v = (1 )2u e-

2v = (1 )2u e+

\1

2

vv = 1

1ee

-+

.

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II. ELECTRIC DIPOLE

A system of two equal and opposite charges fixed at a small distance constitutes adipole. If l is the distance between the charges + q and – q, then dipole moment isdefined as :

rP = q

r

l .

It is a vector quantity and its direction is from negative to positive charge.

Potential due to an electric dipole : Consider a dipole AB of dipole moment, q=rr

P l .

We want to calculate the electric potential at a point P, at a distance r from the centre ofthe dipole. Let line r makes an angle q with the line of dipole.The electric potential due to the charges of the dipole at P

Vp =0

14

q qPB PA

æ ö é ù-ç ÷ ê úp Îè ø ë û

From the figure, cos2

PA r= + ql

and cos ,2

PB r= - ql

assuming that l << r.

\ Vp =0

1cos cos4

2 2

q q

r r

é ù-æ ö ê úæ ö æ öç ÷ ê ú- q + qp Îè ø ç ÷ ç ÷è ø è øê úë û

l l

=20 2 2

1 cos4

cos4

q

r

æ ö qç ÷p Îè ø é ù

- qê úë û

l

l

As l << r, hence 2

2cos4

ql

can be neglected and putting q P=l , we get

VP =20

1 cos4

Pr

æ ö qç ÷p Îè ø

. …(1)

In vector notation it can be written as :

VP =30

1 .4

P rr

æ öç ÷p Îè ø

r r

.

Electric field due to an electric dipole : The electric field at point P varies with r and q

both, so we can not get rE from the differentiation of V at once. The components of r

Ein two perpendicular directions are ; the radial component Er, and transverse componentEq. Thus

Er =Vr

¶-

¶ = 2

0

1 cos4

Pr r

qé ù¶- ê úp Î¶ ë û

=30

1 2 cos4

Pr


… (i)

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and Eq = 20

1 cos1 14

PVr r r

qé ù¶ ¶- = - ê úp Î¶q ¶q ë û

= 30

1 sin4

Pr


… (ii)

\ E = 2 2rE Eq+

or E =2

30

1 3cos 14

Pr

æ ö q +ç ÷p Îè ø

… (2)

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

tan a =tan

2q

or a = 1 tantan2

- qæ öç ÷è ø

… (3)

Here a is the angle made by the resultant field rE with the line of rr . The direction of

rE

from the direction of dipolemoment is q + a.Special cases :1. End -on position : At the axis of the dipole, q = 0, and so

E = 30

1 24

Pr

æ öç ÷p Îè ø

and V = 20

14

Pr

æ öç ÷p Îè ø

2. Broad side -on position : At the equatorial line of the dipole, q = 90° and so

E = 30

14

Pr

æ öç ÷p Îè ø

and V = 0.

A DIPOLE IN AN ELECTRIC FIELD

Consider an electric dipole placed in an uniform field of intensity Er

. The ends of thedipole experience equal and opposite forces, each of magnitude F = Eq.Thus, because of uniform field, the net force on the dipole becomes zero and so thecentre of mass of the dipole does not move. However, the forces on the charged ends doproduce a net torque t

r on the dipole about its centre of mass. The magnitude of thistorque

t = [magnitude of either force] × [ distance between lines of action of the forces]

= sinF ´ ql

= sin ( )sinEq E q´ q = ql l

or t = sinPE qIn vector notation, it can be written as :

tr = P E´

r r … (1)Work done by the agent to increase the angle from q1 to q2 :

P:

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The torque exerted by the agent to increase the angle

agentt = sinPE q

Work done, agentW =2

1

agentdq

q

t qò

=2

1

sinPE dq

q

q qò

= 21

cosPE qq- q

or agentW = 1 2(cos cos )PE q - q … (2)

III. MOTION OF CHARGED PARTICLE IN UNIFORM MAGNETICFIELD

The path of charged particle in magnetic field depends on the angle q between vr and B

r.

Depending on different values of q, the possible cases are :Case 1 : When q is 0° or 180°:

For q = 0° or 180°, the force on the moving charge ( )sin 0 180 0,F qvB or= =o o

and therefore particle goes undeviated along a straight path.Case 2 : When q = 90°:

(i) When particle is projected from inside the field, it experiences a force whichalways perpendicular to the velocity and so its path will be circular. Thenecessary centripetal force is provided by the magnetic force. If r be theradius of the path, then

2mvr

= sin 90qvB o

or r =mvqB

or we can write

r =mvqB

^ ...(1)

or r =v

q Bm

æ öç ÷è ø

Let ,qm

= a is called specific charge,

\ The equation (1) can be written in the form :

r =vBa

The K.E. of the particle K =2

2Pm

or P = 2mKIf charged particle is accelerated by potential V, then

K = qV

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\ r =22 mqVP mK

qB qB qB= = ...(2)

Time period :

T =Length of path

speed

=2

2mv

r qBv v

p ´p

=

or T =2 mqBp

Also linear frequency of rotation

f =1

,2qB

T m=

p

and angular frequency w = 2 Bqfm

p = .

Note: 1.Time period T, f and w are independent of v.

2.The velocity at any instant can be written as

vr = $

x yv i v j+$

(ii) When particle is projected out side the field: If the length of the magneticfield is enough, then the angle with which the charged particle emerges outwill equal to the angle with which it enters into the field. Thus we have twocases:

(a)Time spend in magnetic field

t = 2T m

qBp

=

PQ = 2r(b)The time spend in magnetic field

t = [ ] [ ]22

T Tq q

=p p

, where T = 2 mqBp

PQ = 2r sinq

IV. THE PRISMWhen two refracting surfaces are inclined at some angle, they constitutea prism. Figure shows a triangular prism. The angle between the inclinedsurfaces is called angle of prism or refracting angle. The angle of commonlyused prism is 60°. Prism can cause deviation as well as dispersion.

Refraction through a prismConsider a monochromatic ray of light incident at an angle i on the face AB of the prism.It gets refraced at an angle r1 into the prism, after this the ray incident on the other face

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AC of the prism at an angle r2, and then finally emerges from this face with an angle e (seefigure). By Snell¢s law

m =1

sinsin

ir

= 2

sinsin

er

. ...(1)

Deviation produced by prismBecause of the inclination between the refracting surfaces, the incident ray and emergingray are not parallel. The angle between the incident ray and emerging ray is called angleof deviation and designated by d. In figure

A RÐ +Ð = 180°

\ RÐ = 180° – A

In DPQR, 1 2R r rÐ + Ð + Ð = 180°

or ( ) 1 2180 A r r- + +o = 180°

\ r1 + r2 = A ...(2)

Angle of deviation, d = SPQ SQPÐ + Ð

= ( ) ( )1 2i r e r- + -

= ( ) ( )1 2i e r r+ - +

= ( )1 e A+ -

\ i + e = A + d ...(3)Deviation produced by small angled prismFrom equation (1), for small angle, we have

m =1 2

i er r

=

\ i = m r1 and e = m r2Now from equation (3), we have

m r1 + m r2 = A + dor m(r1 + r2) = A + dor mA = A + d\ d = (m – 1)A ...(4)

There are two values for angle of incidence for same angle of deviationWhen a ray is incident at an angle i, it emerges at an angle e, with a deviation angle d. Ifthe ray is incident at an angle e, then it will emerge at an angle i having same angle ofdeviation. Thus there are two angles of incidence for same angle of deviation. These arei1 = i and i2 = e.

Minimum deviationWe know that

i + e = A + d

Principal section of prism

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\ d = (i + e) – AFrom the above equation, we can say that angle of deviation depends on angle ofincidence. Experiments show that with the increase in angle of incidence, the angle ofdeviation first decreases, passes through minimum and then increases. Thus for a certainvalue of the angle of incidence (i1 = i2), the light passing through prism suffers minimumdeviation. The angle of deviation at this position is called the minimum angle of deviation(dm). Figure shows the minimum deviation and graph shows the variation of angle ofdeviation with angle of incidence.In minimum deviation position, d = dm

i = eand so r1 = r2 = r (say)From equations (2) and (3), we get

r =2A

and .2

mAi

+ d=

If m is the refractive index of material of the prism, then by Snell¢s law

m =sinsin

ir

or m =sin

2

sin2

mA

A

+ dæ öç ÷è ø . ...(5)

This is called prism formula.Maximum deviationWe know that, angle of deviation

d = (i + e) – A.The deviation angle will be maximum, when either of i or e is maximum. Thus for

i = 90°,dmax = (90° + e) – A. ...(i)

At face AB of the prism,

m =1

sin 90sin r

o

\ sin r1 =1m

or r1 = 1 1sin C- æ ö=ç ÷è mø

We have r1 + r2 = A\ r2 = A – r1 = A – C

Now for face AC, m = ( )sin

sine

A C-

or sin e = m sin(A – C)

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or e = ( )1sin sin A C- m -é ùë û ...(ii)

Thus dmax = ( )190 sin sin A C A-+ m - -é ùë ûo . ...(6)

Condition of no emergenceA ray of light will not emerge out from the prism, if it gets totally reflected from the otherface of the prism, even for angle of incidence on first face is 90°. Thus angle of incidenceon second face should be greater than critical angle. i.e.,

r2 > C.

For 190 , .i r C® ®o Thus for no emergence from any face of the prism, angles

r1 + r2 = A,\ A > 2C ...(7)

So, a ray of light will not emerge out from the prism, if A > 2C.Totally reflecting prismThe critical angle for glass-air interface is 42°. Thus if we make a prism in such a way, thatlight ray incident into it at an angle greater than critical, then it becomes totally reflectingprism. Such a prism may be right angled isosceles (45° – 90° – 45°). They can be used todeviate rays through 90° or 180°.

Erecting prismThis is also the right-angled isosceles prism. In this case rays of light should be parallelto the hypotenuse. By doing so the rays invert themselves and an inverted objectappears as erect.

V. PHOTO-ELECTRIC EFFECT

When light of certain frequency is incident on a metal surface, electrons are ejected fromthe metal. This phenomenon is called photoelectric effect (PEE). Electrons ejected fromthe metal are called photoelectrons. The photoelectric effect was first observed byHeinrich Hertz in 1887.Work functionWe know that metals have large number of free electrons. These electrons move freelyinside the metal but can not come out from it due to attraction of the positive ions. Someenergy is needed to liberate the electrons from the bondage of the attraction of the ions.The minimum energy required to liberate the electrons from the metal surface, is calledwork function, and is represented by W0.Work functions of some photometals

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Metal Work function (ev) Metal Work function (eV)

Cesium 1.9 Calcium 3.2 Potassium 2.2 Copper 4.5

Sodium 2.3 Silver 4.7 Lithium 2.5 Platinum 5.6

Experimental set-up of PEEThe experimental set-up to study the photoelectric effect is shown infig.When monochromatic light of frequency greater than f0 is incidenton the cathode, photoelectrons are emitted from it and they movetowards anode A. Initially, the space between the cathode and theanode contains a number of electrons making up electron cloud. Thisnegative charge repels the fresh electrons coming from the cathode.The electrons of maximum kinetic energy are able to reach the anodeand constitutes a photocurrent. If anode is made positive with respectto the cathode, the emitted electrons are attracted by the anode andthe photoelectric current increases. With the increase in anodepotential, the photoelectric current increases and becomes maximum. Thereafter currentwill not increase with the increase in anode potential. This maximum value of current iscalled the saturation current (is). This will happen when all the emitted electrons by thecathode in any time interval are attracted by the anode . Fig. shows the photoelectriccurrent i with anode potential V.

Stopping potentialWhen anode is given negative potential with respect to the cathode, the photoelectriccurrent decreases. For a particular value of anode potential, the photoelectric currentbecomes zero. The minimum negative anode potential at which photoelectric currentbecomes zero is called stopping or cut off potential V0. To stop the photoelectric current,we must ensure that even the fastest electron will not reach the anode. Thus stoppingpotential is related to the maximum kinetic energy of the ejected electrons. If V0 is thestopping potential, then

Kmax = eV0. ...(4)

Characteristics of pee

1. Effect of intensity of incident lightWhen the intensity of the light increases, more number of photons strike with thephotometal and thereby liberate more number of electrons. Because of this, photocurrent increases. As the frequency of the incident light is same, so maximumkinetic energy and hence slopping potential remains same.

2. Effect of frequency of incident lightWhen the intensity of incident light is kept constant and its frequency increases,the number of photons remains same but their kinetic energy increases. Thereforethe emitted electrons are same in number but of greater kinetic energy and hencestopping potential also increases. Fig. shows the variation of photocurrent withfrequency of light f.

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3. Effect of photometalWhen intensity and frequency of incident light are kept constant and photo-metal is changed, the stopping potential V0 versus frequency f are parallel straightlines. This shows that the slope V0/f is same for all metals and is equal to universalconstant (h). If the graph is plotted between Kmax and f, then there is straight line.Slope of which gives the value of h/e (Fig.).

4. Effect of timeMetal starts emitting electrons as soon as light is incident on it and so there is notime lag between incident light and emitted electrons.

Einstein's explanation of pee

Einstein forwarded the Plank¢s quantum theory to explain photoelectric effect. Accordingto him light is made of small energy bundles, called photons. The energy of photon isproportional to the frequency f. That is

E = hf,where h is a universal constant, called Plank¢s constant. He made the followingassumptions :1. The photoelectric effect is the result of collisions between photons of incident

light and free electrons of the metal.2. The electrons of metal are bound with the nucleus by attractive forces. The

minimum energy required to liberate an electron from this binding is called workfunction W0.

3. The incident photon interacts with a single electron and spend energy in twoparts :(i) in liberating the electron from the metal surface,(ii) and imparting kinetic energy to emitted electrons. Thus if hf is the energy of

incident photons, thenh f = W0 + Kmax ...(i)

As f =cl and W0 =

0

hc ,l

\hcl

= 2max

0

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hc mv+l

. ...(ii)

Above equation is known as Einstein photo-electric equation. It should be

remembered that photoelectric effect will occur only if o .l £ l

4. The efficiency of photoelectric effect is less than 1%, i.e., only less than 1% ofphotons are capable of ejecting electrons from the metal surface. The rest 99% ofthe photon energy will convert into thermal energy.

5. If V0 is the stopping potential, then Kmax = eV0 and so

hf = W0 + eV0

orhfe = 0W

e + V0

or V0 = 0W h f.e e

æ ö- + ç ÷è ø

...(iii)

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The equation (iii) is a straight line between V0 and f, whose slope is ,he

æ öç ÷è ø

which

is a universal constant.

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