Physics Disha

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I. HEAD-ON ELASTIC COLLISION

If colliding bodies before and after collision remain in a line, the collision is said to be head-on collision. This will happen when bodies move along the line joining their geometric centres.

Consider two bodies of masses m₁ and m₂ moving with velocities ur₁ andu ur r₂ ₁> ( ur₂) along the same straight line. Let after collision their velocities become vr₁ and vr₂ in the same initial direction. Then

according to conservation of linear momentum, we have

1 1 2 2

mur +m ur = m_{1 1}rv +m_{2 2}rv .

Since all the colliding bodies before and after collision remain in the same line, so we can drop 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

\ 1₂m u1 12+1₂m u2 22 = 1₂m v1 12+1₂m v2 22 …(iii)

or m₁(u₁2_{– v}

12) = m2(v22 – u22)

or m₁(u₁ + v₁)(u₁ – v₁) = m₂(v₂+ u₂) (v₂– u₂) (iv) Dividing equation (iv) by (ii), we get

u₁ + v₁ = v₂ + u₂

or u₁-u₂ = v₂-v₁. …(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, 1 v = 1 2 1 2 2 1 2 1 2 2 m m m u u m m m m æ - ö ₊æ ö ç ₊ ÷ ç ₊ ÷ è ø è ø …(vi) and v₂ = 2 1 2 1 1 1 2 1 2 2 m m m u u m m m m æ - 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 get

v₁ = u₂ and v₂ = u₁

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

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(ii) If m₁ = m₂= m and u₂ = 0, then

v₁ = 0 and v₂ = u₁.

(iii) When a light body collides with a massive stationary body. Here m₁ << m₂ and u₂ = 0

\ v₁ = – u₁ and v₂ _{; 0.}

Hence when a light body collides with a massive stationary body, the light body rebounds after the collision with an equal speed while the massive body remains at rest.

(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 projectile to the target

DK = decrease in K.E. of projectile = 1₂m u1 12-1₂m v1 12.

Fractional decrease in K.E.

K K D = 2 2 1 1 1 1 1 1 2 2 2 1 1 1 2 m u m v m u -or K K D = 2 1 1 1 v u æ ö - ç ÷_{è ø} . …(viii)

Perfectly inelastic collision in 1-D

Consider two bodies of masses m₁ and m₂ moving with velocities u₁ and u₂ along a straight line. They make perfectly inelastic collision. Let after collision, their common velocity becomes v, then by conservation of momentum, we have

m₁u₁ + m₂u₂ = (m₁ + m₂)v \ v = 1 1 2 2 1 2 m u m u m m é + ù ê ₊ ú ë û.

The loss of K.E. in collision

K D = 1 12 2 22 1 2 2 1 1 1 ( ) 2m u 2m u 2 m m v æ ₊ ö_- ₊ ç ÷ è ø = _èæç₂1m u1 12+₂1m u2 22ö÷_ø-1₂(m1+m2) 2 1 1 2 2 1 2 m u m u m m æ + ö ç ₊ ÷ è ø = 1 2 1 2 2 1 2 1 ( ) 2 m m u u m m æ ö -ç ₊ ÷ è ø .

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General analysis of 1-D collision

Newton's experimental law : Coefficient of restitution It is defined as;

e = velocity of separation_{velocity of approach}

= 2 1 1 2 v v u u -or e = 2 1 1 2 2 1 1 2 . v v v v u 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 e£1. (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 the collision 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 velocities become v₁ and v₂ respectively. Then we have,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ … (i) and e = 1 2 1 2 v v u u -- . … (ii) Solving equations (i) and (ii), we get

1 v = 1 2 2 1 2 1 2 1 2 (1 ) m em e m u u m m m m æ - ö ₊æ + ö ç ₊ ÷ ç ₊ ÷ è ø è ø … (iii) and v₂ = 2 1 2 1 1 1 2 1 2 (1 ) m em e m u u m m m m æ - ö ₊æ + ö ç ₊ ÷ ç ₊ ÷ è ø è ø …(iv)

Special case

If m₁ = m₂ = m and u₁ = u, u₂ = 0, then mu = mv₁ + mv₂ and e = 1 2 0 v v u -After solving above equations, we get

1 v = (1 ) 2 u e -2 v = (1 ) 2 u e + \ 1 2 v v = 1₁-₊e_e.

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II. E

LECTRIC DIPOLE

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

r

P = qrl .

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, Pr=qr_{l .} We want to calculate the electric potential at a point P, at a distance r from the centre of the 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 V_p = 0 1 4 q q PB PA æ ö é _- ù ç _{p Î} ÷ ê ú è ø ë û

From the figure, cos 2 PA r= +l q and cos , 2 PB r= -l q assuming that l << r. \ V_p = 0 1 cos cos 4 2 2 q q r r é _- ù æ _{ö ê} _ú æ ö æ ö ç _{p Î} _{÷ ê} _- _q ₊ _q _ú è ø ç_ê_è ÷_ø ç_è ÷_ø_ú ë û l l = 2 0 ₂ ₂ 1 cos 4 cos 4 q r æ ö q ç _{p Î} ÷ è ø é ù - q ê ú ë û l l As _{l << r, hence} 2cos2 4 q

l _{can be neglected and putting q}_l₌_P_{, we get} V_P = 2 0 1 cos 4 P r æ ö q ç _{p Î} ÷ è ø . …(1)

In vector notation it can be written as : V_P ₌ 3 0 1 . 4 P r r æ ö ç _{p Î} ÷ è 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 r_{E from the differentiation of V at once. The components of}Er

in two perpendicular directions are ; the radial component E_r, and transverse component E_q. Thus E_r = V r ¶ -¶ = 0 2 1 cos 4 P r r q é ù ¶ - _ê _ú p Î ¶ ë û = 3 0 1 2 cos 4 P r æ ö q ç _{p Î} ÷ è ø … (i)

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5 and E_q = ₂ 0 1 cos 1 1 4 P V r r r q é ù ¶ ¶ - = - _ê _ú p Î ¶q _{¶q ë} _û = ₃ 0 1 sin 4 P r æ ö q ç _{p Î} ÷ è ø … (ii) \ E = 2 2 r E +E_q or E = 2 3 0 1 3cos 1 4 P r æ ö q + ç _{p Î} ÷ è ø … (2)

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

tana = tan₂q or a = _tan 1 tan 2 - æ qö ç ÷ è ø … (3)

Here a is the angle made by the resultant field rE with the line of rr . The direction of Er 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 = ₃ 0 1 2 4 P r æ ö ç _{p Î} ÷ è ø and V = ₂ 0 1 4 P r æ ö ç _{p Î} ÷ è ø

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

E = ₃ 0 1 4 P r æ ö ç _{p Î} ÷ è 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 the dipole 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 the centre of mass of the dipole does not move. However, the forces on the charged ends do produce a net torque tr on the dipole about its centre of mass. The magnitude of this torque

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

= Eq´_lsinq = E q( )sin_l q

or t = PEsinq

In vector notation, it can be written as :

tr = P Er 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

agent

t = PEsinq

Work done, Wagent =

2 1 agentd q q t q

ò

= 2 1 sin PE d q q q q

ò

= 2 1 cos PE- qq_q

or Wagent = PE(cosq -1 cosq2) … (2)

III. MOTION OF CHARGED PARTICLE IN UNIFORM MAGNETIC

FIELD

The path of charged particle in magnetic field depends on the angle q between vr and Br. 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 F=qvBsin 0

(

oor180o

)

=0, 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 which always perpendicular to the velocity and so its path will be circular. The necessary centripetal force is provided by the magnetic force. If r be the radius of the path, then

2 mv r = qvBsin 90o or r = mv_qB or we can write r = mv_qB^ ...(1) or r = _qv B m æ ö ç ÷ è ø Let q ,

m = a is called specific charge,

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

r = v

B a The K.E. of the particle K =

2

2 P m

or P = _2mK

If charged particle is accelerated by potential V, then K = qV

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7 \ r = P 2mK 2mqV qB= qB = qB ...(2) Time period : T = Length of path speed = 2 2 mv r qB v v p ´ p ₌ or T = 2 m_qBp

Also linear frequency of rotation

f = 1 ,

2 qB T = pm and angular frequency w = 2 f Bq m

p = .

Note:

1.Time period T, f and w are independent of v. 2.The velocity at any instant can be written as

vr = v i v jx$+ y$

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

(a)Time spend in magnetic field t = T₂ =p_qBm PQ = 2r

(b)The time spend in magnetic field t = 2

[ ]

2 T T q q = p p , where T = 2 m qB p PQ = 2r sinq

IV. THE PRISM

When two refracting surfaces are inclined at some angle, they constitute a prism. Figure shows a triangular prism. The angle between the inclined surfaces is called angle of prism or refracting angle. The angle of commonly used prism is 60°. Prism can cause deviation as well as dispersion.

Refraction through a prism

Consider 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 (see figure). By Snell¢s law

m = 1 sin sin i r = ₂ sin sin e r . ...(1)

Deviation produced by prism

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

A R Ð + Ð = 180° \ ÐR = 180° – A In DPQR, Ð + Ð + ÐR r₁ r₂ = 180° or

(

180o-A

)

+ +r1 r2 = 180° \ r₁ + r₂ = A ...(2) Angle of deviation, d = ÐSPQ+ ÐSQP =

(

i r-₁

) (

+ -e r₂

)

=

(

i e+ -

) (

r₁+r₂

)

=

(

1 e+ -

)

A \ i + e = A + d ...(3)

Deviation produced by small angled prism

From equation (1), for small angle, we have

m =

1 2

i e

r =r

\ i = m r₁ and e = m r₂

Now from equation (3), we have

m r₁ + m r₂ = A + d or m(r₁ + r₂) = A + d

or mA = A + d

\ d = (m – 1)A ...(4)

There are two values for angle of incidence for same angle of deviation

When a ray is incident at an angle i, it emerges at an angle e, with a deviation angle d. If the ray is incident at an angle e, then it will emerge at an angle i having same angle of deviation. Thus there are two angles of incidence for same angle of deviation. These are i₁ = i and i₂ = e.

Minimum deviation

We know that

i + e = A + d

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\ d = (i + e) – A

From the above equation, we can say that angle of deviation depends on angle of incidence. Experiments show that with the increase in angle of incidence, the angle of deviation first decreases, passes through minimum and then increases. Thus for a certain value of the angle of incidence (i₁ = i₂), the light passing through prism suffers minimum deviation. The angle of deviation at this position is called the minimum angle of deviation (d_m). Figure shows the minimum deviation and graph shows the variation of angle of deviation with angle of incidence.

In minimum deviation position, d = d_m

i = e

and so r₁ = r₂= r (say)

From equations (2) and (3), we get

r = 2 A and . 2 m A i= + d If m is the refractive index of material of the prism, then by Snell¢s law

m = sin sin i r or m = sin 2 sin 2 m A A + d æ ö ç ÷ è ø_. _...(5)

This is called prism formula.

Maximum deviation

We 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°,

d_max = (90° + e) – A. ...(i)

At face AB of the prism,

m = 1 sin 90 sin r o \ sin r₁ = _m1 or r₁ = sin-1æ ö_{ç ÷}1 =C è mø We have r₁ + r₂ = A \ r₂ = A – r₁ = A – C

Now for face AC, m = _sin

₍

sin_{A C}_-e

₎

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or e = sin-1é_ëmsin A C

(

-

)

ù_û ...(ii) Thus d_max = 90o+sin-1é_ëmsin A C

(

-

)

ù_û-A. ...(6)

Condition of no emergence

A ray of light will not emerge out from the prism, if it gets totally reflected from the other face of the prism, even for angle of incidence on first face is 90°. Thus angle of incidence on second face should be greater than critical angle. i.e.,

r₂ > C.

For i®90 ,o r₁®C. 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 prism

The critical angle for glass-air interface is 42°. Thus if we make a prism in such a way, that light ray incident into it at an angle greater than critical, then it becomes totally reflecting prism. Such a prism may be right angled isosceles (45° – 90° – 45°). They can be used to deviate rays through 90° or 180°.

Erecting prism

This is also the right-angled isosceles prism. In this case rays of light should be parallel to the hypotenuse. By doing so the rays invert themselves and an inverted object appears as erect.

V. PHOTO-ELECTRIC EFFECT

When light of certain frequency is incident on a metal surface, electrons are ejected from

the metal. This phenomenon is called photoelectric effect (PEE). Electrons ejected from the metal are called photoelectrons. The photoelectric effect was first observed by Heinrich Hertz in 1887.

Work function

We know that metals have large number of free electrons. These electrons move freely inside the metal but can not come out from it due to attraction of the positive ions. Some energy 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 called work function, and is represented by W₀.

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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 PEE

The experimental set-up to study the photoelectric effect is shown in fig.

When monochromatic light of frequency greater than f₀ is incident on the cathode, photoelectrons are emitted from it and they move towards anode A. Initially, the space between the cathode and the anode contains a number of electrons making up electron cloud. This negative charge repels the fresh electrons coming from the cathode. The electrons of maximum kinetic energy are able to reach the anode and constitutes a photocurrent. If anode is made positive with respect to the cathode, the emitted electrons are attracted by the anode and the photoelectric current increases. With the increase in anode

potential, the photoelectric current increases and becomes maximum. Thereafter current will not increase with the increase in anode potential. This maximum value of current is called the saturation current (i_s). This will happen when all the emitted electrons by the cathode in any time interval are attracted by the anode . Fig. shows the photoelectric current i with anode potential V.

Stopping potential

When anode is given negative potential with respect to the cathode, the photoelectric current decreases. For a particular value of anode potential, the photoelectric current becomes zero. The minimum negative anode potential at which photoelectric current becomes 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 stopping potential is related to the maximum kinetic energy of the ejected electrons. If V₀ is the stopping potential, then

K_max = eV₀. ...(4)

Characteristics of pee

1. Effect of intensity of incident light

When the intensity of the light increases, more number of photons strike with the photometal and thereby liberate more number of electrons. Because of this, photo current increases. As the frequency of the incident light is same, so maximum kinetic energy and hence slopping potential remains same.

2. Effect of frequency of incident light

When the intensity of incident light is kept constant and its frequency increases, the number of photons remains same but their kinetic energy increases. Therefore the emitted electrons are same in number but of greater kinetic energy and hence stopping potential also increases. Fig. shows the variation of photocurrent with frequency of light f.

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3. Effect of photometal

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

4. Effect of time

Metal starts emitting electrons as soon as light is incident on it and so there is no time lag between incident light and emitted electrons.

Einstein's explanation of pee

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

E = hf,

where h is a universal constant, called Plank¢s constant. He made the following assumptions :

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 work function W₀.

3. The incident photon interacts with a single electron and spend energy in two parts :

(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, then

h f = W₀ + K_max ...(i) As f = _{l and W}c ₀ = 0 hc , l \ hc_l = 2_max 0 1 2 hc mv + l . ...(ii)

Above equation is known as Einstein photo-electric equation. It should be remembered that photoelectric effect will occur only if l £ l_o.

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

5. If V₀ is the stopping potential, then K_max = eV₀ and so hf = W₀ + eV₀ or hf e = 0 W e + V0 or V₀ = W0 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 æ öç ÷_{è ø}h_e , which is a universal constant.