 ## CBSE Class XI Supplementary Textual Material in Physics

Contents:

• Unit 1: Errors in measurement
• Unit 4:
1. Motion in a vertical circle
2. Elastic and inelastic collision in two dimensions
3. Non-Conservative forces
• Unit 7:
1. Poisson’s ratio
2. Elastic energy
3. Critical velocity
4. Relation between Cp and Cv
• Unit 10: Speed of a wave motion

CBSE Class 11 Supplementary Textual Material in Physics is given below.

ERRORS IN MEASUREMENT

UNIT – I

The term ‘Error’ in Physics, has a meaning quite different from ‘mistake’. An experimentalist can commit a mistake due to carelessness or ‘casualness’ on his her part. We do not expect such a behavior from any serious experimentalist. ‘Mistakes’, therefore, do not get any consideration, or quantitative classification, in the observations of a ‘well planned’ and ‘well executed’ experiment.

No matter how well planned, and carefully executed our experiment is, we still cannot avoid ‘errors’ in our observations. This is because no measurement is ever perfect. Errors can arise from causes like a built-in fault in the ‘design’ or
‘graduations’ of a measuring instrument, or a faulty way of carrying out the measurement on the part of the observer. The more important errors, in Physics, however, are what are known as (i) random errors; and (ii) the errors caused by the limitations of the measuring devices used in a given situation. The ‘limitation’, of any measuring instrument, is quantitatively specified through its ‘least count’ – the minimum magnitude of the relevant quantity that it can measure.

Random errors are errors whiCh cannot be associated with any systematic or constant cause or with any definite ‘law of action’. These errors, are usually assumed to follow the well known ‘Gaussian law of Normal Distribution.

In simple words, this law implies that the probability of an error {+Δ x) in a measurement Is the same as the probability of an error (-Δ x)in that very measurement. Also, in a carefully carried out experiment, small magnitudes of ‘error’ are more likely than larger magnitudes of ‘error. The ‘Normal Distribution law’ is graphically represented by a curve of the type shown here (Figure (i)) Fig. (i) Illustrating Gaussian law of Normal Distribution

A very significant ‘conclusion’, from this law, is that the ‘arithmetic mean’ of a large number of observations, is likely to be much closer to its ‘true value’ than any of the individual observations,

It is for this reason that we are atways advised to take a targe number of observations, and use their arithmetic mean, for doing, our ‘calculations’ or for drawing our ‘conclusions’ or inferences’,

The ‘least counts’, of the measuring Instruments, used in an experiment, playa very significant role in the ‘precision’ associated with that experiment. Scientists are, therefore, constantly striving to design instf 1. Ments And measuring techniques, that have better (Smaller) values for their ‘least court

We can appreciate the importance of ‘least count’ by taking the simple example of the ‘measurement of a length’, When we use a meter scale for this purpose, we can rely on this measurement only up to a ‘mrn’. This is because the least count of a meter scale is 1mm only, However, the use of a (simple) vernier caliper, pushes up this reliability to (1 / 10)th of a mm or o.1mm while the use of (usual) ‘screw gauge’ would take this reliability to (1 / 10)th of a mm or 0.01mm,

Most of our ‘experiments require us to use the measured values, of a number of different physical quantities, and put then in the appropriate ‘formula’ , to calculate the required quantity. We then calculate the ‘percentage reliability, or ‘maximum error’, in our final result:

1. by associating a ‘relative error’- equal to the ratio of the least count (Of the measuring instrument used) to the measured value. with each of the quantities involved in our formula.

2. by using the standard ‘routes’ for finding the error in a ‘sum or difference’, ‘product or quotient’, or ‘power’ . of different quantities. involved in a given formula.

We illustrate these ideas – for calculating the maximum error-through a few examples.

Example 1: Suppose we use a phySical balance to measure the mass or an object and find the mean value of our observations to be 156.28 g.

Since we are somewhat uncertain, about the measurement, due to our instrument’s imperfections. we need to express this in our resun. Let the least count of physical balance be 0.lg: This implies that the uncertainty. of any measurement made with this instrument. is ±0.1g. Therefore, we would report the mass of this object to be (156.3g ± O.lg). This implies that we can only say that the mass of the object is somewhere between 156.29 and 156.4g.

Example 2 : It is required to find the volume of a rectangular block. A vernier caliper is used to measure the length, width and height of the block. The measured values are found to be 1.37cm, 4.11cm, and 2.S6cm, respectively.

Solution: The measured (nominal) volume of the block is. therefore.

V = l * w * h

= (1.37 x 4.11 x 2.56)cm3.

= 14.41cm3.

However. each of these measurements has an uncertainty of ± 0.01em, the least count of the vernier caliper. We can say thai the values of length, width, and height should be written as

l = (1.37 cm ± 0.01 cm)

w = (4.11 cm ± 0.01 cm)

h = (2.56 cm ± 0.01 cm)

We thus find that the lower limit, of the volume of the block, is given by

Vmin = 1.36 cm x 4.1O cm x 2.55 cm

= 14.22 cm3

This is 0.19cm3 lower than the (nominal) measured value.

The upper limit can also be calculated:

Vmax = 1.38cm x4.12cm x 2.57cm

= 14.61 cm3

This is 0.20cm3 higher than the measured value.

AS a practical rule, we choose the higher of these two deviations (from the measured value.) as the uncertainty, in our result. We, therefore, should report the volume of the block as (14.41cm3 ± O.20cm3).

Example 3 : In an experiment, on determining the density of a rectangular block, the dimensions of the block are measured with a vernier caliper (with a least count of 0.01cm) and its mass is measured with a beam balance of least count O.lg. How do we report our result for the density of Ihe block?

Solution: Let the measured values be:

Mass of block (m) = 39.39

Length of block (t) = 5.12cm

Breadth of block (b) = 2.S6cm

ThiCkness of block (I) = O.37cm

The density of the block is given by

ρ = mass / volume = m / l * b * t

= = 39.39 / 5.12 cm x 2.56 cm x 0.37 cm g = 8.1037 cm-3

Now uncertainty in m = ± 0.01g

uncertainty in l = ± 0.01 cm

uncertainty in b = ± 0.01 cm

uncertainty in t = ± 0.01 cm

Maximum relative error, in the density value is, therefore, given by

Δρ / ρ = Δl / l + Δb / b + Δt / t = Δm + m

= 0.01 / 5.12 + 0.01 / 2.56 + 0.01 / 0.37 + 0.1 / 39.3

= 0.0019 + 0.0039 + 0.027 + 0.0024

= 0.0358

Hence Δρ = 0.0358 * 8.10379 cm-3 = ~ 0.3g cm-3

We cannot, therefore, report the calculated value of ρ (= 8.1037 gm-3) up to the fourth decimal place. Since Δρ = 0.3g cm-3 the value of ρ can be regarded as accurate up to the first decimal place only. Hence the value of ρ must be rounded off as 8.1 g cm-3 and the result of measurements should be reported as

ρ = (8.1 ± 0.3) g cm-3.

A careful look, at the calculations done above, the main contribution to this (large) error in the measurement of ρ, is contributed by the (large) relative error (0.027) in the measurement of t, the smallest of the quantities measured. Hence the precision of the reported value of ρ could be increased by measuring t with an instrument having a least count smaller than 0.0l cm. Thus if a micrometer screw gauge (least count = 0.001 cm). (rather) than a vernier caliper were to be used. for measuring t. we would be reporting our result for ρ with a considerably lower degree of uncertainty. Experimentalists keep such facts in mind While designing their ‘plan’ for carrying out different measurements in a given experiment.

EXERCISES

1. The radius of a sphere is measured as (2.1 ± 0.5)cm

Calculate its surface area with error limits.

[Ans. {(55.4 ± 26.4)cm2})

2. The voltage across a lamp is (6.0 ± 0.1) volt and the current passing through tt is (4.0 ± 0.2) ampere. Find the power consumed by the lamp.

[Ans. {(24.0 ± 1.6) watt})

3. The length and breadth of a rectangular blOck are 25.2cm and 16.8cm, which have both been measured to an accuracy of 0.1em. Find the area of the rectangular block.

[Ans. {(423.4 ± 4.2) cm2))

4. A force of (2500 + 5) N is applied over an area 01 (0.32 ± 0.02) m2. Calculate the pressure exerted over the area.

[Ans. {(7812.S ± 503.9) N / m2})

5. To find the value of ‘g’, by using a simple pendulum, the following observations were made:

Length of the thread t = (100 ± 0.1) cm

Time period of oscillation T = (2 ± 0.1) s

Calculate the maximum permissible error in measurement of ‘g’. Which quantity should be measured more accurately & why?

6. For a glass prism, of refracting angle 60°, the minimum angle of deviation, Dm, is found to be 36°, with a maximum error of 1.05°. when a beam of parallel light is incident on the prism. Find the range of experimental value of refractive index ‘μ’ it is known that the refractive Index ‘μ’ of the material of the prism is given by:

μ = sin (A + Dm / 2) / sin (A / 2)

[Ans. (1.46 ≤ μ ≤ 1.51, with a mean value of 1.49)]

7. The radius of curvature of a concave mirror, measured by a spherometer, is given by

R = l2 / 6h + h / 2

The value of l and hare 4.0cm and 0.065cm respectively, where ( is measured by a meter scale and h by a spherometer. Find the relative error in the measurement of R.

[Ans. (0.08)]

8. In Searle’s experiment, the diameter of the wire, as measured by a screw gauge, of least count 0.001cm, is O.50O cm. The length, measured by a scale of least count 0.1cm, is 110.0cm. When a weight of 40N is suspended from the wire, its extension is measured to be 0.12Scm by a micrometer of least count 0,001cm. Find the Young’s modulus of the material of the wire from this data.

[Ans. {(2.2 x 1011 ± 10.758 x 109) N/m2)]

9. A small error in the measurement of the quantity having the highest power (in a given formula), will contribute maximum percentage error in the value of the physical quantity to whom ~ is related, Explain why?

10. The two specific heat capacities of a gas are measured as Cp = (12,28 ± 0.2) units and Cv = (3.97 + 0.3) units. Find the value of the gas constant R.

[Ans. {(8.31 ± 0.5) units}].

MOTION IN A VERTICAL CIRCLE

UNIT – IV

Consider a particle P suspended in a vertical plane, by a massless, inextensible siring from a fixed point O. In equilibrium, the siring is vertical with P vertically below the point of suspension 0, as shown in Figure (i) (a) Let the particle P be imparted an initial velocity ϑ1, in a horizontal direction, as shown in Figure (i) (b). Under the lesion in the string, the particle starts moving along a vertical Circular path of radius equal to the length of the string. The point of suspension O is the center of this circle. It turns out that the initial velocity, ϑ1 has to be more than a certain minimum critical value so that the particle may describe a vertical circular motion around palm O.

The motion of a particle in a vertical circle, differs from that in a horizontal circle. In a horizontal circular motion, the force of gravity plays no role in the motion of The particle. However, in a vertical circular motion, gravity plays a very important role. It is easy to realize that a vertical circular motion has to be a non-uniform circular motion. In this case. the velocity of the particle varies both in magnitude and direction. In other words, In a vertical circular motion even Ihe speed of me particle does not remain constant. As The particle moves up the circle, from its lowest position P, its speed continuously decreases liII it reaches the highest point of its circular path. This is due to the work done against the force of gravity. When the particle moves down the circle. re. from R → Q’ → P, Its speed would keep on increasing, This is because of the work done. by the force of gravity, on the particle,

To obtain the basic characteristics of a vertical circular motion, consider an instantaneous position of particle, say at L. In this position, let the string make an angle a with the vertical line oP. as shown in figure (II) The forces acting on the particle (of mass rn) at this position, L, are

(i) its weight = mg ; acting vertically downwards

(ii) the tension; T; in the string acting along LO.

The Instantaneous velocity ϑ of the particle is along the direction of the tangent to the circle at L The corresponding instantaneous centripetal force, force, on the particle, equals mϑ2 / r where r (= length of string I) is the radius of the particle’s circular path. This force must act along Lo. We must, therefore, have

mϑ2 / r = T – mg cos θ

∴ T = mϑ2 / r + mg cos θ (i)

We can take Ihe horizontal direction, at the lowes I point P, as the position 01 zero gravitational potential energy, Now, as per the law of conservation of energy;

Total energy at P = Total energy at L

∴ 1 / 2 mϑ21 + O = 1 / 2 mϑ2 + mgh (ii)

where MP = H, is the vertical height by which particle has risen above P. From right angled Triangle OML,

OM = OL cos θ = r cos θ

∴ M P = h = OP – OM

= r – r cos θ

= r (1 – cos θ ) (iii)

From Eqns. (ii) and (iii), we get

ϑ21 + ϑ2 + 2gr (1 – cos θ) (iv)

We now substitute the value of ϑ2, from Eqn (iv), in Eqn (i). Hence

T = m / r [ϑ21 – 2gr (1 – cos θ) ] + mg cos θ

= mϑ21 / r – 2mg (1 – cos θ) + mg cos θ

= mϑ21 / r – 2mg + 3mg cos θ (v)

This relation gives the lens ion T. in the string as a function of θ. We now use this relation to see the details of the particle when it is at the (i) lowest (ii) mid – way (horizontal) and (iii) highest position of its circular path.

When the particle is at the lowest point P of ~s vertical circular path, we have θ = 0° .
The tension Tp in Ihe siring in this posmon, from Eqn (V) ,is

Tp = mϑ21 / r + – 2mg + 3mg cos 0o

= mϑ21 / r + mg (vi) Consider next the case. when the particle Is In position Q. where the string is momentarily) in its horizontal position. Clearly θ = π / 2 here. Let ϑ2 be the instantaneous velocity of the particle here. Let TQ be the instantaneous tension in the sIring here. Using Eqn (v), we have

TQ = mϑ21 / r – 2mg + 3mg cos (π / 2)

= mϑ21 / r – 2mg (vii)

The change in the tension, as the particle moves from P to Q. equals (Tp – TQ). We have (Tp – TQ)

= (mϑ21 / r + mg) – (mϑ21 / r + 2mg) = 3mg We next consider the partide P to be at the highest point R of its circular path, Let ϑ3 be the instantaneous velocity of the particle here, In this position, θ = π. If TR denotes the tension in string in this position, we have, from eqn(v),

TR = mϑ21 – 2 mg + 3 mg cos (π)

= mϑ21 – 5 mg (viii) Hence the change in the tension in the string, as the particle moves from P to R, along the vertical circle. is (Tp – TR) = (mϑ21 / r + mg) – (mϑ21 / r + 5mg)= 6mg

It is thus seen that, the tension in the string is maximum when the partide is at lowest point P and is minimum at the highest point R of its vertical circular path. This is so because at the highest point, a part of the centripetal force, needed to keep the particle moving In its circular path, is provided by the weight (mg) of the particle.

From Eqn (viii), It Is easy to realize that TR can be (a) positive: (b) negative or (c) zero depending. on the value of ϑ1. If TR becomes a negative number, the string would get slackened, and the particle will be unable to continue moving along its vertical circular path. It will fall down before it is able to complete ils circular path. Hence for completing the vertical circle, the minimum value of TR has to be zero, We, therefore, have

(TR)min = m (ϑ1)2min / r – 5 mg = 0

∴ (ϑ1)min = √5gr (ix)

Using Eqn (iv) the minimum speed, which the particle must have at the highest point R, So that it is able to complete the vertical Circle, is given by

1)2min = (ϑ3)2min + 2 gr (1 – cos π)

∴ 5gr = (ϑ3)2min + 4 gr

or (ϑ3)min = √gr (x)

When the particle completes Its motion along the vertical circle it is referred to as “looping the loop”. For this to be possible, the minimum speed at the lowest point, must be √5gr.

The values of the tension in the string, when the particle is just able to do ‘looping the loop’, correspond to ϑ1 = (ϑ1)min = √5 gr

Hence, in this case,

Tp = 6 mg [From Eqn (vi)]

and TR = mg [From Eqn (viii)]

The results obtained above (for ‘looping the loop’) are put to many practical applications. We list below some of these applications.

(i) The pilot of an air – craft can successfully loop a vertical circle (of rap ius r), if the velocity of the as-craft, at the lowest point of its vertical circle, is more than √5 gr

(ii) Consider a bucket full of water being rotated in a vertical circle. The water. in the bucket, would not spill-over (even when the bucket is at its highest position along The vertical circle, t.e. when it is upside down), if the starting speed of the bucket, at the lowest point of tts path, is more than √5 gr. Under these conditions, the centrifugal force, on the water, inside the bucket, is more than the weight mg of the water. Hence the water does not spill over. If, however. the staining speed, at the lowest point of the vertical circular path, is less than √5 gr, water will spill over when the bucket is upside down, i.e. at the highest point of its circular path. tt is thus obvious that the “trick” really lies in whirling the bucket “fast enough”.

(iiil A circus acrobat, performing in the “circle of death”, speeds up his motor cycle, inside the circular cage. before going into a vertical loop. When he acquire a speed more than √5 gr at the lowest point of Ihe intended vertical circular path, he would not fall down, even when he is “upside down’ (i.e. at the highest point of his circular path). This is again because, in such a case, the centerifugal force on the motor cyclist, when he is “up-side-down, is more than his weight.

Example 1 : A small stone. of mass 200 g. is tied to one end of a string of length 80 cm. Holding the other end in hand. the stone is whirled Into a vertical circle. What is the minimum speed, that needs to be imparted, at the lowest point of the circular path, so that the stone is just able to complete the vertical circle? What would be the tension in the string at the lowest point of circular path? (Take g ~ 10ms-2)

Solution: We know that

ϑmin = minimum speed needed at the lowest point, so that particle is just able to complete the vertical circle = √5 gl

Hence ϑmin = √5 * 10 * 0.8 ms-1

~ 6.32 ms-1

Also T1 = Tension in the string at the lowest point of as circular path.

= mϑ2min / l + mg = 6 mg = 6 * O.2 * l0 N = 12N

Example 2 : A massless string, of length 1.2m, has a breaking strength of 2kgwt. A stone of mass OAkg. tied to one end of the string is made to move in a vertical Circle, by holding the other end In the hand. Can the particle describe the vertical circle? (Take g ~ 1Oms-2)

Solution : We are given that

Tmax = maximum tension in the string so that it does not break

= 2 Kgwt = 2 * 10N = 20N
Let T1 be the tension in the string when the stone is in its lowest position of its circular path. We know that T1 = mϑ21 / r + mg.

T1 would have its minimum value when ϑ1 equals its minimum value (= √5 gl), needed by the stone, to complete its vertical circular path.

Hence (T1)min = mϑ2min / l + mg = 6 mg

=6 * O.4 * 10

= 24 N

We thus see that (T1)min is more than the breaking strength of the string. Hence the particle cannot describe the vertical circle.

Example 3 : A small stone. of mass O.2Kg.tied to a massless. inextensible string, is rotated in a vertical circle of radius 2m. if the particle is just able to complete the vertical Circle, what is its speed at the highest point of its circular path? How would this speed get effected if the mass of the stone is increased by 50%? (Take g ~ 1Om s-2)

Solution:

Let ϑ1 be the speed of the stone at the lowest point of its vertical circle. Since the stone is just able to complete the vertical circle. we have

ϑ1 = √5 gr

= √5 * 10 * 2 ms-1 = 10 ms-1

Let ϑ2 be the speed of the stone, at the highest point, on its circular path. Then

ϑ22 = ϑ21 – 4gr

= (10)2 – 4 * 10 * 2

= 100 – 80 = 20

∴ ϑ2 = √20 ms-1 ~ 4.47 ms-1

It is thus seen that the value of ϑ2, does not depend on the mass (m) of the stone. Hence ϑ2 would remains the same when the mass of the stone increases by 50%.

Example 4 : A particle, of, mass 150g, is attaChed to one end of a massless, inextensible string. It is made to descri.be a vertical circle of radius 1m. When the string is making an angle of 48.2° with the vertical. hs instantaneous speed is 2 ms-1. What is the tension in the string In this position? Would this particle be able to complete its circular path?

(Take g ~ 10ms-2)

Solution:

The tension T, in the string, when it makes an angle θ, with the vertical, is given by

T = mϑ2 / l + mg cos θ

where ϑ is the instantaneous speed of the particle.

Here ϑ = 2 ms-1, l = 1 m, m = 0.15 kg, and θ = 48.2°

∴ T = 0.15 * (2)2 / l + (0.15 * 10 * cos 48.2°)

= (0.6 + 1.5 * 0.67) N

~ 1.6 N

LeI ϑ1 be the speed of the particle at the lowest point of its circular path. Then

ϑ21 = ϑ2 + 2gr (1 – cos θ)

= (2)2 + 2 * 10 * 1 * (1 – cos 48.2°)

= (4 + 20) * (1 – 0.67)

~ (4 + 6.6) = 10.6

∴ ϑ1 = √10.6 ms-1 ~ 3.25 ms-1

The minimum value of ϑ1, so that the particle is able to complete its vertical circle, is √5 gr

∴ (ϑ1)min = √5 * 10 * 1 ms-1 ~ 7.07 ms-1

The value of ϑ1 obtained above, is tess than this minimum speed. The particle on the given case, would not be able to complete its vertical circular path.

Example 5 : A bucket, containing 4 kg of water, is tied to a rope of length 2.5m and rotated in a vertical circle in such a way that tile water in it just does not spill over when the bucket is in its ‘upside down’ position. What is the speed of bucket at the

(a) highest and (b) lowest point of its circular path? (Take g ~ 10 ms-2)

Solution :

Let ϑ1 be the speed of bucket at the lowest point of “s circular path. Then

ϑ1 = √5 gr

= √5 * 10 * 2.5 ms-1

= √125 ms-1 ~ 11.18 ms-1

Let ϑ2 be the speed of the bucket at the highest point of its circular path. Then

ϑ2 = √gr

= √10 * 2.5 ms-1

= 5 ms-1

Example 6 : The figure here shows a smooth ‘Looping-the-loop’ track. A particle, of mass m, is released from point A, as shown. If H = 3r, would the partide ‘loop the loop’?

What is the force on the circular track When the particle is at point (i) B (ii) C ? Solution:

Let ϑB be the speed acquired by the partide at the (lowest) point B. From law of conservation of energy, we have

Total energy at A = Total energy at B

∴ ( 0 + mgH) = 1 / 2 ϑB2 + 0

∴ ϑB = √2gh = √2g * 3r = √6gr

The minimum, speed, needed by the particle at B, so that it can ‘loop the loop’ is √5gr.Since ϑB is more than √5gr ; the particle would ‘loop the loop’.

The forces, acting on particle at B, are as shown here, Let N1 be the force exerted on the particle by the track. According to Newton’s third law, The force exerted by the particle on the track, is equal and opposite to N1, Now

N1 = mg + m ϑB2 / r

= mg + m * 6gr / r = 7 mg Hence the force, exerted, by the particle, on the track, equals 7 mg, directed vertically downwards.

The forces, acting on the particle in position C, are as shown in the figure here. The speed, ϑc, of the people, at C, is given by ϑc2 = ϑB2 – 4gr = 6gr – 4gr = 2gr

∴ N2 = mϑc2 / r – mg

= 2mg – mg = mg

Hence the particle exerts a force mg. directed radially outwards. on the track, at points EXERCISES

1. A stone of mass O.2 kg is tied to one end of a string of length 80 cm. Holding the other end. the stone is whirled into a vertical circle. What Is the minimum speed of the stone at the lowest point so that it just completes Ihe circle. What is the tension in string at the lowest point of the circular path? (g = 10 ms-2).

[Ans: 6.32ms-1 12N]

2. A particle, of mass 100 g, is moving in a vertical circle of radius 2 m. The particle is just ‘looping the loop’. What is the speed of particle and the tension in string at the highest point of the circular path? (g = 10 ms-2).

[Ans. 4.47ms-1, zero]

3. A particle, of mass 0.2 kg, attached to a massless string is moving in a vertical circle of radius 1.2 m. It is imparted a speed of 8 -1 at the lowest point of its circular path. Does the particle complete the vertical circle? What is the Change in tension in the string when the particle moves from the position, where the string is vertical, to the position where the string is horizontal?

[Ans, Yes, 6N]

4. A particle, of mass 200 g, is whirled into a vertical circle of radius 80 cm using a massless stling. The speed of partide. when the string makes an angle of BOo with the vertical line, is 1.5 -1. What is the tension in the string in this position?

[Ans. 1.56 N]

ELASTIC AND INELASTIC COLLISIONS IN TWO DIMENSIONS

UNIT – IV

Consider two particles A and B moving in a plane. If These two particles collide, and still continue moving In the same plane. the Collision is referred to as a two dimensional (or an oblique) collision. A collision, between two billiard balls is an example of such a two dimensional collision. To analyze the basic details of a two dimensional collision. consider a system of two particles A and B, moving, as shown, before and after the collision.

Since the forces, they exert on each other (during Collision)are internal forces, and there are no other forces, (i.e.there is no extemal force), the linear momentum. of the system is conserved. When the cOllision is elastic. the kinetic energy of the system is also conserved. (The collision is inelastic if kinetic energy of system is not conserved).

To illustrate. how calculations can be carried out, we now, consider a simple two dimensional elastic collision, taking place as shown below. Figure ((ii) (a)) shows a particle A, of mass m1 moving along x-axis, in the x-y plane, with an initial speed u. Particle B, of mass m2, is initially at rest. When particle A collides with B, the two particles move with speeds ϑ1 and ϑ2 in the x-y plane, after
the collision (Figure (ii) (b))

After the collision, let particle A move in a direction inclined at an angle θ, with its initial direction of motion. Angle θ is known as the angle of scattering,

After the collision, lei particle B move along a direction, making an angle φ, with the initial direction of motion of A. Angle Φ is known as the angle of recall,

Knowing m1 m2, and u , we have to do the needed calculations. The law of conservation of linear momentum, used for the x and y components separately, gives us the two equations:

m1 u + O = m1ϑ1 cos θ + m2 ϑ2 cos φ (i)

and 0 = m1ϑ1 sin θ – m2 ϑ2 sin φ (ii)

We have assumed the collision to be perfectly elastic. Hence total K. E. before coUision = total K.E. after collision

∴ 1 / 2 m1u2 + 0 = 1 / 2 m1ϑ21 + 1 / 2 m1ϑ22 (iii)

or m1u2 = m1ϑ21 + m1ϑ22

We thus have three equations in all but need to find four unknown parameters. A complete solution, is therefore, NOT POSSIBLE, However ff the value 01 anyone of the four unknowns, i.e. ϑ1, ϑ2, θ or φ is given, the remaining three can be calculated, using Eqns (i), (ii) and (iii),

[For an inelastic two dimensional collision, we only have two equations i.e Eqn (i) and (ii), Hence, In this case, if two of the four unknowns, say θ and φ are given, we can calculate the remaining two unknowns (i e ϑ1 and ϑ1) using Eqns (i) and (ii)].

Special case: We now consider the special case of two dimensional collision of two particles of equal mass, Eqns (i), (ii) and (iii), in this case, reduce to

u = ϑ1 cos θ + ϑ2 cos φ (iv)

o = ϑ1 sin θ + ϑ2 sin φ (v)

u2 = ϑ21 + ϑ22 (vi)

From Eqns (iv) and (vi)

1 cos θ + ϑ2 cos φ)2 = ϑ21 + ϑ22

∴ ϑ21 cos2 θ + ϑ22 cos2 φ + 2 ϑ1 ϑ2 cos θ cos φ = ϑ21 + ϑ22

or 2 ϑ1 ϑ2 cos θ cos φ = ϑ21 ( 1 – cos2 θ) + ϑ22 ( 1 – cos2 φ) (vii)

= ϑ21 sin2 θ + ϑ22 sin2 φ

Using Eqn (v), we can rewrite Eqn (vii) as

2 ϑ1 ϑ2 cos θ cos φ = 2 ϑ21 sin2 θ

or cos θ = (ϑ12) sin2 θ / cos φ (viii)

Now cos (θ + φ) = cos θ cos φ – sin θ sin φ

= (ϑ12) sin2 θ / cos φ cos φ – ϑ1 / ϑ2 sin2 θ ( from Eqn (V) and (vii)

= ϑ1 / ϑ2 sin2 θ – ϑ1 / ϑ2 sin2 θ

= 0

∴ θ + φ = π / 2

We thus see that in the special case, of a perfectly elastic (two-dimensional) conision, between two particles, of the same mass, the two particles move along mutually perpendicular directions after the collision. This is illustrated in the figure below. Example 1: A and e are two particles having the same, mass m. A is moving along x-axis with a speed of 10 m-1, and B is at rest. After undergoing a perfectly elastic collision, with S, particle A gets scattered through an angle of 30°. What is the direction of motion of e. and the speeds of A and e, after this collision?

Solution: Figure (iv), (a) and (b). show !he particles A and B, before and after Ihe collision. Since A and B have the same mass, and the collision is perfectly elastic, we would have

θ + φ = 90° (i)

∴ φ = 90° – 30° = 60°

Using law of conservation of linear momentum, we get

(i) for the x-components,

u = 10 = ϑ1 cos 30° + ϑ2 cos 60°

or 10 = √3 / 2 ϑ1 + ϑ2 / 2

∴ 20 = √3 ϑ1 + ϑ2 (ii)

and (ii) for the y – components,

o = ϑ1 sin 30° – ϑ2 sin 60°

∴ ϑ1 1 / 2 = √3 / 2 ϑ2 or ϑ1 = √3 ϑ2 (iii)

From Eqn (ii) and (iii), We get

20 = 3 ϑ2 + ϑ2 or ϑ2 = 5 ms-1

and ϑ1 = √3 ϑ2

= 1.732 * 5 ms-1 = 8.66 ms-1

Example 2: Two particles. A and B. of masses m and 2 m, are moving along the X and y- axis, respectively, with the same speed e,

They collide, at the origin, and coalesce into one body, after the collision. What is the velocity of this coalesced mass? What is the loss of energy during this collision?

Solution: Figures (v) (a), and (v) (b). show the two particles before and after the collision. Let V be the speed of the combined mass and let the direction of V be making an angle α with the positive x-axis. after the collision. Using law of conservation of linear momentum. we have

For the x – components,

m ϑ = 3m V cos α (i)

and for the y – components

2mϑ = 3m V sin α (ii)

From Eqns (i) and (ii), we get

tan α = 2mϑ / mϑ = 2

∴ α = tan – 1 (2) ~ 63.4° (iii)

Also V2 = (ϑ / 3)2 + (2ϑ / 3)2 = 5 / 9 ϑ2

∴ V = (√5 / 3)ϑ (iv)

Now ki = Total K.E. before collision = 1 / 2 mϑ2 + 1 / 2 (2 m) ϑ2 = 3mϑ2 / 2

Now kf = Total K.E. after collision = 1 / 2 (3m)V2 = (3 / 2) (5 / 9) mϑ2 = 5 / 6 mϑ2

Hence ΔK = Loss of kinetic energy during the collision = ki – kf

= (3 / 2 – 5 / 6) mϑ2 = 2 / 3 mϑ2

EXERCISES

1. A billiard ball A moving with an initial speed of 1 ms-1, undergoes a perfectly elastic collision with another identical ball B at rest. A is scattered through an angle of 30°. What is the angle of recoil of B ? What is the speed of ball A after the collision?

[Ans.60°; √3 / 2 ms-1]

2. Two identical balls, A and B, undergo a perfectly elastic two dimensional collision. Initially A is moving with a speed of 10 ms-1 and B is at rest. Due to collision, A is scattered through an angle of 30°. What are the speeds of A and B after the collision?

[Ans.ϑA =5 √3 ms-1, / ϑB = 5 ms-1]

3. A and S are two identical balls. A, moving with a speed of 6 ms-1, along the positive x – axis. undergoes a collision with B, initially at rest. After collision, each ball moves along directions making angles of ± 30° with the x – axis. What are the speeds of A and B after the collision? Is this collision perfectly elastic?

[Ans.ϑA = ϑB = 2 √3 ms-1, No]

NON – CONSERVATIVE FORCES

UNIT – IV A force is non-conservative; if work done by, or against the force, in moving from one point A, in space, to another point B DEPENDS ON THE PATH FOLLOWED IN MOVING FROM A to B. In Figure (i) (a), let W1 W2 and W3 denote the works done in moving a body. from A to B, along three different paths 1, 2 and 3 respectively.

For a non-conservative force W1 ≠ W2 ≠ W3

Figure (i) (b) shows a particle moving along a closed path A -> 1 -> B -> 2 -> A. Let W, be the work done along the path A -> 1 -> B and let W1 be the work done along the path B -> 2 -> A, For a non-conservative force |W1| ≠ |W2|. Therefore, In such a case, the net work done, along the closed path is not-zero, Expressed mathematically. i.e. for a non-conservative force, the work done along a closed path is not zero.

Two of the common examples, of non-conservative forces, are

(i) Force of friction
(ii) Viscous force

Non- Conservative forces are usually velocity dependent. Consider a particle moving from A to B. on a horizontal, rough surface The force of friction coming into play, is a non-ccnservative force. Let W be the work done against the force of friction in moving from A to B. When the body moves from B to A. W would again be the work done against the force of friction. The net work dona, against friction. in the round trip is therefore, 2W. If this work is being done at the cost of kinetic energy. the loss of Kinetic energy. in the round IJip. is numerically equat to 2W.

Consider next. a body having both a kinetic energy (K), and a potential energy (U). moving in a non-conservative force field. The total energy E (= K + U). of the body, does not remain constant. Let E, and E, represent the total values of the initial and finat energy. If W is the work done, against the non-conservative force. we would have

Ei – Ef = W

The work done. by non-conservative forces. appears in some other form of energy like heat. sound. light etc. When we take into account all forms of energy, the general law of conservation of energy stili holds good both for conservative as well as non-conservative forces.

Example 1 : A particle, of mass O.1 kg, has an initial speed of 4ms-1 at a point A on a rough horizontal road. The coefficient of friction. between the object and the road is 0.15. The particle moves to a point B, at a distance of 2m from A. What is the speed of the particle at B?

(Take g ~ 1O ms-2)

## CBSE Class XII Supplementary Textual Material in Physics

Contents:

• Unit 6: Combination of Lens and a Mirror
• Unit 9: Characteristics of a Light Emitting Diode (LED)

CBSE Class 12 Supplementary Textual Material in Physics is given below.

UNIT – VI

COMBINATION OF A LENS AND A MIRROR

Consider a coaxial arrangement of a lens and a mirror. Let an object be placed in front of the lens. The incident rays, from the object, first undergo refraction at lens and are then incident on the mirror. To obtain the position of the image due to the combination, we can proceed as follows:

Using refraction formula, we can calculate where the image would have been formed, had there been only the lens. We next take this image formed. as a real, (or virtual), object for the mirror. Using the mirror formula, we can then locate the position of its final image formed by the mirror. This final position, would be the position of the image due to the combined effect of refraction by the lens and reflection by the mirror. This procedure is illustrated through the following examples.

Example 1 : A convex lens, of focal length 20 cm, has a point object placed on its principle axis at distance of 40 cm from it. A plane mirror is placed 30 cm behind the convex lens. Locate the position of image formed by this combination. We first consider the effect of the lens. For the lens, we have

u = – 40cm and f = + 20cm

Using the lens formula, we get

1 / ϑ – 1 / (-40) = 1 / 20 &therfore; ϑ = +40 cm

Had there been the lens only the image would have been formed at Q1. The plane mirror M is at a distance of 30cm from the lens L We can, therefore, think of a Q1 as a virtual object, located at a distance of 10 cm, behind the plane mirror M. The plane mirror therefore forms a real image (of this virtual object Q1) at Q, 10cm in front of it. This is show in the Figure (i)

Example 2 : A convex lens is placed in contact with a plane mirror. An axial point object, at a distance of 20cm from this combination, has its image coinciding with itself. What is the focal length of the convex lens? Solution :

Figure (ii) (a) shows a convex lens L in contact with a plane mirror M. P is the point object, kept in front of this combination at a distance of 20cm. from it. Since the image of the object is coinciding with the object itself, the rays from the object, after refraction from the lens, should fall normally on the mirror M. so that they retrace their path and form an image coinciding with the object itself. This will be so, if the incident rays from P form a parallel beam perpendicular to M, after refraction from the lens. For clarity, M has been shown at a finite distance from L, in figure (ii) (b). For lens L, since the rays from P. form a parallel beam after refraction, P must be at the focus of the lens. Hence focal length of the lens is 20cm.

Example 3 : A convex lens. and a convex mirror, (of radius of curvature 20cm) are placed coaxially with the convex mirror placed at a distance of 30cm from the lens. For a point object, at a distance of 25cm from the lens. the final image; due to this combination, coincides with the object itself. What is the focal length of the convex lens? Solution:

The final image, formed by the combination, is coinciding with the object itself. This implies that the rays, from the object, are retracing their path, after refraction from the lens and reflection from the mirror. The (refracted) rays are. therefore, falling normally on the mirror.

It follows that the rays A B, and A’ B’, when produced, are meeting at the centre of curvature. C of the mirror. Hence O2C = 20cm, the radius of curvature of the mirror.

From the figure (iii), we then see that for the convex lens u = -25cm and ϑ = + (30 + 20)cm = + 50cm. If f is the focal length of the lens, we have

1 / 50 – 1 / (-25) = 1 / f

&therefore; 1 / f = 1 + 2 /50

&therefore; f = 50 / 3 cm = 16.67 cm

Example 4 : A convex lens, of focal length 20 cm, is placed co-axially with a convex mirror of radius of curvature 20 cm. The two are kept 15cm apart from each other. A point object is placed 60cm in front of the convex lens. Find the position of the image formed by this combination.

Solution: The ray diagram, for the image formed by the combination, is shown below in Figure.(iv) = – 60cm and f = +20cm

Hence, using the lens formula, we get

1 / ϑ1 = 1 / (-60) = 1 / 20

&therefore; 1 / ϑ1 = 1 / 20 = – 1 / 60

&therefore; ϑ1 = 30 cm

Had there been only the lens L, the image of P would have been formed at Q1, which acts as a virtual object for the convex mirror.

&therefore; Q’ Q1 = distance of this virtual object (Q1) from convex mirror = OQ1 — O’Q1 = (30-15)cm = 15 cm

Hence for the convex mirror,

u2 = +15cm and R =+20cm.

Using the mirror formula. we get

1 / ϑ2 +1 / u2 = 2 / R,

i.e, 1 / ϑ2 + 1 / 15 = 2 / 20

&therefore; 1 / ϑ2 = 1 / 10 – 1 / 15

&therefore; ϑ2 = +30 cm

Hence the final image (a) formed is a virtual image formed at a distance of 30cm behind the convex mirror.

Example 5 : A convex lens. of focal length 20cm, and a concave mirror, of focal length 10 cm, are placed co-axially 50cm apart from each other. An incident beam parallel to its principal axis, is incident on the convex lens. Locate the position of the (final) image formed due to this combination.

Solution : The incident beam, on lens L, is parallel to its principal axis. Hence the lens forms an image Q1 at its focal point, i. e, at a distance 0 Q1 (= 20cm) from the lens. This image, Q1, now acts as a real object for the concave mirror. For the mirror, we then have: u = -30cm, and f = -10cm,

Hence using mirror formula, we get

1 / ϑ + 1 (-30) = 1 / (-10)

&therefore; 1 / ϑ = 1 / 30 – 1 / 10

ϑ = – 15 cm

The lens — mirror combination, therefore, forms a real image Q at a distance of 15cm from M. The ray diagram is as shown in figure (v) below. EXERCISES

1. A point object is placed 60 cm in front of a convex lens of focal length 30cm.

A plane mirror is placed 10cm behind the convex lens. Where is the image formed by this system?

[Ans. At the optical center of the convex lens.]

2. A convex lens. of focal length 15 cm, and a concave mirror, of radius of curvature 20cm. are placed co-axially 10 cm apart. An object is placed in front of the convex lens so that there is no parallax between the object and its image formed by the combination. Find the position of the object.

(Ans. At a distance of 30 cm from the lens]

3. Figure. (vi) show a plane mirror M placed at a distance of 10cm from a concave lens L. A point object is placed at a distance of 60 cm from the lens. The image formed, due to refraction by the lens and reflection by the mirror, is 30cm behind the mirror. What is the focal length of this lens?

(Ans. – 30cm] UNIT – IX

THE I-V CHARACTERISTICS OF A LIGHT EMITTING DIODE (LED)

The light emitting diode, represented by either of the two symbols shown here, is basically the same as a conventional p-n junction diode. Its actual shape is also shown here. The shorter, of its two leads, corresponds to its n (or cathode side) while the longer lead corresponds to its p (or anode side). The general shape, of the I-V characteristics of a LED, is similar to that of a conventional, p-n junction diode. However, the *barrier potential’ changes slightly with the colour.

The colour of the light emitted, by a given LED, depends on its band-gap energy. The energy of the photons emitted, is equal to (or slightly) less than this band gap energy. The other main characteristic, of the emitted light. its intensity. is determined by the forward current conducted by the junction. LED’s usually use a low voltage D C supply for their operation. A given LED has a ‘safe value’ of the forward current that it can carry. This value is around 5 mA for the usual simple LED’s and can go up to 30 mA, or more, for LED’s needed for providing a high brightness output light. In practice, it is usual to have a (suitable) series resistor, connected to the LED, so that the forward current is limited to within its ‘safe value’