Problem Based Learning in Basic Physics - X

Problem Based Learning in Basic Physics - X

 

                                                       A. K .Mody

C-14 Beverly Hills, Lam Road, Devlali, Nashik 422401, Maharashtra

                                                                       H. C. Pradhan

HBCSE, TIFR, V. N. Purav Marg, Mankhurd

Mumbai – 400 088

[In this article- tenth in the series of articles we present problems for a problem based learning course based on system with continuously varying parameter.]

In this article 10^th in the PBL series we consider various examples from different areas of physics where we have systems in which one of the parameter varies continuously and to estimate any effect we need to take differential element and perform integration separately (diefferent from that which is already performed in arriving at standard text book formula).

1.      A block of mass m sliding on a frictionless horizontal surface experiences an air resistance proportional to its velocity. If it has initial speed v₀, find an expression for its velocity as a function of time.

Solution:

Acceleration

                        

Integrating,  

2.      A car starts from rest and start moving while its engine continues to deliver constant power P. Find the distance travelled by the car in time t.

Solution:

In this case, since constant power is delivered, acceleration is not constant.

Power delivered                           

;     integrating which, we get

;     integrating which, we get

3.      A cylindrical solid of mass 10 ^{– 2} kg and cross-sectional area 10 ^{– 4} m² is moving parallel to its axis (the x-axis) with a uniform speed of 10³ m/s in the positive direction. At t = 0, its front face passes the plane x = 0. The region to the right of this plane is filled with dust particle of uniform density 10 ^{– 3} kg/m³. When a dust particle collides with the face of the cylinder, it sticks to its surface. Assuming that the dimension of the cylinder practically remains unchanged and the dust sticks only to the front face of the x-coordinate of the front of the cylinder. Find the x-coordinate of the front of the cylinder at t = 150 s.                 [JEE 1993]

Solution:

In this case we have an object (cylinder) whose mass keeps on increasing (variable mass problem) as  m = m₀ +ρAx  and equation will be   .  If we assume mass of the cylinder as m₀ and velovity at t = 0 as v₀, then integration yields .

The same would also follow from conservation of total momentum.

Further integration yields,  .

Upon substitution of parameters gives x = 10⁵ m   at t = 150 s.

4.      A uniform chain of length L is kept on a smooth frictionless horizontal table with small length hanging from the edge. This makes chain slide down from the table. Find the speed when it just loses contact with the table.

Solution:

Let initially small length hanging from the table. Thus gravity pulling it down with the acceleration         

                       

                       

                       

Integrating we get  

When chain just leaves the table, its vertical length x →L

                           is the speed when it just loses contact with the table.

5.      Calculate the gravitational potential energy of a uniform spherical distribution of mass M and radius R.

Solution:

When the a mass dm comes from infinity to the surface of sphere of mass m and radius r, it potential energy is

When mass  accumulates and grows from zero to M and radius from zero to R, potential energy becomes

Since mass distribution is uniform, density

This gives   and integral becomes

Which upon integration gives

[Calculation of electrostatic potential energy of a charged sphere is  identical to the one discussed here for gravitational potential energy.]

6.      For a particle performing SHM with time period T, calculate time required to travel from a/2 to a, where a is the amplitude.

Solution:

For a particle performing SHM,      velocity         where x is displacement                         and ,

Thus,      which gives 

 

7.      A uniform rope of mass 0.4 kg and length 2.45 m hangs from a ceiling.

(i)     Find an expression for increase in length of the rope.

(ii)   Find the speed of the transverse wave in the rope at a point 0.5 m distance from the lower end.

(iii)  Calculate the time taken by a transverse wave to travel the full length of the rope. Derive the formula used. [Take g = 9.8 m/s², and Y = 2×10¹¹N/m² ].  [(ii) and (iii) Roorkee 1991]

Solution:

Consider an element of length dx at a distance x from the lower end of the rope.

The stress experienced by this element is given by ρxg where ρ is the density of the rope.

If length this element increases by dΔx, then strain = .  Here Y is the Young’s modulus of the material of the rope.

 increase in length of this element

 increase in length of the rope  

where  M is the mass of the rope and A is its area of cross-section.

(ii)  velocity of the wave at a distance x from the lower end                    here T is the tension at a distance x from the lower end and m is linear density of the rope.

(iii)  time taken by wave to travel the rope

8.      A string tied between x = 0 and x = l vibrates in fundamental mode. The amplitude A, tension T and mass per unit length µ is given. Find the total energy of the string. [JEE 2003]

Solution:

The amplitude at any point x is given by a = A sinkx where

Where ,  mass of string element of length dx is dm = µdx

Energy of this element while oscillating is

Thus total energy of the string

Which upon integration yields

9.      The capacitance of a parallel plate capacitor with plate are A and separation d is C. The space between two wedges of dielectric constants K₁ and K₂, respectively. Find the capacitance of the resulting capacitor. [JEE 1996]

Solution:

Let A = bL,   where b is the width of the capacitor into the plane of the paper.

Consider a strip of length dx at a distance x from one end as shown below.

The corresponding thicknesses are y and (d – y). Notice here that

Each of these strips have capacitance  and

The series combination gives capacitance of the entire strip dC given by  =

       

The capacitance C of the entire capacitance therefore can be obtained by integrating dC.

10.  A ray of light travelling in air is incident at grazing angle (incident angle 90^o) on a long rectangular slab of a transparent medium of thickness t = 1.0 m (see fig. below). The point of incidence is the origin A(0, 0). The medium has a variable index of refraction µ(y) given by , where k = 1.0 (metre) ^{– 3/2}.

The refractive index of air is 1.0

(A) Obtain the relation between the slope of the trajectory of the ray at a point B(x, y) in the medium and the incident angle at that point.

(B)  Obtain the equation of the trajectory y(x) of the ray in the medium.

(C)  Determine the coordinates (x₁, y₁) of  the point P, where the ray intersects the upper surface of the slab boundary.

(D) Indicate the path of the ray subsequently. [JEE 1995]

Solution:

Note here that refractive index µ depends on co-ordinate y only.

(A)  For a series of interfaces, Snell’s law is µ₁ sinθ₁ = µ₂ sinθ₂ =….= µ_n sinθ_n

Thus at the point of incidence  and point B, the relation is µ₁ sinθ₁ = µ_y sinθ_y and θ₁ ≈ 90^o.

(B) Which (referring to figure gives µ(y) sinθ_y = 1    and slope = dy/dx = tan (90 - θ_y) = cot θ_y 

        which upon integration yields .

(C)  Point P is at y₁  = 1.0 m which gives x₁ = 4.0 m

(D) At point A and P µ_A sinθ_A = µ_P sinθ_P where θ_A ≈ 90^o where as both µ_A = µ_P = 1 as subsequently at P ray enters air. This gives θ_P ≈ 90^o

References:

1. JEE - Joint Entrance Examination for admission to IIT

2.      Roorke : Roorke University Entrance exam (Now IIT-Roorke)

Problem Based Learning in Basic Physics - IX

Problem Based Learning in Basic Physics - IX

 

                                                       A. K .Mody

C-14 Beverly Hills, Lam Road, Devlali, Nashik 422401, Maharashtra

H. C. Pradhan

HBCSE, TIFR, V. N. Purav Marg, Mankhurd

Mumbai – 400 088

[In this article- ninth in the series of articles we present problems for a problem based learning course based on order of magnitude, approximations and errors.]

In this article 9^th in the PBL series we consider various examples from different areas of physics where very often we make approximations. We are trying to discuss meaning of these approximations.

Approximations with Functions:

 

 

For all of the above approximations to be valid, we must have  x << 1.

The limit of approximation is defined by the value of the first term which is ignored w.r.t the actual value.  For example in  the term x²/2 if smaller than the accuracy we are looking for then the term and higher order terms can be ignored.

Consider the following examples from Physics to understand the importance of the above mentioned approximations and order of magnitude calculations.

1.      We say sin θ ≈ θ  for small angle where θ is in radians. Discuss up to what value of θ is this approximation valid? Check for same value of θ, if the approximation is valid for                                cos θ ≈ 1 – θ²/2! .

Discussion:

Let us consider θ = 30^o = π/6 radian = 0.523598 for which sin θ = 0.5 .  Thus for at 30^odifference between sin θ and θ  is about 4%.  Thus depending on the accuracy we are looking for we should say sin θ ≈ θ  within that many percent. In this case 2^nd term in the expansion become θ³/6 = 0.024 which along with higher order terms can be ignored if 4% difference is acceptable. Otherwise this term is included and we can check for the next term. For example you can convince yourself that at 10^o, sin θ ≈ θ  the accuracy is within 0.5%.

Now at 30^o, cos θ  = 0.866025  and 1 – θ²/2 = 0.862922 . Here the error would be about 0.35%. Next term θ⁴/4 = 0.0031.  Check similar difference for 10^o.

2.      Assuming earth to be a perfect sphere of radius 6400 km and Mt. Everest to have height of    8 km, find the greatest distance on earth surface from where Mt. Everest will be visible.

Discussion:

Consider the diagram shown in the figure below.

Here h is the height of the mountain and R the radius of the earth. If a tangent is drawn from the peak of the mountain to the earth’s surface, then S represents the maximum distance on the earth’s surface from where mountain would be visible (assuming earth’s surface to be perfect sphere).

Referring to the figure :   . Considering h << R we can get cos θ ≈ 1 – h/R

Comparing this with approximation cos θ ≈ 1 – θ²/2, we get θ² = 2h/R = 16/6400 = 1/400

i.e. θ = 1/20 radian or 9/π degree. This gives S = R θ  = 6400/20 = 320 km.

Note that here we used approximation  where x = h/R.  Here x² = (1/800)² and higher order terms can justifiably neglected.

3.      Calculation of g:

When we calculate g near earth surface, We take  where M = 6×10²⁴kg and                      R = 6.4×10⁶ m. Why do we ignore effect of Sun and Moon near earth surface, inspite of the fact that Sun is massive?

Let us check the contribution of each near earth’s surface.

For Earth  SI units

For Sun near earth  SI units

For Moon near earth  SI units , using the data given in example 4 below. Thus, contribution of the Sun is 4 orders and that of moon is 6 orders smaller than that due to earth. Hence for practical purposes we ignore the effect due to the sun and that due to the moon. However, when it comes to force on large masses as earth and its ocean, even tiny effect can prove to make significant difference and due to size of the earth can cause tides in the ocean, (See Example 4 below).

4.      What causes tides:

We want to understand tides in the ocean based on Newton’s law of gravitation. Light take 500 sec to travel from the Sun to the Earth and diameter of earth is 6400 km. From this information, find the difference in gravitational pull (acceleration) experienced by a body when it is on the near side of the Sun to that when it is on far side of the Sun.                             [Given: G = 6.67×10 ^{– 11} Nm²/kg² and mass of the Sun = 2×10 ³⁰ kg]

Solution:

   and    

Which gives Δg = 1.011×10 ^{– 6} cm/s².

Considering mass of the oceans, which is a fraction of earth, this generates a huge difference in the force causing tides.

Now from mass of the moon = 7.3483×10²² kg and earth-moon distance = 384400 km calculate similar differential g, (say Δg) at earth due to moon. Considering relative position of earth Sun and moon, can you estimate magnitude that causes tidal effect on earth on a new moon day, full moon day and at 1^st quarter or 3^rd quarter? Here assume that sun-earth-moon are along the same line on a new moon and full moon day (not in the same order as you can understand). What can happen if they align on the same line?

Take radius of Sun to be 7×10⁸ m and that of a black hole to be , where c is the speed of light. Take average height of a person to be 1.5 m. Calculate differential g between head and toe of an average height person on the surface of the earth, sun and black hole of one solar mass.

This tidal gravity on earth acts on the ocean which is a huge mass (that of ocean), say 100^th of that of earth. Can you calculate this tidal (differential) force on this mass on earth?

Estimate this tidal (differential) force that a normal human (average mass 50 kg) would experience (difference of g between head and toe) on the surface of (i) Sun (ii) 1 solar mass black hole. [Use calculator and preserve at least 8 digits after decimal for all your calculations].

5.      Why water vapour at 100^oC burns the skin whereas at room temperature does not:

Discussion:

Average kinetic energy of a gas molecule is considered to be of the order of k_BT where                      k_B = 1.38×10 ^{– 23} J/K is the Boltzmann constant and T is the absolute temperature. Accordingly, water molecule (of water vapour) at room temperature (300 K) has kinetic energy k_BT ≈ 1/40 eV whereas at 400 K, k_BT ≈ 1/30 eV. Can you explain why at 300 K water vapor is harmless whereas at 400 K it feels very hot and can burn the skin? [Hint: the distribution of number of molecules ΔN  in the energy range between E and E + ΔE is given by Maxwell-Boltzmann distribution function . For simplicity you may assume energy at E  ≈ 1 eV (energy just sufficient to cause burn) in the interval     ΔE ≈ k_BT.] [You should get about 10⁴ times more molecules having energy 1 eV at 400 K compared to that at 300 K].

6.      Estimation of size of an atom:

This means mass of one aluminum atom is  where N_A = 6.02×10²³/mole is the Avogadro number. This means, volume occupied by one atom (assuming cubic arrangement) is  This gives a = 2.554×10 ^{– 10} m = 2.554 A^o.  Here a can be treated as diameter of the atom.

Actual crystal structure of Aluminum is FCC that gives 4 atoms per cubic cell. Which modifies the calculation to get a = 4.05 A^o. The separation of atoms in FCC is   which is diameter of the atom.

Thus Aluminum atom radius is estimated to be 1.43 A^o.  We can safely consider typical atomic radius to be of the order of 1 A^o.

7.      What can exist inside the nucleus?

Consider an Aluminum atom which has atomic weight A = 27. Radius of a nucleus is given by R = R₀A^1/3.  Where R₀ = 1.22 Fermi.  [1 Fermi = 10 ^-15 m]. This gives R_Al ≈ 4 fm.             Any particle, proton, neutron or electron if inside the nucleus will have uncertainty in position of the order of 4 fm.

From uncertainty principle, this gives .

For electron this gives speed greater than that of light. Which means electron will have energy E = pc = 5×10 ^{– 11} J ≈ 47 MeV , whereas for proton and neutron . Electrons observed in β-decay have much less energy and thus cannot exist inside the nucleus as independent particle.  Proton and neutron on the other hand can exist inside the nucleus.

8.      Discussion of SHM

In case of simple pendulum, the differential equation is solved for small amplitude taking approximation sin θ ≈ θ .  Thus, expression for period obtain would be approximate within those appropriate limit.

In case of spring we estimate  assuming spring to behave as elastic for the extension and compressions produced. For a perfectly elastic spring potential energy , where as in reality spring can neither be compressed by indefinite amount nor extended. Thus the V must have higher order terms and should have expression of the type  . Here values of a and b are very very small compared to k and thus corresponding terms become important only when x is large. Time period is also accurate within this approximation.

9.      Infinity in Physics: Examples from Electrostatics and Magnetostatics:

Discussion:

In electrostatic we deal with infinite line of charge, infinite charge plane, and in magnetism we deal with infinite straight wire carrying current. In real life we do not have such infinite objects. Then why do we derive such expressions?

In each of the above mentioned case, we find an expression for field at some distance from these infinite objects.

Consider infinite plane. For any point outside this plane, we can have two elements exactly at the same distance from the foot of the perpendicular, on the plane contributing to the field. The field contributions from these two elements have two components each. The components parallel to the plane cancel each other and the one perpendicular to the plane add up and survive. Thus field will always be perpendicular to the plane. What if we have finite size plate? Even in this case, if we are sufficiently close to the plane surface (distance very small compared to the dimensions of the plate), the result of infinite plane applies. In case of parallel plate capacitor, if the separation between plates is too small compared to the size of the plate, field lines are all perpendicular except near edges. These effects would be not negligible if plate separation is comparable to the size of the plates.

We encourage readers to construct similar line of arguments for linear charge distribution and straight wire carrying current.

10.  Infinity in Exponential equations:

Now, consider famous equations like

(i)       from radioactive decay

(ii)     for charging of a capacitor in an RC circuit

(iii)  for growth of current in an LR circuit

Here  have dimensions of time and are called time constants.  In each case we can write time constant as τ and exponential term can be represented as e  ^{- t/τ}.

We say that as t  → ∞,  e  ^{- t/τ}  → 0. What does infinite time mean here?

Consider values of e  ^{- t/τ}   at t = τ, 2τ, 3τ, 4τ, 5τ…

e ^{– 1} = 0.3679,    e ^{– 2} = 0.1353,    e ^{– 3} =            0.0498,  e ^{– 4} = 0.0183,   e ^{– 5} = 0.0067.                     Thus t  → 5τ   is good enough for infinity at 0.7% accuracy and system would stabilize for all practical purpose as the measuring instrument may not be sensitive enough to record 0.7% variation. Otherwise we need to consider higher values of t.

Note: Even when we say e^x → 1  as x → 0, (since ), the value of x determines the limit of this approximation.

References:

1.      Physics Textbook for class XI and XII,  NCERT New Delhi (2006)

2.      French A. P. , Vibrations and Waves, W.W. Norton & Co Inc, NY (1971)

3.      Halliday, Resnik and Krane, Physics, 5^th Ed. John Wiley & Sons (2002)