To Speed up your Internet Connection

Learn from yesterday, live for today, hope for tomorrow. The important thing is to not stop questioning.

– Albert Einstein

Many among you might have often experienced the difficulty in getting your favourite sites opened up without undue delay. Your AP Physics Exams are fast approaching and you don’t have time to waste. Do you have any difficulty in getting this site opened up as you would legitimately expect it to with the type of the internet connection you have? Often the domain name servers (DNS) may be creating the problem. The openDNS may help you in this context. If you suspect your DNS addresses you may try the very useful service offered by openDNS. You will find a useful post on this at ideagold

Prepare well for AP Physics Exam - Master the Fundamentals

Life is like riding a bicycle. To keep your balance you must keep moving.

– Albert Einstein

If a building is to be strong, its foundation must be strong. The foundation on which you develop your concepts has to be fool proof. I write this just to make you aware of the importance of basic things in physics. One who is well versed in fundamental principles of physics can easily solve problems in physics which, at the first glance, may appear to be tough.

One of my students once brought a problem to me. The problem was related to black holes even though there was no mention of black holes in the syllabus prescribed for the course. He even expressed the doubt that the problem got included in the question paper by mistake. This was the problem:

A black hole is an object with an enormous concentration of mass in a small radius. Its gravitational field is so strong that even light cannot escape from it. To what approximate radius should the earth be compressed so that it becomes a black hole? (Mass of earth = 5.98×10²⁴ kg. Gravitational constant = 6.67×10^–11 m³ kg^–1s^–2)

(a) 100 km

(b) 1 km

(c) 1 cm

(d) 0.01 mm

(e) 0.0001 mm

I asked the student whether he had studied the concept of escape velocity. He answered with enthusiasm that the escape velocity is the minimum velocity with which a body is to be projected so that it will escape from the gravitational pull of the earth (or or any planet or star). He even remembered the value of escape velocity:

V_esc = √(2GM/R)

where G is the constant of gravitation M is the mass and R is the radius (of the planet or star).

[The above expression is obtained by putting the total energy (sum of the kinetic energy and the gravitational potential energy) of the projected body equal to zero : ½ mV² – GMm/R = 0].

“So I have to equate the speed of light to the esacpe velocity. Am I right?,” he asked me.

“Absolutely. If the object is to become a black hole, the escape velocity has to become the velocity of light c,” I answered.

Therefore we have

c =√(2GM/R) from which

R = 2GM/c² = (2×6.67×10^–11×5.98×10²⁴)/(3×10⁸)²

This will will work out to about 0.9 cm. The correct option therefore is (c).

[You cannot strictly treat a photon as an ordinary material particle since its rest mass is zero. The entire energy of the photon is kinetic and you have to obtain the escape velocity from the equation, mc² = GMm/R so that c =√(GM/R). The order of magnitude of the radius is still given by option (c)].

My student then showed me a graph (shown in the adjoining figure) which was related to another problem:

A time varying force F acts on a body of mass 2 kg initially at rest. The nature of variation of the force is shown by the semicircular curve (Fig.). If no other forces act on the body, what will be its velocity at the instant t = 2 s?

(a) 6.28 ms^–1

(b) 12.56 ms^–1

(c) 25.12 ms^–1

(d) 7.07 ms^–1

(e) 3.14 ms^–1

“Sir, I think I can work out this problem using the concept of impulse,” my student said.

“You are on the right path,” I answered.

“Sir, I know that the area under the force-time graph gives the impulse received by the body and the impulse received is equal to the change in the momentum. But my problem is with the semi circle. How will I find the area of this semi circle?”

“Probably the question setter had the idea of distracting students like you,” I told him with a smile. “The shape of the curve is semicircular, but it is really a ‘semi ellipse’ since the scales on the two axes are different. Even the quantities are different. One is time while the other is force. So the area is not the area in the real sense”.

“So the impulse received during two seconds is equal to a quarter of the area of the ellipse with semi major axis (a) equal to 4 newton and semi minor axis (b) equal to 2 seconds. Am I right?,” my student asked.

“Absolutely,” I answered. “The impulse received is πab/4 = (π×4×2)/4. This works out to 2π”.

“I got the answer,” my student said. “Since the body started from rest the change in momentum of the body is equal to its momentum itself and is equal to the impulse 2π. Since the mass of the body is 2 kg, its velocity at the instant t = 2 s is 2π/2 = 3.14 ms^–1, very nearly”.

“You are right,” I said. “I’ll ask you another question involving the calculation of area to clear your doubt thoroughly. Here is the question”:

The PV diagram of one cyclic of operation on an ideal gas is shown in figure. What is the net amount of work done by the gas during the cycle?

(a) 2πP₀ V₀

(b) – 2πP₀ V₀

(c) 6πP₀ V₀

(d) – 6πP₀ V₀

(e) – 12πP₀ V₀

The net amount of work (W) involved in the cyclic process is the area enclosed by the curve and is given by

W = πab = π×2V₀×3P₀ = 6πP₀ V₀

The process shown by the conventional PV diagram is anticlockwise which indicates that more work is done on the system while compressing the gas. The net amount of work done by the gas is therefore negative so that the correct option is – 6πP₀ V₀.

[The equations to be remembered in thermodynamics can be found here]

AP Physics B– Atomic Physics and Quantum Effects– Multiple Choice Practice Questions

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The essential points you need to remember in atomic physics and quantum effects (especially to make you strong in working out multiple choice questions in the stipulated time) were given in the post dated 23^rd February 2008. Some typical multiple choice questions (with solution) were discussed in the posts dated 25^th February 2008 and 1^st March 2008. Two free response questions (with model answers) in this section were discussed in the post dated 8^th March 2008. You can access all these posts by clicking on the label ‘atomic physics’ below this post.

Today we will discuss some more multiple choice questions in this section:

(1) The photoelectric work function of a metal surface is Φ. The thermionic work function of the same metal surface will be

(a) less than Φ

(b) equal to Φ

(c) greater than Φ

(d) dependent on the method of heating used

(e) independent of the photoelectric work function

The difference between photoelectric emission and thermionic emission is in the method of supplying energy to the electrons in the metal to make them come out of the metal. The minimum energy required for the electrons to escape from the surface of a given metal is a constant. This means that the photoelectric work function of a given metal surface is equal to its thermionic work function [Option (b)].

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(2) The adjoining figure shows four energy levels of an atom. The wave lengths emitted due to transitions 1, 2, 3 and 4 (fig.) are λ₁, λ₂, λ₃ and λ₄ and the corresponding frequencies are ν₁, ν₂, ν₃ and ν₄ respectively. Then

(a) λ₁ = λ₂ + λ₃ + λ₄

(b) λ₄ = λ₁ + λ₂ + λ₃

(c) ν₁ = ν₂ + ν₃ + ν₄

(d) ν₄ = ν₁ + ν₂ + ν₃

(e) λ₄ = λ₁ λ₂ λ₃ /(λ₁ + λ₂ + λ₃)

From Bohr’s frequency condition we have

hν₄ = hν₁ + hν₂ + hν₃ where h is Planck’s constant.

Therefore ν₄ = ν₁ + ν₂ + ν₃

(3) In the above question the emitted wave lengths are related as

(a) λ₁ = λ₄ – (λ₂ + λ₃)

(b) λ₄ = λ₁ – (λ₂ + λ₃)

(c) 1/λ₁ = 1/(λ₂ + λ₃ + λ₄)

(d) 1/λ₄ = 1/(λ₁ + λ₂ + λ₃)

(e) λ₄ = λ₁λ₂λ₃ /(λ₁λ₂ + λ₂λ₃ + λ₃λ₁)

As shown above, ν₄ = ν₁ + ν₂ + ν₃

Or, c/λ₄ = c/λ₁ + c/λ₂ + c/λ₃ where c is the speed of light.

Therefore, 1/λ₄ = 1/λ₁ + 1/λ₂ + 1/λ₃ = (λ₂λ₃ + λ₃λ₁ + λ₁λ₂) /λ₁λ₂λ₃ from which

λ₄ = λ₁λ₂λ₃ /(λ₁λ₂ + λ₂λ₃ + λ₃λ₁)

(4) What is the change in the angular momentum of the electron in the hydrogen atom when it undergoes a transition to emit the H_α line in the Balmer series? (Planck’s constant = h).

(a) h/2π

(b) h/π

(c) 2h/π

(d) 3h/2π

(e) zero

When the electron is in the orbit of quantum number n its angular momentum is nh/2π. The H_α line results due to the transition from the 3^rd orbit to the 2^nd orbit. The angular momentum therefore changes from 3h/2π to 2h/2π so that the change in the angular momentum is h/2π.

(5) In an X-ray tube the accelerating voltage is V volt. If the electronic charge is e coulomb, Planck’s constant is h joule second and the speed of light in free space is c ms^–1, the range of wave lengths possible in the continuous X-ray spectrum is

(a) zero to infinity

(b) zero to hc/eV

(c) zero to eV/hc

(d) hc/eV to infinity

(e) eV/hc to infinity

The minimum wave length λ_min of the X-rays produced is given by

hc/λ_min = eV

We have equated the entire energy eV of the electron to the energy hν (= hc/λ) of the photon. This gives λ_min = hc/eV.

All wave lengths above this value are possible so that the correct option is (d).

You will find some useful questions in this section here

Answer to AP Physics C Free Response Practice Question Involving Angular Momentum

In the post dated 4^th March 2009, I had given a free response practice question involving the concepts of moment of inertia, angular momentum and elastic collision for the benefit of AP Physics C aspirants. As promised, I give below the answer along with the question:

An object of mass m collides elastically with the lower end B of a thin uniform rod AB of mass 3m and length L suspended vertically using a frictionless hinge at its upper end A (Fig.) so that it can rotate in a vertical plane. The only external force present is that of earth’s gravity. Just before collision the object was moving horizontally with speed v. Now answer the following questions:

(a) Given that the moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its middle and perpendicular to its length is ML²/12, determine the moment of inertia of the rod AB about the hinge.

(b) Calculate the horizontal speed of the object of mass m after the collision.

(c) Suppose the colliding object were of the same mass (3m) as that of the rod. Calculate the distance d from the hinge at which the colliding object (moving with the horizontal speed v) should hit the rod so that its kinetic energy is fully transferred to the rod.

(d) If the collision at the position obtained in part (c) is such that the colliding object gets attached to the rod, calculate (in terms of the given parameters) the angular velocity with which the rod starts moving after the collision

(a) By parallel axis theorem the moment of inertia I of the rod about the hinge is given by

I = ML²/12 + M(L/2)² since the distance of the parallel axis is L/2.

Here M = 3m so that I = (3mL²/12) + 3mL²/4 = 3mL²/3 = mL²

(b) Since the angular momentum as well as kinetic energy are conserved in elastic collisions we have, on equating the angular momenta,

mvL =mv₁L + I ω

where ω is the angular velocity of the rod and v₁ is the horizontal speed of the object of mass m immediately after the collision.

[There is external force (gravity) and hence we will be justified only if we equate angular momenta just before and just after the collision].

Since I = mL² the above equation becomes

mvL =mv₁L + mL²ω

Therefore, v = v₁ + Lω from which ω = (v – v₁)/L

Equating the kinetic energies immediately before and after the collision, we have

½ mv² = ½ mv₁² + ½ Iω²

Substituting for I and ω we have

½ mv² = ½ mv₁² + ½ mL² [(v – v₁)/L]²

Therefore, v² = v₁² +(v – v₁)²

This gives v₁² = vv₁.

Since v₁ cannot be equal to v the above equation is satisfied only if v₁ = 0

[So the colliding object will have its horizontal momentum killed by the collision and it will fall vertically downwards].

(c) The situation is shown in the adjoining figure. Since the colliding object transfers the entire kinetic energy to the rod, its velocity immediately after the collision is zero. The collision is again elastic. Let ω₁ be the angular velocity with which the rod starts rotating in this case.

Equating angular momenta just before and just after the collision, we have

(3m)vd = Iω₁ = mL²ω₁ so that ω₁ = 3vd/L²

Equating kinetic energies just before and just after the collision, we have

½ ×(3m)v² = ½ Iω₁² = ½ × mL² (3vd/L²)²

Or, 3d²/L² = 1 from which d = L/√3

(d) Since the colliding object gets attached to the rod, this is a case of inelastic collision in which case also angular momentum is conserved.

[Remember that kinetic energy is not conserved in inelastic collisions].

Equating angular momenta just before and just after the collision, we have

(3m)vL/√3 =[3m(L/√3)² + mL²]ω₂ where ω₂ is the angular velocity with which the rod starts rotating in this case.

[The quantity in the square bracket on the right hand side is the total moment of inertia of the system after the object gets attached to the rod].

Therefore, √3 v = 2Lω₂ from which ω₂ = √3 v/2L.

AP Physics C Free Response Practice Question Involving Angular Momentum

Do not worry about your problems with mathematics; I assure you mine are far greater.

– Albert Einstein

Today I give below a free response practice question involving the concepts of moment of inertia, angular momentum, elastic collision and inelastic collision for the benefit of AP Physics C aspirants:

An object of mass m collides elastically with the lower end B of a thin uniform rod AB of mass 3m and length L suspended vertically using a frictionless hinge at its upper end A (Fig.) so that it can rotate in a vertical plane. The only external force present is that of earth’s gravity. Just before collision the object was moving horizontally with speed v. Now answer the following questions:

(a) Given that the moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its middle and perpendicular to its length is ML²/12, determine the moment of inertia of the rod AB about the hinge.

(b) Calculate the horizontal speed of the object of mass m after the collision.

(c) Suppose the colliding object were of the same mass (3m) as that of the rod. Calculate the distance d from the hinge at which the colliding object (moving with the horizontal speed v) should hit the rod so that its kinetic energy is fully transferred to the rod.

(d) If the collision at the position obtained in part (c) is such that the colliding object gets attached to the rod, calculate (in terms of the given parameters) the angular velocity with which the rod starts moving after the collision

The above question carries 15 points and you have 15 minutes at your disposal. Try to answer this question. I’ll be back soon with a model answer for your benefit.

Meanwhile find some useful multiple choice questions (with solution) in this section here.

You can find other posts involving rotation on this site (including equations to be remembered) by clicking on the label ‘rotation’ below this post.