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]