AP Physics Resources

AP Physics C - Answer to Free Response Practice Question on Electrostaics

2012-01-14T07:33:00.000-08:00

"I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend".

–Sir Isaac Newton

In the post dated 11^th January 2012, a free-response question for practice was given to you. As promised, I give below a model answer. The question also is given for convenience:

A thin circular nonconducting disc of radius ‘a’ resting horizontally on the ground has positive charges sprayed on its top surface so that it has a uniform surface charge density σ. The centre of the disc is O and the axis of the disc is vertical (Fig.). There are no other electric charges in the vicinity for producing any appreciable interaction on the charges on the disc and the acceleration ue to gravity at the place is g. Now answer the following questions:

(a) Derive an expression for the electric potential at a point on the axis of the disc at distance x from its centre.

(b) What is the electric potential at the centre of the disc?

(c) A positively charged particle of mass m and charge q with specific charge (q/m) of 8ε₀g/σ is released from rest from a point P on the axis of the disc. The distance of point P from the centre of the disc is d. If the particle just reaches the centre O of the disc, determine the value of d.

(d) Explain qualitatively the nature of motion of the charged particle after its release from the point P stated in part (c) above.

[This question was asked in a different form in IIT entrance examination earlier].

(a) The disc can be imagine to be made of a large number of concentric rings of radii ranging from zero to a. To derive an expression for the potential at a point P (Fig.) on the axis of the disc at distance x from the centre, consider one such ring of radius r and width dr as shown in the figure. The total charge on the ring is 2πrdrσ since the area of the top surface of the ring is 2πrdr.

All charges on the ring are at the same distance √(r² + x²) from the point P so that the electric potential at P due to the charges on the ring is (1/4πε₀){2πrdrσ/√(r² + x²)} where ε₀ is the permittivity of free space.

The electric potential at P due to the charges on the entire disc [_(P)] is obtained by adding the contributions of all the rings forming the entire disc. Therefore we have

V_(P) = ₀∫^a [(1/4πε₀){2πrdrσ/√(r² + x²)}]dr

Or, V_(P) = (σ/2ε₀)₀∫^a[r/√(r² + x²)] dr

Since the value of the integral is √(r² + x²) and the upper and lower limits (of r) are a and zero, we get

V_(P) = (σ/2ε₀)[√(a² + x²) – x]

(b) The electric potential at the centre of the disc [V_(O)] is obtained by putting x = 0 in the above general expression for the potential on the axis.

Therefore, V_(O) = σa/2ε₀

(c) When the particle falls from height d and just comes to rest (momentarily) at O, as is evident from the question, it gains electric potential energy at the cost of gravitational potential energy. Therefore, by the law of conservation of energy we have

mgd =q[V_(O) – V_(P)] where V_(P) here is the electric potential energy when the point P is at height d and is equal to (σ/2ε₀){√(a² + d²) – d}. Therefore we have

mgd =q[(σa/2ε₀) – (σ/2ε₀){√(a² + d²) – d}]

Or, mgd =(qσ/2ε₀) [a –{√(a² + d²) – d}]

Therefore, d =(q/m)(σ/2ε₀g) [a –{√(a² + d²) – d}]

Since the specific charge q/m =8ε₀g/σ, as given in the question, we get

d =(8ε₀g/σ)(σ/2ε₀g) [a –{√(a² + d²) – d}] = 4[a –{√(a² + d²) – d}]

Or, 4√(a² + d²) = 3d + 4a

Squaring, 16a² + 16d² = 9d² + 16a² + 24ad

Or, 7d² = 24ad from which d = 24a/7

(d) Initially the particle moves down with an acceleration because of the gravitational pull mg even though the electrostatic repulsive force due to the like charges on the disc opposes the motion. As the particle approaches the disc, the electric force increases where as the gravitational force remains constant. But the particle overshoots its final equilibrium position because of inertia. The downward motion of the particle gets decelerated because of the increasing electric repulsive force and it comes to rest momentarily at the central point O. The particle then moves upwards because of the electric repulsive force which is greater in magnitude here, compared to the gravitational pull. The particle oscillates about its final equilibrium position and finally comes to rest at the final equilibrium position which is at a height from the point O.

AP Physics C - Free Response Practice Question on Electrostaics

2012-01-11T08:11:00.000-08:00

“There is no substitute for hard work”.

– Thomas A. Edison

I give below a free response practice question on electrostatics for the benefit of AP Physics C aspirants. The question is meant for checking your understanding and the capacity of application of the concepts of electric potential, electric potential energy and the law of conservation of energy. Here is the question:

(a) Derive an expression for the electric potential at a point on the axis of the disc at distance x from its centre.

(b) What is the electric potential at the centre of the disc?

(d) Explain qualitatively the nature of motion of the charged particle after its release from the point P stated in part (c) above.

Try to answer this question. This carries 15 points and you can take 15 minutes to answer it. I’ll be back soon with a model answer for you so that you can compare your answer with it and get an idea of your strength and weakness.

AP Physics B & C - Multiple Choice Practice Questions on Magnetic Fields due to Current Carrying Conductors

2011-12-30T03:52:00.000-08:00

A few multiple choice questions (for practice) related to magnetic fields produced by current carrying wires are given below. You may solve them yourself and check your answers by referring to the solution given below the set of questions.

Question No.1:

A plane square loop of wire (Fig.) carrying a current is oriented with its plane horizontal. On viewing from above, the current in the loop flows in clockwise direction. If the magnitude of the magnetic flux density at the centre of the loop due to each side is B, the resultant magnetic flux density at the centre of the loop is

(a) directed horizontally leftwards and has magnitude 2B

(b) directed horizontally rightwards and has magnitude 2B

(d) directed vertically downwards and has magnitude 4B

(e) zero

Question No.2:

A straight infinitely long wire carrying a current I is given a 90º bend at the position O (Fig.) near its middle. What is the magnetic flux density at the point P (shown in the figure) at distance d from the bend?

(a) μ₀I/2πd, directed normally into the plane of the figure, away from the reader

(b) μ₀I/4πd, directed normally into the plane of the figure, away from the reader

(d) μ₀I/4πd, directed normal to the plane of the figure, towards the reader

(e) Zero

Question No.3:

Two coplanar concentric circular coils P and Q of 20 turns each carry currents of 1A and 2 A respectively in opposite directions. If their radii are 10 cm and 20 cm respectively, what is the magnitude of the resultant magnetic flux density at their common centre?

(a) 10μ₀

(b) 20μ₀

(d) 100μ₀

(e) Zero

Question No.4:

If the coils in question no.3 carry the same current of 2A (in opposite directions), what will be the magnitude of the resultant magnetic flux density at their common centre?

(a) 10μ₀

(b) 20μ₀

(d) 200μ₀

(e) Zero

The above questions are meant for AP Physics B as well as AP Physics C.

The following questions (5 and 6) are meant specifically for AP Physics C.

Question No.5:

An infinitely long coaxial cable has an inner central cylindrical conductor of radius a and an outer conducting cylindrical pipe of inner radius b and outer radius c (A portion of the coaxial cable is shown in the adjoining figure). It carries equal and opposite currents of magnitude I on the inner an outer conductors. What is the magnitude of the magnetic flux density at a point P outsie the coaxial cable at distance r from the axis?

(a) Zero

(b) (μ₀I/2πr)[(c²–r²) /(c²–b²)]

(d) μ₀I/2πr

(e) (μ₀I/2πr)[(c²–b²) /(c²–a²)]

Question No.6:

In the case of the coaxial cable of question no.5 above, what is the magnitude of the magnetic flux density at a point P in between the central conductor and the outer pipe, if the distance of the point P from the axis is r?

(a) Zero

(b) μ₀I/2πr

(d) (μ₀I/2πr)[(c²–r²) /(c²–b²)]

(e) (μ₀I/2πr)[(c²–b²) /(c²–a²)]

Answers to the above questions are given below:

Answer to Question No.1:

The magnetic fields due to all the four sides of the loop act vertically downwards at the centre of the loop and they add up to produce a resultant field of magnitude 4B [Option (d)].

Answer to Question No.2:

The magnetic field at P due to the horizontal portion of the conductor is zero since the point P is lying on the straight line indicating the direction of flow of the current.

[The magnitude B of the magnetic field at a point P due to a finite length of straight conductor is generally given by

B = (μ₀I/4πr) (sinΦ₁– sinΦ₂) where Φ₁and Φ₂ are shown in the adjoining figure

The straight lines joining the point P to the ends of the conductor make the same angles (Φ₁= Φ₂ = π/2) so that sinΦ₁– sinΦ₂= 0. Thus B = 0].

The vertical portion of the conductor in the problem produces a magnetic field of magnitude μ₀I/4πr directed normally into the plane of the figure [Option (b)].

[B = (μ₀I/4πr) (sinΦ₁– sinΦ₂) = (μ₀I/4πr) (sin π/2– sin 0) = μ₀I/4πr]

Answer to Question No.3:

The magnetic flux ensity at the centre of a circular current carrying coil is directed along the axis and has magnitude μ₀nI/2a where μ₀is the permeabitity of free space, n is the number of turns in the coil and a is the radius of the coil.

Since the currents in the coils are in opposite directions, the magnetic fields are in opposite directions and the magnitude B of the resultant magnetic field at the common centre of the coils is given by

B = μ₀n₁I₁/2a₁ – μ₀n₂I₂/2a₂ = (μ₀×20×1)/(2×0.1) – (μ₀×20×2)/(2×0.2) = 0

Answer to Question No.4:

The resultant magnetic field at the common centre of the coils is given by

B = μ₀n₁I/2a₁ – μ₀n₂I/2a₂ = (μ₀×20×2)/(2×0.1) – (μ₀×20×2)/(2×0.2) = 100μ₀

Answer to Question No.5:

This question can be worked out easily using Ampere’s circuital law which states that the line integral of magnetic flux density B over any closed curve is equal to µ₀times the total current I passing through the surface enclosed by the closed curve. This is stated mathematically as

∫B. dℓ = µ₀I (The integration is over the closed path)

[Ampere’s circuital law as modified by Maxwell to accommodate the displacement current flowing through dielectrics and free space is

∫B. dℓ = µ₀[I + ε₀ (dφ_E/dt)], where ε₀ (dφ_E/dt) is the displacement current resulting from the rate of change of electric flux φ_E. ε₀ is the permittivity of free space].

We imagine a circle of radius r, with its centre at the axis of the coaxial cable, as the closed curve for the integration. Since this circular path encloses two equal currents in opposite directions, the total current I passing through the surface enclosed by the closed curve is zero. Therefore the magnitude B of the magnetic flux density at a point P outsie the coaxial cable must be zero.

Answer to Question No.6:

In orer to find the magnetic flux density at a point P in between the central conductor and the outer pipe, we imagine a circle of radius r, with its centre at the axis of the coaxial cable. Since this circular path encloses the entire current I passing through the central conductor, we have (from Ampere’s circuital law)

∫B. dℓ = µ₀I where B is the magnetic flux density at point P at distance r . The direction of the magnetic field coincides with the circular path of integration since the magnetic field lines due to a straight conductor are in the form of concentric circles. The line integral on the left hand side of the above equation therefore simplifies to B×2πr an we have

B×2πr = µ₀I

Therefore B = µ₀I /2πr

The correct option is (b).

AP Physics B & C - Multiple Choice Practice Questions on Electromagnetic Induction

2011-11-21T22:09:00.000-08:00

“Maturity is often more absurd than youth and very frequently is more unjust to youth.”

– Thomas A. Edison

Michael Faraday’s discovery of electromagnetic induction was a turning point in the history of mankind. When he made the first public announcement that the relative motion between a magnet and a coil of wire could cause the flow of a feeble electric current through the coil, he had to face this question: “But what is the use?” Faraday countered this with another question: “What is the use of a new born baby?”

The baby has grown rapidly to become a very healthy youth who will remain so for many more decades!

The phenomenon responsible for the generation of electric power for feeding the modern world still continues to be electromagnetic induction.

Questions on electromagnetic induction are generally interesting. Click on the label ‘electromagnetic induction’ below this post; you will find all posts on electromagnetic induction published so far on this site.

Today we will discuss a few more multiple choice practice questions in this section.

(1) Earth’s resultant magnetic field at California has magnitude B tesla and it makes an angle θ with the horizontal. Assuming that there are no other magnetic fields, what will be the voltage induced between the tips of the wings of an airplane of wing-span L flying horizontally with speed v?

(a) BLv

(b) BLv sin θ

(d) BLv/sin θ

(e) BLv/cos θ

Since the airplane is flying horizontally it can ‘cut’ the vertical magnetic field lines to generate a motional emf V given by

V = B_verticalLv where B_vertical is the vertical component of earth’s magnetic field at the place.

With reference to the adjoining figure we have

B_vertical = B sin θ

Therefore, the voltage induced between the tips of the wings of the airplane is BLv sin θ.

(2) A plane square loop of thin copper wire has 100 turns. Each side of the loop is 10 cm long and the loop is oriented with its plane making an angle of 30º with a uniform magnetic field of flux density 0.4 tesla. If the loop is rotated in 0.5 second so as to orient its plane at right angles to the magnetic field, what will be the magnitude of the average emf induced in the loop?

(a) 0.1 volt

(b) 0.2 volt

(d) 0.8 volt

(e) 2 volt

The induced emf V is given by

V = – dФ/dt where dФ is the change in the total magnetic flux linked with the coil and dt is the time taken for the flux change.

[The negative sign is the consequence of Lenz’s law by which the induced emf has to oppose the change of flux dФ].

Since we are required just to find the magnitude of the induced voltage, we may ignore the negative sign

Since the coil has N (=100) turns, the total flux linked with the coil is Nφ where φ is the flux per turn given by

φ = BAcos θ where B = 0.4 tesla and A = area of the square loop = (0.1)² m² = 0.01 m²

The angle θ is the angle between the magnetic field and the area vector.

[Remember that the area vector is directed perpendicular to the plane of the coil].

Since the plane of the coil makes an angle of 30º with a magnetic field, the area vector makes an angle of 60º with the magnetic field.

The initial magnetic flux linkage is NBAcos 60º = 100×0.4×0.01×(1/2) = 0.2 weber.

Since the area vector and the magnetic field are finally parallel (or anti-parallel), the final flux linkage is NBAcos 0º = 100×0.4×0.01 = 0.4

The change of flux dФ = 0.4 – 0.2 = 0.2

Therefore, induced emf = (Change of flux) /(Time) = 0.2/0.5 = 0.4 volt.

(3) Suppose that the resistance (R) of the loop in the above question is 10 Ω. What will be the induced current in the loop if the loop is kept stationary and the magnetic field is steadily reduced to zero in a time of 40 millisecond?

(a) 0.2 A

(b) 0.5 A

(d) 1.5 A

(e) 2 A

The initial magnetic flux linked with the loop (as shown above) is NBAcos 60º = 100×0.4×0.01×(1/2) = 0.2 weber.

When the magnetic field is reduced to zero, the magnetic flux is reduced to zero. Therefore the change of magnetic flux is 0.2 weber. The emf V induced in the loop is given by

V = (Change of flux) /(Time) = 0.2/(40×10^–3) volt = 5 volt.

The current induced in the loop is V/R = 5/10 A = 0.5 A [Option (b)]

The following question is meant specifically for AP Physics C aspirants:

(4) A straight conductor of length L and mass M can slide down along a pair of long, smooth, conducting vertical rails P and Q of negligible resistance (Fig.). A resistor of resistance R is connected between the ends of the rails as shown in the figure. A uniform magnetic field of flux density B acts perpendicularly into to the plane containing the rails and the sliding conductor. The terminal velocity of fall of the rod is

(a) MgR/LB

(b) mgL/B²R²

(d) mgB/L²R

(e) mgR/B²L²

When the rod slides down under gravity, the magnetic flux linked with the closed circuit comprising the rod, rails and the resistor R changes and a current is induced in the circuit. The induced emf is the motional emf BLv where v is the velocity of the rod. The induced current I in the circuit is BLv/R.

By Lenz’s law the induced current has to oppose the motion of the rod. It is the magnetic force ILB which brings in this opposition. When the velocity of the rod increases, the opposing magnetic force also increases. When the magnitudes of the gravitational force (weight Mg of the rod) and the opposing magnetic force become equal, the rod moves with a constant (terminal) velocity.

Therefore, we have

ILB = Mg

Substituting for I we have (BLv/R)LB = Mg

Or, B²L²v/R = Mg

This gives v = MgR/B²L²

Now, let me ask you a question:

If the direction of the magnetic field in the above question is reversed, will the rod still attain a terminal velocity? Think of it and arrive at the answer ‘YES’.

You will find a few more questions (with solution) in this section here.

AP Physics B & C - Multiple Choice Practice Questions on Work, Energy and Power

2011-10-23T09:29:00.000-07:00

“The best thinking has been done in solitude. The worst has been done in turmoil.”

– Thomas A. Edison

Today we will discuss a few multiple choice practice questions involving work, energy and power. Questions in this section were discussed earlier on this site. You can access them by clicking on the label ‘work energy power’ or by trying a search for the required word using the search box provided on this page.

(1) A sphere A of mass m moves with speed v and kinetic energy E along the positive x-direction and collides inelastically with another identical sphere B at rest. After the collision sphere A moves with kinetic energy E/3 along the positive y-direction. What is the speed of sphere B after the collision?

(a) v/2

(b) v(√3)/2

(d) 2v/√3

(e) 4v/√3

We have E = ½ mv²

Since the kinetic energy is directly proportional to the square of speed, the speed of A after the collision is v/√3.

Momentum is conserved in all collisions (elastic as well as inelastic). The momentum of sphere A before the collision is mv and is directed along the positive x-direction. This is the total momentum of the system since the initial momentum of sphere B is zero.

The momentum mv is represented by the vector OP₁ in the adjoining figure. The momentum of sphere A after the collision is mv/√3 and is directed along the positive y-direction. This is represented by the vector OP₂. The momentum of sphere B after the collision must be given by the vector OP₃ since the parallelogram OP₂P₁P₃ is drawn such that its diagonal OP₁ represents the final momentum, mv of the system

Evidently │OP₃‌│²= │OP₁│²+ │OP₂│²

Or, │OP₃‌│² = (mv)² + (mv/√3)²

But │OP₃‌│ = mv_f where v_f is the speed of sphere B after the collision.

Thus we have

(mv_f)² = (mv)² + (mv/√3)²

This gives v_f = √(v² + v²/3) = 2v/√3

(2) A particle moves in a plane under the action of a force of constant magnitude. If the direction of the force is always at right angles to the direction of motion of the particle, it follows that

(a) the momentum of the particle is constant

(b) the velocity of the particle is constant

(d) the acceleration of the particle is constant

(e) the particle moves along a straight line with increasing speed

A force of constant magnitude acting always at right angles to to the direction of motion of the particle will supply centripetal force required for uniform circular motion. In uniform circular motion the speed of the particle is constant.

[The velocity is not constant since the direction changes continuously. The momentum and the acceleration also are not constant for the same reason].

Since the speed is constant, the kinetic energy of the particle is constant [Option (c)].

(3) A ball of mass 0.5 kg is thrown vertically downwards from the edge of a building of height 65 m. The initial speed of the ball is 20 ms^–1. If the ball strikes the ground with a speed of 40 m/s, the energy lost by the ball because of air resistance is (gravitational acceleration, g = 10 ms^–2)

(a) 10 J

(b) 25 J

(d) 40 J

(e) 45 J

The initial energy of the ball is E_i = mgH + ½ mv_i ²

The first term is the gravitational potential energy of the ball of mass m at height H and the second term is the initial kinetic energy of the ball having initial speed v_i. Substituting proper values, we have,

E_i = 0.5×10×65 + ½ ×0.5×20² = 325 + 100 = 425 J.

The final energy of the ball (at the ground level) having velocity v_f is E_f = ½ mv_f²

Substituting proper values, we have,

E_f = ½×0.5×40² = 400 J.

Therefore, the energy lost by the ball because of air resistance is E_i – E_f = 425 J – 400 J = 25 J.

(4) Two blocks A and B of masses m and 3m respectively (Fig.) travel in opposite directions with the same speed v along a smooth surface. They collide head-on and stick together. Mechanical energy lost during the collision is

(a) ½ mv²

(b) mv²

(d) 2 mv²

(e) 3 mv²

Total momentum of the system before collision is 3mv – mv = 2 mv

[We have taken the leftward momentum as positive and that’s why the total momentum is positive. If you take the leftward momentum as negative, the total momentum will be negative. But this won’t affect your final answer and the energy lost during the collision will be positive].

Since the momentum of the system is conserved, the total momentum after the collision will be 2 mv itself. If the common velocity of A and B (which stick together) after collision is v_f, we have,

(m + 3m)v_f = 2 mv

Therefore v_f = v/2

The initial kinetic energy of the system is ½ mv² + ½ (3m)v² = 2 mv²

Final kinetic energy of the system is ½ (4m)v_f² = ½ (4m)(v/2)² = ½ mv²

Therefore, the loss of kinetic energy during the collision = 2 mv²– ½ mv² = (³/₂) mv²

(5) An electric motor creates a tension of 1000 N in a hoist cable and reels it at the rate of 2.5 ms^–1. The power output of the motor is

(a) 250W

(b) 400 W

(d) 2500 kW

(e) 400 kW

Since the point of application of the force of 1000 N moves through 2.5 m per second, the work done per second (which is the power output) is 1000×2.5 watt = 2500 watt = 2.5 kW

AP Physics B - Multiple Choice Practice Questions on Geometric Optics

2011-10-09T06:35:00.000-07:00

"I know not with what weapons World War III will be fought, but World War IV will be fought with sticks and stones."

–Albert Einstein

Questions on optics were discussed on earlier occasions on this site. You can access those questions by clicking on the label ‘optics’ below this post. Generally simple questions will be asked from this section. But to answer them correctly you need to have a thorough knowledge of basic things in this section.

Today we will discuss a few more multiple choice practice questions on geometric optics (geometrical optics):

(1) A thin converging lens produces an inverted real image of an object on a screen. The size of the image is the same as that of the object. When the upper half of the lens is covered by an opaque sheet of paper,

(a) the image disappears

(b) the lower half of the image disappears

(d) full image with reduced intensity is obtained

(e) full image having half the size of the object is obtained

Even a small portion of the lens can produce the full image of the object. But the brightness (intensity) of the image will be reduced since half the aperture of the lens admits light. The correct option is (d).

(2) A lens behaves as a converging lens in air and in a liquid of refractive index n₁. But it behaves as a diverging lens in liquids of refractive indices n₂ and n₃. The refractive index of the material of the lens is

(a) greater than both n₂and n₃

(b) less than n₁ but greater than n₂

(d) greater than n₁but less than both n₂ and n₃

(e) less than n₁but greater than both n₂ and n₃

The refractive index of the material of the lens must be less than both n₂ and n₃ for the convergent lens to become divergent. Since the lens continues to be convergent in the medium of refractive index n₁, the refractive index of the material of the lens must be.greater than n₁. This follows from lens maker’s equation,

1/f = [(n’/n) – 1](1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, n’ is the refractive index of the material of the lens, n is the refractive index of the medium in which the lens is placed and R₁ and R₂ are the radii of curvature of its surfaces. The quantity (1/R₁ – 1/R₂) is positive in the case of a lens that is converging in air. The ratio n’/n has to be greater than 1 for the focal length to be positive (and hence the lens to be converging) when the lens is placed in any other medium.

The correct option is (d).

[A simple ray diagram will lead to the answer to the above question. The adjoining figure shows the manner in which a lens which is converging in air becomes diverging in a medium of refractive index greater than that of the lens. At the first surface of the lens, the ray is refracted away from the normal where as at the second surface the ray is refracted towards the normal. The net effect is to bend the ray further away from the principal axis of the lens].

(3) A thin monochromatic beam of light proceeding through glass is incident at an angle i at glass-air interface. The beam is partially reflected and partially refracted at the interface such that the angle between the reflected and refracted beams is 90º, as shown in the figure. If the angles of reflection and refraction are r and r’ respectively, what is the critical angle for the glass-air interface?.

(a) sin^–1(r/r’)

(b) sin^–1(r–r’)

(d) sin^–1(cos r)

(e) sin^–1(tan r’)

The critical angle C and refractive index n of glass with respect to air is given by

n = 1/sin C ………….(i)

here n = sin r’/sin i = sin r’/sin r since i = r

Since it is given that the angle between the reflected and refracted rays is 90º, we have (with reference to the figure) r’ = (90º – r)

Therefore, n = sin r’/sin r = sin(90º – r)/sin r =cos r/sin r = 1/tan r

Substituting in Eq (i), we have

1/tan r = 1/sinC

Or, sin C = tan r so that C = sin^–1(tan r)

(4) A transparent slab with parallel faces has its refractive index increasing linearly with distance from end A to end B (Fig.). Parallel rays of light are incident normally on this slab as shown. What will happen to these rays? (Ignore reflection at the face of the slab).

(a) They will become convergent after refraction

(b) They will become divergent after refraction

(d) They will bend towards B consequent on refraction

(e) They will proceed undeviated.

Since the angle of incidence is zero (normal incidence), the angle of refraction also is zero irrespective of the refractive index. Therefore the rays will proceed undeviated.

(5) The figure shows five lenses a, b, c, d, and e made of glass. When placed in air which lens/lenses will act as converging lens/lenses?

(a) None

(b) a, c and d

(d) a and d

(e) c and d

Lens (d) is a planoconvex lens and is converging. Lens (c) has one surface convex and the other concave. But the convex surface has smaller radius of curvature. Therefore, the divergence produced because of the concave surface is more than compensated by the convergence produced by the convex surface. It can therefore function as a converging lens.

Lens (a) also has convex and concave faces. But the radius of curvature of the concave face is smaller so that the convergence produced by the convex face is more than compensated by the divergence produced by the concave face. It will act as a diverging lens.

Lenses (b) and (e) are evidently diverging lenses.

So (c) and (d) are the converging lenses [Option (e)].

AP Physics B – Answer to Free Response Practice Question on Fluid Mechanics

2011-10-03T02:38:00.001-07:00

He who joyfully marches in rank and file has already earned my contempt. He has been given a large brain by mistake, since for him the spinal cord would suffice.

–Albert Einstein

In the post dated 30^th September 2011 a free response practice question on fluid mechanics was given to you. As promised, I give below a model answer along with the question:

The adjoining figure shows an empty thin walled cubical vessel of side 0.4 m and mass 6.4 kg floating on kerosene contained in a large tank (not shown). Density of kerosene is 800 kgm^–3where as the density of water is 1000 kgm^–3. You may take the acceleration due to gravity as 10 ms^–2. Now, answer the following questions:

(a) Calculate the magnitudes of the force of buoyancy and the force of gravitaty acting on the empty vessel and state their directions.

(b) Calculate the height x of the portion of the empty vessel that is submerged in kerosene.

(c) Water is slowly poured into the vessel so that an additional 0.2 m of the height of the vessel is submerged in kerosene. Calculate the volume of water added to obtain this condition.

(a) The force of buoyancy acts vertically upwards while the force of gravity (which is the weight of the vessel) acts vertically downwards. They have the same magnitude since the vessel is in equilibriumn. Since the mass (m) of the empty vessel is 6.4 kg, its weight (mg) is 6.4×10 = 64 newton.

Therefore the force of buoyancy and the force of gravitaty have the same magnitude of 64 N.

(b) The weight of a floating body is equal to the weight of the displaced liquid (in accordance with the law of floatation). Therefore we have

Axρg = mg where A is the area of the base of the vessel, x is the height of the portion of the empty vessel that is submerged in kerosene, ρ is the density of kerosene and g is the gravitational acceleration.

This gives x = m/Aρ = 6.4/(0.4²×800) = 0.05 m

(c) When an additional 0.2 m of the vessel is submerged in kerosene, the additional volume of kerosene displaced by the vessel is 0.2 A and the weight of this volume of kerosene is 0.2 Aρg. This must be equal to the weight of the water added so that we have

0.2 Aρg = Vdg where V is the volume of water added.and d is the density of water.

This gives V = 0.2 Aρ/d = (0.2×0.4²×800)/1000 = 0.0256 m³

[Suppose you were asked to determine the pressure exerted by water at the bottom of the vessel when the condition mentioned in part (c) in the above question is attained. You will then divide the weight of the water by the area of the bottom of the vessel, remembering that the thrust at the bottom is produced by the weight of the water column in the vessel and pressure is thrust per unit area.

Thus presuure at the bottom = 0.2 Aρg/A =0.2 ρg = 0.2×800×10 = 1600 pascal

(This is the same as Vdg/A)].

Now, go over here to find some more refreshing questions in this section.

AP Physics B - Free Response Practice Question on Fluid Mechanics

2011-09-30T01:43:00.000-07:00

“Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time.”

– Thomas A. Edison

Today I will give you a free response practice question on fluid mechanics. The question is meant for AP Physics B aspirants. But it will be useful for AP Physics C aspirants also as it high lights basic principles in hydrostatics. Generally questions meant for AP Physics C will be tougher than those for AP Physics B. So the question I give below can serve only as a part of a more difficult question in their case. Here is the question:

(a) Calculate the magnitudes of the force of buoyancy and the force of gravity acting on the empty vessel and state their directions.

(b) Calculate the height x of the portion of the empty vessel that is submerged in kerosene.

(c) Water is slowly poured into the vessel so that an additional 0.2 m of the height of the vessel is submerged in kerosene. Calculate the volume of water added to obtain this condition.

Try to answer the above question which carries 10 points. You have about 11 minutes for answering it. I’ll be back shortly with a model answer for your benefit.

You will find a useful post in this section here.

AP Physics B & C – Practice Questions (MCQ) Involving Kinematics and Elastic Collision

2011-09-19T02:47:00.000-07:00

“Men often become what they believe themselves to be. If I believe I cannot do something, it makes me incapable of doing it. But when I believe I can, then I acquire the ability to do it even if I didn’t have it in the beginning.”

– Mahatma Gandhi

Today we will discuss a few questions (MCQ) involving kinematics and elastic collision. The first four questions are relevant to AP Physics B as well as AP Physics C while the last question is relevant to AP Physics C.

(1) A particle moves from point A to point B (Fig.) in 2 seconds, covering three quarters of a circle of radius 1 m. What is the magnitude of the average velocity of the particle?

(a) 0.5 ms^–1

(b) 1 ms^–1

(d) 1/√2 ms^–1

(e) 2√2 ms^–1

The displacement of the particle during 2 seconds is equal to the length of the straight line AB. Since OA and OB have the same length of 1 m, AB = √2 m (length of the hypotenuse of the right angled triangle AOB.

Therefore average velocity = (√2)/2 = 1/√2 ms^–1

(2) A small object initially at rest starts sliding down from point P (Fig.) on a perfectly smooth inclined plane of inclination (θ) 30º and collides normally and elastically with the surface A of a large fixed block. If the distance PA (measured along the incline) is 2.5 m, what is the time taken by the object to traverse this distance? (g = 10 ms^–2)

(a) 0.25 s

(b) 0.5 s

(d) 1.25 s

(e) 1.5 s

The motion of the object down the plane is uniformly accelerated and you can use the equation,

s = ut + ½ at² with usual notations.

Here displacement s = 2.5 m, u = 0 and a = g sinθ = 10 sin30º = 5 ms^–2, which is the component of gravitational acceleration down the incline. Therefore we have

2.5 = 0 + ½ ×5 × t²

This gives t = 1 s.

(3) In the above question, after starting from the point P, the minimum time required for the object to return to P is

(a) 0.5 s

(b) 1 s

(d) 2 s

(e) 2.5 s

Because of the elastic collision with the block, the velocity of the small object gets reversed. It travels up the incline for 1 seccond covering the distance of 2.5 metre and momentarily comes to rest. The times required for the trips down the inclined plane and up the inclined plane are equal since the acceleration is g sinθ throughout the motion. Therefore, after starting from the point P, the minimum time required for the object to return to P is

1 s +1 s = 2 s.

(4) In question No.2 suppose the inclined plane is not perfectly smooth, but offers a small frictional resistance. The object slides downwards from point P and collides with the block elastically after time t₁. It then slides upwards and momentarily comes to rest after an additional time t₂. Which one among the following statements is correct?

(a) t₁ is less than 1 s

(b) t₁ = t₂ = 1 s

(d) t₁ is less than t₂

(e) t₁ is greater than t₂

During the downward trip the acceleration has magnitude less than g sinθ since the frictional force opposes the motion of the object. In solving question No.2 we have found that the time for the downward trip is 1 second when the downward acceleration has magnitude g sinθ, appropriate to the case of a perfectly smooth incline. Since the magnitude of the downward acceleration is reduced in the case of an inclined plane that offers frictional resistance, the time required for the downward trip is increased.

During the upward trip (after colliding with the block) the deceleration has magnitude greater than g sinθ since the frictional force as well as gravity oppose the motion of the object. The object therefore comes to rest in a shorter time.

Therefore t₁ is greater than t₂ [Option (e)].

[When you project a ball up, the time of ascent will be equal to time of descent only if the air resistance is negligible. If the air resistance is not negligible, you will find that the time of ascent is less than the time of descent].

The following question is specifically meant for AP Physics C aspirants:

(4) A small object initially at rest at point P (Fig.) on a perfectly smooth inclined plane of inclination (θ) 30º starts sliding down under gravity and collides normally and elastically with the surface A of a large block that is projected up the incline. Assume that the mass of the small object is negligible compared to the mass of the block. If the distance PA (measured along the incline) and the velocity of the block up the incline at the instant of collision are 2.5 m and 2 ms^–1 respectively, what will be the velocity of the small object immediately after the collision? (g = 10 ms^–2)

(a) 5 ms^–1

(b) 7 ms^–1

(d) 3 ms^–1

(e) 2 ms^–1

In the case of an elastic collision the relative velocity after the collision is equal and opposite to the relative velocity before the collision:

u₁– u₂= –(v₁– v₂)…………(i)

At the instant of collision the large block moves up the incline with velocity 2 ms^–1. (Let us take this direction as positive). Or, u₁= 2 ms^–1.

The velocity of the small object at the moment of collision is down the incline and hence negative. Its magnitude is 5 ms^–1 as is obtained from the equation v² = u² + 2as:

v² = 0² + 2 g sinθ × 2.5 = 2×10 sin30º × 2.5 = 25 from which v = 5 ms^–1

Therefore, u₂ = – 5 ms^–1

The relative velocity before collision is u₁– u₂ = 2 – (–5)

The relative velocity after collision is (v₁– v₂) = 2 – v₂where v₂ is the velocity of the small object just after the collision. (The velocity of the large block after collision is unchanged since its mass is large compared to the mass of the small object. Or, v₁ = u₁)

Therefore, from Eq (i) we have

2 – (–5) = –(2– v₂)

This gives v₂ = 9 ms^–1.

[You can obtain v₂ by solving the following equations highlighting the conservation of momentum and kinetic energy in the case of elastic collisions:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂………………………..(i)

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²…………..(ii)

Equations (i) and(ii) can be solved for the velocities v₁ and v₂ of the block and the small object respectively after the collision. You will get

v₁ = [(m₁– m₂)u₁ + 2m₂u₂] /(m₁+m₂) and

v₂ = [(m₂– m₁)u₂ + 2m₁u₁] /(m₁+m₂)

Here m₁ >> m₂, u₁ = 2 ms^–1 and u₂ = – 5 ms^–1so that

v₁ ≈ u₁ = 2 ms^–1 and

v₁ ≈ –u₂ + 2 u₁ = – (– 5) + (2×2) = 9 ms^–1

* * * * * * * * * * * * * * * * * * * * * * *

If you would like just arguments (without using lengthy mathematical steps, you may proceed like this (after obtaining the velocity of the object just before collision as –5 ms^–1):

Before collision the block has velocity 2 ms^–1 where as the small object has velocity –5 ms^–1 (relative to the ground). If the block is taken to be at rest for convenience, you have to imagine that the small object is moving towards the block with a velocity of –7 ms^–1. We are in fact using a frame of reference in which the block is at rest and are finding the velocities of the block and the small object in this frame by adding a velocity of –2 ms^–1to both:

2–2 = 0 and –5–2 = –7.

Just after the elastic collision, the velocity of the object becomes 7 ms^–1 relative to the block which we kept at rest for the convenience of argument. Our frame of reference is to be brought back to the ground. For this we add a velocity of +2 ms^–1to the block and the small object and obtain the velocity of the block as 2 ms^–1 (0+2 = 2) and the velocity of the small object as 9 ms^–1 (7+2=9)].

AP Physics C – Circular Motion and Rotation – Multiple Choice Questions Based on Some Interesting Aspects of Rolling

2011-08-29T04:51:00.000-07:00

“I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent.”

– Mahatma Gandhi

Rolling motion is very much related to our daily life since the wheels used in transportation move between places by rolling. All of you know that the invention of wheels for transportation was a turning point in the history of mankind. It is therefore important that in your AP Physics course you are required to understand the special features of rolling motion.

Even though you come across rolling bodies very commonly in your daily life, many among you might be unaware of some of the interesting aspects of rolling. For instance, some of you might not have noted that a smooth sphere cannot roll on a smooth surface, whether the surface is horizontal or inclined. For a body to roll along a surface, there has to be friction between the body and the surface. If friction is absent, the body will just slide along the surface. If the friction is insufficient, the body will roll as well as slide (slip) along the surface. In pure rolling there will be no slipping.

This post is meant for making you aware of similar aspects of rolling, by working out a few multiple choice practice questions. Here are the questions:

(1) The adjoining figure shows a disc of radius R rolling without slipping along a horizontal surface. Assume that the centre of mass C of the disc is displaced along the positive x-direction. If the angular velocity of the disc is ω, what is the velocity V_A of the point A of the disc at the instant it is in contact with the horizontal surface?

(a) ωR directed along the positive x-direction

(b) 2ωR directed along the positive x-direction

(d) 2ωR directed along the negative x-direction

(e) Zero

Since the disc is rolling without slipping, the contact point A of the disc (to be more precise, the line of contact through A) must be at rest.

The translational velocity V_cm of the centre of mass C of the disc is ωR which is directed parallel to the horizontal surface. Since the disc is a rigid body all points of the disc must move forward (along the positive x-direction) with this translational velocity. But all points of the disc have an additional linear velocity V_r because of the rotation of the disc about its central axis passing through C.

We have V_r= ωr where r is the distance of the point from the centre of mass C. In the case of the contact point A of the disc the distance from the centre of mass is R so that V_r= ωR and is directed along the negative x-direction.

Thus the contact point has the common translational velocity V_cm = ωR directed along the positive x-direction and the linear velocity (due to rotation about C) V_r = ωR directed along the negative x-direction, with the result that it is at rest. The correct option is (e).

(2) What is the velocity V_B of the topmost point B of the disc in question No.1?

(a) ωR directed along the positive x-direction

(b) 2ωR directed along the positive x-direction

(d) ωR directed along the negative x-direction

(e) 2ωR directed along the negative x-direction

The topmost point B of the disc has translational velocity V_cm = ωR directed along the positive x-direction as in the case of all other points of the disc. The additional linear velocity on account of the rotation is V_r = ωR. This too is directed along the positive x-direction in the case of the topmost point A. Therefore, the resultant velocity of the topmost point is 2ωR directed along the positive x-direction [Option (b)].

(3) What is the velocity V_D of the point D (at the left edge) of the disc in question No.1?

(a) ωR directed along the positive x-direction

(b) (√2)Rω directed along the positive x-direction

(d) ωR directed vertically upwards

(e) (√2)ωR directed at 45º to the horizontal

The velocity of the point D has two parts (as explained in the solution of question No.1):

(i) The translational velocity V_cm = ωR directed horizontally (along the positive x-direction).

(ii) The additional linear velocity on account of the rotation given by V_r = ωR, directed vertically upwards.

The resultant velocity of the point D is (√2)ωR, directed at an angle of 45º as shown in the adjoining figure [Option (e)].

[Note that the two velocities of equal magnitude ωR at right angles give a resultant velocity of magnitude (√2)ωR inclined equally (at 45º in this case) to both].

(4) Consider any arbitrary point P on a thin disc disc rolling on any surface. If the centre of mass is C and the point of contact with the surface is A at any instant, the instantaneous velocity of the point P is directed

(a) along the radius through P

(b) parallel to the surface

(d) along the line CP

(e) perpendicular to the line CP

You can forget about the velocities V_cm and V_r we considered in solving the previous questions and imagine that the disc is rotating about a parallel axis through the point of contact A (or, in other words, rotating about the line of contact) with angular velocity ω. You can easily obtain the velocities of the points we discussed above. Try yourself!

You will easily arrive at the answer to the present problem: [Perpendicular to the line AP given in option (c)].

[The magnitude of the velocity of point P will be ωr’ where r’ is the distance AP].

(5) A solid sphere and a hollow sphere made of two different metals, but of the same mass and radius are given to you. Which one of the following methods will be suitable for identifying them?

(a) Arrange two simple pendulums of equal length with the hollow sphere and the solid sphere as bobs and compare their periods of oscillation.

(b) Supply equal charges to them and compare the electric potentials on them.

(d) Allow them to roll down from the top of the same inclined plane and compare the times taken to reach the bottom of the incline.

(e) Compare their apparent loss of weight when fully submerged under water.

In all cases except (d) the quantities measured will not distinguish the hollow sphere from the solid sphere. The time for rolling down the incline will be greater for the hollow sphere since its moment of inertia is greater. It has more laziness (inertia) to roll!

[The acceleration a of a body rolling down an incline of angle θ (with respect to the horizontal) is given by

a = (g sin θ)/[1 + (k²/R²)] where g is the acceleration due to gravity, R is the radius of the rolling body and k is the radius of gyration defined by I = Mk² where I is the moment of inertia of the body about its central axis (axis of rolling) and M is the mass of the body.

For a solid sphere k²/R² = 2/5 since I = (2/5)MR².

For a thin hollow sphere k²/R² = 2/3 since I = (2/3)MR²

For a thick hollow sphere of outer radius R and inner radius R₁ the moment of inertia about the central axis is (2/5)M[(R⁵ – R₁⁵)/(R³ – R₁³)].

The quantity k²/R² is certainly more than 2/5 in this case also. Since k²/R² appears in the denominator of the expression for acceleration, the solid sphere has greater acceleration and it reaches the bottom of the incline earlier.

In fact in the case of regularly shaped bodies such as ring, disc, cylinder, hollow sphere, solid sphere etc., made of material of uniform density, the solid sphere has the maximum acceleration while rolling down an incline and a thin ring has the least acceleration (since it has k²/R² = 1).

If you have a solid sphere with non uniform density such that there is more concentration of material near the centre, it will accelerate faster than a solid sphere made of a material of uniform density].

AP Physics B - Wave Motion (including sound) - Multiple Choice Practice Questions on Sound

2011-08-06T03:14:00.000-07:00

Questions on sound were discussed earlier on this site. You can access all posts related sound by clicking on the label ‘sound’ below this post.

Today we will discuss a few more multiple choice practice questions on sound, relevant to AP Physics B aspirants.

(1) A source producing sound of wave length 0.6 m is moving away from a stationary listener with speed V/6 where V is the speed of sound in air. The wave length of sound heard by the listener is

(a) 0.5 m

(b) 0.54 m

(d) 0.7m

(e) 0.8 m

You might have come across questions of the above type under Doppler effect.

If the real frequency of the source of sound is n we have

V/n = λ where λ is the wavelength of the sound as measured by the stationary listener when the source is stationary. Therefore we have

V/n = 0.6 ……………..(i)

When the source moves away from the listener at speed V/6 the n sound waves produced per second will occupy a distance V + V/6 so that the wave length as measured by the listener becomes (V + V/6)/n. Therefore we have

(V + V/6)/n = λ₁ ……….(ii)

where λ₁ is the new wave length.

Dividing Eq (ii) by Eq (i) we obtain

1 + 1/6 = λ₁/0.6

Or, 7/6 = λ₁/0.6 from which λ₁ = 0.7 m.

(2) Two steel wires A and B of the same length are vibrating under the same tension. If the first overtone of wire A has the same frequency as the fundamental note of wire B, the area of cross section of wire A is

(a) twice that of B

(b) three times that of B

(d) half that of B

(e) one third that of B

The frequency of vibration (f) of a string of length ℓ and linear density m stretched under tension T is given by

f = (s/2ℓ) √(T/m) where s = 1,2,3,4 etc. For the fundamental mode s = 1. For the 2^nd mode (or 1^st overtone) s = 2. For the 3^rd mode (or 2^nd overtone) s = 3 and so on.

The linear density is mass per unit length and is equal to aρ where ‘a’ is the area of cross section and ρ is the density of the material of the string.

Since the first overtone of wire A has the same frequency as the fundamental note of wire B, we have

(2/2ℓ) √(T/m₁) = (1/2ℓ) √(T/m₂) where m₁ is the linear density of A and m₂ is the linear density of B.

From mthe above equation it follows that m₁= 4 m₂.

Since the wires are of the same material, the area of cross section of A must be 4 times that of B.

[Since the changes are confined to the mode of vibration and the linear density, you should be able to write 2/√m₁ = 1/√m₂ to arrive at m₁= 4 m₂ and the final answer a₁ = 4 a₂]

(3) Two sound notes of wave lengths λ₁ and λ₂ in air produce n beats per second. If λ₁ < λ₂ the speed of sound in air is

(a) λ₁λ₂n /(λ₂ + λ₁)

(b) (λ₁+ λ₂)n

(d) λ₁λ₂n / λ₁

(e) λ₁λ₂n /(λ₂ – λ₁)

The frequency f₁ of the sound of wave length λ₁ is given by

f₁= v/λ₁ where v is the speed of sound in air.

The frequency f₂of the sound of wave length λ₂ is given by

f₂= v/λ₂

Since λ₁ < λ₂we have f₁ > f₂

Since the number of beats produced per second is n, we have

f₁ – f₂ = n

Therefore, v/λ₁ – v/λ₂ = n

Or, v (1/λ₁ – 1/λ₂) = n

This gives v = λ₁λ₂n /(λ₂ – λ₁)

(4) A glass tube is open at both ends. The minimum frequency of a tuning fork which vibrates in resonance with the air column in this tube is f. If this tube is held vertically with half its length submerged in water, what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube?

(a) f/3

(b) f/2

(d) 2f

(e) 3f

The air column will vibrate with the minimum frequency in the fundamental mode. When both ends of the glass tube are open, there are anti-nodes at the ends and the adjacent node in the middle so that the length L of the tube is equal to λ/2 where λ is the wave length of the fundamental note. Thus we have

L = λ/2

Or λ = 2L

[Remember that the distance between consecutive anti-nodes (or, consecutive nodes) is equal to λ/2].

When half the length of the glass tube is immersed in water, it becomes a closed pipe of length L/2. In the fundamental mode of vibration of the air column in the tube in this case there is a node at the closed end (water surface) and the adjacent anti-node at the open end as shown in the figure. Therefore we have

L/2 = λ₁ /4 where λ₁ is the wave length of the resonating sound.

Or λ₁ = 2L

In both cases the wave length of the resonating sound is the same. Therefore the resonant frequency is the same as f [Option (c)].

(5) In the above question what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube if a quarter of the tube is submerged in water?

(a) 2f/3

(b) 3f/4

(d) 2f

(e) 3f

Initially, when both ends are open, the resonating wavelength as shown above is given by

λ = 2L

When a quarter of the glass tube is submerged in water, we have

3L/4 = λ₁/4

Therefore λ₁ = 3L

From the above equations λ/λ₁ = 2/3

Since the frequency is inversely proportional to the wave length we have

λ/λ₁ = f₁/f

Therefore f₁/f = 2/3 from which f₁= 2f/3