AP Physics B & C - Newton’s Laws - Multiple Choice Practice Questions

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

 – Sir Isaac Newton

 
Let us discuss a few multiple choice practice questions involving Newton’s laws of motion.

(1) Coefficient of static friction between a horizontal conveyor belt and a 3 kg packet placed on it is 0.2. What is the maximum permissible acceleration of the conveyor belt so that the packet will not slip?

(a) 0.2 ms^–2

(b) 0.6 ms^–2

(c) 2 ms^–2

(d) 3 ms^–2

(e) 6 ms^–2

It is the frictional force between the packet and the conveyor belt that keeps the packet stationary with respect to the belt when the belt moves. The maximum value of this frictional force is μ_smg where μ_s is the coefficient of static friction, m is the mass of the packet and g is the acceleration due to gravity.

If the maximum permissible acceleration of the belt is a_max, we have

            ma_max = μ_smg

Therefore, a_max = μ_sg = 0.2×10 = 2 ms^–2

[Note that the mass of the packet is given just to serve the purpose of a distraction]

(2) An empty open box of mass M (Fig.) is sliding along a smooth horizontal surface with constant velocity ‘v₁’. An object of mass M/4 is dropped vertically into the box and the box continues to move forward with velocity v₂. After some time another object of mass 3M/4 is dropped vertically into the box and then the box moves forward with velocity v₃. Which one among the following is correct?

(a) v₁ = v₂ = v₃

(b) v₂ =  v₁/4

(c) v₂ = 3 v₁/4

(d) v₃ =  v₁/4

(e) v₃ =  v₁/2

The objects dropped vertically into the box will not contribute any horizontal momentum to the box. But they can change the velocity of the box obeying the law of conservation of momentum. The momentum of the system throughout its motion is equal to the initial momentum Mv₁ of the box and we have

Mv₁ = [M + (M/4) + (3M/4)]v₃   

Therefore v₃ =  v₁/2

Therefore, the correct option is (e).

[We have Mv₁ = [M + (M/4)]v₂ and this yields v₂ = 4v₁/5. But we don’t have this answer in the given options]

The following questions are specifically meant for AP Physics C aspirants: 

(3) A bullet of mass m voving horizontally with speed v strikes a wooden block of mass M which rests on a horizontal surface. After the collision the block and the bullet move together and come to rest after moving through a istanc d. The coefficient of kinetic friction between the block and the horizontal surface is

(a) m²v²/[2gd(M + m)²]

(b) mv²/[2gd(M + m)]

(c) m²/[2(M + m)²]

(d) (M + m)v²/2gdM

(e) 2m²v²/[gd(M + m)²]

The common velocity (of the bullet and wooden block) after the impact (v_f) is given by the law of conservation of momentum:

mv + M×0 = (M + m)v_f

Therefore, v_f = mv/(M + m)

The kinetic energy possessed by the the bullet and wooden block, just after the impact, is utilized in doing work against friction while moving through the distance d.

Therefore we have

            ½ (M + m)v_f² = μ_kRd where μ_k is the coefficient of kinetic friction and R is the normal force which is the weight of the bullet and the wooden block.

Substituting for v_f  and R, we have

            ½ (M + m) [mv/(M + m)]² = μ_k(M + m)gd

This gives μ_k = m²v²/[2gd(M + m)²]

(4) A balloon with its payload has mass 40 kg. It is found to descend with acceleration equal to g/4 where g is the acceleration due to gravity. Keeping the volume of the balloon unchanged, the payload is reduced by M kg. The balloon is then found to ascend with acceleration of the same magnitude g/4. What is the value of M? (Ignore viscous drag force)

(a) 4 kg

(b) 6 kg

(c) 8 kg

(d) 16 kg

(e) 20 kg

If the acceleration due to gravity is g we have

            (40g – F)/40 = g/4 where F is the upthrust on the balloon.

Or, g – (F/40) = g/4  from which F = 30g   

If the total mass of the balloon while ascending is x we have

             (30g – xg)/x = g/4

Or, (30/x) – 1 = ¼

Or, 30/x = 5/4

Therefore x = 24 kg

The reduction (M) in the mass of the payload is given by

             M = 40 kg – 24 kg = 16 kg

AP Physics B Electrostatics – Answer to Free Response Practice Question

“There is no democracy in physics. We can't say that some second-rate guy has as much right to an opinion as Fermi.”

– Luis Walter Alvarez

A free response practice question meant for testing your knowledge, understanding and capacity for application of basic principles in electrostatics was given to you in the post dated 21^st September 2012. As promised, I give below a model answer. The question also is given below for your convenience.

Two parallel flat metal plates P₁ and P₂ , each of area 0.16 m², are arranged horizontally (Fig.) in air with a separation of 1.5×10^–2 m. [Separation is shown exaggerated in the figure for convenience].  A constant potential difference of 60 V is established between the plates so that the electric field in the region between the plates is directed vertically downwards as indicated in the figure. Now, answer the following questions:

(a) If the potential of plate P₁ is +10 V, what is the potential of plate P₂? Justify your answer.

(b) Determine the magnitude of the electric field between the plates.

(c) Determine the capacitance of the parallel plate capacitor formed by the plates.

(d) Calculate the increase in the energy stored in the capacitor (formed by the plates P₁ and P₂) when additional charges equal to ten electronic charges.are added to it.  

(e) A charged oil droplet of effective weight 3.2×10^–14 newton is found to remain stationary in the space between the plates. Determine the charge on the oil droplet.

(a) The electric field is pointed vertically downwards. This means that the upper plate P₁ is at higher potential. Since the potential of P₁ is given as 10 V and the potential difference between P₁ and P₂ is given as 60 V, we have

             10 – V₂ = 60 V where V₂ is the potential of plate P₂.

This gives V₂ = – 50 V

(b) The magnitude (E) of the electric field between the plates is given by

             E = V/d where V is the potential difference and d is the separation between the plates.

Therefore, E = 60/(1.5×10^–2) = 4000 NC^–1

(c) The capacitance (C) of a parallel plate capacitor with air (or free space) as dielectric is given by

             C = ε₀A/d where ε₀ is the permittivity of free space (or air, very nearly), A is the area of the plates and d is the separation between the plates.

Therefore, C.= (8.85×10^–12×0.16)/(1.5) = 9.44×10^–11 F

(d) Ten electronic charges is a very small amount of charge which is equal to 10×1.6×10^–19 coulomb. The potential difference between the plates can therefore be assumed to be unchanged. Therefore, the increase in the energy of the capacitor is V(∆q) = 60×1.6×10^–18 joule = 9.6×10^–17 joule.

(e) Since the charged droplet is stationery, the effective weight of the droplet acting vertically downwards is balanced by the electric force on it. The electric force is therefore acting vertically upwards. This means that the charge on the droplet is negative.

Equating the magnitudes of the electric force qE and the effective weight W, we have

             q×4000 = 3.2×10^–14

Therefore q = (3.2×10^–14)/4000 = 8×10^–18 coulomb.

Since the charge is negative, the correct answer is –8×10^–18 coulomb.

[If you were asked to find the number of electrons gained (or lost) by the droplet, you will answer that the droplet has gained 50 electrons since (8×10^–18)/(1.6×10^–19) = 50].

AP Physics B Electrostatics – A Free Response Practice Question

“This time, like all times, is a very good one if we but know what to do with it”

– Ralph Waldo Emerson

Today I give you a free response practice question meant for testing your knowledge, understanding and capacity for application of basic principles in electrostatics:

Two parallel flat metal plates P₁ and P₂ , each of area 0.16 m², are arranged horizontally (Fig.) in air with a separation of 1.5×10^–2 m. [Separation is shown exaggerated in the figure for convenience]. A constant potential difference of 60 V is established between the plates so that the electric field in the region between the plates is directed vertically downwards as indicated in the figure. Now, answer the following questions:

(a) If the potential of plate P₁ is +10 V, what is the potential of plate P₂? Justify your answer.

(b) Determine the magnitude of the electric field between the plates.

(c) Determine the capacitance of the parallel plate capacitor formed by the plates.

(d) Calculate the increase in the energy stored in the capacitor (formed by the plates P₁ and P₂) when additional charges equal to ten electronic charges.are added to it.  

(e) A charged oil droplet of effective weight 3.2×10^–14 newton is found to remain stationary in the space between the plates. Determine the charge on the oil droplet.

The above question carries 15 points. You may take up to 17 minutes for answering it. Try to answer it yourself within the stipulated time. I’ll come back shortly with a model answer for your benefit.

AP Physics B & C Multiple Choice Practice Questions on Electrostatics

“I never did a day's work in my life. It was all fun.”

– Thomas A. Edison

Questions in the section ‘electrostatics’ were discussed on many occasions earlier on this site. You may access them by clicking on the label ‘electrostatics’ below this post. Since the number of posts displayed per page is limited, you will have to use the ‘older posts’ button to access all the posts.

All posts in this section can equally well be accessed by trying a search for ‘electrostatics’ using the search box provided on this page.

Today we will discuss a few more multiple choice practice questions on electrostatics:

(1) Point charges +q and –3q are arranged at points A and B on the x-axis (Fig.). At which point (or points) is the electric field due to this system of charges zero?

(a) At two points on the x axis, one point to the right of the charge –3q  and the other to the left of the charge +q.

(b) Somewhere on the x axis to the left of the charge +q.

(c) Somewhere on the x axis to the right of the charge –3q.

(d) Somewhere on the x axis, in between the two charges.

(e) The electric field cannot be zero anywhere.

The electric field lines due to a positive charge proceed radially outwards (diverge) from the charge where as the electric field lines due to a negative charge proceed radially inwards (converge) to the charge. At points on the x-axis lying in between A and B the fields due to these charges are directed along the positive x-direction. Therefore, the fields add up and the resultant field cannot be zero. But at points on the x-axis to the left of the charge +q, the fields are in opposite directions (along the negative x-direction due to the charge +q and along the positive x-direction due to the charge –3q) so that they can cancel at a point where the magnitudes of the fields due to the charges are the same. So the correct option is (b).

[At points on the x-axis to the right of the charge –3q, the fields are in opposite directions (along the positive x-direction due to the charge +q and along the negative x-direction due to the charge –3q. But the two fields cannot cancel anywhere here since the magnitude of the field due to the charge –3q is greater than that due to  the charge +q (since the charge –3q is nearer as far as points on the right of B are concerned]

(2) A large flat plate is positively charged so that it has uniform surface charge density σ. If the electric field near the central region of the plate at a distance of 1 cm from the plate is 9 NC^–1, what will be the electric field at a distance of 3 cm from the plate?

(a) 81 NC^–1

(b) 27 NC^–1

(c) 9 NC^–1

(d) 3 NC^–1

(e) 1 NC^–1

The electric field due to the surface charge on the plate is normal to the surface of the plate. Since the plate is large, the electric field near the central region of the plate is uniform (with the electric field lines proceeding normal to the flat surface and therefore parallel). This means that the electric field at a point is independent of the distance of the point from the plate so that electric field at 3 cm from the plate is 9 NC^–1 itself.

[For AP Physics C aspirants: 

To find the electric field near the surface of the charged plate, we can apply Gauss theorem and accordingly imagine a Gaussian surface shaped as a rectangular parallelepiped of cross section area A (Fig.). The electric field due to the surface charge on the plate is normal to the surface of the plate. Therefore the electric field is directed normal to the end faces of the rectangular parallelepiped so that the total electric flux through the closed suface (rectangular parallelepiped) is 2EA where E is the magnitude of the electric field. (The flux through the side surfaces of the parallelepiped is zero since the electric field is parallel to these surfaces). The total charge enclosed by the closed surface is σA. Therefore, by Gauss theorem we have

             2EA = σA/ε₀ where ε₀ is the permittivity of free space.

Therefore E = σ/2ε₀, which is independent of distance]

(3) Charges +q, –q and –q are placed at the vertices of an equilateral triangle ABC as indicated in the adjoining figure. What is the direction of the net electric field at the central point P?

(a) Along AP

(b) Along PA

(c) Along CP

(d) Along BP

(e) There is no field at P

The direction of the electric fiel at P is the direction of the force acting on a test positive charge placed at P. The charge +q will exert a repulsive force on the test positive charge. This is irected along AP. The charges –q and –q at B and C will exert attractive forces along PB and PC respectively. These two forces being of equal magnitudes, their resultant will be directed along AP. Therefore the resultant electric field due to all the three charges will be along AP [Option (a)]. 

(4) Points 1, 2 and 3 lie on the axis of an electric dipole AB. Point 4 lie on the equatorial line (perpendicular bisector of AB) of the dipole. Out of the following choices which one correctly gives the directions of the electric fields at the points 1, 2, 3 and 4?        

When you place a positive test charge at point (1) the net force on it is repulsive since the charge +q of the dipole is nearer than the charge –q. The direction of the force (and the electric field) is therefore leftwards.

At point (2) the charge +q exerts a repulsive force on the positive test charge where as the charge –q exerts an attractive force on it. Thus both forces act rightwards an hence the electric field at point (2) acts rightwards.

At point (3) the net force on the positive test charge is attractive since the charge –q of the dipole is nearer than the charge +q. The electric field at point (3) is therefore leftwards.

At point (4) the positive test charge  is repelled by the charge +q and attracted by the charge –q. these forces are equal in magnitude and are inclined equally upwards and downwards respectively.  Their resultant acts rightwards and hence the electric field at point (4) is rightwards.

The above facts are correctly given in option (e).

If you have a mental picture of the distribution of the electric field lines due to an electric dipole (See the figure below), you will be able to answer the above question in no time].

 

(5) In the above question, in the case of the points (1), (2), (3) and (4), where do you find zero electric potential?

(a) Nowhere

(b) At point (1)

(c) At points (1), (2) and (3)

(d) At points (2) and (4)

(e) At point (2)

Points (2) and (4) are equidistant from the charges +q and –q. These equal and opposite charges produce positive and negative potentials of equal values so that they cancel each other, giving rise to zero potential at points (2) and (4). 

Now see these questions with solution.

AP Physics B – Some Interesting Multiple Choice Practice Questions on Atomic and Nuclear Physics

“Non-violence leads to the highest ethics, which is the goal of all evolution. Until we stop harming all other living beings, we are still savages.”

– Thomas A. Edison

The sections ‘atomic physics and quantum effects’ and ‘nuclear physics’ in the AP Physics B syllabus will appear to be interesting to most of the AP Physics B aspirants. Today we will discuss a few practice questions (multiple choice) in these sections:

(1) The de Broglie wave length of a particle with kinetic energy E is λ. If its kinetic energy is increased  to 4E, its de Broglie wave length will be

(a) λ/4

(b) λ/2

(c) λ

(d) 2λ

(e) 4λ

Since the kinetic energy is directly proportional to the square of momentum, the momentum of the particle is doubled when its kinetic energy is quadrupled.

[Note that kinetic energy E = p²/2m where p is the momentum and m is the mass].

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum.

Therefore, when p is doubled, λ is halved [Option (b)].

(2) Particles A and B have masses m and 4m respectively but they carry the same charge. When they are accelerated by the same voltage, their de Broglie wave lengths are in the ratio

(a) 1 : 1

(b) 2 : 1

(c) 4 : 1

(d) 1 : 4

(e) 1 : 8

Let V represent the common accelerating voltage and q represent the common charge of the particles. If p₁ and p₂ are the momenta of the particles A and B, on equating their kinetic energies, we have

             p₁²/2m = p₂²/(2×4m)

[Remember that the kinetic energy of a particle of charge q accelerated by a voltage V is qV. The kinetic energies of A and B are equal since they have the same charge and they are accelerated by the same voltage]

The above equation gives

             p₁/ p₂ = ½

Since the de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum, the ratio of the de Broglie wave lengths of A and B is given by

              λ₁/ λ₂ = p₂/ p₁ = 2, as given in option (b).

(3) A metallic surface is found to emit photo-electrons when monochromatic light rays of frequencies n₁ and n₂ (n₂ > n₁) are incident on it. If the maximum values of kinetic energy of the photo-electrons emitted in the two cases are in the ratio 1 : 3, the threshold frequency of the metallic surface is

(a) (3n₁ – n₂)/ 2

(b) (2n₁ – n₂)/ 2

(c) (3n₁ – n₂)/ 3

(d) (n₂ – n₁)/ 3

(e) (n₂ – n₁)/ 2

If the threshold frequency is n₀ and the maximum values of kinetic energy in the two cases are E₁ and E₂ respectively, we have

             hn₁ = hn₀ + E₁ and

             hn₂ = hn₀ + E₂

Therefore, E₁/E₂ = (n₁ – n₀)/(n₂ – n₀)

Since the ratio is 1 : 3 we have

             (n₁ – n₀)/(n₂ – n₀) = 1/3

Or, 3n₁ – 3n₀ = n₂ – n₀

This gives n₀ = (3n₁ – n₂)/ 2, as given in option (a).

(4) The energy that must be added to an electron to reduce its de Broglie wave length from 2 nm to 1 nm is

(a) half the initial energy

(b) equal to the initial energy

(c) twice the initial energy

(d) thrice the initial energy

(e) four times the initial energy

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum.

Since the de Broglie wave length of the electron is to be reduced to half the initial value (from 2 nm to 1 nm), the momentum of the electron is to be doubled. But when the momentum is doubled, its kinetic energy becomes four times the initial value. Therefore the energy that must be added is three times the initial energy [Option (d)].

(5) An alpha particle of mass m and speed v  proceeds directly towards a heavy nucleus of charge Ze. The distance of closest approach of the alpha particle is directly proportional to

(a) 1/Ze²

(b) 1/Ze

(c) v

(d) m

(e) 1/v²

The alpha particle has to move towards the nucleus with difficulty, doing work against the electrostatic repulsive force. When the alpha particle reaches the distance of closest approach, the entire kinetic energy gets converted into electrostatic potential energy. Therefore we have

             ½ mv² = (1/4πε₀)(2Ze²/r) where r is the distance of closest approach.

[Remember that the charge on the alpha particle is 2e].

This gives r = Ze²/πε₀mv²

This shows that r is directly proportional to 1/v².