AP Physics B – Nuclear Physics- Multiple Choice Practice Questions

Let us discuss some multiple choice questions on nuclear physics which will benefit AP Physics B aspirants:

(1) The ratio of the nuclear radius of ₅₂Te¹²⁵ to that of ₁₃Al²⁷ is

(a) 4

(b) 125/17

(c) ¼

(d) 5/3

(e) 3/5

The nuclear radius of an atom of mass number A is given by

R =R₀A^1/3 where R₀ = 1.2×10^–15 m.

The required ratio is therefore (125/27)^1/3 which is equal to 5/3.

(2) A nucleus of ₉₂U²³⁸ gets converted into a ₉₁Pa²³⁴ nucleus. The particles emitted during this decay are

(a) one α-particle and one positron

(b) one α-particle and one electron

(c) one α-particle and one antineutrino

(d) one α-particle and one neutrino

(e) one α-particle and two β-particles

The mass number decreases by 4 and the atomic number decreases by 1 in the above decay. When an α-particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2. The atomic number can be increased by one from this condition only if an electron also is emitted. The correct option therefore is (b).

(3) The de Broglie wave length of an α-particle of mass m emitted by a nucleus of mass M initially at rest is λ. The de Broglie wave length of the nucleus immediately after the α-emission is

(a) λ

(b) λ (M– m)/m

(c) λm/(M–m)

(d) λ (m/M)²

(e) λ (M/m)²

The nucleus has a recoil momentum on emitting the α-particle. Since the parent nucleus is initially at rest, its recoil momentum has the same magnitude as that of the α-particle but the direction is opposite in accordance with the law of conservation of momentum. The de Broglie wave length λ of the α-particle is given by

λ = h/p where h is Planck’s constant.

Since the recoil momentum of the nucleus has the same magnitude p the de Broglie wave length of the nucleus immediately after the α-emission is λ itself [Option (a)].

(4) Complete the following relation representing one possible fission process:

₀n¹ + ₉₂U²³⁵→ ₃₈Sr⁹⁰ + ----

(a) ₅₄Xe¹⁴⁵

(b) ₅₄Xe¹⁴⁵ + 3 ₀n¹

(c) ₅₄Xe¹⁴³ + 3 ₀n¹

(d) ₅₄Xe¹⁴² + ₀n¹

(e) ₅₄Xe¹⁴² + 3 ₀n¹

The total mass number and the total atomic number on the two sides will match only if the relation is completed with the terms given in option (c).

(5) The mass m of any particle of rest mass m₀ at speed v is given by Einstein’s relativistic relation,

m = m₀/[1– (v²/c²)]^1/2 where c is the speed of light in free space.

The rest energy of an electron is 0.511 MeV. The increase in the energy of the electron when it is accelerated from rest to 80% of the speed of light in free space is (very nearly)

(a) 0.341 MeV

(b) 0.405 MeV

(c) 0.511 Mev

(d) 0.852 Mev

(e) 0.916 Mev

We have m₀c² = 0.511 Mev for the electron.

The total energy at a speed of 0.8 c is m₀c²/[1– (v²/c²)]^1/2 = m₀c²/(0.36)^1/2 since v = 0.8 c.

But m₀c² = 0.511 Mev so that the total energy is 0.511/0.6 = 0.852 MeV nearly.

The increase in the energy is 0.852 – 0.511 = 0.341 MeV.

Nuclear Physics for AP Physics B- Equations to be Remembered

Nuclear physics is supposed to occupy just 3% of the total content of the syllabus for the AP Physics B (2009) Examination. However you need to remember the following things in this section for answering the questions within the stipulated time:

(1) Mass number (A) is the number of nucleons (neutrons & protons) and is given by

A = Z + N

where Z is the atomic number (number of protons in the nucleus) and N is the neutron number.

Isotopes have same Z but different A.

Isotones have same N.

Isobars have same A but different Z

Isomers have same A and same Z but different nuclear energy states

(2) Nuclear radius (size) R is given by

R =R₀A^1/3

where R₀ = 1.2×10^–15 m. The volume of the nucleus which is proportional to R³ is therefore proportional to the mass number A. This means that the density of the nucleus is constant (approximately 2.3×10¹⁷ kgm^–3) and is independent of A.

(3) Einstein’s mass-energy equivalence relation is

E = mc²

where E is the energy (in joules) obtained on annihilating a mass m kg and c is the speed of light in free space.

(4) Mass defect ∆M is the difference between the mass of the nucleus and the total mass of its constituents(nucleons):

∆M = [Zm_p + (A–Z)m_n] – M

where m_p is the mass of the proton, m_n is the mass of the neutron and M is the mass of the nucleus.

The binding energy E_b of the nucleus is given by

E_b = ∆M c²

The binding energy per nucleon is E_b/A

(5) Three types of radioactive decay occur in nature.They are:

(i) α-decay in which an α-particle (helium nucleus ₂He⁴) is emitted;

(ii) β-decay in which electrons or positrons are emitted;

(ii) γ-decay in which γ-rays (high energy photons) are emitted

When an α-particle is emitted by a nucleus its mass number decreases by 4 and its atomic number decreases by 2.

When an electron is emitted by a nucleus (β^– decay) its mass number is unchanged but its atomic number increases by 1.

When a positron is emitted by a nucleus (β⁺ decay) its mass number is unchanged but its atomic number decreases by 1.

When a γ-ray photon is emitted by a nucleus its mass number and atomic number remain unchanged.

In β^– decay a neutron within the nucleus gets transformed into a proton in accordance with the relation,

n→ p + e^– + ν where n, p, e^– and ν represent the neutron, proton, electron and the antineutrino respectively.

In β⁺ decay a proton within the nucleus gets transformed into a neutron in accordance with the relation,

p→ n + e⁺ + ν where e⁺ and ν represent the positron and the neutrino respectively.

(6) In nuclear fission reaction a heavy nucleus such as ₉₂U²³⁵ or ₉₄Pu²³⁹ absorbs a slow neutron and breaks into two fragments of nearly equal size, releasing large amount of energy (typically about 200 Mev per fission). Two or three neutrons also are produced in the reaction. A typical example of fission reaction is

₀n¹ + ₉₂U²³⁵→ ₉₂U²³⁶→ ₅₆Ba¹⁴¹ + ₃₆Kr⁹² + 3 ₀n¹

Fragments other than barium and krypton can be produced in the reaction.

(7) In nuclear fusion reaction two light nuclei combine to form a single larger nucleus with release of energy. A common example of nuclear fusion reaction is the one given below in which two deuterons combine to form the light isotope of helium:

₁H² + ₁H² → ₂He³ + ₀n¹ + 3.27 Mev

You should remember that the total mass number on the left hand side should be the same as that on the right hand side of the equations representing the nuclear reaction. Similarly the charge numbers also should match.

Questions for high lighting the above facts and the law of conservation of energy and momentum are often asked in degree entrance examinations.

Answers to Fre-response Pracice Questions on AP Physics B&C Work, Energy & Power

In the post dated 14-12-2008, two free-response question for practice was given to you.

As promised, I give below the answers along with the questions:

(1) ABCDEF is a track which is straight and horizontal in the region BCDE but curved in the regions AB and EF. The track is smooth everywhere except over a length 1.5 m in the horizontal region CD where the coefficient of kinetic friction is 0.2. A block of mass m₁ = 2 kg is released from rest from position A which is at a height of 3.2 m. It slides down and collides with another stationary block of mass m₂ = 3 kg placed near position C (fig.). The velocity of the 3 kg mass m₂ immediately after the collision is 4 ms^–1. Now answer the following:

(a) Determine the velocity of the 2 kg mass m₁ immediately after the collision.

(b) State whether this collision is elastic or inelastic. Justify your answer.

(c) Calculate the kinetic energy lost by the 3 kg mass during its forward trip along the rough track CD.

(d) Determine the maximum height ‘x’ up to which the 3 kg mass will rise along the smooth track EF.

(e) If the 2 kg mass m₁ is removed (without disturbing any other thing) immediately after the collision, determine the distance from C where the 3 kg mass will finally come to rest.

(a) The speed u₁ of the mass m₁ just before it collides with the mass m₂ is given by

m₁gh = ½ m₁ u₁², on equating the loss of gravitational potential energy to the gain of kinetic energy.

Therefore, u₁ = √(2gh) = √(2×10×3.2) = 8 ms^–1.

The initial momentum of the system (of masses m₁ and m₂) is (m₁u₁ + 0) and the final momentum is (m₁v₁ + m₂v₂) where v₁ and v₂ are the velocities of masses m₁ and m₂ respectively after the collision.

Since the momentum is conserved, m₁u₁ = m₁v₁ + m₂v₂.

Substituting known values, 2×8 = 2 v₁ + 3×4 from which v₁ = 2 ms^–1.

(b) This collision is inelastic since the kinetic energy is not conserved as shown below:

Initial kinetic energy = ½ m₁u₁² + 0 = ½ ×2×8² = 64 J

Final kinetic energy = ½ m₁v₁² + ½ m₂v₂² = ½ (2×2² + 3×4²) = (4+24) J = 28 J, which is less than the initial kinetic energy.

(c) The kinetic energy lost by the 3 kg mass during its forward trip along the rough track CD is μ_km₂gd = 0.2×3×10×1.5 = 9 J (since the coefficient of kinetic friction (μ_k) is 0.2 and the distance moved (d) is 1.5 m.

(d) The kinetic energy remaining in the 3 kg mass after its forward trip along CD is (24 J – 9 J) = 15 J. The maximum height ‘x’ up to which the 3 kg mass will rise along the smooth track EF is given by

m₂gx = 15 J

Therefore, x = 15/(3×10) = 0.5 m.

(e) Since the mass m₁ is removed after the first collision there is no possibility for further collisions. The mass m₂ comes down and moves backwards from D to E losing another 9 J of energy. The kinetic energy remaining in it is now 6 J only and rises up along the smooth track CBA and returns after reaching the maximum possible height. It will stop at a point distant ‘s’ from C after losing its energy in doing work against traction. s is evidently given by

μ_km₂gs =6 J

Therefore s = 6/(0.2×3×10) = 1 m.

(2) An object A of mass 5 kg moving along the positive x-direction has displacement x given by x = 0.2 t + 0.18 t². The force acting on the object ceases after 5 seconds and the object moves with uniform velocity. Another object B of mass 10 kg moving in the positive y-direction has kinetic energy equal to the potential energy of a spring of force constant 3000 Nm^–1 compressed through 0.1 m from its natural length. The objects A and B moving with their respective constant velocities collide at the origin O and stick together.

Now answer the following questions:

(a) Determine the momentum and kinetic energy of object A just before collision.

(b) Determine the momentum and kinetic energy of object B just before collision

(c) Calculate the velocity of the combined mass (after A and B stick together).

(d) Calculate the change in kinetic energy of the system because of the collision.

(e) Is this collision elastic or inelastic? Justify your answer.

(a) The velocity v_A of object A will be constant after 5 s since the force ceases to act. This is obtained by substituting t = 5 s in dx/dt:

Therefore, v_A = 0.2 + 2×0.18t = 0.2 + 2×0.18×5 = 2 ms^–1

Momentum of object A is P_A = m_Av_A = 5×2 = 10 kg ms^–1

Kinetic energy of object A is K_A = P_A²/2m_A = 100/10 = 10 J

[Or, K_A = ½ m_A v_A², if you would prefer this form].

(b) Potential energy of a compressed spring is ½ kx² where k is the spring constant and x is the compression. Substituting appropriate values, the energy is ½ ×3000×(0.1)² = 15 J. Since the kinetic energy K_B of the object B is equal to the above energy, we have

K_B = 15 J.

Therefore, P_B²/2m_B = 15 from which the momentum P_B of object B is obtained as

P_B = √(2×10×15) = 10√3 kg ms^–1

(c) The momentum vectors P_A and P_B are shown in the figure. The resultant momentum P gives the momentum of the combined mass and is given by

P = √(P_A² + P_B²) = √(100+300) = 20 kg ms^–1

The velocity ‘v’ of the combined mass is given by

v = P/(m_A+m_B) = 20/15 = 1.333 ms^–1

The velocity is directed as shown, making an angle θ with the x-axis, given by

tanθ = P_B/P_A = (10√3)/10 = √3 so that θ = 60º

(d) Kinetic energy K of the combined mass after the collision is given by

K = P²/2(m_A+m_B) = 20²/2(5+10) = 13.333 J

The initlal kinetic energy of the system is K_A + K_B = 10+15 = 25 J.

Thefefore, the change in kinetic energy of the system because of the collision is 13.333 –25 = –11.667 J

[Negative sign shows that the kinetic energy has decreased].

(e) The collision is inelastic since the kinetic energy is not conserved.

Fre-response Pracice Questions- AP Physics B&C- Work, Energy & Power

Two free-response questions involving work and energy are given below. The first question is meant for AP Physics B and the second question is meant for AP Physics C aspirants. If you are preparing for AP Physics C exam, you should be able to answer the questions meant for AP Physics B as well.

(1) ABCDEF is a track which is straight and horizontal in the region BCDE but curved in the regions AB and EF. The track is smooth everywhere except over a length 1.5 m in the horizontal region CD where the coefficient of kinetic friction is 0.2. A block of mass m₁ = 2 kg is released from rest from position A which is at a height of 3.2 m. It slides down and collides with another stationary block of mass m₂ = 3 kg placed near position C (fig.). The velocity of the 3 kg mass m₂ immediately after the collision is 4 ms^–1. Now answer the following:

(a) Determine the velocity of the 2 kg mass m₁ immediately after the collision.

(b) State whether this collision is elastic or inelastic. Justify your answer.

(c) Calculate the kinetic energy lost by the 3 kg mass during its forward trip along the rough track CD.

(d) Determine the maximum height ‘x’ up to which the 3 kg mass will rise along the smooth track EF.

(e) If the 2 kg mass m₁ is removed (without disturbing any other thing) immediately after the collision, determine the distance from C where the 3 kg mass will finally come to rest.

The above question carries 15 points and you have about 17 minutes for answering it.

(2) An object A of mass 5 kg moving along the positive x-direction has displacement x given by x = 0.2 t + 0.18 t². The force acting on the object ceases after 5 seconds and the object moves with uniform velocity. Another object B of mass 10 kg moving in the positive y-direction has kinetic energy equal to the potential energy of a spring of force constant 3000 Nm^–1 compressed through 0.1 m from its natural length. The objects A and B moving with their respective constant velocities collide at the origin O and stick together.

Now answer the following questions:

(a) Determine the momentum and kinetic energy of object A just before collision.

(b) Determine the momentum and kinetic energy of object B just before collision

(c) Calculate the velocity of the combined mass (after A and B stick together).

(d) Calculate the change in kinetic energy of the system because of the collision.

(e) Is this collision elastic or inelastic? Justify your answer.

[The above question is worth 15 points and the time at your disposal is about 15 minutes].

Try to answer these questions. I’ll be back soon with model answers for your benefit.

AP Physics B & C- Additional Practice Questions (MCQ) on Work, Energy & Power

This post has been a bit delayed because of a ‘boot.com’ virus infection in my system. Fortunately I could disinfect the system without much delay.

A few multiple choice questions on work energy and power were discussed in the last post. Here are a few more multiple choice questions in this section:

(1) A particle of mass 2 kg moving along the positive x-direction has a constant momentum of 6 kg ms^–1. If a constant force of 2 N is applied on the particle for 1 s along the negative x-direction, the change in the kinetic energy of the particle is

(a) decrease of 4 J

(b) increase of 4 J

(c) decrease of 8 J

(d) increase of 4 J

(e) decrease of 9 J

The initial kinetic energy of the particle = p²/2m = 6²/(2×2) = 9 J.

The force F of 2 N actng on the particle for the time t equal to 2 s imparts a momentum Ft equal to 4 kg ms^–1 along the negative direction. Since the initial momentum of the particle is along the positive x-direction, the resultant momentum is 6 – 4 = 2 kg ms^–1.

Therefore, the final kinetic energy of the particle = 1²/(2×2) = 1 J.

The change in the kinetic energy of the particle is 1 – 9 = – 8 J

The kinetic energy of the particle is thus decreased by 8 J.

(2) An elevator motor creates a tension of 5000 N in a hoisting cable and reels it at 0.8 ms^–1. If the efficiency of the elevator system is 80%, the power input to the motor is

(a) 2 kW

(b) 4 kW

(c) 6 kW

(d) 8kW

(e) 10 kW

The power output (P_out) of the system is Fv = 5000×0.8 = 4000 W.

If the power input is P_in we have

efficiency h = P_out/ P_in

Therefore, 80/100 = 4000/P_in so that P_in = 5000 W = 5 kW.

(3) A liquid of density ρ is being continuously pumped through a pipe of area of cross section a. If the speed of the liquid through the pipe is v, the time rate at which kinetic energy is being imparted to the liquid is

(a) av²ρ/2

(b) av³ρ/2

(c) avρ/2

(d) av³ρ

(e) av²ρ

The mass of liquid flowing per second is avρ. Therefore, the time rate at which kinetic energy is imparted to the liquid is ½ (avρ)v² = av³ρ/2

The above questions will be beneficial for AP Physics B as well as C. The following questions are for AP Physics C aspirants:

(4) A machine delivering constant power moves an object along a straight line. The displacement of the object in time t is proportional to

(a) t^–2

(b) t³

(c) t^1/2

(d) t^3/2

(e) t^–3

Since the power is the product of force and velocity we have

Fv = K where K is a constant.

If m is the mas and v is the velocity of the object, the above equation can be written as

M(dv/dt)v = K so that vdv = (K/m)dt

Intrgrating, v²/2 = (K/m)t

Therefore v α t^1/2.

If ‘s’ is the displacement, v = ds/dt so that (ds/dt) α t^1/2.

Integrating, s α t^3/2 [Option (d)].

(5) The velocity v, momentum p and the kinetic energy E of a particle are related as

(a) p = dv/dE

(b) p = dE/dv

(c) p = (dE/dv)^1/2

(d) p = d²E/dv²

(e) p = (dE/dv)²

If the mass of the particle is m we have E = ½ mv²

Differentiating, dE/dv = mv = p.