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Question 607: [Nuclear Physics]

When a neutron is captured by uranium-235 nucleus, outcome may be represented by the nuclear equation shown below.

²³⁵₉₂ U +          ¹₀n       - - - >   ⁹⁵₄₂ Mo             +  ¹³⁹₅₇ La        +          x¹₀n      +  7 ⁰_–1 e

(a)

(i) Use equation to determine the value of x.

(ii) State name of the particle represented by the symbol ⁰_–1 e.

(b) Some data for the nuclei in the reaction are given in Fig.1.

mass / u           binding energy per nucleon / MeV

uranium-235 (²³⁵₉₂ U)                          235.123

molybdenum-95 (⁹⁵₄₂ Mo)                   94.945                                     8.09

lanthanum-139 (¹³⁹₅₇ La)                     138.955                                   7.92

proton (¹₁ p)                                         1.007

neutron (¹₀ n)                                       1.009

Use data from Fig.1 to

(i) determine binding energy, in u, of a nucleus of uranium-235,

(ii) show that binding energy per nucleon of a nucleus of uranium-235 is 7.18 MeV.

(c) Kinetic energy of the neutron before the reaction is negligible.

Use data from (b) to calculate total energy, in MeV, released in this reaction.

Reference: Past Exam Paper – November 2012 Paper 43 Q8

Solution 607:

(a)

(i) {Consider the relative atomic masses: 235 + 1 = 95 + 139 + x(1) + 0}

Value of x = 2

(ii) EITHER Beta particle or electron

(b)

(i)

{A nucleus of uranium-235 contains 92 protons and 235 – 92 = 143 neutrons. From the equation E = mc², a mass m is equivalent to an energy E. So, there is an amount of energy due to each proton and neutron which are bound to the nucleus. In the calculation below, the c² has been eliminated since it is present in all the terms.}

Mass of separate nucleons = {(92 × 1.007) + (143 × 1.009)} u = 236.931 u

{The uranium-235 has a mass of 235.123u. So, the difference in mass between 236.931u and 235.123u gives the binding energy in terms of u.}

Binding energy = 236.931 u – 235.123 u = 1.808 u

(ii)

{From the list of data given in the question paper, u = 1.66 × 10^–27 kg. The binding energy in terms of u has been calculated above.}

Energy E = mc² = 1.808 × (1.66 × 10^–27) × (3.0 × 10⁸)² = 2.7 × 10^–10 J

{Binding energy per nucleon in Joules= (2.7 × 10^–10) / 235

Binding energy per nucleon in in eV= (2.7 × 10^–10) / (235 × 1.6 × 10^–19)

Binding energy per nucleon in in MeV= (2.7 × 10^–10) / (235 × 1.6 × 10^–19 × 1 × 10⁶)}

Binding energy per nucleon = (2.7 × 10^–10) / (235 × 1.6 × 10^–13) = 7.18 MeV

(c)

{Energy released = energy of products – energy of reactants.

Mo contains 95 nucleons, each with energy 8.09 MeV and La contains 139 nucleons, each with energy 7.92 MeV. The individual neutron and electron are not binded, so we do not considered energies due to these 2.}

Energy released = (95×8.09) + (139×7.92) – (235×7.18) = 1869.43 – 1687.3 = 182 MeV

Question 608: [Work, Energy and Power]

(a) Explain what is meant by work done.

(b) A boy on board B slides down a slope, as shown in Fig.1. 

Angle of the slope to horizontal is 30°. The total resistive force F acting on B is constant.

(i) State a word equation that links work done by force F on B to changes in potential and kinetic energy

(ii) The boy on board B moves with velocity v down slope. Variation with time t of v is shown in Fig.2. 

Total mass of B is 75 kg.

For B, from t = 0 to t = 2.5 s,

1. show that distance moved down slope is 9.3 m,

2. calculate gain in kinetic energy

3. calculate loss in potential energy:

4. calculate resistive force F:

Reference: Past Exam Paper – June 2014 Paper 23 Q3

Solution 608:

(a) Work done is the product of force and the distance moved in the direction of the force OR the product of force and displacement in the direction of the force

(b)

(i) Work done by F = Decrease in GPE – Gain in KE

(ii)

1.

Distance moved = area under the line = 0.5(7.4 x 2.5) = 9.3m (9.25m)

OR

From graph, acceleration a (= gradient) = 7.4 / 2.5 (= 2.96))

From equation of motion: (7.4)2 = 0 + 2(2.96)(s) giving s = 9.3 (9.25)m

2. Gain in KE = ½mv² = 0.5 (75) (7.4)² = 2100J

3.

{Potential energy depends on height. So, the loss in potential energy can be obtained if the decrease in height is known.}

Loss in potential energy = mgh

(Distance moved along slope = 9.3m)

Height, h = 9.3sin(30)

Loss in potential energy = 75(9.81)(9.3sin(30)) = 3400J

4.

Work done by F = energy loss

Resistive force F = (3421 - 2054) / 9.3 = 150 (147) N

Question 609: [Electrostatic]

Two small charged spheres A and B are situated in vacuum. Distance between centres of spheres is 12.0cm, as shown in Fig.1.

Charge on each sphere may be assumed to be point charge at the centre of the sphere.

Point P is a movable point that lies on line joining centres of  the spheres and is distance x from the centre of sphere A.

Variation with distance x of electric field strength E at point P shown in Fig.2.

(a) State evidence provided by Fig.2 for statements that

(i) the spheres are conductors

(ii) the charges on spheres are either both positive or negative

(b)

(i) State relation between electric field strength E and potential gradient at a point

(ii) Use Fig.2 to explain distance x at which the rate of change of potential with distance is

1. maximum

2. minimum

Reference: Past Exam Paper – November 2011 Paper 41 & 42 Q4

Solution 609:

(a)

(i) Zero field (strength) inside the spheres

(ii)

Either The field strength is zero

Or the fields are in opposite directions

At a point between the spheres

(b)

(i) The electric field strength is (-) the potential gradient (not V/x)

(ii)

1. The electric field strength has a maximum value at x = 11.4cm

(peaks at 1.4 and 11.4 not equal à greater at 11.4)

2.

The electric field strength is zero

Either at x = 7.9cm (± 0.3cm)

Or at x = 0 to 1.4cm                or 11.4cm to 12cm

Question 610: [Hooke’s law]

Graph shows the effect of applying a force of up to 5 N to a spring.

What is the total increase in length produced by a 7 N force, assuming spring obeys Hooke’s law?

A 4.2 cm                     B 5.6 cm                      C 15.2 cm                    D 19.6 cm

Reference: Past Exam Paper – November 2011 Paper 12 Q22

Solution 610:

Answer: A.

Hooke’s law: F = kx.

When force F = 5N, extension x = 14 – 11 = 3cm

So, spring constant k = F / x = 5 / 3 = 1.67Ncm^-1

When force F = 7N, extension x = F / k = 7 / 1.67 = 4.2cm