Thursday, April 23, 2015

Physics 9702 Doubts | Help Page 120

  • Physics 9702 Doubts | Help Page 120

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
(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:
(i) {Consider the relative atomic masses: 235 + 1 = 95 + 139 + x(1) + 0}
Value of x = 2

(ii) EITHER Beta particle or electron

{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

{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

{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

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

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

{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

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

(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:
(i) Zero field (strength) inside the spheres

Either The field strength is zero
Or the fields are in opposite directions
At a point between the spheres

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

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)

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


  1. November 2012 Paper 43 Q7 (b)(i). I used mg=qV/d and v^2=u^2+2as and lambda=h/p and got the same answer. Is my approach/concept wrong? And also could you please explain further part (ii)? I don't quite get the question. Thank you.

    1. See solution 705 at

      This method is wrong.
      1. It is not mentioned that the acceleration is constant.
      2. At very high velocities, even the mass can changed (this is not done in A-levels)
      3. What is d? Is it given?
      4. No information was said about the direction of the force. What if it was in the same direction as the weight?

      5. Most important, physics can be classified into Newtonian mechanics (deals with macroscopic objects) and Quantum mechanics (deals with microscopic objects). We can't mixed the formulae ...

      6. ...

  2. Hi. I do not understand question 609 (a) (ii). Why is the field strength zero at a point between spheres? And how do I know the fields are in opposite directions, and how does this provide evidence that both the spheres are +vely/-vely charged? Can you provide a detailed explanation please? Help appreciated. Thank you!

    1. From the graph, E = 0 indicates that the field strength is zero at some point (the graph crosses the x-axis).

      From the graph, it is seen that the electric field strength E has both positive and negative values. This indicates that the fields are in opposite directions.

      The direction of the electric force on a positive charge is from positive to negative. If one of the spheres was positive and the other negative, then the direction of the electric force on a positive in the field would be in only one direction - that is, towards the negative charge. However, since the graph has both positive and negative values for E, this means that the direction of the field changes with distance.

      Now, since the spheres are either both positive or negative, the field due to each of the sphere would oppose the other. At some point, the electric force due to each of them would be equal. This is where the electric field strength is zero.

      If the spheres had unlike charges, the force would always be in only one direction and not zero.

  3. Question 607 (b)(i). I know it stated 'in terms of u', but I'm confused, isn't 1.808u simply the mass defect? How can that be answered as 'the binding energy'? Mass defect is simply the mass isn't it? Please forgive my ignorance. TQVM

    1. If the mass defect is m, then the energy associated with it is
      E = m c^2

      (c^2) is also a constant. So, instead of doing the calculation it is easier to state the mass defect in terms of u.

      But of course, as done as the next part of the question, the have the actual binding energy, we need to multiply to mass defect by c^2 to obtain the correct value of energy.

      Writing in terms of u is just an easy way to work. Large values are avoided.

  4. could you please explain solution 608 (b)(i)

    1. This is the law of conservation of energy.
      As the boy slides down, its PE is being converted to KE. However, some work is done against friction.

      Decrease in PE = Gain in KE + Work done against friction

      Re-arrange to obtain as above.

  5. can you please explain question 609 part b ii .why is the rate of change of potential maximum ehen electric field has extreme negative value ?

    1. as given in the previous part, E is proportional to the potential gradient. SO, the change in potential is maximum when E has a maximum value.

      Note that the sign only indicates the direction / charge while the value indicates the magnitude of the quantity.

      The potential is zero when the field is zero.

  6. From quest 608) why can't we use F=ma to calculate the resistive force .
    And how do u derive the word equation. Is there any hint in this quest regarding to the equation

    1. we would need to know the deceleration caused by the resistive force


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