# 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.

^{235}

_{92}U +

^{1}

_{0}n - - - >

^{95}

_{42}Mo +

^{139}

_{57}La + x

^{1}

_{0}n + 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 (

^{235}_{92}U) 235.123
molybdenum-95 (

^{95}_{42}Mo) 94.945 8.09
lanthanum-139 (

^{139}_{57}La) 138.955 7.92
proton (

^{1}_{1}p) 1.007
neutron (

^{1}_{0}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 = mc

^{2}, 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 c^{2}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 = mc

^{2}= 1.808 × (1.66 × 10^{–27}) × (3.0 × 10^{8})^{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 × 10^{6})}
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 = ½mv

^{2}= 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 Ã 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

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.

ReplyDeleteSee solution 705 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-143.html

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. ...

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!

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

DeleteFrom 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.

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

ReplyDeleteIf the mass defect is m, then the energy associated with it is

DeleteE = 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.

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

ReplyDeleteThis is the law of conservation of energy.

DeleteAs 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.

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 ?

ReplyDeleteas 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.

DeleteNote 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.

From quest 608) why can't we use F=ma to calculate the resistive force .

ReplyDeleteAnd how do u derive the word equation. Is there any hint in this quest regarding to the equation

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

Delete