# Physics 9702 Doubts | Help Page 203

__Question 980: [Radioactivity]__
The grid shows a number of nuclides
arranged according to the number of protons and the number of neutrons in each.

A nucleus of the nuclide

^{8}_{3}Li decays by emitting a β-particle.
What is the resulting nuclide?

**Reference:**

*Past Exam Paper – June 2014 Paper 11 Q40*

__Solution 980:__**Answer: A.**

Beta particle is

^{0}_{-1}β.^{8}

_{3}Li - - - >

^{A}

_{Z}X +

^{0}

_{-1}β

Let the resulting nuclide of the

^{8}_{3}Li nucleus β-decay be^{A}_{Z}X.
The total nucleon number and proton
number should be equal on both sides of the equation.

Nucleon number, A = 8 – 0 = 8

Proton number, Z = 3 – (-1) = 4.

Number of neutrons, N = 8 – 4 = 4

__Question 981: [Nuclear Physics > Fission]__**(a)**The variation with nucleon number A of the binding energy per nucleon BE of nuclei is shown in Fig. 8.1.

On Fig. 8.1, mark the approximate
positions of

(i) iron-56 (label this point Fe),

(ii) zirconium-97 (label this point
Zr),

(iii) hydrogen-2 (label this point
H).

**(b)**(i) State what is meant by

*nuclear fission*.

(ii) By reference to Fig. 8.1,
explain how fission is energetically possible.

**Reference:**

*Past Exam Paper – June 2011 Paper 41 Q8*

__Solution 981:__**(a)**

(i) Fe should be shown near the peak

{The numbers (e.g. 56, 97,
2) given the nucleon number of the elements. Thus, these values can be used to
determine the position along the x-axis.

Iron-56 is among the most
stable nuclei (the most stable is said to be nickel-62). So, iron-56 needs to
be near the peak. Thus, this position represents A = 56 along the x-axis.}

(ii) Zr should be shown about
half-way along the plateau

{We don’t know much about
this stability here – but we use its nucleon number, relatively to Fe, to
locate a position for Zr.}

(iii) H should be shown at less than
0.4 of maximum height

{Hydrogen should be should
close to the left end of the graph.}

**(b)**

(i) In nuclear fission, a heavy /
large nucleus breaks up / splits into two nuclei / fragments of approximately
equal mass

(ii)

{Since B

_{E}is the binding energy per nucleon,}
Binding energy of a nucleus = B

_{E}× A
The binding energy of the parent
nucleus is less than the sum of binding energies of the fragments

__Question 982: [Work, Energy and Power]__
A motor is used to move bricks
vertically upwards, as shown in Fig. 5.1.

Fig. 5.1

The bricks start from rest and
accelerate for 2.0 s. The bricks then travel at a constant speed of 0.64 m s

^{−1}for 25 s. Finally the bricks are brought to rest in a further 3.0 s.
The total mass of the bricks is 25 kg.

**(a)**Determine the change in kinetic energy of the bricks

(i) in the first 2.0 s,

(ii) in the next 25 s,

(iii) in the final 3.0 s.

**(b)**The bricks are in a container. The weight of the container and bricks is 350 N.

Calculate, for the lifting of the
bricks and container when travelling at constant speed,

(i) the gain in potential energy,

(ii) the power required.

**Reference:**

*Past Exam Paper – November 2014 Paper 21 Q5*

__Solution 982:__**(a)**

(i) Change in kinetic energy = ½ mv

^{2}= 0.5 × 25 × (0.64)^{2}= 5.1(2) J
{The bricks start from
rest (speed = 0, initial KE = ½ mv

^{2}= 0) and accelerate for 2.0s until the speed finally becomes 0.64 m s^{−1}.}
(ii) Change in kinetic energy = Zero

{In the next 25s, the
bricks travel at constant speed. Since there is no change in speed, the

**change**in KE is zero.}
(iii) Change in kinetic energy = (–)
5.1(2) J

{In the final 3.0s, the
bricks are brought to rest – that is the speed changes from 0.64 m s

^{−1}to zero. The magnitude of the change in KE would thus be the same as in (i). However, the change is a decrease in KE here. The negative sign accounts for this.}**(b)**

(i) PE = mgh = 350 × 0.64 × 25 =
5600 J

{Total weight, mg = 350 N.

The constant motion occurs
at a speed of 0.64 m s

^{−1}for 25 s. The distance moved upwards during this time = height gained, h = Speed × Time = (0.64 × 25) m.
(ii)

Power P = Fv OR = gain in PE / t, EP / t OR = work done / t, W / t

P = 350 × 0.64 OR 5600 / 25

__Question 983: [Kinematics]__
A moving body undergoes uniform
acceleration while travelling in a straight line between points X, Y and Z. The
distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and from
Y to Z is 6.0 s.

What is the acceleration of the
body?

A 0.37 m s

^{–2}B 0.49 m s^{–2}C 0.56 m s^{–2}D 1.1 m s^{–2}**Reference:**

*Past Exam Paper – November 2010 Paper 12 Q8*

__Solution 983:__**Answer: A.**

The time taken to travel the same
distances of XY and YZ are different. This means that the same of the moving
body is greater during YZ than during XY.

Since acceleration is uniform we can
use the equations of motion. There is a single value for the acceleration a
from X to Z.

2 sets of data are given, so we may
need to solve the equations simultaneously for the acceleration a.

__Consider the 1__

^{st}stage: XY.
Initial velocity u = ???, acceleration
a = needs to be found, s = distance XY = 40m and time t = 12s.

s = ut + ½ at

^{2}
40 = 12u + ½ a(12)

^{2}…………….. (1)__Now consider both stages together: XZ.__

(We cannot consider only
YZ only because the initial speed would not be the same as the ‘u’ above. We
need to use quantities that relates both situation as much as possible.)

Initial velocity = u, acceleration =
a, s = distance XZ = 40 + 40 = 80 and total time t = 12 + 6 = 18s.

80 = 18u + ½ a(18)

^{2}…………….. (2)
The initial velocity is not required
and can easily be eliminated.

From equation (1), u = (10 – 18a) /
3

Replacing ‘u’ in equation (2) and
solving gives a = 0.37ms

^{-2}
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