AS CHALLENGE PAPER 2016

SOLUTIONS

Marking

The mark scheme is prescriptive, but markers must make some allowances for alternative
answers. It is not possible to provide notes of alternative solutions which students devise since
there is no opportunity to mark a selection of students’ work before final publication. Hence
alternative valid physics should be given full credit. If in doubt, email the BPhO office.

A positive view should be taken for awarding marks where good physics ideas are rewarded.

A value quoted at the end of a section must have the units included. Candidates lose a mark
the first time that they fail to include a unit, but not on subsequent occasions except where it
is a specific part of the question.

The paper is not a test of significant figures. Significant figures are related to the number of
figures given in the question. A single mark is lost the first time that there is a gross
inconsistency (more than 2 sf out) in the final answer to a question. Almost all the answers
can be given correctly to 2 sf.

ecf: this is allowed in numerical sections provided that unreasonable answers are not being
obtained.

owtte: “or words to that effect” – is the key physics idea present and used?

1
Section A: Multiple Choice
Question 1. C
Question 2. D
Question 3. D
Question 4. C
Question 5. C
Question 6. A

There is 1 mark for each correct answer.

Maximum 6 marks

Multiple Choice Solutions

Qu. 1 Energy change = work done = force x displacement = MLT ×L=M

Qu. 2 Since the liquid is incompressible, what goes in of volume each second must go through the

smaller pipe. So the flow rate remains at 6 m s

Qu. 3 However, the fluid must speed up in order to get through the small tube. The cross sectional
area is th of the larger tube area, so the speed must be 9 times greater.

cos 45° =
Qu. 4 (5 & 6) Resolving the forces on the join in the string vertically

sin 45° =
And horizontally

And dividing
tan 45° =
And with tan 45° = 1 then =

2
Section B: Written Answers
Question 7.

Area = (rectangle) ! +
a) Distance (or displacement) = area under graph

(triangle) ½ base x height or × ! × # [= (& − )/!]! = #!

# × ! is the height, the gradient times the base.
Must be clear how the terms are obtained.
[3]
b) This is a subtlety of constant acceleration. The falling object increases in speed by the

acceleration, , is the same (acceleration is change of speed ÷ time). So the height ℎ

same amount as the rising object loses speed, as the travel time is the same and the

is covered in time ! at speed &. Hence ℎ = &!.

The argument must be quite clear and without obfuscation as the answer is given.
OR by calculation: [convention ℎ ↑, / ↓, ↓, & ↑, so (ℎ − /) ↑, etc.]
&3 , &4 are the speeds of the top/bottom particles when they meet
&3 = ! and &4 = & − !
So &3 + &4 = &
ℎ = (& − &4 )! + &3 !
Hence ℎ = &!
Alternatively
A starting point – rising particle: ℎ − / = &! − !
Reasoning – and with /= !
ℎ − / = &! − /
ℎ = &! (no mark for the result)
[3]
c) Falling object /= !
And ℎ = &!
6
/=
7
Therefore
[2]
d)
20 ms
one crossing ! axis and
Parallel lines with

! 5s
one through origin
(may be inverted)
0

! (5 8) and below the time

Two points at the same

30 ms
axis.

&3 = − ! = −50 ms
50 ms
&4 = & − ! = −30 ms

[3]
Total 11

3
Question 8.

Diagram

<
E

Spin of the Earth

Informative diagram to illustrate the angle between Sun and Venus

=.>
sin < = .=
or < = 46°
idea that planets remain in position whilst Earth spins in a few hours
?@°
time before sunrise = @=° × 24 = 3 hours 4 minutes ≈ 3 hours
[4]
Total 4

Question 9.

Slower ball dropped

from a lower height
Faster ball dropped
from a greater height
so travelling faster
Diagram

The balls fall with constant acceleration from two different heights.
When they bounce off the 45o slope, their speeds then remain constant
As there is no component of gravity/no force on them
The elastic collisions maintain their speeds that they gained in the fall
Travelling at different constant speeds, the faster ball (dropped from the greater
height) will collide with the slower ball.

[5]

4
The calculation of the horizontal distance 8 travelled is N.B.(ℎ, ℎE ) are the starting heights
Note that this calculation is not required.

vertically ℎ = !I and horizontally 8 = &!K (constant speed x time)

6 Q
!K + !I = !3L3MN = O +
P 7
So that
6E Q
!3L3MN = O +
P 7E
Similarly
& =2 ℎ
& E = 2 ℎ′
And with

2ℎ 8 2ℎ′ 8
S + =S +
T2 ℎ T2 ℎ′

6U VQ 6VQ
Multiply through by T2 =
√6E √6
to give

Rearranging gives 8 = 2√ℎℎ′

Total 5

Question 10.
a) Since the resistivity is much less than copper, the cube will need to be much larger.

N
1.68 × 10 Y = 0.53
N
Z = 3.2 × 10> m
[2]

b) A semiconductor has many fewer free electrons (by a factor of 108 or more)
[1]

ℰ
[=
]V]^
c)
[1]

`
d) _=
a

` M
Z = _2b# = 2b# = cb a
a
[2]

fN f M fM
de = = cπ = c
jZ
e) (use of d = k )
ga ga a ai

n ℰ
l = mL [ = mL
[2]

` a o]Vpq s
f) three terms
ri
Correct denominator term
[2]
mt 2u j#c = mt 2j#c
1 ℰu3 ℰu2
g) For d≈0 l= [1]

5
 ℰ
dv ≪ d l = mL
a]
h)
i.e. Independent of the geometry of the coil, but does depend on u [1]

i)

dv ≪ d Small d ≈ 0
l

u
graphs
Labels to state which is which
[2]

Total 14
Question 11.
Rounding errors are not penalized
Any order is allowed
Data really to 2 sf but avoiding multiple rounding errors

5 W electrical → 62.5 W thermal

Thermal power with 5.5 MeV per decay and 1.6 MeV = 1.6 × 10
….is 8.8 × 10
J

Results in 7.1 × 10 decays per second

J per decay

2.8 x 109 seconds

{| 3}~
Number of decaying atoms (molecules) _L =
Half life in seconds used in calculation

=.@
= 2.8 x 1023

n
No of moles calculated n| = 0.47 moles
•

Calculation of mass of PuO2: 238 + 16 + 16 = 270 g/mole

So 127 g of pure PuO2
80% factor as not all Pu is radioactive (wrong isotope) 159 g

Density calculation 15.9 cm3 ≈ 16 cm3

Individual intermediate values may not be calculated, but if the subsequent value is obtained
then credit should be given.
ECF allowed. If the right calculation given with the wrong numbers then credit given unless
numbers are silly (e.g. 1000 atoms used, say)
[10]
Total 10
END OF SOLUTIONS