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BPo 对
些特定
过程有要求吗 ?
g
e
:
次
British Physics Olympiad 2018-19
程必须写求根
公式
mm
BPhO Round 1
Section 1
.
16th November 2018
This question paper must not be taken out of the exam room
Instructions
Time: 1 hour 20 minutes for this section.
Questions: Students may attempt any parts of Section 1, but are not expected to complete all parts.
Working: Working, calculations, explanations and diagrams, properly laid out, must be shown for full
credit. The final answer alone is not sufficient. Writing must be clear.
Marks: A maximum of 50 marks can be awarded for Section 1. There is a total of 87 marks allocated
to the problems of Question 1 which makes up the whole of Section 1.
Instructions: You are allowed any standard exam board data/formula sheet.
Calculators: Any standard calculator may be used, but calculators cannot be programmable and must
not have symbolic algebra capability.
Solutions: Answers and calculations are to be written on loose paper or in examination booklets.
Graph paper and formula sheets should also be made available. Students should ensure that their name
and their school/college are clearly written on each and every answer sheet. Number each question
clearly.
Setting the paper: There are two options for sitting BPhO Round 1:
a. Section 1 and Section 2 may be sat in one session of 2 hours 40 minutes plus 5 minutes reading time
(for Section 2). Section 1 should be collected in after 1 hour 20 minutes and then Section 2 given out.
高
b. Section 1 and Section 2 may be sat in two sessions on separate occasions, with 1 hour 20 minutes
plus 5 minutes reading time allocated for Section 2. If the paper is taken in two sessions on separate
occasions, Section 1 must be collected in after the first session and Section 2 handed out at the
beginning of the second session.
一
二
Q
Important Constants
Constant
Symbol
Value
Speed of light in free space
c
3.00 ⇥ 108 m s
Elementary charge
e
1.60 ⇥ 10
19 C
Planck constant
h
6.63 ⇥ 10
34 J s
Mass of electron
me
9.11 ⇥ 10
31 kg
Mass of proton
mp
1.67 ⇥ 10
27 kg
atomic mass unit
1
u
1.661 ⇥ 1027 kg
Gravitational constant
G
6.67 ⇥ 10
Acceleration of free fall at Earth’s surface
g
9.81 m s
Permittivity of free space
"0
8.85 ⇥ 10
Permeability of free space
µ0
4⇡ ⇥ 10
Avogadro constant
NA
6.02 ⇥ 1023 mol
Mass of Sun
MS
1.99 ⇥ 1030 kg
Radius of Earth
RE
6.37 ⇥ 106 m
(1u is equivalent to 931.5 MeV)
T(K) = T(
C)
+ 273
Volume of a sphere = 43 ⇡r3
For small angles, sin ✓ ⇡ tan ✓ ⇡ ✓
For x ⌧ 1,
(1 + x)n ⇡ 1 + nx
11 m3 kg 1 s 2
2
12 F m 1
7Hm 1
1
a) The Milky Way galaxy has a period of rotation of 240 ⇥ 106 years. The Sun is 26 light years from
the centre of the galaxy. How fast is the Sun moving with respect to the centre of the galaxy, given
in units of m s 1 ?
A light year is the distance that light travels in one year of 365.25 days.
V
T
=
r
w σ
26
=
2
10
×
240
=
x
106
J 65 × 2
×
x
360
π
的
2
器
0…
:
, nx
8
0 …
7
=
,
x
.
365
46
×
,
或 x 24
5 )七
v
=
wr
6
3600
×
X 号 X
2017
= . 3
X
8
10
108 [3]
46
wums
xh
x
”
0
b) A smooth sphere of radius 6.0 cm is suspended from a thread of length 9.0 cm attached to a smooth
wall as shown in shown in Figure 1. If the mass of the sphere is 0.5 kg, calculate the tension, T ,
in the thread.
a
t
-62
-
Txsn
Jy
-
瓷
sm θ=
θ=
司
:
5
0 .
9
T
T
=
0
.
0
↓
g
5 × 9 81 ÷
.
义
~
1
n
a
5
Figure g1
0
弘
.
[3]
B 5 N
c) The displacement of an object is determined by the following function:
s = 2t3
9t2 + 12t + 4
where s is the displacement in metres, and t the time elapsed in seconds. Determine
r|
=
'
s
=
urtree
6E
3
-
t
18 E
5t
-
1
6t ε ,
+ 12
8t
(i) the times when the object comes to rest,
vao
2
t
412
+ 2
(ii) the time when the acceleration is zero,
0
≈
(iii) the object’s velocity when its acceleration is zero,
20
(iv) the object’s accelerations when its velocity is zero.
tz = 2
1 ss
=
i
L 5
t
When
U = Gx ) 5 - 18 x 1 5 ti 2
i 5
omg
=
ii
a
=
whon
七
r
'
=
12
E
-
18
irs
a 20
是
180
t
=
1
.
5
when t
wthen
=
t
.
.
G
1
=
2
G
12
=
=
-
.
18
12 x
2
=
-
s
18
-
6
-
=
ons
6
-
[4]
-
2
ams
-
z
d) The distance in which a train can be stopped is given by:
s = av + bv 2
where s is the stopping distance, v the initial velocity, and a and b are constants. When moving at
40 km hr 1 , the train can be stopped in 100 m, and at 80 km hr 1 it can be stopped in 280 m.
Find the o
greatest speed such that the train can be stopped in 500 m.
40
lkmhor
sa
kuhnt
r
-
0
"
=
g
mst
算
号
wos
t b
280
Dx
ga
2
③ ③
-
200
+
2 X
a
+
号
b
第
ga
13
“
)
=
100
=
D
20000
b =
8
筑江
②
=20 D
和哈筑
1
2了
5
+
号
L
,
v
500
-
=
[4]
0
戚x
☆∞ )
“
So
≥
噩
☆
31
=
U
1
>
Thenefome
,
二
门
一
门
己
比
二
一
Question 1
.
V2
8
=
-
48
.
5
Va
the
velocity
is
31 8 mst
,
4o
222
~
⑪a
200
θ
:
,
品
e) Two planes set out at the same time from an aerodrome. The first flies north at 360 km h 1 , the
second south-east at 300 km h 1 . After 40 minutes they both turn and fly towards each other.
Calculate
(i) the bearing, and
beaiing
he
angle
τ
”
Cls
E→ ,
的o
)
20
o
㉕
c
=
2
zhcos 0
fros
1350
切
Wly
L
)
t
=
毙
☆
:
62
α~
[7]
→
Ekabefore )
定
=
4
:
e,
…
szoirxwo
q
σ
2
(i) Assuming the mass of the neutron is equal to half that of the deuteron, find the ratio of the
final speed of the deuteron to the initial speed of the∞
neutron.
2
空
当成当G
.
b
引
(ii) What percentage of the initial kinetic energy is transferred to the deuteron?
,
年 m Gv
=
京
+ 21 B
n V
2
{ mbo
1
=
的r
化了了
?
“
^
:
µ mv ( Gz+
2
2
B
l
e
)
(iii) How many such collisions would be needed to slow the neutron down from 10 MeV to
”
0.01 eV?
≤
产
( )
106
÷λ
宁
⑮
部
189 .
9
p
[6]
:
下
≤
{ mP
^
! … … nig
mxB
0
—
B
点
(
63
G1
lowf
τ
=
4 bt 4 b 42
-
"
-
-
0
43
^以
1
-
B
之b
:
=
==
=
LGLN)
1
\ ×
m
0元
x
01
.
I02
1
-
88
-
0
ba
=
登
10
六
9
∞
,
多
≥
it
0
-
9 43
,
reswires
to
tGlaes
g) A uniform chain of mass per unit length, µ, is suspended from one end above a table, with the
oratio
lower end just touching the surface. The chain is released, falls and comes to rest on the table
without bouncing.
obe
sre,
定
G
a
系
,
(i) Determine an expression, in terms of µ and the gravitational field strength g, for the reaction
force exerted by the table on the chain as a function of time, t.
Hint: you might consider F in the form F = mt v.
(ii) In terms of the total weight W of the chain, what is the maximum reaction force exerted by
the table, and at what time during the fall does this occur?
iL
v
at
:
= gt so
=
F
[6]
fgtculg
h) A small particle of mass m can slide without friction round the inside of a cylindrical hole of
radius r, in a rectangular shaped object of mass M . The rectangular object is held between rigid
walls by small wheels so that it can slide up and down without friction, as shown in Figure 2. If
the small particle m is initially at rest at the bottom of the cylindrical hole, and is then given an
impulse to give it a speed v, what is the minimum speed v needed to just lift the rectangular mass
M off the ground?
”
mg +
Mg
m
=
卡
rdmg
==
当
mgzr +
mv
v
=
4
-
gr
m
mxUg
-
+
Mg
rugrtgr
心
心
二
二
立
一
比
」
十
立
弓
立
《二
二
十
比
十
「、
一
v
=
=
ng
=
2
mg
M
+
4
r
=
mv 2
m
gryngr
.4
Figure 2
mg
so
[5]
ve
gr 筑 gr
+
2
-
.
-
2z
1
年 haz
Eicafters
u
x
20 s
theantrai
Thafm
t
:
toonth
α
sm
40
:
=
45
f) A neutron moving through heavy water strikes an isolated and stationary deuteron (the nucleus of
an isotope of hydrogen) head-on in an elastic collision.
街H
m
iL
=
<
f
o
P
→
→
soa
24
'
9
-
(ii) the distance
of the meeting point from the aerodrome.
= 30 x 皆 =
× 皆 = 240 km
5 =
3
十
「
"
s
s 3 ix
广
言
方
一一
一
一一
门
工
i) Two resistors and two cells are connected in the circuit shown in Figure 3. One cell has an
e.m.f. of 2.0 V and an internal resistance of 1.0 ⌦, the other an e.m.f. of 1.5 V and an internal
resistance of 0.5 ⌦. The resistors are connected in series and the point between them is at earth,
i.e. zero potential. Calculate
DV
(i) the current through the cells,
=
IR
=
0
.
368
x
8
(ii) the potential difference across each cell, and
(iii) the potential, relative to earth, at points A and B.
1r
D
R
7
=
VJ
=
资
ii
Vs
③
0
2 +
1
IR
=
0
Z及
=
54 5 f
.
5
.
0.
:
0
-
1
368 x
o
.
5
.
②
368 H
.
0
=
.
①
9 5r
=
.
368 x
.
3
3 5
=
器
:
=
U
1+
D
.
5
368 U
=
0
Figure 3
186 L
.
[4]
j) A thick-bottomed, cylindrical glass beaker is placed on a bench. Water and oil are poured into the
beaker and form discrete layers, as shown in Figure 4. The bottom of the beaker is 1.8cm thick,
the water is 1.2 cm deep, and the oil layer is 0.8 cm deep.
(i) Draw a diagram showing the path of a ray at a small angle to the normal, travelling from the
underside of the beaker and being refracted through the layers.
(ii) Assuming the angles of deviation of the ray are small, calculate the apparent vertical
displacement of the lab bench when viewed from above.
The refractive indices are 1.5, 1.3 and 1.1 for the glass, water and oil respectively.
(
"i
1
=
1 5 sin
.
1
I
.
θ
g
2+
=
1
1
.
SPOWE
3
.
.
)
sim
Do
=
840 8
.
=
.
3
38
observer's
eye
.
sin θ
w
1
.
1
soh
lxsin
.
θ0
8
!
θ
3.
oil
water
glass
,
Figure 4
[7]
k) A person might reasonably expect to jump a height of 1 m on Earth. On a planet with a density
two thirds that of Earth, and radius twice that of the Earth, to what height might the person jump?
Assume that they supply the same energy to make the jump
,
F
=
的
G
箭
=
Px
→
=
P
π2
r
G
“
F
F
V
器
3
:
哥
:
6P
6
0
=
3
F
[4]
α
m
301
告
≈
年
m
,
=
2.
95
…
m
=
号P 2
PV
bg
=
4
ogbh
“
F
0
=
g
4b
4b
,
p3
0号
gu
Pg gbxabih
正 egb h
ogb 2 . 号 ebh
g
b
=
.
bih + p
4
2
4
β
,
Figure 5
h
⑤
+
红
=
2
λ
,
b
-
+
4 x
可
酱
接
2
SoP
=
.
[4]
copper
,
m) A volume of 80 cm3 of water in a copper calorimeter of mass 150 g takes 12 minutes to cool from
40 C to 15 C in a cold room. The same volume of ethanol of density 0.8 g cm 3 takes 8 minutes
to cool also from 40 C to 15 C in the same calorimeter in the same circumstances. Calculate the
specific heat capacity of ethanol.
The specific heat capacity of copper = 400 J kg 1 C 1 and of water = 4200 J kg 1 C 1 .
The density of water, ⇢w = 1.0 g cm 3 .
13 75 w
= 毙器
08
=
m
OV
[5]
SO ×
80
=
~
E
=
0
.
1
=
08 × 400
xL
40 -
14
g
=
0
kg
.
) + 150× L 0 -
3
x
400 xL
=
P
40 -
{
15)
:
.
=
9900
Ee
=
80
13
× 0
.
.
75
8
x
-
×
10
8 xt
3
×
0
6600
-
CX
,
40 -
15 )
n) A steel girder is planted securely between two sides of a ravine in order to provide a bridge. The
total cross-sectional area of the girder is 30 cm2 , and the length of the girder is 4.0 m. Installed
at a temperature of 5 C, the temperature now rises to 20 C. Calculate the force exerted by the
计时了
Ih 2
girder due to the change in temperature, assuming the ends do not move.
( )
Young modulus of steel = 2.0 ⇥ 1011 Pa
差不多写到
Linear expansivity of steel (fractional expansion per unit temperature rise) = 1.2 ⇥ 10 7 C 1 at
写完包括前
6
5 C.
20 x
空的题
. 2 x 103
1 8 xc
1
[4]
,
o
mim
n
,
so
JF
=
~
-
=
0
.
*
“
A
x …
'
=
2x 0
"
0
xsox .
*
o) A narrow beam of monochromatic light falls on a diffraction grating of 1200 lines mm 1 , and
two diffracted beams of successive orders are observed at 14 and 73 to the normal, both of them
on the same side of the normal. The incident beam of light is not along the normal to the grating.
(i) Sketch a diagram to show the path difference between rays passing through adjacent slits, for
a ray incident on the diffraction grating at angle ✓1 , and for the corresponding ray emerging
from the grating at angle ✓2 , with respect to the normal.
(ii) Derive an equation relating the angles ✓1 and ✓2 to the order of diffraction, n, and the
wavelength, .
Determine:
(iii) The wavelength of the light used.
(iv) The angle of incidence of the beam on the grating.
(v) The angle of diffraction of a third transmitted beam.
面
4
二
二
二
一
一
一一一
二
l) A pond containing water of density ⇢ is covered to a depth b by oil of density 23 ⇢. A long wood
block of square cross section 4b ⇥ 4b, with the same density as the oil, floats in the pond, as shown
in Figure 5. What fraction of the wood block is immersed below the surface oil level?
[6]
+
160 × 10交
40 × L 40
=
-
15 )
6000
Ceb
190
JkgYa
」
σσ
p) Two identical spherical glass containers are joined by a narrow tube, whose volume is negligible
compared to the spheres. The spheres contain
θ air at 100 C. One of the spheres is then heated by
50 C whilst the other is cooled by 50 C. This produces a small change in pressure, from Pinitial
to Pfinal , of the air in the system. What common temperature of the two spheres could produce
the same final pressure Pfinal ?
[4]
prn算等
o Pinitial
q) A simple pendulum consists of a small mass on the end of a light, inextensible string, as shown in
Figure 6. It swings from an initial angle ✓ = 14 , for which it would have a period ?
T0 , but it hits
a wall elastically, which is at angle = 7 to the vertical. What is the new period of oscillation in
terms of T0 ?
:
BfGmal
.
are small angles such that sin ✓ ⇡ ✓ and sin
(✓,
⇡ ).
下
wall
.
 
Figure 6
[4]
r) Four charges are placed at the corners of a square of side 10 cm, as shown in Figure 7.
A = +10 ⇥ 10
9
B = +8 ⇥ 10
9
6 ⇥ 10
9
C=
D=
12 ⇥ 10
C
C
9
C
C
(i) Calculate the magnitude and direction of the electric field strength at the centre of the square.
(ii) Calculate the work done taking an electron from the centre to the mid-point of side CD.
loc
如
的
量
A
~
2
x 10
①
E
-
=
5
x 10
2
B
瓷
=
…
8
:
:
99
.
O
x 09
xtr + s
…
C
D
Figure 7
[7]
END OF SECTION 1
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