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2013泛珠三角及中华名校物理奥林匹克邀请赛答案


Part-I Q1 (9 points) Rotational inertia of the beam 杆的转动惯量的一般表达式: I (l1 , l2 ) ? ? a) I ? I (0, L) ?
l2 l1

m 3 2 l m / Ldl ? l . 3L l1
2

l

m 2 L 3 C

onservation of angular momentum 角动量守恒: Lmv0 3v I ? ? mL2? ? mLv0 ? ? ? ? 0. 2 I ? L m 4L Energy lost 动能损失:
?E ? 1 2 ?1 2 1 2 2? 2 mv0 ? ? I ? ? mL ? ? ? mv0 / 8 . (3 points) 2 2 ?2 ?
2

1 7 3 1 7 ?L? ? 1 1 ? mL2 b) I ? I (? L, L) ? mL2 , or I ? mL2 ? m ? ? ? ? ? ? mL2 ? 12 4 ? ? 12 16 ? 48 4 4 48 ?
Conservation of momentum and angular momentum 角动量、动量守恒:

?mv0 ? 2mv? , ? 2 ? L ?L? ? I ? ? m ? ? ? ? m v0 4 ?4? ?
Energy lost 动能损失:

6v0 ? , ?? ? . ? ? 5L ? ?v ? v / 2 0 ?

?E ?

2 ? 1 2 ?1 2 1 ?L? 2 2 mv0 ? ? I ? ? m ? ? ? ? mv?2 ? ? mv0 /10 . (3 points) ?2 ? 2 2 ?4? ? ?

1 1 1 c) I ? I (? L, L) ? mL2 2 2 12 Conservation of momentum, angular momentum and energy 角动量、动量、能量守恒:
? ?mv0 ? mv1 ? mv2 ? L L ? ? I ? ? m v2 ? m v0 2 2 ? ?1 2 1 2 1 2 1 2 ? 2 I ? ? 2 mv2 ? 2 mv1 ? 2 mv0 ?
Q2 (10 points) a) m1r1 ? G
m1m2 mm (r2 ? r1 ), m2 r2 ? G 1 2 3 (r1 ? r2 ). (1 point) 3 | r1 ? r2 | | r1 ? r2 |
r2 ? m1 r ? rc . m1 ? m2

12v0 ? ?? ? 5L ? 2v ? ? ?v1 ? 0 5 ? 3v0 ? ?v2 ? 5 ?

(3 points)

b) r1 ?

m2 r ? rc , m1 ? m2

1

?(m1 ? m2 )rc ? 0 ? rc ? 0 ? m1m2 m1 ? m2 ? m1m2 ? m ? m r ? ?G | r |3 r ? r ? ?G | r |3 r 2 ? 1
c) rc ? 0 . (1 point)

(2 points)

d) The equation for r is the same as a uniform circular motion, r 满足的方程与匀速圆周运动满足 的方程一致. Therefore 因此,

G (m1 ? m2 ) G (m1 ? m2 ) ? ? 2a ? ? ? . a2 a 3/2

r (t ) ? a cos(?t ) x0 ? a sin(?t ) y0 . (3 points)
e) m1 ? m2 ? m ,
2? a 3/2 T? 2Gm ? 2T 2Gm ? ? a?? ? 2 ? 4? ?
1/3

? 9 ?108 ? 6.7 ?10?11 ? 4 ?1030 ? ?? ? 4? 2 ? ? . (3 points)

1/3

? 6.116 ?1027

?

?

1/3

? 1.8 ?109 m

Q3 (12 points) a) q ' ? ?QR / x, x ' ? R 2 / x.

Qq ' 1 Q 2 Rx f ?? ? 4?? 0 ( x ? x ') 2 4?? 0 ( x 2 ? R 2 ) 2 1 W ? ? f ( x)dx ?
0 L

L2Q 2 . 8?? 0 R( R 2 ? L2 ) 1

(3 points) 不用积分扣 2 分。

b) Since the outside charge does not create any field inside the conductor sphere, the work done here is no different from the one in a):外电荷对导体内部无影响,所以结果与 a)一样。

L2 e 2 W? . 8?? 0 R( R 2 ? L2 ) 1

(2 points)

c) The static electric field on the charge q2 ? LQ / R outside the conducting sphere is equivalent to a
? field generated by three point charges: q1 ? Q at (0,0) , image charge q2 ? ?q2 R / x at
?? ? (l ? ? R 2 / l , 0) , and q2 ? ?q2 at (0, 0) . Therefore ? 球外 q2 ? LQ / R 受的静电力来自以下三个点电荷: q1 ? Q 在 (0, 0) , 镜像电荷 q2 ? ? q2 R / x 在

?? ? (l ? ? R 2 / l , 0) , 以及 q2 ? ?q2 在 (0, 0) 。

f ( x) ?

? q (q ? q?? ) ? 1 ? q2 q2 1 ? 2 12 2 ? ? ? 2 4? ò0 ? ( x ? x ') x ? 4? ò0

? q 2 xR q ( xq ? Rq2 ) ? ? 22 2 2 ? 2 13 ? ? . (2 points) x ? (x ? R ) ?

And the work done by electric field is then 电场做功 2

?

R2 / L

?

f ( x)dx ?

1 8? ò0

2 ? L2 q2 Lq2 ( Lq2 ? 2 Rq1 ) ? ? 3 2 ? ? R3 ?R ?L R ?

?

1 8? ò0

? L4Q 2 L2 (2 R 2 ? L2 )Q 2 ? 1 L2 ( L4 ? 2 L2 R 2 ? 2 R 4 )Q 2 ? ? 5 ?? 3 2 R5 R 5 ( R 2 ? L2 ) ?R ?R L ? 8? ò0

(3 points)

d) W ? Win c) (2 points) Q4 (9 points) r r cos ? a) ?t ? ? v c

v?

r sin ? cv sin ? ? ?t c ? v cos ?

?
(2 points)

r

v?

0.9 ? 0.707c ? 1.75c . (1 point) 1 ? 0.9 ? 0.707
vs v sin ? ? 0 ? vs ? v cos ? ? 0 vs ? v cos ?

b) v ?

(2 points)

Thus v ? vs / cos ? . (1 point) c) Let the emission angle of the beads be ? ' . Then to ensure the net velocity of the beads is along the Y-direction, we have 令小球的发射角为 ? ' ,为使球的合速度沿 Y-方向,须有
' v sin ? ? vb sin ? ' , vb ? v cos ? ? vb cos ? ' ? v cos ? ? vb 1 ? ? v / vb ? ? v cos ? 2

v
v
? b

r r cos ? . ?t ? ? ? v vb

vb

v?

v sin ? (v cos ? ? vb 1 ? ? v / vb ? )
2

vb 1 ? ? v / vb ?

2

?0

(2 points)

So it will not happen 不会发生. (1 point) Q5 (a) The initial number of mole of air in the tire is 轮胎内原有气体摩尔量 ni ?
PV0 i RTa

The final number of mole of air in the tire is 打完气轮胎内有气体摩尔量 n f ?

Pf V0 RTa

So number of mole of air pumped into the tire is 因 此 , 打 进 轮 胎 的 气 体 摩 尔 量 为

3

n f ? ni ?

?P

f

? Pi ?V0 RTa



Inside the compressor, this amount of air has volume 在压缩机里,此摩尔量的气体的体积为 ′ = ( ? ) ? = 0 (1 point)

The work done by the compressor is hence 压缩机做功 Wc ? PV ' ? ? Pf ? Pi ?V0 c

The internal energy of the gas now in the tire increases by 轮胎内气体内能增加为

?E ? (n f ? ni )CV Ta ? Wc ?

R (n ? n )T ? ? Pf ? Pi ?V0 ? ?1 f i a

(1 point)

The maximum temperature is then 因此这时的温度 ( ? )0 = + = + = [ ? ( ? 1) ] 0 ? 1 The maximum pressure is 气压 = = [ ? ( ? 1) ] = ? ( ? 1) 0

The minimum Pc required is 最小 Pc 必须满足

Pc ? Pmax ? ? Pf ? ? ? ? 1? Pi
(b) The total number of stokes is 打气总次数 N ?
n f ? ni PVp / RTa a ?

(2 points)

?P

f

? Pi ?V0 PVp a

. (1 point)

During the j-th stroke, in the adiabatic compression inside the pump from Pa to Pj 第 j 个斯托克循环绝 热压缩使压强从 Pa 变为 Pj = ′ where V’ is the volume of the air inside the pump after the adiabatic compression. V’:压缩后气体体积 The internal energy of the air at this moment is 此时气体的内能 1 1 ′ = ( ) ? 1 ? 1 The amount of work done to inject this amount of air into the tire is 打气所做的总功 ′ = ( ) Hence the change in internal energy of the air inside the tire during the j-th stroke is 内能的变化
?P 1 ? ? ?U ? PV ' ? PV ' ? PV ' ? Pj ? a j j j ? ?1 ? ?1 ? ? 1 ? Pj ? ?? ? V p (2 points) ? ?
1

1

1

4

On the other hand, 另一方面 1 ?1 1 1 ? = ? ( )= ( ? )0 = 0 ? = ?1 ? 1 ? 1 ? 1 ? 1 So 因此 1? 1 ? = 0 Replacing the finite-difference equation by differential equation and integrate 改用微分形式表示 1? 1 = 0 ∫




1?



= ∫
0



1 0

We have 我们得到 ( ? ) = ? =
1 1 1 1

1 0 ? )
?

1? (

Pmax

1 ? ? ? Pa ? ? Pf ? Pi ? ?1 ? ? Pi ? ? Pi ? Pa ? ? ? ? ? ? ?

(3 points)

5

Part-II Q1 (20 points) a) 2? rB ? ?0 J ? r 2 ? B ? ?0 Jr / 2

?0 I 2 B 2? rdr ? Energy per length is 单位长度能量 W ? ?0 ? 8? 0
1
R 2

Compare with 相比较 W ?

? 1 2 LI ? L ? 0 (H/m) (1 point) 2 4?

b) Newton’s second law gives us the differential equation 牛顿第二定律引出的方程
? ?? ? = ? ? ? ?

(1 point)

c) ? ? ?? ? = ?0 ? ? ? ? ?? ? = ?0 ? ? ?e? E0 (2 points) x? ?1 ? i?? ? m d) The current density is given by 电流密度 J ? ?nex ?
I e 2? n V D ?1 ? i?? ? m , so V ? 2 ? I 2 ?r ?r e2? n ?1 ? i?? ? m D e 2? nE0 ?1 ? i?? ? m

(2 points)

e)

(2 points)
D m (1 point) ? r 2 e2? n

f) The real part of the impedance represents the resistance 阻抗实部 R ?

The imaginary part represents 虚部 ? LI ? g) LI ? R? ? 2.0 ?10?9 ( H ) . LFaraday ? h)

D m? D m , so LI ? 2 2 . (1 point) 2 2 ?r e n ?r e n

?0 D ? 1.0 ?10?7 ?10?6 ? 1.0 ?10?13 ( H ) 4?

(2 points)

? ?1 =

=

= , ?1 = ?1 { +1 + = + ?1 = ?1

6

?i?C ?Vm?1 ? 2Vm ? Vm?1 ? ?
i)

Vm ? 0 (5 points) ? 2 LC

? + ? 2 +

1 2

=0

??
j) Yes 是(1 point)

1 4 LC

ka ? ? ? sin ? 2 ? ?

?1

(2 points)

Q2 (30 points) a) Largest 最大: compressed 压缩; Smallest 最小: stretched 伸展, or vice versa 反之亦然 (1 point) b) : Joule/s → N ? m/s → kg ? m2 s ?3 : Bb kg 2 m4 s?6 ? = ?5 , = ? = ?5 2 4 6 (1 point) (1 point)

c) with 1 + 2 = ? 3 = Kinetic energy: M ? (1 + 2 ) 2 1 2 = 1 2 1 = 2 2 2 2

M 1M 2 , M ? M1 ? M 2 M1 ? M 2

KE ?

1 1 1 MGM 1 2 M1? 2 R12 ? M 2? 2 R2 ? ? (2? )2/3 M (GM )2/3 T ?2/ 3 2 2 2 R 2

(2 points)

d) Neutron binary 双中子星系统

dEk 1 ? ? (2? ) 2/3 M (GM ) 2/3 T ?5/3 , (1 point) dT 3 dEk dEk dT (1 point) LG ? ? dt dT dt

?

5 1 8 5 ? ? dT ? 3G 3 c ?5 M1M 2 ( M1 ? M 2 ) 3 (2? ) 3 T 3 (1 point) dt

7

?

1 5 ? ? dT ? 1.4 ?10?67 ? 4 ?1060 ? (2) 3 (3 ?104 ) 3 ? 1.4 ?10?67 ? 4 ?1060 ? 0.79 ? 3.4 ?10 ?8 ? 1.5 ?10 ?14 (1 point) dt

e) For Sun-Earth system,日-地系统 with ? = 1s

?

5 ? dT ? 1.4 ?10?67 ? 2 ?1030 ? 6 ?1024 (365.24 ? 86400) 3 ? 1.7 ?10?12 ? 3.2 ?10?13 ? 5.4 ?10?25 (1 point) dt

t?

dt 1.8 ?1025 1.8 ?1025 ? 1.8 ?1025 s ? ? y ? 5.7 ?1018 y . (1 point) 6 dT 365.24 ? 86400 3.155 ?10

f) =

2

(1 point)

g) Rs ? GMc ?2 , or Rs c 2G ?1 ? M . (1 point) Also v ? ? R / 2 . (1 point)
c5 ? v ? c5 LG ? Gc M R ? ? 16Gc c G R R v ? 16 ? ? ? 16 G ?c? G
?5 2 4 6 ?5 4 ?2 2 s ?2 6 s 6

(1 point) 数量级正确即可

35 ? 16 ?1040 ?1011 ? 5.9 ?1053W (Order of magnitude only) 6.7

h) 0 = 2, ′ =

?

≈ 1

(2 points)
2

i) (′? )2 = 1?′2 with ′ = 104 light year = 9.46 × 1019 , =

= 1.49 × 103

? ' ? ? Rs / R ' ? 1.5 ?1.5 ?103 ?10?29 / 0.95 ? 2.4 ?10?22 . (3 points)
j) 1 = 2 = 6m, 1 = = 1067Hz (2 points)
1



k) 1st resonance: 第一个共振态
1 1 ?kT ? m? 2 x 2 ? k BT ? x ? ? B 2 ? 2 2 ? m? ?
1/2

? 42 ?1.4 ?10?24 ? ?? ? 3 ? 1.1?10 ?

1/2

1 ? 3.3 ?10?17 m (4 points) 3 2? ?1.07 ?10

l) The change in the length due to gravitational wave should be at least in the same order of magnitude as 10?15 m 重力波所引起的长度改变量至少为10?15 m量级

RL x Rs 9 ?103 2.7 ?1020 ? ?D? s ? ? 2.7 ?1020 m ? ? 2.8 ?104 Ly L D x 3.3 ?10?17 9.5 ?1015

(2 points)

10?6 LG 5.9 ?1053 ?10?6 ? ? 6.9 ?105 (W / m2 ) . We would be toast. (2 points) m) I ? 2 2 40 4? D 4? ? 2.7 ?10
8


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