第8届上海交通大学大学生物理竞赛试题及答案

60
1
F
2
1
3
0
4m
1, 2 a2
R 2r
R 2 a2 a1 R 2 r 1
1 22
a2 a1 r
1
I 2 4R 2 / 2 8mr 2 I1 mr 2 / 2
M=6m a1
F Ffr 6ma2 , (2) 4mg N1 N 0, (3) Ffr N 2 4ma1, (4) N1 mg, (5) N 2 ma1, (6)
1
a1
g cos 1 sin
g 23
10
2g cos
2g
1 (1 sin )r (2 3)r
1
a2
3g cos 1 sin
3g (2 3)r
9
2
F 23mg cos 1 sin
23g (2 3)
l
l 2
1 2 (a2
a1 )t 2
l 2g cos 1 sin
2g 23
0
a1
g, a2
2g, 1
2g ; r
2m
V0
2 d2 E0
2m dx 2
Asin(kx )
x 0, a
0
ka n , n 1,2,...
nx
n (x)
A sin a
En
n2h 2 , n 1,2... 8ma 2
A 2/a
n (x,t)
2 / a sin n x e iEnt / ,0 x a a
n 1,2,...,
2
n
a
x
nx
* n
1.306kJ / K
:T=373K
0.184 kJ/K
S mC p (T T0 ) 1.122kJ / K T
(b)
1.306kJ/K
1
S1
mC p (T1 T0 ) T1
0.648kJ / K
2
S2
mC p (T2 T1 ) T2
0.561kJ / K
S 1.306 0.648 0.561 0.097kJ / K
dx
a/2
0
( x x)2
a 0
n (x
a)2 2
* n
dx
a2
6
(1
)
12 n 2 2
(3)
P 1/ a
a
x P xdx a / 2
0
( x x)2
a
P (x
a )2 dx
a2
0
2
12
n
(4)
1/2
1
3
(5)
21 2 0
1
0
1 21
3 20
2 / a ( 1 sin 2 x e iE2t / 2a
7
arccos(1/3)
A EC AF AH
0
B
EC F
O
O
AHO AH=AOcos(A),
AH A
F OH
OE=OC,F CE
OF
EC
B
OFC
F
CO
来自百度文库
AF
O’
A
FO’
AF AO’=3FO’, cos(FO’A)=1/3
FO’
EO
FO’C= EOA,
F O’ CE CO
arccos(1/3)
8
S
v sin l
2(U l U r )
) v2
m
2(U l U r ) v 2 m
3
1 2
1y 11 1 uv f
f y
y=kx+b (x<0)
1 11
x x' f
y' x'
y
x
y' y x' (kx b) x'
x
x
y' (k b( 1 1 )x' x' f
y' (k b )x' b f
y' (k b )x' b f
46
N1r cos N 2r sin I1 1, (7) Ffr R N 2R sin N1R cos I 2 2. (8)
Ffr ma1 4ma1 5ma1 (9) 56789
mgr cos
N2ma1r sin
mr 2 1 , (10) 2
10ma1r 2ma1r sin 2mgr cos 4mr 2 2. (11)
3 sin x e iE1t / ) 2a
a
x
nx
* n
dx
0
2a x
1 sin 2 2 x
a0 4
a
A B cos( t)
3 sin 2 x 4a
3 sin 2 x sin x (ei(E1 E2 )t / 4 aa
e i(E1 E2 )t / ) dx
AB
(E2 E1 ) /
(E2 E1 ) / h
6
4.186 kJ /(kg K )
0o C 1.0 kg
100o C
b
0o C 1.0 kg
100o C
a 100o C 50o C
S
12
2 dQ
S S2 S1
( 1T
)
(a) : Q C p (T T0 )
dS dQ mC p dT
T
T
S
373 mC p dT 273 T
mC
p
ln( 373 ) 273
3 cos( ) 4
l 1, m 1: Y1, 1( , )
3 sin( )e i 8
n 2,l 1: R2,1(r)
r
1 ( 1 )3/2 ( r )e 2 1
3 21
1
1 0.053(nm)
500
1
22
2m x 2 V0
i t
(x,t) (x)T (t) ( x)e iEt /
2
[
2 V (x)] E
p z
z
m=0
m=1 m=-1
21
2 3
M
I
I’
4 5
500
31
1900 2
500
Matrix
E2 )t ] sin 2 x cos x exp[ i(E1
a 0 4a a 8a a
aa
aa
E2 )t ]}dx
iC iD exp[i[E1 E2 ]t ] iF exp[ i(E1 E2 )t ]
i(C D cos t F cos t) (F D) sin t CDF
6 xy
(E1 E2 ) /
v
a,b;
a’,b’.
a,a’
b,b’
l sin
l sin
l
l
l0 l sin 2 2 cos2 l
l0x l cos l0 y l sin
x
1
l'
l sin 2
cos 2
2
l
l cos l'x
l' y l sin
x
2
b,b’
a,a’
y
l sin
l sin
10 F
M=6kg
r R=2r m=1kg
1 k
m0 dm kmv dt
(m dm)(v dv) mv udm mgdt
dm v m dv mg
dt
dt
dm kmv dt
kmv2 m dv mg dt
dv g kv2 dt
0
0
u0
2
Ur
v
l
l
l
vr
r
Ul
l
mv sin l mvr sin r
1 mv2 2
Ul
1 2
mvr2
Ur
vr
2
arcsin(
L0
L0
k
L0
E0
L
L
E 0 L0 k S
0
L0
E F
S’
S' Es d S' E0S'
0
0 E0
E’
1
1
E' 20
2 E0
1 E E0 E' 2 E0
1 F S 2 E0 dS
F
1 2
0 E0 2 S
F k(l l0 ) l
l
0 E0 2 S 2k
l0
9
v l
x l sin
x
vv
x
y
y
y
x
x
l
x
E2 E1
a
p
n (i
0
x n )dx
i2 a (1 sin 2 x eiE2t / a 02 a
3 sin x eiE1t / )(1 2 cos 2 x e iE2t /
2a
2a a
3 cos x e iE1t / )dx 2a a
i2
a
{
sin 4 x
3 sin 2 x sin x cos 2 x exp[i(E1
F 23mg, N1 N 2 mg;
Ffr 5mg, N 5mg;
0
0
s
1
1
V
, x 0,x a 0, 0 x a
1
2
n
3
x (x x )2
2
4
5
t=0
1a [0 2 ]
1
3
21 2 0
x
p
6 m
n=2,l=1
n
l
(r, , ) Yl,m ( , )Rn,l (r)
l 1, m 0 : Y1,0 ( , )
2
4 a,T
76
a0 ,T0
a3 a03 T 2 T0 2
76 years
a3 1year
a0
a 17.94Au
1Au 1.49 1011 m
a 2.67 1012 m
8
Au
5
xy
r
e
F Ce r r
F 0rˆ c ˆ r
d s drrˆ rd ˆ W
W F d s 0 cd c
l
W 2 c 0, F F