SỞ GD& ĐT NGHỆ AN KỲ THI CHỌN HỌC SINH GIỎI CẤP TỈNH LỚP 12
NĂM HỌC 2010 - 2011
HƯỚNG DẪN CHẤM ĐỀ THI CHÍNH THỨC
VẬT LÍ LỚP 12 THPT – BẢNG A
Hướng dẫn chấm gồm 05 trang
Câu NỘI DUNG Điểm
1.1a
(1,5đ)
x A c t
ω ϕ
= +
K
rad s
m
ω
= =
x cm Ac cm
t
v A cm
ϕ ϕ π
ϕ
= − = − =
= → →
= = =
x c t cm
π
= +
0,5
0,5
0,5
1.1b
(1,5đ)
!"#$%&'cm ()(#*(+#,-.()(/0(/12#$%&
#*34560
t t T
−
= +
78t
2
#9560#$%&'cm
#*3
!%)(:560#$%&'cm#*
32;<..7=(>/(16?
560@*/-#A(#$%&'cm#*
37=(>/0(
B
C ' CM OM t
ω π π π
= = − =
'
D
t s
π
→ =
E560#$%&'cm#*3#9
' DF
G
D D
t s
π π π
= + =
0,5
0,5
0,5
1.2
(1,5đ)
+ HA((0)24 !IJ(K7#$
%@-404
2'
mg
l m
K
µ
∆ = =
!"(!IJ(K7.</(79(L/(/6(K72-/7
.M#A(#$%&2'00!IJ#9@,)N7:OI
2#A(7)09
#$%P2'00!IJ#9@,.MN7:OI
Q.<#/@M9R#S2OS(M04(T(4/
0U#*>/N#9VW79@V
%
2'
m
mg
x m
K
µ
∆ = =
XW((K7Y(L/#*3G3787>/!IJI
=(L/
)#*32).<:#/@M9R#SS(
[ ]
G
% % % %
C C
m m m m
mvKA K l
mg A A x A x A x A x l
µ
∆
− + =
= + − ∆ + − ∆ + − ∆ + − ∆ − ∆
G
2D' v m s→ =
0,25
0,5
0,25
0,5
2a
(1,5đ)
Z#[(.(K4060()(\2J]4
79
#9
0,5
^'
^
_
_
_
O
C
1
C
2
x
d d
π π
ϕ
λ
∆ = − +
78
C
C
v
cm
f
λ
= = =
!4#9(T(4-/
G
d d k d d k
π π
ϕ π λ
λ
∆ = − + = → − = −
/(\J,
D``D
G
AB d d k AB k
λ
− < − = − < → = −
!,4\J(C60(T(4
!4#9(T(6/
G
d d k d d k
π π
ϕ π λ
λ
∆ = − + = + → − = +
/(4\J
D``D
G
AB d d k AB k
λ
− < − = + < → = −
!,4\J(C60(T(6/
0,25
0,25
0,25
0,25
2b
(1,0đ)
!460/(\J()(/60a04x2([/5(K
#9
d d x− =
Z60/(4\J3,M0P
G G
d d x k x k
λ
λ
− = = + → = +
78
D``Dk = −
E
%
0
C
D F2Cb'
G
C
2Cb'
G
m
x cm
x cm
= + =
= + =
0,25
0,25
0,5
2c
(1,5đ)
YS.4()(\2J]4
79
#9
G G
M
u c d d c t d d mm
π π π π
ω
λ λ
= − + + + +
a60
79
L//(0=#.\2J#90,/60,
AM BM AM BM b+ = + =
c/.(K
79
#9
C
C G G
G2'
C G G
M
M
M
M
b
u c c t
u
u
b
u c c t
π π π π
ω
λ
π π π π
ω
λ
= + + +
→ = −
= + + +
!4560
M M
u mm u mm= → = −
0,5
0,75
0,25
3
(4đ)
de*W
'f Hz=
"
AM AB MB
U U U= +
(3fU
AB
7/.78U
MB
,4\J16(3
g79I271U
AM
7/.U
MB
g79(//*(M0H271U
AM
7/.U
MB
(//*(M0H79<[I271U
AM
S(.U
MB
(/(M0([h/*79[h/*g271(#[(.]U
AB
79
U
MB
#9(i
E24\J(6(3(/(M0([h/*2T(M0H79<[I
deMR.j(3<[2k(3(/(M02H
e
'f Hz=
2"
` C
C MB r L L C L C
U V U U U U U Z Z= = + = → < → <
l"1R*W#,>/)'HzZ
L
RZ
C
M02-#A(Z
L
= Z
C
$[
[/<084(T(42m#9R*W#,>/)'HzIR2)
E21MR9@:#4
deMR.j(3(/(M02H79.k(3<I
e
'f Hz=
2([
C
C
C
C
C
AM r L L
r
AB r L C
U V
U V
U U U U V
U V
U U U U
=
=
= + = → =
=
= + − =
C
' C
' C
' C 2' C
'
'
C
L
Z
C F
Z L H
r
r
π
π
−
= Ω
=
→ = Ω → =
= Ω
= Ω
El"#A(
'f Hz=
%M(h2n
0%
opg,-/Rq#,>/)'Hz
nM0M0P
.j(3(/(M0(
' ` 2' C r L H
π
= Ω =
79.k(3
<
C
' C C F
−
=
0,5
0,25
0,25
0,25
0,25
0,25
0,5
0,5
0,75
0,5
4a
(2đ)
!,/(T(K"/1O
2'
R
f cm
n
= = =
− −
Zr1M()(]779M#9l2(
l d d
d ld lf
f d d
= +
→ − + =
= +
'
Z6.'([02(*L/1[
G G s l lf l f cm∆ = − ≥ → ≥ =
2
"/o%M1
Gd d f cm= = =
E@*/M()(701M
C
C F
C
l cm= + =
−
e:((/6"/1O%72-7:O"/1O()(7Gcm1M
0,25
0,25
0,5
()(lM0-):l
min
=scm2/:("/1O%,01M()(
lR-
C b
C b '
l cm
+
= + + =
−
>/P5:((/6(KM
0 0
F s ' s '' s l l l l cm= − + − = − + − =
0,5
0,5
4b
(2đ)
ct4M
O L
AB A B A B→ →
!(
C
C D
C
d cm d cm= → = =
−
eM()(]"/1O#9x2#4(
D
F
d x d x
d x
= − = −
= −
D
u>/[#9M2,\
J
#97MW78"/1OH79\
J
(v(L/
\
J
23(M\
J
S((L/\J7\
J
S((L/\J2=/
W.4M>/[
F
G G C
C D
f d x
k k k x cm
f d d x
−
= = − = − → = − → =
− − −
!x oCcm79D(
C ` D d cm d cm= − =
2/
CD
D
C D
d d
f cm
d d
−
= = = −
+ − +
0,25
0,5
0,5
0,25
0,5
5(2đ)
w
!
w
o
DUu
AB
→=
0
x
2(
y
D
2
z
` UCqUuuuu
MMBAM
=→====→
z
MB
uTt G
<<
x
0
y
→
U
,D
1
0
x
/
w
I
.
z
,
w
>/I
7
y
/
y
1.
z
,
w
>/D
1
w
( 2(
z
UCqq
=+−
b
!
w
o!G
=+→=
MBAMAB
uuu
s`1,
z
w
.7
y
{
y
w
o!G
w
(
>
+
=
<
+
−=
CC
UC
u
CC
UC
u
Mb
AM
F,L/@:("0
c/o!G
x
(,
z
w
x
w
2,
y
/@
w
(|
z
02(
z
$>/<#9t
"2,
((
tUCCtICCtUCCqCqC
tUCCuCCuCCtUuu
MBAMMBAM
ωωϕωωω
ωω
−=++−⇔−=+→
=+→=+
C C U
I
q q c t a
C C U
C C
i t
q q c t a
C C
ω
ω
ω
ω
ω
ϕ
=
= +
−
+
→ → = →
= +
+
=
(
d
(
AM
MB
C U
q a
u t
C C C C
C U
q a
u t
C C C C
ω
ω
= = +
+
→
= = +
+
0,25
0,25
0,25
0,5
0,25
0,25
!
w
o!Gd
x
0
}
F,
w
(
=
+
=
+
−
C
a
CC
UC
C
a
CC
UC
7
y
d(
( )
+
+
+
=
−
+
=
(
(
CC
UC
t
CC
UC
u
t
CC
UC
u
Mb
AM
ω
ω
"
`
AM MB
u u t≤ ≥ ∀
,1Y:
L/@:("0
0,25
6
(1,5đ))
+ !8(-O(5$[(4>/
51O(K7$$,79.*($#4(K
.O8
C
C
C
C
C
I I I
I I
r R
I
I I
I r R
π
π
π
π
π
+ =
=
+
→
= =
=
+
+ $|(|@V100=(K#T(?)(<#,
7$|(|@V7800=(Ki#T(
dZ6O00=#T(?)(<#,7$2%=0787$|
^7$3"Ws($[(5n2(4S((L/10tt2
(00=#T(?
M I B r c
π α
=
^7$|3#97$|t07$$879;7$$,78(5
n
(4=(L/(v(L/10tt>/451O7$$,
=.*7$|82(00=#T(?
2' M I B r c
π α
= −
d0=(Ki#T((K1/
G
p
M r g r r g r
π ρ α ρ α
= +
ZL/1[(|@V
G 2'
p
M M M r g r B r I I c
π ρ α π α
= + → + = −
→
2' C s
G s C
− +
= =
+ + +
B I I B I
g g
π π π
α
π ρ π π ρ
0,25
0,25
0,25
0,25
0,25
0,25
Chú ý : Nếu học sinh giải cách khác đúng vẫn cho điểm tối đa
A
B
C
1
C
2
M
D
1
D
2
H.2
B
ur
α
n
P
r
n
n