奥林匹克物理竞赛之力学解题方法
DESCRIPTION
[email protected]. 奥林匹克物理竞赛之力学解题方法. 整体法 隔离法 等效法 对称法 图像法 类比法 递推法 微元法 极限法. 一、整体法. 1 .方法简介 : 从整体或全过程去把握物理现象的本质和规律的方法。 层次深、理论性强,运用价值高。 变繁为简、变难为易。. 2 .赛题精讲. 例 1 . 如图所示,人和车的质量分别为 m 和 M ,人用水平力 F 拉绳子,图中两端绳子均处于水平方向,不计滑轮质量及摩擦,若人和车保持相对静止,且水平地面是光滑的,则车的加速度为多少?. 点评:五说题意. - PowerPoint PPT PresentationTRANSCRIPT
PowerPoint Presentation2
ANT BNT
CNT DNT
[email protected]
0.61N
1.?
81.0kg20m5.0m1.0sg=10m/s2
A10N·s B20N·s C30N·s D40N·s
[email protected]
M
1ABm1m2ABμ1Aμ2FABBF
T
[email protected]
[email protected]
B
C
[email protected]
[email protected]
.m=150gABCD123∠ABC=∠BCD=1200ABAI=4.2NsC
1.
2.
3.
[email protected]
CDCv
1
[email protected]
[email protected]
7.ABCD Aq R=0.2mCD2ABABABABC g10m/s2
(1) A
x2d-Q
P
2
[email protected]
[email protected]
[email protected]
“”
[email protected]
6OABCOAxOCyPxyQx1y1OA=BC=25dmAB=OC=12dmPx=10dmy=8dmQQABBCCOOAP
[email protected]
1.
P1108P2108P31032P46032
Q
QP4
D
2.
[email protected]
DP3
[email protected]
2
[email protected]
2.AL=5mM=5kgAs=3mBm=2kgABμ1=0.1ABμ2=0.2FABBg=10m/s2
1F
2F
Ba1t1s1a2t2s2V
[email protected]
F’ATAAa’1 V’a’2Tt1-T
[email protected]
[email protected]
4.ABCAB=BCABa1BCa2B
Aa1> a2 Ba1= a2
Ca1< a2 D
[email protected]
t
v
VA
VC
VB
0
[email protected]
1.56
1.51
1.50
1.53
1.54
1.59
0.20
0.25
0.30
0.35
0.40
0.45
[email protected]
210.050.53(0.191.09)“ ”1
2
[email protected]
1
1
1
1
[email protected]
1AOBα=10BOAAv=5m/sAOθ=600OA=10mOBOA
1AOAA
1°29°28°…0°0°A60
2
OOB1°2°3°……ABBC′C′D′D′E′……BCCDDE……. 0°90°60°RtAMOAM
[email protected]
[email protected]
[email protected]
tLst.1/ka
[email protected]
415°2.5km/h4.0km/h2.0km/h
[email protected]
2
1at2a2t3a…ntn+1a
1nt
2nt
1.
Δta1a2a3…an
[email protected]
[email protected]
6AiBi(i=12…6mA1A6A1B1A6B6
12…n12:
23
1
[email protected]
9θ=450L=2.00mμ=0.2011
[email protected]
12m=2kgM=3kgμ=0.50v0=3m/s
v1v2v3…vn…s1s2s3…sn…. (n-1)n
[email protected]
M
Mnn
[email protected]
140%
2
n
n
[email protected]
1.
2.
[email protected]
1.
v’
u
1212121
[email protected]
[email protected]
[email protected]
4LMm
1
L
1.
2.
[email protected]
α
[email protected]
12OO′ABA′B′O′A′OO′v∠ A′ O′O=αO
tO′C′OCC′CO′OC′D′CD
αEC′≈ED′EC≈ED
OD+O′D′≈OO′CC′
OO′CC′=O′C′
[email protected]
mg=kx
[email protected]
[email protected]
[email protected]
5mlMμμ’μ’ >> μ L
x
8m=50kgμ=0.3θ=300
1Fh1=2L/5LF
2h2=4L/5
[email protected]
[email protected]
1
1μ
[email protected]
BμO
O
OO
1AA
[email protected]
m
M
F
a
ANT BNT
CNT DNT
[email protected]
0.61N
1.?
81.0kg20m5.0m1.0sg=10m/s2
A10N·s B20N·s C30N·s D40N·s
[email protected]
M
1ABm1m2ABμ1Aμ2FABBF
T
[email protected]
[email protected]
B
C
[email protected]
[email protected]
.m=150gABCD123∠ABC=∠BCD=1200ABAI=4.2NsC
1.
2.
3.
[email protected]
CDCv
1
[email protected]
[email protected]
7.ABCD Aq R=0.2mCD2ABABABABC g10m/s2
(1) A
x2d-Q
P
2
[email protected]
[email protected]
[email protected]
“”
[email protected]
6OABCOAxOCyPxyQx1y1OA=BC=25dmAB=OC=12dmPx=10dmy=8dmQQABBCCOOAP
[email protected]
1.
P1108P2108P31032P46032
Q
QP4
D
2.
[email protected]
DP3
[email protected]
2
[email protected]
2.AL=5mM=5kgAs=3mBm=2kgABμ1=0.1ABμ2=0.2FABBg=10m/s2
1F
2F
Ba1t1s1a2t2s2V
[email protected]
F’ATAAa’1 V’a’2Tt1-T
[email protected]
[email protected]
4.ABCAB=BCABa1BCa2B
Aa1> a2 Ba1= a2
Ca1< a2 D
[email protected]
t
v
VA
VC
VB
0
[email protected]
1.56
1.51
1.50
1.53
1.54
1.59
0.20
0.25
0.30
0.35
0.40
0.45
[email protected]
210.050.53(0.191.09)“ ”1
2
[email protected]
1
1
1
1
[email protected]
1AOBα=10BOAAv=5m/sAOθ=600OA=10mOBOA
1AOAA
1°29°28°…0°0°A60
2
OOB1°2°3°……ABBC′C′D′D′E′……BCCDDE……. 0°90°60°RtAMOAM
[email protected]
[email protected]
[email protected]
tLst.1/ka
[email protected]
415°2.5km/h4.0km/h2.0km/h
[email protected]
2
1at2a2t3a…ntn+1a
1nt
2nt
1.
Δta1a2a3…an
[email protected]
[email protected]
6AiBi(i=12…6mA1A6A1B1A6B6
12…n12:
23
1
[email protected]
9θ=450L=2.00mμ=0.2011
[email protected]
12m=2kgM=3kgμ=0.50v0=3m/s
v1v2v3…vn…s1s2s3…sn…. (n-1)n
[email protected]
M
Mnn
[email protected]
140%
2
n
n
[email protected]
1.
2.
[email protected]
1.
v’
u
1212121
[email protected]
[email protected]
[email protected]
4LMm
1
L
1.
2.
[email protected]
α
[email protected]
12OO′ABA′B′O′A′OO′v∠ A′ O′O=αO
tO′C′OCC′CO′OC′D′CD
αEC′≈ED′EC≈ED
OD+O′D′≈OO′CC′
OO′CC′=O′C′
[email protected]
mg=kx
[email protected]
[email protected]
[email protected]
5mlMμμ’μ’ >> μ L
x
8m=50kgμ=0.3θ=300
1Fh1=2L/5LF
2h2=4L/5
[email protected]
[email protected]
1
1μ
[email protected]
BμO
O
OO
1AA
[email protected]
m
M
F
a