完整,高中物理3動量好題集錦

p

=

mvI

=

m

m

H

g

Hg

mg

mg

α

mg

α

I

I

m

m

,I

,I

m

I=Δp

F

m

Δ

p=F=mg

Δp

mg

mg

F

, F

mg

m

FI+

I=0,

I=

>>

F>>mg

M

m

Fμ

m

M

/

g

Mg

g

MgM

m

/μgM

m

M

M

m

,

M

m

g

M

m

m

h

hΔ

g=10m/sF

mgΔ=0

Fm

m

m

m

m

m

m

m

p

+

p

=

p/

+

p/ Δp+Δp=0Δp=

Δp

m

m

1/ 2/ m

m

和

m

mm

m

,

mm

m

m

m

m m

m

m

m

m

m

m

m

M

m

mv

M

m

M

mg

M

mg

Mv

m M

m

m/s

m/s

m/s

A>B,

；碰后

m m m m

m

m

m

m

m m

m

M

d

mv

mv

M

m

d

=d

mv

mv

M

M

m

mv

M

mmv

Mmd

d

M

M

md

Mmv

m

d

/

,

M

m

,

d

/2

m

M

mE

d d m M

m

<<dMm

M

m ΔEK=

d

d

ΔEK

m

M

L

Lll，則：mv=Mv=l

l

l+l=Ll

mM

m

L

l

lM

M

m

m=mm+m=

m+

m

M

u

m

M-m

mu

M

m,

mu動量易錯題

ABBeq

\o\ac(△,I=)

F

= eq

\o\ac(△,P)

F eq

\o\ac(△,P)

h

h

量變化快，所以掉在水泥地上杯子受到的合力大，沖力也大，所以杯子所以

【小結(jié)】 判斷這一類問題，應(yīng)從作用力大小判斷入手，再由動量

F

F

F

F

F

FIF=

F

=

F

F

=

=

=10m/s

,

h

V

eq

\o\ac(△,m)

A B

=

=

eq

\o\ac(△,P)

=

=P

eq

\o\ac(△,P)

g

eq

\o\ac(△,P)B

Ab

B

b

b

b

A

b

B

A+mBAAmBBBA

A+mB

=

mAAmBB

A

B

A

A,

A

A

mAAA+mBBAA

b

兩塊在水平飛行的同時，豎直方向做自由落體運動即做平拋運

=

b

ABB

b

FA=FB

mE

AB

m

=

=

I

木=

=

I車

=

=

,I

合=P

mg

Mg

N=mg

AB

TT

=

V,

=

I

木=

=

=Mg

eq

\o\ac(△,P)

F

FF

m

m

V

=

=

=

+M

eq

\o\ac(△,t)

:(Teq

\o\ac(△,g))

=

(Teq

\o\ac(△,g)) L

=

T

n

ml,

n

=

=

=

d+s

l

+

A

B

A

B

A

A

B

AB

B

A+mBBmAAmBmA

=

mB

2mAmBmBA

B

A

A

A

B

A

A

A

A

B

B

A

BB

AB

B

B

A

A

ABB

A

B

AB

B

A

B

B

M+mu=

M+mu=M+m+mv

m

u

m

A

M+m+mv+moB

+mvMu=M+mA

A

A

M+m

=

M+m+mvB

A

=

+mv

B

B

=30

h

BA

B

B

A

A

B

V

A

h

A

V=mv

B

m

A

A

A

V

V

m

u

eq

\o\ac(△,v)V+ eq

\o\ac(△,v))mu eq

\o\ac(△,v)

),+

u

+ eq

\o\ac(△,v))+

eq

\o\ac(△,v)+ eq

\o\ac(△,v))eq

\o\ac(△,0)

eq

\o\ac(△,M) 0=M eq

\o\ac(△,v) eq

\o\ac(△,v))

m

u

mumu

m

,

AV

VV

V

Vl

L

m

V

+M

ABA

B

A

BB

A

AB

AB

mAmB

m

A

AB

ABA

A

=

A

=

mAAAB

mv=m+mBBB

A

A

B

A

AB

AB

AB

mAmB

m

A

AB

ABA

AB

t=mA+mBu

u

AB

u

=

6m/sB

A

A

A

VA=u=6m/sA

B

=mAA+m+mBBB=

AB

A=6m/sB

A

A

B

6m/sA

AV=mv+mA+mBu

B

mv+mBu=m+mB

B

2m/s

mM+m=0+mv=0

4m/s

4m/s

M+m4m/s4m/s

M+m=

=m+M

AB

A

m

B

m

B

A

6m/s

B

B

A

A

B

m

m

m

m/

A

動量是一個瞬時狀態(tài)量，動量守恒指的是系統(tǒng)任一瞬時的動量守恒，列方程m

m

m

m

M

m

m/

m