Conservación de la cantidad de movimiento problemas.pdf

�� E� E� �� D�� D��

��

��

��

0624604

��

��. F� F��

�� E� �� 60�. E� �� 25�/� � �� 5��2, �� , ��, �� , �� . ��. D�� D�� F �� . C�� 1000 ��/�3.

��

2

�� E� �� ,, �� , �� :

∑

F x =

∑

Fs x +

∑

FBx =

d

∫

dt VC

ρudV +

∫ SC





ρ u V idA

(1.1)

�� . �� (1) � �� (2) �� .

��

E� �� (�C) ��:

F x = F / F =

d

∫

dt VC

ρudV +

∫





ρu V idA +

SC1

∫





ρ u V idA

(1.2)

SC 2

�� , ��, �� , �� :

d

∫

ρ udV = 0 dt VC

∫





ρ u V idA = − ρ V1 (V1 A )

SC 1

(1.3)

(1.4)

3

∫





ρ u V idA = ρV2 cos θ (V2 A )

(1.5)

SC 2

�� V1 = V2 = V , �� (1.2) �� :

F x = FM / F = − ρV (V A ) + ρV cos θ (V A )

(1.6)

2

F x = FM / F = ρV A ( cos θ − 1 )

(1.7)

A�� , �� , �� , �� :

FF / M = − F M / F

FF / M = − ρV 2 A ( cos θ − 1 )

(1.8) (1.9)

A�� , �� :

FF / M − F = 0

(1.10)

F = FF / M = − ρV 2 A ( cos θ − 1 )

(1.11)

�� : N F = 156.25

(1.12)

4

�� D�� , �� . �� , �� (�� ) � �� , �� . (��: A�� )

��

�� D�� , �� , �� (�: ��, �: ��) �� , �� , �� . E� �� .

��

5

�� :

0=



d

∫

dt VC



∑F = ∑F +∑F





ρ V i dA

(1.13)

SC



s

∫

ρ dV +

B

=

d



 



∫ ρVdV + ∫ ρ V V idA

dt VC

SC

(1.14)

A�� :

0=

∫ ρ dV + ∫ ρ V idA

∂t VC





∫ ρV idA + ∫ SC 1





∂

(1.1)

SC





∫

ρV idA +

SC 2





ρ V idA = 0

(1.2)

SC 3

ρVA1 + ρVA2 − ρ VA = 0

(1.3)

Q1 + Q2 − Q = 0

(1.4)

A�� :

FT = 0 =

d

∫ u ρ dV + ∫

dt VC

T





uT ρ V idA

SC

(1.5)

C�� , �� :

∫





uT ρV idA +

SC 1

∫ SC 2





uT ρV idA +

∫





uT ρ V idA = 0

(1.6)

SC 3

VQ1 − VQ2 − V cosθ Q = 0

(1.7)

C�� (1.4) � (1.7) �� , ��:

V ( Q − Q2 ) − VQ2 − V cos θ Q = 0

(1.8)

6

Q2 =

Q1 =

1 2 1 2

Q ( 1 − cos θ )

(1.9)

Q ( 1 + cos θ )

(1.10)

�� :

F N =





∫ u ρ V idA

(1.11)

T

SC1

F N = −V sinθ Q

(1.12)

�� .

�� A�� 180�, �� 2� � �� 50�. E� �� 150�� 30��. ��

� 5�/� � ��

�� , �� . �� .��

��

�� :

7

∑

F x =

∑

Fs x +

∑

FBx =

d

∫

dt VC

ρ udV +

∫ SC





ρ u V i dA

(1.13)

�� :

��

�� , � �� :

Fsuj y − W f = 0

(1.14)

Fsuj y = W f

(1.15)

Fsuj y =

π 4

2

D L ρ g

(1.16)

�� :

Fsuj y = 25.25 N

(1.17)

A��, �� (1.13) ��:

Fs x +

∑

FBx =

d

∫

dt VC

 F c / f + ( P1 A1 − P2 A2 )  + F B

x

ρ udV +

=

∫ SC

∫ ρu SC1

(1.18)

 i V dA + 





ρ u V i dA

∫

 i V A 

ρ u

(1.19)

SC 2

E� �� , �� (1.19) �� :

8

Fc / f + ( − P1 A1 − P2 A2 )  = ρV A + ρ V A

(1.20)

Fc / f =  2 ρ V + ( P1 + P2 )  A

(1.21)

2

2

�� : 2

A�� , �� , �� :

Fsuj x = Fc / f =  2 ρ V + ( P1 + P2 )  A 2

(1.22)

�� :

Fsuj x = Fc / f = 648.59 N

(1.23)

�� C�� , �� , � �� . �� , �� 180� �� . E� �� . D�� (�� , �� ), �� , � �� (��, �� ) �� .

��

9

�� E� �� , �� :

0=

∑

F x =

Fs x +

∂

∫

∂t VC

∑

∫

ρ dV +

FBx =





ρ Vrel idA

(1.24)

SC

d

∫

dt VC

ρ u rel dV +

∫



ρ u relVrel id

SC

(1.25)

��

�� :

Vrel = V − V a

(1.26)

A�� , �� :

0=

∫

 i Vrel dA + 

SC 1

 i Vrel dA + 

∫ SC 2

∫

 i Vrel dA +

SC 3



∫

 i Vrel dA 

(1.27)

SC 4

0 = −Vrel A1 + Vrel A2 + Vrel A3 + Vrel A3

(1.28)

A� �� :

A3 =

A1 − A2 2

�� :

(1.29)

10

∫

Fa / f =





ρ urelV rel idA +

SC 1

∫





∫

ρ u relVrel idA +

SC 2





ρu relVrel idA +

SC 3

∫



ρ

SC 4



i relVrel dA

(1.30)

�� :

F a / f = − ρVrel 2 A1 + ρVrel 2 A2 − ρVrel 2 A3 − ρ Vrel 2 A3

(1.31)

Fa / f = ρ Vrel 2 ( A2 − A1 − 2 A3 )

(1.32)

�� , �� , �� , �� :

suj

= Fa / f = ρ (V − Va )

2

( A2 − A1 − 2 A3 )

(1.33)

C�� , �� , �� .

�� D=5�� 5 ��/�. E� �� , �� . E� �� 0.�� 0��. D��, �� , �� . D�� ρ ��000��/��.

��

��

11

�� E� �� , �� :

Q = VA

(1.34)

�� :

V e =

4Q

(1.35)

π D 2

A� �� :

Ve = 2.55m / s

(1.36)

�� , �� :

As =

V s =

π 2

Re

(1.37)

2Q

(1.38)

π R e

Vs = 6.37 m / s

(1.39)

A�� :

∑



F =

∑



Fs +

∑



FB =

d

∫

dt VC



ρVdV +

∫ SC

��





ρ VV idA

(1.40)

12

A�� (1.40) �� (�C) � �� , �� :

��

F x B / F =

∫





Vx ρ V idA

(1.41)

SC s

E� �� (�� ) �� : θ V x = V s cos

(1.42)

π /4

F x B / F =

∫

Vs eRdθ Vs cos θρ

(1.43)

−π /4

N F x B / F = 28.66

(1.44)

�� : θ V y = V s sin

F y B / F =

∫



(1.45) 

Vy ρ V idA

(1.46)

Vs sin θρ Vs eRdθ

(1.47)

SC s

π /4

F y B / F =

∫

−π /4

F y B / F = 0 N

(1.48)

��

F z B / F =

∫





VZ ρ V idA

(1.49)

SC e

Q F z B / F = Ve ρ

(1.50)

13

F z B / F = 12.73N

(1.51)

A�� , �� : 

(

)

Fsuj = 28.66iˆ + 0 ˆj + 12.73kˆ N

(1.52)

�� D=10�� 0��, �� . C�� 0�� (��). E� �� , �� /� � � ��/�. �� 1000��/�3, ��: �� , �� .

��

�� :

0=

∂





∫ ρ d V + ∫ ρ V idA

∂t VC

SC

(1.53)

14

∑



F =

∑



Fs +

∑



FB =

d

∫



dt VC

ρVdV +

∫





ρ VV idA

SC

(1.54)

�� , � �� , �� (�) � �� (��). E� �� C�.

��

E� �� :

V II ( x ) =

(V2 − V 1 ) L

x + V1

(1.55)

E�� . A��, �� :

∫

0=





ρV idA +

SC I

∫





ρ V idA

(1.56)

SCII

 (V2 − V 1 )  x V e dx = 0 + 1 ∫0  L 

L

−V I AI +

(1.57)

D�� :

V I =

 (V2 − V 1 )  x V e dx +  1 L π D 2 ∫0   4

L

(1.58)

�� :

15

V I = 8.117 m / s

(1.59)

A�� :

��

∑

F x =

∑

Fsx +

∑

FBx =

d

∫

dt VC



∫

ρuVdV +



ρuV idA +

SCI

F D / F x + PI AI =

∫





ρ uV idA

(1.60)

SCII





∫ ρ V V idA

(1.61)

I

SCI

F D / F x = − PI AI − ρ VI 2 AI 2 F D / F x = ( − PI − ρ VI )

π 4

(1.62)

D

2

(1.63)

N F D / F x = −674.53

(1.64)

D� �� (�� ) �� . D� �� :

��

∑

F y =

∑

Fsy +

∑

FBy =

d

∫

dt VC

F D / F y =

ρ vVdV +

∫





ρ vV i dA +

SCI

∫



∫





ρ vV i dA

(1.65)

SCII



ρ VII V idA

(1.66)

SCII

F D / F y

2

 (V2 − V 1 )  x + V1  e dx = − ρ ∫  L  0  L

(1.67)

N F D / F y = −572.5

(1.68)

A�� : 

(

)

Fsuj = −673.53iˆ − 572.5 ˆj N

(1.69)

16

�� (�� . �� )

�� , �� 300�/�. �� 3000�/�. D�� , ��: a.

�� .

b. ��

1.8��/�.

��

�� A�� :

0=

∂

∫

∂t VC

ρ dV +

∫





ρ V i dA

(1.70)

SC

�� , �� , �� , �� , �� :

dM VC dt

ɺ = 0 +m

dM VC dt

ɺ = −m

(1.71)

(1.72)

E� �� , �� :

17

M

∫ dM

t

VC

∫

ɺ = − mdt

M o

(1.73)

0

ɺ M = M o − mt

(1.74)

E�� . A�� , �� , �� : −

∫

d

a X ρ dV =

∫

dt VC

VC

∫

u x ρ dV +





u x ρ V idA

SC

(1.75)

E� �� (C�� ) �� , �� . �� : −

∫

a X ρ dV =

VC

−

SC

dU VC

dU VC dt dU VC dt



(1.76)

ɺ M VC = − u gas m

(1.77)

ɺ ) = −ugas m ɺ ( M o − mt

(1.78)

dt −

∫



u x ρ V idA

ɺ m

= u gas

dUVC = u gas

(1.79)

dt

(1.80)

ɺ ) ( M o − mt ɺ m ɺ ) ( M o − mt

E�� , ��: U

∫ dU

U0

t

VC

=

∫u 0

ɺ m gas

ɺ ) ( M o − mt

dt

 M − mt ɺ  U − U 0 = −u gas ln  o  M o  

(1.81)

(1.82)

18

 M − mt ɺ  U = U 0 − u gas ln  o   M o 

(1.83)

E�� . D� �� , �� :

1−

ɺ mt

M o

ɺ mt

M o

U −U 0    u gas 

−

=e

U −U 0    u gas 

(1.84)

(1.85)

−

= 1− e

�� , �� 1.8��/� ��: ɺ mt

M o

= 0.393

(1.86)

�� (�� . �� )

E� �� . E� �� , �� . �� ( �), D��: a.

�� .

b. �� .

19

��

�� A�� :

0=

∂

∫

∂t VC

ρ dV +

∫





ρ V i dA

(1.87)

SC

E� �� , �� , �� :

dM VC

− ρ (V − U ) A = 0

dt

dM VC dt

(1.88)

= ρ (V − U ) A

(1.89)

E�� , �� . A�� : −

∫ a ρ dV X

VC

=

d

∫ u ρ dV + ∫

dt VC

x





u x ρ V idA

SC

(1.90)

E� �� (�� ) �� . �� : −

∫

VC

a X ρ dV =

∫ SC





u x ρ V idA

(1.91)

20

−

dU VC dt dU VC dt

2

M VC = − (V − U ) ρ A

(1.92)

2

M VC = (V − U ) ρ A

(1.93)

�� (1.12) �� :

1

dU

(V − U )

dt

M VC = (V − U ) ρ A =

1

(V − U )

dU =

1 M VC

dM VC dt

dM VC

(1.94)

(1.95)

�� : U

1

∫ (V − U )

M

dU =

0

1

∫ M

M 0

dM VC

(1.96)

VC

 M   V  = ln    V U M −    0

ln 

 V   V U −  

M = M 0 

(1.97)

(1.98)

�� . E�� (1.93) ��:

dU VC dt

2  V  = (V − U ) ρ A V − U 

M 0  1

(V − U )

3

dU VC =

ρ A M 0V

dt

(1.99)

(1.100)

�� :

21

U

t

1

∫ (V − U )

3

dUVC =

0

1

1

 2 2  (V − U ) 

ρ A

∫ M V dt 0

−

1 

(1.101)

0

ρ A

t = V 2  M 0V 

(1.102)

D�� :

    1   U = V 1 − 1/ 2    1 + 2 ρ VA t      M 0  

(1.103)

�� 100�� , �� , �� 2000�/� �� 5��/� ��. E� �� , ��: �.

E� �� 50%.

�. �� . �.

�� .

C�� =9.81 �/�2 � �� , �� .

22

��

�� D�� , �� :

0=

Fsy + FBy −

∫

∂





∫ ρ dV + ∫ ρ V idA

∂t VC

(1.104)

SC

a y ρ dV =

VC

d

∫

dt VC

v y ρdV +

∫ SC





v y ρ V idA

(1.105)

�� :

0=

∂

∫

∂t VC

∫

ρ dV +





ρ V i dA

(1.106)

SC

E� �� :

dM VC dt

ɺ e = 0 +m

dM VC dt

ɺ e = −m

(1.107)

(1.108)

�� : M

∫ dM

M o

t

VC

∫

ɺ e dt =− m

(1.109)

0

ɺ e t M ( t ) = M o − m

(1.110)

23

�� (1.110) ��:

M 0 = 100kg ɺ e = 5kg / s m

(1.111)

A�� :

Fsy + FBy −

∫

a y ρ dV =

VC

d

∫

dt VC

v y ρ dV +

∫ SC





v y ρ V idA

(1.112)

E� �� (�� ) � �� , �� :

F By −

∫

∫

a y ρ dV =

VC





v y ρ V idA

SC

ɺ e − M ( t )g − a( t )M ( t ) = −Ve m

a( t ) =

ɺe Ve m

M ( t )

(1.113)

−g

(1.114)

(1.115)

�� (1.110) �� (1.115) �� :

a( t ) =

ɺe Ve m ɺ et M 0 − m

−g

(1.116)

�� :

Ve = 2000 m / s

(1.117)

�� (1.116) �� :

dV dt

=

ɺe Ve m

−g

ɺ et M0 − m

 Ve mɺ e  − g  dt dV =   M 0 − mɺ et  �� :

(1.118)

(1.119)

24

 Ve mɺ e  dV =  − g  dt ɺ et M 0 − m  0 

(1.120)

 M 0  V (t ) = Ve ln  −g t ɺ M m t −  0 e 

(1.121)

V(t )

∫ 0

t

∫

E�� . A�� 50%, �� :

M (t ) = 0.5 M 0

(1.122)

�� (1.110) �� :

t =

( M 0 − 0.5M 0 ) ɺe m

(1.123)

�� :

t = 10 s

(1.124)

C�� (1.116) � �� (1.121): 2 a(t) = 190.19m / s

V (t ) = 1288.19m / s

(1.125)

�� 0�� 0��0�/�, �� 0.�� /�, �� . D�� .��. D�� .

25

��

�� E� �� :

F sx + F Bx −

∫

a x ρ dV =

VC

d

∫

dt VC

u x ρ dV +



∫



u x ρ V idA

SC

(1.126)

E� �� . �� .

��

A�� :

∫

a x ρ dV =

VC

−a x M = (

∫





u x ρV idA +

SC1

∫





u x ρ V idA

SC2

] − U ) ρ [ −(V − U ) A] + − (V + U )  ρ [− (V + U ) A a x = (V − U ) − (V + U )  2



2

(1.127)

A ρ

 M

(1.128)

(1.129)

26

a x = −4VU

A ρ

(1.130)

M

C�� , �� , �� , �� :

dU a x = dt dU dt dU U U

∫

U 0

dU

= −4VU

= −4V

(1.131)

A ρ

A ρ

M

t

(1.133)

dt

A ρ

∫

(1.132)

M

= − 4V

dt

(1.134)

U  A ρ t  = −4V U M  0

(1.135)

U

M

0

ln 

−4V

U = U 0e

A ρ t M

(1.136)

�� (1.136) �� (1.130) �� :

 −4V A M ρ t  A ρ a x = −4V U 0e  M  

(1.137)

�� (1.136) � (1.137) �� :

U (1.5 s ) = 2.59m / s a(1.5 s ) = −2.33m / s

2

�� .

(1.138)

Conservación de la cantidad de movimiento problemas.pdf

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