1. No category

GENERAL PHYSICS PH 221-3A (Dr. S. Mirov)
Test 4 (12/03/07)
Sample
Key
STUDENT NAME: ________________________ STUDENT id #: ___________________________
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ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.
NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)
Important Formulas:
1
1.
Motion along a straight line with a constant acceleration
vaver. speed = [dist. taken]/[time trav.]=S/t;
vaver.vel. = Δx/Δt;
vins =dx/Δt;
aaver.= Δvaver. vel./Δt;
a = dv/Δt;
v = vo + at; x= 1/2(vo+v)t; x = vot + 1/2 at2; v2 = vo2 + 2ax (if xo=0 at to=0)
2.
Free fall motion (with positive direction ↑)
g = 9.80 m/s2;
y = vaver. t
vaver.= (v+vo)/2;
v = vo - gt; y = vo t - 1/2 g t2; v2 = vo2 – 2gy (if yo=0 at to=0)
3
3.
Motion in a plane
4.
Projectile motion (with positive direction ↑)
5.
Uniform circular Motion
vx = vo cosθ;
vy = vo sinθ;
x = vox t+ 1/2 ax t2; y = voy t + 1/2 ay t2; vx = vox + at; vy = voy + at;
vx = vox = vo cosθ;
x = vox t;
xmax = (2 vo2 sinθ cosθ)/g = (vo2 sin2θ)/g for yin = yfin;
vy = voy - gt = vo sinθ - gt;
y = voy t - 1/2 gt2;
a=v2/r,
T=2πr/v
6. Relative motion
r
r
r
v PA = v PB + v BA
r
r
a PA = a PB
7.
Component method of vector addition
1
A =
A x2 + A y2 ; θ = tan-1 ⏐Ay /Ax⏐;
r
r
The scalar product A a ⋅ b = a b c o s φ
r
r
a ⋅ b = ( a x iˆ + a y ˆj + a z kˆ ) ⋅ ( b x iˆ + b y ˆj + b z kˆ )
r r
a ⋅ b = a xb x + a yb y + a zbz
r
r
The vector product a × b = ( a x iˆ + a y ˆj + a z kˆ ) × ( b x iˆ + b y ˆj + b z kˆ )
A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2;
iˆ
r
r
r
r
a × b = −b × a = ax
bx
ˆj
a
b
y
y
kˆ
a
a z = iˆ
b
bz
y
a
y
bz
= ( a y b z − b y a z ) iˆ + ( a z b x − b z a x ) ˆj + ( a x b
y
z
ax
bx
− ˆj
− bxa
y
ax
az
+ kˆ
bx
bz
a
y
b
y
=
) kˆ
1.
Second Newton’s Law
ma=Fnet ;
2.
Kinetic friction fk =μkN;
3
3.
St ti friction
Static
f i ti
fs =μsN;
N
4.
Universal Law of Gravitation: F=GMm/r2; G=6.67x10-11 Nm2/kg2;
5.
Drag coefficient
6.
Terminal speed
7.
Centripetal force: Fc=mv2/r
8.
Speed of the satellite in a circular orbit: v2=GME/r
9.
The work done by a constant force acting on an object:
10. Kinetic energy:
D =
vt =
K =
1
C ρ Av
2
2
2m g
C ρ A
1
m v
2
W
r r
= F d cosφ = F ⋅ d
2
11. Total mechanical energy: E=K+U
12. The work-energy theorem: W=Kf-Ko; Wnc=ΔK+ΔU=Ef-Eo
13. The principle of conservation of mechanical energy: when Wnc=0, Ef=Eo
14. Work done by the gravitational force:
W
g
= m gd cosφ
2
1.
Work done in Lifting and Lowering the object:
Δ K
=
K
f
− K
= W
i
F
a
+ W
= − k x
; if
g
=
f
K
Spring Force:
3.
Work done by a spring force:
4.
Work done by a variable force:
W
1
k x
2
=
s
x
W
; P
Δ t
Pavg =
Power:
6.
Potential energy:
7.
d W
d t
=
Δ U
; W
2
i
−
= − W
a
g
= −W ; Δ U
2
o
; if x
i
=
x ; W
s
= −
1
k x
2
2
F ( x )d x
∫
= −
x
x
f
r
r
F ⋅ v
=
F ( x )d x
i
= m g ( y
f
−
y i ) = m g Δ y ; if
y
i
= 0 a n d U
1
k x
2
8.
Elastic potential Energy:
U ( x ) =
9.
Potential energy curves:
F ( x ) = −
10.
Work done on a system by an external force:
F ric tio n is n o t in v o lv e d W
i
= 0; U ( y ) = m g y
2
d U ( x )
; K ( x ) =
d x
= Δ E
m e c
= Δ K
E
Δ E
th
=
− U ( x )
m e c
+ Δ U
W h e n k in e tic fric tio n fo rc e a c ts w ith in th e s y s te m
= Δ E
W
m e c
+ Δ E
th
+ Δ E
in t
fkd
Conservation of energy:
Pavg =
Δ E
; P
Δ t
12.
Power:
13.
Center of mass:
14.
f
Gravitational Potential Energy:
Δ U
11.
= 0 a n d x
i
F v c o s φ
=
; P
1
k x
2
f
∫
=
W
x
5.
i
( H o o k 's l a w )
2.
x
K
r
rc o m
=
W
= Δ E
= Δ E
m e c
+ Δ E
fo r is o la te d s y s te m
=
1
M
d E
d t
i =1
+ Δ E
in t
m e c
+ Δ E
th
= 0
;
n
∑
th
(W = 0 ) Δ E
m
i
r
ri
Newtons’ Second Law for a system of particles:
r
F net =
r
M a
c o m
3
1.
2.
r
r
r
dP
r
Linear Momentum and Newton’s Second law for a system of particles: P = M v c o m a n d F n e t =
dt
r
J =
Collision and impulse:
∫
t
r
F ( t ) d t ; J = F a v g Δ t ; when a stream of bodies with mass m and
f
ti
F avg = −
speed v, collides with a body whose position is fixed
r
p
Impulse-Linear Momentum Theorem:
r
− p
f
n
n
Δ m
Δ p = −
Δ v
m Δ v = −
Δ t
Δ t
Δ t
r
= J
i
r
r
Pi = P f f o r c l o s e d , i s o l a t e d s y s t e m
3.
Law of Conservation of Linear momentum:
4.
Inelastic collision in one dimension:
5.
Motion of the Center of Mass: The center of mass of a closed, isolated system of two colliding bodies is
r
r
p 1i + p
r
= p1
2 i
f
r
+ p
2 f
not affected by a collision.
6.
Elastic Collision in One Dimension:
7.
Collision in Two Dimensions:
8.
Variable-mass system:
9.
Angular Position:
v
− vi = v
10. Angular Displacement:
α
m
m
=
− m
+ m
1
1
= p1
2 ix
fx
2
v1i ;
v
2 f
2
+ p
2 fx
;
=
m
p 1 iy + p
2m1
1 + m
2 iy
v1i
2
= p1
fy
+ p
2 fy
ln
rel
M
M
i
( s e c o n d ro c k e t e q u a tio n )
f
(ra d ia n m e a s u re )
Δ θ = θ
11. Angular velocity and speed:
12. Angular acceleration:
f
= M a (firs t ro c k e t e q u a tio n )
rel
S
r
θ =
p 1 ix + p
R v
f
v1
avg
ω
=
2
− θ
1
(p o s itiv e fo r c o u n te r c lo c k w is e ro ta tio n )
=
Δ θ
;
Δ t
Δ ω
;
Δ t
α =
avg
ω =
dθ
dt
(p o s itiv e fo r c o u n te rc lo c k w is e ro ta tio n )
d ω
dt
4
ω = ω
1.
angular acceleration:
o
=
θ − θ
o
= ω ot +
2
= ω
θ − θ
o
+ ω )t
o
1
ω t2
2
+ 2 α (θ − θ o )
2
o
= ω t −
1
ω t
2
v = ω r;
a
= α r;
t
a
r
1
Iω
2
=
I =
∫
2
r 2d m
; I =
∑
2
r;
T
=
2π r
2π
=
v
ω
m i ri 2 f o r b o d y a s a s y s t e m o f d i s c r e t e p a r t i c l e s ;
f o r a b o d y w ith c o n tin u o u s ly d is trib u te d m a s s .
= I
+ M h
4.
The parallel axes theorem: I
5.
Torque:
6.
Newton’s second law in angular form:
τ
7.
Work and Rotational Kinetic Energy:
W
P =
dW
dt
com
2
τ = r F t = r⊥ F = r F s i n φ
; Δ K
= K
v
Rolling bodies:
f
− K
i
2
f
−
net
= Iα
=
1
Iω
2
θ
∫θ
2
i
f
τ dθ ; W
= τ (θ
f
− θ i ) fo r τ = c o n st;
i
= W
w o rk e n e rg y th e o re m fo r ro ta tin g b o d ie s
= ω R
com
a
com
a
com
=
=
Torque as a vector:
1
Iω
2
=
1
I com ω
2
= α R
K
9.
v 2
= ω
r
=
Rotational Kinetic Energy and Rotational Inertia:
K
8.
2
Linear and angular variables related:
s = θ r;
3.
1
(ω
2
θ − θ
ω
2.
+ α t
o
r
τ
2
+
1
m v
2
g s in θ
1 + I com / M R
r
r
= r × F ;
2
2
com
fo r ro llin g s m o o th ly d o w n th e ra m p
τ = r F s in φ = r F
⊥
= r⊥ F
5
1.
2.
r
r
r
r
r
l = r × p = m (r × v );
Angular Momentum of a particle:
l = r m v s in φ = r p
r
Newton’s Second law in Angular Form: τ
n et
=
r
L =
3.
5.
6.
n
∑
i=1
Angular momentum of a system of particles:
r
τ
4.
= rm v
⊥
⊥
= r⊥ p = r⊥ m v
r
d l
d t
L = Iω
r
r
Conservation of Angular Momentum: L i = L
n et ext
=
r
li
r
d L
d t
Angular Momentum of a Rigid Body:
Static equilibrium:
if a ll th e fo rc e s lie in x y p la n e
F
n et,x
= 0;
F
n et, y
= 0;
τ
= 0
net,z
s tre s s = m o d u lu s × s tra in
Elastic Moduli:
8.
Tension and Compression:
9.
Shearing:
= G
10. Hydraulic Stress:
Δ L
, G
L
p = B
11. Simple harmonic motion:
12. The Linear Oscillator:
13. Pendulums:
(is o la te d s y s te m )
r
r
F net = 0 ; τ net = 0
7.
F
A
f
F
A
= E
Δ L
, E i s t h e Y o u n g 's m o d u l u s
L
is th e s h e a r m o d u lu s
Δ V
V
, B is th e b u lk m o d u lu s
x = x
ω =
m
c o s(ω t + φ ); v = − ω x
k
, T
m
= 2π
m
s in (ω t + φ ); a = − ω
2
x
m
c o s (ω t + φ )
m
k
T
= 2π
I
, to rs io n p e n d u lu m
k
T
= 2π
L
, s im p le p e n d u lu m
g
T
= 2π
I
, p h y s ic a l p e n d u lu m
m g h
6
1.
Damped Harmonic Motion:
2.
Sinusoidal waves:
x (t) =
x
m
− b t
e
2 m
k
m
c o s (ω ' t + φ ), ω ' =
2π
ω
−
b 2
4 m
1
T
, v =
3.
Wave speed on stretched string:
4.
Average power transmitted by a sinusoidal wave on a stretched string:
Pavg =
5.
Interference of waves:
6.
Standing waves:
7.
Resonance:
8.
Sound waves:
f
=
y
v
λ
B
λ
Interference:
Δ L
φ =
λ
2π
= (2 m
Sound level in decibels:
12.
Standing wave patterns in pipes:
13.
f
=
v
v
λ
Beats:
β
+ 1 )π
1
ρ vω
2
11.
λ
=
f
=
ω
k
=
λ
1
μ vω
2
2
2
fo r m
f
2
m
s
2
m
= (1 0 d B ) lo g
, I =
I
, I
Io
= 0 ,1 , 2 , 3 ..., d e s tr u c tiv e in te r f e r e n c e
Ps
4π r
o
2
= 1 0
− 1 2
W
/ m
2
n v
, n = 1 , 3 , 5 , ..., f o r p ip e c lo s e d a t o n e e n d a n d o p e n e d a t th e o th e r
4 L
f1 −
y
= 0 ,1 , 2 , 3 ..., c o n s tr u c tiv e in te r f e r e n c e
=
=
m
1
1
φ ] s in ( k x − ω t +
φ )
2
2
n v
, n = 1 , 2 , 3 , ..., f o r p ip e o p e n e d f r o m
2 L
b ea t
− b t
= λ f
T
=
f
e
s in k x ] c o s ω t
= m (2π ) fo r m
Sound Intensity:
=
co s
m
2π
10.
f
2π
2
m
v
, f o r n = 1 , 2 , 3 , ...
2 L
P
, I =
A
I =
,
1
k x
2
,
ρ
Δ L
m
λ
, E (t) ≈
τ
μ
v =
y '( x , t ) = [ 2 y
= n
v =
s in ( k x − ω t ), k =
m
y '( x , t ) = [ 2 y
φ =
9.
y ( x , t) =
2
b o th e n d s
2
7
1.
The Doppler effect:
f ' = f (1 ±
vR
),
vs
vR the speed of the receiver; vs the speed of the
sound;(vs=331m/s);
+ for
f receiver
i
approaching
hi stationary
i
emitter,
i
- for receiver moving away from the stationary emitter;
⎛
⎞
⎜
1 ⎟
f' = f ⎜
⎟,
v
E
⎜1m
⎟
vs ⎠
⎝
vE the speed of the emitter, vs the speed of the sound,
- for emitter approaching stationary receiver,
+ for emitter moving away from the stationary receiver;
f '= f
vs ± vR
general Doppler Effect
vs m vE
8
9
2.
Your grandfather clock’s pendulum has a length of 0.9930 m. if the clock
loses half a minute per day, how should you adjust the length of the
pendulum?
For a simple pendulum T = 2π
L
g
Suppose
pp
clock's ppendulum oscillates "n" times in a day.
y
⇒ nT1 = (24 ⋅ 3600 − 30) = 86370s
after the adjustment of the pendulum's length
nT2 = (24 ⋅ 3600) = 86400 s
L2
g
T2
Take ratio
=
=
T1
L
2π 1
g
2π
L2 86400
=
L1 86370
864002 ⋅ 0.9930
⇒ L2 =
= 0.9937
2
86370
⇒ L2 − L1 = 0.9937
0 9937 − 00.9930
9930 = 7 ⋅ 10−4 m
10
11
4.
12
5.
13
6.
14