Formula/Review Sheet by Qiu PHY 2020 for Mid-Term Exam 3
v
x
t
[average speed = (distance)/(time)]
v
x x  x o 

t
t
a
v
t
Velocity is a vector! Speed is the magnitude of the velocity vector.
[instantaneous velocity = (change in distance)/(change in time)]
[average acceleration = (change in velocity)/(time)]
Acceleration is a vector!
v
a
t
[instantaneous acceleration = (change in velocity)/(change in time)]
When given a graph of distance versus time, the slope is the velocity.
When given a graph of velocity versus time, the slope is the acceleration.
Workhorse Equations (for constant acceleration)
(v  v o )
a
t
a
v  v
x  xo   o
t
 2 
g  9.8
Ax  Bx  C  0
F  ma
1
x  x o  v o t  at 2
2
m
ft
 32 2
2
s
s
2
v x  v cos 
 B  B 2  4AC
x
2A
v y  v sin 
v  v 2x  v 2y
[Force = (mass)(acceleration)]
Force is a vector!
v 2  v o2
2x  x o 
tan  
unit is N = Newton
vy
vx
Frictional Force is opposite to direction of motion.
In equilibrium, the sum of all forces acting on a body is zero.
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Formula/Review Sheet by Qiu PHY 2020 for Mid-Term Exam 3
Projectile Motion
x-direction
y-direction
ax  0
a y  g
v xo  v o cos  o
v x  v xo (constant!)
v yo  v o sin o
v y  v o sin o  gt
x  x o  ( v o cos  o ) t
y  y o  ( v o sin  o ) t  gt 2
1
2
Trajectory:
y  (tan  o ) x 
Max. Height:
h
v o2 sin 2  0
2g
Range:
R
v o2 sin 2 o
g
Eo  Ef
g
x2
2
2 v cos  o
Conservation of Energy
2
o
(PE) o  (KE) o  W  (PE) f  (KE) f
W  Fd
[Work = (Force) (distance)] (unit is J, Joule)
Force and distance are vectors and their dot product gives the scalar Work.
So, use only the component of the Force that is in the direction of the movement.
W  (F cos ) d
KE(linear) =
P
W
t
1
m v2
2
W  KE  (KE) f  (KE) o
[Power = (work)/(time)]
Spring (with spring constant k):
p  mv
F
p
t
PE 
F  kx
PE or U (Potential Energy) of gravity:
P  F v  (F cos ) v
(unit is W, Watt)
U  mgh
[momentum = (mass) (velocity)]
1
k x2
2
Momentum is a vector!
[Force = (change in momentum)/(change in time)] (recall Force is a vector)
“I or J” Impulse (if force applied in a short period of time):
2
IFt
Formula/Review Sheet by Qiu PHY 2020 for Mid-Term Exam 3
G = 6.67  1011 N m2 kg2
NA = 6.022  1023 particles = 1 mole
R = 8.314 J/(mole K)
 = 5.67  108 W m2 K4
1 bar = 1  105 Pa
1 Pa = 1 N/m2
1 cal = 4.186 J
Conservation of Momentum: no external forces and no internal work done, then
 po   pf .
Totally Elastic Collision: KE conserved and Momentum conserved.
1
1
1
1
m1 v12o  m 2 v 22 o  m1 v12f  m 2 v 22f
2
2
2
2




m1 v1o  m 2 v 2o  m1 v1f  m 2 v 2f
Totally Inelastic Collision (two bodies stick together): KE is NOT conserved ; Momentum is.



m1 v1o  m 2 v 2o  (m1  m 2 ) V
RECALL Linear vs. Rotational Motion! RECALL Conservation of Energy and Momentum
contains Linear and Rotational parts! RECALL Bodies in Equilibrium have net Force of zero
and net Torque of zero.
m m
2 R
Period (T) is a time: T 
Gravitational Force: F  G 1 2
v
R2
 mi x i
 mi yi
M   mi
Center of Mass:
X
Y
 mi
 mi
m
Density:  
Pressure is force per unit area.
V
1
P  gh  v 2  const .
Bernouilli:
and
A1v1  A 2 v 2
2
F f

Archimedes: FB  liquid Vdisplaced g
Pascal:
A a
Ideal Gas Law: PV  nRT
Temperature Scales:
°F = (9/5) °C + 32
°C = (5/9) [°F  32]
°K= °C + 273
Heat Capacity/Specific Heat:
Q
C
C = m c (here, c is “specific heat”) Q  m c (Tf  To )
T
Recall: heat is needed at a phase transition! Latent Heat of Fusion (solid-liquid) or Latent Heat
of Vaporization (Evaporation). The latent heats at these transitions are “normally” independent
of direction of warming/cooling through the transition. Units of latent heat is cal/g or J/g.
Q
Q
T
 kA
(unit is J/s = W) Conduction: H 
(k = thermal conductivity).
t
t
L
Q
Q
 h A (T) 5 / 4
 A e  T4
Convection: H 
Radiation: H 
t
t
(e is the emissivity, 0  e  1, higher for a dark, rough surfaces, good absorber is good emitter)
( is the Stefan-Boltzmann constant)
Heat Flow: H 
A  2 A o T
V  3 Vo T   Vo T
Thermal Expansion (): L   L o T
Apparent (liquid in a container): apparent  liquid  3 container
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Formula/Review Sheet by Meisel PHY 2020 for Mid-Term Exam 3, v. 14 November 2005
Linear Motion
Rotational Motion
Displacement
x
Velocity
v
Acceleration
a
Angular Displacement

x
t
Angular Velocity


t
v
t
Angular Acceleration


t
Constant acceleration equations
v  vo  at
1
vo  v 
2
1
x  x o  vo t  at 2
2
vaverage 
a
Constant angular acceleration equations
  o  t
1
average  o  
2
1
  o  o t  t 2
2
v2  vo2
2x  x o 
Mass
m
Momentum
p  mv
Force
F
Power
PFv
Newton’s 2nd Law
 F  ma
Equilibrium
Fnet  0

2  o2
2  o 
 mi ri2
Moment of Inertia
I
Angular Momentum
LI
Torque
  F   I
Power
P
Newton’s 2nd Law
Equilibrium
  I 
net  0
Rotational Kinetic Energy
Con. of Linear Momentum
1
KE  m v 2
2
 pi o   p j f
Con. of Angular Momentum
1
I 2
2
 Li o   L j f
Work
WFd
Work
W
Kinetic Energy
Conservation of Energy holds for both linear and rotational motion!
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KE 
Formula/Review Sheet by Meisel PHY 2020 for Mid-Term Exam 3, v. 14 November 2005
Efficiency () is, in general, (benefit)/(cost).
q q
F  k 1 2 where k = 9  109 (N m2)/C2 and negative sign is “attractive”
r2
T
( work output )
Carnot heat engine:  
 1 C
(heat input )
TH
Electron (negative) & Proton (positive) of the unit of charge 1.6  1019 C
Carnot refrigerator: COP (coefficient of performance)
F
q
(total is sum of all fields, “lines” + to )
k
q
r2
q
Potential (Point): Vp  k
Potential Difference (Volt): VAB  VA  VB
r
W  QV
Electric Field (E):
TC
(heat output )
COP =

( work input ) (TH  TC )
Entropy (S):
S  0.
Frequency (f): f 
1
T
Angular frequency ():   2 f
I
Spring Motion: Recall: F = m a =  k x (negative sign means “restoring” force)
1
1
Speed of a wave in a string with tension (T) and  (mass/length): v 
R’s in parallel:
2L
n
4L
Closed Pipe (n = 1,3,5,…):  n 
n
1

1
RT

1
R1

1
R2

1
R3
 ...
8
T

I
Io
nv
2L
nv
fn 
4L
v  vO
f beat  f1  f 2 Doppler shift: f O 
fS
v  vS
n 

The speed of light in vacuum: c  3  10 m/s.
(where IO is 1  1012, i.e. “faintest” resolved sound).
Open Pipe (n = 1,2,3,…):
L
A
V2
R
R’s in series: RT = R1 + R2 + R3 + …
Speed of sound in air at 1 bar: v  331  0.6 T (in m/s and T in C).
Sound level in dB:   10 log
P  I V  I2 R 
V=IR
R 
k
E  kx 2  mv 2  kA 2
v
A2  x 2
2
2
2
m

k
m
a
x
T  2
(for a simple pendulum)
T  2
m
g
k
vf
2L
nv
fn 
Standing Waves: (ends are fixed)  n 
n
2L
1
Q
t
E
fn 
(along line O and S, sign: toward/away!)
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