Physics 121
Exam Formula Sheet
You may take this sheet into the Testing Center. You are allowed to
put additional handwritten information on this sheet if you so desire.
Conversion Factors
Mass
1 kg = 1000 g, 1 u = 1.66  10-27 kg
Length
1 m = 100 cm = 3.28 ft = 39.37 in, 1 mi = 1.61 km = 5280 ft = 1609 m
Time
1 day = 86 400 s, 1 yr = 3.156  107s
Speed
60 mi/h = 88.0 ft/s = 26.83 m/s, 100 km/h = 27.78 m/s = 62.14 mi/h
Angle
1 rev = 360° = 2π rad
Force
1 N = 0.225 lb, 1 lb = 4.45 N
Energy
1 eV = 1.60  10-19 J, kcal = 1 Cal = 103 cal = 4.19 kJ,
1kWh = 3.6  106J = 3.6 MJ
speed of light
gravitational constant
For Earth:
free-fall acceleration
mass
mean radius
tera- T 1012
milli- m 10-3
c
G
Physical Constants
2.998  108 m/s
6.670  10-11 Nm2/kg2
9.80 m/s2 = 32.15 ft/s2
5.98  1024 kg
6370 km
g
ME
RE
Metric Prefixes
giga- G 109 mega- M 106
micro- μ 10-6 nano- n 10-9
kilo- k 103
pico- p 10-12
Chapter 2 - Motion in One Dimension
Position: x(t)  x-coordinate at time t, y(t)  y-coordinate at time t, etc.
Average speed: average speed  total distance moved/Δt (Δt = elapsed time)
Displacement: Δx  xfinal – x initial
Average velocity: <v>  Δx/Δt

t
Velocity: v  dx/dt = slope on x(t) plot. Hence Δx = v (t ) dt
t0
Acceleration: a  dv/dt = d2v/dt2 = slope on v(t) plot.
Hence Δv =
t
 a(t )dt
Kinematic relationships valid iff a = constant:
v = v0 + at, r = r0 + v0t + ½ at2, r = r0 + ½ (v + v0)t,
vv = v0v0 + 2aΔr
For free fall with a = (0, -g), v0 = (vx0, vy0) = (v0 cos θ, v0 sin θ)
ax = 0, ay = -g, vx = vx0, vy = vy0 - gt, x = x0 + vx0t, y = y0 + vy0t - ½ gt2
Projectile motion over level terrain with negligible air resistance
Trajectory: y = (tan θ0)x – [g / (2v02cos2θ0)]x2 (a parabola)
Range: R = (v02/g) sin 2θ0, Maximum height: h = v02sin2θ0 / 2g
Speed v as a function of time: v2 = vx2 + vy2 = vx02 + (vy0-gt)2
Speed v as a function of y: v2 = v02 – 2g (y – y0)
Acceleration in Uniform Circular Motion:
ar = v2/ r, directed toward the center of the circle
Radial (centripetal) and Tangential Acceleration: a = ar + at, with
ar = v2/ r, directed toward the center of the circle,
at = dv/dt, directed toward the direction of motion
Relative Motion: Relative velocity:
vBA = vBC + vCA, and, by induction, vBA = vBC + vCD + vDE + . . . + vXA
Relative acceleration: aBA = aBC, if aCA = 0
Vector Differentiation:
Rectangular form:
dV dvx ˆ dv y ˆ dvz ˆ

i
j
k  (a x , a y , a z )
dt
dt
dt
dt
Polar form with v = (v, θ):
dv dv
d
 uˆ ||  v
uˆ 
dt dt
dt
Chapters 5 and 6 – The Laws of Motion
Newton’s First Law – The First Law of Motion:
In the absence of a force (a free object) moves with a = 0, i.e., if at rest, it
remains at rest. If moving, it continues to move in a straight line at a
constant speed. This is a law describing an inertial reference frame.
If it appears to be violated, the observer’s reference frame is not in an
inertial reference frame.
Definitions:
dv
Acceleration: a 
.
dt
Force: F, a push or pull, the cause of any true acceleration, a vector
quantity. In SI units forces are measured in Newtons (N); in British
units forces are measure in pounds (lbs).
Inertial mass: m, a measure of how an object responds to a force, a scalar
quantity. In SI units masses are measured in kilograms (kg); in British
units masses are measured in slugs.
Newton’s Second Law – The Second Law of Motion:
In an inertial reference frame:
1. Accelerations are caused by forces.
2. a = F/m, or F = ma. The force in this relationship is the net force
acting upon the accelerating body, i.e., it is the sum of all forces
acting upon that body, F  Fnet 
t0
Kinematic relationships valid iff a = constant:
v = v0 + at, x = x0 + v0t + ½ at2, x = x0 + ½ (v + v0)t, v2 = v02 + 2a (x - x0)
Free fall:
a = -g (assuming that displacement is taken as positive upward and
ignoring air resistance and other smaller effects)
Chapter 3 - Vectors
Vector notation: A is a vector, A is its magnitude (which includes units) and
û is a unit vector in the direction of A. Hence A  Aû. The
direction of û is given by the right-hand rule.
Vector components in two dimensions: A = Axî+ Ayĵ = (Ax, Ay)
r is a position vector in two dimensions: r = xî+ yĵ = (x, y), r = (x2 + y2)½
Dot (scalar) product: AB  AB cos θ (θ is the angle between A and B)
Also A  B  Ax Bx  Ay By  Az Bz .
Cross (vector) product: AB  AB sin θ û ( û is a unit vector normal to
the plane of A and B in the direction given by the right-hand rule. )
ˆi
ˆj
kˆ
Also A  B  Ax
Ay
Az .
Bx
By
Bz
Chapter 4 - Motion in Two [Three] Dimensions
Definitions:
Position: r  (x, y, [z]) Displacement: Δr  (Δx, Δy, [Δz])
Velocity v  dr/dt = (dx/dt, dy/dt, [dz/dt]) = (vx, vy, [vz])
Acceleration: a  dv/dt = (dvx/dt, dvy/dt, [dvz/dt]) = (ax, ay, [az])
F .
i
i
3. a is in the same direction as F, always.
Newton’s Third Law – The Third Law of Motion:
If body A exerts a force on body B, then body B exerts a force, equal in
magnitude, but opposite in direction, on body A, i.e.., FAB =  FBA, where
FAB is the force exerted on body B by body A and FBA is the force exerted
on body A by body B. This law is sometimes called the Law of Action
and Reaction. This is a somewhat misleading title because it implicitly
implies a cause-effect relation between the two forces which are associated
with any interaction. In reality, neither force of a force pair is more
fundamental than the other and neither should be viewed as the cause of
the other. All forces occur in pairs. There are no isolated forces.
Fundamental Forces: There are only four kinds of forces:
1. Gravity: all objects with mass attract all other objects with mass.
2. Electromagnetic Force: all forces experienced at the macroscopic
level, except gravity, are electromagnetic, including tension,
compression, friction, buoyancy and viscous drag.
3. Nuclear Strong Force: this is the force that binds atomic nuclei
together by overcoming the repulsive electromagnetic forces exerted
by the protons on each other. This is a very short-range force.
4. Nuclear Weak Force: this is also a short-range force manifest in
certainly nuclear reactions, including the emission of beta radiation.
Chapter 7 – Energy of a System
Work-Kinetic Energy Theorem: for any object ΔK = Wnet 
W . W is
i
i
the work done on the object by the ith force acting upon it.
i
Definitions:
ΔK  Kfinal Kinitial 
Wi 

(The integral forms of these equations are the impulse-momentum theorems.)
Center of mass:
½ mvf2 ½ mvi2
N
rf
Fi  dr , Wi is the work done by any force Fi in a displacement
ri
rf
F
ri
net  dr 
 W    F  dr, where F
rf
i
i
i ri
net 
i
F.
i
i
i i for discrete particles.

Chapter 10 – Rigid-body, Fixed-axis Rotation
Dynamics:
Fx dx.
1
e.g., WH   k ( xv2  xi2 ) for FH  kx ˆi. (Hooke's law) .
2
3. W = 0, if F  r .
Power:
lim W
P
 F  v.
t  o t
Chapter 8 - Conservation of Energy
If a physical quantity X is conserved for an isolated system, then
dX
X  0 or X final  X initial or X  constant or
 0.
dt
Conservation of Energy:
dE
For an isolated system: E  0 or E final  Einitial or
 0.
dt
Definitions:
int  kineticenergy  potentialenergy  int ernalenergy
functionof rest mass, temperatu
re, physical
1
1
mv2  mv  v, U int  mc2 
state, chemical state and nuclear structure
2
2
Conservative force: is a force which does not change the total mechanical
energy of a system when it does work on the system. i.e., K  U  0 as
a consequence of work done on a system by such a force. For a
conservative force the associated potential energy can be defined by
r
U ˆ for a one- dimensional
U  Wc  Fc  dr and thereforeFc  U (r )  
i
force.
r
x

Potential Energy:
U g  mgy for Fg  -mgˆj. (near - earth gravity),
1 2
kx for FH  kx ˆi (Hooke's law) ,
2
k
k
U   for F  
rˆ (long - range gravity and electrostatic forces)
r
r2
UH 
For gravity k = GMm for electrostatic force k = kCQ.
U
Force from a radial potential energy: Fr  
rˆ .
r
Chapter 9 – Linear Momentum and Collisions
Conservation of Momentum:
dP
For an isolated system (Fext = 0) Pf  Pi , or P  0, or
 0.
dt
Definition of momentum:
pp
For a particle p  mv  K 
,
2m
For a system ptotal = Σ pi = Σ mivi = Mvcm, where M  Σ mi
mv mv
M
m
i i
i
i
i
i i
.)
i
(Warning: Ktotal is not usually equal to
2
p total
2M
unless the system consists of a
Angular variables:  [rad] 
nd
2 Law of Motion and the Impulse-Momentum Theorem:
dp
particle:

F  Fnet  p 
Fdt 
I.
dt
dP
system:
 Ma CM 
Fext  P 
Fext dt 
dt

i Iα, where τ i  ri  Fi
 

I
d d 2
s
s
d
, 

,  
,
dt
r
r
dt
dt 2
2
Angular units: 1 rev = 360° = 2π rad
1 
For constant angular velocity, f  
, with f [cyc/s],T [s], [rad/s]
T 2
Rotational Inertia:
General definition: I 
mr
2
i  (particles), 
i
r
2
 dm (object)
Parallel-axis theorem: I = Icm+ Md 2
Specific uniform bodies:
Solid sphere about a diameter: I 
Sph. shell about a diameter: I 
2
3
2
5
MR
MR
2
2
Hoop or cyl shell about sym axis: I  MR
Hollow cyl about sym axis: I 
1
4
Solid cylinder about symmetry axis: I 
Solid cyl abt centered diam: I 
2
Rectangular plate abt  axis through center: I 
1
4
MR
2
Thin rod, abt centered  axis: I 
2
2
M ( R1  R 2 )
1
12

1
12
“, abt  axis through an end: I 
M (a
2
1
2
1
12
MR
2
ML
MR
1
3
2
2
MR
2
2
b )
Chapter 11 – Torque and Angular Momentum
Conservation of Angular Momentum:
dL
For an isolated system (τext = 0) L f  L i , or L  0, or
 0.
dt
Second Law of Motion:
Definition of torque: τ  r  F
dL
particle:

τ  τ net  L 
τ dt.
dt
dL
system:

τ ext  L 
τ ext dt.
dt
(The integral forms of these equations are the angular impulse-momentum
theorems.)
Definition of angular momentum:
L  r  p  r  mv for a particle.


L
L
i


i  (R  P)  L CM  (R  MV)  L CM for a system
L  Iω for a rigid body rotatingabout a principalaxis
Rolling (for a object of cylindrical symmetry, rolling without slipping):
1
1
2
vcenter  R and K  I CM  2  MvCM
2
2
Chapter 12 – Static Equilibrium, Elasticity
Two equilibrium conditions:
F  0 , (2)  τ  0
For coplanar forces: (1a)  F
 0, (1b)  F
(2a) 
0.
Ext
Ext
x, Ext
z ,Ext
ext .
θ
v
 r 2  v
r
The translational kinematic and dynamic relationships of other chapters
hold for these variables if we identify t↔t, x↔θ, v↔ω, a↔α, m↔I, F↔τ,
p↔L, K↔K, k↔κ, W↔W, e.g., Newton’s 2nd Law: F = ma ↔ τ = Iα
s  r , v  r , a tan  r , a radial 
Vector format: (1)
single particle or object.)

i
Kinematics:
K
(Hence vCM 
τ
Kinetic energy: K 
xf
xi
K  U  U
 r dm for continuoussystems.
1 2
I for a rotatingrigid body
2
1
1
2 for a body rolling without slipping
 I CM  2  MvCM
2
2
e.g., Wg  mg( y f  yi ) for Fg  mg ˆj
2. One-dimensional motion: W 
M rCM 
Equation of motion: τ net 
Special Cases:
1. F = constant: w  F  d  F  r ,
E
m r
i 1
from ri to r f. If Fi is (is not) conservative, the integral and therefore Wi
does not (does) depend upon the path taken in evaluating the integral.
Wnet 
M rCM 
y,Ext  0,
s
r
s
If (1) and (2) are true are true about one inertial reference point, they are
true for all inertial reference points.
Center of Gravity:
N
MgCG rCG 
mi gi ri for discrete particles.

i 1
Mg CG rCG 
 r gdm for continuoussystems.
Note that if g is constant across the mass distribution (or varies negligibly)
then rCG = (or  ) rCM.
Chapter 13  Universal Gravitation
Law of Universal Gravitation:
m m
F21  G 1 2 (rˆ12 ) where F21 is the force exerted by m2 on m1, r̂12 is a
r2
unit vector directed from m2 towards m1, i.e., the force is attractive, and
G  6.674 1011
N.m2
2
kg

2
N.m2
 1010
is theuniversalgravitational constant.
3
kg2
This expression is valid for point masses and also for symmetric spheres (r
being the center-to-center separation). Where more than two masses are
present, the net force on any particular mass is obtained by simple
superposition, i.e., Fj,net  Fj1  Fj2  Fj3  . . . . 

i,i  j
Fji .
Astrophysical consequences:
g of earth
Kepler’s Laws of planetary motion: (1) elliptical (conic section) orbits,
 4 2 
dA
a 3  K a 3
(2)
 constant , or L = constant, (3) T 2  
S
 GM S 
dt


Motion in binary star systems
Large-scale cosmic structure
Gravitational potential energy:
Gm1m2
. Note that U (r = ) = 0.
Ug  
r
Bound and unbound orbits. vesc 
2GM
. Superposition applies.
R
Gravitational field:
Fg
g
[m/s2  N/kg] .
m
Gravitaional mass versus inertial mass:
mia = mgg, or a 
mg
mi
g
.
Any departure of a from g (in a vacuum) would imply the two are different.
Experiment verifies that any difference must be <1 part in 10 11.