PHY 3221: Mechanics I
Fall Term 2010
Exam 1, September 29 2010
• This is a closed book exam lasting 50 minutes.
• There are three problems worth a total of 20 pts. Begin each problem on a fresh sheet of
paper. Use only one side of the paper. Avoid microscopic handwriting.
• Put your name, the problem number and the page number in the upper right-hand corner
of each sheet.
• To receive partial credit you must explain what you are doing. Carefully labelled figures
are important! Randomly scrawled equations are not helpful.
• Draw a box around important results (or at least results which you think might be important).
• Good luck!
1
Problem 1. Conservative forces. [4 pts] Consider the following force: Fx = y, Fy = x,
Fz = z. Is this a conservative force? If so, find the potential energy U(x, y, z) associated with
it.
Problem 2. Spring launcher. [10 pts.] A block of mass m is launched using a spring
of spring constant k which has been compressed from its equilibrium configuration by a displacement ∆x (see figure). When the spring is released, the block slides first on a smooth
(frictionless) level surface, then goes up a ramp in the form of a wedge of height h and inclination angle θ = 45◦ (see figure). The ramp’s inclined surface is rough, with a coefficient of
kinetic friction µk .
~vtop
equilibrium
~g
rough, µk
∆x
k
smooth
h
θ = 45◦
R
m
Figure 1: An illustration for the spring launcher problem.
(a) [5 pts] Find the speed vtop of the block at the top of the ramp.
(b) [5 pts] The block flies off the ramp at an angle θ = 45◦ and lands a distance R from the
base of the ramp. Ignoring air resistance, find R in terms of vtop , h and g.
2
Problem 3. Diving Gators. [6 pts] You are contracted by the UF Athletic Association
to calculate the minimum depth of the swimming pool used by the Gator diving team. Assume
that the diving platform is a distance h above the surface of the pool. Upon leaving the
platform, divers typically jump straight up with velocity vh (ignore its horizontal component).
(a) [3 pts] Find the velocity v0 of the diver at the moment of impact onto the water surface,
in terms of h, vh and g. Ignore air resistance.
(b) [3 pts] Assume that upon entering the water of the pool, the diver begins to experience
a retarding force linearly proportional to the velocity, i.e. Fr = mkv, where m is the mass of
the diver and k is a known constant. Let M be the mass of the water displaced by the body of
the diver. For full credit [3 pts], you can assume that the density of the human body is equal
to the density of water, i.e. m = M. Find the distance D that the diver will travel in the water
before she stops. (You can neglect the size of the diver and treat her as a point particle.) Write
your answer for D in terms of v0 , k and g.
(c) [3 pts] For extra credit, do part (b) under the more realistic assumption of M > m, i.e.
that the body of the diver floats (does not sink) in the pool. Write your answer for D in terms
of v0 , m, M, k and g.
3
Formula sheet
A·(B × C) = B·(C × A) = C·(A × B) ≡ ABC
A×(B × C) = (A · C)B − (A · B)C
(A × B) · (C × D) = A · [B × (C × D)]
= A · [(B · D)C − (B · C)D]
= (A · C)(B · D) − (A · D)(B · C)
(A × B) × (C × D) = [(A × B) · D] C − [(A × B) · C] D
= (ABD)C − (ABC)D = (ACD)B − (BCD)A
v = ṙ er + r θ̇ eθ + r sin θφ̇ eφ
a =
+ 2ṙ φ̇ sin θ + 2r θ̇φ̇ cos θ + r θ̈ sin θ eφ
v = ṙ er + r φ̇ eφ + ż ez
a =
r̈ − r θ̇2 − r φ̇2 sin2 θ er + 2ṙ θ̇ + r θ̈ − r φ̇2 sin θ cos θ eθ
r̈ − r φ̇2 er + r φ̈ + 2ṙ φ̇ eφ + z̈ ez
X
εijk εlmk = δil δjm − δim δjl
k
X
εijk εljk = 2 δil
j,k
X
εijk εijk = 6
i,j,k
Time averages over one period T :
1
hsin ωti =
T
2
hcos2 ωti =
Tidal force
Z
t+T
dt sin2 ωt =
t
1 Z t+T
1
dt cos2 ωt =
T t
2
2GmMm r cos θ
D3
GmMm r sin θ
= −
D3
FT x =
FT y
1
2
Rocket motion
Fext = mv̇ + uṁ
4
Simple harmonic oscillator:
mẍ + kx = 0
x(t) = A sin(ω0 t − δ)
x(t) = A cos(ω0 t − φ)
s
2π
=
ω0 = 2πν0 =
τ0
k
m
Damped oscillator:
b
ẍ + 2β ẋ + ω02 x = 0, 2β =
m
√ 2 2
√ 2 2 x(t) = e−βt A1 e β −ω0 t + A2 e− β −ω0 t
Underdamped motion
x(t) = Ae−βt cos(ω1 t − δ),
ω1 =
q
ω02 − β 2
Critically damped motion
x(t) = (A + Bt)e−βt
Overdamped motion
i
h
x(t) = e−βt A1 eω2 t + A2 e−ω2 t ,
Driven oscillator
ω2 =
q
β 2 − ω02
F0
ẍ + 2β ẋ + ω02 x = A cos ωt, A =
m
√ 2 2
√ 2 2 xc (t) = e−βt A1 e β −ω0 t + A2 e− β −ω0 t
A
cos(ωt − δ)
xp (t) = q
(ω02 − ω 2 )2 + 4ω 2β 2
δ = tan
−1
q
2ωβ
2
ω0 − ω 2
!
ω02 − 2β 2
ωR
Q=
2β
ωR =
RLC circuit
VL = L
Gauss’s law
Z
S
dI
dt
VR = RI
~n · ~g da = −4πG
5
VC =
Z
V
ρ dv
q
C