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!
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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.
~ × F~ :
Solution To check if the force is conservative, calculate ∇
~ × F~ )x =
(∇
∂Fy ∂Fz
−
= 0 − 0 = 0.
∂z
∂y
∂Fz ∂Fx
−
= 0 − 0 = 0.
∂x
∂z
~ × F~ )z = ∂Fx − ∂Fy = 1 − 1 = 0.
(∇
∂y
∂x
~ × F~ )y =
(∇
~ × F~ = 0 and the force is conservative. It can be represented as
Therefore ∇
F~ = −∇U(x, y, z)
∂U
=⇒ U(x, y, z) = −yx + C1 (y, z)
∂x
∂U
∂C1
∂C1
Fy = x = −
=x−
=⇒
= 0 =⇒ C1 (y, z) = C2 (z) =⇒ U(x, y, z) = −yx + C2 (z)
∂y
∂y
∂y
Fx = y = −
Fz = z = −
∂C2
∂C2
1
∂U
=−
=⇒
= −z =⇒ C2 (z) = − z 2 + const
∂z
∂z
∂z
2
The end result is
1
U(x, y, z) = −xy − z 2 + const
2
Check: the negative of the gradient of this function should reproduce the given force.
2
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.
Solution
(a) The easiest method is to use energy conservation: the initial energy stored in the spring
1
Ei = k(∆x)2
2
is converted into gravitational potential energy
U = mgh,
kinetic energy of the block
1 2
T = mvtop
2
and heat loss due to the work Wf done by the friction force Ff . The heat loss is
|Wf | =
where
L=
Z
L
0
Ff dL
h
sin θ
3
is the length of the ramp. The friction force is constant:
Ff = µk N = µk mg cos θ
and its work is
|Wf | =
Z
L
0
◦
µk mg cos θdL = µk mg cos θ
Z
L
0
dL = µk mgL cos θ = µk mgh cot θ = µk mgh
because cot 45 = 1. Putting these together,
Ei = T + U + |Wf |
we get
1
1 2
1 2
k(∆x)2 = mgh + mvtop
+ µk mgh = (µk + 1)mgh + mvtop
2
2
2
from where
vtop =
s
k
(∆x)2 − 2(µk + 1)gh
m
Alternatively you can first find the velocity at the bottom of the ramp, then use Newton’s
law with the forces of gravity and friction to compute the velocity at the top of the ramp.
(b) This is a problem on projectile motion. Choosing the origin at the base of the ramp,
and x axis pointing to the right and y-axis pointing up, the initial conditions are
x(0) = 0,
y(0) = h,
ẋ(0) = vtop cos θ
ẏ(0) = vtop sin θ
The equation of motion in the horizontal direction is
mẍ = 0
with solution (accounting for the initial conditions)
x(t) = vtop t cos θ
The equation of motion for y is
mÿ = −mg
with solution (accounting for the initial conditions)
1
y(t) = h + vtop t sin θ − gt2 .
2
The time T when the particle falls on the ground is found from
y(T ) = 0
and is given by
T =
vtop sin θ +
q
2
vtop
sin2 θ + 2gh
g
(the other root gives negative T , i.e. corresponds to going backwards in time). The range then
is
q
q
2
2
vtop cos θ vtop sin θ + vtop
sin2 θ + 2gh
vtop (vtop + vtop
+ 4gh)
R = x(T ) =
=
g
2g
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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.
Solution
(a) Since there is no air resistance, energy is conserved:
1
1
mgh + mvh2 = 0 + mv02
2
2
from where
v0 =
q
vh2 + 2gh
(b) The vertical axis (say, z) can be chosen pointing up or down, it doesn’t matter. Let’s
choose it to point up, with origin at the pool’s surface. The equation of motion for z is then
mz̈ = −mg + Mg − mk ẋ
where F~g = −mg~e3 is gravity, F~A = Mg~e3 is the buoyant (Archimedes) force and F~r = −mk~v is
the retarding force. Because we assume m = M in this part of the problem, gravity is precisely
cancelled by the buoyant force and we get
z̈ = −k ż
which can be immediately integrated:
ż = −kz + C
The constant C is determined from the initial condition ż(z = 0) = −v0 , which gives C = −v0 :
ż(z) = −v0 − kz
The diver will stop at a distance z = −D at which ż(−D) = 0:
v0
0 = −v0 − k(−D) =⇒ D = .
k
(c) Left for homework.
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