Contemporary Physics I – HW 6
HW 6
Due November 16, 2016
As always, please answer all questions clearly and concisely. While you need not transcribe the question
completely, it should be clear from your answer alone what you are talking about.
You are strongly encouraged to discuss the homework with your classmates, but you must complete the
written homework by yourself, and of course, the material you submit must be your own.
Remember, show all of your work!
1. A 4 kg block is attached to a spring with a spring constant k = 200N/m, and is stretched an amount
0.2m.
(a) Sketch the potential energy curve for the spring. Make sure that the x=axis goes from at least
-0.2m to 0.2m.
(b) How much potential energy is stored in the spring when it is stretched?
1
1 2
kx = (200 N/m)(0.2 m)2
2
2
U =4J
U=
(c) The block is then released from rest. Where (what value of x) will the block have its maximum
kinetic energy? How fast will it go at that moment?
It will have maximum kinetic energy right before it is pulled back in the other direction, at 0.2m.
Ui + Ki + W = Uf + Kf
1 2
1
kx = mv 2
2
2
r
k
v=
x
m
s
200 N/m
(0.2 m)
=
4 kg
v = 1.41 m/s
2. A very complicated molecule has an energy diagram as shown below:
where the x-axis represents the separation between the two atoms. Your answers to the following will
be approximate, but in each case, I’ve tried to make it so that correct answer(s) will be integers or
half-integers.
(a) What separations are the equilibria of the system? Are they stable or unstable?
Equilibria points occur at extrema of the system, maximums (unstable) and minimums (stable).
From the image it’s clear at which values of r we have extrema.
r ≈ 1Å (stable)
r ≈ 3Å (unstable)
r ≈ 4Å (stable)
(b) Suppose the atoms are separated by 2Å and are at rest, describe what will happen to the molecule.
The separation will oscillate from ≈ 2Å to ≈ 0.5Å, eventually settling to ≈ 1Å.
(c) The atoms are now separated by 1Å and are given 3eV of kinetic energy. What is the total energy
of the system?
ET = EK + EU
= 3 eV + (−5 eV )
ET = −2 eV
(d) Under those circumstances, what are the range of motion of the atoms?
By investigating the diagram we can see that at -2eV, it can move from ≈ 0.5Å to ≈ 2.5Å.
(e) Under the circumstances described by part c) (and since), how much energy is required to disassociate the molecule?
Dissociating a molecule means that r → ∞. Looking at the graph shows that as r increases
Ux → 0. Since we are at -2eV we would then need more 2eV to reach 0.
(f) Estimating from the curve at r = 2Å, what is the approximate force on an atom? (Hint: 1eV /Å =
1.6 × 10−9 N ).
dU
dx
−2 eV − (−4 eV )
=−
1Å
= (−2 eV /Å) · (1.6 × 10−9 N/(eV /Å))
F =−
F = −3.2 × 10−9 N
3. I’m going to make some not entirely accurate statements about air. In order to avoid confusion, I’ll
indicate true statements with a “T.” The other statements should be treated as true for the purposes
of this problem.
• (T) Air is made up predominantly of molecular nitrogen, which has a molecular mass of approximately 28. (28 × the mass of a proton).
• (T) At sea level, typical air density is approximately 1.3kg/m3 .
• (T) Room temperature is about 300K.
• (T) An ideal monatomic gas has an energy of:
E=
3
kb T
2
per atom and an equation of state (pressure relation)
P V = N kb T
where N is the number of atoms and V is the volume of the gas.
• You may treat air as being made entirely of molecular nitrogen, and you should treat the molecular
nitrogen as a monatomic gas.
(a) What is the number density of air molecules?
n=
=
ρ
m
1.3 kg/m3
28 · (1.7 × 1027 kg)
n = 2.73 × 1025 m3
(b) What is the typical kinetic energy of an air molecule?
3
kb T
2
3
= (1.38 × 10−23 J/K)(300 K)
2
EK =
EK = 6.21 × 10−21 J
(c) How fast is a typical air molecule moving?
1
mv 2
2
r
2K
v=
m
s
2 · (6.21 × 10−21 J)
=
28 · (1.7 × 10−27 kg)
K=
v = 510.81 m/s
(d) Incidentally, it can be shown that the sound speed of a monatomic gas is:
r
kb T
cs =
m
where m is the mass of an atom (molecule). What is the sound speed of air, and how does that
compare to the result from the previous part?
r
kb T
cs =
m
s
1.38 × 10−23 J/K)
=
28 · (1.7 × 10−27 kg)
cs = 294.91 m/s
(e) What is air pressure at room temperature?
P V = N kB T
N
P = kB T
V
= nkB T
= (2.73 × 1025 m−3 )(1.38 × 10−23 J/K)(300K)
P = 1.13 × 105 N/m2
4. 6.P.62 Calculate the speed of a satellite in a circular orbit near the Earth (just above the atmosphere).
If the mass of the satellite is 200kg, what is the minimum energy required to move the satellite from
this near-Earth orbit to very far away from the Earth?
Let’s assume the distance of the satellite from the centre of Earth is approximately the same as the
radius of Earth. We can balance the gravitational force with the centripetal force.
GME m
mv 2
=
2
RE
RE
r
GME
v=
R
r E
(6.67 × 10−11 m3 /kg/s2 )(5.97 × 1024 kg)
=
6.37 × 106 m
v = 7.92 × 103 m/s
W = ∆E
W = (Kf − Ki ) + (Uf − Ui )
1
GME m
= 0 − mv 2 + 0 − −
2
RE
1
GME
GME m
=− m
+
2
RE
RE
GME m
=
2RE
(6.67 × 10−11 m3 /kg/s2 )(5.97 × 1024 kg)(200 kg)
=
2(6.37 × 106 m)
W = 6.28 × 109 J
5. 6.P.69 (a,d only) A pendulum consists of a very light but stiff rod of length L hanging from a nearly
frictionless axle, with a mass m at the end of the rod.
(a) Calculate the gravitational potential energy as a function of the angle θ, measured from the vertical.
L−h
L
L cos θ = L − h
cos θ =
h = L(1 − cos θ)
U = mgh
U = mgL(1 − cos θ)
(b) Suppose that you hit the stationary hanging mass so it has an initial speed vi . What is the
minimum initial speed needed for the pendulum to go over the top (θ = 180◦ )?
From conservation of energy...
K=U
1
mv 2 = mgL(1 − cos θ)
2
p
v = 2gL(1 − cos θ)
p
= 2gL[1 − cos (180◦ )]
p
= 4gL
p
v = 2 gL
6. 7.P.23 A relaxed spring of length 0.15 m stands vertically on the floor; its stiffness is 1000 N/m. You
release a block of mass 0.4 kg from rest, with the bottom of the block 0.8 m above the floor and straight
above the spring. How long is the spring when the block comes momentarily to rest on the compressed
spring?
∆L = Lf − Li = 0.8 m − 0.15 m
= 0.65 m
1
U = mgL = k(x0 − x)2
r 2
2mgL
→ x0 = x −
k
s
2 · (0.4 kg) · (9.8 m/s2 ) · (0.65 m)
= 0.15 m −
1000 N/m
x0 = 0.078 m
7. 7.P.36 During three hours one winter afternoon, when the outside temperature was 0◦ C (32◦ F),
a house heated by electricity was kept at 20◦ C (68◦ F) with the expenditure of 45 kWh (killowatthours) of electric energy. What was the average energy leakage in joules per second through the walls
of the house to the environment (the outside air and ground)? The rate at which energy is transferred
between two systems is often proportional to their temperature difference. Assuming this to hold in
this case, if the house temperature had been kept at 25◦ C (77◦ F), how many kWh of electricity would
have been consumed?
Energy leakage =
45 kW h
= 15 kW
3h
= 1.5 × 104 J/s
T2
E2 =
E1
T1
◦ 25 C
=
· 45 kW h
20◦ C
E2 = 56.3 kW h