Physics 8
Spring 2012
TA:
NAME:
Physics 8 Spring 2012
Midterm 1 Solutions
For the midterm, you may use one sheet of notes with whatever you want to put on it, front
and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please
ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR
FINAL ANSWERS! You have the full length of the class. If you attach any additional
scratch work, then make sure that your name is on every sheet of your work. Good luck!
1. Wile E. Coyote (Carnivorus hungribilus) is chasing the Roadrunner (Speedibus cantcatchmi ) yet again. While running down the road, they come to a deep gorge, 15.0 m
straight across and 100 m deep. The Roadrunner launches himself across the gorge at
a launch angle of 15◦ above the horizontal, and lands with 1.5 m to spare.
(a) What was the Roadrunner’s launch speed?
(b) Wile E. Coyote launches himself across the gorge with the same initial speed, but
at a different launch angle. To his horror, he is short of the other lip by 0.50 m.
What was his launch angle? (Assume that it was less than 15◦ .)
————————————————————————————————————
Solution
For a given launch speed and angle, the range of a projectile is given by
v02
R=
sin(2θ).
g
(a) The roadrunner jumps at an angle of θ = 15◦ , and reaches a range R = 16.5
meters. So, solving for the velocity gives
s
r
Rg
16.5 × 9.8
= 18 m/s.
v0 =
=
sin(2θ)
sin 30◦
(b) The poor coyote only jumps 14.5 meters. Solving for the angle gives
1 −1 14.5 × 9.8
1 −1 Rg
θ = sin
= 13◦ .
= sin
2
2
2
v0
2
18
1
2. When visiting Yosemite, campers staying in Curry Village are warned by park rangers
to store food in lockers outside the cabins to keep bears from getting into your supplies
(and never leave food in your car - bears have broken in to get at a single candy bar
wrapper ). However, bear lockers are not available when camping in the wilderness and
the next best alternative is to hang up your pack. Consider a 26 kilogram pack seen
the two circumstances in the figures below.
(a) In the first case, the ropes were of different lengths and the pack hangs at the
angles seen. Calculate the tension in the two ropes.
(b) The first situation hung too low, making the pack accessible to the bear. So,
you’ve shortened the longer rope to hang the pack as seen in the second case.
Now what is the tension in the ropes?
————————————————————————————————————
Solution
(a) To determine the tensions, we simply need to look at the forces acting on the
pack. In the first case, we have the tension from the first rope, T~1 , pulling the
pack up to the left at an angle of θ1 = 71◦ with respect to the horizontal, and the
tension, T~2 , pulling the pack up to the right at an angle of θ2 = 28◦ . Furthermore,
there is also the weight of the pack pulling down on it. So, we just need to look
at the force components (we’ll plug in the numbers at the end),
ΣFx =
−T1 cos θ1 + T2 cos θ2
= 0
ΣFy = T1 sin θ1 + T2 sin θ2 − mg = 0.
The first equation tells us that T1 cos θ1 = T2 cos θ2 ⇒ T1 =
into the second equation, and factoring out the T2 , gives
(cos θ2 × tan θ1 + sin θ2 ) T2 = mg ⇒ T2 =
2
cos θ2
T.
cos θ1 2
Plugging this
mg
.
(cos θ2 × tan θ1 + sin θ2 )
Plugging in the numbers gives
T2 =
26 × 9.8
mg
=
= 84 Newtons.
◦
(cos θ2 × tan θ1 + sin θ2 )
cos (28 ) tan (71◦ ) + sin (28◦ )
Then, since T1 =
cos θ2
T,
cos θ1 2
T1 =
we find
cos θ2
cos (28◦ )
T2 =
× 84 = 228 Newtons.
cos θ1
cos (71◦ )
(b) Now we do everything over again, simply with different angles. Letting θ1 = θ2 ≡
θ, we have
ΣFx =
−T1 cos θ + T2 cos θ
= 0
ΣFy = T1 sin θ + T2 sin θ − mg = 0.
The first equation tells us that T1 = T2 , meaning that the tensions are the same
in both ropes, as we should expect for this symmetric situation. The second
equation tells us that
2T sin θ = mg ⇒ T =
26 × 9.8
mg
= 340 Newtons,
=
2 sin θ
2 sin (22◦ )
which is a bigger tension than the first case, since the pack is pulled tighter.
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3. Consider two displacement vectors, ~r1 = 16î + 31ĵ meters and ~r2 = 21î + 10ĵ + 14k̂
meters.
(a) What is the length of each vector, r1 = |~r1 |, and r2 = |~r2 |?
(b) What is the angle, θ, between the two vectors, ~r1 and ~r2 ?
(c) How much work does a force F~ = 67î + 23ĵ + 55k̂ Newtons do as it acts on a body
moving in a straight line from ~r1 to ~r2 (i.e., along ∆~r = ~r2 − ~r1 )?
————————————————————————————————————
Solution
(a) The length of the first vector is
√
r1 = 162 + 312 = 34.9 meters,
while
r2 =
√
212 + 102 + 142 = 27.1 meters.
(b) The angle between the two vectors can be found using the dot product. We have
~r1 · ~r2 = r1 r2 cos θ = r1x r2x + r1y r2y + r1z r2z ⇒ cos θ =
r1x r2x + r1y r2y + r1z r2z
.
r1 r2
So,
cos θ =
r1x r2x + r1y r2y + r1z r2z
16 × 21 + 31 × 10 + 0 × 14
= 0.682,
=
r1 r2
34.9 × 27.1
which gives θ = cos−1 (0.682) = 47◦ .
(c) The work done by a constant force, F~ is just W = F~ · ∆~r. Now,
∆~r = ~r2 − ~r1 = (21 − 16) î + (10 − 31) ĵ + (14 − 0) k̂ = 5î − 21ĵ + 14k̂.
So, the work is
~
W = F
· ∆~r
= 67î + 23ĵ + 55k̂ · 5î − 21ĵ + 14k̂
= 67 × 5 − 23 × 21 + 55 × 14
= 622 Joules.
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4. A 190 gram block is launched by compressing a spring of constant k = 200 N/m by
15 cm. The spring is mounted horizontally,
and the surface directly under it is frictionless. But, beyond the equilibrium position
of the spring end, the surface has frictional
coefficient µk = 0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise, as shown in the figure to
the right.
(a) If there was no frictional part, how high up the ramp would the block slide?
(b) If we now include the frictional part, how high up the ramp does the block slide?
(c) After the block reaches the top of the ramp, it slides back down, over the frictional
surface, compressing the spring. The spring then pushes the block back to do the
same trip over, again. However, each during each trip, the block slows down
when passing over the rough patch. How many times does the block slide over
the frictional surface? (Note: you’ll get a decimal, meaning that it doesn’t make
it all the way. Just give the whole number actual completed trips.)
(d) How far along the rough patch (measured from the left-hand-side) does the block
ultimately stop?
————————————————————————————————————
Solution
(a) Compressing the spring gives the block an initial spring potential energy, Ei =
1
kx2 . Once released, the spring pushes the block, speeding it up. Once the spring
2
gets to its equilibrium point, the block comes free and slides with kinetic energy,
1
mv 2 . Energy conservation sets 12 kx2 = 12 mv 2 . Now, if there is no friction, then
2
the block would simply slide along at the same speed until it reached the upward
curve. Then, as it climbs the block slows down, converting its kinetic energy into
gravitational potential energy, mgh, where h is its maximum height. Once again,
if there is no friction, then 12 mv 2 = mgh, as required by conservation of energy.
So,
1 2 1 2
kx2
200 (0.15)2
kx = mv = mgh ⇒ h =
=
= 1.21 meters.
2
2
2mg
2 (0.19 × 9.8)
(b) Now, when we include the friction then the block should not go as high. It still
has the same initial potential energy, Ei = 21 kx2 , and so, therefore the same initial
kinetic energy, but it loses speed (kinetic energy) doing work against the frictional
force. So, since W = ∆KE, then the kinetic energy after it crosses over the rough
patch is
1 2 1 2
1
mvf = mvi + W = kx2 − µk mgD,
2
2
2
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where we have set the initial kinetic energy equal to the initial spring potential
energy by conservation, and also W = −Ff D = −µk mgD is the work done by
friction. Now, the block will use this final kinetic energy to slide up the ramp,
changing this energy to gravitational potential energy, mgh. So, altogether we
find
kx2
1 2
− µk D.
mgh = kx − µk mgD ⇒ h =
2
2mg
Notice that the height is smaller than the frictionless case, as we expected. Plugging in the numbers gives
h=
kx2
200 (0.15)2
− µk D =
− 0.27 × 0.85 = 0.98 meters.
2mg
2 (0.19 × 9.8)
(c) Every time the block passes over the rough patch, it loses an energy µk mgD. So,
it will pass over the patch enough times until all the energy has been used up.
So, the total number of times it moves over the patch will be
N=
1
kx2
kx2
Ei
= 2
=
.
W
µk mgD
2µk mgD
With the numbers, we find
N=
kx2
200 (0.15)2
=
= 5.27 times.
2µk mgD
2 (0.27 × 0.190 × 9.8 × 0.85)
So, the block makes five complete trips over the rough patch, and starts heading
back.
(d) The block made an odd number of total trips, meaning that it ended up over on
the ramp side, and then starts heading back to the spring, but only makes it back
27% of the way from the right-hand side. So, if the whole distance is 85 cm, then
the block ends up a distance from the left-hand side of
d = (1 − 0.27) × 85 = 62 centimeters.
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Extra Credit Question!!
The following is worth 10 extra credit points!
where and σ are constants,
and r is the distance between the molecules. The
potential energy is plotted in
the figure to the right. The
vertical axis is in units of ,
while the horizontal axis is
in units of σ.
Energy
The potential energy between a pair of neutral
atoms or molecules is very
well-approximated by the
Lennard-Jones
Potential,
given by the expression
σ 12 σ 6
,
P E(r) = 4
−
r
r
Molcular Bond Energy
8
7
6
5
4
3
2
1
0
0.75
-1
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
-2
-3
-4
Distance
(a) Why does the potential energy approach zero as the distance gets bigger?
(b) At what separation distance, in terms of σ and , is the potential energy zero?
(c) At approximately what distance is the system in equilibrium? What is the potential energy at that distance? (Express your answers in terms of σ and .)
(d) How much energy would you need to add to the system at equilibrium in order
to break the molecular bonds holding it together? Why?
(e) How much energy is released in the breaking of those molecular bonds? Why?
Note - no calculation is needed to answer these problems!
————————————————————————————————————
Solution
(a) As the two molecules get further apart, the attractive force between them gets
weaker and weaker. When the are very far apart, they hardly interact at all they are basically free molecules. The potential energy of a free particle is zero,
since potential energy depends on the interaction between multiple particles.
(b) We can just read the value off from the graph. We see that the potential energy
crosses the x axis when x = 1, which means that r = σ. We can see this from the
equation, too: setting r = σ gives P E(σ) = 0.
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3.25
(c) The system is in equilibrium when the net force on it is zero. Since the force is
the slope of the potential energy graph, this happens when the slope is zero. The
potential energy graph has zero slope when it’s at it’s minimum point. Checking
the graph, we see that this happens right around x ≈ 1.15, or r ≈ 1.15σ. We could
d
(P E(r)) = 0, which gives r = 21/6 σ ≈ 1.12σ,
check the exact answer by finding dr
and so we were close on our guess. The energy at this distance can just be read
off the graph, giving y = −3, or P E = −3.
(d) In order to break the molecular bonds apart, we’d need to raise the energy to
zero. At equilibrium the energy is P E = −3, and so we’d need to add +3 units
of energy.
(e) There is no energy released in breaking these molecular bonds - we had to add
the energy to break these bonds. Energy is never released in the breaking of
bonds! One can obtain energy by breaking a less stable bond, then forming a
more stable bond. The more stable bond has a more negative potential energy
(a deeper potential “well”). The difference in energy between the initial and final
states is released to the environment.
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