Download Midterm Exam 3

Physics 130
General Physics
Fall 2012
Midterm Exam 3
November 29, 2012
Name:
Instructions
1. This examination is closed book and closed notes. All your belongings
except a pen or pencil and a calculator should be put away and your
bookbag should be placed on the floor.
2. You will find one page of useful formulae on the last page of the exam.
3. Please show all your work in the space provided on each page. If you
need more space, feel free to use the back side of each page.
4. Academic dishonesty (i.e., copying or cheating in any way) will
result in a zero for the exam, and may cause you to fail the
class.
In order to receive maximum credit,
each solution should have:
1.
2.
3.
4.
5.
6.
7.
8.
9.
A labeled picture or diagram, if appropriate.
A list of given variables.
A list of the unknown quantities (i.e., what you are being asked to find).
One or more free-body or force-interaction diagrams, as appropriate,
with labeled 1D or 2D coordinate axes.
Algebraic expression for the net force along each dimension, as appropriate.
Algebraic expression for the conservation of energy or momentum equations, as appropriate.
An algebraic solution of the unknown variables in terms of the
known variables.
A final numerical solution, including units, with a box around it.
An answer to additional questions posed in the problem, if any.
1
Physics 130
General Physics
Fall 2012
1. A 10 m long glider with a mass of 680 kg (including the passengers) is gliding horizontally
through the air at 30 m/s when a 60 kg skydiver drops out by releasing his grip on the
glider. Using the appropriate conservation law, determine the glider’s velocity just after
the skydiver lets go.
Solution:
This is a conservation of momentum problem. At the start, both the glider and the
skydiver are traveling at the same speed.
pf x “ pix
mg pvf qg ` ms pvf qs “ pmg ` ms qpvi q
(1)
(2)
Immediately after release, the skydiver’s horizontal velocity is still 30 ms. Solving
for the final velocity of the glider,
pmg ` ms qpvi q ´ ms pvf qs
mg
p680kgqp30m{sq ´ 60kgp30m{sqs
“
620kg
“ 30m{s
pvf qg “
(3)
pvf qg
(4)
pvf qg
(5)
The skydiver’s motion in the vertical direction has no influence on the glider’s horizontal motion.
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Physics 130
General Physics
Fall 2012
2. A 20 g ball of clay traveling east at 2.0 m{s collides with a 30 g ball of clay traveling
30˝ south of west at 1.0 m{s. What are the speed and the direction of the resulting 50 g
ball of clay?
Solution:
This is an inelastic two-dimensional conservation of momentum problem. Applying
conservation of momentum in the x-direction, the final momentum in the x-direction
is
pf x “ pix “ m1 pvix q1 ` m2 pvix q2
“ p0.020 kgqp2.0 m{sq ´ p0.030 kgqp1.0 m{sq cos 30˝
“ 0.0140 kg m{s
(1)
(2)
(3)
Applying conservation of momentum in the y-direction, the final momentum in the
y-direction is
pf y “ piy “ m1 pv ´ iyq1 ` m2 pviy q2
“ p0.020 kgqp0 m{sq ´ p0.030 kgqp1.0 m{sq sin 30˝
“ ´0.0150 kg m{s
(4)
(5)
(6)
Next, we find the magnitude of the final momentum by combining the x- and ycomponents using the Pythagorean theorem:
a
pf x ` pf y
pf “
(7)
a
p0.0140 kg m{sq2 ` p´0.0150 kg m{sq2
(8)
“
“ 0.0205 kg m{s
(9)
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Physics 130
General Physics
Fall 2012
Using the final momentum, we can solve for the final speed:
pf “ pm1 ` m2 qvf
pf
ñ vf “
m1 ` m2
0.0205 kg m{s
“
0.020 kg ` 0.030 kg
“ 0.41 m{s
(10)
(11)
(12)
(13)
(14)
The direction is given by
ˆ
˙
pf y
θ “ tan
p
ˆ fx
˙
´0.0150 kg m{s
´1
“ tan
0.0140 kg m{s
˝
“ ´47
´1
4
(15)
(16)
(17)
Physics 130
General Physics
Fall 2012
3. Most geologists believe that the dinosaurs became extinct 65 million years ago when a
large comet or asteroid struck the earth, throwing up so much dust that the sun was
blocked out for a period of many months. Suppose an asteroid with a diameter of 2.0 km
and a mass of 1.0 ˆ 1013 kg hits the earth with an impact speed of 4.0 ˆ 104 m{s.
(a) What is the earth’s recoil speed after such a collision? (Hint: Use a reference frame
in which the earth is initially at rest.)
(b) What percentage is this of the earth’s speed around the sun?
Solution:
(a) This is an inelastic collision, and therefore a conservation of momentum problem.
At the start, the asteroid has a speed of 4.0 ˆ 104 m{s while the earth is not
moving.
pix “ pf x
Mearth pvi qe ` ma pvi qa “ pMearth ` ma qpvf q
(1)
(2)
(3)
Solving for the final velocity of the earth and asteroid system,
vf “
Mearth pvi qe ` ma pvi qa
pMearth ` ma q
5
(4)
Physics 130
General Physics
0 kg m{s ` p1.0 ˆ 1013 kgqp4.0 ˆ 104 m{sq
p5.98 ˆ 1024 kg ` 1.0 ˆ 1013 kgq
“ 6.7 ˆ 10´8 m{s
“
Fall 2012
(5)
(6)
(b) The speed of the earth going around the sun is
vE “
2πrearth´sun
2πp1.50 ˆ 1011 mq
“
“ 3.0 ˆ 104 m{s
T
3.15 ˆ 107 s
(7)
(8)
Therefore,
vf {vE “ p6.7 ˆ 10´8 m{sq{p3.0 ˆ 104 m{sq
“ 2.2 ˆ 10´12
“ 2.2 ˆ 10´10 %.
6
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
4. Fred (mass 60 kg) is running with the football at a speed of 6.0 m{s when he is met
head-on by Brutus (mass 120 kg), who is moving at 4.0 m{s. Brutus grabs Fred in a
tight grip, and they fall to the ground. Which way do they slide, and how far? The
coefficient of kinetic friction between football uniforms and Astroturf is 0.30.
Solution:
This is an inelastic collision in which Fred and Brutus are moving toward each other.
Assume that Brutus is moving in the positive x-direction and Fred is moving in the
negative x-direction. Because momentum is conserved, we have
pix “ pf x
mB pvi qB ´ mF pvi qF “ pmB ` mF qpvf q
(1)
(2)
Solving for the final velocity of Brutus and Fred,
mB pvi qB ´ mF pvi qF
pmB ` mF q
p120 kgqp4.0 m{sq ´ p60 kgqp6.0 m{sq
“
p120 kg ` 60 kgq
“ 0.667 m{s
vf “
(3)
(4)
(5)
Now that we know that Brutus and Fred moved to the right after the collision, we
can find the distance that they traveled. Friction slows them down, and therefore
the force of friction must act in the negative x-direction. The net force is equal to
the total mass (the mass of both players) times the acceleration.
ÿ
Fx “ ´fk “ pmB ` mF qa
(6)
´µn “ pmB ` mF qa
´µpmB ` mF qg “ pmB ` mF qa
ñ a “ ´µg
“ ´p0.30q ˆ p9.8 m{s2 q
“ ´2.94 m{s2
7
(7)
(8)
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
Finally, we can use the kinematic equation to find the distance traveled with vf “ 0:
vf2 “ vi2 ` 2ad
ñd “ ´
vi2
2a
´p0.667 m{s2 q2
“
2 ˆ p´2.94 m{s2 q
“ 0.076 m
“ 7.6 cm
8
(12)
(13)
(14)
(15)
(16)
Physics 130
General Physics
Fall 2012
5. A two-stage rocket is traveling at 1200 m{s with respect to the earth when the first
stage runs out of fuel. Explosive bolts release the first stage and push it backward with
a speed of 35 m{s relative to the second stage. The first stage is three times as massive
as the second stage. What is the speed of the second stage after the separation?
Solution:
The rocket is traveling at an initial speed before the explosion. The two parts separate
after the explosion. To find the speed of the second stage, we’ll use the conservation
of momentum equation.
pf
pm1 ` m2 qvi
p3m2 ` m2 qvi
4m2 vi
4vi
“
“
“
“
“
pi
m1 v1f ` m2 v2f
3m2 v1f ` m2 v2f “ m2 p3v1f ` v2f q
m2 p3v1f ` v2f q
3v1f ` v2f
(1)
(2)
(3)
(4)
(5)
The fact that the first stage is pushed backward at 35 m{s relative to the second can
be written
v1f “ ´35 m{s ` v2f
4vi “ 3p´35 m{s ` v2f q ` v2f
4vi “ ´105 m{s ` 4v2f
(6)
(7)
(8)
Solving for the velocity of the second stage v2f
v2f “
4vi ` 105 m{s
4
9
(9)
Physics 130
General Physics
4 ˆ 1200 m{s ` 105 m{s
4
“ 1.2 km{s
“
10
Fall 2012
(10)
(11)
Physics 130
General Physics
Fall 2012
6. A 500 g rubber ball is dropped from a height of 10 m and undergoes a perfectly elastic
collision with the earth.
(a) For an elastic collision, what quantities are conserved?
(b) Write out the conservation of momentum equation for this problem.
(c) Write out the conservation of energy equation for this problem.
(d) Which quantities are not known in these equations?
(e) Explain how to obtain the formulas for these unknown quantities.
(f) What is the earth’s velocity after the collision? Assume the earth was at rest just
before the collision.
(g) How many years would it take the earth to move 1.0 mm at this speed?
Solution:
(a) In an elastic collision, both energy and momentum are conserved.
(b) The conservation of momentum equation is
pi “ pf
mb vbi ` me vei “ mb vbf ` me vef
mb vbi ` 0 “ mb vbf ` mevef
(1)
(2)
(3)
(c) The conservation of energy equation is
Ki ` Ui “ Kf ` Uf
1
1
1
1
mb pvbi q2 ` me pvei q2 “
mb pvbf q2 ` me pvef q2
2
2
2
2
1
1
1
mb pvbi q2 ` 0 “
mb pvbf q2 ` me pvef q2
2
2
2
2
2
mb pvbi q ` 0 “ mb pvbf q ` me pvef q2
(4)
(5)
(6)
(7)
(d) The quantities that are not known in these equations are vbf and vef .
(e) We’re given the mass of the rubber ball and it’s initial velocity. We also know
the mass of the earth. Therefore, we have two unknowns and two equations.
The unknowns are vbf and vef . To find the final velocity of the earth, we can
11
Physics 130
General Physics
Fall 2012
solve for vbf in the momentum equation and plug this into the energy equation to
eliminate vbf . This will give us the equation for vef . Once we have an expression
for vef , we can plug this into the momentum equation to get an expression for
vbf .
(f) We’ll assume that the earth is at rest right before the collision. To find the
speed of the rubber ball right before the collision, we can use kinematics.
pvf B q2 “ ppviB q2 ` 2a∆h “ 2p9.8m{s2 qp10mq
b
2p9.8m{s2 qp10mq “ 14.0m{s
vf B “
(8)
(9)
By continuing with the derivations above, we find that the equations for vbf are
and vef
mb vbi ´ me vef
mb
2mb vbi
“
me ` mb
vbf “
(10)
vef
(11)
We found that the velocity of the ball right before it collided with the earth was
14.0m{s. This is our initial velocity of the ball in our conservation of momentum
and energy equations. So, vbi “ 14.0m{s. Plugging this into the above equation,
we solve for vef .
vef “
2p0.500kgqp14.0m{sq
“ 2.3 ˆ 10´24 m{s
24
5.98 ˆ 10 kg
(12)
(g) The time it would take the earth to move 1.0mm at this speed is given by the
kinematics equation.
1
xf “ xi ` vi ∆t ` a∆t2 “ 0 ` vi ∆t ` 0
2
xf
0.001m
∆t “
“ 4.3 ˆ 1020 sec “ 1.36 ˆ 1013 years
“
vi
2.3 ˆ 10´24 m{s
12
(13)
(14)
Physics 130
General Physics
Fall 2012
7. A package of mass m is release from rest at a warehouse loading dock and slides down
a 3.0 m high frictionless chute to a waiting truck. Unfortunately, the truck driver went
on a break without having removed the previous package, of mass 2m, from the bottom
of the chute.
(a) Suppose the packages stick together. What is their common speed after the collision?
(b) For an elastic collision, what quantities are conserved?
(c) Write out the conservation of momentum equation for this problem.
(d) Write out the conservation of energy equation for this problem.
(e) Which quantities are not known in these equations?
(f) Explain how to obtain the formulas for these unknown quantities.
(g) If the collision between the packages is perfectly elastic, to what height does the
package of mass m rebound?
Solution:
(a) From conservation of energy, we can find the speed of the package at the bottom
of the chute.
Ki ` Ui “ Kf ` Uf
1 2
mv
0 ` mgh “
2
b
a
v “
2gh “ 2p9.8m{s2 qp3.0mq “ 7.668m{s
(1)
(2)
(3)
If the packages stick together, the collision is inelastic and conservation of momentum is conserved.
pi “ pf
m1 v1i ` m2 v2i “ pm1 ` m2 qvf
mv1i ` 0 “ 3mvf
13
(4)
(5)
(6)
Physics 130
General Physics
Fall 2012
Cancelling the m’s
v1i “ 3vf
(7)
Solving for vf , we find
vf “
7.668m{s
v1i
“
“ 2.56m{s
3
3
(8)
(9)
The packages move together with a speed of 2.56m{s.
(b) In an elastic collision, both energy and momentum are conserved.
(c) In this conservation of momentum equation, we’ll use the velocity calculated
above for the initial velocity before the collision. Also, the package at the
bottom is at rest before the collision. The conservation of momentum equation
gives
pi
m1 v1i ` m2 v2i
m1 v1i ` 0
mv1i
v1i
v1f
“
“
“
“
“
“
pf
m1 v1f ` m2 v2f
m1 v1f ` m2 v2f
mv1f ` p2mqv2f
v1f ` 2v2f
v1i ´ 2v2f
(10)
(11)
(12)
(13)
(14)
(15)
(d) The conservation of energy equation is
Ki ` Ui “ Kf ` Uf
1
1
1
1
m1 pv1i q2 ` m2 pv2i q2 “
m1 pv1f q2 ` m2 pv2f q2
2
2
2
2
(16)
(17)
(18)
(e) The quantities that are not known in these equations are v1f and v2f .
(f) We have two unknowns and two equations. The unknowns are v1f and v2f .
To find the final velocity of the second package, we can solve for v1f in the
momentum equation and plug this into the energy equation to eliminate v1f .
This will give us the equation for v2f . Once we have an expression for v2f , we
can plug this into the momentum equation to get an expression for v1f .
From the derivations above, the equation relating the final speed of the first
mass to it’s initial speed is
v1f “
m ´ 2m
´1
v1i “
p7.668m{sq “ ´2.56m{s
m ` 2m
3
14
(19)
(20)
Physics 130
General Physics
Fall 2012
The package rebounds and goes back up the ramp. To determine how high the
package goes, we can use the conservation of energy.
Ki ` Ui “ Kf ` Uf
1 2
mv ` 0 “ 0 ` mgh
2
v2
p´2.56m{sq2
h “
“
“ 0.33 m
2g
2p9.8m{s2 q
15
(21)
(22)
(23)
Physics 130
General Physics
Fall 2012
8. A roller coaster car on the frictionless track shown below starts from rest at height h.
The track’s valley and hill consist of circular-shaped segments of radius R.
(a) What is the maximum height hmax from which the car can start so as not to fly off
the track when going over the hill? Give your answer as a multiple of R. Hint First
find the maximum speed for going over the hill.
(b) Evaluate hmax for a roller coaster that has R “ 10 m.
Solution:
(a) When the car is in the valley and on the hill it can be considered to be in circular
motion. We start at the top of the hill and determine the maximum speed so
that the car stays on the track. We can use the equations for circular motion.
The acceleration is given by
v2
a“´ ,
(1)
r
where the minus sign means that the centripetal acceleration points inward,
toward the center of the circular track.
The free-body diagram for the car at the top of the hill illustrates the two forces
acting on the car. Using Newton’s second law, we know
ÿ
F “ ma “ ´m
FN ´ Fg “ ´m
v2
R
v2
R
(2)
(3)
When the normal force is greater than zero, the car is still touching the track.
Therefore, by setting the normal force equal to zero we can find the maximum
speed that the car can have and still stay on the track.
Fg
16
2
vmax
“ m
R
(4)
Physics 130
General Physics
v2
mg “ m max
a R
ñ vmax “
gR
Fall 2012
(5)
(6)
Now that we know the maximum velocity at the top of the hill, we can use
conservation of energy to find the starting height.
Kf ` Uf ` Ki ` Ui
1 2
1 2
mvmax ` mgyf “
mv ` mgyi
2
2 i
1 2
mv
` mgyf “ 0 ` mgyi
2 max
(7)
(8)
(9)
We’ll use yf “ R for the height at the top of the hill and factor the mass term,
1 2
v
` gR “ 0 ` gyi
2 max
v2
yi “ max ` R
2g
We can now plug in our results for vmax from above
?
p gRq2
yi “
`R
2g
R
`R
“
2
3R
“
2
(10)
(11)
(12)
(13)
(14)
(b) Plugging into the above equation, hmax for a roller coaster that has R “ 10 m
we get
yi “
3R
3 ˆ 10 m
“
“ 15 m
2
2
17
(15)
Physics 130
General Physics
Fall 2012
9. A sled starts from rest at the top of the frictionless, hemispherical, snow-covered hill
shown below.
(a) Find an expression for the sled’s speed when it is at angle φ.
(b) Use Newton’s laws to find the maximum speed the sled can have at angle φ without
leaving the surface.
(c) At what angle φmax does the sled “fly off” the hill?
Solution:
(a) Find sled’s speed When the sled is at a position that relates to angle φ, the
height is the R cos φ. Using conservation of energy to find the speed,
Kf ` Uf “ Ki ` Ui
1
1
mpvf q2 ` mgyf “
mpvi q2 ` mgyi
2
2
1
2
pvf q ` gR cos φ “ 0 ` gyi
2
pvf q2 “ 2gR ´ 2gR cos φ “ 2gRp1 ´ cos φq
a
vf “ 2gRp1 ´ cos φq
(1)
(2)
(3)
(4)
(5)
(b) Find max speed
ÿ
F “ ma “ ´m
v2
FN ´ Fg cos φ “ ´m
R
18
v2
R
(6)
(7)
Physics 130
General Physics
Fall 2012
The critical speed is just as the normal force equals zero. Factoring the negatives
and the masses, and setting Fg “ mg and FN “ 0,
pvc q2
aR
“
gR cos φ
g cos φ “
(8)
vc
(9)
(c) Find the angle when the sled gets air! We can set the speeds from the two
previous parts equal to each other to find φ.
a
a
gR cos φ “
2gRp1 ´ cos φq
(10)
gR cos φ “ 2gRp1 ´ cos φq
(11)
cos φ “ 2 ´ 2 cos φ
(12)
3 cos φ “ 2
(13)
2
φ “ cos´1 p q Ñ φ “ 48˝
(14)
3
19
Physics 130
General Physics
Fall 2012
10. Bob can throw a 500 g rock with a speed of 30 m/s. He moves his hand forward 1.0 m
while doing so.
(a) How much work does Bob do on the rock?
(b) How much force, assumed to be constant, does Bob apply to the rock?
(c) What is Bob’s maximum power output as he throws the rock?
Solution:
(a) Calculate work
W “ ∆K
1
1
mpvf q2 ´ mpvi q2
W “
2
2
1
W “
mpvf q2 ´ 0
2
1
0.5kgp30m{sq2
W “
2
W “ 225 J
(1)
(2)
(3)
(4)
(5)
(b) Calculate force
W “ F ¨ ∆r “ F ∆r
W
225 J
F “
“
∆r
1m
F “ 225 N
(6)
(7)
(8)
(c) Calculate power
P “ F ¨ v “ Fv
P “ 225 Np30m{sq
P “ 6750 W
20
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
11. Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose
gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling resistance as the truck
tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes
upward at 6.0˝ and the coefficient of rolling friction is 0.40. Use work and energy to find
the length of a ramp that will stop a 15, 000 kg truck that enters the ramp at 35 m/s
(« 75 mph).
Solution:
We’ll identify the truck and the loose gravel as the system. We need the gravel inside
the system because friction increases the temperature of the truck and the gravel.
We will also use the model of kinetic friction and the conservation of energy equation.
Kf ` Uf ` ∆Eth “ Ki ` Ui ` Wext
0 ` Uf ` ∆Eth “ Ki ` 0 ` 0
(1)
(2)
The thermal energy created by friction is
∆Eth “
“
“
“
fk p∆xq
pµk FN qp∆xq
pµk mg cos θqp∆xq
pµk mg cos θqp∆xq
(3)
(4)
(5)
(6)
Using geometry, the final gravitationl potential energy of the truck is
Uf “ mgyf
“ mgp∆xq sin θ
(7)
(8)
Finally, the initial kinetic energy is simply
Ki “
21
1 2
mv
2 i
(9)
Physics 130
General Physics
Fall 2012
Plugging everything into the conservation of energy equation, we get
1 2
mv
2 i
1 2
g∆xpsin θ ` µk cos θq “
v
2 i
mgp∆xq sin θ ` µk mg cos θp∆xq “
(10)
(11)
vi2
(12)
2gpsin θ ` µk cos θq
p35 m{sq2
(13)
“
2 ˆ p9.8 m{s2 q ˆ psin 6˝ ` 0.4 ˆ cos 6˝ q
“ 124 m
(14)
“ 0.12 km.
(15)
ñ ∆x “
22
Physics 130
General Physics
Fall 2012
12. An 8.0 kg crate is pulled 5.0 m up a 30˝ incline by a rope angled 18˝ above the incline.
The tension in the rope is 120 N, and the crate’s coefficient of kinetic friction on the
incline is 0.25.
(a) How much work is done by tension, by gravity, and by the normal force?
(b) What is the increase in thermal energy of the crate and incline?
Solution:
(a) Find work done by tension, gravity and the normal
WT “ T ¨ ∆r “
“
“
Wg “ Fg ¨ ∆r “
“
“
WN “ 0 J
T ∆x cosp18˝ q
p120 Nq ˆ p5.0 mq ˆ pcos 18˝ q
570J
mg∆x cos 120˝
p6 kgq ˆ p9.8 m{s2 q ˆ p5.0 mq ˆ pcos 120˝ q
´196 J
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(b) Find the increase in thermal energy
∆Eth “ fk ¨ ∆r “ Fx ∆x “ µk Fn ∆x
To find the normal force,
ÿ
Fy “ 0
(8)
(9)
(10)
Fn ´ Fg cos 30˝ ` T sin 18˝ “ 0
ñ Fn “ Fg cosp30˝ q ´ T sin 18˝
“ p8.0 kgq ˆ p9.8 m{s2 q ˆ pcos 30˝ q ´ p120 Nq ˆ psin 18˝ q
“ 30.814 N
23
(11)
(12)
(13)
(14)
Physics 130
General Physics
Fall 2012
Plugging this result into the equation for the thermal energy, we find
∆Eth “ µk Fn ∆x
“ p0.25q ˆ p30.814 Nq ˆ p5.0 mq
“ 38.5 J
24
(15)
(16)
(17)
(18)
Physics 130
General Physics
Fall 2012
13. A 5.0 kg cat leaps from the floor to the top of a 95 cm high table. If the cat pushes
against the floor for 0.20 s to accomplish this feat, what was her average power output
during the pushoff period?
Solution:
The average power output during the push-off period is equal to the work done by
the cat divided by the time the cat applied the force. Since the force on the floor by
the cat is equal in magnitude to the force on the cat by the floor, work done by the
cat can be found using the workkinetic-energy theorem during the push-off period:
Wnet “ Wfloor “ ∆K. We do not need to explicitly calculate Wcat , since we know
that the cat’s kinetic energy is transformed into its potential energy during the leap.
In other words,
∆Ug “ mgpy2 ´ y1 q
“ p5.0 kgqp9.8 m{s2 qp0.95 mq
“ 46.55 J.
(1)
(2)
(3)
Thus, the average power output during the push-off period is
Wnet
t
46.55 J
“
0.20 s
“ 0.23 kW.
P “
25
(4)
(5)
(6)
Physics 130
General Physics
Fall 2012
14. In a hydroelectric dam, water falls 25 m and then spins a turbine to generate electricity.
(a) What is ∆U of 1.0 kg of water?
(b) Suppose the dam is 80% efficient at converting the water’s potential energy to
electrical energy. How many kilograms of water must pass through the turbines
each second to generate 50 MW of electricity?
Solution:
(a) The change in the potential energy of 1.0 kg of water falling 25 m is
∆Ug “ ´mgh
“ ´p1.0 kgqp9.8 m{s2 qp25 mq
“ ´250 J.
(1)
(2)
(3)
(b) The dam is required to produce 50 MW “ 50 ˆ 106 W of power every second,
or in other words 50 ˆ 106 J of energy per second. If the dam is 80% efficient at
converting the water’s potential energy into electrical energy, then every kilogram
of water that falls produces 0.8 ˆ 250 J “ 200 J of electrical energy. Therefore, the
total mass of water needed every second is
50 ˆ 106 J ˆ
1.0 kg
“ 2.5 ˆ 105 kg.
200 J
This is a typical value for a small hydroelectric dam.
26
(4)
Physics 130
General Physics
Fall 2012
15. You need to determine the density of a ceramic statue. If you suspend it from a spring
scale, the scale reads 28.4 N. If you then lower the statue into a tub of water, so that it
is completely submerged, the scale reads 17.0 N. What is the statue’s density?
Solution:
The buoyant force, FB , on the can is given by Archimedes’ principle.
The free-body diagram sketched above shows that the gravitational force of the
statue, FG,statue is balanced by the spring force, Fsp , and the buoyant force, FB
exerted by the water. Since the statue is in static equilibrium, we know that the
net force is zero. The rest of the solution follows through a series of substitutions,
recalling that the density of water is ρw “ 1000 kg m´3 :
FB ` Fsp ´ FG,statue “ 0
ñ FB “ FG,statue ´ Fsp
ρw Vstatue g “ FG,statue ´ Fsp
FG,statue ´ Fsp
Vstatue “
ρw g
FG,statue ´ Fsp
mstatue
“
ρstatue
ρw g
(1)
(2)
(3)
(4)
(5)
(6)
ρstatue
ρw g
ñ
“
mstatue
FG,statue ´ Fsp
ρw mstatue g
ρstatue “
FG,statue ´ Fsp
ρw FG,statue
ρstatue “
FG,statue ´ Fsp
p1000 kg m´3 q ˆ p28.4 Nq
“
p28.4 N ´ 17.0 Nq
“ 2490 kg m´3 .
27
(7)
(8)
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
16. A 355 mL soda can is 6.2 cm in diameter and has a mass of 20 g. Such a soda can half
full of water is floating upright in water. What length of the can is above the water
level?
Solution:
The buoyant force, FB , on the can is given by Archimedes’ principle.
Let the length of the can above the water level be d, the total length of the can be
L, and the cross-sectional area of the can be A. The can is in static equilibrium, so
from the free-body diagram sketched above we have
ÿ
Fy “ FB ´ FG,can ´ FG,water “ 0
(1)
ρwater ApL ´ dqg “ pmcan ` mwater qg
pmcan ` mwater q
L´d “
Aρwater
pmcan ` mwater q
ñd “ L´
Aρwater
(2)
(3)
(4)
Recalling that one liter equals 10´3 m3 and the density of water is ρwater “ 1000 kg m´3 ,
the mass of the water in the can is
ˆ
˙
Vcan
mwater “ ρwater
(5)
2
ˆ
˙
355 ˆ 10´6 m3
´3
“ p1000 kg m q ˆ
(6)
2
“ 0.1775 kg.
(7)
To find the length of the can, L, we have:
Vcan “ AL
355 ˆ 10´6 m3
ñL “
πp0.031 mq2
“ 0.1176 m.
28
(8)
(9)
(10)
Physics 130
General Physics
Fall 2012
Finally, substituting everything into equation (4), we get:
d “ 0.1176 m ´
“ 0.0522 m
“ 5.2 cm.
p0.020 kg ` 0.1775 kgq
π ˆ p0.031 mq2 ˆ p1000 kg m´3 q
29
(11)
(12)
(13)
Physics 130
General Physics
Fall 2012
17. Water flowing out of a 16 mm diameter faucet fills a 2.0 L bottle in 10 s. At what
distance below the faucet has the water stream narrowed to 10 mm in diameter?
Solution:
We treat the water as an ideal fluid obeying Bernoulli’s equation. The pressure at
point 1 is p1 and the pressure at point 1 is p2 . Both p1 and p2 are atmospheric
pressure. The velocity and the area at point 1 are v1 and A1 , and at point 2 they
are v2 and A2 . Let d be the distance of point 2 below point 1.
First, we use the time it takes to fill a 2.0 L bottle to compute the flow rate, Q, the
rate at which water is flowing out of the faucet, remembering that one liter equals
10´3 m3 :
p2.0 Lq ˆ p10´3 m3 {Lq
Q “
10 s
“ 2.0 ˆ 10´4 m3 {s.
(1)
(2)
Next, we use this result to find the velocity, v1 , with which the water leaves the
faucet through the D1 “ 16 ˆ 10´3 m diameter aperture (at point 1):
Q “ v1 A1
Q
ñ v1 “
A1
Q
“
πpD1 {2q2
2.0 ˆ 10´4 m3 {s
“
π ˆ p8 ˆ 10´3 mq2
“ 1.0 m{s.
30
(3)
(4)
(5)
(6)
(7)
Physics 130
General Physics
Fall 2012
Now, from the continuity equation we have
v2 A2 “ v1 A1
A1
ñ v2 “ v1
A2
πpD1 {2q2
“ v1
πpD2 {2q2
ˆ ˙2
D1
“ v1
D2
ˆ
˙2
16 ˆ 10´3 m
“ 1.0 m{s
10 ˆ 10´3 m
“ 2.56 m{s.
(8)
(9)
(10)
(11)
(12)
(13)
Finally, we turn to Bernoulli’s equation to find the height, d:
1
1
p1 ` ρv12 ` ρgy1 “ p2 ` ρv22 ` ρgy2
2
2
1 2
ρgpy1 ´ y2 q “
ρpv2 ´ v12 q
2
1 2
pv2 ´ v12 q
gd “
2
pv22 ´ v12 q
d “
2g
p2.56 m{sq2 ´ p1.0 m{sq2
“
2 ˆ 9.8 m{s2
“ 0.283 m
« 28 cm.
31
(14)
(15)
(16)
(17)
(18)
(19)
(20)
Physics 130
General Physics
Fall 2012
18. A hurricane wind blows across a 6.0 m ˆ15.0 m flat roof at a speed of 130 km/hr.
(a) Is the air pressure above the roof higher or lower than the pressure inside the house?
Explain.
(b) What is the pressure difference?
(c) How much force is exerted on the roof? If the roof cannot withstand this much
force, will it “blow in” or “blow out”?
Solution:
(a) The pressure above the roof is lower due to the higher velocity of the air.
(b) Bernoulli’s equation with yinside « youtside is
1
pinside “ poutside ` ρair v 2
2
1
ρair v 2
ñ ∆p “
2
ˆ
˙2
1 hr
1
130 km 1000 m
3
ˆ p1.28 kg{m q ˆ
ˆ
ˆ
“
2
hr
km
3600 s
“ 835 Pa.
(1)
(2)
(3)
(4)
The pressure difference is 0.83 kPa.
(c) The force on the roof is
Froof “ p∆pqA
“ p835 Paq ˆ p6.0 m ˆ 15.0 mq
“ 7.5 ˆ 104 N.
(5)
(6)
(7)
The roof will blow up, because the pressure inside the house is greater than the
pressure on the top of the roof.
32
Physics 130
General Physics
Fall 2012
19. A gas is compressed from 600 cm´3 to 200 cm´3 at a constant pressure of 400 kPa. At
the same time, 100 J of heat energy is transferred out of the gas. What is the change in
thermal energy of the gas during this process?
Solution:
This is a first law of thermodynamics problem involving an isobaric (constant pressure) process. We know that W ą 0 because the gas is compressed, which transfers
energy into the system. Also, 100 J of heat is transferred out of the gas. The first
law of thermodynamics gives
∆Eth “
“
“
“
W `Q
´p∆V ` Q
´p4.0 ˆ 105 Paq ˆ p200 ´ 600q ˆ 10´6 m´3 ´ 100 J
60 J.
Therefore, the thermal energy increases by 60 J.
33
(1)
(2)
(3)
(4)
Physics 130
General Physics
Fall 2012
20. A 100 g ice cube at ´10˝ C is placed in an aluminum cup whose initial temperature is
70˝ C. The system comes to an equilibrium temperature of 20˝ C. What is the mass of
the cup?
Solution:
There are two interacting systems: aluminum and ice. The system comes to thermal
equilibrium in four steps: (1) the ice temperature increases from ´10 ˝ C to ´0 ˝ C;
(2) the ice becomes water at 0 ˝ C; (3) the water temperature increases from 0 ˝ C to
20 ˝ C; and (4) the cup temperature decreases from 70 ˝ C to 20 ˝ C.
Since the aluminum and ice form a closed system, we have
Q “ Q1 ` Q2 ` Q3 ` Q4 “ 0.
(1)
Mice cice ∆T
p0.100 kgq ˆ r2090 J{pkg Kqs ˆ p10 Kq
2090 J.
Mice Lf
p0.100 kgq ˆ p3.33 ˆ 105 J{kgq
33, 300 J.
Mice cwater ∆T
p0.100 kgq ˆ r4190 J{pkg Kqs ˆ p20 Kq
8380 J.
MAl cAl ∆T
MAl ˆ r900 J{pkg Kqs ˆ p´50 Kq
´p45, 000 J{kgqMAl .
(2)
Each term is as follows:
Q1 “
“
“
Q2 “
“
“
Q3 “
“
“
Q4 “
“
“
(3)
(4)
(5)
Inserting everything into equation (1), we get
43, 770 J ´ p45, 000 J{kgqMAl “ 0
ñ MAl “ 0.97 kg.
34
(6)
Physics 130
General Physics
Kinematics and Mechanics
Energy
1
xf “ xi ` vxi t ` ax t2
2
vxf “ vxi ` ax t
Kf ` Ugf “ Ki ` Ugi
∆K ` ∆U ` ∆Eth “ ∆Emech ` ∆Eth “ ∆Esys “ Wext
2
2
vxf
“ vxi
` 2ax pxf ´ xi q
1
yf “ yi ` vyi t ` ay t2
2
vyf “ vyi ` ay t
1
θf “ θi ` ωi ∆t ` α∆t2
2
ωf “ ωi ` α∆t
2
ωf “ ωi ` 2α∆θ
s “ rθ
c “ 2πr
vt “ ωr
v2
“ ω2r
r
2πr
v“
T
fk “ µk FN
F~AonB “ ´F~BonA
Momentum
p~i “ p~f
p~ “ m~v
m1 ´ m2
pvix q1
m1 ` m2
2m1
pvix q1
pvfx q2 “
m1 ` m2
pvfx q1 “
P “ ∆E {∆t
Fluids and Thermal Energy
F
P“
A
m
ρ“
V
Q “ vA
Q “ MLf for freezing/melting
Q “ Mc∆T
2
v
F~net “ ΣF~r “ ma “ m
r
Fg “ mg
0 ă fs ă“ µs FN
~ ¨ ~v
P “F
v1 A 1 “ v2 A 2
1
1
p1 ` ρv12 ` ρgy1 “ p2 ` ρv22 ` ρgy2
2
2
Qnet “ Q1 ` Q2 ` ... “ 0
ar “
Forces
F~net “ ΣF~ “ ma
Kf ` Uf ` ∆Eth “ Ki ` Ui ` Wext
1
K “ mv 2
2
U “ mgy
∆Eth “ fk ∆s
~ ¨ ~d
W “F
2
2
vyf
“ vyi
` 2ay pyf ´ yi q
2
Fall 2012
Constants
g “ 9.8 m{s2
Mearth “ 5.98 ˆ 1024 kg
rearth´sun “ 1.50 ˆ 1011 m
ρwater “ 1000 kg{m 3
ρair “ 1 .28 kg{m 3
J
cice “ 2090
kg K
J
cwater “ 4190
kg K
J
cAl “ 900
kg K
J
Lf “ 333 , 000
ice to water
kg
35