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.
10.
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).
A force-interaction diagram, if appropriate.
One or more free-body 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 ms when a 60 kg skydiver drops out by releasing his grip on the
glider. What is 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.
mg pvf qg
pf x
ms pvf qs
pix
p mg
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,
pvf qg pmg msqpmviq mspvf qs
g
pv q p680kgqp30m{sq 60kgp30m{sqs
f g
pvf qg 620kg
30m{s
(3)
(4)
(5)
The skydivers motion in the vertical direction has no influence on the gliders horizontal motion. Notice that we did not need to invoke conservation of momentum
to solve this problem. Because there are no external horizontal forces acting on either the skydiver or the glider, neither will change their horizontal speed when the
skydiver lets go!
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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:
Applying conservation of momentum in the x-direction, find the final momentum in
the x-direction
pf x
pf x
pix m1pv ixq1 m2pvixq2
p0.02kgqp2.0m{sq p0.03kgqp1.0m{sq cosp30q 0.0140kgms
(1)
(2)
(3)
Applying conservation of momentum in the y-direction, find the final momentum in
the y-direction
pf y
pf y
piy m1pv iyq1 m2pviy q2
p0.02kgqp0m{sq p0.03kgqp1.0m{sq sinp30q 0.0150kgm{s
(4)
(5)
(6)
Find the magnitude of the final momentum using the pythagorean theorem to combine the x- and y-components
pf
pf
a
pf x pf y
a
p0.0140kgm{sq2 p0.0150kgm{sq2 0.0205kgm{s
(7)
(8)
(9)
We know that
pf
pm1
m2 qvf
3
0.0205kgm{s
(10)
(11)
Physics 130
General Physics
Fall 2012
Solving for the final speed,
vf
vf
vf
pf
m1 m2
0.0205kgm{s
0.02kg 0.03kg
.41m{s
(12)
(13)
(14)
(15)
The direction is given by
θ
θ
tan1
47
pf y
pf x
{s m{s
0.014kgm
0.0150kg
(16)
(17)
(18)
4
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 1013 kg hits the earth with an impact speed of 4.0 104 m{s.
(a) What is the earth’s recoil speed after such a collision? (Use a reference frame in
which the earth was initially at rest.)
(b) What percentage is this of the earth’s speed around the sun? (Use the astronomical
data provided.)
Solution:
(a) This is a conservation of momentum problem. At the start, the asteroid has a
speed of 4.0 104 m{swhile the earth is not moving.
me pvi qe
pix
ma pvi qa
pf x
pme
ma qpvf q
(1)
(2)
(3)
Solving for the final velocity of the earth and asteroid system,
vf
me pvi qe
pme
ma pvi qa
ma q
5
(4)
Physics 130
General Physics
vf
vf
Fall 2012
0 kgm{s p1.0 1013 kgqp4.0 104 m{sq
p1.0 1013kg 5.98 1024kgq
6.8 108 m{s
(5)
(6)
(b) The speed of the earth going around the sun is
vE
2πr
T
1.50 10 mq
4
2πp3.15
107s 3.0 10 m{s
11
6
(7)
(8)
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. Momentum is conserved so that
pf x
mB pvi qB mF pvi qF pmB
pix
mF qpvf q
(1)
(2)
(3)
Solving for the final velocity of Brutus and Fred,
vf
vf
vf
mB pvi qB mF pvi qF
pmB mF q
p120kgqp4.0m{sq p60kgqp6.0m{sq
p180kgq
0.667m{s
(4)
(5)
(6)
(7)
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. The force of
friction is in the negative x-direction. The sum of the forces is equal to the mass
times the acceleration.
¸
Fx
fk
a
a
mT a
µFn µpmB mF qg pmB
µg p0.30qp9.8m{s2q
2.94m{s2
7
mF qa
(8)
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
Using the kinematic equation,
pvf q2 pviq2 2ad 2ad
pvf q2 p0.667m{s2q2 0.076m
d 2a
p2qp2.94m{s2q
d 7.6cm
8
(12)
(13)
(14)
(15)
Physics 130
General Physics
Fall 2012
5. A neutron is an electrically neutral subatomic particle with a mass just slightly greater
than that of a proton. A free neutron is radioactive and decays after a few minutes
into other subatomic particles. In one experiment, a neutron at rest was observed to
decay into a proton (mass 1.67 1027 kg) and an electron (mass 9.11 1031 kg). The
proton and electron were shot out back-to-back. The proton speed was measured to be
1.0 105 m{s and the electron speed was 3.0 107 m{s. No other decay products were
detected.
(a) Was momentum conserved in the decay of this neutron?
(b) If a neutrino was emitted in the above neutron decay, in which direction did it
travel? Explain your reasoning.
(c) How much momentum did this neutrino ”carry away” with it?
Solution:
(a) The initial momentum is pi
afar the decay is
pf
pf
pf
0 since the neutron is not moving. The momentum
meve mpvp
(1)
31
7
27
5
p9.11 10 kgqp3.0 10 m{sq p1.67 10 kgqp1.0 10 m{sq (2)
1.4 1022kg m{s
(3)
(4)
Momentum is not conserved.
(b) Because the initial momentum is zero,
pe pp pn 0
pn 1.4 1022 kg m{s
(c) The neutrino must ”carry away” 1.4 1022 kg m{s.
9
(5)
(6)
(7)
Physics 130
General Physics
Fall 2012
6. 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.
pm1
p3m2
pf
m2 qvi
m2 qvi
4m2 vi
4vi
pi
m1 v1f m2 v2f
3m2 v1f m2 v2f
m2 p3v1f v2f q
3v1f v2f
m2p3v1f
v2f q
(1)
(2)
(3)
(4)
(5)
(6)
The fact that the first stage is pushed backward at 35 m{s relative to the second can
be written
v1f
4vi
4vi
35m{s v2f
3p35m{s v2f q
105m{s 4v2f
10
v2f
(7)
(8)
(9)
(10)
Physics 130
General Physics
Fall 2012
Solving for the velocity of the second stage v2f
v2f
v2f
v2f
105m{s
4
4 1200m{s 105m{s
4
1.2 km{s
4vi
11
(11)
(12)
(13)
(14)
Physics 130
General Physics
Fall 2012
7. 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
mb vbi me vei
mb vbi 0
pf
mb vbf
mb vbf
me vef
mevef
(1)
(2)
(3)
(c) The conservation of energy equation is
1
me pvei q2 2
Ki
1
mb pvbi q2
2
1
mb pvbi q2
2
mb pvbi q2
Ui
0
0
Kf Uf
1
1
mb pvbf q2
me pvef q2
2
2
1
1
mb pvbf q2
me pvef q2
2
2
mb pvbf q2 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 asteroid 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
12
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 ppbviB q2 2a∆h 2p9.8m{s2qp10mq
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
vbf
vef
mb vbi me vef
mb
2mb vbi
me mb
(10)
(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
qp14.0m{sq 2.3 1024m{s
2p0.500kg
5.98 1024 kg
(12)
(g) The time it would take the earth to move 1.0mm at this speed is given by the
kinematics equation.
xf
xi
∆t
xf
vi
1
a∆t2 0 vi ∆t 0
2
0.001m
4.3 1020 sec 1.36 1013 years
2.3 1024 m{s
vi ∆t
13
(13)
(14)
Physics 130
General Physics
Fall 2012
8. 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
0
mgh
v
Ui
Kf Uf
1 2
mv
2
a
2gh (1)
(2)
b
2p9.8m{s2 qp3.0mq 7.668m{s
(3)
If the packages stick together, the collision is inelastic and conservation of momentum is conserved.
pi
m1 v1i m2 v2i
mv1i 0
14
pf
p m1
3mvf
m2 qvf
(4)
(5)
(6)
Physics 130
General Physics
Fall 2012
Cancelling the m’s
v1i
3vf
(7)
Solving for vf , we find
vf
v1i
3
{s 2.56m{s
7.668m
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)
Kf Uf
1
1
m1 pv1f q2
m2 pv2f q2
2
2
1
1
2
m1 pv1f q
me 2v2f q2
2
2
mpv1f q2 2mpv2f q2
pv1f q2 2pv2f q2
(16)
(d) The conservation of energy equation is
1
m2 pv2i q2 2
Ki
Ui
1
m1 pv1i q2
2
1
m1 pv1i q2 0
2
mpv1i q2
pv1iq2
(17)
(18)
(19)
(20)
(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 .
15
Physics 130
General Physics
Fall 2012
From the derivations above, the equation relating the final speed of the first
mass to it’s initial speed is
v1f
2m v 1 p7.668m{sq 2.56m{s
m
1i
m 2m
3
(21)
(22)
The package rebounds and goes back up the ramp. To determine how high the
package goes, we can use the conservation of energy.
0
h
Ki
Ui
1 2
mv
2
Kf
Uf
(23)
0
mgh
(24)
v2
2g
2.56m{sq
p
0.33 m
2p9.8m{s2 q
16
2
(25)
Physics 130
General Physics
Fall 2012
9. 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
(1)
a
r
The free-body diagram for the car at the top of the hill, illustrates the two forces
acting on the car. Using Newton’s law, we know,
¸
FN
F
Fg ma m
v2
R
v2
m R
(2)
(3)
To stay on the track, the normal force has to be greater than zero. If we set the
normal force equal to zero, this is the maximum speed of the car that allows it
to stay on the track.
Fg
mg
v2
v2
R
v2
m
R
gR Ñ v
m
17
(4)
a
gR
(5)
(6)
Physics 130
General Physics
Fall 2012
Now that we know the maximum height at the top of the hill, we can use the
conservation of energy to find the starting height.
Kf Uf
1
mpvf q2 mgyf
2
1
mpvf q2 mgyf
2
We’ll use yf
Ki U i
1
mpvi q2
2
0
(7)
mgyi
mgyi
(8)
(9)
R for the height at the top of the hill and factor the mass term,
1
pvf q2
2
gR
yi
0
gyi
2
pv2gf q
(10)
R
(11)
We can now plug in our results for vmax from above
yi
yi
?gRq2
p
2g
R
2
R
R
(12)
3R
2
(13)
(14)
(b) Plugging into the above equation, hmax for a roller coaster that has R 10 m
yi
3R
2
18
3 210m 15m
(15)
Physics 130
General Physics
Fall 2012
10. A pendulum is formed from a small ball of mass m on a string of length L. As the
figure shows, a peg is height h L{3 above the pendulum’s lowest point. From what
minimum angle θ must the pendulum be released in order for the ball to go over the
top of the peg without the string going slack?
Solution: This is a two part problem. First, we need to find the velocity for the ball
to go over the peg without the string going slack. Then we need to find the potential
energy
to
match
that
velocity.
(a) Find velocity of ball so that string doesn’t go slack.
We’ll set our origin on the peg. This means that once the string touches the
peg, the ball will be moving in circular motion with a radius of L{3.
When the ball is let go, it moves below the peg and then moves in circular
motion. When it reaches the top of the circle and is directly above the peg, we
can use Newton’s second law and the fact that the ball is moving in circular
motion.
¸
F
T Fg ma m
v2
m R
v2
R
(1)
(2)
(3)
19
Physics 130
General Physics
Fall 2012
The critical speed is just as the tension equals zero. Factoring the negatives and
the masses, and setting Fg mg and T 0,
2
m pvRcq
a
pvcq gR
0
mg
Setting R L{3,
pvcq c
(4)
(5)
gL
3
(6)
(b) Find the angle to release the ball. Now that we know the velocity for the string
to remain taut, we’ll use the conservation of energy to find the height to release
the ball.
Kf
Uf
1
mpvf q2 mgyf
2
1
p
vc q2 gyf
2
Plugging in yf
Ki U i
1
mpvi q2
2
0
gyi
(7)
mgyi
(8)
(9)
L{3 and pvcq2 gL{3,
gL gL
6
3
L L
6
3
gyi
(10)
yi
(11)
yi
L
2
(12)
This result tells us that we need to release the ball at a height of L/2 above the
peg.
To find the angle, consider the third figure in the three-part figure above. We
need to find the distance of the ball below the point where the string is attached.
We can use the string length and result above.
hL
L
2
L3 L6
(13)
To find the angle, we can use cosine,
cos θ
θ
20
L
6
1
L
6
80.4
(14)
(15)
Physics 130
General Physics
Fall 2012
11. 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
pvf q2
Uf
1
mpvf q2 mgyf
2
1
p
vf q2 gR cos φ
2
2gR 2gR cos φ 2gRp1 cos φq
vf
(b) Find max speed
Ki Ui
1
mpvi q2
2
0
a
F
Fg cos φ 21
ma m
v2
m R
gyi
(1)
mgyi
(2)
(3)
(4)
2gRp1 cos φq
¸
FN
(5)
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,
g cos φ
vc
2
pvRcq
a
gR cos φ
(8)
(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
gR cos φ 2gRp1 cos φq
cos φ 2 2 cos φ
3 cos φ 2
2
φ cos1 p q Ñ φ 48
3
22
(10)
(11)
(12)
(13)
(14)
Physics 130
General Physics
Fall 2012
12. 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
W
W
W
W
∆K
1
1
mpvf q2 mpvi q2
2
2
1
mpvf q2 0
2
1
0.5kgp30m{sq2
2
225 J
(1)
(2)
(3)
(4)
(5)
(b) Calculate force
F
F ∆r F ∆r
W
225 J
∆r
1m
225 N
P
P
P
F v Fv
225 Np30m{sq
6750 W
W
F
(6)
(7)
(8)
(c) Calculate power
23
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
13. 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. Usw 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
0
The
Plugging
thermal
this
into
Uf
Uf
Uf
∆Eth
∆Eth
Ki
Ki
Ui Wext
0 0
energy
created
∆Eth
∆Eth
∆Eth
∆Eth
fk p∆xq
pµk FN qp∆xq
pµk mgcosθqp∆xq
pµk mg cos θqp∆xq
the
conservation
µk mg cos θp∆xq
∆x
∆x
(1)
(2)
(3)
by
friction
is
(4)
(5)
(6)
(7)
of
energy
Ki
Ki Uf
µk mg cos θ
1
mpvi q2 mgyf q
2
µk mg cos θ
equation,
(8)
(9)
(10)
(11)
24
Physics 130
General Physics
Fall 2012
14. 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
Wg
WN
T ∆r T ∆x cos 18 p120Nqp5.0mqpcos 18 570J
Fg ∆r mg∆x cos 120 p6kgqp9.8m{s2 qp5.0mqpcos 120
0J
(1)
196J
(2)
(3)
(b) Find the increase in thermal energy
∆Eth
fk ∆r
Fx∆x µk Fn∆x
(4)
(5)
To find the normal force,
¸
Fn Fg cos 30
Fy
0
T sin 18
Fn
Fn
(
0
Fg cos 30 T sin 18
30.814N
(
p8.0kgqp9.8m{s2q cos 30 p120Nqpsin 18(
(
Plugging this result into the equation for the thermal energy, we find
∆Eth
µk Fn ∆x p0.25qp30.814Nqp5.0mq 38.5J
25
(10)
(11)
Physics 130
General Physics
Fall 2012
15. The spring shown in the figure below is compressed 50 cm and used to launch a 100 kg
physics student. The track is frictionless until it starts up the incline. The student’s
coefficient of kinetic friction on the 30 incline is 0.15.
(a) What is the student’s speed just after losing contact with the spring?
(b) How far up the incline does the student go?
Solution:
(a) This is a conservation of energy problem. We want to know the initial velocity
immediately after the spring is no longer in contact with the student.
Kf
1
mpvf q2
2
Ugf Usf ∆Eth
1
mgyf
k p∆xf q2 0J
2
26
Ki Ugi
1
mpvi q2
2
Usi
Wext
1
mgyi
k p∆xi q2
2
(1)
0J(2)
Physics 130
General Physics
1
mpvf q2
2
v
Fall 2012
1
k p∆sq2
2
c
k∆x
14.14m{s
m
(3)
(4)
(b) Using the conservation of energy equation again, we can how far the student
travels up the incline. .Let’s define the y-height when the student reaches the
highest point as yf p∆sq sin 30 .
Kf
1
mpvf q2
2
Ugf Usf ∆Eth
1
mgyf
k p∆xf q2 0J
2
mg p∆sq sin 30
µk Fn ∆s
Ki Ugi Usi Wext
1
1
mpvi q2 mgyi
k p∆xi q2
2
2
1
mgyi
p∆xq2
2
(5)
0J(6)
(7)
(8)
From the sum of the forces in the y-direction on the incline, the normal force is
Fn mg cos 30 . Plugging this in,
mg p∆sq sin 30
µk mgcos30 ∆s
∆sqmg psin 30
µk cos 30 q
∆s
∆s
27
1
p∆xq2
2
1
mgyi
p∆xq2
2
mgyi 12 k p∆xq2
mg psin 30 µk cos 30 q
32.1m
mgyi
(9)
(10)
(11)
(12)
Physics 130
General Physics
Fall 2012
16. 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
Thus,
the
average
mgpy2 y1q
p5.0 kgqp9.8 m{s2qp0.95 mq
46.55 J.
power
output
P
28
during
Wnet
t
46.55 J
0.20 s
0.23 kW.
the
(1)
(2)
(3)
push-off
period
is
(4)
(5)
(6)
Physics 130
General Physics
Fall 2012
17. 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{s2qp25 mq
250 J.
(1)
(2)
(3)
(b) The dam is required to produce 50 MW 50 106 W of power every second, or in
other words 50 106 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 0.8 250 J 200 J of electrical energy. Therefore, the total mass of water
needed
every
second
is
50 106 J This
is
a
typical
1.0 kg
200 J
value
2.5 105 kg.
for
29
a
small
(4)
hydroelectric
dam.
Physics 130
General Physics
Fall 2012
18. (a) The 70 kg student in the figure below balances a 1200 kg elephant on a hydraulic
lift. What is the diameter of the piston the student is standing on?
(b) When a second student joins the first, the piston sinks 35 cm. What is the second
student’s mass? Assume that the density of oil is 900 kg{m3 .
Solution:
To
solve
this
problem
we
F2
need
the
hydraulic
AA2 F1 ρghA2.
lift
equation,
(1)
1
(a) The hydraulic lift is in equilibrium and the pistons on the left and the right are
at the same level. Therefore, h 0 and the hydraulic lift equation simplifies to
Fleft
Aleft
ñ FπrG,student
2
student
ñ rstudent Fright
Aright
FG,elephant
2
πrelephant
d
student
melephant
70 kg
1200 kg
0.242 m
24 cm.
30
(3)
FG,student
relephant
FG,elephant
cm
d
(2)
(4)
relephant
(5)
p1.0 mq
(6)
(7)
(8)
Physics 130
General Physics
Fall 2012
(b) With the second student added, the pistons are no longer at the same height.
Therefore,
the
hydraulic
lift
equation
becomes
Fright
Aright
AFleft ρgh.
Noting that the force on the piston on the left is Fleft
the
second
student’s
ñ
melephant g
Aright
ms 70 kg
Aleft
ms
Substituting
ma
pms
70 kg
ms
1200 kg
π p1.0 mq2
58 kg
(9)
left
70 kgqg
Aleft
pms
70 kgqg, we can solve for
mass,
ms :
ρgh
(10)
melephant
ρh
Aright
melephant
ρh Aleft
Aright
melephant
ρh Aleft 70 kg
Aright
everything,
we
p900 kg{m qp0.35 mq πp0.242 mq2 70 kg
(11)
(12)
(13)
get:
3
(14)
31
Physics 130
General Physics
Fall 2012
19. 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
ñ FB
ρw Vstatue g
Vstatue
mstatue
ρstatue
ρstatue
mstatue
0
FG,statue Fsp
FG,statue Fsp
FG,statue Fsp
ρw g
FG,statue Fsp
ρw g
(1)
(2)
(3)
(4)
(5)
(6)
ñ
ρstatue
ρstatue
32
ρw g
FG,statue Fsp
ρw mstatue g
FG,statue Fsp
ρw FG,statue
FG,statue Fsp
p1000 kg m3q p28.4 Nq
p28.4 N 17.0 Nq
2490 kg m3 .
(7)
(8)
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
20. 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
¸
Fy
FB FG,can FG,water 0
ρwater ApL dqg pmcan mwater qg
pmcan mwaterq
Ld Aρwater
ñ d L pmcan mwaterq
(1)
(2)
(3)
(4)
Aρwater
Recalling that one liter equals 103 m3 and the density of water is ρwater
the
mass
of
the
water
in
the
mwater
To
find
the
ρwater
Vcan
2
1000 kg m3,
can
(5)
355 106 m3
3
p1000 kg m q (6)
2
0.1775 kg.
length
Vcan
ñL
of
(7)
the
can,
AL
355 106 m3
π p0.031 mq2
0.1176 m.
33
is
L,
we
have:
(8)
(9)
(10)
Physics 130
Finally,
General Physics
substituting
d
everything
0.1176 m into
Fall 2012
equation
(4),
p0.020 kg 0.1775 kgq
π p0.031 mq2 p1000 kg m3 q
0.0522 m
5.2 cm.
we
get:
(11)
(12)
(13)
34
Physics 130
General Physics
Fall 2012
21. 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 103 m3 :
Q
3 3
p2.0 Lq 10p10s m {Lq
2.0 104 m3{s.
(1)
(2)
Next, we use this result to find the velocity, v1 , with which the water leaves the faucet
through the D1
16 103 m diameter aperture (at point 1):
Q
ñ v1
v1 A1
Q
A1
Q
π pD1 {2q2
2.0 104 m3 {s
π p8 103 mq2
1.0 m{s.
35
(3)
(4)
(5)
(6)
(7)
Physics 130
General Physics
Now,
from
the
v2 A2
ñ v2
continuity
Finally,
we
turn
p1
to
equation
we
v1 A1
A1
v1
A2
π pD1 {2q2
v1
π pD2 {2q2
D1
D2
v1
1.0 m{s
2.56 m{s.
Bernoulli’s
(9)
(10)
ρg py1 y2 q
gd
d
16 103 m
10 103 m
2
(12)
1 2
ρv
2 2
to
find
ρgy2
1 2
ρpv2 v12 q
2
1 2
p
v2 v12 q
2
pv22 v12q
2g
p2.56 m{sq2 p1.0 m{sq2
2 9.8 m{s2
0.283 m
28 cm.
36
(11)
(13)
equation
p2
have
(8)
2
ρgy1
1 2
ρv
2 1
Fall 2012
the
height,
d:
(14)
(15)
(16)
(17)
(18)
(19)
(20)
Physics 130
General Physics
22. A hurricane wind blows across a 6.0 m
Fall 2012
15.0 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
pinside
ñ ∆p The
(c)
pressure
The
poutside
(1)
1
ρair v 2
2
2
1
130 1000 m
4
p1.28 kg{m q 2
3600 s
835 Pa.
force
Froof
1
ρair v 2
2
difference
on
is
the
p∆pqA
p835 Paq p6.0 m 15.0 mq
7.5 104 N.
0.83
roof
(2)
(3)
(4)
kPa.
is
(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.
37
Physics 130
General Physics
Kinematics and Mechanics
xi vxit 12 axt2
vxf vxi ax t
2
vxf
vxi2 2axpxf xiq
1 2
ay t
yf yi vyi t
2
vyf vyi ay t
2
vyi2 2ay pyf yiq
vyf
1
θf θi ωi ∆t
α∆t2
2
ωf ωi α∆t
ωf 2 ωi 2 2α∆θ
s rθ
c 2πr
vt ωr
v2
ar ω2r
r
xf
v
2πr
T
Forces
F~net ΣF~
ma
v2
F~net ΣF~r ma m
r
Fg mg
0 fs µs FN
f k µk F N
F~AonB F~BonA
Momentum
p~f
p~ m~v
m1 m2
pvfx q1 m
pvix q1
m
p~i
1
2
pvfx q2 m 2m1m pvix q1
1
2
Energy
38
Fall 2012
Physics 130
Kf
Ugf
∆K
General Physics
Ki
∆U
Ugi
∆Eth
∆Emech
∆Eth
∆Esys Wext
21 mv 2
U mgy
∆Eth fk ∆s
~ ~d
W F
~ ~v
P F
∆E P∆t
K
Fluids and Thermal Energy
F
P
A
m
ρ
V
Q vA
v2 A2
v1 A1
p1
Qnet
1 2
ρv
2 1
Q1
ρgy1
Q2
p2
... 0
1 2
ρv
2 2
ρgy2
MLf for freezing
Q Mc∆T
Q
Constants
m
g 9.8 2
s
ρoil
900 mkg3
ρwater
1 , 000 mkg3
ci
J
2090 kgK
cw
J
4190 kgK
cAl
J
900 kgK
Lf
333 , 000 kgJ
ice to water
39
Fall 2012