203-1-1371 'א 1 הקיסיפ ןילדג לאכימ 'פורפ :םיצרמ 13.07.14 'א דעומ

‫פיסיקה ‪ 1‬א' ‪203-1-1371‬‬
‫מרצים‪ :‬פרופ' מיכאל גדלין‬
‫מועד א' ‪13.07.14‬‬
‫• משך המבחן ‪ 3‬שעות‬
‫• חומר עזר‪ :‬דף נוסחאות מצורף‪ ,‬מחשבון אסור‬
‫• בשאלות פתוחות יש לרשום פתרון באמצעות אותיות בלבד‪ ,‬להגיע‬
‫לנוסחה סופית ולהציב מספרים רק בה‬
‫• בשאלות עם מספרים חובה להגיע למספר סופי )בקירוב(‬
‫• בשאלות אמריקאיות רק תשובות סופיות )בטופס( נבדקות‬
‫• שאלות פתוחות יש לפתור במחברת‬
‫• אסור לכתוב בעפרון‪ ,‬אסור להשתמש בצבע אדום‬
‫בהצלחה !‬
‫חלק א' ‪ -‬שאלות אמריקאיות )כל שאלה ‪ 4 -‬נק'(‬
‫נא לסמן תשובות בטבלה זו‬
‫לכל שאלה רק תשובה אחת נכונה‬
‫‪No. A B C D E‬‬
‫‪1‬‬
‫‪2‬‬
‫‪3‬‬
‫‪4‬‬
‫‪5‬‬
‫‪6‬‬
‫‪7‬‬
‫‪8‬‬
‫‪9‬‬
‫‪10‬‬
‫‪1‬‬
‫‪ (1‬מרכז המסה של מערכת חלקיקים נע במהירות קבועה אם‬
‫‪E‬‬
‫מרכז המסה‬
‫נמצא במרכז‬
‫הגיאומטרי‬
‫של המערכת‬
‫‪D‬‬
‫התפלגות‬
‫החלקיקים‬
‫סימטרית‬
‫ביחס למרכז‬
‫המסה‬
‫‪C‬‬
‫מהירות‬
‫מרכז המסה‬
‫בהתחלה‬
‫הנה אפס‬
‫‪B‬‬
‫סכום הכוחות‬
‫החיצוניים‬
‫אשר‬
‫על‬
‫פועלים‬
‫החלקיקים‬
‫הנו אפס‬
‫‪A‬‬
‫סכום הכוחות‬
‫שהחלקיקים‬
‫מפעילים זה‬
‫על זה הנו‬
‫אפס‬
‫‪ (2‬שני ילדים בעלי מסות ‪ 40 kg‬ו ‪ 60 kg‬עומדים על משטח אופקי ללא חיכוך‬
‫ומחזיקים במוט שאורכו ‪ 10 m‬בשני הקצוות‪ .‬הילדים מתחילים למשוך את המוט כל‬
‫אחד לכיוונו ומתקרבים זה לזה‪ .‬כאשר הם נפגשים הילד בעל מסה ‪ 60 kg‬עובר מרחק‬
‫‪E‬‬
‫תלוי בכוחות‬
‫שהם‬
‫מפעילים‬
‫‪D‬‬
‫‪10 m‬‬
‫‪C‬‬
‫‪6m‬‬
‫‪B‬‬
‫‪5m‬‬
‫‪A‬‬
‫‪4m‬‬
‫‪ (3‬מסה תלויה בקצה של קפיץ מושלם ומתנדנדתלמעלה ומטה עם זמן מחזור ‪.T‬‬
‫אם מגדילים את התנופה )אמפליטודה( פי‪-‬שניים‪ ,‬זמן המחזור יהיה‬
‫‪E‬‬
‫‪4T‬‬
‫‪C‬‬
‫‪2T‬‬
‫‪D‬‬
‫‪T /2‬‬
‫‪B‬‬
‫‪1.5T‬‬
‫‪A‬‬
‫‪T‬‬
‫‪ (4‬חלקיק מבצע תנועה הרמונית פשוטה המתוארת ע"י הביטוי )‪x = 2 cos(50t‬‬
‫כאשר ‪ x‬במטרים ו ‪ t‬בשניות‪ .‬מהירותו המרבית ב ‪ m/s‬הנה‬
‫‪E‬‬
‫תשובה‬
‫אף‬
‫איננה נכונה‬
‫‪C‬‬
‫‪100‬‬
‫‪D‬‬
‫‪200‬‬
‫‪B‬‬
‫)‪100 cos(50t‬‬
‫‪A‬‬
‫)‪100 sin(50t‬‬
‫‪ (5‬שני גלילים זהים בעלי מומנט התמד ‪ I = 0.5M R2‬מתגלגלים על רצפה אופקית‬
‫ללא החלקה במהירויות שוות ואחר כך מעלה במישור משופע‪ .‬דיסקה ‪ A‬מתגלגלת‬
‫במישור משופע ללא החלקה‪ .‬דיסקה ‪ B‬מתגלגלת במישור משופע עם אותה זווית‬
‫אבל ללא חיכוך‪ .‬שיא הגובה שאליו מגיעה דיסקה ‪ A‬הנו ‪ .12 cm‬מהו שיא הגובה‬
‫אליו מגיעה דיסקה ‪? B‬‬
‫‪E‬‬
‫‪6 cm‬‬
‫‪D‬‬
‫‪8 cm‬‬
‫‪C‬‬
‫‪12 cm‬‬
‫‪B‬‬
‫‪18 cm‬‬
‫‪A‬‬
‫‪24 cm‬‬
‫‪ (6‬מחליק על קרח בעל מומנט התמד ‪ I0‬מסתובב במהירות זוויתית ‪ .ω0‬הוא מקרב‬
‫ידיים לגופו ומגדיל את המהירות הזוויתית ל ‪ .4ω0‬מהו מומנט ההתמד החדש שלו ?‬
‫‪A‬‬
‫‪B‬‬
‫‪C‬‬
‫‪D‬‬
‫‪E‬‬
‫‪I0‬‬
‫‪I0 /2‬‬
‫‪2I0‬‬
‫‪I0 /4‬‬
‫‪4I0‬‬
‫‪2‬‬
‫‪ (7‬גוף שמסתו ‪ 2 kg‬נע במהירות ‪ .3 m/s‬כוח שגודלו ‪ 4 N‬מתחיל לפעול על הגוף‬
‫ברגע ‪ t1‬מסוים בכיוון התנועה ומפסיק לפעול ברגע ‪ .t2‬במשך זמן זה הגוף עובר מרחק‬
‫‪ .5 m‬העבודה של הכוח הנה‬
‫‪E‬‬
‫‪38 J‬‬
‫‪C‬‬
‫‪18 J‬‬
‫‪D‬‬
‫‪20 J‬‬
‫‪B‬‬
‫‪15 J‬‬
‫‪A‬‬
‫‪12 J‬‬
‫‪ (8‬בן אדם דוחף גוף‪ ,‬שמשקלו במנוחה על הקרקע הנו ‪ ,80 N‬במעלה מדרון בעל‬
‫זווית ◦‪ 30‬עם האופק‪ .‬הגוף עובר מרחק ‪ 5 m‬לאורך המדרון ומהירותו יורדת בקצב‬
‫‪ .1.5 m/s2‬העבודה שעושה האדם הנה‬
‫‪E‬‬
‫‪260 J‬‬
‫‪D‬‬
‫‪200 J‬‬
‫‪C‬‬
‫‪140 J‬‬
‫‪B‬‬
‫‪61 J‬‬
‫‪A‬‬
‫‪−200 J‬‬
‫‪ (9‬אבן נזרקת מצוק שגובהו ‪ 60 m‬במהירות שהרכיב האנכי שלה הנו ‪ 20 m/s‬כלפי‬
‫מעלה‪ .‬כמה זמן תישאר האבן באוויר ? ‪g = 10 m/s2‬‬
‫‪E‬‬
‫‪8s‬‬
‫‪C‬‬
‫‪6s‬‬
‫‪D‬‬
‫‪7s‬‬
‫‪B‬‬
‫‪5s‬‬
‫‪A‬‬
‫‪4s‬‬
‫‪ (10‬גוף מחובר לקצה חוט ומסתובב במישור אופקי במעגל בעל רדיוס ‪1.5 m‬‬
‫במהירות זוויתית קבועה‪ .‬אם הוא עושה שני סיבובים שלמים תוך שנייה‪ ,‬התאוצה‬
‫שלו הנה‬
‫‪E‬‬
‫‪2400 m/s2‬‬
‫‪D‬‬
‫‪240 m/s2‬‬
‫‪C‬‬
‫‪24 m/s2‬‬
‫‪3‬‬
‫‪B‬‬
‫‪2.4 m/s2‬‬
‫‪A‬‬
‫‪0.24 m/s2‬‬
frictional force on the Cheerios box is 2.0 N,
its central axis, is set spinning counterclockthe
force
the
Wheaties
box is 4.0
ock
2frictional
(mass 1.0
kg) on
is at
rest
on a frictionless
he second disk, with rotational inertia 6.60
Faxis,
isend
12 N, what
is counterclockwise
the magnitude
the
force
an unstretched
springof
lthe
is setofspinning
atof
900spring
the end
Cheerios
‫סעיפים‬
‫ אין‬,'‫נק‬
erom
other
of What
thebox?
spring
is
fixed
to20
a ‫שאלה‬
wall. ‫ כל‬,‫ שאלות פתוחות‬- '‫חלק ב‬
uple
together.
(a)
is their
angular
speed
traveling
at speed
4.0 m/s,clockwise
collides with
ead
the second
disk isvset!spinning
mstick
re their
(b) angular
speed
and (c)
direction
blocks
together.
When
the
blocks moouple
together?
m
t distance
is the spring
compressed?
F
the rotational inertia of the block – rod – bullet system about point
! M/4.00 on its outer edge, at radius R2. By how much does
••30
A16cat
2.0
incline
angle
"box
!
Fig.kinetic
6-23
Problems
and
22.breadbox
A?
(b) If
the
angularof
speed
of the
system
about
A be
justpulled
after impact
ncrease the
energy
of the
–kg
ring
system if theon
cata frictionless
••21
An
initially
stationary
of sand
is to
across a
isfloor
4.5 rad/s,
what to
is the
bullet’s
speed
just before
is Rconnected,
by a cord that runs over
a by
pulley,
a light
spring
the inner edge, at 40°
radius
1?
means
of
a cable
in which
the impact?
tension should not
:
: spring
of
constant
k !radius
120
N/m,
in1100
Fig.
The
box0.60
is
••61
The
uniform
rod
(length
exceed
N.8-41.
The
coefficient
of static friction between the box
horizontal
mass
0.10 kg
and
mas shown
Fig.
6-24, a vinyl
forcerecord
a block
weighing
45 0.10
N. The
P actsofon
Rotation axis
m,
mass
1.0
kg) Assume
in0.35.
Fig. 11-54
rotates
reely
about
a
vertical
axis
through
its
center
with
an
angureleased
from
rest
when
the
spring
is
unstretched.
that
the
and
the
floor
is
(a)
What
should be the angle between the
nitially at rest on a plane inclined at angle u ! 15° to the
D
in
the
plane
of
the
figure
about
an
dl.of
4.7
rad/s.
The
rotational
inertia
of
the
record
about
its
cable and
thespeed
horizontal
in box
order to pull the greatest possible
pulley of
is the
massless
frictionless.
is the
of the
The positive direction
x axis isand
up the
plane. The (a) What
#4
2
Rod
axis
through
one
end,
with
a
rotaHALLIDAY
REVISED
otation
is 5.0 between
" 10
kgblock
$it
mhas
.A
wad
of wet
of mass
amount(b)
of sand,
and
whatthe
is the weight of the sand and box
when
moved
10 putty
cm
the incline?
How
far(b)
down
ts of friction
and
plane
are
m
0.50
and
s !down
tional
of 0.12 kg $ m2. As the
vertically onto the record from above and sticks to
in thatinertia
situation?
.drops
In unit-vector
notation,
what
is
the
frictional
force
on
θ
incline
from its point of release does rod
the swings
box slide
before
momenthrough
its lowest
posi: angular speed of the record imof the record. What is the
from the plane when
Pstopping,
is ‫הסטטי‬
(a) ("5.0
N)
î1
, (b) ("8.0
N)î , (c) tion,
••22 it collides
In‫עם‬
Fig.
6-23,
a
sled
is
held
on
an
inclined
plane
by
a
cord
‫החיכוך‬
‫מקדם‬
.‫האופק‬
θ
‫זווית‬
‫בעל‬
‫מדרון‬
‫על‬
‫נמצא‬
m
‫שמסתו‬
‫( גוף‬1
tarily
and
what
are
the
magnitude
and
(d)
direction
with a 0.20 kg putty
ly after the putty sticks to it?
Fig. 8-44 Problem 33.
"15 N)î ?
pulling
directly
up
the
plane.
The
sled is to be on the verge of movwad
that
sticks
to
the
end
of
the
rod.
(up
or
down
the
incline)
of
the
box’s
acceleration
at
the
instant
the
‫את‬leaves
‫משחררים‬
k ‫קפיץ‬
‫קבוע‬
‫בעל‬
‫לקפיץ‬
‫מחובר‬
‫ הגוף‬.µFs required
= µk =of µthe,‫והקינטי שווים‬
In a long C
jump, an athlete
the ground .with
an
the
plane.
In Fig.
the magnitude
Ifing
theup
rod’s
angular
speed
just6-28,
before
ngular momentumbox
thatmomentarily
tends to rotatestops?
her body ?
forward,
cord’s
force
on
the
sled
is
plotted
versus
a
range
of
values for
the
‫הגוף‬
‫של‬
‫המקסימלית‬
‫המהירות‬
‫מהי‬
.‫רפוי‬
‫במצב‬
‫כאשר‬on‫הגוף‬
collision is 2.4 rad/s, what is the angu•••34
A
boy is ‫הקפיץ‬
initially
seated
the top of a hemispherical ic
x
UM
ing to ruin her Planding. To counter thisWtendency, she rocoefficient
static
ms between sled and plane: F1 ! 2.0 N,
lar
speed ofofthe
rodfriction
– putty system
!
mound
of
radius
R
13.8
m.
He
begins
to slide down the ice, wit
r outstretched arms to “take up” the angular momentum
F2 ! 5.0 N, and
u is the plane inclined?
2 ! 0.50. At what angle
immediately
aftermcollision?
Fig. 11-54 Problem 61.
a
negligible
initial
speed
(Fig.
8-45).
Approximate
the ice as bein
18).
In
0.700
s,
one
arm
sweeps
through
0.500
rev
and
the
θ
In Fig. 9-67, block 1 of mass m1 slides from rest along a
mDuring
F a jump to his
•••62
m sweeps
through
1.000
rev.2.50
Treat
each
arm
ascollides
a thin rod
of
frictionless.
At
what
height
does
the
boy
lose
contact
with the ice
!
ss
ramp from
height
h
m
and
then
with
Fig. 6-24 Problem 17.
partner, an aerialist
F2 is to make a quadruple somersault lasting a
and2,length
rotating
end.
the athy kg
block
which0.60
has m,
mass
m2 ! around
2.00m1. one
After
theIncollision,
1
time t ! 1.87 s. For the first and last quarter-revolution, he is in the
ference
what
is the
magnitude
total friction
angular
ides intoframe,
a region
where
the
coefficientofofthe
kinetic
extended orientation shown in Fig. 11-55, with rotational inertia
um
of
the
arms
around
the
common
rotation
axis
through
0 testify
and comes
a stopwitness
in distance
d within
that an
region.
u
as antoexpert
in a case
involving
acciF1
I1 ! 19.9 kg $ m2 around
his center of mass (the dot). During the
lders?
he
value
ofslid
distance
d ifrear
the of
collision
is (a) elastic
and (b)at
hich
car A
into the
car B, which
was stopped
rest of the flight he is in a tight tuck, with rotational inertia I2 !
uniform
disk
of
mass
10m
and
radius
3.0r
can
rotate
freely
y
inelastic?
µs
ht along a road headed down a hill (Fig. 6-25).
find
3.93 kg $ m2. What must
be his angular speed
his center
µ2 v 2 around
0
2 You
slope
fixed
merry-go-round.
A smaller
uniform
of center
the hilllike
is ua!p
12.0°, that the cars
were separated
R
of
mass
during
the
tuck?
1
θ
mass
r lies
top ofof
the
larger
disk,
ce d m
! and
24.0radius
m when
theondriver
car
A put
theconcentric
car into a
Fig.
6-28
Problem
22.
nitially
the
two disksanti-brake-lock
rotate togethersystem),
with an and
angular
cked any
automatic
that vethe
Fig.
8-41
20 A
rad/s.
Then
a slight
disturbance
smaller
diskProblem 30.
car
at the
onset
of braking
was v0causes
! 18.0the
m/s.
With
what
ω2
Tuck
Frictionless
µk outer edge of the
outward
larger
disk, untilkinetic
the
car A hitacross
car B the
if the
coefficient
friction
was (a)
2
2 ‫ את‬of‫)להזניח‬
‫גובה‬
‫מאותו‬
‫זמנית‬
‫בו‬
‫משוחררים‬
‫קטן‬
‫כדור‬
‫( כדור גדול ומעליו‬2
disk catches on the outer ILW
edge of the larger disk. Afterward,
••23 When the three blocksI 2in Fig.
A surface
blockcovered
withwithmass
road surface) and••31
(b) 0.10 (road
wet
disks again rotate together (without
further
sliding).
(a)
What
‫ההתנגשויות‬
‫ כל‬.(‫נקודתיים‬
‫גופים‬
‫הם‬
‫כאילו‬
‫לכדורים‬
‫ להתייחס‬,‫הרדיוסים של הכדורים‬
6-29 are
released
from
rest, they
ac! 2.00
m about
is ofplaced
9-67
Problem
68.
heir angular Fig.
velocity
the kg
center
the largeragainst
disk? (b) a
Parabolic
celerate
with‫שהכדור‬
am
magnitude
of 0.500
Fig.
8-45
Problem 34.
?
‫הרצפה‬
‫על‬
‫יישאר‬
‫הגדול‬
‫כדי‬
‫המסות‬
‫צריך להיות יחס‬
‫מה‬
.‫אלסטיות‬
spring
on energy
a frictionless
incline
the ratio K/K0 of the
new kinetic
of the two-disk
sysm/s2. Block 1 has mass M, block 2 1 path of
A
aerialist
3
v
he system’s
initial
kinetic
energy?
" !0 30.0° (Fig. 8-42). has 2M, and block 3 has 2M. What is
A small ball ofwith
mass angle
m
electric motor has rotational inertia I !
out its vcentral axis. The motor is used to
n1
of the space2probe in which it is mounted.
Fig. 6-27 Problem 20.
nted along the central axis of the probe; the
lig.inertia
! 12 kg " m58. about this axis.
9-62 I Problem
of revolutions of the rotor required to turn
tionary
box ofaxis.
sand is to be pulled across a
about
its
central
3, block 1 (mass 2.0 kg) is moving rightward
a cable in which the tension should not
eel is rotating
freely
at angular
speedat800
(mass
5.0 kg)
is
moving
rightward
3.0
oefficient
of static
friction
between
them/s.
box
ose
rotational
inertia
is
negligible.
A
second
less,
and ashould
spring be
with
a angle
springbetween
constantthe
of
(a)
What
the
and
with
twice
the
rotational
inertia
of
the
(The
block
is
not
attached
to
the
in the
d horizontal
above a platform
larger ball
of shape of a circular disk rotates
the coefficient of
k kinetic friction be- •••35 In Fig. 8-40, a block of mass m ! 3.20 kg slides from rest
When
the
blocks
collide,
the
compresB2.
tionless
asepvertical
axle
through
thethe
center
ofgreatest
!ock
0.63 kgbearing
(with
a about
slight
tween
block
2possible
and the table?
spring.)
The
spring,
with
spring
ontal
in
order
to
pull
distance
a 23.
frictionless incline at angle " ! 30.0° where
d
Fig. 6-29d down
Problem
θ
ed
to
the
same
shaft.
(a)
is
aand
mass
of 150
a19.6
radius
ofWhat
2.0 m,
and
a the angular
sThe
withplatform
the baseball
basω
θhas constant
! instant
k kg,
N/cm,
isthe
com••24 A 4.10
kg block
is pushed runs into a spring of spring constant 431 N/m. When the block mo
Baseball
maximum
at
the
blocks
have
the
ω
al
inertia
ofand
300the
kg $two
m about
the axis of rotation. A 60 kg
Fig.
9-68a),
(b)
what
isareathe
weight
ofcen-rethe
sand
box
Fig.
31.
along
a 8-42
floorand
by Problem
a constant
applied mentarily
pressed
20.0
cm shaft
and
then
v
combination
ofthe
the
and
two
wheels?
18.
stops, it has compressed the spring by 21.0 cm. What ar
walks
slowlyFig.
from6-25
the rimProblem
of
platform
toward
the
simultaneously
from
I horizontal and has a
force that is
leased.
(a)
What
is
the
elastic
pomaximum
compression.
e angular
of the
(a)I distance d and (b) the distance between the point of the fir
h
! 1.8 m.speed
(Assume
thesystem
ra- is 1.5 rad/s when the student
magnitude
of 40.0 N. Figure 6-30
he original rotational kinetic energy
is lost?
1
2
1
s
v (m/s)
1
1
theball
rim,iswhat
is the
angularenergy
she is
0.50 m from spring? (b) What
tential
of the
compressed
the change
:speed when
ch
negligible
relative
Releaseis speed
y
gives the block’s
v versus block – spring contact and the point where the block’s speed
12 N horizontal force F
er?
If the larger ball rebounds
in
the
gravitational
potential
energy
of
the
block
–
Earth
Basketball
Catch
time t as the block moves system
along an x greatest?
a block weighing 5.0 N
from11-52
the floor
thenblock moves from
F
gure
anand
overhead
as
the
the release
point
to
its
highest
point
on
Fig.
11-55
Problem
62.
axis
on
the
floor.
The
scale
of
the
figvertical
wallis(Fig.
6-26).
The
x
•••36
Two children are playing a game in which they try to h
elastically
a ball
thin rebounds
uniform rod
of length
ure’sisvertical
axis is set
by vfrom
incline? (c) How far along the incline
the highest
point
t of static frictionthe
between
s ! 5.0
0
0.5
1.0 with a marble fired from a spring-loade
arger
ball,
value
of m
and
mass
Mwhat
rotating
horizona
small
box
on the floor
Rotation
m/s. What is the coefficient of
and the block is the
0.60,
and
Ball
release
t (s)
•••63
In Fig. 11-56, a 30 kg
axis
the larger
ball rad/s
stopping
ngular
speed 20.0
about point?
gun
that
is
mounted
on a table. The target box is horizontal di
kinetic friction between the block
ficient of kinetic friction is (a) Before
(b) After
stands on table
the edge
of aonestaollides
the small
ball?
hroughwith
its center.
A particle
Fig.
11-52
Problem
59. on child
••32
In Fig.
8-43,
a chain
is held
aand
frictionless
with
v m from
tance
D
!
2.20
the edge of the table; see Fig. 8-46. Bobb
Fig. 6-30
Problem
24.
the
floor?
ume that the block
is not
Fig.
6-26
Problem
19.
tionary merry-go-round of radius
height
the attached
small ballto Fig. 9-68 Problem 69.
M/3.00 does
initially
fourth of its length hanging over the 2.0
edge.
If
the
chain
has
length
Child
m. The rotational inertia of the
his(Fig.
9-68b)?
ejected
from the rod and travels along a path that is per‫אופקי‬
‫משטח‬
‫להסתובב על‬
‫יכול‬
,m ‫ ומסתו‬l ‫ שאורכו‬,‫( מוט דק אחיד‬3
merry-go-round
about its
rotation
ar
to the
rodpuck
at the1 instant
of(‫חיכוך‬
ejection.
If‫)ללא‬
the
speed
Fig.
9-69,
of mass
m1 ! 0.20
kg particle’s
is sent sliding
axis is 150 kg $ m2. The child catches
0rictionless
m/s greater
than
the
speed
of
‫בקצהו‬
‫פוגע‬
‫כאשר‬
‫במנוחה‬
‫נמצא‬
‫המוט‬
.‫שלו‬
‫סביב ציר אנכי העובר דרך מרכז‬
lab bench,
undergo a one-dimensional
A elass towhat
1 1.0 kg thrown by ‫המסה‬
a ball of mass
a
end
juststationary
after ejection,
is 2 then slides off the
on
with
puck 2. Puck
bench 0
.θ ‫ מהירותו והמוט הנה‬friend.
‫כיוון‬Just
‫בין‬before
‫והזווית‬
the ball‫האופקי‬
is caught, ‫ במישור‬φ‫ הקליע נע‬.M ‫קליע שמסתו‬
eaofdistance
vp?
d from the base of the bench. Puck 1 rebounds
:
it
has
a
horizontal
velocity
of magv
ncollision
Fig. 11-53,
1.0 g off
bullet
is
.ωedge
‫הנה‬
‫המערכת‬
‫של‬
‫המהירות הזוויתית‬
‫הקליע נשאר במוט ואחרי ההתנגשות‬
andaslides
the ‫מהי‬
opposite
of the
bench,
nitude 12 m/s, at angle f ! 37° with
Rod
Fig. 11-56 Problem 63.
odistance
a 0.50 kg
attached
tothe bench. What is the
2d block
from the
base of
mass
?
a line tangent to the outer edge of ‫מהירות הקליע רגע לפני ההתנגשות‬
a 0.60Be
m careful
nonuniform
rod of
?of(Hint:
with signs.)
the merry-go-round, as shown. What is the angular speed of the
50 kg. The block – rod – bullet
D
merry-go-round just after the ball is caught?
hen rotates in the plane of the
massismheld
lies on
of a uniform
aofsled
onthe
an rim
inclined
plane disk
by a cord
rotate freely
about
its be
center
like verge
a merryen plane.
The sled
is to
on the
of mov1
2
cockroach
and
disk
rotate
together
with
an
Fig. 6-28, the magnitude F required of the
60
rad/s.
Then versus
the cockroach
halfway
ed is
plotted
a rangewalks
of values
for the
sk.
(a)
What
then
is
the
angular
velocity
of
ig. 9-63
Problem
59.and plane: F ! 2.0 N,
iction
m between
sled
stem? (b) What is the ratio K/K of the new
0.50.
Attowhat
angle u is the
plane(c)
inclined?
ystem
its
energy?
What
ollisions
ininitial
One kinetic
Dimension
nge in the kiAxisinto block B (mass 2.4
k A (mass 1.6 kg) slides
1
2
Fig. of
8-43three
Problem
32.
surface.
directions
velocities
be•••64
A ballerina
begins a tour jeté (Fig. 11-19a) with an-Fig. 8-46
bout
a fixed axis atThe
A. The roBlock
gular
speed
and
a
rotational
inertia consisting of two parts:
'
inertia
of
the
rod
alone
about
thin0.060rod
of the Bullet
! 1.44 kg$m for her leg extended outward at angle % ! 90.0&
he
indicated; the Icorresponding
at A iscollision
kg $ m . Treatare
i
2
leg
2
Problem 60.
to her body and I
then is Fig. 11-53
ssa particle.
4.00 (a)
kgWhat
can
2d
d
θ
Fig. 9-69 Problem 70.
plane about
a
tsCollisions
center.
The
in Two
Dimensions
In Fig. 9-21, projectile particle 1 is an alpha particle and
3.00 g bullet
rticle 2 is an oxygen nucleus. The alpha particle isµscats
ngle
u ! 64.0°
and
the oxygen nucleus
µ2recoils with speed
n m/s
plane
is
fired
and at angle u ! 51.0°. Fig.
In atomic11-50
mass units, theProblem 53.
he alpha
is 4.00 u and the mass of the oxygen nuod.
Asparticle
viewed
.0 u. What are the (a) final and (b) initial speeds of the alle?
path6-28
makes angle
u ! 60.0°
Fig.
Problem
22. with the rod (Fig.
l B, moving in the positive direction of an x axis at speed
ges
in the rod and the angular velocity of the
with stationary ball A at the origin. A and B have differs.
After
collision, the
B moves
in the negative direction
atelytheafter
collision,
what ofis the bullet’s
s at speed v/2. (a) In what direction does A move? (b)
ct?
2
5
trunk
! 0.660 kg$m2 for the rest of her body (pri-
1
2
e1 blocks
Fig.
shows inan
m rest, they ac-
4
Problem 36.
No. A B C D E
1
X
2 X
3 X
4
X
5
X
6
X
7
X
8
X
9
X
10
X
5
‫פתרונות‬
‫חלק א׳‬
‫‪ 1.‬לפי הגדרת מרכז המסה‪ ,‬מהירות שלו שווה לתנע של המערכת חלקי מסת‬
‫המערכת‪ .‬מהירות מרכז המסה קבועה אם תנע המערכת קבוע‪ ,‬ז״א כאשר סכום‬
‫הכוחות החיצוניים הנו אפס‪.‬‬
‫‪ 2.‬אין כוחות חיצוניים‪ ,‬מהירות מרכז המסה קבועה ושווה אפס )כי כך היה‬
‫בהתלה(‪.‬‬
‫‪ 3.‬זמן המחזור של תנודות הרמוניות לא תלוי באמפליטודה‪.‬‬
‫‪ 4.‬מהיאות מקסימלית שווה לאמפליטודה כפול תדירות‪.‬‬
‫‪ 5.‬על גליל ‪ A‬פועל כוח חיכולך סטטי‪ ,‬אשר לא מבצע עבודה‪ .‬הגליל עוצר כאשר‬
‫כל האנרגיה הקינטית שלו הופכת לאנרגיה פוטנציאלית‪ .‬על גליל ‪ B‬לא פועל כוח‬
‫החיכוך‪ ,‬לכן אין מומנט כוח אשר ישנה את התנועה הסיבובית שלו סביב מרכז המסה‪.‬‬
‫רק האנרגיה הקינטית של תנועה קווית הופכת לפואהציאלית‪.‬‬
‫‪ 6.‬תנע זוויתי נשמר כי אין מומנט כוח חיצוני‪.‬‬
‫‪W = F l 7.‬‬
‫‪8.‬‬
‫‪W = Ef − Ei‬‬
‫‪m 2‬‬
‫) ‪(vf − vi2‬‬
‫‪2‬‬
‫‪Ef − Ef = mg(hf − fi ) +‬‬
‫‪vf2 − vi2 = 2al,‬‬
‫!‪a < 0‬‬
‫‪hf − hi = l sin θ‬‬
‫‪9.‬‬
‫‪h + v0 t − gt2 /2 = 0‬‬
‫‪10.‬‬
‫‪a = ω2R‬‬
‫‪ω = 2π/T‬‬
‫חלק ב'‬
‫‪1.‬‬
‫המהירות מקסימלית בנקודה שבה התאוצה היא אפס‪ .‬נסמן ב ‪ x‬את המרחק‬
‫שעובר הגוף עד נקודה זו‪ ,‬אז‬
‫‪kx + µmg cos θ = mg sin θ‬‬
‫‪mg‬‬
‫=‪x‬‬
‫)‪(sin θ − µ cos θ‬‬
‫‪k‬‬
‫)‪(1‬‬
‫)‪(2‬‬
‫אם‬
‫)‪(3‬‬
‫‪µ ≥ tan θ‬‬
‫‪6‬‬
‫הגוף לא יזוז בכלל‪ .‬אחרת‪ ,‬ממשפט עבודה ואנרגיה‬
‫‪Ef − Ei = W‬‬
‫‪mv 2 kx2‬‬
‫‪+‬‬
‫‪− mgx sin θ = −µmg cos θx‬‬
‫‪2‬‬
‫‪2‬‬
‫‪mg‬‬
‫)‪v = √ (sin θ − µ cos θ‬‬
‫‪k‬‬
‫)‪(4‬‬
‫)‪(5‬‬
‫)‪(6‬‬
‫‪ 2.‬זאת אינני התנגשות אחת אלא שתי התנגשויות‪ .‬הראשונה‪ :‬הכדור התחתון‬
‫מתנגש עם הרצפה ומהירותו מתהפכת מ ‪) v‬כלפי מטה( ל‪ −v -‬כלפי מעלה‪ .‬בהתנגשות‬
‫השנייה מתנגשים שני הכדורים ) ‪ m‬הוא העליון‪ M ,‬הוא התחתון( ‪:‬‬
‫)‪(7‬‬
‫)‪(8‬‬
‫)‪(9‬‬
‫)‪(10‬‬
‫)‪(11‬‬
‫)‪(12‬‬
‫‪mv − M v = mu‬‬
‫‪mv 2 M v 2‬‬
‫‪mu2‬‬
‫‪+‬‬
‫=‬
‫‪2‬‬
‫‪2‬‬
‫‪2‬‬
‫‪M‬‬
‫‪u = (1 − )v‬‬
‫‪m‬‬
‫‪M 2‬‬
‫‪2‬‬
‫‪v + v = u2‬‬
‫‪m‬‬
‫‪M‬‬
‫‪M‬‬
‫‪1+‬‬
‫‪= (1 − )2‬‬
‫‪m‬‬
‫‪m‬‬
‫‪M‬‬
‫‪=3‬‬
‫‪m‬‬
‫‪.3‬רק תנע זוויתי נשמר כי אין מומנט כוח חיצוני )כוח חיצוני ישנו ‪ -‬בציר(‬
‫)‪(13‬‬
‫)‪(14‬‬
‫‪mv(l/2) sin θ = Iω‬‬
‫‪I = M l2 /12 + m(l/2)2‬‬
‫‪7‬‬