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Unformatted text preview: ActivPhysics can help with these problems:
m Activities in Section 3, Projectile Motion Section +1: Velocity and Acceleration . A skater is gliding along the ice at 2.4 m/s, when
4. she undergoes an acceleration of magnitude "1.1 m/s2 for 3.0 8. At the end of that time she is
moving at 5.7 m/s. What must be the angle
between the acceleration vector and the initial ‘ velocity vector? tion
‘ . constant acceleration, Equation 43 shows that the
' rs v0, aAt and v form a triangle as shown. The
of cosines gives '02 = 03 + (aAt)2 — ZuoaAtx
ms(180° — 90). When the given magnitudes are
ﬂistituted, one can solve for 00: (5.7 rn/s)2 =
4 m/s)2 + (1.1 m/32)2(3.0 s)2 + 2(2.4 m/s)x
m/sz)(3.0 5) cos 60, or cos 00 = 1.00 (exactly), and
0°. Since v0 and a are colinear, the change in 17 51M:
T7 90 Problem 1 Solution. Problem 2. In the preceding problem, what would have been
' ' , the magnitude of the skater’s ﬁnal velocity if the
‘ acceleration had been perpendicular to her initial
velocity? Solution
When v0 and aAt are perpendicular, Equation 4—3 and
the Phythagorean theorem imply v = «no: + (aAt)2 = {(2.4 m/s)2 + (1.1 m/s2)(3.0 s)2 = 4.08 m/s. VHAPTER 4 MOTION IN MORE THAN ONE
DIMENSION Problem An Object is moving in the x direction at 1.3 m/s
when it is subjected to an acceleration given by
a=0.52j m/s . What is its velocity vector after 4.4 s of acceleration? Solution
From Equation 43, v =vo + aAt : (1.30 m/s)i+
(0.523 m/52)(4.4s) = (1.30i+2.293) m/s. Problem 4. An airliner is ﬂying 'at a velocity of 2601‘ m/s, when
a wind gust gives it an acceleration of 0.38i+ 0.72j m/s2 for a period of 24 s. (a) What is its
velocity at the end of that time? (b) By what angle
has it been deﬂected from its original course? Solution (a) Equation 4—3 gives v = 2601‘ m/s + (0.38i +
0.723)(m/52)(24 s) = (269i+ 17.33) m/s. (b) Since Va
is along the xaxis, the angular deﬂection is just
tan“l(vy/v,) = tan—1(17.3/269) = 3.67°. Section 42: Constant Acceleration Problem @The position of an object as a function of time is
given by r = (3.2t + 1.8t2)i+ (1.7t — 2.4t2)j m,
where tis the time in seconds. What are the
magnitude and direction of the acceleration? Solution One can always ﬁnd the acceleration by taking the
second derivative of the position, a(t) = d2r(t)/dt2.
However, collecting terms with the same power of t,
one can write the position in meters as r(t) = (3.2i + 1.7j)t + (1.8‘i — 2.4j)t2. Comparison with
Equation 44 shows that this represents motion with
constant acceleration equal to twice the coefﬁcient of
the t2 term, or a = (3.6‘i—4.8j) m/sz. The magnitude
and direction of a are «(3.6)2 + (~48)? m/s2 =
4.49 m/s2, and tan—1(4.8/3.6) = 307° (cow from
the xaxis, in the fourth quadrant) or —53.1" (CW
from the :caxis). 50 CHAPTER 4 Problem 6. An airplane heads northeastward down a runway,
accelerating from rest at the rate of 2.1 m/sz.
Express the plane’s velocity and position at t = 30 s
in unit vector notation, using a coordinate system
with z—axis eastward and y—axis northward, and
with origin at the start of the plane’s takeoff roll. Solution Since the acceleration is constant and the airplane
starts from rest (v0 = 0) at the origin (to = 0),
Equations 43 and 44 give v(t)=at and r(t) = %at2.
Both vectors are in the NE direction, parallel to a = (2.1 rn/sz)(i‘cos45° +jsin 45°) = (1.48 m/sz)x
n+3). Thus, v(30 s) = (1.48 m/s2)(30 s)(i‘+j) =
(44.5 m/s)(i+j), and r(30 s) = %(1.48 m/s)(30 s)2x
(i+j) = (668 m) (3+3). (Note: (Hp/ﬂ is a unit
vector in the NE direction.) Problem An asteroid is heading toward Earth at a steady 21 kin/s. To save their planet, astronauts strap a
giant rocket to the asteroid, giving it an
acceleration of 0.035 km/s2 at right angles to its
original motion. If the rocket ﬁring lasts 250 s, (a) by what angle does the direction of the
asteroid’s motion change? (b) How far does it move
during the ﬁring? Solution Av  an: ((1) FIGURE 4—2a Problem 7 Solution. (a) The change in velocity, Av =aAt, is at right
angles to the initial velocity, v0, so they form the legs
of a right triangle with hypotenuse v (see Fig. 4124;).
The angle between v and Va is 0 = tan—lﬂAvl —:—
Ivol) = tan‘1(0.035 km/s)(250 s)/(21 kin/s) = 22.6°.
(b) For constant acceleration, the displacement is (see
Equation 44) Ar = r — r0 = vot + éatz. Again, the
two vectors vet and ﬁatz form the legs of a right
triangle, whose hypotenuse, Ar, has magnitude ,/(21 km/s)2(250 S)? + ($110035 km/sz)2(250 s)4 =
5.36x 103 km. (Note: this is the asteroid’s displacement during the rocket ﬁring, not the
pathlength of its trajectory.) Problem 8. An object is moving initially in the a: direction at
4.5 m/s, when an acceleration is applied in the
y direction for a period of 18 s. If it moves equal
distances in the .1: and y directions during this time,
what is the magnitude of its acceleration? Solution The :1: component of the displacement is due only to
the initial velocity, Arr: = vOIAt. The y component is
just due to the acceleration, Ag 2 éaAt2 = Arr. Thus, a = 2110, /At = 2(4.5 m/s)/18 s = 0.5 m/sz. Problem .A hockey puck is moving at 14.5 m/s when a stick imparts a constant acceleration of 78.2 m/s2 at a
90.0" angle to the original direction of motion. If
the acceleration lasts 0.120 3, what is the magnitude
of the puck’s displacement during this time? Solution Take the z~aads in the direction of the initial velocity,
v0 = 14.5 m/s, and the y—axis in the direction of the
acceleration, a = 78.23 m/sz. The displacement during the 0.120 s interval of constant acceleration is
(Equation 44) Ar: 1 — r0 = vot + %at2 = (14.5 m/s)(0.120 s)i+ 315(782 m/s’)(0.120 S)?“ =
(1.74i+0.563j In. This has magnitude «(1.74)2 + (0.563)2 m = 1.83 m (and makes an angle
of tan“1(0.563/ 1.74) = 17.9° with the direction of the
initial velocity). Problem 10. Repeat the preceding problem, except that now
the acceleration makes a 65.0° angle with the
original direction of motion. Solution Now the acceleration is (78.2 m/s2)icos 650° + jsin 65.0°) and Ar = (14.5 m/s)(0.120 s)i+ §(33.03+ 70.9j)(m/s2) (0.120 s)2 = (1.98‘i +0.5103) m.
This has magnitude 2.04 m (and makes an angle of
14.5° with the initial velocity). Problem 11. A particle leaves the origin with initial velocity
v0 = 111‘ + 143 m/s. It undergoes a constant
acceleration given by a: — 1.2i+0.26j‘ m/s2.
(a) When does the particle cross the yaxis? (b) What is its y—coordinate at the time? (c) How
fast is it moving, and in what direction, at that
time? Solution I (a) Since the particle leaves from the origin (to = 0),
its position is r(t) = vat + §at2. It crosses the y—axis
when z(t) = vozt + %a.,t2 = O, or t = «21103 /a,, =
—2(11 m/s)/(—1.2 m/sz) = 18.3 s. (b) w) = toy: +
gay? = [14 m/s + §(o.26 m/sz)(18.3 mass 3) =
300 m. (c) v(t) =vo +at = (11i+ 143) m/s + (1.2’i+0.263 18.3 m/s = (~11i+18.83) m/s. Then
lv(t)l = (~11)2 + (18.8)2 m/s = 21.8 m/s and
9, = tan—1(18.8/(—11))= 120°. Problem @ particle starts from the origin with initial
, velocity v0 = mi and constant acceleration a: aj. Show that the particle’s distance from the origin
and its direction relative to the zaxis are given by d = “/03 + §¢22t2 and 9 = tan1(at/2'uo). Solution Rom Equation 44, the particle’s displacement from
its original position is Ar: r — r0 = Vat + éat2 = (vo‘i + %at j)t. This has magnitude t‘lvﬁ + («éat)2 and
direction (CCW from the :raxis) 0 = tan“1(at/2vo). Problem 13. Figure 427 shoWs a cathoderay tube, used to
display electrical signals in oscilloscopes and other
scientiﬁc instruments. Electrons are accelerated by
the electron gun, then move down the center of
the tube at 2.0x 109 cm/s. In the 4.2cm—long
deﬂecting region they undergo an acceleration
directed perpendicular to the long axis of the
tube. The acceleration “steers” them to a
particular spot on the screen, where they produce
a visible glow. (a) What acceleration is nwded to
deﬂect the electrons through 15°, as shown in the
ﬁgure? (b) What is the shape of an electron’s path
in the deﬂecting region? Solution (a) With z~y axes as drawn on Fig. 427, the electrons
emerge from the deﬂecting region with velocity
v=voi+atj, after a time t = :r/vo, Where a: = 4.2 cm
and '00 = 2x 109 cm/s. The angle of deﬂection
(direction of v) is 15°, so tan 15° = vy/v, = at/vo =
arr/113. Thus, a = vgtan 15°/a: = 2.55x1017 cm/s2
(when values are substituted). (b) Since the CHAPTER 4 51 acceleration is assumed constant, the electron
trajectory is parabolic in the deﬂecting region. FIGURE 427 Problem 13 Solution. Section 43: Projectile Motion Problem 14. You toss an apple horizontally at 8.7 m/s from a
height of 2.6 m. Simultaneously, you drop a peach
from the same height. How long does each take to
reach the ground? Solution The time of flight for either projectile can be
determined from the vertical component of the
motion, which is the same for both, since coy = 0. Thus, Equation 48 gives t = (/2(y  yo)/ = (/2(2.6 m)/(9.8 m/s2) = 0.728 s. Problem 15. A carpenter tosses a shingle off a 8.8m—high roof,
giving it an initially horizontal velocity of 11 m/s.
(a) How long does it take to reach the ground?
(b) How far does it move horizontally in this time? ‘Uo = (llm/s)i I Problem 15 Solution. 52 CHAPTER 4 Solution
(a) The shingle reaches the ground when y(t) = 0 =
yo — égtjl 01' _ En: lm=
13‘ g (9.8 m/sz) 1'345' (b) The horizontal displacement is x = not =
(11 m/s)(1.34 s) = 14.7 m. Problem 16. An arrow ﬁred horizontally at 41 m/s travels 23 m
horizontally before it hits the ground. E'om what height was it ﬁred? Solution From a:  :co = vat, one ﬁnds the time of ﬂight t: 23 m/(41 m/s) = 0.561 s, and from yo — y = égtz
(recall that 120,, = 0) one ﬁnds the height yo — y =
%(9.8 m/s2)(0.561 s)2 = 1.54 m. Problem @ A kid ﬁres water horizontally from a squirt gun
held 1.6 m above the ground. It hits another kid
2.1 m away square in the back, at a point 0.93 m
above the ground (see Fig. 428). What was the
initial speed of the water? FIGURE 4—28 Problem 17. Solution Since the water was ﬁred horizontally (v03, = 0), the
time it takes to fall from yo = 1.6m to y z: 0.93m is given by Equation 48, t = (/2010 — ill/9 == (/2(1.6 — 0.93) m/(9.8 m/s2) = 0.370 3. Its initial speed, '00 = v03, can be found from Equation 47,
110 = (1: — zo)/t = 2.1 m/0.370 s = 5.68 m/s. Problem 18. In a chase scene, a movie Stuntman is supposed to
run right off the ﬂat roof of one city building and land on another roof 1.9 m lower. If the gap
between the buildings is wide, how fast must he
run? Solution The horizontal and vertical distances covered by
the stuntman are a: — so = vot and ya — y = £th
(since 120, = 1:0, and my = 0). Eliminating t, one ﬁnds 00 = (.7: — mo)\/g/2(yo — y) = (4.5 m)“ (9.8 m/sz)/2(1.9 m) = 7.23 m/s. (Note that
Equation 49 with 00 = 0 and yo = 0 provides an
equivalent solution.) Problem 19. Ink droplets in an inkjet printer are ejected
horizontally at 12 m/s, and travel a horizontal
distance of 1.0 mm to the paper. How far do they
fall in this interval? Solution From a: — $0 = vast, the time of ﬂight can be found.
Substitution into yo — y = §gt2 (recall that 001, = 0) yields 90 — y = $9“?  mop/”ii = gas m/SZXIO—a m)2/(12 m/s)2 = 3.40x108 m =
34 nm for the distance fallen, practically negligible.
Note that this analysis is equivalent to using
Equation 49 with 90 = 0. Problem 20. Protons in a. particle accelerator drop 1.2 pm over
the 1.7km length of the accelerator. What is their
approximate average speed? Solution The horizontal and vertical distances in projectile
motion (range and drop) are related by the
trajectory equation (Equation 49). With 00 = 0,
y = ga:2/2v§, or 120 = :c‘/(—g)/2y = (1.7x103 m) (—9.8 m/s2)/2(—1.2x10‘6 m) ='
3.44x106 m/s. Since 11,, = —gt = —g(z — 2:0)/vo =
—4.85x10‘3 m/s is negligible compared to this, 110 is
hardly different from the average speed. Problem 21. You’re standing on the ground 3.0 m from the wall
of a building, and you want to throw a package
from your 1.54.2m shoulder level to someone in a
secondﬂoor window above the ground. At what
speed and angle should you throw it so it just
barely reaches the window? 54 CHAPTER 4 the vertical component of the velocity (negative t = hog/g = 2(6.64 kin/s) sin 45” / (9.8 m/sz) =
downward) for the given drop, 11;, = _ “3y __ 29y = 958 s = 16.0 min. (c) At a 20° launch angle, .
= 2 . O =
.. r———_——_———_Q8.1 m /s)2 $112150 _ 2 (9.8 m/s2)(—5.9 m) = 00 l/ (4500 km)(0.0098 km/s )/ s1n40 8.28 km/s.
—14.6 m/s . Problem
Problem A rescue airplane is ﬂying horizontally at speed on
_ . . at an altitude h above the ocean, attempting to
26. Compare the travel times for the prejectiles drop a package of medical supplies to a shipwreck
lauPched at 30° and 60° II} F 1a 415, Poth 0_f victim in a lifeboat. At what lineofsight angle a
meh have the same starting and ending pomts. (Fig. 4—29) should the pilot release the package? Solution The package, when dropped, is a horizontally launched
projectile, like that in Problem 15 (the expressions for
t and a: are the same). The lineofsight angle is a =
tan1(h/x), where :r = vo‘/2h/g is the horizontal
distance at drop time. Thus, a = tan‘1 gh/2vo. V0: V0? N FIGURE 4—15 Problem 26. Solution The time of ﬂight (t > 0) for a projectile trajectory
beginning and ending at the same height (y(t) = ya)
can be found from Equation 48, y(t) — yo = 0 = voyt —
égt2 or t = 2vOy/g = 200 sinQO/g. In Fig. 415, on =
50 m/s, so two = 2(50 m/s) sin 30°/(9.8 m/sz) = 5.10 S, and tsoo = ﬂ t3oo = 8.84 8. FIGURE 429 Problem 28 Solution. Problem A submarinelaunched missile has a range of
4500 km. (a) What launch speed is needed for this range when the launch angle is 45°? (b) What is Problem
the total ﬂight time? (c) What would be the 29. At a circus, a human cannonball is shot from a
minimum launch speed at a 20° launch angle, used cannon at 35 km/h at an angle of 40°. If he leaves the cannon 1.0 m 013" the ground, and lands in a to “depress” the trajectory so as to foil 3
net 2.0 m 011’ the ground, how long is he in the air? spacebased antimissile defense? Solution The time of ﬂight can be calculated from
Equation 48, where the trajectory begins at Solution (a) Assuming Equation 410 applies (i.e., the
trajectory begins and ends at the same height, or y t = y , one ﬁnds 11 : ﬁg 531129 ___ yo = 1.0 m and t = 0, ends at y(t) = 2.0 m, and
H—‘—oT—./‘: you = vosinﬂo = (35 m/3.6 s) sin 40° = 6.25 m/s. The
(4500 km)(0.0098 km/s )/ 3m 90 = 6.64 km/s. equation is a quadratic, §gt2 _ "01,1 + (y __ yo) ___ 0’ (b) The time of ﬂight is the positive solution of   _ 2 _ _ _
Equation 48 when y(t)  yo = 0 = ”out — égtz. Thus With solutions t _ [”0” i— 009 29 (y yo)]/g ’— [6.25 m/s i ,/ (6.25 m/s)2 — 2(9.8 m/52)(2 m — 1 m)]~:— (9.8 m/sg) = 0.188 s or 1.09 s. The trajectory crosses
the height 2.0 in twice, once going up, at the smaller
time of ﬂight, and once going down into the net, at the
larger time of ﬂight. The latter is the answer to the
question asked here. Problem 30. Preparing to give an injection, a physician ejects a
drop of medicine to ensure there’s no air in the
syringe. The syringe is pointing upward, at 20° to
the vertical, with the tip of the needle 45.0 cm
above a tabletop. The drop leaves the needle at
1.30 m/s. (a) What maximum height does it
reach? (b) How long is it before the drop hits the
tabletop? (c) How far does the drop travel
horizontally? Solution (a) Take the tip of the syringe to be at yo = 45.0 cm
above the tabletop at y = 0. At its maximum height,
just the vertical component of the drop’s velocity is
instantaneously zero, so '03 = 0 = v33, — Zg(ymmt — yo),
or gym“ — yo = nay/2g. The vertical component of the
initial velocity is boy 2 (1.30 m/s) cos 20°, which gives
ymax — yo = 7.61 cm. This is the maximum height
above the tip, corresponding to 52.6 cm above the
tabletop. (b) The time of ﬂight is the positive solution
of the equation y(t) = 0 = yo + voyt — égtz, or t = (coy + Mug” + Zgyo)/g. Substituting the numbers from part (a), we obtain t.= 0.452 s. (c) The horizontal
distance traveled is a: — :co = volt =
(1.30 m/s)(sin 20°)(0.452 s) = 20.1 cm. Problem fyou can hit a golf ball 180 m on Earth, how far
can you hit it on the moon? (Your answer is an
underestimate, because the distance on Earth is
restricted by air resistance as well as by a larger 9.) Solution
For given W), the horizontal range is inversely
proportional to 9. With surface gravities from Appendix E, we ﬁnd zmoon = (gEarth/gmoon)$EsIth =
(9.81 /1.62)(180 m) = 1090 m. Problem
32. Prove that a projectile launched on level ground
reaches its maximum height midway along its
trajectory. CHAPTER 4 55 Solution From the trajectory (Equation 49), the maximum
height occurs when dy/dx = tanB — gz/vg cos2 00 = 0,
or :1: = 0% sin 00 cos 9/9, which is midway, or half of the
horizontal range (Equation 4.10). This result can be
derived in other ways; for example, the maximum
height is reached at time vow/g (when 11,, = 0), whereas
the total time of ﬂight is twice this (the solution of O = voyt  égtz). Problem 33. A projectile launched at an angle 00 to the
horizontal reaches a maximum height h. Show
that its horizontal range is 4h / tan 00. Solution The intermediate expression for the horizontal range
(when the initial and ﬁnal heights are equal) is x = 2123 sin 00 cos 90/9 = vavoy/g (see the equation
before Equation 4—10). The components of the initial
velocity are related by coy / 1103 = tan 4%. The
maximum height, h = gmax  yo, can be found from
Equation 211 (when 2),, = 0) or 113,, = 29h. Then 2 =
ZonUoz/g = 20%(voy/ tan 90)/g == 2(2gh)/g tan 00 :
4h/ tan 00. (This result reﬂects a classical geometrical
property of the parabola, namely, that the latus ‘ rectum is four times the distance from vertex to focus.) Problem 34. You’re 5.0 m from the lefthand wall of the house
shown in Fig. 430, and you want to throw a ball to a friend 5.0 m from the righthand wall.
(a) What is the minimum speed that will allow the ball to clear the roof? (b) At what angle
should it be thrown? Assume the throw and catch
both occur 1.0 m above the ground. FIGURE 430 Problem 34. Solution Since the trajectory is symmetrical (begins and ends
at the same height), one can use the result of the 56 CHAPTER 4 previous problem with h = ymax — yo = 6 m — 1 m=5 in and horizontal rangez =5 m+6 m+ 5 m = 16 m. Then (b) 00 = tan“1(4h/a:) = 51.3°, and
(a) v0 = coy/sinﬂo = W/sinﬂo = 12.7 m/s. Problem 35. A circular fountain has jets of water directed from
the circumference inward at an angle of 45°. Each
jet reaches a maximum height of 2.2 m. (a) If all
the jets converge in the center of the circle and at
their initial height, what is the radius of the
fountain? (b) If one of the jets is aimed at 10" too
low, how far short of the center does it fall? Solution (a) The radius is the horizontal range, 1‘ = rig/g
(Equation 410 with 9 = 45°). The maximum height is
h = vgy/2g = v3/4g (Equation 211 with 0,, = 0 and
210,, = 120 cos 45° = ooh/2). Therefore, 1' = (4gh)/g ==
4h = 4(2.2 m) = 8.8 m. (b) If one jet is directed at 35°
with the same initial speed (113 = rg), it would fall
short by 1' — 2:, where :r: is given by Equation 4—10.
Therefore, r — :c = r — (cg/g) sin(2 x 35°) = (8.8 m)(l — sin 70°) = 0.531 m. Problem 35 Solution. Problem 36. When the Olympics were held in Mexico City in
1968, many sports fans feared that the high
altitude would result in poor performances due to
reduced oxygen. To their surprise, new records
were set in track and ﬁeld events, probably as a
result of lowered air resistance and a decrease in g
to 9.786 m/s2, both ultimately associated with the
high altitude. In particular, Robert Beamon set a
new world record of 8.90 m in the long jump.
Photographs suggest that Beamon started his
jump at a 25° angle to the horizontal. If he had
jumped at sea level, Where g = 9.81 m/sz, at the
same angle and initial speed as in Mexico City,
how far would Beamon have gone? Neglect air
resistance in both cases (although its effe...
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 Spring '08
 WORMER
 Physics

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