Post on 25-Jun-2018
1
Motion in Two Dimensions
SOLUTIONS TO PROBLEMS Section 3.1 The Position, Velocity, and Acceleration Vectors P3.1
x m( )0
!3 000!1 270!4 270 m
y m( )!3 600
01 270!2 330 m
(a)
Net displacement = x2 + y2
= 4.87 km at 28.6° S of W
FIG. P3.1
(b) Average speed =
20.0 m s( ) 180 s( ) + 25.0 m s( ) 120 s( ) + 30.0 m s( ) 60.0 s( )180 s + 120 s + 60.0 s
= 23.3 m s
(c) Average velocity =
4.87 ! 103 m360 s
= 13.5 m s along rR
2 Motion in Two Dimensions
Section 3.2 Two-Dimensional Motion with Constant Acceleration P3.3
rv i = 4.00i + 1.00 j( ) m s and
rv 20.0( ) = 20.0i ! 5.00 j( ) m s
(a) ax =
! vx
! t=
20.0 " 4.0020.0
m s2 = 0.800 m s2
ay =
! vy
! t="5.00 " 1.00
20.0 m s2 = "0.300 m s2
(b) ! = tan"1 "0.300
0.800#$%
&'(= "20.6° = 339° from + x axis
(c) At t = 25.0 s
x f = xi + vxit +12
axt2 = 10.0 + 4.00 25.0( ) + 12
0.800( ) 25.0( )2 = 360 m
y f = yi + vyit +12
ayt2 = !4.00 + 1.00 25.0( ) + 12!0.300( ) 25.0( )2 = !72.7 m
vxf = vxi + axt = 4 + 0.8 25( ) = 24 m svyf = vyi + ayt = 1 ! 0.3 25( ) = !6.5 m s
" = tan!1 vy
vx
#
$%&
'(= tan!1 !6.50
24.0#$%
&'(= !15.2°
P3.5
ra = 3.00 j m s2 ; rv i = 5.00i m s ;
rri = 0i + 0 j
(a)
rrf =rri +
rv it +12
ra t2 = 5.00ti + 12
3.00t2 j!"#
$%&
m
rv f =
rv i +ra t = 5.00i + 3.00tj( ) m s
(b) t = 2.00 s , rrf = 5.00 2.00( ) i + 1
23.00( ) 2.00( )2 j = 10.0i + 6.00 j( ) m
so x f = 10.0 m , y f = 6.00 m
rv f = 5.00i + 3.00 2.00( ) j = 5.00i + 6.00 j( ) m s
v f =rv f = vxf
2 + vyf2 = 5.00( )2 + 6.00( )2 = 7.81 m s
Chapter 3 3
P3.6 (a) For the x-component of the motion we have x f = xi + vxit +
12
axt2 .
0.01 m = 0 + 1.80 ! 107 m s( )t + 12
8 ! 1014 m s2( )t2
4 ! 1014 m s2( )t2 + 1.80 ! 107 m s( )t " 10"2 m = 0
t ="1.80 ! 107 m s ± 1.8 ! 107 m s( )2 " 4 4 ! 1014 m s2( ) "10"2 m( )
2 4 ! 1014 m s2( )="1.8 ! 107 ± 1.84 ! 107 m s
8 ! 1014 m s2
We choose the + sign to represent the physical situation
t = 4.39 ! 105 m s
8 ! 1014 m s2 = 5.49 ! 10"10 s .
Here
y f = yi + vyit +
12
ayt2 = 0 + 0 + 12
1.6 ! 1015 m s2( ) 5.49 ! 10"10 s( )2 = 2.41 ! 10"4 m .
So,
rrf = 10.0 i + 0.241 j( ) mm .
(b)
rv f =rv i +
rat = 1.80 ! 107 m s i + 8 ! 1014 m s2 i + 1.6 ! 1015 m s2 j( ) 5.49 ! 10"10 s( )= 1.80 ! 107 m s( ) i + 4.39 ! 105 m s( ) i + 8.78 ! 105 m s( ) j
= 1.84 ! 107 m s( ) i + 8.78 ! 105 m s( ) j
(c) rv f = 1.84 ! 107 m s( )2 + 8.78 ! 105 m s( )2 = 1.85 ! 107 m s
(d) ! = tan"1 vy
vx
#
$%&
'(= tan"1 8.78 ) 105
1.84 ) 107#
$%&
'(= 2.73°
4 Motion in Two Dimensions
Section 3.3 Projectile Motion
*P3.9 At the maximum height vy = 0 , and the time to reach this height is found from
vyf = viy + ayt as t =
vyf ! viy
ay=
0 ! viy
!g=
viy
g.
The vertical displacement that has occurred during this time is
!y( )max = vy ,avgt =
vyf + viy
2"
#$%
&'t =
0 + viy
2"
#$%
&'viy
g"
#$%
&'=
viy2
2g.
Thus, if !y( )max = 12 ft
1 m3.281 ft
"#$
%&'= 3.66 m , then
viy = 2g !y( )max = 2 9.80 m s2( ) 3.66 m( ) = 8.47 m s , and if the angle of projection is ! = 45° , the launch speed is
vi =
viy
sin!=
8.47 m ssin 45°
= 12.0 m s .
P3.11 Take the origin at the mouth of the cannon.
x f = vxi t 2 000 m = 1 000 m s( )cos!it
Therefore, t = 2.00 s
cos!i
y f = vyi t + 1
2ay t2 :
800 m = 1 000 m s( )sin!i t + 1
2"9.80 m s2( )t2
800 m = 1 000 m s( )sin!i
2.00 scos!i
"
#$%
&'(
12
9.80 m s2( ) 2.00 scos !i
"
#$%
&'
2
800 m cos2 !i( ) = 2 000 m sin!i cos!i( ) " 19.6 m
19.6 m + 800 m cos2 !i( ) = 2 000 m 1 " cos2 !i cos!i( )
384 + (31360)cos2 !i + (640000)cos4 !i = (4 000000)cos2 !i " (4 000000)cos4 !i
4 640000cos4 !i " 3 968 640cos2 !i + 384 = 0
cos2 !i =
3 968 640 ± (3 968 640)2 " 4(4 640000)(384)9 280000
cos!i = 0.925 or 0.00984
!i = 22.4° or 89.4° (Both solutions are valid.)
Chapter 3 5 P3.12 (a)
x f = vxi t = 8.00cos 20.0° 3.00( ) = 22.6 m
(b) Taking y positive downwards, and the final point just before ground impact,
y f = vyit +12
g t2
y f = 8.00sin 20.0° 3.00( ) + 12
9.80( ) 3.00( )2 = 52.3 m .
(c) Now take the final point 10 m below the window.
10.0 = 8.00 sin 20.0°( )t + 1
29.80( )t2
4.90t2 + 2.74t ! 10.0 = 0
t = !2.74 ± 2.74( )2 + 1969.80
= 1.18 s
P3.13 Consider the motion from original zero height to maximum height h:
vyf2 = vyi
2 + 2ay y f ! yi( ) gives 0 = vyi2 ! 2g h ! 0( ) or
vyi = 2gh Now consider the motion from the original point to half the maximum height:
vyf2 = vyi
2 + 2ay y f ! yi( ) gives vyh
2 = 2gh + 2 !g( ) 12
h ! 0"#$
%&'
so
vyh = gh
At maximum height, the speed is vx =
12
vx2 + vyh
2 =12
vx2 + gh
Solving, vx =
gh3
Now the projection angle is !i = tan"1 vyi
vx= tan"1 2gh
gh/3= tan"1 6 = 67.8° .
6 Motion in Two Dimensions
P3.15 (a) We use the trajectory equation: y f = x f tan!i "
gx f2
2vi2 cos2 !i
. With x f = 36.0 m ,
vi = 20.0 m s , and ! = 53.0° we find
y f = 36.0 m( ) tan 53.0° !
9.80 m s2( ) 36.0 m( )2
2 20.0 m s( )2 cos2 53.0°( )= 3.94 m . The ball clears the bar by
3.94 ! 3.05( ) m = 0.889 m . (b) The time the ball takes to reach the maximum height is
t1 =
vi sin!ig
=20.0 m s( ) sin53.0°( )
9.80 m s2 = 1.63 s .
The time to travel 36.0 m horizontally is t2 =
x f
vix
t2 =
36.0 m(20.0 m s ) cos 53.0°( )
= 2.99 s .
Since t2 > t1
the ball clears the goal on its way down .
P3.16 The horizontal component of displacement is
x f = vxit = vi cos!i( )t . Therefore, the time required to
reach the building a distance d away is t = d
vi cos!i. At this time, the altitude of the water is
y f = vyit +
12
ayt2 = vi sin!id
vi cos!i
"
#$%
&'(
g2
dvi cos!i
"
#$%
&'
2
.
Therefore the water strikes the building at a height h above ground level of
h = y f = d tan!i "
gd2
2vi2 cos2 !i
.
Section 3.4 The Particle in Uniform Circular Motion
P3.23 ac =
v2
r=
20.0 m s( )2
1.06 m= 377 m s2
The mass is unnecessary information.
Chapter 3 7
P3.25 r = 0.500 m ;
vt =2! rT
=2! 0.500 m( )
60.0 s200 rev
= 10.47 m s = 10.5 m s
a = v2
R=
10.47( )2
0.5= 219 m s2 inward
P3.27 The satellite is in free fall. Its acceleration is due to gravity and is by effect a centripetal acceleration.
ac = g
so v2
r= g . Solving for the velocity, v = rg = 6, 400 + 600( ) 103 m( ) 8.21 m s2( ) = 7.58 ! 103 m s .
v = 2!r
T and
T =2! r
v=
2! 7,000 " 103 m( )7.58 " 103 m s
= 5.80 " 103 s
T = 5.80 " 103 s 1 min60 s
#$%
&'(= 96.7 min .
Section 3.5 Tangential and Radial Acceleration P3.28 (b) We do part (b) first. The tangential speed is described by v f = vi + att
0.7 m s = 0 + at 1.75 s( ) so at = 0.400 m s2 forward
(a) Now at t = 1.25 s , v f = vi + att = 0 + 0.4 m s2( )1.25 s
v f = 0.5 m s
so ac =
v2
r=
0.5 m s( )2
0.2 m= 1.25 m s2 toward the center
(c)
ra = rar +rat = 0.4 m s2 forward + 1.25 m s2 inward
ra = 0.42 + 1.252 forward and inward at ! = tan"1 1.250.4
#$%
&'(
ra = 1.31 m s2 forward and 72.3° inward
8 Motion in Two Dimensions
P3.30 (a) See figure to the right. (b) The components of the 20.2 and the 22.5 m s2 along the rope together
constitute the centripetal acceleration:
ac = 22.5 m s2( )cos 90.0° ! 36.9°( ) + 20.2 m s2( )cos 36.9° = 29.7 m s2
(c) ac =
v2
r so v = acr = 29.7 m s2 1.50 m( ) = 6.67 m s tangent to circle
rv = 6.67 m s at 36.9° above the horizontal
FIG. P3.30
Section 3.6 Relative Velocity
P3.33 Total time in still water t = d
v=
2 0001.20
= 1.67 ! 103 s . Total time = time upstream plus time downstream:
tup =1 000
(1.20 ! 0.500)= 1.43 " 103 s
tdown =1 000
1.20 + 0.500= 588 s.
Therefore, ttotal = 1.43 ! 103 + 588 = 2.02 ! 103 s, 21.0% more than if the water were still. .
*P3.34
rvce = the velocity of the car relative to the earth.
rvwc = the velocity of the water relative to the car.
rvwe = the velocity of the water relative to the earth.
These velocities are related as shown in the diagram at the right.
(a) Since
rvwe is vertical, vwc sin 60.0° = vce = 50.0 km h or
rvwc = 57.7 km h at 60.0° west of vertical .
(b) Since
rvce has zero vertical component,
FIG. P3.34
vwe = vwc cos 60.0° = 57.7 km h( )cos 60.0° = 28.9 km h downward .
Chapter 3 9
Additional Problems P3.47
x f = vixt = vit cos 40.0°
Thus, when x f = 10.0 m , t = 10.0 m
vi cos 40.0°.
At this time, y f should be 3.05 m ! 2.00 m = 1.05 m .
Thus, 1.05 m =
vi sin 40.0°( )10.0 mvi cos 40.0°
+12!9.80 m s2( ) 10.0 m
vi cos 40.0°"
#$
%
&'
2
.
From this, vi = 10.7 m s .
10 Motion in Two Dimensions
P3.50 Measure heights above the level ground. The elevation yb of the ball follows
yb = R + 0 ! 1
2gt2
with x = vi t so yb = R !
gx2
2vi2 .
(a) The elevation yr of points on the rock is described by
yr2 + x2 = R2 .
We will have yb = yr at x = 0 , but for all other x we require the ball to be above the rock
surface as in yb > yr . Then yb2 + x2 > R2
R !gx2
2vi2
"
#$%
&'
2
+ x2 > R2
R2 !gx2R
vi2 +
g2x4
4vi4 + x2 > R2
g2x4
4vi4 + x2 >
gx2Rvi
2 .
If this inequality is satisfied for x approaching zero, it will be true for all x. If the ball’s
parabolic trajectory has large enough radius of curvature at the start, the ball will clear the
whole rock: 1 > gR
vi2
vi > gR .
(b) With vi = gR and yb = 0 , we have 0 = R !
gx2
2gR
or x = R 2 . The distance from the rock’s base is
x ! R = 2 ! 1( )R .