Download Chapter 10 Solution Problem Solution Problem Solution

Chapter 10 Solution
Problem
1.
A 28-kg child sits at one end of a 3.5-m-long seesaw. Where should her 65-kg father sit so the center of mass will be at
the center of the seesaw?
Solution
Take the x-axis along the seesaw in the direction of the father, with origin at the center. The center of mass of the child and
her father is at the origin, so xcm = 0 = mc x c + m f x f , where the masses are given, and xc = !(3.5 m)=2 (half the length of
the seesaw in the negative x direction). Thus, x f = !mc x c=m f = (28=65)(1.75 m) = 75.4 cm from the center.
Problem
3.
Four trucks, with masses indicated in Fig. 10-23, are on a rectangular barge of mass 35 Mg whose center of mass is at its
center. The trucks’ individual centers of mass are located 25 m apart on the barge’s long dimension and 10 m apart on
the short dimension, as shown. Where is the center of mass of the entire system? Express in relation to the truck at lower
left.
FIGURE 10-23 Top view of four trucks on a barge, with truck masses given.
Dots mark individual trucks’ centers of mass (Problem 3).
Solution
As explained in the text (see Figs. 10-8 and 9), in order to find the center of mass of this system, the trucks and the barge can
be treated as point masses located at their centers of mass. With x-y axes as shown superimposed on Fig. 10-23,
xcm = (18 Mg)(0) + (23 Mg)( 25 m) + (19 Mg)( 25 m) + (11 Mg)( 0) + (35 Mg)(12.5 m) = 14.0 m,
(18 + 23 + 19 + 11 + 35) Mg
(19 Mg + 11 Mg)(10 m) + (35 Mg)( 5 m)
ycm =
= 4.48 m
106 Mg
230 CHAPTER 10
Problem
5.
Three equal masses lie at the corners of an equilateral triangle of side l. Where is the center of mass?
Solution
Take x-y coordinates with origin at the center of one side as shown. From the symmetry (for every mass at x, there is an equal
mass at ! x) xcm = 0. Since y = 0 for the two masses on the x-axis, and y = l sin60° = l 3=2 for the other mass,
Equation 10-2 gives ycm = m(l 3=2)=3m = l=2 3 = 0.289 l .
Problem 5 Solution.
Problem
9.
Find the center of mass of a pentagon of side a with one triangle missing, as shown in Fig. 10-24. Hint: See
Example 10-3, and treat the pentagon as a group of triangles.
Solution
Choose coordinates as shown. From symmetry, xcm = 0. If the fifth isosceles triangle (with the same assumed uniform
density) were present, the center of mass of the whole pentagon would be at the origin, so 0 = (my5 + 4mycm )=5m , where
ycm gives the position of the center of mass of the figure we want to find, and y5 is the position of the center of mass of the
fifth triangle. Of course, the mass of the figure is four times the mass of the triangle. In Example 10-3, the center of mass
of an isosceles triangle is calculated, so y5 = ! 23 l, and from the geometry of a pentagon, tan36° = 12 a=l. Therefore,
ycm = ! 14 y5 =
1
6
l=
1
1 2 a cot 36°
= 0.115a.
FIGURE 10-24 Problem 9 Solution.
CHAPTER 10 231
Problem
10. A solid cube of side a has a density that varies linearly from zero at the bottom to !0 at the top, as suggested in
Fig. 10-25; that is, ! = ! 0 ( y=a). Find expressions for (a) the cube’s mass and (b) the y coordinate of its center of mass.
Hint: Use a square mass element of thickness dy, as shown in the figure.
Solution
Take the y-axis through the center of each square mass element with y measured vertically from the bottom. Then the center
2
2
of mass of the cube is on the y-axis. (a) A mass element has volume a dy and mass dm = ! a dy = ! 0 ay dy. The total mass
of the cube is
M =
!
a
0
dm =
!
a
0
" 0 ay dy =
1
" a3 .
2 0
(b) The y component of Equation 10-5 gives the height of the center of mass,
ycm =
1
M
"
a
0
y dm =
2
!0 a
3
"
a
0
!0 ay 2 dy =
FIGURE 10-25 Problem 10 Solution.
2
a.
3
232 CHAPTER 10
Problem
18. Find the center of mass of the uniform, solid cone of height h and constant density ! shown in Fig. 10-28. Hint:
Integrate over disk-shaped mass elements of thickness dy, as shown in the figure.
Solution
Choose the y-axis along the axis of the cone, and the origin at the center of the base. For mass elements, take disks at
2
height y, of radius r = R(1 ! y=h) , parallel to the base. Then dm = !" r dy, where ! is the density of the cone, and
M =
1
3
!" R2 h is its mass. The CM is along the axis (from symmetry) at a height
ycm =
=
1
M
3
h
$
h
0
$
y dm =
3
!"R2 h
$
h
0
y !" R2 (1 # y=h)2 dy
2 y2
y3 (
3 % h2 2 h2 h2 (
1
y
#
+
dy
=
#
+ * = h.
'&
2*
'
h
h& 2
3
4)
4
h )
h%
0
FIGURE 10-28 Problem 18 Solution.
Problem
19. A popcorn kernel in a hot pan bursts into two pieces, with masses of 91 mg and 64 mg. The more massive piece moves
horizontally at 47 cm/s. Describe the motion of the second piece.
Solution
Suppose the popcorn kernel was initially at rest, and not subject to a net external force while bursting apart. The
momentum of the two pieces will still be zero immediately afterwards (conservation of momentum during the break-up).
Then 0 = m1 v1 + m2 v 2 , where v1 and v 2 are the final velocities of the two pieces, so v2 = !( m1=m2 )v1 .
If v1 = (47 cm/s) î is the final velocity of the more massive piece, then v2 = !(91 mg=64 mg)( 47 cm/s) î = !(66.8 cm/s) î;
i.e., the less massive piece moves with speed 66.8 cm/s in the opposite direction to the more massive piece.
CHAPTER 10 233
Problem
21. A firecracker, initially at rest, explodes into two fragments. The first, of mass 14 g, moves in the positive x direction at
48 m/s. The second moves at 32 m/s. Find its mass and the direction of its motion.
Solution
The momentum of the firecracker/fragments is approximately conserved during a short time interval around the explosion (as
for the popcorn kernel in Problem 19), so 0 = m1 v1 + m2 v 2 , or m2 v 2 = !(14 g)( 48 î m/s) . The direction of the second piece
is opposite to that of the first, or v2 = !32î m/s. Then m2 = (14 g)( 48=32) = 21 g.
Problem
23. A 680-g wood block is at rest on a frictionless table, when a 27-g bullet is fired into it. If the block with the embedded
bullet moves off at 19 m/s, what was the original speed of the bullet?
Solution
Since the table’s surface is frictionless, there are no horizontal external forces on the block and bullet, and the component of
their combined momentum, in the direction of the bullet’s initial horizontal velocity v 0 , is conserved. As in Example 10-6,
mv0 = (m + M )V, so v 0 = (680 + 27)(19 m/s)=27 = 498 m/s.
Problem
4
24. A plutonium-239 nucleus at rest decays into a uranium-235 nucleus by emitting an alpha particle ( He) with kinetic
energy of 5.15 MeV. What is the speed of the uranium nucleus?
Solution
Since the plutonium decays at rest, the conservation of momentum involves only one-dimensional motion, mPu239 v Pu2 3 9 =
0 = mU 2 3 5v U 2 3 5 + m! v ! . The momentum of the alpha particle is m! v ! =
2m! K ! , where K ! is its kinetic energy. (Since
2
K ! = 5.15 MeV ¿ m! c = 3.73 Gev , relativity can be ignored.) Thus,
vU 2 3 5 = !
2m" K "
mU 2 3 5
1/2
$ 2(4 u)(5.15 MeV)(1.60 # 10 !3 J/MeV) '
= !&
)
&%
(235 u)2 (1.66 # 10 !2 7 kg/u)
)(
= !2.68 # 105 m/s.
(The minus sign means that the velocity of the uranium nucleus is opposite to that of the ! particle; the speed is the
magnitude of the velocity.)
234 CHAPTER 10
Problem
29. An 11,000-kg freight car rests against a spring bumper at the end of a railroad track. The spring has constant
k = 3.2 ! 105 N/m . The car is hit by a second car of 9400-kg mass moving at 8.5 m/s, and the two cars couple together.
(a) What is the maximum compression of the spring? (b) What is the speed of the two cars together when they
rebound from the spring?
Solution
(a) The motion of the center of mass of the two freight cars (on an assumed horizontal, frictionless track) is determined by the
only horizontal external force, that of the spring. If this is conservative, the potential energy of the spring at maximum
2
compression equals the kinetic energy of the center of mass prior to contact with the spring; i.e., 12 kx2max = 12 MVcm
. Now
xcm = (m1 x1 + m2 x2 )=M , so Vcm = m2 v 2 =M , since the first car is initially at rest. Thus, Vcm = (9, 400 kg)(8.5 m/s)=
5
(11, 000 + 9, 400) kg = 3.92 m/s, and xmax = Vcm M=k = (3.92 m/s) (20, 400 kg)=(3.2 ! 10 N/m) = 98.9 cm.
(b) When the cars rebound, they are coupled together and both have the same velocity as their center of mass. Since the
2
spring is ideal (by assumption), its maximum potential energy, 12 k xmax , is transformed back into kinetic energy of the cars,
2
MVcm
, so the rebound speed equals the initial Vcm, or 3.92 m/s. (Reconsider this problem after reading Chapter 11,
especially Example 11-2.)
1
2
CHAPTER 10 235
Problem
32. A car of mass M is initially at rest on a frictionless surface. A jet of water carrying mass at the rate dmydt and moving
horizontally at speed v0 strikes the rear window of the car, which makes a 45° angle with the horizontal; the water
bounces off at the same relative speed with which it hit the window, as shown in Fig. 10-30. (a) Find an expression for
the initial acceleration of the car. (b) What is the maximum speed reached by the car?
FIGURE 10-30 Problem 32.
Solution
(a) The force exerted on the car by the water is the negative of the rate of change of momentum of the water,
Fc = !( d p=dt ) w . If v i and vf are the initial and final velocities of the water for the interval dt, then
(d p=dt ) w = (dm=dt )( v f ! v i ), where vi = v0 î . Since the water is deflected 90° without loss of relative speed, we should
evaluate the change in its velocity in the rest frame of the car, vf ! vi = (vf ! vc ) ! (vi ! vc ) = v"f ! v"i, where vc = vc î is
the velocity of the car, v!i = (v0 " vc )î = v r î , v !f = v r ˆj, and vr is the relative velocity. Thus, v f ! v i = v r ( ˆj ! î). If Fc is
the only force with a horizontal component acting on the car, then
ax =
dv c
1
1 ) # dm &
,
v # dm &
=
F !î =
"%
( v ( ˆj " î) . ! î = r %
(.
$ dt '
dt
M c
M +* $ dt ' r
M
-
Initially, the relative speed is just v0 (vr = v0 ! vc and vc = 0), so the initial acceleration is (v 0 =M )(dm=dt). (b) The water
ceases to exert a force on the car when vr = 0, or vc = v0 (which is the car’s final speed).
Problem 32 Solution.
236 CHAPTER 10
Problem
35. A biologist fires 20-g rubber bullets at a rhinoceros that is charging at 1.9 m/s. If the gun fires 5 bullets per second, with
a speed of 1600 m/s, and if the biologist fires for 13 s to stop the rhino in its tracks, what is the mass of the rhino?
Assume the bullets drop vertically after striking the rhino, and neglect forces exerted by the rhino’s feet.
Solution
The problem describes a one-dimensional collision between a rhino and 5 ! 13 = 65 rubber bullets, fired in the opposite
direction to the rhino’s charge (assumed to be horizontal). If the external horizontal forces are negligible, the horizontal
momentum of the rhino and bullets is conserved. Since the bullets are supposed to drop vertically and the rhino is stopped,
the final total horizontal momentum is zero, so initially 65(0.02 kg)(1600 m/s) − m (1.9 m/s) = 0, or the rhino’s mass is
m = 1.09 metric tons. (Note that 1.9 m/s corresponds to a 14:07 min/mi pace, so a healthy biologist could outrun this
rhino’s charge.)
Problem
38. Eighty percent of a rocket’s initial mass is fuel. If the fuel is exhausted at 2.5 km/s relative to the rocket, what is the
terminal speed of the rocket?
Solution
The mass of the rocket is 20% of the initial mass, so from Equation 10-11, the terminal speed (starting from rest) is
v f = v i + v ex ln( M i=M f ) = (2.5 km/s) ln 5 = 4.02 km/s.
(Any external gravitational influence has been ignored.)
Problem
42. A 150-g trick baseball is thrown at 60 km/h. It explodes in flight into two pieces, with a 38-g piece continuing straight
ahead at 85 km/h. How much energy do the pieces gain in the explosion?
Solution
The velocities of the baseball and the two pieces into which it explodes are all along the same line, so the conservation of
momentum (before and after) takes the form Pi = (m1 + m2 )v 0 = Pf = m1v 1 + m2 v 2 . Using the given values, we find
v 2 = [( 150 g)( 60 m=3.6 s) ! (38 g)( 85 m=3.6 s)] =(112 g) = 14.3 m/s. The difference in kinetic energy, Kf ! K i (which is
1
2
2
2
2
[(0.112 kg)(14.3 m/s) + (0.038 kg)(23.6 m/s) ] ! 12 (0.150 kg)(16.7 m/s) = 22.1 J !
20.8 J = 1.23 J. (Since K i = K cm , this is also K int,f .)
released in the explosion) is
Problem
50. A 60-kg astronaut floating in space simultaneously tosses away a 14 kg oxygen tank and a 5.8-kg camera. The tank
moves in the x direction at 1.6 m/s, and the astronaut recoils at 0.85 m/s in a direction 200° counterclockwise from the xaxis. Find the velocity of the camera.
Solution
In a coordinate system floating with the astronaut and her/his equipment before they are separated, the conservation of
momentum requires that m1v1 + m2 v2 + m3 v3 = 0 (1, 2, 3 representing astronaut, tank, and camera). Since the masses and
two velocities are given,
v 3x = !
(60 kg)(0.85 m/s) cos 200° + (14 kg)(1.6 m/s)
= 4.40 m/s,
5.8 kg
and v 3 y = !(60 kg)(0.85 m/s) sin200°=(5.8 kg) = 3.01 m/s, or v 3 = 5.33 m/s at 34.3° CCW from the x-axis.
CHAPTER 10 237
Problem
53. Firefighters spray water horizontally at the rate of 41 kg/s from a nozzle mounted on a 12,000-kg fire truck. The water
speed is 28 m/s relative to the truck. (a) Neglecting friction, what is the initial acceleration of the truck? (b) If the 12,000kg truck mass includes 2,400 kg of water, how fast will the truck be moving when the water is exhausted? Hint: Think
about rockets.
Solution
The fire truck behaves like a water rocket with thrust given by Equation 10-10b: Fthrust = ! v ex (dM =dt) = !(28 m/s) "
(!41 kg/s) = 1.15 kN. (a) If the thrust is the only significant horizontal force, the truck’s initial acceleration is ai =
Fthrust=M i = 1.15 kN=12,000 kg = 9.57 cm/s2 . (b) If the truck starts from rest, Equation 10-11, with vi = 0, gives vf =
v ex ln( M i=M f ) = (28 m/s) ln(12, 000=(12, 000 ! 2, 400)) = 6.25 m/s.
Problem
!
60. The triangle of Fig. 10-33 has a density that varies in proportion to y , where ! is a constant. Show that its center of
mass is located at y = (! + 2)h=(! + 3).
Solution
!
From symmetry, xcm = 0 . For mass elements, choose strips with area (w=h)y dy and density cy , as shown, where c is a
! +1
proportionality constant which cancels out in the final answer. Thus, dm = (cw=h) y
h
ycm =
!
!
0
y dm
h
0
=
dm
!
!
h
0
h
0
y " + 2 dy
y" +1dy
=
dy , and
h" + 3=(" + 3) # " + 2 &
=%
( h.
h" +2 =(" + 2) $ " + 3'
FIGURE 10-33 Problem 60 Solution.