CHAPTER 14
PROBLEM 14.1
An airline employee tosses two suitcases, of mass 15 kg and
20 kg, respectively, onto a 25-kg baggage carrier in rapid
succession. Knowing that the carrier is initially at rest and that
the employee imparts a 3-m/s horizontal velocity to the 15-kg
suitcase and a 2-m/s horizontal velocity to the 20-kg suitcase,
determine the final velocity of the baggage carrier if the first
suitcase tossed onto the carrier is (a) the 15-kg suitcase, (b) the
20-kg suitcase.
Solution
There are no horizontal external forces acting during the impacts. The
baggage carrier is to coast between impacts.
(a)
15-kg suitcase tossed on carrier first:
Let v1 be the common velocity of suitcase A and the carrier after
the first impact and v2 be the common velocity of the two suitcases
and the carrier of the second impact.
Conservation of momentum:
(15)(3) = ( 40)v1
v1 = 1.125 m/s
20-kg suitcase tossed next:
(20)(2) + ( 40)(1.125) = 60 v2
(b)
v 2 = 1.417 m/s
b
20-kg suitcase tossed on carrier first:
Let v1¢ be the common velocity of suitcase B and the carrier after
the first impact and v2¢ be the common velocity of all after the
second impact.
Conservation of momentum:
(20)(2) = ( 45)v1¢
v1¢ = 0.8889 m/s
15-kg suitcase tossed next:
(15)(3) + ( 45)(0.8889) = 60v2¢
v 2¢ = 1.417 m/s
b
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PROBLEM 14.2
An airline employee tosses two suitcases in rapid succession, with
a horizontal velocity of 2.4 m/s, onto a 25-kg baggage carrier, which
is initially at rest. (a) Knowing that the final velocity of the baggage
carrier is 1.2 m/s and that the first suitcase the employee tosses onto
the carrier has a mass of 15 kg, determine the mass of the other
suitcase. (b) What would be the final velocity of the carrier if the
employee reversed the order in which he tosses the suitcases?
Solution
There are no horizontal external forces acting during the impacts. The baggage carrier is to coast between
impacts.
(a)
Conservation of momentum:
2.4( mB + 15) = ( mB + 40)vfinal
(1)
Let vfinal = 1.2 m/s:
2.4 mB + 36 = 1.2mB + 48
(b)
mB = 10.00 kg b
Conservation of momentum equation is still valid.
Let mB = 10.00 kg in Eq. (1):
2.4(10 + 15) = (10 + 40)vfinal v final = 1.200 m/s
b
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4
PROBLEM 14.3
A 90-kg man and a 60 kg woman stand side by side at the
same end of a 150-kg boat ready to dive, each with a 5-m/s
velocity relative to the boat. Determine the velocity of the
boat after they have both dived, if (a) the woman dives first,
(b) the man dives first.
Solution
(a)
Woman dives first.
Conservation of momentum:
60(5 - v1 ) - (150 + 90)v1 = 0
v1 =
300
= 1 m/s
300
Man dives next. Conservation of momentum:
-240 v1 = -150 v2 + 90(5 - v2 )
v2 =
(b)
240v1 + 450
= 2.875 m/s
240
v 2 = 2.88 m/s
b
v 2¢ = 2.93 m/s
b
Man dives first.
Conservation of momentum:
90(5 - v1¢) - 210 v1¢ = 0
v1¢ =
450
= 1.5 m/s
300
Woman dives next. Conservation of momentum:
-210 v1¢ = -150 v2¢ + 60(5 - v2¢ )
v2¢ =
210v1¢ + 300
= 2.9286 m/s
210
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5
PROBLEM 14.4
A 90-kg man and a 60-kg woman stand at opposite ends of a
150-kg boat, ready to dive, each with a 5-m/s velocity relative
to the boat. Determine the velocity of the boat after they have
both dived, if (a) the woman dives first, (b) the man dives first.
Solution
(a)
Woman dives first.
Conservation of momentum:
-60(5 - v1 ) + 240v1 = 0
v1 =
300
= 1 m/s
300
Man dives next. Conservation of momentum:
240v1 = -150v2 + 90(5 - v2 )
v2 =
(b)
450 + 240 v1
= 0.875 m/s
240
v 2 = 0.875 m/s
b
Man dives first.
Conservation of momentum:
90(5 - v1¢) - 210 v1¢ = 0
v1¢ =
450
= 1.5 m/s
300
Woman dives next. Conservation of momentum:
-210 v1¢ = 150 v2¢ - 60(5 - v2¢ )
v2¢ = -210v1¢ = 210v2¢ - 300
v2¢ =
-210v1¢ + 300 -210 ¥ 1.5 + 300
=
= -0.0714 m/s
210
210
v 2¢ = -0.0714 m/s
b
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PROBLEM 14.5
A bullet is fired with a horizontal velocity of 450 m/s through
a 3-kg block A and becomes embedded in a 2.5-kg block B.
Knowing that blocks A and B start moving with velocities of
1.5 m/s and 2.7 m/s, respectively, determine (a) the weight of
the bullet, (b) its velocity as it travels from block A to block B.
Solution
The masses are m for the bullet and mA and mB for the blocks.
(a)
The bullet passes through block A and embeds in block B. Momentum is conserved.
Initial momentum:
mv0 + mA (0) + mB (0) = mv0
Final momentum:
mv B + mA v A + mB v B
Equating,
mv0 = mv B + mA v A + mB v B
m=
mA v A + mB v B (3)(1.5) + (2.5)(2.7)
=
= 0.0252 kg
v0 - v B
( 450 - 2.7)
m = 25.2 g b
(b)
The bullet passes through block A. Momentum is conserved.
Initial momentum:
mv0 + mA (0) = mv0
Final momentum:
mv1 + mA v A
Equating,
mv0 = mv1 + mA v A
v1 =
mv0 - mA v A
m v
3(1.5)
= v0 - A A = 450 = 271.07 m/s
m
0.02515
m
v1 = 271 m/s
b
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7
PROBLEM 14.6
A 45,000 kg boxcar A is moving in a railroad switchyard with a velocity of 10 km/h toward cars B and C, which
are both at rest with their brakes off at a short distance from each other. Car B is a 25000 kg flatcar supporting
a 30,000 kg container, and car C is a 40,000 kg boxcar. As the cars hit each other, they get automatically and
tightly coupled. Determine the velocity of car A immediately after each of the two couplings, assuming that
the container (a) does not slide on the flatcar, (b) slides after the first coupling but hits a stop before the second
coupling occurs, (c) slides and hits the stop only after the second coupling has occurred.
Solution
Each term of the conservation of momentum equation is mass times velocity. As long as the same units are used
in all terms, any unit may be used for mass and for velocity. We use kg for mass and km/h for velocity.
Conservation of momentum:
(a)
Container does not slide
450, 000 = 100, 000v1 = 140, 000v2
(b)
b
v 2 = 3.21 km/h
b
v1 = 6.43 km/h
b
v 2 = 3.21 km/h
b
v1 = 6.43 km/h
b
v 2 = 4.09 km/h
b
Container slides after 1st coupling, stops before 2nd
( 45, 000)(10) = (70, 000)v1 = (140, 000)v2 (c)
v1 = 4.50 km/h
Container slides and stops only after 2nd coupling
( 45000)(10) = (70, 000)v1 = (110, 000)v2
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8
PROBLEM 14.7
At an amusement park, there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg,
and 35 kg, respectively. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity
v B = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8.
Determine the final velocity of each car, after all impacts, assuming (a) cars A and C hit car B at the same time,
(b) car A hits car B before car C does.
Solution
Assume that each car with its rider may be treated as a particle. The masses are:
mA = 200 + 40 = 240 kg,
mB = 200 + 60 = 260 kg,
mC = 200 + 35 = 235 kg.
Assume velocities are positive to the right. The initial velocities are:
v A = 2 m/s v B = 0 vC = -1.5 m/s
Let v A¢ , v B¢ , and vC¢ be the final velocities.
(a)
Cars A and C hit B at the same time. Conservation of momentum for all three cars.
mA v A + mB v B + mC vC = mA v A¢ + mB v B¢ + mC vC¢
(240)(2) + 0 + (235)( -1.5) = 240 v A¢ + 260v B¢ + 235vC¢ (1)
Coefficient of restitution for cars A and B.
v B¢ - v A¢ = e( v A - v B ) = (0.8)(2 - 0) = 1.6
(2)
Coefficient of restitution for cars B and C.
vC¢ - v B¢ = e( v B - vC ) = (0.8)[0 - ( -1.5)] = 1.2
(3)
Solving Eqs. (1), (2), and (3) simultaneously,
v A¢ = -1.288 m/s v B¢ = 0.312 m/s vC¢ = 1.512 m/s
vA¢ = 1.288 m/s
b
vB¢ = 0.312 m/s
b
vC¢ = 1.512 m/s
b
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9
Problem 14.7 (continued)
(b)
Car A hits Car B before C does.
First impact. Car A hits car B. Let v A¢ and v B¢ be the velocities after this impact. Conservation of
momentum for cars A and B.
mA v A + mB v B = mA v A¢ + mB v B¢
(240)(2) + 0 = 240 v A¢ + 260 v B¢ (4)
Coefficient of restitution for cars A and B.
v B¢ - v A¢ = e( v A - v B ) = (0.8)(2 - 0) = 1.6
(5)
Solving Eqs. (4) and (5) simultaneously,
v A¢ = 0.128 m/s, v B¢ = 1.728 m/s
v ¢A = 0.128 m/s
v ¢B = 1.728 m/s
Second impact. Cars B and C hit. Let v B¢¢ and vC¢¢ be the velocities after this impact. Conservation of
momentum for cars B and C.
mB v B¢ + mC vC = mB v B¢¢ + mC vC¢¢
(260)(1.728) + (235)( -1.5) = 260v B¢¢ + 235vC¢¢ (6)
Coefficient of restitution for cars B and C.
vC¢¢ - v B¢¢ = e( v B¢ - vC) = (0.8)[1.728 - ( -1.5)] = 2.5824
(7)
Solving Eqs. (6) and (7) simultaneously,
v B¢¢ = -1.03047 m/s vC¢¢ = 1.55193 m/s
v ¢¢B = 1.03047 m/s
vC¢¢ = 1.55193 m/s
Third impact. Cars A and B hit again. Let v A¢¢¢ and v B¢¢¢ be the velocities after this impact. Conservation of
momentum for cars A and B.
mA v A¢ + mB v B¢¢ = mA v A¢¢¢+ mB v B¢¢¢
(240)(0.128) + (260)( -1.03047) = 2440vA¢¢¢+ 260v B¢¢¢ (8)
Coefficient of restitution for cars A and B.
v B¢¢¢- v A¢¢¢= e( v A¢ - v B¢¢ ) = (0.8)[0.128 - ( -1.03047)] = 0.926776
Solving Eqs. (8) and (9) simultaneously,
(9)
v A¢¢¢= -0.95633 m/s
v B¢¢¢ = -0.02955 m/s
v ¢¢¢
A = 0.95633 m/s
v ¢¢¢
B = 0.02955 m/s
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Problem 14.7 (continued)
There are no more impacts. The final velocities are:
v ¢¢¢
A = 0.956 m/s
b
v ¢¢¢
B = 0.0296 m/s
b
vC¢¢ = 1.552 m/s
b
We may check our results by considering conservation of momentum of all three cars over all three
impacts.
mA v A + mB v B + mC vC = (240)(2) + 0 + (235)( -1.5)
= 127.5 kg ◊ m/s
mA v A¢¢¢+ mB v B¢¢¢+ mC vC¢¢ = (240)( -0.95633) + (260)( -0.02955) + (2355)(1.55193)
= 127.50 kg ◊ m/s.
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11
PROBLEM 14.8
At an amusement park, there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg,
and 35 kg, respectively. Car A is moving to the right with a velocity vA = 2 m/s when it hits stationary car B.
The coefficient of restitution between each car is 0.8. Determine the velocity of car C so that after car B collides
with car C, the velocity of car B is zero.
Solution
Assume that each car with its rider may be treated as a particle. The masses are:
mA = 200 + 40 = 240 kg
mB = 200 + 60 = 260 kg
mC = 200 + 35 = 235 kg
Assume velocities are positive to the right. The initial velocities are:
v A = 2 m/s, v B = 0, vC = ?
First impact. Car A hits car B. Let v A¢ and v B¢ be the velocities after this impact. Conservation of momentum for
cars A and B.
mA v A + mB v B = mA v A¢ + mB v B¢
(240)(2) + 0 = 240 v A¢ + 260 v B¢ (1)
Coefficient of restitution for cars A and B.
v B¢ - v A¢ = e( v A - v B ) = (0.8)(2 - 0) = 1.6
Solving Eqs. (1) and (2) simultaneously,
(2)
v A¢ = 0.128 m/s
v B¢ = 1.728 m/s
v ¢A = 0.128 m/s
v ¢B = 1.728 m/s
Second impact. Cars B and C hit. Let v B¢¢ and vC¢¢ be the velocities after this impact. v B¢¢ = 0. Coefficient of
restitution for cars B and C.
vC¢¢ - v B¢¢ = e( v B¢ - vC ) = (0.8)(1.728 - vC )
vC¢¢ = 1.3824 - 0.8vC
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Problem 14.8 (continued)
Conservation of momentum for cars B and C.
mB v B¢ + mC vC = mB v B¢¢ + mC vC¢¢
(260)(1.728) + 235vC = (260)(0) + (235)(1.3824 - 0.8vC )
(235)(1.8)vC = (235)(1.3824) - (260)(1.728)
vC = -0.294 m/s
vC = 0.294 m/s
b
Note: There will be another impact between cars A and B.
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13
PROBLEM 14.9
A system consists of three particles A, B, and C. We know that
mA = 3 kg, mB = 4 kg, and mC = 5 kg, and that the velocities of
the particles expressed in m/s are, respectively, vA = -4i + 4j + 6k,
vB = -6i + 8j + 4k, and vC = 2i - 6 j - 4k. Determine the angular
momentum HO of the system about O.
Solution
Linear momentum of each particle ( kg ◊ m/s):
mA v A = -12i + 12 j + 18k
mB v B = -24i + 32 j + 16k
mC vC = 10i - 30 j - 20k
Position vectors, (meters):
rA = 1.2 i + 1.5k , rB = 0.9i + 1.2 j + 1.2k , rC = 2.4 j + 1.8k
Angular momentum about O ( kg ◊ m2 /s):
HO = rA ¥ mA vA + rB ¥ mB v B + rC ¥ mC vC
i
j k
i
j k
i
j
k
= 1.2 0 1.5 + 0.9 1.2 1.2 + 0 2.4 1.8
-12 12 18
-24 32 16
10 -30 -20
= ( -18i - 39.6 j + 14.4k) + ( -19.2i - 43.2 j + 57.6k ) + (6i + 18 j - 24k )
= -31.2i - 64.8 j + 48.0k
HO = - (31.2 kg ◊ m2 /s)i - (64.8 kg ◊ m2 /s) j + ( 48.0 kg ◊ m2/s)k b
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14
PROBLEM 14.10
For the system of particles of Problem 14.9, determine (a) the
position vector r of the mass center G of the system, (b) the linear
momentum mv of the system, (c) the angular momentum HG of the
system about G. Also verify that the answers to this problem and to
Problem 14.9 satisfy the equation given in Problem 14.27.
PROBLEM 14.9 A system consists of three particles A, B, and C. We
know that mA = 3 kg, mB = 4 kg, and mC = 5 kg and that the velocities
of the particles expressed in m/s are, respectively, vA = -4i + 4j + 6k,
v B = -6i + 8 j + 4k , and vC = 2i - 6 j - 4k. Determine the angular
momentum HO of the system about O.
Solution
Position vectors, (meters):
rA = 1.2i + 1.5k , rB = 0.9i + 1.2 j + 1.2k , rC = 2.4 j + 1.8k
(a)
Mass center:
( mA + mB + mC ) r = mArA + mB rB + mC rC
12 r = (3)(1.2i + 1.5k ) + ( 4)(0.9i + 1.2 j + 1.2k ) + (5)(2.4 j + 1.8k )
r = 0.6i + 1.4 j + 1.525k
r = (0.600 m)i + (1.400 m) j + (1.525 m)k b
Linear momentum of each particle ( kg ◊ m/s):
mA v A = -12i + 12 j + 18k
mB v B = -24i + 32 j + 16k
mC vC = 10i - 30 j - 20k
(b)
Linear momentum of the system ( kg ◊ m/s):
mv = mA v A + mB v B + mC vC = -26i + 14 j + 14k
mv = -(26.0 kg ◊ m/s)i + (14.00 kg ◊ m/s) j + (14.00 kg ◊ m/s)k b
Position vectors relative to the mass center (meters).
rA¢ = rA - r = 0.6i - 1.4 j - 0.025k
rB¢ = rB - r = 0.3i - 0.2 j - 0.325k
rC¢ = rC - r = -0.6i + 1.0 j + 0.275k
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15
Problem 14.10 (continued)
(c)
Angular momentum about G ( kg ◊ m2 /s):
HG = rA¢ ¥ mA v A + rB¢ ¥ mB v B + rC¢ ¥ mC vC
i
j
k
i
j
k
i
j
k
= 0.6 -1.4 -0.025 + 0.3 -0.2 -0.325 + - 0.6 1.0 0.275
-12 12
18
-24 32
16
10 -30 -20
= ( -24.9i - 10.5 j - 9.6k) + (7.2i + 3.0 j + 4.8k ) + ( -11.75i - 9.25 j + 8.0k )
= -29.45i - 16.75 j + 3.2k
HG = - (29 .5 kg ◊ m2 /s)i - (16.75 kg ◊ m2 /s) j + (3.20 kg ◊ m2 /s)k b
i
j
k
r ¥ mv = 0.6 1.4 1.525 = -1.75i - 48.05 j + 44.8k
-26 14
14
HG + r ¥ mv = - (31.2 kg ◊ m2 /s)i - (64.8 kg ◊ m 2 /s) j + ( 48.0 kg ◊ m 2 /s)k
Angular momentum about O ( kg ◊ m2/s):
HO = rA ¥ mA v A + rB ¥ mB v B + rC ¥ mC vC
i
j k
i
j k
i
j
k
= 1.2 0 1.5 + 0.9 1.2 1.2 + 0 2.4 1.8
-12 12 18
-24 32 16
10 -30 -20
= ( -18i - 39.6 j + 14.4k) + ( -19.2i - 43.2 j + 57.6k ) + (6i + 18 j - 24k )
= -(31.2 kg ◊ m2 /s)i - (64.8 kg ◊ m2 /s) j + ( 48.0 kg ◊ m2 /s)k
Note that
H O = HG + r ¥ m v .
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16
PROBLEM 14.11
A system consists of three particles A, B, and C. We know that
mA = 3 kg, mB = 4 kg, and mC = 5 kg and that the velocities of the
particles expressed in m/s are, respectively, vA = -4i + 4j + 6k,
vB = vxi + vyj + 4k, and vC = 2i - 6j - 4k. Determine (a) the
components vx and vy of the velocity of particle B for which the
angular momentum HO of the system about O is parallel to the z axis,
(b) the corresponding value of HO.
Solution
HO = rA ¥ mA vA + rB ¥ mB v B + rC ¥ mC vC
i
j
k
i
j
k
i
j
k
= (3 kg) 1.2 m
0
1.5 m + 4 0.9 1.2 1.2 + 5 0 2.4 1.8
-4 m/s 4 m/s 6 m/s
vx v y 4
2 -6 -4
= -18i - 39.6 j + 14.4k + (19.2 - 4.8v y )i + ( 4.8v x - 14.4) j + (3.6v y - 4.8v x )k + 6i + 18 j - 24k
HO = (7.2 - 4.8vy )i + ( -36 + 4.8v x ) j + ( -9.6 + 3.6v y - 4.8v x )k
(a)
For HO to be parallel to the z axis:
H x = 7.2 - 4.8v y = 0
H y = -36 + 4.8v x = 0
v x = 7.50 m/s, v y = 1.500 m/s b
(b)
HO = H z k = ( -9.6 + 3.6 ¥ 1.500 - 4.8 ¥ 7.50)k
HO = -( 40.2 kg ◊ m2 /s)k b
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17
PROBLEM 14.12
For the system of particles of Problem 14.11, determine (a) the
components vx and vy of the velocity of particle B for which
the angular momentum HO of the system about O is parallel to the
y-axis, (b) the corresponding value of HO.
PROBLEM 14.11 A system consists of three particles A, B,
and C. We know that mA = 3 kg, mB = 4 kg, and mC = 5 kg
and that the velocities of the particles expressed in m/s
are, respectively, v A = - 4i + 4j + 6k, v B = v x i + v y j + 4k, and
vC = - 2i - 6j - 4k. Determine (a) the components vx and vy of the
velocity of particle B for which the angular momentum HO of
the system about O is parallel to the z axis, (b) the corresponding
value of HO.
Solution
HO = rA ¥ mA vA + rB ¥ mB vB + rC ¥ mC vC
i
j
k
i
j
k
i
j
k
= (3 kg) 1.2 m
0
1.5 m + 4 0.9 1.2 1.2 + 5 0 2.4 1.8
-4 m/s 4 m/s 6 m/s
vx v y 4
2 -6 -4
= -18i - 39.6 j + 14.4k + (19.2 - 4.8vy )i + ( 4.8vx - 14.4) j + (3.6v y - 4.8vx )k + 6i + 18 j - 24k
HO = (7.2 - 4.8v y )i + ( -36 + 4.8v x ) j + ( -9.6 + 3.6v y - 4.8v x )k
(a)
For HO to be parallel to the y axis:
H x = 7.2 - 4.8 v y = 0
H z = -9.6 + 3.6 v y - 4.8v x = 0
v y = 1.500 m/s
-9.6 + 3.6(1.500) - 4.8v x = 0
v x = -0.875 m/s
v x = -0.875 m/s, v y = 1.500 m/s b
(b)
HO = H y j = [-36 + 4.8 ¥ ( -0.875)]j
HO = -( 40.2 kg ◊ m2 /s) j b
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18
PROBLEM 14.13
A system consists of three particles A, B, and C. We know that mA = 2.5 kg,
mB = 2 kg, and mC = 1.5 kg, and that the velocities of the particles expressed
in m/s are, respectively, v A = i + 1.5j - k, v B = v x i + v y j + v z k, and
vC = -1.5i - j + 0.5k. Determine (a) the components vx and vy of the
velocity of particle B for which the angular momentum HO of the system
about O is parallel to the x axis, (b) the value of HO.
Solution
i
HO = Sri ¥ mvi = Smi xi
( vi ) x
i
j
k
i
= 2.5 0 2.5 2 + 2 2
1 1.5 -1
vx
j
yi
( vi ) y
k
zi
( vi ) z
j k
i
j
k
2 1.5 + 1.5 4
3
0
v y vz
-1.5 -1 0.5
= [2.5( -2.5 - 3) + 2(2v2 - 1.5v y ) + 1.5(1.5 - 0)]i
+[2.5(2 - 0) + 2(1.5v x - 2v z + 1.5(0.2)]j
+[2.5(0 - 2.5) + 2(2v y - 2v x ) + 1.5( -4 + 4.5)]k
HO = [( 4v z - 3v y - 11.5)i + (3v x - 4v z + 2) j + ( -4v x + 4v y - 5.5)k ]
(a)
For HO to be parallel to the x axis, we must have H y = H z = 0:
H z = 0 : - 4 v x + 4 v y - 5.5 = 0
H y = 0 : (3v x - 4 v z + 2) = 0 (b)
(1)
vx =
vz =
1
( 4v y - 5.5) m/s b
4
1
(3v y - 2.125) m/s b
4
Substitute into Eq. (1):
1 Ï3
1
¸
v z = (3v x + 2) = Ì ( 4v y - 5.5) + 2 ˝
4
4 Ó4
˛
HO = ( 4v z - 3v y - 11.5)i = (3v y - 2.125 - 3v y - 11.5)i = 13.625 i N ◊ m ◊ s
HO = ( -13.63 N ◊ m ◊ s)i b
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19
PROBLEM 14.14
For the system of particles of Problem 14.13, determine (a) the components
vx and vz of the velocity of particle B for which the angular momentum HO
of the system about O is parallel to the z axis, (b) the value of HO.
PROBLEM 14.13 A system consists of three particles A, B, and C. We
know that mA = 2.5 kg, mB = 2kg, and mC = 1.5 kg and that the velocities of
the particles expressed in m/s are, respectively, vA = i + 1.5 j - k , vB = vxi +
vyj + vzk, and vC = -1.5i - j + 0.5k. Determine (a) the components vx and
vz of the velocity of particle B for which the angular momentum HO of the
system about O is parallel to the x axis, (b) the value of HO.
Solution
i
HO = Sri ¥ mvi = Smi xi
( vi ) x
i
j
k
i
= 2.5 0 2.5 2 + 2 2
1 1.5 -1
vx
j
yi
( vi ) y
k
zi
( vi ) z
j k
i
j
k
2 1.5 + 1.5 4
3
0
v y vz
-1.5 -1 0.5
= [2.5( -2.5 - 3) + 2(2v2 - 1.5v y ) + 1.5(1.5 - 0)]i
+[2.5(2 - 0) + 2(1.5v x - 2v z + 1.5(0.2)]j
+[2.5(0 - 2.5) + 2(2v y - 2v x ) + 1.5( -4 + 4.5)]k
HO = [( 4v z - 3v y - 11.5)i + (3v x - 4v z + 2) j + ( -4v x + 4v y - 5.5)k ]
(a)
(1)
For HO to be parallel to the z axis, we must have H x = H y = 0:
H x = 0 : 4v z - 3v y - 11.5 = 0 vz =
1
(3v y + 11.5) m/s b
4
H y = 0 : 3v x - 4 v z + 2 = 0
3v x - (3v y + 11.5) = 0
1
v x = (3v y + 11.5)
3
(b)
1
v x = (3v y + 11.5) m/s b
3
Substituting into Eq. (1):
HO =
28 ˆ
-4
Ê
(3v y + 11.5) + (3v y + 11.5) - 5.5 Á - v y - ˜ k Ë
3
3¯
28 ˆ
Ê
HO = - Á v y + ˜ N ◊ m ◊ s b
Ë
3¯
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20
PROBLEM 14.15
A 450-kg space vehicle traveling with a velocity v0 = (360 m/s)i passes through the origin O at t = 0. Explosive
charges then separate the vehicle into three Parts A, B, and C, weighing, respectively, 225 kg, 150 kg, and 75 kg.
Knowing that at t = 4 s, the positions of Parts A and B are observed to be A (1200 m, –300 m, –600 m) and
B (1300 m, 3500 m, 800 m), determine the corresponding position of Part C. Neglect the effect of gravity.
Solution
Motion of mass center:
Since there is no external force,
r = v 0 t = (360 m/s)( 4 s) = (1440 m)i
Equation (14.12):
m r = Smi ri :
( 450)(1440i ) = 225(1200i - 300 j - 600k )
+ 150(1300i + 350 j + 800k )
+ 75rC
75rC = (648000 - 270000 - 195000)i
+ (67500 - 52500) j
+ (135000 - 120000)k
75rC = (183000i + 15000 j + 15000)k
rC = (2440 m)i + (200 m) j + (200 m)k b
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21
PROBLEM 14.16
A 15-kg projectile is passing through the origin O with a velocity v0 = (40 m/s)i when it explodes into two
fragments A and B, of weight 6 kg and 9 kg, respectively. Knowing that 3 s later the position of fragment A is
(100 m, 10 m,–20 m), determine the position of fragment B at the same instant. Assume ay = -g = - 9.81 m/s2 and
neglect air resistance.
Solution
Motion of mass center:
It moves as if projectile had not exploded.
r = v0 ti -
1 2
gt j
2
1
= ( 40 m/s)(3 s)i - (9.81 m/s2 )(3 s)2 j
2
= (120 m)i - ( 44.145 m)j
Equation (14.12):
m r = Smi ri :
m r = mArA + mB rB
15(120i - 44.145 j) = 6(100i + 10 j - 20k ) + 9rB
9rB = (15 ¥ 120 - 6 ¥ 100)i - (15 ¥ 44.145 + 6 ¥ 10) j + (6 ¥ 20)k
= 1200i - 722.175 j + 120k
rB = (133.3 m)i - (80.2 m) j + (13.33 m)k b
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22
PROBLEM 14.17
A small airplane of mass 1500 kg and a helicopter of mass
3000 kg flying at an altitude of 1200 m are observed to collide
directly above a tower located at O in a wooded area. Four
minutes earlier, the helicopter had been sighted 8.4 km due west
of the tower and the airplane 16 km west and 12 km north of the
tower. As a result of the collision, the helicopter was split into
two pieces, H1 and H2, of mass m1 = 1000 kg and m2 = 2000 kg,
respectively; the airplane remained in one piece as it fell to the
ground. Knowing that the two fragments of the helicopter were
located at Points H1 (500 m, -100 m) and H2 (600 m, - 500 m),
respectively, and assuming that all pieces hit the ground at the
same time, determine the coordinates of the Point A where the
wreckage of the airplane will be found.
Solution
Motion of mass center G.
vH =
At collision:
(8400 m)
i = (35.00 m/s)i
4(60 s)
(16,000 m)i - (12,000 m) j
4(60 s)
= (66.67 m/s)i - (50 m/s) j
vA =
Velocity of mass center:
( mH + mA ) v = mH v H + mA v A
4500 v = 3000(35.00i ) + 1500(66.67i - 50 j)
v = ( 45.556 m/s)i - (16.667 m/s) j
Vertical motion of G:
h=
1 2
gt
2
t=
2(1200 m)
2h
=
= 15.641 s
g
9.81 m/s2
Position of G at time of ground impact:
r = vt = ( 45.556i - 16.667 j)(15.641)
r = (712.55 m)i - (260.69 m) j
(1)
From Eq. (14.12):
( mH + mA ) r = mH1 rH1 + mH 2 rH 2 + mArA
(2)
4500(712.55i - 260.69 j) = 1000(500i - 100 j) + 2000(600i - 500 j) + 1500rA
1.5rA = ( 4.5 ¥ 712.55 - 500 - 2 ¥ 600)i + ( -4.5 ¥ 260.69 + 100 + 2 ¥ 500) j
rA = (1004 m)i - ( 48.7 m) j b
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23
PROBLEM 14.18
In Problem 14.17, knowing that the wreckage of the small airplane
was found at Point A (1200 m, 80 m) and the 1000-kg fragment of
the helicopter at Point H1 (400 m, –200 m), and assuming that all
pieces hit the ground at the same time, determine the coordinates
of the Point H2 where the other fragment of the helicopter will be
found.
Solution
Motion of mass center G.
vH =
At collision:
(8400 m)
i = (35.00 m/s)i
4(60 s)
(16,000 m)i - (12,000 m) j
4(60 s)
= (66.67 m/s)i - (50 m/s) j
vA =
Velocity of mass center:
( mH + mA ) v = mH v H + mA v A
4500 v = 3000(35.00i ) + 1500(66.67i - 50 j)
v = ( 45.556 m/s)i - (16.667 m/s) j
Vertical motion of G:
h=
1 2
gt
2
t=
2(1200 m)
2h
=
= 15.641 s
g
9.81 m/s2
Position of G at time of ground impact:
r = vt = ( 45.556i - 16.667 j)(15.641)
r = (712.55 m)i - (260.69 m) j
(1)
From Eq. (14.12):
( mH + mA ) r = mH1 rH1 + mH 2 rH 2 + mArA
(2)
Substituting data:
4500(712.55i - 260.69 j) = 1000( 400i - 200 j) + 2000rH 2 + 1500(1200i + 80 j)
2rH 2 = ( 4.5 ¥ 712.55 - 400 - 1.5 ¥ 1200)i
+ ( -4.5 ¥ 260.69 + 200 - 1.5 ¥ 80) j
rH 2 = (503 m)i - (547 m) j b
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24
PROBLEM 14.19
Car A was traveling east at high speed when it
collided at Point O with car B, which was traveling
north at 72 km/h. Car C, which was traveling west
at 90 km/h, was 10 m east and 3 m north of Point O
at the time of the collision. Because the pavement
was wet, the driver of car C could not prevent his car
from sliding into the other two cars, and the three
cars, stuck together, kept sliding until they hit the
utility pole P. Knowing that the masses of cars A,
B, and C are, respectively, 1500 kg, 1300 kg, and
1200 kg, and neglecting the forces exerted on the cars
by the wet pavement, solve the problems indicated.
Knowing that the coordinates of the utility pole
are xP = 18 m and yP = 13.9 m, determine (a) the
time elapsed from the first collision to the stop at P,
(b) the speed of car A.
Solution
Let t be the time elapsed since the first collision. No external forces in the xy plane act on the system consisting
of cars A, B, and C during the impacts with one another. The mass center of the system moves at the velocity
it had before the collision.
Initial mass center r0 :
( mA + mB + mC )( x0 i + y0 j) = mC ( xC i + yC j)
x0 = 0.3xC = (0.3)(10) = 3 m,
y0 = 0.3 yC = (0.3)(3) = 0.9 m
Given velocities:
vA = v Ai,
vB = (20 m/s) j,
vC = -(25 m/s)i
Velocity of mass center:
( mA + mB + mC )v = mA vA + mB vB + mC vC
v = 0.375vA + 0.325vB + 0.3vC
Since the collided cars hit the pole at
rP = xP i + yP j
x P i + yP j = x0 i + y0 j + vt
Resolve into components.
x: xP = x0 + 0.375v At P - 0.3vC t P (1)
y: yP = y0 + 0.325v B t P (2)
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25
Problem 14.19 (continued)
Data:
xP = 18 m, yP = 13.9 m
( a) From (2),
13.9 = 0.9 + (0.325)(20)t P (b) From (1),
t P = 2.00 s b
18 = 3 + (0.375)vA (2.00) - (0.3)(25)(2.00)
vA = 40 m/s
vA = 144.0 km/h b
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26
PROBLEM 14.20
Car A was traveling east at high speed when it collided
at Point O with car B, which was traveling north at
72 km/h. Car C, which was traveling west at 90 km/h,
was 10 m east and 3 m north of Point O at the time of
the collision. Because the pavement was wet, the driver
of car C could not prevent his car from sliding into the
other two cars, and the three cars, stuck together, kept
sliding until they hit the utility pole P. Knowing that the
masses of cars A, B, and C are, respectively, 1500 kg,
1300 kg, and 1200 kg, and neglecting the forces exerted
on the cars by the wet pavement, solve the problems
indicated.
Knowing that the speed of car A was 129.6 km/h and
that the time elapsed from the first collision to the stop
at P was 2.4 s, determine the coordinates of the utility
pole P.
Solution
Let t be the time elapsed since the first collision. No external forces in the xy plane act on the system consisting
of cars A, B, and C during the impacts with one another. The mass center of the system moves at the velocity
it had before the collision.
Initial mass center r0 :
( mA + mB + mC )( x0 i + y0 j) = mC ( xC i + yC j)
x0 = 0.3xC = (0.3)(10) = 3 m,
y0 = 0.3 yC = (0.3)(3) = 0.9 m
Given velocities:
vA = vAi, v B = (20 m/s) j,
vC = -(25 m/s)i
Velocity of mass center:
( mA + mB + mC ) v = mA v A + mB v B + mC vC
v = 0.375vA + 0.325vB + 0.3vC
Since the collided cars hit the pole at
rP = xP i + yP j
x P i + yP j = x0 i + y0 j + vt
Resolve into components.
x: xP = x0 + 0.375v At P - 0.3vC t P (1)
y: yP = y0 + 0.325v B t P (2)
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27
Problem 14.20 (continued)
Data:
vA = 129.6 km/h = 36 m/s,
tP = 2.4 s
From (1),
xP = 3 + (0.375)(36)(2.4) - (0.3)(25)(2.4) = 17.4 m
From (2),
yP = 0.9 + (0.325)(20)(2.4) = 16.5 m
xP = 17.40 m b
yP = 16.50 m b
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28
PROBLEM 14.21
In a game of pool, ball A is moving with a velocity v0 when it strikes
balls B and C, which are at rest and aligned as shown. Knowing that
after the collision the three balls move in the directions indicated,
and that v0 = 4 m/s and vC = 2 m/s, determine the magnitude of the
velocity of (a) ball A, (b) ball B.
Solution
Conservation of linear momentum. In x direction:
m( 4 m/s) cos 45∞ = mv A sin 4.3∞ + mvB sin 37.4∞
+ m(2 m/s) cos 30∞
0.07498vA + 0.60738vB = 1.09638
(1)
In y direction:
Solving:
m( 4 m/s)sin 45∞ = mv A cos 4.3∞ - mvB cos 37.4∞
+ m(2 m/s)sin 30∞
0.99719vA - 0.79441vB = 1.82843
(2)
vA = 2.98 m/s b
v B = 1.437 m/s b
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29
PROBLEM 14.22
In a game of pool, ball A is moving with a velocity v0 when it strikes
balls B and C, which are at rest and aligned as shown. Knowing that
after the collision the three balls move in the directions indicated
and that v0 = 4 m/s and vC = 2 m/s, determine the magnitude of the
velocity of (a) ball A, (b) ball B.
Solution
Conservation of linear momentum. In x direction:
m( 4 m/s) cos 30∞ = mv A sin 7.4∞ + mvB sin 49.3∞
+ m(2 m/s) cos 45∞
0.12880vA + 0.75813vB = 2.04989
(1)
In y direction:
m( 4 m/s)sin 30∞ = mvA cos 7.4∞ - mvB cos 49.3∞
+ m(2 m/s)sin 45∞
Solving: 0.99167vA - 0.65210vB = 0.58579
(2)
vA = 2.13 m/s b
v B = 2.34 m/s b
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30
PROBLEM 14.23
An expert archer demonstrates his ability by hitting tennis balls thrown by an assistant. A 58-g tennis ball has
a velocity of (10 m/s) i – (2 m/s) j and is 10 m above the ground when it is hit by a 40-g arrow traveling with a
velocity of (50 m/s) j + (70 m/s) k where j is directed upwards. Determine the position P where the ball and
arrow will hit the ground relative to Point O located directly under the point of impact.
Solution
Assume that the ball and arrow move together after the hit.
Conservation of momentum of ball and arrow during the hit.
mA vA + mB v B = ( mA + mB ) v
(0.040)(50 j + 70k ) + (0.058)(10i - 2 j) = (0.040 + 0.058) v
v = (5.9184 m/s)i + (19.2245 m/s) j + (28.5714 m/s)k
After the hit, the ball and arrow move as a projectile.
1 2
gt
2
1
y = 10 + 19.2245 t - (9.8) t 2
2
y = y0 + ( v y )0 t -
Vertical motion.
y = 0 at ground.
2
-4.9 t + 19.2245 t + 10 = 0
Solve for t.
After rejecting the negative root,
t = 4.3884 s
Horizontal motion.
x = x0 + ( v x )0 t
z = z0 + ( v2 )0 t
x = 0 + (5.9184)( 4.3884)
= 26.0 m
z = 0 + (28.5714)( 4.3884)
= 125.4 m
rp = (26.0 m)i + (125.4 m)k b
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PROBLEM 14.24
In a scattering experiment, an alpha particle A is projected with
the velocity u 0 = -(600 m/s)i + (750 m/s) j - (800 m/s)k into
a stream of oxygen nuclei moving with the common velocity
v 0 = (600 m/s) j . After colliding successively with nuclei
B and C, particle A is observed to move along the path defined
by the PointsA1 (280, 240, 120) and A2(360, 320, 160), while
nuclei B and C are observed to move along paths defined,
respectively, by B1(147, 220, 130), B2(114, 290, 120), and by
C1(240, 232, 90) and C2 (240, 280, 75). All paths are along
straight lines and all coordinates are expressed in millimeters.
Knowing that the mass of an oxygen nucleus is four times that
of an alpha particle, determine the speed of each of the three
particles after the collisions.
Solution
Position vectors (mm):
Unit vectors:
uuuuur
A1 A2 = 80i + 80 j + 40k
uuuuur
B1 B2 = -33i + 70 j - 10k
uuuuur
C1C2 = 48 j - 15k
( A1 A2 ) = 120
( B1 B2 ) = 78.032
(C1C2 ) = 50.289
Along A1 A2 ,
l A = 0.66667i + 0.66667 j + 0.33333k
Along B1 B2 ,
l B = -0.42290i + 0.89707 j - 0.12815k
Along C1C2 ,
lC = 0.95448 j - 0.29828k
Velocity vectors after the collisions:
vA = vA l A
vB = vB l B
vC = vC lC
Conservation of momentum:
mu 0 + 4mv 0 + 4mv 0 = mvA + 4mvB + 4mvC
Divide by m and substitute data.
( -600i + 750 j - 800k ) + 2400 j + 2400 j = vA l A + 4vB l B + 4vC lC
Resolving into components,
i: - 600 = 0.66667vA - 1.69160v B
j: 5550 = 0.66667vA + 3.58828v B + 3.81792vC
k : - 800 = 0.33333vA - 0.51260v B - 1.19312vC
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32
Problem 14.24 (continued)
Solving the three equations simultaneously,
vA = 919.26 m/s
vB = 716.98 m/s
vC = 619.30 m/s
vA = 919 m/s b
v B = 717 m/s b
vC = 619 m/s b
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33
PROBLEM 14.25
An 6-kg shell moving with a velocity
v 0 =(12 m/s)i - (10 m/s)j - (400 m/s)k
explodes at Point D into three fragments
A, B, and C weighing, respectively, 2.5-kg,
2-kg, and 1.5-kg. Knowing that the fragments
hit the vertical wall at the points indicated,
determine the speed of each fragment
immediately after the explosion.
Solution
rD = 6 k
Position vectors (m):
Unit vectors:
rA = -2.5i
rA/D = -2.5i - 6k
rA/D = 6.5
rB = 6i + 3j
rB/D = 6i + 3j - 6k
rB/D = 9
rC = -4.5 j
rC/D = -4.5 j - 6k
rC/D = 7.5
Along rA/D ,
Along rB/D ,
Along rC/D ,
1
( -2.5i - 6k )
6.5
1
l B = (6i + 3j - 6k )
9
1
lC =
( -4.5 j - 6k )
7.5
lA =
Assume that elevation changes due to gravity may be neglected. Then, the velocity vectors after the explosion
have the directions of the unit vectors.
vA = vA l A
vB = v B l B
vC = vC lC
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34
Problem 14.25 (continued)
Conservation of momentum:
mv 0 = mA vA + mB vB + mC vC
Êv ˆ
Êv ˆ
6(12i - 10 j - 400k ) = 2.5 Á A ˜ ( -2.5i - 6k ) + 2 Á B ˜ (6i + 3j - 6k )
Ë 6.5 ¯
Ë 9¯
Êv ˆ
+ 1.5 Á C ˜ ( -4.5 j - 6k )
Ë 7.5 ¯
Resolve into components.
25
4
v A + vB
26
3
2
9
-60 = v B - vC
3
10
4
6
30
-2400 = - v A - v B - vC
13
3
5
72 = -
Solving,
vA = 563 m/s b
vB = 460 m/s b
vC = 407 m/s b
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35
PROBLEM 14.26
An 6-kg shell moving with a velocity
v 0 = (12 m/s)i - (10 m/s)j - (400 m/s)k
explodes at Point D into three fragments A, B,
and C weighing, respectively, 2.5-kg, 2-kg,
and 1.5-kg. Knowing that the fragments hit the
vertical wall at the points indicated, determine
the speed of each fragment immediately after
the explosion.
Solution
rD = 6 k
Position vectors (m):
Unit vectors:
rA = -2.5i
rA/D = -2.5i - 6k
rA/D = 6.5
rB = 6i + 3j
rB/D = 6i + 3j - 6k
rB/D = 9
rC = -4.5 j
rC/D = -4.5 j - 6k
rC/D = 7.5
Along rA/D ,
Along rB/D ,
Along rC/D ,
1
( -2.5i - 6k )
6.5
1
l B = (6i + 3j - 6k )
9
1
lC =
( -4.5 j - 6k )
7.5
lA =
Assume that elevation changes due to gravity may be neglected. Then, the velocity vectors after the explosion
have the directions of the unit vectors.
vA = vA l A
vB = v B l B
vC = vC lC
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36
Problem 14.26 (continued)
Conservation of momentum:
mv 0 = mA vA + mB vB + mC vC
Êv ˆ
Êv ˆ
6(12i - 10 j - 400k ) = 2 Á A ˜ ( -2.5i - 6k ) + 1.5 Á B ˜ (6i + 3j - 6k )
Ë 6.5 ¯
Ë 9¯
Êv ˆ
+ 2.5 Á C ˜ ( -4.5 j - 6k )
Ë 7.5 ¯
Resolve into components.
10
v A + vB
13
1
3
-60 = v B - vC
2
2
24
-2400 = - v A - v B - 2vC
13
72 = -
Solving,
vA = 703 m/s b
vB = 613 m/s b
vC = 244 m/s b
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37
PROBLEM 14.27
Derive the relation
H O = r ¥ m v + HG
between the angular momenta HO and HG defined in Eqs. (14.7) and (14.24), respectively. The vectors r and
v define, respectively, the position and velocity of the mass center G of the system of particles relative to the
newtonian frame of reference Oxyz, and m represents the total mass of the system.
Solution
n
From Eq. (14.7),
HO = Â (ri ¥ mi vi )
i =1
n
= Â ÈÎ( r + ri¢) ¥ mi vi ˘˚
i =1
n
n
i =1
i =1
= r ¥ Â ( mi vi ) + Â ( ri¢¥ mi vi )
= r ¥ m v + HG
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38
PROBLEM 14.28
Show that Eq. (14.23) may be derived directly from Eq. (14.11) by substituting for HO the expression given in
Problem 14.27.
Solution
n
HO = Â (ri ¥ mi vi )
From Eq. (14.7),
i =1
n
= Â ÈÎ( r + ri¢) ¥ mi vi ˘˚
i =1
n
n
i =1
i =1
= r ¥ Â ( mi vi ) + Â ( ri¢¥ mi vi )
= r ¥ m v + HG
Differentiating,
& O = r& ¥ mv + r ¥ mv& + H
&G
H
Using Eq. (14.11),
&G
SMO = r& ¥ mv + r ¥ mv& + H
&G
= v ¥ mv + r ¥ m a + H
Ê n ˆ
& = 0 + r ¥ Á Â Fi ˜ + H
G
Ë i =1 ¯
(1)
n
n
Ê n ˆ
M
M
r
=
+
¥
 O  G  ÁË Â Fi ˜¯
i =1
i =1
i =1
i =1
n
But
(
)
Subtracting r ¥ S in=1 Fi from each side of Eq. (1) gives
&G
SMG = H
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39
PROBLEM 14.29
Consider the frame of reference Ax ¢y ¢z ¢ in translation with respect to the
newtonian frame of reference Oxyz. We define the angular momentum H¢A of a
system of n particles about A as the sum
n
HA¢ = Â ri¢ ¥ mi vi¢
(1)
i =1
of the moments about A of the momenta mi vi¢ of the particles in their motion
relative to the frame Ax ¢y ¢z ¢. Denoting by H A the sum
n
HA = Â ri¢ ¥ mi vi
(2)
i =1
of the moments about A of the momenta mi vi of the particles in their
motion relative to the newtonian frame Oxyz, show that HA = HA¢ at a given
instant if, and only if, one of the following conditions is satisfied at that
instant: (a) A has zero velocity with respect to the frame Oxyz, (b) A coincides
with the mass center G of the system, (c) the velocity vA relative to Oxyz is
directed along the line AG.
Solution
vi = vA + vi¢
n
HA = Â ri¢ ¥ mi vi
i =1
n
= Â ri¢ ¥ mi ( vA + vi¢ )
i =1
n
n
= Â (ri¢ ¥ mi vA ) + Â ri¢ ¥ mi vi
i =1
n
i =1
= Â ( mi ri¢) ¥ vA + HA¢
i =1
n
= Â mi (ri - rA ) ¥ vA + HA¢
i =1
= m( r - rA ) ¥ vA + HA¢
HA = HA¢ if, and only if,
m(r - rA ) ¥ vA = 0
This condition is satisfied if
or
or
( a)
( b)
(c)
vA = 0
r = rA
vA is parallel to r - rA.
Point A has zero velocity.
Point A coincides with the mass center.
Velocity vA is directed along line AG.
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40
PROBLEM 14.30
& A¢ , where HA¢ is defined by Eq. (1) of
Show that the relation SMA = H
Problem 14.29 and where SM A represents the sum of the moments about A of
the external forces acting on the system of particles, is valid if, and only if,
one of the following conditions is satisfied: (a) the frame Ax ¢y ¢z ¢ is itself a
newtonian frame of reference, (b) A coincides with the mass center G,
(c) the acceleration aA of A relative to Oxyz is directed along the line AG.
Solution
n
From equation (1),
H ¢A = Â (ri¢ ¥ mi vi¢ )
i =1
n
HA¢ = Â [(ri - rA ) ¥ mi ( vi - vA )]
i =1
Differentiate with respect to time.
n
n
i =1
i =1
& ¢ = Â [(r& - r& ) ¥ m ( v - v )] + Â [(r - r ) ¥ m ( v& - v& )]
H
A
i
A
i i
A
i
A
i i
A
But
r&i = vi
v& i = ai
r&A = vA
and
v&A = a A
Hence,
& A¢ = 0 + Â [(ri - rA ) ¥ mi (ai - aA )]
H
n
i =1
n
= Â [(ri - rA ) ¥ ( Fi - mi aA )]
i =1
n
n
i =1
i =1
= Â [((ri - rA ) ¥ Fi ] - Â [mi (ri - rA )] ¥ aA
= MA - m( r - rA ) ¥ aA
& A¢ = MA
H
if, and only if,
m( r - rA ) ¥ aA = 0
This condition is satisfied if
or
or
( a)
( b)
(c)
aA = 0
r = rA
aA is parallel to r - rA.
The frame is newtonian.
Pointt A coincides with the mass center.
Acceleration aA is directed along line AG.
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41
PROBLEM 14.31
Assuming that the airline employee of Problem 14.1 first tosses
the 15-kg suitcase on the baggage carrier, determine the energy
lost (a) as the first suitcase hits the carrier, (b) as the second
suitcase hits the carrier.
PROBLEM 14.1 An airline employee tosses two suitcases,
of mass 15 kg and 20 kg, respectively, onto a 25-kg baggage
carrier in rapid succession. Knowing that the carrier is initially
at rest and that the employee imparts a 3-m/s horizontal
velocity to the 15-kg suitcase and a 2-m/s horizontal velocity
to the 20-kg suitcase, determine the final velocity of the
baggage carrier if the first suitcase tossed onto the carrier is
(a) the 15-kg suitcase, (b) the 20-kg suitcase.
Solution
There are no horizontal external forces acting during the impacts. The baggage carrier is to coast between
impacts.
Let v1 be the common velocity of suitcase A and the carrier after the first impact and v2 be the common velocity
of the two suitcases and the carrier of the second impact.
(a)
15-kg suitcase tossed on carrier first:
Conservation of momentum:
(15)(3) = ( 40)v1
v1 = 1.125 m/s
Initial kinetic energy:
Final kinetic energy:
1
mA v02
2
1
= (15 kg)(3 m/s)2
2
= 67.5
T0 =
1
T1 = ( mA + mC )v12
2
1
= ( 40 kg)(1.125 m/s2 )
2
= 25.3125 J
T0 - T1 = 42.1875 J
energy lost = 42.2 J b
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42
Problem 14.31 (continued)
(b)
20-kg suitcase tossed next:
(20)(2) + ( 40)(1.125) = 60 v2
v 2 = 1.417 m/s
Initial kinetic energy:
1
T0¢ = T1 + mB v02
2
1
= 25.3125 J + (20 kg)(2 m/s)2
2
= 65.3125 J
Final kinetic energy:
1
T2 = ( mA + mB + mC )v22
2
1
= (60 kg)(1.417 m/s)2
2
= 60.2083 J
T0¢ - T2 = 5.10 J
energy lost = 5.10 J b
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43
PROBLEM 14.32
Determine the energy loss as a result of the series of collisions described in Problem 14.7.
Solution
Summary of results from Problem 14.7:
(a)
mA = 240 kg,
mB = 260 kg,
mC = 235 kg
vA = 2 m/s
vB = 0,
vC = 1.5 m/s
Cars A and C hit B at the same time.
vA¢ = 1.288 m/s
,
vB¢ = 0.312 m/s
vC¢ = 1.512 m/s
,
1
1
1
mA v 2A + mB v B2 + mC vC2
2
2
2
1
1
T1 = (240)(2)2 + 0 + (235)(1.5)2
2
2
= 744.375 J
Initial kinetic energy:
T1 =
Final kinetic energy:
T2 =
1
1
1
mA ( v A¢ )2 + mB ( v B¢ )2 + mC ( vC¢ )2
2
2
2
1
1
1
= (240)(1.288)2 + (260)(0.312)2 + (235)(1.512)2
2
2
2
= 480.35 J
T1 - T2 = 264.03 J
(b)
.
energy loss = 264 J b
Car A hits car B. Then B and C hit. Then A and B hit. The final velocities (from Problem 14.7) are:
v A¢¢¢= 0.956 m/s
Initial kinetic energy:
T1 = 744.375 J
Final kinetic energy:
T2¢ =
,
v B¢¢¢ = 0.0296 m/s
,
vC¢¢ = 1.552 m/s
1
1
1
mA ( v A¢¢¢) + mB ( v B¢¢¢)2 + mC ( vC¢¢ )2
2
2
2
1
1
1
T2¢ = (240)(0.956)2 + (260)(0.0296)2 + (235)(1.552)2
2
2
2
= 392.8 J
T1 - T2¢ = 351.57 J
energy loss = 352 J b
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44
PROBLEM 14.33
In Problem 14.3, determine the work done by the
woman and by the man as each dives from the boat,
assuming that the woman dives first.
Solution
Woman dives first.
Conservation of momentum:
60(5 - v1 ) - (150 + 90)v1 = 0
v1 = 1 m/s
5 - v1 = 4 m/s
T0 = 0
Kinetic energy before dive:
1
1
T1 = (150 + 90)(1)2 + (60)( 4)2
2
2
= 600 J
Kinetic energy after dive:
Work of woman:
T1 - T0 = 600 J
T1 - T0 = 600 J b
Man dives next. Conservation of momentum:
-240 v1 = -150 v2 + 90(5 - v2 )
240 v1 + 450
= 2.875 m/s
v2 =
240
5 - 2.875 = 2.125 m/s
Kinetic energy before dive:
Kinetic energy after dive:
Work of man:
1
T1¢= (150 + 90)(1)2
2
= 120 J
1
1
T2¢ = (150)(2.875)2 + (90)(2.125)2
2
2
= 823.125 J
T2¢ - T1¢= 703.125 J
T2¢ - T1¢= 703 J b
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45
PROBLEM 14.34
In Problem 14.5, determine the energy lost as the bullet (a) passes
through block A, (b) becomes embedded in block B.
Solution
The masses are m for the bullet and mA and mB for the blocks.
(a)
The bullet passes through block A and embeds in block B. Momentum is conserved.
Initial momentum:
mv0 + mA (0) + mB (0) = mv0
Final momentum:
mv B + mA v A + mB v B
Equating,
mv0 = mv B + mA v A + mB v B
m=
mA v A + mB v B (3)(1.5) + (2.5)(2.7)
=
= 0.0252 kg
v0 - v B
( 450 - 2.7)
m = 25.2 g b
(b)
The bullet passes through block A. Momentum is conserved.
Initial momentum:
mv0 + mA (0) = mv0
Final momentum:
mv1 + mA v A
Equating,
mv0 = mv1 + mA v A
v1 =
mv0 - mA v A
m v
3(1.5)
= v0 - A A = 450 = 271.07 m/s
m
0.02515
m
v1 = 271 m/s
The masses are:
b
m = 0.0252 kg
mA = 3 kg
mB = 2.5 kg
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46
Problem 14.34 (continued)
(a)
Bullet passes through block A. Kinetic energies.
1 2
mv0
2
1
= (0.0252)( 450)2
2
= 2551.5 J
Before:
T0 =
After:
T1 =
Lost:
1 2 1
mv1 + mA v 2A
2
2
1
1
= (0.0252)(271.07)2 + (3)(1.5)2
2
2
= 929.21 J
T0 - T1 = (2551.5 - 929.21)
= 1622.29 J
(b)
energy lost = 1622 J b
Bullet becomes embedded in block B. Kinetic energies.
Before:
After:
Lost:
1 2
mv1
2
1
= (0.0252)(271.07)2
2
= 925.835 J
T2 =
1
T3 = ( m + mB )v B2
2
1
= (0.0252 + 2.5)(2.7)2
2
= 9.2043 J
T2 - T3 = 925.835 - 9.2043 J
= 916.631 J
energy lost = 917 J b
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47
PROBLEM 14.35
Two automobiles A and B, of mass mA and mB , respectively, are traveling in opposite directions when they
collide head on. The impact is assumed perfectly plastic, and it is further assumed that the energy absorbed
by each automobile is equal to its loss of kinetic energy with respect to a moving frame of reference attached
to the mass center of the two-vehicle system. Denoting by EA and EB, respectively, the energy absorbed by
automobile A and by automobile B, (a) show that EA /E B = mB /mA , that is, the amount of energy absorbed
by each vehicle is inversely proportional to its mass, (b) compute EA and EB , knowing that mA = 1600 kg and
mB = 900 kg and that the speeds of A and B are, respectively, 90 km/h and 60 km/h.
Solution
Velocity of mass center:
( mA + mB ) v = mA vA + mB vB
m v + mB vB
v= A A
mA + mB
Velocities relative to the mass center:
mA vA + mB vB mB ( vA + vB )
=
mA + mB
mA + mB
m v + mB vB mA ( vA + vB )
v ¢B = vB - v = vB - A A
=
mA + mB
mA + mB
vA¢ = vA - v = vA -
Energies:
(a)
(b)
EA =
m m2 ( v + v B ) ◊ ( vA + vB )
1
mA vA¢ ◊ vA¢ = A B A
2
2( mA + mB )2
EB =
m2 m ( v + vB ) ◊ ( vA + vB )
1
mB vB¢ ◊ vB¢ = A B A
2
2( mA + mB )2
EA mB
=
b
E B mA
Ratio:
vA = 90 km/h = 25 m/s
vB = 60 km/h = 16.667 m/s
vA + vB = 41.667 m/s
EA =
(1600)(900)2 ( 41.667)2
= 180.0 ¥ 103 J
2
(2)(2500)
EA = 180.0 kJ b
EB =
(1600)2 (900)( 41.667)2
= 320 ¥ 103 J
2
(2)(2500)
EB = 320 kJ b
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48
PROBLEM 14.36
It is assumed that each of the two automobiles involved in the collision described in Problem 14.35 had been
designed to safely withstand a test in which it crashed into a solid, immovable wall at the speed v0. The severity
of the collision of Problem 14.35 may then be measured for each vehicle by the ratio of the energy absorbed in
the collision to the energy it absorbed in the test. On that basis, show that the collision described in Problem 14.35
is ( mA /mB )2 times more severe for automobile B than for automobile A.
Solution
Velocity of mass center:
( mA + mB ) v = mA vA + mB vB
m v + mB vB
v= A A
mA + mB
Velocities relative to the mass center:
mA vA + mB vB mB ( vA + vB )
=
mA + mB
mA + mB
m v + mB vB
m (v + v )
vB¢ = v B - v = vB - A A
= A A B
mA + mB
mA + mB
v A¢ = vA - v = v A -
Energies:
Energies from tests:
Severities:
EA =
m m2 ( v + vB ) ◊ ( vA + vB )
1
mA vA¢ ◊ vA¢ = A B A
2
2( mA + mB )2
EB =
m2 m ( v + vB ) ◊ ( vA + vB )
1
mB vB¢ ◊ vB¢ = A B A
2
2( mA + mB )2
( E A )0 =
1
1
mA v02 , ( E B )0 = mB v02
2
2
SA =
EA
m2 ( v + vB ) ◊ ( vA + vB )
= B A
( E A )0
( mA + mB )2 v02
SB =
EB
m2 ( v + vB ) ◊ ( vA + vB )
= A A
( E B )0
( mA + mB )2 v02
Ratio:
SA mB2
=
b
SB m2A
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49
PROBLEM 14.37
Solve Sample Problem 14.4, assuming that cart A is given an initial horizontal velocity v0 while ball B is at rest.
Solution
(a)
Velocity of B at maximum elevation. At maximum elevation, ball B is at rest relative to cart A. vB = vA
Use impulse-momentum principle.
x components:
mA v0 + 0 = mA vA + mB v B
= ( mA + mB )v B
(b)
vB =
mA v0
mA + mB
h=
v02
mA
b
mA + mB 2 g
b
Conservation of energy:
1
mA v02 , V1 = 0
2
1
1
T2 = mA v 2A + mB v B2
2
2
1
= ( mA + mB )v B2
2
m2A v02
=
2( mA + mB )
V2 = mB gh
T2 + V2 = T1 + V1
T1 =
m2A v02
1
+ mB gh = mA v02
2( mA + mB )
2
h=
m2A v02 ˘
1 È
2
ÍmA v0 ˙
mA + mB ˙˚
2mB g ÍÎ
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50
PROBLEM 14.38
In a game of pool, ball A is moving with the velocity v 0 = v0 i
when it strikes balls B and C, which are at rest side by side.
Assuming frictionless surfaces and perfectly elastic impact
(i.e., conservation of energy), determine the final velocity of
each ball, assuming that the path of A is (a) perfectly centered
and that A strikes B and C simultaneously, (b) not perfectly
centered and that A strikes B slightly before it strikes C.
Solution
(a)
A strikes B and C simultaneously.
During the impact, the contact impulses make 30∞ angles with the velocity v0.
Thus,
v B = v B (cos 30∞i + sin 30∞ j)
vC = vC (cos 30∞i - sin 30∞ j)
By symmetry,
vA = vAi
Conservation of momentum:
mv 0 = mv A + mv B + mvC
y component:
0 = 0 + mv B sin 30∞ - mvC sin 30∞
x component:
mv0 = mvA + mv B cos 30∞ + mvC cos 30∞
vC = v B
v0 - vA
2
=
( v0 - vA )
cos 30∞
3
v -v
v B = vC = 0 A
3
v B + vC =
Conservation of energy:
1 2 1 2 1 2 1 2
mv0 = mvA + mv B + mvC
2
2
2
2
2
v02 = v 2A + ( v0 - vA )2
3
2
v02 - v 2A = ( v0 - vA )( v0 + vA ) = ( v0 - vA )2
3
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51
Problem 14.38 (continued)
2
1
5
v0 + vA = ( v0 - vA )
v0 = - v A
3
3
3
6
2 3
vB = vC =
v0 =
v0
5
5 3
1
vA = - v0
5
vA = 0.200v0
b
vB = 0.693v0
30∞ b
vC = 0.693v0
30∞ b
(b)
A strikes B before it strikes C.
First impact: A strikes B.
During the impact, the contact impulse makes a 30∞ angle with the velocity v 0 .
Thus,
v B = v B (cos 30∞i + sin 30∞ j)
Conservation of momentum:
mv 0 = mvA + mvB
y component:
0 = m( vA¢ ) y + mv B sin 30∞ ( vA¢ ) y = - vB sin 30∞
x component:
v0 = m( vA¢ ) x + mv B cos 30∞ ( vA¢ ) x = v0 - v B cos 30∞
Conservation of energy:
1 2 1
1
1
mv0 = m( vA¢ )2x + m( vA¢ )2y + mv B2
2
2
2
2
1
1
1
= m( v0 - v B cos 30∞)2 + ( v B sin 30∞)2 + v B2
2
2
2
1
= m v02 - 2v0 v B + v B2 cos2 30∞ + v B2 sin 2 30∞ + v B2
2
(
)
3
1
v0 , ( v A¢ ) x = v0 sin 2 30∞ = v0 ,
2
4
3
( vA¢ ) y = - v0 cos 30∞ sin 30∞ = v0
4
v B = v0 cos 30∞ =
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52
Problem 14.38 (continued)
Second impact: A strikes C.
During the impact, the contact impulse makes a 30∞angle with the velocity v 0 .
Thus,
vC = vC (cos 30∞i - sin 30∞ j)
Conservation of momentum:
mvA¢ = mvA + mvC
m( v A¢ ) x = m( vA ) x + mvC cos 30∞,
x component:
( vA ) x = ( v A¢ ) x - vC cos 30∞ =
1
v0 - vC cos 30∞
4
m( v A¢ ) y = m( vA ) y - mvC sin 30∞
y component:
( vA ) y = ( vA¢ ) y + vC sin 30∞ = -
3
v0 + vC sin 30∞
4
Conservation of energy:
1
1
1
1
1
m( v A¢ )2x + m( v A¢ )2y = m( v A )2x + m( v A )2y + mvC2
2
2
2
2
2
2
2
˘
ˆ
1 È 1 2 3 2 ˘ 1 ÈÊ 1
3
ˆ Ê
v0 + vC sin 30∞˜ + vC2 ˙
m Í v0 + v0 ˙ = m ÍÁ v0 - vC cos 30∞˜ + Á ¯ Ë 4
2 Î16
16 ˚ 2 ÍÎË 4
˙
¯
˚
1 È1
1
= m Í v02 - v0 vC cos 30∞ + vC2 cos2 30∞
2 Î16
2
+
˘
3 2
3
v0 vC sin 30∞ + vC2 sin 2 30∞ + vC2 ˙
v0 16
2
˚
Ê1
ˆ
3
0 = - v0 vC Á cos 30∞ +
sin 30∞˜ + 2vC2
2
Ë2
¯
Ê1
ˆ
3
3
v0
vC = v0 Á cos 30∞ +
sin 30∞˜ =
4
4
Ë4
¯
(v A ) x =
1
3
1
v0 v0 cos 30∞ = - v0
4
4
8
(vA ) y = -
3
3
3
v0 +
v0 sin 30∞ = v0
4
4
8
vA = 0.250v0
60° b
vB = 0.866v0
30° b
vC = 0.433v0
30° b
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53
PROBLEM 14.39
In a game of pool, ball A is moving with a velocity v 0 of magnitude
v0 = 5 m/s when it strikes balls B and C, which are at rest and aligned
as shown. Knowing that after the collision the three balls move in the
directions indicated and assuming frictionless surfaces and perfectly
elastic impact (i.e., conservation of energy), determine the magnitudes
of the velocities vA, vB, and vC.
Solution
Velocity vectors:
v 0 = v0 (cos 45∞ i + sin 45∞ j)
v0 = 5 m/s
vA = vA j
vB = v B (sin 30∞ i - cos 30∞ j)
vC = vC (cos 30∞ i + sin 30∞ j)
Conservation of momentum:
mv 0 = mvA + mvB + mvC
Divide by m and resolve into components.
i: v0 cos 45∞ = v B sin 30∞ + vC cos 30∞
j: v0 sin 45∞ = vA - v B cos 30∞ + vC sin 30∞
Solving for v B and vC ,
v B = -0.25882v0 + 0.86603vA
vC = 0.96593v0 - 0.5vA
Conservation of energy:
1 2 1 2 1 2 1 2
mv0 = mvA + mv B + mvC
2
2
2
2
Divide by 12 m and substitute for v B and vC .
v02 = vA2 + ( -0.25882 v0 + 0.86603vA )2
+ (0.96593 v0 - 0.5vA )2
= 2v 2A + v02 - 1.41422 v0 vA
vA = 0.70711 v0 = 3.5355 m/s
vA = 3.54 m/s b
v B = 0.35355 v0 = 1.7677 m/s
v B = 1.768 m/s b
vC = 0.61237 v0 = 3.06185 m/s
vC = 3.06 m/s b
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54
PROBLEM 14.40
In a game of pool, ball A is moving with a velocity v 0 of magnitude
v0 = 5 m/s when it strikes balls B and C, which are at rest and aligned
as shown. Knowing that after the collision the three balls move in
the directions indicated and assuming frictionless surfaces and
perfectly elastic impact (that is, conservation of energy), determine
the magnitudes of the velocities vA, vB, and vC.
Solution
Velocity vectors:
v 0 = v0 (cos 30∞ i + sin 30∞ j)
v0 = 5 m/s
vA = vA j
vB = v B (sin 45∞ i - cos 45∞ j)
vC = vC (cos 45∞ i + sin 45∞ j)
Conservation of momentum:
mv 0 = mvA + mvB + mvC
Divide by m and resolve into components.
i: v0 cos 30∞ = v B sin 45∞ + vC cos 45∞
j: v0 sin 30∞ = vA - v B cos 45∞ + vC sin 45∞
Solving for v B and vC ,
v B = 0.25882v0 + 0.70711v A
vC = 0.96593v0 - 0.70711v A
Conservation of energy:
1 2 1 2 1 2 1 2
mv0 = mv A + mv B + mvC
2
2
2
2
Divide by m and substitute for v B and vC .
v02 = v 2A + (0.25882v0 + 0.70711vA )2
+ (0.96593v0 - 0.70711vA )2
= v02 - v0 vA + 2v 2A
vA = 0.5v0 = 2.500 m/s
vA = 2.50 m/s b
v B = 0.61237v0 = 3.06185 m/s
v B = 3.06 m/s b
vC = 0.61237v0 = 3.06185 m/s
vC = 3.06 m/s b
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55
PROBLEM 14.41
Two hemispheres are held together by a cord which maintains a spring under
compression (the spring is not attached to the hemispheres). The potential
energy of the compressed spring is 120 J and the assembly has an initial
velocity v 0 of magnitude v0 = 8 m/s. Knowing that the cord is severed when
q = 30∞, causing the hemispheres to fly apart, determine the resulting velocity
of each hemisphere.
Solution
Use a frame of reference moving with the mass center.
Conservation of momentum:
0 = -mA v A¢ + mB v B¢
m
v A¢ = B v B¢
mA
Conservation of energy:
V=
1
1
mA ( v A¢ )2 + mB ( v B¢ )2
2
2
2
Data:
=
Êm
ˆ
1
1
mA Á B v B¢ ˜ + mB ( v B¢ )2
2
2
Ë mA ¯
=
mB ( mA + mB )
( v B¢ )2
2 mA
v B¢ =
2mAV
mB ( mA + mB )
mA = 2.5 kg mB = 1.5 kg
V = 120 J
v B¢ =
(2)(2.5)(120)
= 10 v B¢ = 10 m/s
(1.5)( 40)
v A¢ =
1.5
(10) = 6 v A¢ = 6 m/s
2.5
30∞
30∞
Velocities of A and B.
vA = [8 m/s
] + [6 m/s
30∞]
vA = 4.11 m/s
46.9∞ b
vB = [8 m/s
] + [10 m/s
30∞]
vB = 17.39 m/s
16.7∞ b
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56
PROBLEM 14.42
Solve Problem 14.41, knowing that the cord is severed when q = 120∞.
PROBLEM 14.41 Two hemispheres are held together by a cord which
maintains a spring under compression (the spring is not attached to the
hemispheres). The potential energy of the compressed spring is 120 J and the
assembly has an initial velocity v 0 of magnitude v0 = 8 m/s. Knowing that the
cord is severed when q = 30∞, causing the hemispheres to fly apart, determine
the resulting velocity of each hemisphere.
Solution
Use a frame of reference moving with the mass center.
Conservation of momentum:
0 = -mA v A¢ + mB v B¢
m
v A¢ = B v B¢
mA
Conservation of energy:
V=
1
1
mA ( v A¢ )2 + mB ( v B¢ )2
2
2
2
Êm
ˆ
1
1
= mA Á B v B¢ ˜ + mB ( v B¢ )2
2
2
Ë mA ¯
Data:
=
mB ( mA + mB )
( v B¢ )2
2 mA
v B¢ =
2mAV
mB ( mA + mB )
mA = 2.5 kg mB = 1.5 kg
V = 120 J
v B¢ =
(2)(2.5)(120)
= 10 v B¢ = 10 m/s
(1.5)( 40)
v A¢ =
1.5
(10) = 6 v A¢ = 6 m/s
2.5
60∞
60∞
Velocities of A and B.
vA = [8 m/s
] + [6 ft/s
60∞]
vA = 12.17 m/s
25.3∞ b
vB = [8 m/s
] + [10 ft/s
60∞]
vB = 9.17 m/s
70.9∞ b
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57
PROBLEM 14.43
A 20-kg block B is suspended from a 2-m cord attached to a 30-kg cart A,
which may roll freely on a frictionless, horizontal track. If the system is
released from rest in the position shown, determine the velocities of A and B
as B passes directly under A.
Solution
Conservation of linear momentum.
Since block and cart are initially at rest,
L0 = 0
Thus, as B passes under A,
L = mA vA + mB vB = 0
+
mA v A + mB v B = 0
vA = -
mB
vB
mA
(1)
Conservation of energy.
Initially,
T0 = 0
V0 = mB gl (1 - cos q )
As B passes under A,
Thus,
1
1
mA v 2A + mB v B2
2
2
V =0
T=
T0 + V0 = T + V : mB gl (1 - cos q ) =
1
1
mA v 2A + mB v B2
2
2
Substituting for vA from (1) and multiplying by 2:
Ê m2 ˆ
2mB gl (1 - cos q ) = mA Á B2 v B2 ˜ + mB v B2
Ë mA ¯
Ê m2
ˆ
m + mA 2
= Á B + mB ˜ v B2 = mB B
vB
mA
Ë mA
¯
vB =
2 mA
gl (1 - cos q )
mA + mB
(2)
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58
Problem 14.43 (continued)
Given data:
mA = 30 kg mB = 20 kg, l = 2m
g = 9.81 m/s2 q = 25∞
2mA
2 ¥ 30
=
= 1.2
mA + mB 30 + 20
From Eq. (2),
From Eq. (1),
v B = (1.2)(9.81)(2)(1 - cos 25∞)
= 1.48522 m/s
vA = -
mB
20
v B = - (1.48522)
mA
30
v B = 1.485 m/s
b
vA = 0.990 m/s
b
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59
PROBLEM 14.44
Three spheres, each of mass m, can slide freely on a frictionless, horizontal surface.
Spheres A and B are attached to an inextensible, inelastic cord of length l and are at
rest in the position shown when sphere B is struck squarely by sphere C, which is
moving to the right with a velocity v 0 . Knowing that the cord is slack when sphere
B is struck by sphere C, and assuming perfectly elastic impact between B and C,
determine (a) the velocity of each sphere immediately after the cord becomes taut,
(b) the fraction of the initial kinetic energy of the system which is dissipated when
the cord becomes taut.
Solution
(a)
Determination of velocities.
Impact of C and B.
Conservation of momentum:
mv 0 = mvC + mv1
vC + v1 = v0
(1)
Conservation of energy (perfectly elastic impact):
1 2 1 2 1 2
mv0 = mvC + mv1
2
2
2
Square Eq. (1):
vC2 + 2vC v1 + v12 = v02
Subtract Eq. (2):
2vC v1 = 0
vC2 + v12 = v02
v1 = 0 corresponds to initial conditions and should be eliminated. Therefore,
From Eq. (1):
(2)
vC = 0 b
v1 = v0
Cord AB becomes taut.
Because cord is inextensible, component of vB along AB must be equal to vA .
Conservation of momentum:
+
mv 0 = 2mv A + mv B /A
y comp:
0 = 2mv A sin 60∞ - mv B /A sin 30∞
v B /A = 2 3vA +
x comp:
(3)
mv0 = 2mv A cos 60∞ + mv B /A cos 30∞
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60
Problem 14.44 (continued)
Dividing by m and substituting for v B /A from Eq. (3):
(
v0 = 2vA (0.5) + 2 3vA
v0 = 4vA
)( 3/2 )
vA = 0.250v0
Carrying into Eq. (3):
v B /A = 2 3(0.250v0 ) = 0.866v0
Thus,
vB = vA + vB /A
= 0.250v0
60∞ + 0.866v0
vA = 0.250v0
60∞ b
vB = 0.901v0
13.9∞ b
30∞
vB = (0.250v0 cos 60∞ + 0.866v0 cos 30∞)i
+ (0.250v0 sin 60∞ - 0.866v0 sin 30∞) j
vB = 0.875v0 i - 0.2165 j
vB = 0.90139v0
(b)
13.90∞
Fraction of kinetic energy lost.
T0 =
1 2
mv0
2
1 2 1 2 1 2
mv A + mv B + mvC
2
2
2
1
1
1
2
= m(0.250v0 ) + (0.90139v0 )2 + m(0)
2
2
2
1
2
= m(0.875)v0
2
Tfinal =
Kinetic energy lost = T0 - Tfinal =
1
Fraction of kinetic energy lost = 8
1
1 1
m(1 - 0.875)v02 = - mv02
2
2 8
b
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61
PROBLEM 14.45
A 360-kg space vehicle traveling with a velocity v 0 = ( 450 m/s) k passes through the origin O. Explosive
charges then separate the vehicle into three parts A, B, and C, with masses of 60 kg, 120 kg, and 180 kg,
respectively. Knowing that shortly thereafter the positions of the three parts are, respectively, A(72, 72, 648),
B(180, 396, 972), and C( -144, - 288, 576), where the coordinates are expressed in meters, that the velocity of
B is vB = (150 m/s)i + (330 m/s) j + (660 m/s) k , and that the x component of the velocity of C is -120 m/s,
determine the velocity of Part A.
Solution
rA = 72i + 72 j + 648 k
rB = 180 i + 396 j + 972 k
rC = -144 i - 288 j + 576 k
Since there are no external forces, linear momentum is conserved.
Position vectors (meters):
( mA + mB + mC )v 0 = mA v A + mB v B + mC vC
vA =
mA + mB + mC
m
m
v0 - B vB - C vC = 6v0 - 2vB - 3vC mA
mA
mA
(1)
= (6)( 450k ) - (2)(150i + 330 j + 660k ) - (3)[-120i + ( vC ) y j + ( vC ) z k ]
= -3( vC ) y j - 3( vC ) z k + 60i - 660 j + 1380 k
( v A ) x = 60, ( vC ) y = -3( vC ) y - 660, ( v A ) z = -3( vC ) z + 1380
Conservation of angular momentum about O:
( HO )2 = ( HO )1
Since the vehicle passes through the origin, ( HO )1 = 0.
( HO )2 = rA ¥ ( mA vA ) + rB ¥ ( mB vB ) + rC ¥ ( mC vC ) = 0
Divide by mA .
m
m
rA ¥ vA + B rB ¥ vB + C rC ¥ vC = rA ¥ vA + 2rB ¥ vB + 3rC ¥ vC
mA
mA
= rA ¥ (6v 0 - 2v B - 3vC ) + 2rB ¥ v B + 3rC ¥ vC
= 3(rC - rA ) ¥ vC + 6rA ¥ v0 + 2(rB - rA ) ¥ vB
= ( -648 i - 1080 j - 216 k ) ¥ vC + ( 432 i + 432 j + 3894 k ) ¥ v0
+ (216i + 648 j + 648k ) ¥ v B = 0
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62
Problem 14.45 (continued)
i
-648
( vC ) x
j
k
i
j
k
i
j
k
-1080 -216 + 432 432 3894 + 216 648 648 = 0
( vC ) y ( vC ) z
0
0
450
150 330 660
Resolve into components.
i:
216(vC ) y - 1080( vC ) z + 194, 400 + 213, 840 = 0
(2)
j:
648( vC ) z - 216( vC ) x - 194, 400 - 45, 360 = 0
(3)
k:
1080( vC ) x - 648( vC ) y + 0 - 25, 920 = 0
(4)
Set
( vC ) x = -120 m/s
From Eq. (4),
( vC ) y = -240 m/s
From Eq. (3),
( vC ) z = 330 m/s
From Eq. (1),
vA = (6)( 450k ) - (2)(150 i + 330 j + 660 k ) - 3( -120 i - 240 j + 330 k )
= 60 i + 60 j + 390 k
vA = (60.0 m/s) i + (60.0 m/s)j + (390 m/s) k b
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63
PROBLEM 14.46
In the scattering experiment of Problem 14.24, it is known that the alpha particle is projected from A0 (300, 0, 300)
and that it collides with the oxygen nucleus C at Q(240, 200, 100), where all coordinates are expressed in
millimeters. Determine the coordinates of Point B0 where the original path of nucleus B intersects the xz plane.
(Hint: Express that the angular momentum of the three particles about Q is conserved.)
PROBLEM 14.24 In this scattering experiment, an alpha particle A is projected with the velocity
u 0 = -(600 m/s)i + (750 m/s) j - (800 m/s)k into a stream of oxygen nuclei moving with the common velocity
v 0 = (600 m/s) j . After colliding successively with nuclei B and C, particle A is observed to move along the
path defined by the Points A1 (280, 240, 120) and A2(360, 320, 160), while nuclei B and C are observed to
move along paths defined, respectively, by B1 (147, 220, 130), B2 (114, 290, 120), and by C1(240, 232, 90) and
C2 (240, 280, 75). All paths are along straight lines and all coordinates are expressed in millimeters. Knowing
that the mass of an oxygen nucleus is four times that of an alpha particle, determine the speed of each of the
three particles after the collisions.
Solution
Conservation of angular momentum about Q:
uuur
uuur
uuur
uuur
uuur
uuur
QA0 ¥ ( mu 0 ) + QB 0 ¥ ( 4mv0 ) + QC 0 ¥ ( 4 mv0 ) = QA1 ¥ ( mvA ) + QB1 ¥ ( 4mvB ) + QC1 ¥ ( 4mvC )
uuur
uuur
uuur
QA0 ¥ ( mu 0 ) + QB 0 ¥ ( 4mv 0) + 0 = 0 + QB1 ¥ ( 4 mvB ) + 0
uuur
where
QA0 = rA0 - rQ = (300i + 300 k ) - (240 i + 200 j + 100 k )
(1)
= (60 mm)i - (200 mm) j + (200 mm)k
uuur
QB 0 = ( Dx )i + ( Dy ) j + ( Dz ) k
uuur
QB1 = rB1 - rQ = (147i + 220 j + 130 k ) - (240i + 200 j + 100 k )
- (93 mm)i + (20 mm)j + (30 mm) k
u 0 = -(600 m/s)i + (750 m/s)j - (800 m/s)k
v 0 = (600 m/s)j
and from the solution to Problem 14.27,
vB = v B l B = (716.98)( -0.42290i + 0.89707 j - 0.12815 k )
= -(303.21 m/s)) i + (643.18 m/s) j - (91.88 m/s) k
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64
Problem 14.46 (continued)
Calculating each term and dividing by m,
i
j
k
uuur
QA0 ¥ u 0 60 -200 200 = 10, 000 i - 72, 000 j - 75, 000 k
-600 750 -800
uuur
QB 0 ¥ ( 4 v 0 ) = [( Dx )i + ( Dy ) j + ( Dz ) k ] ¥ (2400 j)
= -2400( Dz )i + 2400( Dx) k
uuur
QB1 ¥ ( 4vB ) =
i
j
k
-93
20
30
= -84, 532 i - 70, 565 j - 215, 006 k
-1212.84 2572.72 -367.52
Collect terms and resolve into components.
i:
k:
Coordinates:
10, 000 - 2400( Dz ) = -84, 532
Dz = 39.388 mm
- 75, 000 + 2400( Dx ) = -215, 006 Dx = -58.336 mm
x B0 = xQ + Dx = 240 - 58.336 x B0 = 181.7 mm b
yB0 = 0 b
z B0 = zQ + Dz = 100 + 39.388 z B0 = 139.4 mm b
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65
PROBLEM 14.47
Two small spheres A and B, weighing 2.5 kg and 1 kg, respectively, are connected by a rigid
rod of negligible weight. The two spheres are resting on a horizontal, frictionless surface
when A is suddenly given the velocity v 0 = (3 m/s)i. Determine (a) the linear momentum of
the system and its angular momentum about its mass center G, (b) the velocities of A and B
after the rod AB has rotated through 180∞.
Solution
Position of mass center.
AG + BG = 200 mm = 0.2 m
AG (2.5 kg) = BG(1 kg) BG = 2.5 AG 3.5 AG = 0.2
BG =
(a)
Linear and angular momentum
AG =
2
m
35
1
m
7
L = mA v0 = (2.5 kg) (3 m/s ) i = 7.5 kg ◊ m/s
uuur
Ê 2 ˆ
HG = GA ¥ mA v 0 = Á m˜ j × (7.50 kg ◊ m/s) i
Ë 35 ¯
L = (7.50 kg ◊ m/s)i b
ˆ
Ê3
= - Á kg ◊ m2 /s˜ k
¯
Ë7
= -(0.42857 kg ◊ m2 /s) k
(b)
HG = -(0.429 kg ◊ m2 /s)k b
Velocities of A and B after 180° rotation
Conservation of linear momentum:
mA v0 = mA v A¢ + mB v B¢
2.5 v0 = 2.5 v A¢ + v B¢
2.5v A¢ + v B¢ = 7.5
(1)
Conservation of angular momentum about G ¢ :
+ r m v = -r m v ¢ + r m v ¢
A A 0
A A A
B B B
Ê 2 ˆ
Ê 2 ˆ
Ê 1ˆ
ÁË m˜¯ (2.5)(3) = - ÁË m˜¯ (2.5)v ¢A + ÁË ˜¯ (1)( v B¢ )
35
35
7
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66
Problem 14.47 (continued)
v A¢ v B¢ 3
+
=
7
7 7
- v A¢ + v B¢ = 3 -
(2)
From Eq. (1):
Solving:
9
30
v ¢A = m/s, v B¢ =
m/s
7
7
vA¢ = (1.286 m/s)i; vB¢ = ( 4.29 m/s)i b
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67
PROBLEM 14.48
Solve Problem 14.47, assuming that it is B which is suddenly given the velocity v 0 = (3 m/s)i.
PROBLEM 14.47 Two small spheres A and B, weighing 2.5 kg and 1kg, respectively, are
connected by a rigid rod of negligible weight. The two spheres are resting on a horizontal,
frictionless surface when A is suddenly given the velocity v 0 = (3 m/s)i. Determine (a) the
linear momentum of the system and its angular momentum about its mass center G, (b) the
velocities of A and B after the rod AB has rotated through 180∞.
Solution
Position of mass center.
AG + BG = 200 mm = 0.2 m
AG (2.5 kg) = BG(1 kg) BG = 2.5 AG 3.5 AG = 0.2
BG =
(a)
AG =
2
m
35
1
m
7
Linear and angular momentum
L = mB v 0 = (1 kg ) (3 m/s)i = (3 kg ◊ m/s)i
uuur
Ê1 ˆ
HG = GB ¥ mB v 0 = - Á m˜ j ¥ (3 kg ◊ m/s)i
Ë7 ¯
L = (3kg ◊ m/s)i b
ˆ
Ê3
= + Á kg ◊ m2 /s˜ k
¯
Ë7
HG = +(0.429 kg ◊ m2 /s)k b
(b)
Velocities of A and B after 180° rotation
Conservation of linear momentum:
mB v0 = mA v A¢ + mB v B¢
1(3) = (2.5)v A¢ + v B¢
2.5v A¢ + v B¢ = 3
(1)
Conservation of angular momentum about G:
+
rB mB v0 = rA mA v A¢ - rB mB v B¢
Ê 1ˆ
Ê 2ˆ
Ê 1ˆ
ÁË ˜¯ (1)(3) = ÁË ˜¯ (2.5)v A¢ - ÁË ˜¯ (1)v B¢
7
35
7
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68
Problem 14.48 (continued)
v A¢ v B¢ 3
=
7
7 7
v A¢ - v B¢ = 3 Solving:
v A¢ =
(2)
12
9
m/s, v B¢ = - m/s
7
7
v ¢A = (1.714 m/s)i; v ¢B = -(1.259 m/s)i b
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69
PROBLEM 14.49
Three identical spheres A, B, and C, which can slide freely on a
frictionless horizontal surface, are connected by means of inextensible,
inelastic cords to a small ring D located at the mass center of the
three spheres (l ¢ = 2l cos q ). The spheres are rotating initially about
ring D, which is at rest, at speeds proportional to their distances from
D. We denote by v0 the original speed of A and B and assume that
q = 30∞. Suddenly, cord CD breaks, causing sphere C to slide away.
Considering the motion of spheres A and B and of ring D after the
other two cords have again become taut, determine (a) the speed of
ring D, (b) the relative speed at which spheres A and B rotate about
D, (c) the percent of energy of the original system which is dissipated
when cords AD and BD again become taut.
Solution
We consider the following two positions of the spheres A and B and the ring D.
Position 1:
Immediately after cord CD breaks linear momentum:
L = (2mv0 cos q )i (1)
Angular momentum about G:
H G = 2(l sin q )( mv0 sin q )k
H G = (2 lmv0 sin 2 q )k
Position 2:
(a)
(2)
After cords AD and BD become taut:
Speed of mass center (now located at D).
Recalling Eq. (1):
L = (2m)v = (2 mv0 cos q )i
vD = v = v0 cosq (b)
v = ( v0 cos q )i
(3)
Relative speed v ¢at which A and B rotate about D.
Angular momentum about G:
H G = (2 mv ¢ )l
Recalling Eq. (2):
2 mv ¢l = 2 lmv0 sin 2 q
v ¢ = v0 sin 2 q
(4)
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70
Problem 14.49 (continued)
(c)
Energy lost:
Considering system of 3 spheres:
Initially,
Ê l¢ˆ
vC = Á ˜ v A = (2 cos q )v0
Ë l¯
Therefore,
T0 =
1 2 1 2 1 2
mv A + mv B + mvC = mv02 (1 + 2 cos2 q )
2
2
2
1
ˆ 1
Ê1
Tf = (2m)vD2 + 2 Á mv ¢ 2 ˜ + mvC2
¯ 2
Ë
2
2
= m ÈÎv02 cos2 q + m( v0 sin 2 q )2 + 2v02 cos2 q ˘˚
= mv02 (3 cos2 q + sin 4 q )
% loss = 100
= 100
= 100
T0 - T f
T0
1 + 2 cos2 q - 3 cos2 q - sin 4 q
1 + 2 cos2 q
sin 2 q cos2 q
1 + 2 cos2 q
(5)
Making q = 30∞ in Eqs. (3), (4), and (5)
(a) 0.866v0 , (b) 0.250 v0 , (c) 7.50% b
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71
PROBLEM 14.50
Solve Problem 14.49, assuming that q = 45∞.
PROBLEM 14.49 Three identical spheres A, B, and C, which can
slide freely on a frictionless horizontal surface, are connected by
means of inextensible, inelastic cords to a small ring D located at
the mass center of the three spheres (l ¢ = 2l cos q ). The spheres are
rotating initially about ring D, which is at rest, at speeds proportional
to their distances from D. We denote by v0 the original speed of A and
B and assume that q = 30∞. Suddenly, cord CD breaks, causing sphere
C to slide away. Considering the motion of spheres A and B and of
ring D after the other two cords have again become taut, determine
(a) the speed of ring D, (b) the relative speed at which spheres A and B
rotate about D, (c) the percent of energy of the original system which
is dissipated when cords AD and BD again become taut.
Solution
We consider the following two positions of the spheres A and B and the ring D.
Position 1:
Immediately after cord CD breaks linear momentum:
L = (2mv0 cos q )i (1)
Angular momentum about G:
H G = 2(l sin q )( mv0 sin q )k
H G = (2 lmv0 sin 2 q )k
Position 2:
(a)
After cords AD and BD become taut:
Speed of mass center (now located at D).
Recalling Eq. (1):
L = (2m)v = (2 mv0 cos q )i
vD = v = v0 cosq (b)
(2)
v = ( v0 cos q )i
(3)
Relative speed v ¢ at which A and B rotate about D.
Angular momentum about G:
H G = (2 mv ¢ )l
Recalling Eq. (2):
2 mv ¢l = 2 lmv0 sin 2 q
v ¢ = v0 sin 2 q
(4)
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72
Problem 14.50 (continued)
(c)
Energy lost:
Considering system of 3 spheres:
Initially,
vC = (l ¢ /l )vA = (2 cos q )v0
Therefore,
T0 =
1 2 1 2 1 2
mv A + mv B + mvC = mv02 (1 + 2 cos2 q )
2
2
2
1
ˆ 1
Ê1
Tf = (2m)vD2 + 2 Á mv ¢ 2 ˜ + mvC2
¯ 2
Ë
2
2
= m ÈÎv02 cos2 q + m( v0 sin 2 q )2 + 2v02 cos2 q ˘˚
= mv02 (3 cos2 q + sin 4 q )
% loss = 100
= 100
= 100
T0 - T f
T0
1 + 2 cos2 q - 3 cos2 q - sin 4 q
1 + 2 cos2 q
sin 2 q cos2 q
1 + 2 cos2 q
(5)
Making q = 45∞ in Eqs. (3), (4), and (5):
(a) 0.707v0 , (b) 0.500 v0 , (c) 12.50% b
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73
PROBLEM 14.51
Two small disks A and B, of mass 3 kg and 1.5 kg,
respectively, may slide on a horizontal, frictionless
surface. They are connected by a cord 600 mm long
and spin counterclockwise about their mass center
G at the rate of 10 rad/s. At t = 0, the coordinates
of G are x0 = 0, y0 = 2 m, and its velocity is
v0 = (1.2 m/s)i + (0.96 m/s) j. Shortly thereafter,
the cord breaks; disk A is then observed to move
along a path parallel to the y axis and, disk B along
a path which intersects the x axis at a distance
b = 7.5 m from O. Determine (a) the velocities of
A and B after the cord breaks, (b) the distance a
from the y-axis to the path of A.
Solution
Initial conditions.
Location of G:
AG BG AG + GB AB 0.6 m
=
=
=
=
mB mA mB + mA
m 4.5 kg
Ê 0.6 ˆ
= 0.2 m
AG = 1.5 Á
Ë 4.5 ˜¯
BG = 0.4 m
Linear momentum:
L0 = m v0 = ( 4.5 kg)(1.2 i + 0.96 j)
= 5.4i + 4.32 j
Angular momentum:
About G:
uuur
uuur
( HG )0 = GA ¥ mA v ¢A + GB ¥ mB v ¢B
= (0.2 m)(3 kg)(0.2 m ¥ 10 rad/s)k
+ (0.4 m)(1.5 kg)(0.4 m ¥ 10 rad/s)k
( HG )0 = (3.6 kg ◊ m2 /s)k
About O: Using formula derived in Problem 14.27,
( H O ) 0 = r ¥ m v 0 + ( HG ) 0
= 2 j ¥ (5.4i + 4.32 j) + 3.6k
= -10.8k + 3.6k = -(7.2 kg ◊ m2 /s)k
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74
Problem 14.51 (continued)
Kinetic energy: Using Eq. (14.29),
1
1
1
1
1
mv02 + S m : vi¢ 2 = mv02 + mA v A¢ 2 + mB v B¢ 2
2
2i
2
2
2
1
1
1
= ( 4.5)[(1.2)2 + (0.96)2 ] + (3)(0.2 ¥ 10)2 + (1.5)( 0.4 ¥ 10)2
2
2
2
= (5.3136 + 6 + 12) = 23.314 J
T0 =
(a)
Conservation of linear momentum:
L0 = L
5.4i + 4.32 j = mA v A + mB v B
= 3( v A j) + 1.5[( v B ) z i + ( v B ) y j]
Equating coefficients of i:
5.4 = 1.5( v B ) x
( v B ) x = 3.6 m/s
Equating coefficients of j:
(1)
4.32 = 3v A + 1.5( v B ) y
( v B ) y = 2.88 - 2vA
(2)
Conservation of energy.
T0 = T : T0 =
1
1
mA v 2A + mB v B2
2
2
1
1
23.314 J = (3) v 2A + (1.5)[( v B )2x + ( v B )2y ]
2
2
( )
Substituting from Eqs. (1) and (2):
23.314 = 1.5v 2A + 0.75(3.6)2 + 0.75(2.89 - 25 A)2
4.5v 2A - 8.64vA - 7.373 = 0
v 2A - 1.92vA - 1.6389 = 0
vA = 0.96 + 1.60 = 2.56 m/s
vA = 2.56 m/s b
and vA = 0.96 - 1.60 = -0.64 m/s (rejected, since vA is shown directed up)
From Eqs. (1) and (2):
( v B ) x = 3.6 m/s
( v B ) y = 2.88 - 2(2.56) = -2.24 m/s
vB = 3.6 i - 2.24 j
(b)
vB = 4.24 m/s
31.9° b
Conservation of angular momentum about O:
( HO )0 = HO : - 7.2k = ai ¥ mA vA + bi ¥ mB vB
- 7.2k = ai ¥ 3(2.56 j) + 7.5i ¥ 1.5(3.6i - 2.24 j)
-7.2k = 7.68ak - 25.2k
a = 2.34 m b
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75
PROBLEM 14.52
Two small disks A and B, of mass 2 kg and 1 kg, respectively,
may slide on a horizontal and frictionless surface. They
are connected by a cord of negligible mass and spin
about their mass center G. At t = 0, G is moving with the
velocity v0 and its coordinates are x0 = 0, y0 = 1.89 m.
Shortly thereafter, the cord breaks and disk A is observed
to move with a velocity v A = (5 m/s) j in a straight line and
at a distance a = 2.56 m from the y axis, while B moves
with a velocity v B = (7.2 m/s)i - ( 4.6 m/s) j along a path
intersecting the x axis at a distance b = 7.48 m from the
origin O. Determine (a) the initial velocity v0 of the mass
center G of the two disks, (b) the length of the cord initially
connecting the two disks, (c) the rate in rad/s at which the
disks were spinning about G.
Solution
Initial conditions.
Location of G:
Linear momentum:
Angular momentum about G:
AG BG AG + GB l
=
=
=
mB mA mB + mA m
m
1
AG = B l = l
3
m
m
2
BG = A l = l
m
3
L0 = mv0 = 3 v0
uuur
uuur
( H G )0 = GA ¥ mA v ¢A + GB ¥ mB v ¢B
Ê2 ˆ
Ê1 ˆ
Ê2 ˆ
Ê1 ˆ
= Á l ˜ (2 kg) Á lw ˜ k + Á l ˜ (1 kg) Á lw ˜ k
Ë3 ¯
Ë3 ¯
Ë3 ¯
Ë3 ¯
2
= l2w k2
3
Kinetic energy: Using Eq. (14.29),
T0 =
1
1
1
1
1
mv02 + S mi vi2 = mv02 + mA v A¢ 2 + mB v B¢ 2
2
2i
2
2
2
2
1
1 Ê1 ˆ
1 Ê2 ˆ
= (3)v02 + (2) Á lw ˜ + (1) Á lw ˜
¯
Ë
2
2
3
2 Ë3 ¯
1
3
T0 = v02 + l 2 w 2
2
3
2
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76
Problem 14.52 (continued)
Conservation of linear momentum:
mv0 = mA v A + mB v B
3v0 = (2)(5 j) + (1)(7.2i - 4.6 j) = 7.2i + 5.4 j
v0 = (2.4 m/s)i + (1.8 m/s) j b
(a)
Conservation of angular momentum about O:
+ (1.89 j) ¥ mv0 + ( HG )0 = ( 2.56i ) ¥ mA vA + ( 7.48 i ) ¥ mB vA
Substituting for v0 , ( HG )0 , vA , vB and masses:
2
(1.89 j) ¥ 3(2.4i + 1.8 j) + l 2 wk = (2.56i ) ¥ 2(5 j) + (7.48i ) ¥ (7.2i - 4.6 j)
3
2 2
-13.608k + l wk = 25.6k - 34.408k
3
2 2
l w = 4.80 l 2 w = 7.20
3
(1)
Conservation of energy:
3 2 1 2 2 1
1
v0 + l w = mA v 2A + mB v B2
2
3
2
2
3
1
1
1
[(2.4)2 + (1.8)2 ] + l 2 w 2 = (2)(5)2 + (1)[( 7.2)2 + ( 4.6)2 ]
2
3
2
2
1
(2)
13.5 + l 2 w 2 = 25 + 36.5 l 2 w 2 = 144.0
3
Dividing Eq. (2) by Eq. (1), member by member:
144.0
w=
= 20.0 rad/s
7.20
Original rate of spin = 20.0 rad/s b
T0 = T :
(c)
Substituting for w into Eq. (1):
l 2 (20.0) = 7.20 l 2 = 0.360 l = 0.600 m
Length of cord = 600 mm b
(b)
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77
PROBLEM 14.53
In a game of billiards, ball A is given an initial velocity v0 along the
longitudinal axis of the table. It hits ball B and then ball C, which
are both at rest. Balls A and C are observed to hit the sides of the
table squarely at A' and C', respectively, and ball B is observed to
hit the side obliquely at B'. Knowing that v0 = 4 m/s , v A = 2 m/s,
and a = 1600 mm , determine (a) the velocities vB and vC of balls
B and C, (b) the Point C' where ball C hits the side of the table.
Assume frictionless surfaces and perfectly elastic impacts (i.e.,
conservation of energy).
Solution
Conservation of linear momentum.
mv0 i = - mv A j + m( v B ) x i + m( v B ) y j + mvC i
Equating coefficients of unit vectors:
i:
mv0 = m( v B ) x + mvC
j:
0 = - mv A + ( v B ) y
( v B ) x + vC = v0 = 4 m/s
(1)
( v B ) y = v A = 2 m/s
(2)
Conservation of angular momentum about corner D.
+ ( 750 mm) v0 = -(1600 mm) v A + (1800 mm)( v B ) y + cvC
(750)( 4) = (1600)(2) + (1800)(2) + cvC
cvC = 2600
(3)
1 2 1 2 1
1
mv0 = mv A + m ÈÎ( v B )2y + ( v B )2y ˘˚ + mvC2
2
2
2
2
Conservation of energy.
Dividing by m, multiplying by 2, and substituting for v0 ; v A , ( v B ) y , their values and ( v B ) x = 4 - vC from Eq. (1):
( 4)2 = (2)2 + ( 4 - vC )2 + (2)2 + vC2
vC2 - 4 vC + 4 = 0, vC = 2 m/s
With vC = 2, Eq. (3) yields C = 1300
(a)
vC = 2 m/s
Therefore:
From Eq. (1):
b
( v B ) x = 4 - 2 = 2 m/s
v B = (2 m/s) i + (2 m/s) j
v B = 2.328 m/s
or
(b)
From Eq. (3): C(7.68) = 322.56
45° b
C = 1300 mm b
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78
PROBLEM 14.54
For the game of billiards of Problem 14.53, it is now assumed
that v0 = 5 m/s, vC = 3 m/s, and c = 1200 mm. Determine (a) the
velocities vA and vB of balls A and B, (b) the point A' where ball
A hits the side of the table.
PROBLEM 14.53 In a game of billiards, ball A is given an
initial velocity v0 along the longitudinal axis of the table. It hits
ball B and then ball C, which are both at rest. Balls A and C
are observed to hit the sides of the table squarely at A' and C',
respectively, and ball B is observed to hit the side obliquely at B'.
Knowing that v0 = 4 m/s, v A = 2 m/s, and a = 160 mm, determine
(a) the velocities vB and vC of balls B and C, (b) the Point C' where
ball C hits the side of the table. Assume frictionless surfaces and
perfectly elastic impacts (i.e., conservation of energy).
Solution
Conservation of linear momentum.
mv0 i = - mv A j + m( v B ) x i + m( v B ) y j + mvC i
i:
mv0 = m( v B ) x + mvC
j:
0 = - mv A + ( v B ) y
( v B ) x = 5 - 3 = 2 m/s (1)
(vB ) y = v A (2)
Conservation of angular momentum about corner D.
+ ( 750 mm) v0 = - av A + (1800 mm)( v B ) y + cvC
Substituting given data and using Eq. (2):
750 (5 m/s) = - av A + 1800v A + 1200 (3)
150 = (1800 - a)v A
(3)
Conservation of energy.
1 2 1 2 1
1
mv0 = mv A + m ÈÎ( v B )2x + ( v B )2y ˘˚ + mvC2
2
2
2
2
Dividing by m, multiplying by 2, and substituting
(5)2 = v 2A + (2)2 + v 2A + (3)2
v 2A = 6
(a)
vA = 2.4495 m/s
vA = 2.45 m/s b
From Eqs. (1) and (2): v B = ( v B ) x i + ( v B ) y j = (2 m/s)i + (2.45 m/s) j
v B = 3.16 m/s
or
(b)
From Eq. (3):
50.8° b
(1800 - a)(2.4495) = 150
a = 1738.7 mm
a = 1739 mm b
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79
PROBLEM 14.55
Three small identical spheres A, B, and C, which can slide on a horizontal, frictionless surface, are attached to
three 200-mm-long strings, which are tied to a ring G. Initially, the spheres rotate clockwise about the ring with
a relative velocity of 0.8 m/s and the ring moves along the x-axis with a velocity v0 = (0.4 m/s)i. Suddenly, the
ring breaks and the three spheres move freely in the xy plane with A and B following paths parallel to the y-axis
at a distance a = 346 mm from each other and C following a path parallel to the x axis. Determine (a) the
velocity of each sphere, (b) the distance d.
Solution
Conservation of linear momentum.
Before break:
L0 = (3 m) v = 3m (0.4i ) = m (1.2 m/s)i
After break:
L = mv A j - mv B j + mvC i
L = L0 :
mvC i + m( v A - v B ) j = m(1.2 m/s)i
vA = v B
Therefore,
(1)
vC = 1.200 m/s v C = 1.200 m/s
(2)
Conservation of angular momentum.
Before break:
+
( H O )0 = 3mlv ¢ = 3m (0.2 m)(0.8 m/s)
= 0.480m
After break:
+
HO = m vA xA
+ mv A ( x A + 0.346)
+ mvC d
H O = ( H O )0 :
0.346mvA + mvC d = 0.480m
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80
Problem 14.55 (continued)
Recalling Eq. (2):
0.346vA + 1.200d = 0.480
d = 0.400 - 0.28833vA
(3)
Conservation of energy.
Before break:
1
Ê1
ˆ
T0 = (3 m)v 2 + 3 Á mv ¢ 2 ˜
Ë2
¯
2
3
3
= m v02 + v ¢ 2 = [(0.4)2 + (0.8)2 ]m = 1.200 m
2
2
(
)
After break:
T=
1 2 1 2 1 2
mv A + mv B + mvC
2
2
2
T = T0 : Substituting for v B from Eq. (1) and vC from Eq. (2),
1 2
Èv A + v 2A + (1.200)2 ˘ = 1.200
˚
2Î
2
v A = 0.480
v A = v B = 0.69282 m/s
(a)
Velocities:
vA = 0.693 m/s ;
(b)
v B = 0.693 m/s ;
vC = 1.200 m/s
b
Distance d:
From Eq. (3):
d = 0.400 - 0.28833(0.69282) = 0.20024 m
d = 200 mm b
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81
PROBLEM 14.56
Three small identical spheres A, B, and C, which can slide on a horizontal, frictionless surface, are attached to
three strings of length l which are tied to a ring G. Initially, the spheres rotate clockwise about the ring which
moves along the x axis with a velocity v0. Suddenly, the ring breaks and the three spheres move freely in the
xy plane. Knowing that vA = (1.039 m/s) j, vC = (1.800 m/s)i, a = 416 mm, and d = 240 mm, determine (a) the
initial velocity of the ring, (b) the length l of the strings, (c) the rate in rad/s at which the spheres were rotating
about G.
Solution
Conservation of linear momentum.
(3m) v = mvA + mvB + mvC
3mv0 i = m (1.039 m/s) j - mv B j + m (1.800 m/s)i
Equating coefficients of unit vectors:
i:
3v0 = 1.800 m/s
j:
0 = 1.039 m/s - vB
v0 = 0.600 m/s
(a)
v B = 1.039 m/s
(1)
b
Conservation of angular momentum.
Before break:
After break:
+
( H O )0 = 3ml 2q&
+
H O = - m v A xA
+ mv A ( x A + 0.416)
+ mvC (0.240)
= m(1.039)(0.416) + m(1.800)(0.240)
= m(0.864224)
( H O )0 = H 0 :
l 2q& = 0.28807 3ml 2q& = m(0.864224)
(2)
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82
Problem 14.56 (continued)
Conservation of energy.
Before break:
1
3
Ê1
ˆ 3
T0 = (3m)v 2 + 3 Á mv ¢ 2 ˜ = mv02 + m(lq& )2
Ë2
¯ 2
2
2
3
3
= m(0.600)2 + ml 2q& 2
2
2
After break:
1 2 1 2 1 2
mv A + mv B + mvC
2
2
2
1
1
= m [(1.039)2 + (1.039)2 + (1.800)2 ] = m (5.399)
2
2
T=
T = T0 :
1
3
3
m(5.399) = m(0.600)2 + ml 2q& 2
2
2
2
l 2q& 2 = 1.4397
(b)
(3)
Dividing Eq. (3) by Eq. (2):
1.4397
q&& =
= 4.9976
0.28807
From Eq. (2):
l2 =
0.28807
4.9976
l = 0.2401 m
l = 240 mm b
(c)
q& = 5.00 rad/s
Rate of rotation:
b
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83
PROBLEM 14.57
A stream of water of cross-sectional area A and velocity v1 strikes a plate which
is held motionless by a force P. Determine the magnitude of P, knowing that
A = 500 mm2, v1 = 25 m/s, and V = 0.
Solution
Mass flow rate. As the fluid moves from Section 1 to Section 2 in time Dt, the mass Dm moved is
Dm = r A( Dl )
dm Dm r A( Dl )
=
=
= r Av1
dt
Dt
Dt
Then
Data:
r = 1000 kg/m3 ,
A = 500 mm2 = 500 ¥ 106 m2 ,
v1 = 25 m/s
dm
= (1000)(500 ¥ 10 -6 )(25) = 12.5 kg/s
dt
Principle of impulse and momentum.
: ( Dm)v1 - P Dt = 0
+
P=
Dm
dm
v=
v
Dt
dt
P = (12.5)(25)
P = 312 N b
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84
PROBLEM 14.58
A stream of water of cross-sectional area A and velocity v1 strikes a plate which
moves to the right with a velocity V. Determine the magnitude of V, knowing
that A = 600 mm2, v1 = 30 m/s, and P = 400 N.
Solution
Consider velocities measured with respect to the plate, which is moving with velocity V. The velocity of the
stream relative to the plate is
u = v1 - V (1)
Mass flow rate. As the fluid moves from Section 1 to Section 2 in time Dt, the mass Dm moved is
Dm = r A( Dl )
dm Dm r A( Dl )
=
=
= r Au dt
Dt
Dt
Then
(2)
Principle of impulse and momentum.
( Dm)u - P ( Dt ) = 0
P=
Dm
dm
u=
u = r Au 2
Dt
dt
u=
P
rA
+
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85
Problem 14.58 (continued)
From Eq. (1),
V = v1 - u = v1 -
P
rA
Data:
P = 400 N,
A = 600 mm2 = 600 ¥ 10 -6 m2
v 1 = 30 m/s,
V = 30 -
r = 1000 kg/m3
400
(1000)(600 ¥ 10 -6 )
V = 4.18 m/s b
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86
PROBLEM 14.59
Tree limbs and branches are being fed at A at the rate of 5 kg/s
into a shredder which spews the resulting wood chips at C with
a velocity of 20 m/s. Determine the horizontal component of
the force exerted by the shredder on the truck hitch at D.
Solution
Eq. (14.38):
( D m) vA + SFD t = ( D m)vC
SF =
Dm
vC = (5 kg/s)(20 m/s
Dt
Force exerted on chips = S F = 100 N
25∞)
25∞
Free body: shredder.
+
SFx = 0 : Fx - (100 N ) cos 25∞ = 0
Fx = 90.6 N
Fx = 90.6 N
On hitch:
b
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87
PROBLEM 14.60
A rotary power plow is used to remove snow from a level
section of railroad track. The plow car is placed ahead of an
engine which propels it at a constant speed of 20 km/h. The
plow car clears 18 ¥ 104 kg of snow per minute, projecting it
in the direction shown with a velocity of 12 m/s relative to the
plow car. Neglecting friction, determine (a) the force exerted
by the engine on the plow car, (b) the lateral force exerted by
the track on the plow car.
Solution
vP = 20 km/h =
Velocity of the plow:
50
m/s
9
Velocity of thrown snow:
ˆ
Ê 50
v s = (12 m/s)(cos 30∞i + sin 30∞ j) + Á
m/s˜ k
¯
Ë 9
Mass flow rate:
dm 18 ¥ 10 4 kg
=
= 3000 kg/s
60 s/min
dt
Let F be the force exerted on the plow and the snow.
Apply impulse-momentum, noting that the snow is initially at rest and that the velocity of the plow is constant.
Neglect gravity.
F( Dt ) = ( Dm)v s
50 ˆ
Ê
Ê dm ˆ
F = Á ˜ vs = (3000) Á12 cos 30∞i + 12 sin 30∞ j + k ˜
Ë
Ë dt ¯
9 ¯
= (31177 N )i + (18000 N ) j + (16666.7 N )k
( a)
Force exerted by engine.
Fz = 16.67 kN b
( b)
Lateral force exerted by track.
Fx = 31.2 kN b
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88
PROBLEM 14.61
Water flows in a continuous sheet from between two plates A
and B with a velocity v of magnitude 30 m/s. The stream is
split into two parts by a smooth horizontal plate C. Knowing
that the rates of flow in each of the two resulting streams are,
respectively, Q1 = 100 L/min and Q2 = 500 L/min, determine
(a) the angle q, (b) the total force exerted by the stream on the
horizontal plate.
Solution
Q1 + Q2 = Q
For steady flow
(1)
Assume that the fluid speed is constant.
v = v(sin q i - cos q j),
Velocity vectors:
v1 = - v i,
v2 = v i
Let Pj be the force that plate C exerts on the fluid.
Impulse-momentum principle:
( Dm)v + Pj( Dt ) = ( Dm)1 v1 + ( Dm)2 v 2
Pj =
( Dm)1
( Dm)2
Dm
v1 +
v2 v
Dt
Dt
Dt
Ê dm ˆ
Ê dm ˆ
Ê dm ˆ
= Á ˜ ( - v i ) + Á ˜ ( v i ) - Á ˜ ( v sin q i - v cos q j)
Ë dt ¯
Ë dt ¯ 2
Ë dt ¯ 1
= - rQ1v i + rQ2 v i - rQv(sin q i - cos q j)
Resolve into components.
Data:
i:
0 = - rQ1v + rQ2 v - rQv sin q
j:
P = rQv cos q Q sin q = Q2 - Q1
(2)
(3)
r = 1000 kg/m3,
v = 30 m/s
Q1 = 100 L/min = 1.6667 ¥ 10 -3 m3 /s
Q2 = 500 L/min = 8.3333 ¥ 103 m3 /s
From Eq. (1),
Q = 600 L/min = 10 ¥ 10 -3 m3 /s
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89
Problem 14.61 (continued)
(a)
From Eq. (2),
Q2 - Q1
= 0.66667
Q
q = 41.810∞
sin q =
q = 41.8∞ b
From Eq. (3),
(b)
P = (1000)(10 ¥ 10 -3 )(30) cos 41.810∞
= 224 N
Force that stream exerts on plate C:
- P j = 224 N b
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90
PROBLEM 14.62
Water flows in a continuous sheet from between two plates A
and B with a velocity v of magnitude 40 m/s. The stream is split
into two parts by a smooth horizontal plate C. Determine the
rates of flow Q1 and Q2 in each of the two resulting streams,
knowing that q = 30∞ and that the total force exerted by the
stream on the horizontal plate is a 500-N vertical force.
Solution
Q1 + Q2 = Q
For steady flow
(1)
Assume that the velocity is constant.
v = v(sin q i - cos q j),
Velocity vectors:
v1 = - vi,
v 2 = vi
Let Pj be the force that plate C exerts on the fluid.
Impulse-momentum principle:
( Dm)v + P j( Dt ) = ( Dm)1 v1 + ( Dm)2 v 2
Pj =
( Dm)1
( Dm)2
Dm
v1 +
v2 v
Dt
Dt
Dt
Ê dm ˆ
Ê dm ˆ
Ê dm ˆ
= Á ˜ ( - v i ) + Á ˜ ( v i ) - Á ˜ ( v sin q i - v cos q j)
Ë dt ¯
Ë dt ¯ 2
Ë dt ¯ 1
= - rQ1v i + rQ2 v i - rQv(sin q i - cos q j)
Resolve into components.
i:
0 = - rQ1v + - rQ2 v - rQv sin q
j: P = rQv cos q Q sin q = Q2 - Q1
(2)
(3)
Data:
r = 1000 kg/m3 ,
From Eq. (3),
Q=
v = 40 m/s
P
r v cos q
500
=
(1000)( 40) cos 30∞
= 14.4334 ¥ 10 -3 m3 /s
= 866.03 L/min
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91
Problem 14.62 (continued)
Solving Eqs. (1) and (2) simultaneously for Q1 and Q2,
1
Q1 = Q(1 - sin q )
2
1
= (866.03)(1 - sin 30∞)
2
1
Q2 = Q(1 + sin q )
2
1
= (866.03)(1 + sin 30∞),
2
Q1 = 217 L/min b
Q2 = 650 L/min b
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92
PROBLEM 14.63
The nozzle discharges water at the rate of
1.3 m3/min. Knowing the velocity of the water
at both A and B has a magnitude of 20 m/s and
neglecting the weight of the vane, determine the
components of the reactions at C and D.
Solution
Volumetric flow rate:
Q = 1.3 m3 /min = 21.667 ¥ 10 -3 m3 /s
Mass density of water:
r = 1000 kg/m3
Mass flow rate:
dm
= rQ = 21.667 kg/s
dt
Assume that the flow speed remains constant.
Principle of impulse and momentum.
+ Moments about D:
[( Dm)V cos 50∞](0.575) - [( Dm)V sin 50∞](0.750) + C ( Dt )(0.750) = ( Dm)V (0.075)
(0.750 sin 50∞ - 0.575 cos 50∞ + 0.075) Dm
V
0.750
Dt
dm
= 0.37324
V
dt
= (0.37324)(21.667 kg/s)(20 m/s)
C=
C = 161.7 N b
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93
Problem 14.63 (continued)
+
Horizontal components:
( Dm)V cos 50∞ + Dx ( Dt ) = ( Dm)V
Dx = (1 - cos 50∞)
Dm
V
Dt
dm
V
dt
= (0.35721)(21.667 kg/s)(20 m/s)
+
= 0.35721
Vertical components:
Dx = 154.8 N
b
-( Dm)V sin 50∞ + C ( Dt ) + D y ( Dt ) = 0
Dm
V -C
Dt
dm
V -C
= 0.76604
dt
= (0.76604)(21.667 kg/s)(20 m/s) - 161.7 N
D y = (sin 50∞)
Dy = 170.2 N b
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94
PROBLEM 14.64
Knowing that the blade AB of Sample Problem 14.7 is in the shape of an arc of circle, show that the resultant
force F exerted by the blade on the stream is applied at the midpoint C of the arc AB. (Hint: First show that the
line of action of F must pass through the center O of the circle.)
Solution
u A = uB = u
Assume
Principle of impulse and momentum.
Moments about O + :
R( Dm) u Ak + rc xF( Dt ) = R( Dm) uB k
rc xF( Dt ) = R( Dm)(uB - u A )k = 0
The line of action of F passes through Point O.
Components + :
( Dm)u A + F( Dt ) sin a = ( Dm)uB cos q
Components
+
F sin a =
:
Dm
u(1 - cos q ) Dt
(1)
0 + F( Dt ) cos a = ( Dm)u sin q
F cos a =
Dividing Eq. (2) by Eq. (1),
tan a =
a=
Dm
u sin q Dt
(2)
2 sin 2 q2
1 - cos q
q
=
= tan
sin q
2 sin q2 cos q2
2
q
2
Thus, Point C lies at the midpoint of arc AB.
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95
PROBLEM 14.65
The stream of water shown flows at a rate of 600 L/min and moves with
a velocity of magnitude 20 m/s at both A and B. The vane is supported
by a pin and bracket at C and by a load cell at D which can exert only
a horizontal force. Neglecting the weight of the vane, determine the
components of the reactions at C and D.
Solution
Ê 1 ˆ
dm
kg ˆ
Ê
= rQ = Á1000 3 ˜ (600 ¥ 10 -3 m3 ) Á
Ë
dt
Ë 60 s ˜¯
m ¯
Mass flow rate:
dm
= 10 kg/s
dt
vA = 20 m/s
Velocity vectors:
v B = 20 m/s
40∞
Apply the impulse-momentum principle.
+
Moments about C :
-30( Dm)v A + 120 D( Dt ) = 160( Dm)v B cos 40∞ + 130v B sin 40∞
1 Ê Dm ˆ
Á
˜ [160 v B cos 40∞ + 130 v B sin 40∞ + 30 v A ]
120 Ë Dt ¯
1
=
(10)[(160)(20 cos 40∞) + (130)(20 sin 40∞) + (30)(20)]
120
D=
= 393.549 N
Dx = 394 N b
Dy = 0 b
+
x components:
C x ( Dt ) + D( Dt ) = ( Dm)v B cos 40∞
+
Ê Dm ˆ
Cx = Á
v cos 40∞ - D = (10)(20 cos 40∞) - 393.549
Ë Dt ˜¯ B
y components:
C x = -240 N b
-( Dm)v A + C y ( Dt ) = ( Dm)v B sin 40∞
Dm
Ê Dm ˆ
Cy = Á
vA +
v B cos 40∞ = (10)(20 + 20 sin 40∞)
˜
Ë Dt ¯
Dt
C y = 329 N b
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96
PROBLEM 14.66
The nozzle shown discharges water at the rate of 800 L/min. Knowing that at
both B and C the stream of water moves with a velocity of magnitude 30 m/s,
and neglecting the weight of the vane, determine the force-couple system
which must be applied at A to hold the vane in place.
Solution
800 ¥ 10 -3
60
1 3
m /s
=
75
Q=
dm
Ê 1 ˆ 40
kg/s
= rQ = 1000 Á ˜ =
Ë 75 ¯ 3
dt
vB = (30 m/s) j
vC = (30 m/s)(sin 40∞ i + cos 40∞ j)
Apply the impulse-momentum principle.
+
x components:
0 + Ax ( Dt ) = ( Dm)(30 sin 40∞)
Ax =
Dm
(30 sin 40∞)
Dt
+
Ê 40 ˆ
= Á ˜ (30 sin 40∞)
Ë 3¯
y components:
Ax = 257 N
( Dm)(30) + Ay ( Dt ) = ( Dm)(30 cos 40∞)
Ay =
Dm
(30)(cos 40∞ - 1)
Dt
Ê 40 ˆ
= Á ˜ (30)(cos 40∞ - 1)
Ë 3¯
= -93.582 N
Ay = 93.6 N
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97
Problem 14.66 (continued)
+
Moments about A:
60( Dm)(30) + M A ( Dt ) = (180)( Dm)(30 cos 40∞)
- (300)( Dm)(30 sin 40∞)
Ê D mˆ
(5400 cos 40∞ - 9000 sin 40∞ - 1800)
MA = Á
Ë D t ˜¯
Ê 40 ˆ
= Á ˜ ( -3448.448)
Ë 3¯
= -45979 N ◊ mm
= -45.979 N ◊ m
MA = 46.0 N ◊ m
A = 274 N
b
20.0° b
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98
PROBLEM 14.67
A high speed jet of air issues from the nozzle A with a
velocity of v A and mass flow rate of 0.36 kg/s. The air
impinges on a vane causing it to rotate to the position
shown. The vane has a mass of 6 kg. Knowing that
the magnitude of the air velocity is equal at A and B,
determine (a) the magnitude of the velocity at A, (b) the
components of the reactions at O.
Solution
Assume that the speed of the air jet is the same at A and B.
v A = vB = v
Apply the principle of impulse and momentum.
(a)
+
Moments about O : (0.190)( Dm)v - (0.250)W ( Dt ) = -(0.250)( Dm)v cos 50∞
- (0.500)( Dm)v sin 50∞
v=
Dt
0.250W
◊
Dm 0.150 cos 50∞ + 0.500 sin 50∞ + 0.190
0.250W dm
0.66944 dt
(0.250)(6)(9.81)
=
(0.66944)(0.36)
=
= 61.058 m/s
vA = 61.1 m/s
b
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99
Problem 14.67 (continued)
(b)
+
x components: ( Dm)v + Rx ( Dt ) = -( Dm)v sin 50∞
Dm
v (1 + sin 50∞)
Dt
= -(0.36)(61.058)(1 + sin 50∞)
= -38.82 N
Rx =
+
y components: 0 + Ry ( Dt ) - W ( Dt ) = -( Dm)v cos 50∞
Dm
v cos 50∞
Dt
= (6)(9.81) - (0.36)(61.058) cos 50∞
= 44.73 N
Ry = W +
R = (38.82)2 + ( 44.73)2
= 59.2 N
tan a =
44.73
38.82
a = 49.0∞ R = 59.2 N
49.0° b
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100
PROBLEM 14.68
Coal is being discharged from a first conveyor
belt at the rate of 120 kg/s. It is received at A
by a second belt which discharges it again at B.
Knowing that v1 = 3 m/s and v 2 = 4.25 m/s and
that the second belt assembly and the coal it
supports have a total mass of 472 kg, determine
the components of the reactions at C and D.
Solution
Velocity before impact at A:
( v A ) x = v1 = 3 m/s
( v A )2y = 2 g (Dy ) = (2)(9.8)(0.545) = 10.685 m2 /s2 tan q =
Slope of belt:
Velocity of coal leaving at B:
2.4 - 1.2
,
2.25
( vA ) y = 3.268 m/s
q = 28.07∞
v2 = 4.25 (cos q i + sin q j)
Apply the impulse-momentum principle.
( Dm)( v A ) x + C x ( Dt ) = ( Dm)v2 cos q
x components:
+
Cx =
Dm
[v2 cos q - ( v A ) x ] = (120)( 4.25 cos 28.07∞ - 3)
Dt
C x = 90.0 N
+
moments about C: ( Dm)[-1.2( v A ) x - 0.75( v A ) y ] + 3.00 D( Dt ) - 1.8W ( Dt ) = ( Dm) [-2.4 v2 cos q + 3v2 sin q ]
Dm
[-(1.2)(3) - (0.75)(3.268)] + 3D - (1.8)(472)(9.8)
Dt
Dm
=
[-(2.4)( 4.25 cos q ) + (3)( 4.25 sin q )]
Dt
D = 2775 + 1.0168
dm
= 2775 + (1.0168)(120) = 2897 N
dt
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101
Problem 14.68 (continued)
+
y components:
( Dm)( - v A ) y + (C y + D - W )( Dt ) = ( Dm)v2 sin q
Cy + D - W =
Dm
(3.268 + 4.25 sin q )
Dt
= (120)(5.268) = 632.2 N
C y = 4625.6 - 2897 + 632.2
C y = 2361 N
C x = 90.0 N, C y = 2360 N b
Dx = 0,
D y = 2900 N b
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102
PROBLEM 14.69
While cruising in level flight at a speed of 900 km/h, a jet plane scoops in air at the rate of 90 kg/s and discharges
it with a velocity of 660 m/s relative to the airplane. Determine the total drag due to air friction on the airplane.
Solution
Flight speed:
v = 900 km/h = 250 m/s
dm
= 90 kg/s
dt
Mass flow rate:
SF = D =
dm
(u - v)
dt
or
D=
dm
( v - u)
dt
where, for a frame of reference moving with the plane, v is the free stream velocity (equal to the air speed) and
u is the relative exhaust velocity.
D = (90)(660 - 250) = 36, 900 N
D = 36.9 kN b
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103
PROBLEM 14.70
The total drag due to air friction on a jet airplane cruising in level flight at a speed of 900 km/h is 40 km.
Knowing that the exhaust velocity is 550 m/s relative to the airplane, determine the rate in kg/s at which the air
must pass through the engine.
Solution
v = 900 km/h = 250 m/s
Flight speed:
SF = D =
dm
( u - v)
dt
or
D=
dm
(u - v )
dt
where, for a frame of reference moving with the plane, v is the free stream velocity (equal to the air speed) and
u is the relative exhaust velocity.
dm
D
4000
400
=
=
=
kg/s = 133.3 kg/s
dt u - v 550 - 250
3
dW
= 133.3 kg/s b
dt
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104
PROBLEM 14.71
The jet engine shown scoops in air at A at a rate of 100 kg/s
and discharges it at B with a velocity of 600 m/s relative to
the airplane. Determine the magnitude and line of action of the
propulsive thrust developed by the engine when the speed of
the airplane is (a) 500 km/h, (b) 1000 km/h.
Solution
Use a frame of reference moving with the plane. Apply the impulse-momentum principle. Let F be the force
that the plane exerts on the air.
x components:
( Dm)u A + F ( Dt ) = ( Dm) uB
+
F=
+
moments about B:
Dm
dm
(uB - u A ) =
(uB - u A )
Dt
dt
(1)
-e( Dm)u A + M B ( Dt ) = 0
MB = e
dm
u A
dt
(2)
Let d be the distance that the line of action is below B.
Fd = M B
Data:
(a)
From Eq. (1),
From Eq. (3),
d=
MB
eu A
=
F
uB - u A
dm
= 100 kg/s, uB = 600 m/s, e = 4 nv
dt
u A = 500 km/h = 1250 /9 m/s
1250 ˆ
Ê
F = 100 Á 600 ˜ = 46111.1 N Ë
9 ¯
Ê 1250 ˆ
( 4) Á
Ë 9 ˜¯
d=
= 1.2048 m
1250 ˆ
Ê
ÁË 600 ˜
9 ¯
u A = 1000 km/h =
(b)
From Eq. (1),
From Eq. (3),
(3)
F = 46100 N b
d = 1.205 m b
2500
m/s
9
2500 ˆ
Ê
F = 100 Á 600 ˜
Ë
9 ¯
Ê 2500 ˆ
( 4) Á
Ë 9 ˜¯
d=
2500 ˆ
Ê
600
ÁË
˜
9 ¯
F = 32200 N b
d = 3.45 m b
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105
PROBLEM 14.72
In order to shorten the distance required for landing, a jet airplane is equipped
with movable vanes, which partially reverse the direction of the air discharged
by each of its engines. Each engine scoops in the air at a rate of 120 kg/s and
discharges it with a velocity of 600 m/s relative to the engine. At an instant when
the speed of the airplane is 270 km/h, determine the reverse thrust provided by
each of the engines.
Solution
Apply the impulse-momentum principle to the moving air. Use a frame of reference that is moving with the
airplane. Let F be the force on the air.
v = 270 km/h = 75 m/s
u = 600 m/s
-( Dm)v + F( Dt ) = 2
( Dm)
u sin 20∞
2
Dm
dm
( v + u sin 20∞) =
( v + u sin 20∞)
Dt
dt
F = (120)(75 + 600 sin 20∞) = 33.6 ¥ 103 N
F=
Force on airplane is - F.
F = 33.6 kN
b
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106
PROBLEM 14.73
A floor fan designed to deliver air at a maximum velocity of
6 m/s in a 400-mm-diameter slipstream is supported by a
200-mm-diameter circular base plate. Knowing that the total
weight of the assembly is 60 N and that its center of gravity is
located directly above the center of the base plate, determine the
maximum height h at which the fan may be operated if it is not to
tip over. Assume r = 1.21 kg/m3 for air and neglect the approach
velocity of the air.
Solution
Calculation of dm
at a section in the airstream:
dt
mass = density ¥ volume = density ¥ area ¥ length
Dm = r A( Dl ) = r Av( Dt )
Dm dm
=
= r Av
Dt
dt
Thrust on the airstream:
dm
( vB - vA )
dt
where vB is the velocity just downstream of the fan and vA is the velocity for upstream. Assume that vA is
negligible.
ˆ
Êp
F = ( r Av )v = r Á D 2 ˜ v 2
¯
Ë4
F=
Êpˆ
F = (1.21 kg/m3 ) Á ˜ (0.400 m)2 (6 m/ss)2 = 5.474 N
Ë 4¯
F = 5.474 N
Force on fan:
F ¢ = - F = 5.474 N
Maximum height h.
1
Ê
ˆ
ÁË e = d = 100 mm = 0.1 m˜¯
2
+
 ME = 0
F ¢h - We = 0
We (60)(0.1)
h=
=
F¢
5.474
h = 1.096 m b
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107
PROBLEM 14.74
The helicopter shown can produce a maximum downward air
speed of 25 m/s in a 10-m-diameter slipstream. Knowing that
the weight of the helicopter and its crew is 18 kN and assuming
r = 1.21 kg/m3 for air, determine the maximum load that the
helicopter can lift while hovering in midair.
Solution
F=
The thrust is
Calculation of
dm
.
dt
dm
(vB - v A )
dt
mass = density ¥ volume = density ¥ area ¥ length
Dm = r AB ( Dl ) = r AB v B ( Dt )
dm
Dm
= r AB v B =
Dt
dt
where AB is the area of the slipstream well below the helicopter and v B is the corresponding velocity in the
slipstream. Well above the blade, v A ª 0.
F = r AB v B2
Hence,
(
)
Êpˆ
= 1.21 kg/m3 Á ˜ (10 m)2 (25 m/s)2
Ë 4¯
= 59, 384.5 N
F = 59, 384.5 N
The force on the helicopter is 59.385 kN
WH = 18 kN
Weight of payload:
WP = WP
Statics:
 Fy = F - WH - WP = 0
+
Weight of helicopter:
WP = F - WH = 59.385 - 18 = 41.385 kN W = 41.4 kNb
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108
PROBLEM 14.75
A jet airliner is cruising at a speed of 1000 km/h with each of its
three engines discharging air with a velocity of 600 m/s relative to
the plane. Determine the speed of the airliner after it has lost the
use of (a) one of its engines, (b) two of its engines. Assume that
the drag due to air friction is proportional to the square of the
speed and that the remaining engines keep operating at the same
rate.
Solution
Let v be the air liner speed and u be the discharge relative velocity.
u = 600 m/s.
dm
(u - v )
dt
Thrust formula for one engine:
F=
Drag formula:
D = kv 2
Three engines working. Cruising speed = v0 = 1000 km/h =
3F - D = 3
2500
m/s
9
dm
(u - v0 ) - kv02 = 0
dt
2
Ê 2500 ˆ
kÁ
Ë 9 ˜¯
dm
= 79.8212k
=
=
2500 ˆ
dt 3(u - v0 )
Ê
3 Á 600 ˜
Ë
9 ¯
kv02
(a)
One engine fails. Two engines working. Cruising speed = v1
2F - D = 2
dm
(u - v1 ) - kv12 = 0
dt
(2)(79.8212k )(600 - v1 ) - kv12 = 0
v12 + 159.642v1 - 95.785 ¥ 103 = 0
v1 = 239.798 m/s
(b)
v1 = 863 km/h b
Two engines fail. One engine working. Cruising speed = v2
F-D=
dm
(u - v2 ) - kv22 = 0
dt
(79.8212k )(600 - v2 ) - kv22 = 0
v22 + 79.8212v2 - 47.893 ¥ 103 = 0
v2 = 182.543 m/s
v2 = 657 km/h b
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109
PROBLEM 14.76
A 16-Mg jet airplane maintains a constant speed of 774 km/h while
climbing at an angle a = 18∞. The airplane scoops in air at a rate of
300 kg/s and discharges it with a velocity of 665 m/s relative to the
airplane. If the pilot changes to a horizontal flight while maintaining
the same engine setting, determine (a) the initial acceleration of the
plane, (b) the maximum horizontal speed that will be attained.
Assume that the drag due to air friction is proportional to the square
of the speed.
Solution
Calculate the propulsive force using velocities relative to the airplane.
F=
dm
(vB - v A )
dt
dm
= 300 kg/s
dt
v A = 774 km/h
= 215 m/s
v B = 665 m/s
F = (300)(665 - 215)
= 135, 000 N
Data:
Since there is no acceleration while the airplane is climbing, the forces are in equilibrium.
+
18∞ SF = 0 :
F - D - mg sin a = 0
F - D = mg sin a = (16, 000)(9.8)sin 18∞ = 48, 454 N
(a)
Initial acceleration of airplane in horizontal flight
ma = F - D : 16, 000 a = 48.454 ¥ 103
Corresponding drag force:
D = 135, 000 - 48, 454
= 86, 546 N
Drag force factor:
D = kv 2A
or
k=
a = 3.03 m/s
18° b
D
v 2A
86, 546
=
(215)2
= 1.87228 N ◊ s2 /m2
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110
Problem 14.76 (continued)
(b)
Maximum speed in horizontal flight
Since the acceleration is zero, the forces are in equilibrium.
F-D=0
dm
dm
dm
( v B - v A ) - kv 2A = 0 kv 2A +
vA vB = 0
dt
dt
dt
1.87228v 2A + 300 v A - (300)(665) = 0
v A = 256.0 m/s
v A = 922 km/h b
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111
PROBLEM 14.77
The wind turbine-generator shown has an output-power rating of 5 kW for a wind
speed of 30 km/h. For the given wind speed, determine (a) the kinetic energy of
the air particles entering the 7.50-m-diameter circle per second, (b) the efficiency
of this energy-conversion system. Assume r = 1.21 kg/m3 for air.
Solution
(a)
Kinetic energy per second.
=
1 Dm 2 1
1
1
v = rQv 2 = r( Av )v 2 = r Av 3
2 Dt
2
2
2
3
Ê 30 ¥ 103 m ˆ
1
p
= (1.21 kg/m3 ) (7.50 m)2 Á
˜ 2
4
Ë 3.6 ¥ 103 s ¯
(b)
dT
= 15.47 kJ/s b
dt
Efficiency.
h=
5 kW
15.47 kJ/s
h = 0.323 b
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112
PROBLEM 14.78
For a certain wind speed, the wind turbine-generator shown produces 28 kW
of electric power and has an efficiency of 0.35 as an energy-conversion system.
Assuming r = 1.21 kg/m3 for air determine (a) the kinetic energy of the air
particles entering the 7.50-m-diameter circle per second, (b) the wind speed.
Solution
(a)
Kinetic energy per second.
= input power =
(b)
output power 28 kW
=
efficiency
0.35
dT
= 80 kJ/s b
dt
Wind speed.
1 Dm 2
v
2 Dt
1
= rQv 2
2
1
= r( Av )v 2
2
1
= r Av 3
2
K. E. per second =
Therefore,
1
p
80 kJ/s = (1.21 kg/m3 ) (7.50 m)2 v 3
4
2
v 3 = 2993.1
v = 14.411 m/s
v = 51.9 km/h b
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113
PROBLEM 14.79
While cruising in level flight at a speed of 900 km/h, a jet airplane scoops in air at a rate of 120 kg/s and
discharges it with a velocity of 650 m/s relative to the airplane. Determine (a) the power actually used to propel
the airplane, (b) the total power developed by the engine, (c) the mechanical efficiency of the airplane.
Solution
dm
= 120 kg/s
dt
u = 650 m/s
v = 900 km/h = 250 m/s
dm
F=
(u - v )
dt
= 120(650 - 250)
= 48, 000 N
Data:
(a)
Power used to propel airplane:
P1 = Fv
= ( 48, 000)(250)
= 12 ¥ 106 W = 12 MW
Propulsion power = 12 MW b
Power of kinetic energy of exhaust:
1
P2 ( Dt ) = ( Dm)(u - v )2
2
1 dm
P2 =
( u - v )2
2 dt
1
= (120)(650 - 250)2
2
= 9.6 ¥ 106 W = 9.6 MW
(b)
Total power:
P = P1 + P2
= 21.6 MW
Total power = 21.6 MW b
(c)
Mechanical efficiency:
P1
12
=
P 21.6
5
= = 0.556
9
Mechanical efficiency = 0.556 b
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114
PROBLEM 14.80
The propeller of a small airplane has a 2-m-diameter slipstream and produces a thrust of 4000 N when the
airplane is at fix on the ground. Assuming r = 1.21 kg/m3 for air, determine (a) the speed of the air in the
slipstream, (b) the volume of air passing through the propeller per second, (c) the kinetic energy imparted per
second to the air in the slipstream.
Solution
Calculation of dm
at a section in the airstream:
dt
mass = density ¥ volume
= density ¥ area ¥ length
Dm = r A( Dl ) = r Av Dt
Dm dm
=
= r Av
Dt
dt
(a)Thrust = dm
( vB - vA ) where vB is the velocity just downstream of propeller and vA is the velocity far
dt
upstream. Assume vA is negligible.
Êp
ˆ
Thrust = ( r Av )v = r Á D 2 ˜ v 2
Ë4 ¯
Ê pˆ
4000 = (1.21) Á ˜ (2)2 v 2
Ë 4¯
= 3.80133v 2
v = 32.4386 m/s
(b)
Q=
(c)
Kinetic energy of mass Dm:
1 dm
p
ˆ
Êp
= Av = Á D 2 ˜ v = (2)2 (32.4386)
¯
Ë
4
4
r dt
v = 32.4 m/s b
Q = 101.9 m3 /s b
1
1
1
DT = ( Dm)v 2 = r A( Dl )v 2 = r Av( Dt )v 3
2
2
2
DT dT
=
Dt
dt
1
= r Av 3
2
1 Ê pD 2 ˆ 3
v
= rÁ
2 Ë 4 ˜¯
1
Êpˆ
= (1.21) Á ˜ (2)2 (32.4386)3
Ë 4¯
2
= 64, 877.1 W
dT
= 64, 900 W b
dt
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115
PROBLEM 14.81
In a Pelton-wheel turbine, a stream of water is deflected by a series of
blades so that the rate at which water is deflected by the blades is equal
to the rate at which water issues from the nozzle ( Dm/Dt = Ar v A ). Using
the same notation as in Sample Problem 14.7, (a) determine the velocity
V of the blades for which maximum power is developed, (b) derive an
expression for the maximum power, (c) derive an expression for the
mechanical efficiency.
Solution
Let u be the velocity of the stream relative to the velocity of the blade.
u = (v - V )
Dm
= r Av A
Dt
Mass flow rate:
Principle of impulse and momentum.
( Dm)u - Ft ( Dt ) = ( Dm)u cos q
+
Dm
u(1 - cos q )
Dt
= r Av A ( v A - V )(1 - cos q )
Ft =
where Ft is the tangential force on the fluid.
The force Ft on the fluid is directed to the left as shown. By Newton’s law of action and reaction, the tangential
force on the blade is Ft to the right.
Output power:
Pout = FV
t
= r Av A ( v A - V )V (1 - cos q )
(a)
V for maximum power output.
dPout
= r A ( v A - 2V ) (1 - cos q ) = 0
dV
1
vA = V b
2
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116
Problem 14.81 (continued)
(b)
Maximum power.
1 ˆÊ1 ˆ
Ê
( Pout ) max = r Av A Á v A - v A ˜ Á v A ˜ (1 - cos q )
Ë
2 ¯Ë2 ¯
( Pout ) max =
1
r Av 3A (1 - cos q ) b
4
Input power = rate of supply of kinetic energy of the stream
1 È1
˘
( Dm)v 2A ˙
Í
Dt Î 2
˚
1 Dm 2
vA
=
2 Dt
1
= pAv 3A
2
Pin =
(c)
Efficiency.
h=
h=
pout
pin
r Av A ( v A - V )V (1 - cos q )
3
1
2 r Av A
Ê
V ˆV
h = 2 Á1 - ˜ (1 - cos q ) b
Ë vA ¯ vA
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117
PROBLEM 14.82
A circular reentrant orifice (also called Borda’s mouthpiece) of
diameter D is placed at a depth h below the surface of a tank. Knowing
that the speed of the issuing stream is v = 2 gh and assuming that
the speed of approach v1 is zero, show that the diameter of the stream
is d = D / 2. (Hint: Consider the section of water indicated, and note
that P is equal to the pressure at a depth h multiplied by the area of
the orifice).
Solution
From hydrostatics, the pressure at Section 1 is p1 = rh = r gh.
The pressure at Section 2 is p2 = 0.
Calculate the mass flow rate using Section 2.
mass = density ¥ volume
= density ¥ area ¥ length
Dm = r A2 ( Dl ) = r A2 v( Dt )
dm Dm
=
= r A2 v
dt
Dt
Apply the impulse-momentum principle to fluid between Sections 1 and 2.
( Dm)v1 + p1 A1 ( Dt ) = ( Dm)v
dm
dm
v1 + p1 A1 =
v
dt
dt
dm
( v - v1 )
p1 A1 =
dt
= r A2 v( v - v1 )
But v1 is negligible,
p1 = r gh and v = 2 gh
r ghA1 = r A2 (2 gh) or
p 2
Êp ˆ
D = 2Á d2˜ Ë4 ¯
4
A1 = 2 A2
d=
D
b
2
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118
PROBLEM 14.83*
The depth of water flowing in a rectangular channel of width b
at a speed v1 and a depth d1 increases to a depth d2 at a hydraulic
jump. Express the rate of flow Q in terms of b, d1, and d2.
Solution
mass = density ¥ volume
= density ¥ area ¥ length
Dm = rbd ( Dl ) = rbdv( Dt )
dm Dm
=
= rbdv
dt
Dt
1 dm
Q=
= bdv
r dt
Mass flow rate:
Q1 = Q2 = Q
Q
Q
v1 =
v2
bd1
bd2
Continuity of flow:
Resultant pressure forces:
p1 = g d1
p2 = g d2
1
1
F1 = p1bd1 = g bd12
2
2
1
1
F2 = p2 bd2 = g bd22
2
2
Apply impulse-momentum principle to water between Sections 1 and 2.
( Dm)v1 + F1 ( Dt ) - F2 ( Dt ) = ( Dm)v2
Dm
( v1 - v2 ) = F2 - F1
Dt
Ê Q
Q ˆ 1
= g b d22 - d12
rQ ◊ Á
˜
Ë bd1 bd2 ¯ 2
(
)
rQ 2 ( d2 - d1 ) 1
= g b( d1 + d2 )( d2 - d1 )
bd1d2
2
Noting that g = r g,
Q=b
1
gd1d2 ( d1 + d2 ) b
2
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119
PROBLEM 14.84*
Determine the rate of flow in the channel of Problem 14.83,
knowing that b = 3.6 m, d1 = 1.2 m and d2 = 1.5 m.
PROBLEM 14.83 The depth of water flowing in a rectangular
channel of width b at a speed v1 and a depth d1 increases to
a depth d2 at a hydraulic jump. Express the rate of flow Q in
terms of b, d1, and d2.
Solution
mass = density ¥ volume
= density ¥ area ¥ length
Dm = rbd ( Dl ) = rbdv( Dt )
dm Dm
=
= rbdv
dt
Dt
1 dm
Q=
= bdv
r dt
Mass flow rate:
Continuity of flow:
Q1 = Q2 = Q
Q
Q
v1 =
v2
bd1
bd2
Resultant pressure forces:
p1 = g d1
p2 = g d2
1
1
F1 = p1bd1 = g bd12
2
2
1
1
F2 = p2 bd2 = g bd22
2
2
Apply impulse-momentum principle to water between Sections 1 and 2.
( Dm)v1 + F1 ( Dt ) - F2 ( Dt ) = ( Dm)v2
Dm
( v1 - v2 ) = F2 - F1
Dt
Ê Q
Q ˆ 1
= g b d22 - d12
rQ ◊ Á
Ë bd1 bd2 ˜¯ 2
(
)
rQ 2 ( d2 - d1 ) 1
= g b( d1 + d2 )( d2 - d1 )
bd1d2
2
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120
Problem 14.84* (continued)
1
gd1d2 ( d1 + d2 )
2
Noting that g = r g,
Q=b
Data:
g = 9.81 m/s2
Q = 3.6
b = 3.6 m, d1 = 1.2 m, d2 = 1.5 m
1
(9.81)(1.2)(1.5)(2.7) 2
Q = 17.58 m3 /s b
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121
PROBLEM 14.85
Gravel falls with practically zero velocity onto a conveyor
belt at the constant rate q = dm/dt . (a) Determine the
magnitude of the force P required to maintain a constant
belt speed v. (b) Show that the kinetic energy acquired
by the gravel in a given time interval is equal to half the
work done in that interval by the force P. Explain what
happens to the other half of the work done by P.
Solution
(a)We apply the impulse-momentum principle to the gravel on the belt and to the mass Dm of gravel hitting
and leaving belt in interval Dt.
x comp.: mv + P Dt = mv + ( Dm)v
+
P=
(b)
Dm
v = qv Dt
P = qv b
Kinetic energy acquired for unit time:
1
DT = ( Dm)v 2
2
DT 1 Dm 2 1 2
=
v = qv Dt 2 Dt
2
(1)
Work done per unit time:
DU P Dx
=
= Pv
Dt
Dt
Recalling the result of Part a:
DU = P ( Dx )
DU
= ( qv )v = qv 2 Dt
(2)
Comparing Eqs. (1) and (2), we conclude that
DT 1 DU
=
Dt 2 Dt
Q.E.D:
The other half of the work of P is dissipated into heat by friction as the gravel slips on the belt
before reaching the speed v.
b
b
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122
PROBLEM 14.86
A chain of length l and mass m falls through a small hole in a
plate. Initially, when y is very small, the chain is at rest. In each
case shown, determine (a) the acceleration of the first link A
as a function of y, (b) the velocity of the chain as the last link
passes through the hole. In Case 1, assume that the individual
links are at rest until they fall through the hole; in Case 2,
assume that at any instant all links have the same speed. Ignore
the effect of friction.
Solution
Let r be the mass per unit length of chain. Assume that the weight of any chain above the hole is supported by
the floor. It and the corresponding upward reaction of the floor are not shown in the diagrams.
Case 1. Apply the impulse-momentum principle to the entire chain.
r yv + r gy Dt = r( y + Dy )( v + Dv )
= r yv + r( Dy )v + r y( Dv ) + r( Dy )( Dv )
( Dy )( Dv )
Dy
Dv
v + ry
+r
r gy = r
Dt
Dt
Dt
Let Dt Æ 0.
Multiply both sides by yv.
Let v =
dy
on left hand side.
dt
Integrate with respect to time.
dy
dv
v + ry
dt
dt
d
= r ( yv )
dt
r gy = r
r gy 2 v = r yv
r gy 2
d
( yv )
dt
dy
d
= r yv ( yv )
dt
dt
r g Ú y 2 dy = r Ú ( yv ) d ( yv )
1
1
r gy 3 = r( yv )2
3
2
Differentiate with respect to time.
2v
or v 2 =
2
gy
3
(1)
dv 2 dy 2
= g
= gv
dt 3 dt 3
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123
Problem 14.86 (continued)
a=
dv 1
= g
dt 3
v2 =
2
gl 3
(a)
(b)
Sety = l in Eq. (1).
a = 0.333g b
v = 0.817 gl b
Case 2. Apply conservation of energy using the floor as the level from which the potential energy is measured.
T1 = 0
V1 = 0
y
1 2
mv V2 = -r gy
2
2
T1 + V1 = T2 + V2
T2 =
0=
Differentiating with respect to y,
2v
1 2 1
mv - r gy 2
2
2
v2 =
r gy 2 gy 2
=
m
l
(2)
dv 2 gy
=
dy
l
dv gy
= dy
l
(a)
Acceleration.
a=v
(b)
Setting y = l in Eq. (2),
v 2 = gl a=
gy
b
l
v = gl b
Note: The impulse-momentum principle may be used to obtain the force that the edge of the hole
exerts on the chain.
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124
PROBLEM 14.87
A chain of length l and mass m lies in a pile on the floor. If its end A is raised
vertically at a constant speed v, express in terms of the length y of chain which
is off the floor at any given instant (a) the magnitude of the force P applied at A,
(b) the reaction of the floor.
Solution
Let r be the mass per unit length of chain. Apply the impulse-momentum to the entire chain. Assume that the
reaction from the floor is equal to the weight of chain still in contact with the floor.
Calculate the floor reaction.
R = r g (l - y )
yˆ
Ê
R = mg Á1 - ˜
Ë
l¯
Apply the impulse-momentum principle.
r yv + P ( Dt ) + R( Dt ) - r gL( Dt ) = r( y + Dy )v
P Dt = r( Dy )v + r gL( Dt ) - R( Dt )
P=r
(a)
Dy
v + r gL - r( L - y ) g
Dt
= r v 2 + r gy Let
(b)
P=
m 2
( v + gy ) b
l
Dy dy
=
=v
Dt dt
yˆ
Ê
R = mg Á1 - ˜ b
Ë
l¯
From above,
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125
PROBLEM 14.88
Solve Problem 14.87, assuming that the chain is being lowered to the floor at a
constant speed v.
Solution
(a)
Let r be the mass per unit length of chain. The force P supports the weight of chain still off the floor.
P = r gy (b)
P=
mgy
b
l
Apply the impulse-momentum principle to the entire chain.
- r yv + P ( Dt ) + R( Dt ) - r gL( Dt ) = - r g ( y + Dy )v
R( Dt ) = r gL( Dt ) - P ( Dt ) - r g (D
Dy )v
Dy
v
R = r gL - r gy - r
Dt
Let Dt Æ 0. Then
Dy dy
=
= -v
Dt dt
R = r g ( L - y ) + r v 2
R=
m
[ g( L - y) + v2 ] b
l
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126
PROBLEM 14.89
A toy car is propelled by water that squirts from an internal
tank at a constant 2 m/s relative to the car. The weight of the
empty car is 200 g and it holds 1 kg of water. Neglecting other
tangential forces, determine the top speed of the car.
Solution
Consider a time interval Dt. Let m be the mass of the car plus the water in the tank at the beginning of the
interval and ( m - Dm) the corresponding mass at the end of the interval. m0 is the initial value of m. Let v be the
velocity of the car. Apply the impulse and momentum principle over the time interval.
+
Horizontal components
:
mv + 0 = ( Dm)( v - U cos 20∞) + ( m - Dm)( v + Dv )
Dm
Dv = U cos 20∞
m - Dm
(1)
Let Dv be replaced by differential dv and Dm be replaced by the small differential -dm, the minus sign meaning
that dm is the infinitesimal increase in m, and neglecting the dm in denominator of (1)
dv = -U cos 20∞
dm
m
Integrating,
v = v0 - U cos 20∞ ln
Since v0 = 0,
v = U cos 20∞ ln
m
m0
m0
m
The velocity is maximum when m = m f , the value of m when all of the water is expelled.
vmax = U cos 20∞ ln
m0
mf
vmax = (2 m/s) cos 20∞ ln
1 + 0.2
0.2
vmax = 3.37m/s b
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127
PROBLEM 14.90
A toy car is propelled by water that squirts from an internal
tank. The weight of the empty car is 200 g and it holds 1 kg of
water. Knowing the top speed of the car is 2.5 m/s, determine
the relative velocity of the water that is being ejected.
Solution
Consider a time interval Dt. Let m be the mass of the car plus the water in the tank at the beginning of the
interval and ( m - Dm) the corresponding mass at the end of the interval. m0 is the initial value of m. Let v be the
velocity of the car. Apply the impulse and momentum principle over the time interval.
+
Horizontal components
:
mv + 0 = ( Dm)( v - U cos 20∞) + ( m - Dm)( v + Dv )
Dm
(1)
Dv = U cos 20∞
m - Dm
Let Dv be replaced by differential dv and Dm be replaced by the small differential -dm, the minus sign meaning
that dm is the infinitesimal increase in m and neglecting dm in denominator of (1)
dv = -U cos 20∞
dm
m
Integrating,
v = v0 - U cos 20∞ ln
Since v0 = 0,
v = U cos 20∞ ln
m
m0
m0
m
The velocity is maximum when m = m f , the value of m when all of the water is expelled.
vmax = U cos 20∞ ln
2.5 m/s = U cos 20∞ ln
m0
mf
1 + 0.2
0.2
U = 1.485 m/s b
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128
PROBLEM 14.91
The main propulsion system of a space shuttle consists of three identical
rocket engines, each of which burns the hydrogen-oxygen propellant at
the rate of 340 kg/s and ejects it with a relative velocity of 3750 m/s.
Determine the total thrust provided by the three engines.
Solution
From Eq. (14.44) for each engine:
dm
u
dt
= (340 kg/s)(3750 m/s)
P=
= 1.275 ¥ 106 N
For the 3 engines:
Total thrust = 3(1.275 ¥ 106 N )
Total thrust = 3.03 MN b
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129
PROBLEM 14.92
The main propulsion system of a space shuttle consists of three identical
rocket engines which provide a total thrust of 6 MN. Determine the rate
at which the hydrogen-oxygen propellant is burned by each of the three
engines, knowing that it is ejected with a relative velocity of 3750 m/s.
Solution
Thrust of each engine:
Eq. (14.44):
1
P = (6 MN ) = 2 ¥ 106 N
3
dm
u
dt
dm
2 ¥ 106 N =
(3750 m/s)
dt
P=
dm 2 ¥ 106 N
=
dt 3750 m/s
dm
= 533 kg/s b
dt
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130
PROBLEM 14.93
A space vehicle describing a circular orbit at a speed of 24 ¥ 103 km/h
releases its front end a capsule which has a gross mass of 600 kg,
including 400 kg of fuel. If the fuel is consumed at the constant rate
of 18 kg/s and ejected with a relative velocity of 3000 m/s, determine
(a) the tangential acceleration of the capsule as the engine is fired,
(b) the maximum speed attained by the capsule.
Solution
Thrust:
dm
u
dt
= (18 kg/s)(3000 m/s)
P=
= 54 ¥ 103 N
( at )0 =
( a)
(b)
P 54 ¥ 103
=
= 90 m/s2
m0
600
( at )0 = 90.0 m/s2 b
Maximum speed is attained when all the fuel is used up.
t1
t1 P
0
0 m
v1 = v0 + Ú at dt = v0 + Ú
= v0 + Ú
( ) dt = v + u m Ê - dm ˆ
dm
t1 u dt
0
dt
m
Úm ÁË
1
0
0
˜
m¯
Ê
m
m ˆ
v1 = v0 + u Á - ln 1 ˜ = v0 + u ln 0
m1
m0 ¯
Ë
Data:
v0 = 24 ¥ 103 km/h
= 6.6667 ¥ 103 m/s
u = 3000 m/s
m0 = 600 kg
m1 = 600 - 400 = 200 kg
v1 = 6.6667 ¥ 103 + 3000 ln
= 9.9625 ¥ 103 m/s
600
200
v1 = 35.9 ¥ 103 km/h b
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131
PROBLEM 14.94
A rocket has a mass of 1200 kg, including 1000 kg of fuel, which is consumed at the rate of 12.5 kg/s and
ejected with a relative velocity of 4000 m/s. Knowing that the rocket is fired vertically from the ground,
determine its acceleration (a) as it is fired, (b) as the last particle of fuel is being consumed.
Solution
From Eq. (14.44) of the text book, the thrust is
dm
u
dt
= (12.5 kg/s)(4000 m/s)
P=
= 50 ¥ 103 kg ◊ m/s2
= 50 ¥ 103 N
S F = ma
P - mg = ma
(a)
a=
P
- g
m
(1)
m = m0 = 960 kg
At the start of firing,
g = 9.81 m/s2
a=
From Eq. (1),
(b)
50 ¥ 103
- 9.81 = 31.86 m/s2
1200
a = 31.9 m/s2 b
As the last particle of fuel is consumed,
m = 1200 - 1000 = 200 kg
From Eq. (1),
a=
g = 9.81 m/s2 (assumed)
54 ¥ 103
- 9.81 = 240.2 m/s2
200
a = 240 m/s2 b
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132
PROBLEM 14.95
A communication satellite weighing 50 kN, including fuel, has been
ejected from a space shuttle describing a low circular orbit around the
earth. After the satellite has slowly drifted to a safe distance from the
shuttle, its engine is fired to increase its velocity by 2500 m/s as a first
step to its transfer to a geosynchronous orbit. Knowing that the fuel is
ejected with a relative velocity of 4000 m/s, determine the weight of fuel
consumed in this maneuver.
Solution
Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time Dt.
: mv = ( m - Dm)( v + Dv ) + ( Dm)( v + Dv - v )
+
= mv + m( Dv ) - ( Dm)v - ( Dm)( Dv ) + ( Dm)v + ( Dm)( Dv ) - ( Dm)v
m( Dv ) - u( Dm) = 0
dm
( Dt )
dt
u dm
Dv dv
=
=Dt dt
m dt
Dm = -
v1
t1 u dm
m1 dm
Úv dv = - Ú0 m dt dt = - Úm m
0
0
v1 - v0 = - u ln
m
m1
= u ln 0
m0
m1
m0
Êv -v ˆ
= exp Á 1 0 ˜
Ë u ¯
m1
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133
Problem 14.95 (continued)
Data:
v1 - v0 = 2500 m/s
u = 4000 m/s
50000
m0 =
kg
9.81
50000
2500
= exp
4000
9.81 m1
= 1.8682
m1 = 2728.14 kg
mfuel = m0 - m1 = 2368.7 kg
weight fuel = 23236.9 N
wfuel = 23200 N b
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134
PROBLEM 14.96
Determine the increase in velocity of the communication satellite of
Problem 14.95 after 12.5 kN of fuel has been consumed.
Solution
Data from Problem 14.95:
m0 =
50000 kg
9.81
m1 = m0 - mfuel =
u = 4000 m/s
(50000 - 12500)
= 3822.63 kg.
9.81
Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time Dt.
+
mv = ( m - Dm)( v + Dv ) + ( Dm)( v + Dv - v )
= mv + m( Dv ) - ( Dm)v - ( Dm)( Dv )
+ ( Dm)v + ( Dm)( Dv ) - ( Dm)v
:
m( Dv ) - u( Dm) = 0
Dm = -
dm
( Dt )
dt
Dv dv
u dm
=
=Dt dt
m dt
v1
t1 u dm
m1
dm
Úv dv = - Ú0 m dt dt = - Úm u m
0
0
v1 - v0 = -u ln
m
m1
= u ln 0
m0
m1
Dv = v1 - v0 = 4000 ln
50000
37500
Dv = 1151 m/s b
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135
PROBLEM 14.97
A 540-kg spacecraft is mounted on top of a rocket with a mass of 19 Mg, including 17.8 Mg of fuel.
Knowing that the fuel is consumed at a rate of 225 kg/s and ejected with a relative velocity of
3600 m/s, determine the maximum speed imparted to the spacecraft if the rocket is fired vertically
from the ground.
Solution
See sample Problem 14.8 for derivation of
v = u ln
Data:
m0
- gt m0 - qt
(1)
u = 3600 m/s q = 225 kg/s, mfuel = 17, 800 kg
m0 = 19, 000 kg + 540 kg = 19, 540 kg
We have
mfuel = qt , 17, 800 kg = (225 kg/s)t
t=
17, 800 kg
= 79.111 s
225 kg/s
Maximum velocity is reached when all fuel has been consumed, that is, when qt = mfuel. Eq. (1) yields
vm = u ln
m0
- gt
m0 - mfuel
19, 540
- (9.81 m/s2 )(79.111 s)
19, 540 - 17, 800
= (3600 m/s) ln 11.230 - 776.1 m/s
= (3600 m/s) ln
= 7930.8 m/s
vm = 7930 m/s b
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136
PROBLEM 14.98
The rocket used to launch the 540-kg spacecraft of Problem 14.97 is redesigned to include two
stages A and B, each of mass 9.5 Mg, including 8.9 Mg of fuel. The fuel is again consumed at
a rate of 225 kg/s and ejected with a relative velocity of 3600 m/s. Knowing that when stage A
expels its last particle of fuel its casing is released and jettisoned, determine (a) the speed of the
rocket at that instant, (b) the maximum speed imparted to the spacecraft.
Solution
Mass of rocket + unspent fuel:
dm
= uq
dt
m = m0 - qt
Corresponding weight force:
W = mg
Acceleration:
a=
P=u
Thrust force.
F P -W P
uq
=
= -g=
-g
m
m
m
m0 - qt
Integrating with respect to time to obtain the velocity,
t
0 m - qt
0
= v0 - u ln
For each stage,
qdt
t
0
v = v0 + Ú adt = v0 + u Ú
- gt
m0 - qt
- gt m0
mfuel = 8900 kg
q = 225 kg/s
For the first stage,
v0 = 0
(a)
v1 = 0 - 3600 ln
(1)
u = 3600 m/s
m fuel
t=
q
=
8900
= 39.556 s
225
m0 = 540 + (2)(9500) = 19, 540 kg
19, 540 - 8900
- (9.81)(39.556) = 1800.1 m/s
19, 540
v1 = 1800 m/s b
For the second stage,
(b)
v0 = 1800.1 m/s,
m0 = 540 + 9500 = 10, 040 kg
v2 = 1800.1 - 3600 ln
10, 040 - 8900
- (9.81)(39.556) = 9244 m/s
10, 040
v2 = 9240 m/s b
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137
PROBLEM 14.99
Determine the altitude reached by the spacecraft of Problem 14.97 when all the fuel of its launching
rocket has been consumed.
Solution
See Sample Problem 14.8 for derivation of
v = u ln
m0
m - qt
- gt = -u ln 0
- gt m0 - qt
m0
(1)
Note that g is assumed to be constant.
dy
Set dt = v in Eq. (1) and integrate with respect to time.
Let
h
t
tÊ
ˆ
m0
h = Ú dy = Ú vdt = Ú Á u ln
- gt ˜ dt
0
0
0Ë
m0 - qt
¯
t
m - qt
1
dt - gt 2
= -u Ú ln 0
0
m0
2
m - qt
m
q
dz = dt
dt = - 0 dz
z= 0
or
m0
m0
q
h=
m0 u z
mu
1
1
z
ln z dz - gt 2 = 0 [( z ln z + z ) ]z - gt 2
Ú
0
q z0
2
q
2
=
m0 u È m0 - qt Ê m0 - qt ˆ m0 Ê m0 ˆ ˘ 1 2
lnn
ln
- 1˜ - 1 ˙ - gt
Í
m0
q ÍÎ m0 ÁË
¯ m0 ÁË m0 ˜¯ ˙˚ 2
=
m0 u ÈÊ
qt ˆ Ê m0 - qt ˆ ˘ 1 2
- 1˜ + 1˙ - gt
ln
ÍÁ1 q ÍÎË m0 ˜¯ ÁË
m0
¯ ˙˚ 2
=
˘
È m - qt ˘ 1 2
m0 u È m0 - qt
- 1˙ - gt
- 1 + 1˙ - ut Íln 0
Íln
m0
q Î
m0
˚
Î
˚ 2
ˆ m - qt 1 2
Êm u
= ut + Á 0 - ut ˜ ln 0
- gt
¯
Ë q
m0
2
È Êm
ˆ
m0 ˘ 1 2
h = u Ít - Á 0 - t ˜ ln
˙ - gt ¯
m0 - qt ˚ 2
Î Ë q
(2)
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138
Problem 14.99 (continued)
Data:
u = 3600 m/s
m0 = 19, 000 + 540 = 19, 540 kg
q = 225 kg/s
mfuel = 17, 800 kg
mfuel 17, 800
=
= 79.111 s
q
225
m0 - qt = 1740 kg
t=
From Eq. (2),
g = 9.81 m/s2
È
ˆ 19, 540 ˘ 1
Ê 19, 540
- (9.81)(79.111)2
h = (3600) Í79.111 - Á
- 79.111˜ ln
¯
Ë 225
1740 ˙˚ 2
Î
= 186, 766 m
h = 186.8 km b
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139
PROBLEM 14.100
For the spacecraft and the two-stage launching rocket of Problem 14.98, determine the altitude at
which (a) stage A of the rocket is released, (b) the fuel of both stages has been consumed.
Solution
dm
= uq
dt
Thrust force.
P=u
Mass of rocket + unspent fuel:
m = m0 - qt
Corresponding weight force:
W = mg
Acceleration:
a=
F P -W P
uq
=
= -g=
-g
m
m
m
m0 - qt
Integrating with respect to time to obtain velocity,
qdt
t
t
0
0 m - qt
0
v = v0 + Ú adt = v0 + u Ú
= v0 - u ln
- gt
m0 - qt
- gt m0
(1)
Integrating again to obtain the displacement,
m0 - qt
1
dt - gt 2
m0
2
m0 - qt
m
q
z=
dz = dt
dt = - 0 dz
m0
m0
q
m0 u z
1 2
s = s0 + v0 t +
ln z dz - gt
q Úz0
2
mu
1
z
= s0 + v0 t + 0 ( z ln z + z ) z - gt 2
0
q
2
t
s = s0 + v0 t - u Ú ln
0
Let
Then
= s0 + v0 t +
m0 u È m0 - qt m0 - qt m0 - qt m0
m
m ˘ 1
+
ln
ln 0 + 0 ˙ - gt 2
Í
q Î m0
m0
m0
m0
m0 m0 ˚ 2
È Êm
ˆ m - qt ˘ 1 2
= s0 + v0 t + u Ít + Á 0 - t ˜ ln 0
˙ - gt ¯
m0 ˚ 2
Î Ë q
(2)
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140
Problem 14.100 (continued)
For each stage,
mfuel = 8900 kg
u = 3600 m/s
q = 225 kg/s
For the first stage,
v0 = 0
t=
mfuel 8900
=
= 39.556 s
q
225
s0 = 0
m0 = 540 + (2)(9500) = 19, 540 kg
From Eq. (1),
v1 = 0 - 3600 ln
19, 540 - 8900
- (9.81)(39.556)
19, 540
= 1800.1 m/s
From Eq. (2),
(a)
È
ˆ 19, 540 - 8900 ˘ 1
Ê 19, 540
s1 = 0 + 0 + 3600 Í39.556 + Á
- 39.556˜ ln
- (9.81)(39.556)2
˙
¯
Ë
225
19, 540
Î
˚ 2
= 31, 249 m
For the second stage,
h1 = 31.2 km b
v0 = 1800.1 m/s s0 = 31, 249 m
m0 = 540 + 9500 = 10, 040 kg
From Eq. (2),
(b)
È
ˆ 10, 040 - 8900 ˘
Ê 10, 040
s2 = 31, 249 + (1800.1)(39.556) + 3600 Í39.556 + Á
- 39.556˜ ln
˙
Ë
¯
225
10, 040
Î
˚
1
- (9.81)(39.556)2
2
= 197, 502 m
h2 = 197.5 km b
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141
PROBLEM 14.101
Determine the distance separating the communication satellite of
Problem 14.95 from the space shuttle 60 s after its engine has been fired,
knowing that the fuel is consumed at a rate of 20 kg/s.
PROBLEM 14.95 A communication satellite weighing 50 kN, including
fuel, has been ejected from a space shuttle describing a low circular orbit
around the earth. After the satellite has slowly drifted to a safe distance
from the shuttle, its engine is fired to increase its velocity by 2500 m/s as
a first step to its transfer to a geosynchronous orbit. Knowing that the fuel
is ejected with a relative velocity of 4000 m/s, determine the weight of
fuel consumed in this maneuver.
Solution
Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time Dt.
+
:
mv = ( m - Dm)( v + Dv ) + ( Dm)( v + Dv - v )
= mv + m( Dv ) - ( Dm)v - ( Dm)( Dv )
+ ( Dm)v + ( Dm)( Dv ) - ( Dm)v
m( Dv ) - u( Dm) = 0
dm
( Dt )
dt
Dv dv
u dm
uq
uq
=
===Dt dt
m dt
m
m0 - qt
Dm = -
v = v0 + Ú
t
uq
0 m - qt
0
t
dt = v0 - u ln ( m0 - qt ) 0
= v0 + u ln ( m0 - qt ) + u ln m0
Ê m - qt ˆ
= v0 - u ln Á 0
Ë m0 ˜¯
(1)
Set dx
= v in Eq. (1) and integrate with respect to time.
dt
t
Ê m - qt ˆ
dt
x = x0 + v0 t + u Ú ln Á 0
0
Ë m0 ˜¯
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142
Problem 14.101 (continued)
Call the last term x¢ and let
m - qt
z= 0
m0
dz = -
x¢ =
q
dt
m0
or
dt = -
m0
dz
q
mu
m0 u z
ln z dz = 0 [( z ln z + z )]zz0
Ú
q
q z0
=
m0 u È m0 - qt Ê m0 - qt ˆ m0 Ê m0 ˆ ˘
-1 ˙
- 1˜ ln
ln
Í
q ÍÎ m0 ÁË
m0
¯ m0 ÁË m0 ˜¯ ˙˚
=
m0 u ÈÊ
qt ˆ Ê m0 - qt ˆ ˘
- 1˜ + 1˙
ln
ÍÁ1 q ÍÎË m0 ˜¯ ÁË
m0
¯ ˙˚
=
È m - qt ˘
˘
m0 u È m0 - qt
- 1 + 1˙ - ut Íln 0
- 1˙
Íln
m0
q Î
m0
Î
˚
˚
ˆ m - qt
Êm u
= ut + Á 0 - ut ˜ ln 0
m0
¯
Ë q
È Êm
ˆ
m0 ˘
= u Ít - Á 0 - t ˜ ln
˙
¯ m0 - qt ˚
Î Ë q
È Êm
ˆ
m0 ˘
x = x0 + v0 t + u Ít - Á 0 - t ˜ ln
˙
¯ m0 - qt ˚
Î Ë q
Data:
x0 = 0
v0 = 0
(2)
q = 20 kg/s
50000
kg, t = 60 sec u = 4000 m/s
9.81
È
˘
Í
˙
ˆ
Ê 50000
50000
˙
- 60˜ ln
x = 0 + 0 + ( 4000) Í60 - Á
¯
Ë (9.81)(20)
Ê 50000
ˆ˙
Í
- (20)(60)˜
(9.81) Á
ÍÎ
Ë 9.81
¯ ˙˚
m0 =
= 30, 774.9 m
x = 30.8 km b
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143
PROBLEM 14.102
For the rocket of Problem 14.94, determine (a) the altitude at which all the fuel has been consumed, (b) the
velocity of the rocket at that time.
PROBLEM 14.94 A rocket has a mass of 1200 kg, including 1000 kg of fuel, which is consumed at the rate of
12.5 kg/s and ejected with a relative velocity of 4000 m/s. Knowing that the rocket is fired vertically from the
ground, determine its acceleration (a) as it is fired, (b) as the last particle of fuel is being consumed.
Solution
See Sample Problem 14.8 for derivation of
v = u ln
m0
m - qt
- gt = -u ln 0
- gt m0 - qt
m0
(1)
Note that g is assumed to be constant.
dy
Set dt = v in Eq. (1) and integrate with respect to time.
h
t
tÊ
ˆ
m0
h = Ú dy = Ú vdt = Ú Á u ln
- gt ˜ dt
0
0
0Ë
m0 - qt
¯
= -u Ú ln
m0 - qt
1
dt - gt 2
m0
2
z=
m0 - qt
m0
dz = -
h=
m0 u z
mu
1
1
ln z dz - gt 2 = 0 [( z ln z + z )]zz0 - gt 2
Ú
q z0
2
q
2
t
0
Let
q
dt
m0
or
dt = -
m0
dz
q
=
m0 u È m0 - qt Ê m0 - qt ˆ m0 Ê m0 ˆ ˘ 1 2
lnn
ln
- 1˜ - 1 ˙ - gt
Í
m0
q ÍÎ m0 ÁË
¯ m0 ÁË m0 ˜¯ ˙˚ 2
=
m0 u ÈÊ
qt ˆ Ê m0 - qt ˆ ˘ 1 2
- 1˜ + 1˙ - gt
ln
ÍÁ1 q ÍÎË m0 ˜¯ ÁË
m0
¯ ˙˚ 2
=
˘
È m - qt ˘ 1 2
m0 u È m0 - qt
- 1˙ - gt
- 1 + 1˙ - ut Íln 0
Íln
m0
q Î
m0
˚
Î
˚ 2
ˆ m - qt 1 2
Êm u
= ut + Á 0 - ut ˜ ln 0
- gt
¯
Ë q
m0
2
È Êm
ˆ
m0 ˘ 1 2
h = u Ít + Á 0 - t ˜ ln
˙ - gt ¯
m0 - qt ˚ 2
Î Ë q
(2)
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144
Problem 14.102 (continued)
m0 = 1200 kg
Data:
qt = mfuel = 1000 kg
q = 12.5 kg/s
u = 4000 m/s
g = 9.81 m/s2
m
1000
= 80 s
t = fuel =
q
12.5
(a)
From Eq. (2),
1200
È
˘ 1
ˆ
Ê 1200
2
h = ( 4000) Í80 - Á
- 80˜ ln
˙ - 2 (9.81)(80)
¯
Ë
12
.
5
1200
1000
Î
˚
= ( 4000)(80 - 16 ln 6) - 31, 392
= 173, 935 m
h = 173.9 km b
(b)
From Eq. (1),
1200 - 1000
- (9.81)(80)
1200
= 4000 ln 6 - 784.8
= 6382 m/s
v = -4000 ln
v = 6.38 km/s b
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145
PROBLEM 14.103
In a jet airplane, the kinetic energy imparted to the exhaust gases is wasted as far as propelling the airplane is
concerned. The useful power is equal to the product of the force available to propel the airplane and the speed
of the airplane. If v is the speed of the airplane and u is the relative speed of the expelled gases, show that the
mechanical efficiency of the airplane is h = 2v/(u + v ). Explain why h = 1 when u = v.
Solution
dm
Let F be the thrust force, and dt be the mass flow rate.
Absolute velocity of exhaust:
ve = u - v
Thrust force:
F=
Power of thrust force:
P1 = Fv =
Power associated with exhaust:
Total power supplied by engine:
dm
(u - v )v
dt
1
1
P2 ( Dt ) = ( Dm)ve2 = ( Dm)(u - v )2
2
2
1 dm
P2 =
( u - v )2
2 dt
P = P1 + P2
P=
=
Mechanical efficiency:
dm
(u - v )
dt
h=
h=
dm È
1
˘
( u - v ) v - ( u - v )2 ˙
dt ÍÎ
2
˚
1 dm 2
(u - v 2 )
2 dt
useful power P1
=
total power
P
2(u - v )v
u2 - v2
h=
2uv
b
(u 2 + v )
h = 1 when u = v. The exhaust, having zero velocity, carries no power away.
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146
PROBLEM 14.104
In a rocket, the kinetic energy imparted to the consumed and ejected fuel is wasted as far as propelling the
rocket is concerned. The useful power is equal to the product of the force available to propel the rocket and the
speed of the rocket. If v is the speed of the rocket and u is the relative speed of the expelled fuel, show that the
mechanical efficiency of the rocket is h = 2uv/(u 2 + v 2 ). Explain why h = 1 when u = v.
Solution
dm
Let F be the thrust force, and dt be the mass flow rate.
Absolute velocity of exhaust:
ve = u - v
Thrust force:
F=
Power of thrust force:
P1 = Fv =
Power associated with exhaust:
Total power supplied by engine:
Mechanical efficiency:
dm
u-v
dt
dm
(u - v )v
dt
1
1
P2 ( Dt ) = ( Dm)ve2 = ( Dm)(u - v )2
2
2
1 dm
P2 =
( u - v )2
2 dt
P = P1 + P2
P=
dm È
1
˘ 1 dm 2
uv - (u - v )2 ˙ =
(u - v 2 )
Í
dt Î
2
˚ 2 dt
h=
useful power P1
=
total power
P
h=
2uv
b
(u + v 2 )
2
h = 1 when u = v. The exhaust, having zero velocity, carries no power away.
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147
PROBLEM 14.105
A 30-g bullet is fired with a velocity of 480 m/s into
block A, which has a mass of 5 kg. The coefficient of
kinetic friction between block A and cart BC is 0.50.
Knowing that the cart has a mass of 4 kg and can roll
freely, determine (a) the final velocity of the cart and
block, (b) the final position of the block on the cart.
Solution
(a)
Conservation of linear momentum.
m0 v0 = ( m0 + mA )v ¢ = ( m0 + mA + mC )vf
(0.030 kg)( 480 m/s) = (5.030 kg)v ¢ = (9.030 kg)vf
0.030
( 480 m/s) = 2.863 m/s
5.030
0.030
( 480 m/s) = 1.5947 m/s
vf =
9.030
v¢ =
(b)
v f = 1.595 m/s b
T ¢ + U = T f : 20.615 - 24.672 x = 11.482
x = 0.370 m b
Work-energy principle.
Just after impact:
Final kinetic energy:
Work of friction force:
1
T ¢ = ( m0 + mA )v ¢ 2
2
1
= (5.030 kg)(2.863 m/s)2
2
= 20.615 J
1
1
Tf = ( m0 + mA + mC ) v 2f
2
2
1
= (9.030 kg)(1.5947 m/s)2
2
= 11.482 J
F = mk N
= m k ( m0 + mA ) g
= 0.50(5.030)(9.81)
= 24.672 N
Work = U = - Fx = -24.672 x
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148
PROBLEM 14.106
An 80-Mg railroad engine A coasting at 6.5 km/h strikes a 20-Mg flatcar C carrying a 30-Mg load B which can
slide along the floor of the car ( m k = 0.25). Knowing that the car was at rest with its brakes released and that
it automatically coupled with the engine upon impact, determine the velocity of the car (a) immediately after
impact, (b) after the load has slid to a stop relative to the car.
Solution
The masses are the engine ( mA = 80 ¥ 103 kg), the load ( mB = 30 ¥ 103 kg), and the flat car ( mC = 20 ¥ 103 kg).
Initial velocities:
( v A )0 = 6.5 km/h
= 1.80556 m/s
( v B )0 = ( vC )0 = 0.
No horizontal external forces act on the system during the impact and while the load is sliding relative to the
flat car. Momentum is conserved.
Initial momentum:
mA ( v A )0 + mB (0) + mC (0) = mA ( v A )0 (1)
(a)Let v' be the common velocity of the engine and flat car immediately after impact. Assume that the
impact takes place before the load has time to acquire velocity.
Momentum immediately after impact:
mA v ¢ + mB (0) + mC v ¢ = ( mA + mC )v ¢
(2)
Equating (1) and (2) and solving for v ¢,
v¢ =
mA ( v A ) 0
mA + mC
(80 ¥ 103 )(1.80556)
(100 ¥ 103 )
= 1.44444 m/s
=
v ¢ = 5.20 km/h
b
(b)Let vf be the common velocity of all three masses after the load has slid to a stop relative to the car.
Corresponding momentum:
mA v f + mB v f + mC v f = ( mA + mB + mC )vf (3)
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149
Problem 14.106 (continued)
Equating (1) and (3) and solving for vf ,
vf =
mA ( v A ) 0
mA + mB + mC
(80 ¥ 103 )(1.80556)
(130 ¥ 103 )
= 1.11111 m/s
=
vf = 4.00 km/h
b
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150
PROBLEM 14.107
Three identical freight cars have the velocities indicated. Assuming that car B is first hit by car A, determine
the velocity of each car after all the collisions have taken place if (a) all three cars get automatically coupled,
(b) cars A and B get automatically coupled while cars B and C bounce off each other with a coefficient of
restitution e = 0.8.
Solution
v A = 10 km/h
vC = 8km/h
All cars are coupled.
Conservation of linear momentum. Components
+
(a)
vB = 0
mA v A + 0 - mC vC = ( mA + mB + mC )vF
vF =
mA v A - mC vC 1
1
1
= v A - vC = (10 - 8)
mA + mB + mC 3
3
3
vF =
2
km/h
3
v A = v B = vC = 0.667 km/h
(b)
b
Cars A and B hit and couple.
Conservation of linear momentum for cars A and B.
mA v A + 0 = ( mA + mB )v AB
v AB =
mA
Ê 1ˆ
v A = Á ˜ (10) = 5 km/h
Ë 2¯
mA + mB
Cars A and B are impacted by car C.
Conservation of linear momentum for unit AB and car C.
( mA + mB )v AB - mC vC = ( mA + mB )v AB
¢ + mC vC¢
1
1
v AB - vC = v AB
¢ + vC¢
2
2
1
5 - (8)) = v AB
¢ + 0.5vC¢
2
(1)
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151
Problem 14.107 (continued)
Coefficient of restitution.
vC¢ - v AB
¢ = e( v AB + vC )
= (0.8)(5 + 8) (2)
Solving Eqs. (1) and (2) simultaneously,
v AB
¢ = -2.8 km/h
vC = 7.6 km/h
v A = v B = 2.8 km/h
vC = 7.6 km/h
b
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152
PROBLEM 14.108
A 4500 kg helicopter A was traveling due east in level flight at a speed of 120 km/h and at an altitude of 750 m
when it was hit by a 6000 kg helicopter B. As a result of the collision, both helicopters lost their lift, and their
entangled wreckage fell to the ground in 12 s at a point located 450 m east and 115.2 m south of the point of
impact. Neglecting air resistance, determine the velocity components of helicopter B just before the collision.
Solution
Let the x axis point east, the y axis point up, and the z axis point south. The origin of the xyz coordinate system
is the point of the collision.
At the time of the collision,
t=0 x=0
y=0 z=0
At the time of impact with the ground,
t = 12 s x = 450 m
y = -750 m
z = 115.2 m
Let v ¢ with components v x¢ , v y¢ , v z¢ be the velocity of the entangled wreckage just after the collision. The
entangled wreckage falls as a projectile.
x = 0 + v x¢ t :
450 = ( v x¢ )(12)
v x¢ = 37.5 m/s
y = 0 + v y¢ t -
1 2
gt :
2
z = 0 + v z¢ t :
1
-750 = ( v y¢ )(12) - (9.81)(12)2
2
v y¢ = -3.64 m/s
115.2 = ( v z¢ )(12)
v z¢ = 9.6 m/s
v ¢ = (37.5 m/s)i - (3.64 m/s) j + (9.6 m/s)k
Let vA and vB be the velocities of helicoptors A and B, respectively, before the collision.
ˆ
Ê 100
vA = ( 120 km/h)i = Á
m/s ˜ i
¯
Ë 3
vB = ?
Conservation of linear momentum before and after the collision.
mA vA + mB vB = ( mA + mB )v ¢
m + mB
m
vB = A
v ¢ - A vA
mB
mB
Data:
mA + mB 4500 + 6000
=
= 1.75
mB
6000
mA 4500
=
= 0.75
mB 6000
Ê 100 ˆ
vB = (1.75)( 37.5 i - 3.64 j + 9.6 k ) - 0.75 Á
i
Ë 3 ˜¯
= ( 40.625 m/s) i - (6.37 m/s) j + (16.8 m/s) k
= (146.25 km/h) i - (22.932 km/h) j + (60.48 km/h) k
Components of vB :
146.3 km/h
60.5 km/h
22.9 km/h
b
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153
PROBLEM 14.109
A 7.5 kg block B is at rest and a spring of constant
k = 15000 N/m is held compressed 75 mm by a cord.
After 2.5-kg block A is placed against the end of the
spring, the cord is cut, causing A and B to move.
Neglecting friction, determine the velocities of
blocks A and B immediately after A leaves B.
Solution
mA = 2.5 kg
mB = 7.5 kg
k = 15000 N/m
e = 75 mm = 0.075 m
h = 150 mm = 0.15 m
Conservation of linear momentum.
+
Horizontal components
:
0 + 0 = mA v A - mB v B
vB =
mA
1
vA = vA
mB
3
Conservation of energy.
State 1.
1 2 1
ke = (15000)(0.075)2 = 42.1875 J
2
2
V1g = 0
V1e =
T1 = 0
State 2.
V2e = 0
V2 g = W A h = (2.5)(9.81)(0.15) = 3.67875 J
T2 =
1
1
mA v 2A + mB v B2
2
2
2
1
1
5
Êv ˆ
= (2.5)v 2A + (7.5) Á A ˜ = v 2A
Ë
¯
2
2
3
3
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154
Problem 14.109 (continued)
T1 + V1 = T2 + V2 :
5
0 + 42.1875 = v 2A + 3.67875
3
v 2A = 23.10525 m2
vA = 4.81 m/s
b
vB = 1.602 m/s
b
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155
PROBLEM 14.110
A 9-kg block B starts from rest and slides down the inclined
surface of a 15-kg wedge A, which is supported by a horizontal
surface. Neglecting friction, determine (a) the velocity of B
relative to A after it has slid 0.6 m down the surface of the
wedge, (b) the corresponding velocity of the wedge.
Solution
We resolve vB into its components: vA
and vB /A
30∞
Impulse-momentum principle.
Smv 0 + SFt = Smv
x component:
C + 0 = mB v B /A cos 30∞ - mA v A - mB v A
+
vA =
mB cos 30∞
(9 kg) cos 30∞
v B /A =
v B /A
mA + v B
15 kg + 9 kg
v A = 0.32476v B /A
(1)
Conservation of energy.
T0 = 0
V0 = mB gh
= (9 kg)(9.81 m/s2 )(0.6 m)sin 30∞
= 26.487 J
1
1
mA v 2A + mB v B2
V =0
2
2
Referring to figures shown above and using the law of cosines:
1
1
T = mA v 2A + mB v 2A + v B2 /A - 2v A v B /A cos 30∞
2
2
T=
(
)
Recalling Eq. (1) and substituting the given data:
1
1
T = (15)(0.32476)2 v B2 /A + (9)[(0.32476)2 + 1 - 2(0.32476) cos 30∞] v B2 /A
2
2
2
= 0.79102 v B /A + 2.04336 v B2 /A = 3.2344 v B2 /A
T + V = T0 + V0 :
3.2344 v B2 /A = 26.487 J
v B /A = 2.8617 m/s
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156
Problem 14.110 (continued)
vB /A = 2.86 m/s
(a)
(b)
30∞ b
From Eq. (1):
v A = 0.32476(2.8617 m/s)
= 0.92936 m/s
vA = 0.929 m/s
b
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157
PROBLEM 14.111
A mass q of sand is discharged per unit time from a conveyor belt moving with
a velocity v0. The sand is deflected by a plate at A so that it falls in a vertical
stream. After falling a distance h, the sand is again deflected by a curved plate
at B. Neglecting the friction between the sand and the plates, determine the
force required to hold in the position shown (a) plate A, (b) plate B.
Solution
(a)When the sand impacts on plate A, it is momentarily brought to rest. Apply the principle of impulse and
momentum to find the force on the sand.
( Dm)v0 + Ax ( Dt ) = 0
x component:
+
Ax = -
Dm
v0 = - qv0
Dt
+
0 + Ay ( Dt ) = 0
y component:
Ay = 0
A = qv0
b
The sand falls vertically. Use conservation of energy for mass element Dm. Let v be the speed at the
curved portion of plate B.
1
T1 + V1 = T2 + V2 : 0 + ( Dm) gh = ( Dm)v 2 + 0
2
v 2 = 2 gh
v = 2 gh
Over the curved portion of plate B, there is negligible change of elevation. Hence, by conservation of
energy, v is both the entrance speed and exit speed of the curved portion of plate B.
(b)
Force exerted through plate B.
Entrance velocity:
v = - 2gh j
Exit velocity:
v ¢ = 2 gh ( - cos 30∞i - sin 30∞ j)
Mass flow rate:
dm Dm
=
=q
dt
Dt
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158
Problem 14.111 (continued)
Principle of impulse and momentum.
( Dm)v + B( Dt ) = ( Dm)v ¢
Ê Dm ˆ
B=Á
( v ¢ - v) = q 2 gh ( - cos 30∞i - sin 30∞ j + j)
Ë Dt ˜¯
3
2 gh
2
1
By = 2 gh (1 - sin 30∞) =
2 gh
2
Bx = - 2 gh cos 30∞ = -
B = 2gh
30∞ b
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159
PROBLEM 14.112
The final component of a conveyor system receives
sand at a rate of 100 kg/s at A and discharges it at B.
The sand is moving horizontally at A and B with a
velocity of magnitude v A = v B = 4.5 m/s. Knowing
that the combined weight of the component and of
the sand it supports is W = 4 kN, determine the
reactions at C and D.
Solution
Apply the impulse-momentum principle.
+
Moments about C:
-(0.9)( D m) v A + 3D ( D t ) - 1.8W ( D t ) = -(1.65)( D m)v B
1.8
1 Dm
W+
(0.9v A - 1.65v B )
3
3 Dt
(1.8)( 4000) 1
=
+ (100)[(0.9)( 4.5) - (1.65)( 4.5)] = 2287.5 N
3
3
D=
D = 2.29 kN b
x components:
( D m) v A + C x ( D t ) = ( D m)v B
+
Ê Dmˆ
Cx = Á
( v B - v A ) = (100)( 4.5 - 4.5) = 0
Ë D t ˜¯
+
y components:
0 + C y ( D t ) + D( D t ) - W ( D t ) = 0
C y = W - D = 4000 - 2287.5 = 1712.5 N
C = 1.712 kN b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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160
PROBLEM 14.113
A garden sprinkler has four rotating arms, each of which consists
of two horizontal straight sections of pipe forming an angle of 120°
with each other. Each arm discharges water at a rate of 20 L/min
with a velocity of 18 m/s relative to the arm. Knowing that the
friction between the moving and stationary parts of the sprinkler
is equivalent to a couple of magnitude M = 0.375 N ◊ m, determine
the constant rate at which the sprinkler rotates.
Solution
The flow through each arm is 20 L/min.
20 L/min 1 min
Q=
¥
= 333.33 ¥ 10 -6 m3 /s
1000 L/m3 60 s
dm
= rQ = (1000 kg/m3 )(333.33 ¥ 10 -6 )
dt
= 0.33333 kg/s
Consider the moment about O exerted on the fluid stream of one arm. Apply the impulse-momentum principle.
Compute moments about O. First, consider the geometry of triangle OAB. Using first the law of cosines,
(OA)2 = 1502 + 1002 - (2)(150)(100) cos120∞
OA = 217.95 mm = 0.21795 m
Law of sines.
sin b sin 120∞
=
100
217.95
b = 23.413∞, a = 60∞ - b = 36.587∞
+ Moments about O:
( Dm)( vO )(0) + M O ( Dt ) = (OA)( Dm)vs sin a - (OA)( Dm)(OA)w
Dm
[(OA)vs sin a - (OA)2 w ]
MO =
Dt
= (0.33333)[(0.21795)(18)sin 36.5877∞ - (0.21795)2 w ]
= 0.77945 - 0.015834w
Moment that the stream exerts on the arm is -M O .
Balance of the friction couple and the four streams
MF - 4 M O = 0
0.375 - 4(0.77945 - 0.015834w ) = 0
w = 43.305 rad/s
w = 414 rpm b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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161
PROBLEM 14.114
The ends of a chain lie in piles at A and C. When given an initial speed v, the
chain keeps moving freely at that speed over the pulley at B. Neglecting friction,
determine the required value of h.
Solution
Let r be the mass per unit length of chain. Consider the impulse-momentum applied to the link being brought
to rest at Point C.
Calculation of Dm.
Dm = r( Dl ) = r v( Dt )
Impulse-momentum principle:
-( Dm)v + C Dt = 0
- rv( Dt )v + C Dt = 0
C = rv 2
Impulse-momentum applied to the moving portion of the chain. Consider only the changes in momentum and
forces contributing to moments about O in the diagram.
+
Moments about O:
( Dm)v + [ r gh - C ]Dt = ( Dm)v
C = r gh
Equating the two expressions for C,
r v 2 = r gh
h=
v2
b
g
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or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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162
PROBLEM 14.115
A railroad car of length L and mass m0 when empty is moving
freely on a horizontal track while being loaded with sand from
a stationary chute at a rate dm/dt = q. Knowing that the car was
approaching the chute at a speed v0 , determine (a) the mass of
the car and its load after the car has cleared the chute, (b) the
speed of the car at that time.
Solution
Consider the conservation of the horizontal component of momentum of the railroad car of mass m0 and the
sand mass qt.
m0 v0 = ( m0 + qt )v
v=
m0 v0
m0 + qt
(1)
mv
dx
=v= 0 0
dt
m0 + qt
x0 = 0
Integrating, using
tL
tL
0
0
L = Ú vdt = Ú
=
x = L when t = t L ,
and
m0 v0
mv
dt = 0 0 [ln ( m0 + qt L ) - ln m0 ]
m0 + qt
q
m0 v0 m0 + qt L
ln
m0
q
ln
m0 + qt L
qL
=
m0
m0 v0
(a)
Final mass of railroad car and sand
(b)
Using Eq. (1),
vL =
m0 + qt L
= e qL/m0 v0
m0
m0 + qt L = m0 e qL/m0 v0 b
m0 v0
mv
= 0 0 e - qL/m0 v0
m0 + qt L
m0
vL = v0 e - qL/m0 v0 b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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163
PROBLEM 14.116
A possible method for reducing the speed of a training plane as
it lands on an aircraft carrier consists in having the tail of the
plane hook into the end of a heavy chain of length l which lies
in a pile below deck. Denoting by m the mass of the plane and
by v0 its speed at touchdown, and assuming no other retarding
force, determine (a) the required mass of the chain if the
speed of the plane is to be reduced to bv0 , where b 1, (b) the
maximum value of the force exerted by the chain on the plane.
Solution
Let m¢ be the mass of chain per unit length and x be the distance traveled in time t. Let v be the velocity of
the plane at time t. Apply the principle of conservation of linear momentum.
+
(a)
mv0 = ( m + m¢ x )v
x-component:
v=
mv0
m + m¢x
bv0 =
mv0
m + m¢ l
(1)
Speed is reduced to bv0 when x = l.
m¢ l =
Mass of chain:
(b)
1- b
mb
b
Maximum force on the plane.
Acceleration of plane:
a=
( - m¢ mv0 )
mv0
m¢ m2 v02
dv
dv
=v
=
◊
=
dt
dx m + m¢ x ( m + m¢ x )2
( m + m ¢ x )2
Force acting on the plane:
F = ma = -
m¢ m3 v02
( m + m ¢ x )3
| Fmax | occurs when x = 0.
| Fmax | = m¢ v02 =
1 - b mv02
l
b
| Fmax | =
(1 - b )mv02
b
bl
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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164