Unit 2 Review, pages 274–281
Knowledge
1. (a)
2. (c)
3. (b)
4. (d)
5. (d)
6. (c)
7. (c)
8. (b)
9. (d)
10. (a)
11. (b)
12. (c)
13. (b)
14. (a)
15. (c)
16. (b)
17. (b)
18. (b)
19. (b)
20. (b)
21. False. Gravity does negative work on a mountaineer as she climbs up the side of a cliff.
22. True
23. True
24. True
25. False. A diver jumps from a diving board. If the work done by air resistance is negligible, at
any given moment until she enters the water the sum of her kinetic energy and her gravitational
potential energy is constant.
26. True
27. False. As you expand a spring, the force necessary to continue pulling it apart increases.
28. True
29. False. The spring constant, k, has no units.
30. True
31. True
32. False. A ball bouncing up and down due to gravity is in damped harmonic motion.
33. False. Gravitational potential energy depends on an object’s elevation, but elastic potential
energy does not depend on an object’s elevation.
34. True
35. False. The amplitude in simple harmonic motion is the maximum displacement of the object
from its equilibrium position.
36. False. A larger spring constant means a stronger pull or push of the spring for a given
displacement.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-16
Understanding
37. No, you would not do the same work to lift a 2 kg box vertically through 1.5 m on the Moon
1
as you would to lift it on Earth. The value of g on the Moon is the value of g on Earth.
6
1
Therefore, the weight of the box is less, and of the work is done.
6
38. No, when a box slides down an inclined plane, increasing in speed, the normal force does not
do any work on the box. The normal force is perpendicular to the displacement of the box. It is
the force of gravity that does work on the box.
39. When a push is applied to an object, and the object undergoes a displacement but its speed
does not increase, you can conclude that the net work done on the object is zero because the
object moves at constant speed. Therefore, another force must have been present to do work
opposite the work done by the push.
40. Given: F = 50 N; ∆d = 20 m
Required: W
Analysis: W = F∆d
Solution: W = F!d
= (50 N)(20 m)
W = 1" 103 J
Statement: The work done by the force is 1 ! 103 J.
41. Answers may vary. Sample answer: The child does work by pumping his legs and raising and
lowering his body to initiate simple harmonic motion. If he supplies energy rhythmically,
resonance is produced and the child swings with an increasing amplitude. As he pumps, chemical
potential energy in his muscles is transformed into kinetic energy, and as he swings higher, he
gains gravitational potential energy. As the child swings downward, gravitational potential
energy is transformed into kinetic energy. As the child swings higher, the child’s gravitational
potential energy at the top of each swing increases, and the amount of kinetic energy it is
transformed into at the bottom of each swing also increases.
42. Given: m = 250 kg; ∆y = 2.1 m; ∆t = 2.4 s; g = 9.8 m/s2
Required: P
Analysis: The work done by the weightlifter is equal to the gravitational potential energy at the
W
mg!y
top of the lift, W = mg∆y. Power is work divided by time, P =
. So, P =
.
!t
!t
mg!y
Solution: P =
!t
(250 kg)(9.8 m/s 2 )(2.1 m)
=
2.4 s
3
P = 2.1 " 10 W
Statement: The power output of the weightlifter is 2.1 ! 103 W .
43. The blocks do not move, so there is no work done. Therefore, the change in energy is zero.
44. (a) Given: Fk = 5.0 N; ∆d = 29 m
Required: Wf
Analysis: W = Fk∆d
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-17
Solution: W = Fk !d
= ("5.0 N)(29 m)
W = "145 J (one extra digit carried)
Statement: The work done by friction is !1.4 " 102 J .
(b) Given: m = 41 kg; Fk = −5.0 N; ∆d = 29 m; Wf = 145 J
Required: vi
Analysis: The kinetic energy of the block before it slows down is equal to the work done by
friction on the block to slow it down to a stop.
1
1
Wk = Ek and Ek = mvi2 , so Wk = mv 2 . Rearrange this equation to isolate vi.
2
2
vi =
2Wk
m
Solution: vi =
=
2Wk
m
2(145 J)
41 kg
vi = 2.7 m/s
Statement: The initial speed of the block was 2.7 m/s.
45. Fuel economy is the ratio of distance travelled per unit of fuel consumed. The mid-sized car
has a lower fuel economy than the compact car because the mass of the mid-sized car is much
more than that of the compact car, so more force, and thus more energy, is required to move the
mid-sized car the same distance.
46. (a) Given: m = 150 kg; ∆y = 25 m; g = 9.8 m/s2
Required: Eg
Analysis: Eg = mg∆y
Solution: Eg = mg!y
= (150 kg)(9.8 m/s 2 )(25 m)
Eg = 3.7 " 104 J
Statement: The gravitational potential energy of the wrecking ball is 3.7 × 104 J.
(b) The kinetic energy is equal to the gravitational potential energy, so it is also 3.7 × 104 J.
47. Given: m = 91 000 kg; v = 980 km/h; ∆y = 12 km = 1.2 × 104 m; g = 9.8 m/s2
Required: ET
1
1
Analysis: ET = Ek + Eg , Eg = mg∆y, and Ek = mv 2 , so ET = mv 2 + mg!y .
2
2
First, convert the speed to metres per second:
1h
km 1000 m
!
!
= 272 m/s (one extra digit carried)
v = 980
3600 s
h
1 km
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-18
1 2
mv + mg!y
2
1
= (91 000 kg)(272 m/s)2 + (91 000 kg)(9.8 m/s 2 )(1.2 " 104 m)
2
ET = 1.4 " 1010 J
Solution: ET =
Statement: The total energy is 1.4 × 1010 J.
48. Given: m = 3.6 g = 3.6 × 10−3 kg; v = 6.5 m/s; g = 9.8 m/s2
Required: ∆y
1
Analysis: Ek = Eg; Eg = mg∆y; Ek = mv 2 ; solve for ∆y:
2
Ek = Eg
1
mv 2 = mg!y
2
v2
!y =
2g
Solution: !y =
=
=
v2
2g
(6.5 m/s)2
2(9.8 m/s 2 )
(6.5)2 m 2 / s 2
2(9.8 m/ s 2 )
!y = 2.2 m
Statement: The marble must be dropped from a height of 2.2 m.
49. (a) Given: m = 110 kg; ∆y = 12 m; g = 9.8 m/s2
Required: !Eg
Analysis: !Eg = "mg!y
Solution: !Eg = "mg!y
= "(110 kg)(9.8 m/s 2 )(12 m)
!Eg = "1.3 # 104 J
Statement: The magnitude of the change in gravitational potential energy is 1.3 × 104 J.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-19
(b) Given: m = 110 kg; ∆y = 12 m; g = 9.8 m/s2
Required: v
1
Analysis: Eg = Ek; Eg = mg∆y; Ek = mv 2 ; solve for v:
2
Eg = Ek
mg!y =
1
mv 2
2
v = 2g!y
Solution: v = 2g!y
= 2(9.8 m/s 2 )(12 m)
v = 15 m/s
Statement: The climber’s speed is 15 m/s. You need to assume that the rope does not stretch, so
there is no elastic potential energy.
50. As the ball rises, the kinetic energy decreases as the velocity decreases. At the top of the
ball’s arc, the kinetic energy and the velocity are zero.
51. The suspension spring of a car has a larger spring constant than a spring on the screen door of
a house because the suspension spring of a car must support a large weight. The spring on a door
is designed to stretch when the door is opened under the action of small forces and compress
when the door is closed, holding it shut, so it has a smaller spring constant.
52. Given: m = 0.021 kg; k = 160 N/m; ∆x = 0.13 m; g = 9.8 m/s2
Required: ∆y
Analysis: The elastic potential energy when the spring is compressed is equal to the gravitational
potential energy at the maximum height of the ball.
1
Ee = Eg; Ee = k(!x)2 ; Eg = mg∆y; solve for Δy:
2
Ee = Eg
1
k(!x)2 = mg!y
2
k(!x)2
!y =
2mg
Solution: !y =
=
k(!x)2
2mg
(160 N/m)(0.13 m)2
2(0.021 kg)(9.8 m/s 2 )
!y = 6.6 m
Statement: The ball rises to a height of 6.6 m.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-20
53. (a) Given: m = 0.43 kg; ∆y = 3.0 m
!
Required: p
Analysis: The kinetic energy when the ball hits the player’s chest the second time is equal to the
gravitational potential energy of the ball at its maximum height. Solve the equation
1
mg!y = mv 2 for v:
2
1
mg!y = mv 2
2
v = 2g!y
Then, use the equation p = mv to determine p.
Solution: v = 2g!y
= 2(9.8 m/s 2 )(3.0 m)
v = 7.668 m/s (two extra digits carried)
p = mv
= (0.43 kg)(7.668 m/s)
p = 3.3 kg ! m/s
Statement: The momentum of the ball is 3.3 kg·m/s [down].
(b) Given: F = 31 N [up]; ∆t = 0.2 s; m = 0.43 kg; vi = 7.668 m/s
Required: vf
Analysis: F∆t = m∆v
Solution: F!t = m!v
F!t = m(vf " vi )
F!t + mvi
m
(31 N)(0.2 s) + (0.43 kg)("7.668 m/s)
=
0.43 kg
vf = 6.8 m/s
Statement: The final velocity is 6.8 m/s [up].
54. Given: m = 78 kg; initial height above water = 69 m; length of cord = 39 m;
final height above water = 6.0 m
Required: impulse of stretching cord
Analysis: Impulse is equal to change in momentum, or ∆p = m∆v. Determine the momentum of
the jumper just before the cord begins to stretch, when the length of the cord is 39 m. Determine
1
the initial velocity, vi, using the conservation of energy: mg(length of cord) = mvi2 . When the
2
cord is stretched as far as possible, the jumper's final velocity, vf, is zero, so the jumper's
momentum is also 0.
vf =
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-21
1
mv 2
2 i
1
(9.8 m/s 2 )(39 m) = vi2
2
Solution: mg(length of cord) =
vi = 2(9.8 m/s 2 )(39 m)
!p = m!v
= m(vf " vi )
vi = 27.65 m/s [down] (two extra digits carried)
= (78 kg)(0 " ("27.65 m/s)]
!p = 2.2 # 103 N $ s
Statement: The impulse is 2.2 ! 103 N " s [up].
55. Answers may vary. Sample answer:
Given: graph of force versus time
Required: impulse
Analysis: Estimate the average force from the graph, and then multiply by the change in time.
Solution: From the graph, estimate the average force to be about 3.0 N. The total time is 40 ms,
or 0.04 s. Thus, the impulse is (3.0 N)(0.04 s) = 0.12 N·s.
Statement: The impulse is 0.12 N·s.
56. Researchers thought that energy and momentum were not conserved in certain radioactive
decays, so Pauli hypothesized that a new particle carried the missing energy and momentum.
This particle became known as the neutrino.
57. When a grapefruit is tossed across a room, if we ignore air friction, the horizontal component
of the linear momentum remains unchanged because no horizontal external force is acting on the
grapefruit.
58. Yes, a wet snowball colliding with a stationary parked car is an example of an inelastic
collision, because the car is stationary and the snowball sticks to the car on impact.
59. Answers may vary. Sample answer:
(a) An example of a collision in everyday life where one of the objects is at rest after the
collision is when the cue ball hits the triangle of billiard balls for the break, and the cue ball stops
and the other balls scatter.
(b) An example of a collision where both objects are at rest after the collision is when a snowball
hits a wall and sticks.
!
!
60. Given: m1 = 1850 kg; vi 1 = 26 m/s [E]; m2 = 1200 kg; vf = 6.5 m/s [E]
!
Required: vi 2
Analysis: The momentum before and after the collision is equal.
!
!
!
m1vi 1 + m2 vi 2 = (m1 + m2 )vf
!
!
(m1 + m2 )vf ! m1vi 1
!
vi 2 =
m2
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-22
!
!
(m1 + m2 )vf ! m1vi 1
!
Solution: vi 2 =
m2
=
(1850 kg + 1200 kg )(6.5 m/s) ! (1850 kg )(26 m/s)
1200 kg
!
vi 2 = !24 m/s
Statement: The initial velocity of car 2 is 24 m/s [W].
!
!
!
61. Given: m1 = 46 kg; v1 = 0 m/s ; a = 3.4 m/s 2 ; ∆t = 2.7 s; m2 = 56 kg; vi 2 = 0 m/s
!
Required: vf
Analysis: Determine the velocity of the first hockey player just before the collision using the
! !
!
appropriate equation of motion, vi 1 = v1 + a!t . Then, use conservation of momentum to
!
determine vf .
!
! !
Solution: vi 1 = v1 + a!t
= 0 m/s + (3.4 m/s 2 )(2.7 s)
!
vi 1 = 9.18 m/s (one extra digit carried)
!
!
!
m1vi 1 + m2 vi 2 = (m1 + m2 )vf
!
!
! m1vi 1 + m2 vi 2
vf =
(m1 + m2 )
(46 kg)(9.18 m/s)
46 kg + 56 kg
!
vf = 4.1 m/s
Statement: The final velocity of the hockey players is 4.1 m/s in the same direction as the initial
velocity of the first player before the collision.
62. (a) Given: m1 = 1.5 kg; m2 = 3.5 kg; vi 1 = 12 m/s; vi 2 = −7.5 m/s
Required: vf
Analysis: Use conservation of momentum and solve for vf.
!
!
!
m1vi 1 + m2 vi 2 = (m1 + m2 )vf
!
!
! m1vi 1 + m2 vi 2
vf =
(m1 + m2 )
!
!
! m1vi 1 + m2 vi 2
Solution: vf =
(m1 + m2 )
=
(1.5 kg)(12 m/s) + (3.5 kg)(!7.5 m/s)
1.5 kg + 3.5 kg
!
vf = !1.65 m/s (one extra digit carried)
Statement: The final velocity of the objects is 1.6 m/s [left].
=
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-23
(b) Given: m1 = 1.5 kg; m2 = 3.5 kg; vi 1 = 12 m/s; vi 2 = −7.5 m/s; vf = −1.65 m/s
Required: ∆Ek
1
1
1
Analysis: !Ek = Ek i " Ek f = m1vi21 + m2 vi22 " (m1 + m2 )vf2
2
2
2
#1
&
1
1
Solution: !Ek = (m1 + m2 )vf2 " % m1vi21 + m2 vi22 (
2
2
$2
'
=
#1
&
1
1
(1.5 kg + 3.5 kg)("1.65 m/s)2 " % (1.5 kg)(12 m/s)2 + (3.5 kg)("7.5 m/s)2 (
2
2
$2
'
!Ek = "2.0 ) 102 J
Statement: The kinetic energy lost in the collision is 2.0 × 102 J.
Analysis and Application
!
63. Given: ∆d = 12 m; F = 480 N [21° above the horizontal]
Required: W
Analysis: W = Fx∆d; Fx = F cos 21°
Solution: W = Fx !d
= F(cos 21°)(!d)
= (480 N)(cos 21°)(12 m)
W = 5.4 " 103 J
Statement: The work done by the child is 5.4 × 103 J.
64. Given: ∆d = 200.0 m; F = 0.25 N
Required: W
Analysis: W = F∆d
Solution: W = F!d
= (0.25 N)(200.0 N)
W = 5.0 " 101 J
Statement: The father does 5.0 × 101 J of work on the stroller.
65. Given: F = 9.3 N; W = 87 J; ∆d = 11 m
Required: θ
Analysis: Rearrange W = Fx∆d to solve for Fx :
W
!d
Then, use Fx = F cos θ to determine θ.
W
Solution: Fx =
!d
87 J
=
11 m
Fx = 7.9 N
Fx =
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-24
Fx = F cos!
cos! =
Fx
F
#F &
! = cos "1 % x (
$ F'
# 7.9 N &
= cos "1 %
$ 9.3 N ('
! = 32°
Statement: The angle between the applied force and the horizontal is 32°.
66. Given: vi = 25 m/s; vf = 0 m/s; a = −8.0 m/s2
Required: ∆d
Analysis: Use the appropriate equation of motion, vf2 = vi2 + 2a!d , rearranged to solve for ∆d.
vf2 = vi2 + 2a!d
!d =
vf2 " vi2
2a
vf2 " vi2
2a
(0 m/s)2 " (25 m/s)2
=
2("8.0 m/s 2 )
Solution: !d =
=
(0)2 m 2 / s 2 " (25)2 m 2 / s 2
2("8.0 m/ s 2 )
!d = 39 m
Statement: The car travels 39 m before it stops.
67. Given: F = 52 N; θ = 13°; ∆d = 3.8 m
Required: W
Analysis: W = Fx∆d; Fx = F cos θ
Solution: W = Fx !d
= F cos" (!d)
= (52 N)(cos 13°)(3.8 m)
W = 1.9 # 10 2 J
Statement: The work done by the applied force is 1.9 ! 102 J .
68. Given: m = 18 kg; ∆y = 5.1 m
Required: W
Analysis: W = mg∆y
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-25
Solution: W = mg!y
= (18 kg)(9.8 m/s 2 )(5.1 m)
W = 9.0 " 102 J
Statement: The work done to lift the box is 9.0 × 102 J.
69. Given: m = 73 kg; vi = 4.2 m/s; vf = 0 m/s; θ = 10°
Required: d
Analysis: First determine the height of the hill, ∆y, using conservation of energy, and then
determine the distance along the hill, d, using trigonometry.
1
!y
mg!y = mvi2 ;
= sin "
2
d
1
Solution: mg!y = mvi2
2
v2
!y = i
2g
=
=
(4.2 m/s)2
2(9.8 m/s 2 )
(4.2)2 m 2 / s 2
2(9.8 m/ s 2 )
!y = 0.90 m
!y
= sin "
d
!y
d=
sin "
0.90 m
=
sin 10°
d = 5.2 m
Statement: The skier slides 5.2 m up the hill before stopping.
70. Given: m = 2.0 kg; ∆y = 10.0 m
Required: v
Analysis: Use conservation of energy:
1
mg!y = mv 2
2
v = 2g!y
Solution: v = 2g!y
= 2(9.8 m/s 2 )(10.0 m)
v = 14 m/s
Statement: The stone’s speed when it hits the ground is 14 m/s.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-26
71. Given: ∆y = 4.0 m; Eg = 2.7 kJ = 2.7 × 103 J
Required: m
Analysis: Eg = mg!y
m=
Solution: m =
=
Eg
g!y
Eg
g!y
2.7 " 103 kg # m 2 / s 2
(9.8 m/ s 2 )(4.0 m )
m = 69 kg
Statement: The mass of the pole vaulter is 69 kg.
72. No, it is not possible for a rubber ball to be dropped from shoulder height and then bounce to
a height above your head because then the potential energy after the bounce would be greater
than the potential energy when the ball is dropped, which violates the law of conservation of
energy.
73. (a) Given: ∆y = 41 m; vi = 22 m/s; θ = 37°
Required: vf
Analysis: Let down be positive. Use conservation of energy to determine the speed just before
the baseball hits the ground.
1
1
mg!y + mvi2 = mvf2
2
2
vf = 2g!y + v 2y
Solution: vf = 2g!y + v 2y
= 2(9.8 m/s 2 )(41 m) + ("22 m/s)2
vf = 36 m/s
Statement: The speed of the ball just before it hits the ground is 36 m/s.
(b) The speed does not change. The angle does not matter when calculating kinetic energy.
74. (a) Given: m = 55 kg; r = ∆y = 4.0 m
Required: Eg top
Analysis: Eg = mg∆y
Solution: Eg = mg!y
= (55 kg)(9.8 m/s 2 )(4.0 m)
= 2156 J (two extra digits carried)
Eg = 2.2 " 103 J
Statement: The gravitational potential energy at the top of the half-pipe is 2.2 ! 103 J .
(b) The kinetic energy at the bottom of the half-pipe is equal to the gravitational potential energy
at the top of the half-pipe, or 2.2 ! 103 J .
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-27
(c) Given: m = 55 kg; ∆y = 2.0 m; ET = 2156 J
Required: Eg; Ek
Analysis: The sum of the gravitational potential energy and the kinetic energy at a height of
2.0 m is equal to the total energy, ET.
Eg = mg∆y; ET = Eg + Ek
Solution: Eg = mg!y
= (55 kg)(9.8 m/s 2 )(2.0 m)
Eg = 1078 J (two extra digits carried)
ET = Eg + Ek
Ek = ET ! Eg
= 2156 J ! 1078 J
= 1078 J
Ek = 1.1 " 103 J
Statement: The kinetic energy and the gravitational potential energy are both 1.1 ! 103 J.
75. A meteorite is moving at a tremendously high velocity. When it hits the ground, it produces a
high pressure, high velocity shock wave. Impact-generated shock waves transfer some of their
kinetic energy as thermal energy to the rocks through which they pass. At very high pressures,
such as in a meteor impact, the resulting higher temperatures transfer enough thermal energy to
cause significant melting.
76. Given: flow rate = 20 kg/s; v = 30 m/s
Required: F
m!v
Analysis: The impulse equation is F∆t = ∆p = m∆v, so F =
. The flow rate is the mass over
!t
m
the change in time, or
, and ∆v = 0 m/s − 30 m/s = −30 m/s. Multiply the flow rate by the
!t
change in velocity to get the force.
m!v
Solution: F =
!t
" m%
= $ ' !v
# !t &
= (20 kg/s)((30 m/s)
F = (6 ) 102 N
Statement: The force is 6 ! 102 N in the opposite direction to the flow of water.
77. Given: m = 39 kg; vi = 3.4 m/s; ∆y = 11 m; ∆x = 11 m; Fk = 22 N
Required: vf
Analysis: The kinetic energy at the bottom of the hill is equal to the gravitational potential
energy plus the kinetic energy at the top of the hill minus the work done by friction. The distance
along the hill, ∆d, can be found using the Pythagorean theorem with ∆x and ∆y.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-28
1 2
1
mvf = mg!y + mvi2 " Fk !d
2
2
2
2
(∆d) = (∆x) + (∆y)2
Solution: (!d)2 = (!x)2 + (!y)2
!d = (!x)2 + (!y)2
= (11 m)2 + (11 m)2
!d = 15.56 m (two extra digits carried)
1 2
1
mvf = mg!y + mvi2 " Fk !d
2
2
2
mvf = 2mg!y + mvi2 " 2Fk !d
vf =
=
2mg!y + mvi2 " 2Fk !d
m
2(39 kg)(9.8 m/s 2 )(11 m) + (39 kg)(3.4 m/s)2 " 2(22 N)(15.56 m)
39 kg
vf = 14 m/s
Statement: Her speed at the bottom is 14 m/s.
78. Water held behind a high dam possesses a lot of gravitational potential energy. The amount
of potential energy is proportional to the height of the dam. When the water in a dam is released
and begins to fall, its gravitational potential energy is transformed into kinetic energy. This
kinetic energy is used to turn turbines which drive an electromagnetic generator, converting
kinetic energy into electrical energy. Eventually, moving much more slowly, the water emerges
at the bottom of the generating station and flows into the river. At this point the energy of the
water is almost entirely kinetic energy.
79. Given: m = 60 g = 0.060 kg; ∆y1 = 2.0 m; ∆y2 = 1.5 m
Required: ∆E
Analysis: The change in energy is the difference between the gravitational potential energy
before the ball is dropped and the gravitational potential energy of the ball at the maximum
height after the first bounce.
∆E = mg∆y2 − mg∆y1
Solution: !E = mg(!y2 " !y1 )
= (0.06 kg)(9.8 m/s 2 )(1.5 m " 2.0 m)
!E = "0.29 J
Statement: The amount of energy lost is 0.29 J.
80. Given: v1 = 2.0 m/s; ∆y1 = 45 m; ∆y2 = 31 m
Required: v2
Analysis: The total energy at the top of the first peak is equal to the total energy at the top of the
second peak. Both include gravitational potential energy and kinetic energy.
1
1
mg!y1 + mv12 = mg!y2 + mv22
2
2
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-29
Solution: mg!y1 +
1
1
mv12 = mg!y2 + mv22
2
2
v2 = 2g!y1 + v12 " 2g!y2
= 2(9.8 m/s 2 )(45 m) + (2.0 m/s)2 " 2(9.8 m/s 2 )(31 m)
v2 = 17 m/s
Statement: The skier’s speed at the second peak is 17 m/s.
81. (a) Given: v = 45 m/s; m = 0.25 kg
Required: ∆y
Analysis: The kinetic energy when the ball is hit is equal to the gravitational potential energy
and the kinetic energy at its maximum height:
1
1
mv 2 = mg!y + mvx2
2
2
1 2
1
v = g!y + (v cos35°)2
2
2
2
2
v " v (cos35°)2
!y =
2g
At its maximum height, the ball only has horizontal velocity, which is equal to the horizontal
velocity when the ball is hit: vx = v cos 35°.
v 2 " v 2 (cos 35°)2
Solution: !y =
2g
=
v 2 (1 " cos 2 35°)
2g
=
(45 m/s)2 (1 " cos 2 35°)
2(9.8 m/s 2 )
!y = 34 m
Statement: The maximum height of the ball is 34 m.
(b) Because air resistance is neglected, the speed of the ball when it hits the ground is equal to its
speed when it was hit by the bat, or 45 m/s.
82. (a) Given: m1 = 1990 kg + 102 kg = 2092 kg; m2 = 1990 kg; vi = 0 m/s; vf = 240 m/s
Required: impulse
Analysis: Impulse is equal to the change in momentum, m2vf − m1vi.
Solution: m2 vf ! m1vi = (1990 kg)(240 m/s) ! 0
"p = 4.8 # 105 N $ s
Statement: The impulse is 4.8 × 105 N·s.
(b) Given: ∆p = 4.8 × 105 kg·m/s; ∆t = 25 s
Required: F
Analysis: F!t = !p
F=
!p
!t
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-30
!p
!t
4.8 " 105 kg # m/s
=
25 s
F = 1.9 " 104 N
Statement: The average force is 1.9 × 104 N [forward].
83. (a) Given: m = 0.2 kg; vi = 40 m/s; vf = −60 m/s
Required: impulse
Analysis: Impulse is equal to the change in momentum, ∆p = m(vf − vi).
Solution: !p = m(vf " vi )
Solution: F =
= (0.2 kg)("60 m/s " 40 m/s)
!p = "20 kg # m/s
Statement: The impulse is 20 kg ! m/s [backward].
(b) The magnitude of the change in momentum is 20 kg·m/s.
84. Given: m1 = 100.0 kg; vi 1 = 5.0 m/s [E]; m2 = 130 kg; vi 2 = 3.0 m/s [W]
Required: vf
Analysis: Use conservation of momentum:
m1vi 1 + m2 vi 2 = (m1 + m2 )vf
vf =
m1vi 1 + m2 vi 2
m1 + m2
Solution: vf =
m1vi 1 + m2 vi 2
m1 + m2
=
(100.0 kg )(5.0 m/s) + (130 kg )(!3.0 m/s)
(100.0 kg + 130 kg )
vf = 0.48 m/s
Statement: The velocity of the hockey players after the collision is 0.48 m/s [E].
85. Given: mn = m; mc = 12m; vi n = 1.0 × 106 m/s; vi c = 0 m/s; perfectly elastic collision
Required: fraction of the neutron’s kinetic energy transferred to the carbon atom
Analysis: Determine the final velocity of the carbon atom, vf c. Use the equation for vf c for vi
! 2mn $
c = 0, vf c = #
& vf n . Then, calculate the fraction of its kinetic energy compared to that of
" mn + mc %
the neutron, using Ek =
1 2
mv .
2
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-31
! 2mn $
Solution: vf c = #
& vi n
" mn + mc %
! 2m $
=#
(1.0 ' 106 m/s)
" m + 12 m &%
vf c = 1.54 ' 105 m/s (one extra digit carried)
1
mc vf2c
Ek c
2
=
Ek n
1
m v2
2 n in
=
12 m (1.54 ! 105 m/s )2
m (1.0 ! 106 m/s )2
Ek c
= 0.28
Ek n
Statement: 28 % of the neutron’s kinetic energy is transferred to the carbon atom.
86. Given: ∆y = 165 m; vi = 180 m/s
Required: vf
Analysis: The total energy of the paintball when it is launched is equal to the total energy when
it lands. The launch energy consists of kinetic energy and gravitational potential energy. The
landing energy consists only of kinetic energy.
1
Eg = mg∆y; Ek = mv 2
2
1
1
Solution: mg!y + mvi2 = mvf2
2
2
vf = 2g!y + vi2
= 2(9.8 m/s 2 )(165 m) + (180 m/s)2
vf = 1.9 " 102 m/s
Statement: The velocity of the paintball when it hits the ground is 1.9 ! 102 m/s .
87. Given: ∆y = 15 m; m = 10.0 kg; vA = 10.0 m/s
Required: ∆yA
Analysis: The total energy at the top of the first hill, which consists only of gravitational
potential energy, is equal to the total energy at the top of hill A, which consists of gravitational
potential energy and kinetic energy.
1
Eg = mg∆y; Ek = mv 2
2
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-32
Solution: mg!y = mg!yA +
1
mv 2
2 A
1
g!y " vA2
2
!yA =
g
1
(9.8 m/s 2 )(15 m) " (10.0 m/s)2
2
=
9.8 m/s 2
!yA = 9.9 m
Statement: The height of hill A is 9.9 m.
88. (a) Given: ∆x = 0.62 m; F = 199 N
Required: k
Analysis: F = k!x
F
k=
!x
F
Solution: k =
!x
199 N
=
0.62 m
k = 3.2 " 102 N/m
Statement: The spring constant is 3.2 ! 102 N/m .
(b) Given: k = 3.2 ! 102 N/m ; ∆x = 0.25 m
Required: F
Analysis: F = k∆x
Solution: F = k!x
= (3.2 " 102 N/m )(0.25 m )
F = 8.0 " 101 N
Statement: The force is 8.0 × 101 N.
(c) Given: k = 3.2 ! 102 N/m ; ∆x = 0.25 m
Required: W
1
Analysis: W = E e = k(!x)2
2
1
Solution: W = k(!x)2
2
1
= (3.2 " 102 N/m)(0.25 m)2
2
W = 1.0 " 101 J
Statement: The work done is 1.0 × 101 J.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-33
(d) Given: k = 3.2 ! 102 N/m ; ∆x = 0.50 m
Required: W
1
Analysis: W = E e = k(!x)2
2
1
Solution: W = k(!x)2
2
1
= (3.2 " 102 N/m)(0.50 m)2
2
W = 4.0 " 101 J
Statement: The work done is 4.0 × 101 J.
89. Answers may vary. Sample answers:
(a) An example in which the damping of vibrations is useful is where the shock absorbers on a
car damp out the vibration of the car when the car hits a bump. Also, some buildings have
damping devices to reduce vibrations caused by the wind or earthquakes.
(b) An example of a situation in which the damping of vibrations would not be useful is if your
house key slipped out of your pocket and landed on a carpet that damped the sound of impact;
you would be more likely to notice that your key had dropped if the sound had not been damped.
90. Given: m1 = 9.1 g = 0.0091 kg; m2 = 98 g = 0.098 kg; ∆d = 8.0 m; µk = 0.60
Required: vi 1
Analysis: The kinetic energy of the ball before the collision is equal to the work done on the ball
by friction.
1
E = mv 2 ; W = Fk∆d; Fk = µkmg
2
1
Solution: m1v12 = Fk !d
2
1
m1v12 = µk (m1 + m2 )g!d
2
v1 =
=
2 µk (m1 + m2 )g!d
m1
2(0.60)(0.0091 kg + 0.098 kg)(9.8 m/s 2 )(8.0 m)
0.0091 kg
v1 = 33 m/s
Statement: The initial speed of the ball is 33 m/s.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-34
91.
Given: m1 = 8.0 × 102 kg; m2 = 560 kg; vf = 15 m/s [N 30° E]
Required: v1; v2
Analysis: According to the law of conservation of momentum, pi x = pf x and pi y = pf y.
Consider the x-direction and y-direction separately.
p = mv
Solution: In the x-direction, only the compact car has momentum before the collision. The
velocity in the x-direction after the collision is vf sin 30°.
m2 v2 = (m1 + m2 )vf sin 30°
v2 =
=
(m1 + m2 )vf sin 30°
m2
(8.0 ! 102 kg + 560 kg)(15 m/s)(sin 30°)
560 kg
v1 = 18 m/s
In the y-direction, only the minivan has momentum before the collision. The velocity in the
y-direction after the collision is vf cos 30°.
m1v1 = (m1 + m2 )vf cos30°
v1 =
=
(m1 + m2 )vf cos30°
m1
(8.0 ! 102 kg + 560 kg)(15 m/s)(cos 30°)
800 kg
v1 = 22 m/s
Statement: The initial velocity of the minivan was 22 m/s [N], and the initial velocity of the
compact car was 18 m/s [E].
92. (a) When you drop a ball toward Earth, the direction of the gravitational force exerted by
Earth on the ball is downward.
(b) When you drop a ball toward Earth, the direction of the gravitational force exerted by the ball
on Earth is upward.
(c) The forces are the same according to Newton’s second law of motion.
(d) If Earth and the ball are initially stationary, after the ball drops Earth moves upward, but the
amount is so infinitesimally small that it is not noticeable.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-35
(e) If somebody in China drops an identical ball at exactly the same instant from the same height,
assuming that China is on the exact opposite side of Earth, the two effects cancel each other out,
so that Earth does not move at all.
93. Given: m1 = 2.2 kg; v1 = 1.2 m/s; m2 = 3.3 kg
Required: v2
Analysis: The momentum of the two chunks must be equal and opposite:
m1v1 = −m2v2
Solution: m1v1 = !m2 v2
v2 = !
=!
m1v1
m2
(2.2 kg )(1.2 m/s)
3.3 kg
v2 = !0.8 m/s
Statement: The speed of the second chunk is 0.8 m/s.
Evaluation
94. (a) Given: m = 4.81 kg; θ = 25.7°; ∆y = 27.3 m
Required: W
Analysis: The work done is equal to the gravitational potential energy of the sled at the top of
the hill:
W = mg∆y
Solution: W = mg!y
= (4.81 kg)(9.8 m/s 2 )(27.3 m)
W = 1.3 " 103 J
Statement: The work done by the child is 1.3 × 103 J.
(b) The work done does not change, because the height of the hill remains the same.
95. The kinetic energy decreases. Friction removes energy from the system and the normal force
does not change the energy.
96. The roller coaster car is carried up a hill by a chain, driven by a motor. Work is done, and
potential energy is stored. Much of this energy becomes kinetic energy as the car goes down the
hill. This kinetic energy is converted back into gravitational potential energy as the car climbs
the next hill. Because the car’s kinetic energy at the bottom of the first hill is less than its
gravitational potential energy at the top of the first hill, the second hill must be shorter than the
first hill (or else the car would not have sufficient energy to make it over the hill; it would stop
and start moving backward). Each subsequent hill must be shorter than the one before it.
(a) As the car approaches the top of a hill, kinetic energy turns into potential energy and the
speed of the car decreases.
(b) As the car goes down the hill, potential energy turns into kinetic energy and the speed
increases.
97. (a) As Earth moves closer to the Sun, its kinetic energy increases and its gravitational
potential energy decreases. An increase in kinetic energy means an increase in speed.
(b) No, Earth does not have more total energy in relation to the Sun when it is closer or further
away; the total energy of the Earth–Sun system is constant throughout Earth’s orbit, in
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-36
accordance with conservation of energy. As Earth’s distance from the Sun decreases, its
gravitational potential energy decreases and its kinetic energy increases. As Earth’s distance
from the Sun increases, its gravitational potential energy increases and its kinetic energy
decreases. In this way, the total energy stays constant.
98. Given: m = 105 kg; k = 7600 N/m
Required: T; f
Analysis: T = 2!
1
m
; f =
k
T
Solution: T = 2!
m
k
105 kg
7600 N/m
T = 0.74 s
= 2!
f =
1
T
1
0.74 s
f = 1.4 Hz
Statement: The period is 0.74 s and the frequency is 1.4 Hz.
99. (a)
=
(b) Given: graph of data
Required: spring constant, k
Analysis: Draw a line that passes as closely through all five points as possible. Then calculate
F
the slope of the line, because k =
.
!x
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-37
Solution:
The points (10, 1.6) and (25, 4) are on the line. Calculate the slope:
k = slope
F ! F1
= 2
x2 ! x1
4 N ! 1.6 N
25 cm ! 10 cm
= 0.16 N/cm
k = 16 N/m
Statement: The spring constant is approximately 16 N/m.
100. Answers may vary. Sample answer: Cars hit bumps in the road that produce vibrations that
need to be damped out. Also, a car’s motor will vibrate differently at different speeds. These
vibrations should also be damped to ensure a smooth ride.
101. Given: m = 61 g = 0.061 kg; v = 53 m/s
Required: W
Analysis: The work done is equal to the final kinetic energy of the ball:
1
W = Ek = mv 2
2
1
Solution: W == mv 2
2
1
= (0.061 kg)(53 m/s)2
2
W = 86 J
Statement: The work done on the ball is 86 J.
102. Some mechanical energy converts to thermal energy, which is really molecular motion. The
molecular momentum is random, in all directions, and therefore does not decrease the overall
momentum.
=
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-38
Reflect on Your Learning
103. Answers may vary. Sample answer: What I found most surprising in this unit was that
neutrinos oscillate between two types, muon and tau neutrinos. What I found most difficult to
understand was which were the problems where I needed to use vectors in solving the problem
and which were the problems where all I needed to use was the magnitude of the vectors. I can
learn more about neutrinos through Internet research, and I can get better at understanding
problems by doing more of them.
104. Work is the result of force acting on an object to move it through a displacement. If this
displacement is vertical, gravitational potential energy is stored. Work is the change in kinetic
energy produced by the force exerted on the object. Thus, when a force is applied, it changes the
kinetic energy of the object. This force changes the velocity because the force is applied for a
given length of time, so it produces a change in the momentum of the object.
105. Answers may vary. Sample answer:
(a) Everyday applications that use Hooke’s law are automobile suspensions, playground toys,
retractable pens, hair elastics, and breathing (lungs).
(b) The limit to which something elastic can stretch is the point at which the material ceases to
obey Hooke’s law. If a material reaches the limit of stress for that material, it will break or
deform instead of returning to its original shape.
106. Answers may vary. Sample answer: An understanding of conservation of energy has lead to
the development of many practical applications such as roller coasters, metronomes, grandfather
clocks, automobile shock absorbers, and competitive sports such as skiing and snowboarding.
Research
107. Answers may vary. Sample answer:
Skid Mark Forensics: An accident reconstruction engineer can estimate the speed of the car just
before the driver hit the brakes by using data from the car’s tires and the road surface. A
frictional force is required to push or slide an object. Heavier objects stick harder to a surface
because of friction. Friction can be modelled using the equation F = µ mg , where, F is the force
of friction, µ is the coefficient of kinetic friction, m is the mass of the car, and g is the
acceleration of gravity.
First, look at the car’s kinetic energy to determine how long the skid mark is for a car travelling
1
at a given initial speed. The equation for kinetic energy is Ek = mv 2 , where, Ek is the kinetic
2
energy, m is the mass of the car, and v is the speed of the car.
As the car skids to a stop, all of the kinetic energy is converted into thermal energy in the tires,
road, and air. Kinetic energy is transformed into thermal energy because of the work of friction,
W, given by the equation W = F!d , or W = µ mg!d .
If the initial kinetic energy, Ek , is equal to the work, W, done by friction to slow down the car
then,
W = Ek
1
mv 2
2
v2
!d =
2µ g
µ mg!d =
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-39
The distance, !d , is very sensitive to speed, v, since it increases as the square of speed. So
doubling the speed will quadruple the distance. It should be noted that the car’s mass does not
influence the skid length. A heavier car has more kinetic energy as well as work done by friction
per metre skid, so the skid length is the same for a heavier or a lighter car.
The skills required to become an accident reconstruction engineer may include:
Analytical and critical thinking skills: Accident reconstruction engineers may not have a chance
to visit the accident site and generally rely on evidence to re-create the accident and work out the
cause. Evidence may include photographs of the accident site; the police accident report and
eyewitness reports; and the vehicle wreckage. Using the evidence available, an accident
reconstruction engineer will be able to provide information on the velocities and positions of
vehicles, skid marks, impact damage to the vehicles, condition of the vehicle (tires in particular),
the weather conditions, and the drivers’ conditions.
Deductive skills and logical thinking: The final positions of the vehicles are usually known and
accident reconstruction engineers may have to work backward from this point to determine what
caused the collision. Furthermore, investigators typically start with a general reconstruction of
the scenario and then fill in the details from the facts available. Therefore, they need to be logical
thinkers as well.
Physics knowledge: Accident reconstruction engineers can apply the laws of physics to
determine the cause of the accident. The law of conservation of energy and the law of
conservation of momentum are often used together to analyze collisions.
Metallurgical skills: Accident reconstruction engineers may examine the vehicle wreckage for
clues to the cause of the collision as well as analyze any fractures or dents in the wreckage.
108. Answers may vary. Sample answer: The pendulum is a nonlinear oscillator and is an
application of the law of conservation of energy. Due to conservation of energy, the total energy
of the system is the sum of the potential and kinetic energy. When the pendulum is at extreme
ends, it has the maximum potential energy and the kinetic energy is zero. Thus, the speed at the
extreme ends is also zero. As the pendulum swings down, it picks up kinetic energy equal to the
potential energy lost because the total energy is conserved. At the bottom of the cycle, the
pendulum has the maximum kinetic energy and its speed is also at a maximum. By letting the
kinetic energy equal the potential energy, the equation can be rearranged and solved for the
speed. (Note: The speed of the pendulum does not depend on mass.)
109. Answers may vary. Sample answer: Solar photovoltaics (PV) is a technology that converts
sunlight into electricity. Solar PV is used for grid-connected electricity for a variety of
residential, commercial, and industrial uses. Solar PV panels can be ground-mounted, installed
on rooftops, or designed into building materials. Colder temperatures increase the efficiency of
solar PV, making it well-suited for Canada’s climate. Solar PV modules can be grouped together
to provide power at any amount, from watts to megawatts. Battery storage can allow solar PV to
operate independently, without requiring utility or generator backup. They should be oriented
between south-east and south-west (due south is best), and they require an unobstructed view of
the Sun year round.
In 2009, the total energy use by all sectors (i.e., residential, commercial/institutional,
industrial, passenger transportation, freight transportation, off-road, and agriculture) was
8 541.61 × 1015 J.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-40
Canada receives 1.2 to 1.8 × 107 J/m2 of sunlight per day. Assuming that the maximum
sunlight is received,
1.8 × 107 J/m2/ day × 365 days /year = 6.57 × 109 J/m2/year (one extra digit carried)
Therefore, Canada receives 6.57 × 109 J/m2/ of sunlight in one year
The area of land needed to provide Canada’s electrical needs for one year is
8541.61 ! 1015 J/ year
9
2
6.57 ! 10 J/m / year
= 1 286 394 216 m 2
Approximately 1.3 × 104 km2 of land is needed to provide all of Canada’s electrical needs in one
year. (Note: 1.8 × 107 J /m2 is the average daily value reported by CanmetENERGY on the map
of the photovoltaic (PV) potential in Canada for a solar panel with a south-facing tilt or latitude.)
110. Answers may vary. Sample answer: Concentrating solar power (CSP) technologies use Suntracking mirrors (heliostats) to focus and concentrate sunlight onto a receiver that is on the top of
a tower. The solar energy is converted into thermal energy which is used to drive a steam turbine
which produces electricity. Some CSP technologies are also experimenting with ways to store
the energy generated, allowing the system to deliver electricity during cloudy weather or at night.
The THEMIS solar tower in the French Pyrenees produces electricity from concentrated
solar energy. It was built in the late 1970s and is being refurbished and upgraded from 2006 to
2013, under the PEGASE project. The proposed tower will provide 4.6 MW of energy using 107
Sun-tracking mirrors.
The Sierra Sun Tower supplies 5 MW of energy to the electrical grid in Southern
California and can satisfy the electrical energy needs of 4000 homes. It is the only full-scale
commercial concentrated solar power plant in the United States. The solar power plant consists
of two towers that span 20 acres. Surrounding the towers are 24 000 mirrors that rotate with the
Sun. The Sun-tracking mirrors reflect the solar rays toward the two towers. The focused solar
energy collected in the towers boils water to produce steam. The steam is sent from the towers to
a turbine and powers a power generator.
111. Answers may vary. Sample answer: Wind energy is powered by the Sun. The Sun heats
Earth unevenly, creating different temperatures at different places and at different times. The
uneven temperature distribution results in wind, since warm air rises and cooler air falls and fills
the void. Wind is the movement of warm and cool air.
Wind is captured using wind turbines. Mechanical power is created when the turbines
turn in the wind. The mechanical power turns a generator to produce electricity. Cables carry the
electrical current to transmission lines which delivery the electricity into homes and businesses.
Turbine blades turn when wind reaches 13 km/h and shut off when the wind reaches
speeds greater than 90 km/h. The average large turbine is about 8 m wide and 90 m tall, and the
turbines are spaced 250 m apart on wind farms. Modern turbines rotate with an average speed
between 18 to 20 rpm.
An average large turbine can generate 1.5 MW of energy. An average small turbine
(15 m tall) can generate 20 to 500 W of energy. A small turbine requires an average annual wind
speed of 14.4 to 16.2 km/h.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-41
In 2007, Canada produced a total of 1846 MW of energy from wind power, compared to
a total of 25.8 MW of energy from solar power. Other renewable energy sources, such as ocean
energy, biomass energy, and hydroelectricity, produced total energies of 20 MW, 1652 MW, and
73 000 MW, respectively.
112. Answers may vary. Sample answer: Torsion tests measure the strength of a material against
maximum bending forces. Products such as shafts, axles, or biomedical catheter tubing may
experience torsional loading in service. Torque refers to the applied bending pressure.
Manufactures will perform torsion tests to check the product quality and ensure the
product will function properly during service. Torsion testing can generally be classified as
failure, proof, or operational testing. Failure testing twists the material until it breaks. Proof
testing determines if a material can resist a certain amount of torque load over a period of time.
Operational testing checks that products such as bottle caps perform as expected when torque is
applied.
To perform a torsion test, the testing apparatus twists the material at quarter-degree
increments and records the torque the material can withstand. A graph is plotted such that strain,
which corresponds to the twist angle measured, is on the x-axis. Stress, which corresponds to the
torque measured, is on the y-axis. The elastic limit of a material is the point beyond which the
material can no longer return to its original state. The elastic limit is represented by the slope on
the stress–strain graph from the start of the test to the proportional limit. Hooke’s law states that
stress is directly proportional to strain until reaching the proportional limit.
113. Answers may vary. Sample answer: Newton’s second law states that force is equal to the
change in momentum with a change in time. Momentum is the product of mass and velocity.
Thus the force can be described as:
m v ! mi vi
F= f f
tf ! ti
If we assume that the mass stays a constant value equal to m as the object travels from the initial
to final velocity, then the equation becomes:
v –v
F= f i
tf – ti
Acceleration is the change in velocity divided by the change in time. Therefore, Newton’s
second law can be expressed as: F = ma
114. Answers may vary. Sample answer: Air bags are safety devices that inflate during a
collision to protect passengers from impact. Momentum is the product of an object’s mass and
velocity. In a collision, the driver’s body goes from high velocity to zero. The sudden change in
velocity results in a large change in the driver’s momentum. Impulse is the product of force and
time that acts on an object to create a change in momentum. A high-speed collision will result in
a large impulse and thus a large force that is required to stop the driver in a short period of time.
Air bags minimize the effect of the force on the driver during a collision by extending the time
required to stop the momentum of the driver. Air bags are connected to sensors that detect any
sudden decrease in acceleration that exceeds a minimum value. When a sudden decrease in
acceleration is detected, the sensor sends an electrical signal to ignite a chemical propellant
inside the air bag. When the propellant is ignited, nitrogen gas is produced and inflates the air
bag. The entire process occurs faster than the blink of an eye. To cushion the head as it moves
forward into the deploying air bag, there are vents in the back of the bag that allow it to slowly
deflate. Transport Canada estimates that between 1999 to 2000, airbags prevented 313 deaths.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-42
115. Answers may vary. Sample answer: Conventional rockets require some type of propellant
(fuel) to operate. However, space travel requiring no fuel can be accomplished by using recycled
kinetic energy. A spacecraft uses recycled kinetic energy via an impulse engine. An impulse
engine uses regenerative braking to convert the kinetic energy from a spacecraft when it lands,
into electricity. The electricity generated is then used to launch a departing partner spacecraft.
Regenerative braking is a highly efficient process commonly used by the electric
automobile industry to convert kinetic energy into electrical energy. The braking system
recaptures some of the car's kinetic energy when the brakes are applied and converts the energy
into electricity, which then can be used to recharge the car's batteries.
The new method of space travel based on recycling the kinetic energy of an arriving
spacecraft may be suited for round-trip, high-volume space travel such as space tourism.
However, recycling stations would be required to provide a power source for the departing
spacecrafts.
116. Answers may vary. Sample answer: Hydroelectricity is generated by extracting energy from
flowing water. A hydroelectric station is built near a sharp incline or waterfall to maximize the
speed gained by the water as it falls. The water flow is directed at the turbine blades, causing
them to spin and turn an electrical generator to generate electricity. Dams are also used to
regulate water flow and electricity generation.
Power plants using fossil fuels include electricity generation by coal, oil, and natural gas.
In 2003, Canada used hydropower for 60 % of it electrical production while production by fossil
fuels was 28 %.
Hydroelectricity production is less efficient than fossil-fuel production because of
seasonal variations, particularly during the winter. Fossil-fuel production has less variation, just
some limitations on smog days.
Hydroelectric power plants have a lower environmental impact than fossil-fuel power
plants, are generally dependable, and are competitive in pricing. However, investments in
hydroelectric power plants are limited by the appropriate sites available in Ontario.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-43
Cost
(cents per
KWh)
Dependability
Dependence on
local resources
Environmental
impact
Hydroelectricity
Pros
Cons
2.5-11.25
Fossil fuel
Pros
Cons
Coal: 5–7
Oil: 8–13
Natural gas: 6–8.5
– can be
used for base
and peak
load needs
– able to run at full
capacity at 85 % of
the time
– can be used for
base and peak load
needs
– low to medium
– some seasonal
variation
– low or no
production
during winter
– high
– no waste
generated
– no air
pollutants
– low
greenhouse
gas
emissions
– changes fish
migration.
– changes flow
patterns
(flooding)
– interferes with
recreational
activity
Natural gas: low
air pollutants,
medium greenhouse
gas emissions,
moderate cooling
water demand.
Oil: moderate
cooling water
demand
Clean coal with
carbon dioxide:
low- medium air
pollutants and green
house gas emissions
– waste generated (by
natural gas, oil, coal
generation).
– moderate/high cooling
water demand (by
natural gas, oil, coal
generation).
Oil: high air pollutants
and greenhouse gas
emissions
Conventional coal: high
air pollutants and green
house gas emissions
Clean coal with carbon
dioxide: increased coal
consumption per MWH
117. Answers may vary. Sample answer: Bioluminescence is light produced by a chemical
reaction from an organism. It mostly occurs in marine life, at anytime and at various locations or
depths in the sea. Bioluminescence is the primary light source in the deep ocean but is rarely
found in fresh water. It may be found on land in glowing fungus (foxfire) on wood, or in
luminous insects such as fireflies.
In the sea, bioluminescent light is typically in the blue wavelength range. Most organisms
emit light between 440 to 479 nm. Some organisms emit light continuously while most emit
flashes periodically, every 0.1 to 10 s. For multicellular species, light is neurally controlled. For
example, some fish control their luminescence by the neurotransmitter noradrenaline. In fireflies,
glutamate triggers luminescence.
Examples of bioluminescence applications include illuminated candy. BioLume, an
American Candy company, added bioluminescent ingredients to lollipops, chewing gum, and
other products. These products will light up when oxygen and an aqueous solution (saliva) is
added.
Another application is a litmus test for toxins. SDIX, an American biotechnology
company, uses a luminescent marine bacterium to assess water quality. The glow of the bacteria
lessens in the presences of heavy metals and pesticides.
Researchers at the Wellcome Trust Sanger Institute are investigating how to combine
genes from bioluminescent firefly and marine bacteria into trees to produce glowing trees.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Energy and Momentum
U2-44