Unit 2 Self-Quiz, pages 272–273 1. 2. 3. 4. 5. 6. 7. 8. 9

Copyright © 2012 Nelson Education Ltd. Unit 2: Energy and Momentum U2-15

Unit 2 Self-Quiz, pages 272–273 1. (c) 2. (a) 3. (d) 4. (b) 5. (d) 6. (a) 7. (a) 8. (b) 9. (d) 10. (a) 11. (d) 12. (b) 13. (c) 14. (a) 15. (b) 16. (c) 17. (d) 18. (a) 19. (a) 20. True 21. True 22. True 23. False. Friction does negative work on a baseball player sliding into home plate. 24. False. The energy contained in gasoline is referred to as chemical energy. 25. False. Energy and power are different words that have different meanings. 26. True 27. True 28. True 29. False. When two objects collide in a perfectly elastic collision, kinetic energy and momentum are conserved. 30. True 31. False. In a perfectly elastic two-body head-on collision, if the two objects have the same mass, the objects collide and travel in the reverse direction at their original speed. 32. True


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.


Understanding 37. No, you would not do the same work to lift a 2 kg box vertically through 1.5 m on the Moon

as you would to lift it on Earth. The value of g on the Moon is

16

the value of g on Earth.

Therefore, the weight of the box is less, and

16

of the work is done.

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

top of the lift, W = mg∆y. Power is work divided by time, P =

W!t

. So, P =

mg!y!t

.

Solution:

P =mg!y!t

=(250 kg)(9.8 m/s2 )(2.1 m)

2.4 sP = 2.1"103 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


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.

Wk = Ek and Ek =

12

mvi2 , so

Wk =

12

mv2 . Rearrange this equation to isolate vi.

vi =

2Wk

m

Solution:

vi =2Wk

m

=2(145 J)

41 kgvi = 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/s2 )(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

Analysis: ET = Ek + Eg ,

Eg = mg∆y, and Ek =

12

mv2 , so ET =

12

mv2 + mg!y .

First, convert the speed to metres per second:

v = 980

kmh

!1000 m

1 km!

1 h3600 s

= 272 m/s (one extra digit carried)


Solution:

ET =12

mv2 + mg!y

=12

(91000 kg)(272 m/s)2 + (91 000 kg)(9.8 m/s2 )(1.2 "104 m)

ET = 1.4 "1010 J

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

Analysis: Ek = Eg; Eg = mg∆y; Ek =

12

mv2 ; solve for ∆y:

Ek = Eg

12

mv2 = mg!y

!y =v2

2g

Solution:

!y =v2

2g

=(6.5 m/s)2

2(9.8 m/s2 )

=(6.5)2 m 2 / s2

2(9.8 m/ s2 )!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/s2 )(12 m)!Eg = "1.3#104 J

Statement: The magnitude of the change in gravitational potential energy is 1.3 × 104 J.


(b) Given: m = 110 kg; ∆y = 12 m; g = 9.8 m/s2

Required: v

Analysis: Eg = Ek; Eg = mg∆y; Ek =

12

mv2 ; solve for v:

Eg = Ek

mg!y =12

mv2

v = 2g!y

Solution:

v = 2g!y

= 2(9.8 m/s2 )(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.

Ee = Eg; Ee =

12

k(!x)2 ; Eg = mg∆y; solve for Δy:

Ee = Eg

12

k(!x)2 = mg!y

!y =k(!x)2

2mg

Solution:

!y =k(!x)2

2mg

=(160 N/m)(0.13 m)2

2(0.021 kg)(9.8 m/s2 )!y = 6.6 m

Statement: The ball rises to a height of 6.6 m.


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

mg!y =

12

mv2 for v:

mg!y =12

mv2

v = 2g!y

Then, use the equation p = mv to determine p. Solution:

v = 2g!y

= 2(9.8 m/s2 )(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!vF!t = m(vf " vi )

vf =F!t + mvi

m

=(31 N)(0.2 s) + (0.43 kg)("7.668 m/s)

0.43 kgvf = 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

the initial velocity, vi, using the conservation of energy: mg(length of cord) = 12

mvi2 . When the

cord is stretched as far as possible, the jumper's final velocity, vf, is zero, so the jumper's momentum is also 0.


Solution:

mg(length of cord) =12

mvi2

(9.8 m/s2 )(39 m) =12

vi2

vi = 2(9.8 m/s2 )(39 m)

vi = 27.65 m/s [down] (two extra digits carried)

!p = m!v= m(vf " vi )= (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.

m1

!vi 1 + m2

!vi 2 = (m1 + m2 )!vf

!vi 2 =(m1 + m2 )!vf ! m1

!vi 1

m2


Solution:

!vi 2 =(m1 + m2 )!vf ! m1

!vi 1

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/s2 ; ∆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/s2 )(2.7 s)!vi 1 = 9.18 m/s (one extra digit carried)

m1

!vi 1 + m2

!vi 2 = (m1 + m2 )!vf

!vf =m1

!vi 1 + m2

!vi 2

(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.

m1

!vi 1 + m2

!vi 2 = (m1 + m2 )!vf

!vf =m1

!vi 1 + m2

!vi 2

(m1 + m2 )

Solution:

!vf =m1

!vi 1 + m2

!vi 2

(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].


(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

Analysis: !Ek = Ek i " Ek f =

12

m1vi 12 +

12

m2vi 22 "

12

(m1 + m2 )vf2

Solution:

!Ek =12

(m1 + m2 )vf2 "

12

m1vi 12 +

12

m2vi 22#

$%&'(

=12

(1.5 kg + 3.5 kg)("1.65 m/s)2 "12

(1.5 kg)(12 m/s)2 +12

(3.5 kg)("7.5 m/s)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 :

Fx =

W!d

Then, use Fx = F cos θ to determine θ.

Solution:

Fx =W!d

=87 J11 m

Fx = 7.9 N


Fx = F cos!

cos! =Fx

F

! = cos"1 Fx

F#

$%&

'(

= cos"1 7.9 N9.3 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, vf

2 = vi2 + 2a!d , rearranged to solve for ∆d.

vf2 = vi

2 + 2a!d

!d =vf

2 " vi2

2a

Solution:

!d =vf

2 " vi2

2a

=(0 m/s)2 " (25 m/s)2

2("8.0 m/s2 )

=(0)2 m 2 / s2 " (25)2 m 2 / s2

2("8.0 m/ s2 )!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 #102 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


Solution:

W = mg!y= (18 kg)(9.8 m/s2 )(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.

mg!y =

12

mvi2 ;

!yd

= sin"

Solution:

mg!y =12

mvi2

!y =vi

2

2g

=(4.2 m/s)2

2(9.8 m/s2 )

=(4.2)2 m 2 / s2

2(9.8 m/ s2 )!y = 0.90 m

!yd

= sin"

d =!y

sin"

=0.90 msin 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:

mg!y =12

mv2

v = 2g!y

Solution:

v = 2g!y

= 2(9.8 m/s2 )(10.0 m)v = 14 m/s

Statement: The stone’s speed when it hits the ground is 14 m/s.


71. Given: ∆y = 4.0 m; Eg = 2.7 kJ = 2.7 × 103 J Required: m Analysis:

Eg = mg!y

m =Eg

g!y

Solution:

m =Eg

g!y

=2.7 "103 kg # m2 / s2

(9.8 m/ s2 )(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.

mg!y + 12

mvi2 =

12

mvf2

vf = 2g!y + vy2

Solution:

vf = 2g!y + vy2

= 2(9.8 m/s2 )(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/s2 )(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 .


(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/s2 )(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

Analysis: The impulse equation is F∆t = ∆p = m∆v, so F =

m!v!t

. The flow rate is the mass over

the change in time, or m!t

, and ∆v = 0 m/s − 30 m/s = −30 m/s. Multiply the flow rate by the

change in velocity to get the force.

Solution:

F =m!v!t

=m!t

"#$

%&'!v

= (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.


12

mvf2 = mg!y + 1

2mvi

2 " Fk!d

(∆d)2 = (∆x)2 + (∆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)

12

mvf2 = mg!y + 1

2mvi

2 " Fk!d

mvf2 = 2mg!y + mvi

2 " 2Fk!d

vf =2mg!y + mvi

2 " 2Fk!dm

=2(39 kg)(9.8 m/s2 )(11 m) + (39 kg)(3.4 m/s)2 " 2(22 N)(15.56 m)

39 kgvf = 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/s2 )(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.

mg!y1 +

12

mv12 = mg!y2 +

12

mv22


Solution:

mg!y1 +12

mv12 = mg!y2 +

12

mv22

v2 = 2g!y1 + v12 " 2g!y2

= 2(9.8 m/s2 )(45 m) + (2.0 m/s)2 " 2(9.8 m/s2 )(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:

12

mv2 = mg!y +12

mvx2

12

v2 = g!y +12

(vcos35°)2

!y =v2 " v2 (cos35°)2

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°.

Solution:

!y =v2 " v2 (cos 35°)2

2g

=v2 (1" cos2 35°)

2g

=(45 m/s)2(1" cos2 35°)

2(9.8 m/s2 )!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:

m2vf ! 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


Solution:

F =!p!t

=4.8 "105 kg #m/s

25 sF = 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 )= (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:

Solution:

vf =m1vi 1 + m2vi 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

c = 0, vf c =

2mn

mn + mc

!

"#$

%&vf n . Then, calculate the fraction of its kinetic energy compared to that of

the neutron, using Ek =

12

mv2 .

m1vi 1 + m2vi 2 = (m1 + m2 )vf

vf =m1vi 1 + m2vi 2

m1 + m2


Solution:

vf c =2mn

mn + mc

!

"#$

%&vi n

=2 m

m +12 m!"#

$%&

(1.0 '106 m/s)

vf c = 1.54 '105 m/s (one extra digit carried)

Ek c

Ek n

=

12

mcvf c2

12

mnvi n2

= 12 m (1.54!105 m/s )2

m (1.0!106 m/s )2

Ek c

Ek n

= 0.28

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.

Eg = mg∆y; Ek =

12

mv2

Solution:

mg!y + 12

mvi2 =

12

mvf2

vf = 2g!y + vi2

= 2(9.8 m/s2 )(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.

Eg = mg∆y; Ek =

12

mv2


Solution:

mg!y = mg!yA +12

mvA2

!yA =g!y " 1

2vA

2

g

=(9.8 m/s2 )(15 m) " 1

2(10.0 m/s)2

9.8 m/s2

!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

k =F!x

Solution:

k =F!x

=199 N0.62 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

Analysis: W = E e =

12

k(!x)2

Solution:

W =12

k(!x)2

=12

(3.2 "102 N/m)(0.25 m)2

W = 1.0 "101 J

Statement: The work done is 1.0 × 101 J.


(d) Given: k = 3.2 !102 N/m ; ∆x = 0.50 m Required: W

Analysis: W = E e =

12

k(!x)2

Solution:

W =12

k(!x)2

=12

(3.2 "102 N/m)(0.50 m)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.

E =

12

mv2 ; W = Fk∆d; Fk = µkmg

Solution:

12

m1v12 = Fk!d

12

m1v12 = µk (m1 + m2 )g!d

v1 =2µk (m1 + m2 )g!d

m1

=2(0.60)(0.0091 kg + 0.098 kg)(9.8 m/s2 )(8.0 m)

0.0091 kgv1 = 33 m/s

Statement: The initial speed of the ball is 33 m/s.


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°.

m2v2 = (m1 + m2 )vf sin30°

v2 =(m1 + m2 )vf sin30°

m2

=(8.0 !102 kg + 560 kg)(15 m/s)(sin 30°)

560 kgv1 = 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 kgv1 = 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.


(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 = !m2v2

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/s2 )(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


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! m

k;

f =

1T

Solution:

T = 2! mk

= 2! 105 kg7600 N/m

T = 0.74 s

f =1T

=1

0.74 sf = 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

the slope of the line, because k =

F!x

.


Solution:

The points (10, 1.6) and (25, 4) are on the line. Calculate the slope:

k = slope

=F2 ! F1

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:

W = Ek =

12

mv2

Solution:

W ==12

mv2

=12

(0.061 kg)(53 m/s)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.


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 = µmg , 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

at a given initial speed. The equation for kinetic energy is Ek =

12

mv2 , where, Ek is the kinetic

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 = µmg!d . If the initial kinetic energy, Ek , is equal to the work, W, done by friction to slow down the car then,

W = Ek

µmg!d =12

mv2

!d =v2

2µg


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.


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

6.57 !109 J/m2 / year= 1 286 394 216 m2

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 Sun-tracking 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.


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:

F =

mf vf ! mivi

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:

F =

vf – vi

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.


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.


Hydroelectricity Fossil fuel Pros Cons Pros Cons

Cost (cents per KWh)

2.5-11.25 Coal: 5–7 Natural gas: 6–8.5

Oil: 8–13

Dependability – can be used for base and peak load needs

– some seasonal variation – low or no production during winter

– able to run at full capacity at 85 % of the time – can be used for base and peak load needs

Dependence on local resources

– high – low to medium

Environmental impact

– 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.

Documents

Unit 2 Self-Quiz, pages 272–273 1. 2. 3. 4. 5. 6. 7. 8. 9