SPH 4U1 – Final Exam Review
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. A football player successfully kicks a field goal through the uprights situated at the south end of the stadium.
What are the directions of the instantaneous velocity and acceleration, respectively, of the football at the peak of
its trajectory?
a. south, south
d. south, up
b. up, south
e. down, down
c. south, down
____
2. An object is thrown vertically upward with a speed of 25 m/s. How much time passes before it comes back down
at 15 m/s? (Air resistance is negligible.)
a. 1.0 s
d. 18 s
b. 4.1 s
e. 27 s
c. 9.8 s
____
3. A toy car is moving at 13 cm/s when it begins accelerating at 1.4 cm/s2. If the acceleration is uniform, what is the
speed of the car after it has travelled a distance of 27 cm?
a. 2.4  102 cm/s
d. 16 cm/s
b. 93 cm/s
e. 1.0  101 cm/s
c. 62 cm/s
____
4. Which of the following is NOT considered to be one of the “fundamental forces”?
a. gravity
d. weak nuclear
b. friction
e. electromagnetic
c. strong nuclear
____
5. With respect to Newton’s third law, the action and reaction forces
a. being equal, imply a “balanced” force situation
b. act on different objects
c. are equal provided the object is at rest
d. are equal provided the object is moving with uniform motion
e. are equal provided the object is NOT at rest or moving with uniform motion
____
6. An object sits at rest on a ramp. As the angle of inclination of the ramp increases, the object suddenly begins to
slide. Which of the following explanations best accounts for the object’s movement?
a. The coefficient of static friction has decreased sufficiently.
b. The force of gravity acting on the object has increased sufficiently.
c. The component of gravity along the ramp has increased sufficiently.
d. The friction has decreased sufficiently while the normal force has remained unchanged.
e. The normal force has increased sufficiently.
____
7. A wheel of diameter 85 cm spins at a rate such that a point on the rim of the wheel has an acceleration of 45 m/s2.
How many rotations does the wheel make in 1.0 min?
a. 1.7  102
d. 6.9
b. 98
e. 0.93
c. 69
____
8. Imagine you are a passenger upside-down at the top of a vertical looping roller coaster. The centripetal force
acting on you at this position
a. is perhaps the least of anywhere in the loop
b. is supplied at least partly by gravity
c. is supplied partly by the seat of the roller coaster
d. is directed vertically downward
e. all of the above
____
9. Which of the following graphs best illustrates the relationship between a satellite’s orbital radius ro and its orbital
speed vo?
a. A
b. B
c. C
d. D
e. E
____ 10. A child pulls a 7.4-kg toboggan 4.2 m across level ground with a 15-N force that is 34 above the horizontal. The
work done is
a. 35 J
d. 3.9  102 J
b. 52 J
e. 2.1  103 J
c. 63 J
____ 11. A farmhand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23 below
the horizontal]. A frictional force of 280 N opposes the motion. The distance the wagon travels is
a. 0.39 m
d. 32 m
b. 1.7 m
e. 1.8  102 m
c. 3.4 m
____ 12. A 75-kg parachutist jumps from a plane at a height of 1.2 km. At the instant he leaves the plane, his gravitational
potential energy compared to the plane is
a. 8.8  102 J
d. –8.8  105 J
3
b. –8.8  10 J
e. 0 J
c. 8.8  105 J
____ 13. A cyclist reaches the bottom of a hill with a speed of 18 m/s. Neglecting air resistance and other friction, to what
maximum height can they coast up the hill without pedalling?
a. 17 m
d. 20 m
b. 18 m
e. 21 m
c. 19 m
____ 14. If the mass of a car is doubled and its speed is cut in half, then the kinetic energy changes by a factor of
a. 0.25
d. 2
b. 0.5
e. 4
c. 1
____ 15. A person running in a race has to pick up a mass equal to her own mass. Assuming she can still do the same
amount of work, her speed will be changed by a factor of
a. 0.25
d. 1
b. 0.50
e. 2
c. 0.71
____ 16. A person runs at a constant speed up a slope that is angled at 12 to the horizontal. At one point, he is 3.2 m
vertically above the bottom of the hill. To double the gravitational energy compared to the bottom of the hill, the
runner must run an additional
a. 0.66 m up the slope
d. 6.4 m up the slope
b. 1.6 m up the slope
e. 15 m up the slope
c. 3.2 m up the slope
____ 17. A crane exerts a force of 4.8  105 N to lift a steel beam 15 m into the air in 18 s. If the time taken was 36 s, the
energy required would change by a factor of
a. 4
d. 0.5
b. 2
e. 0.25
c. 1
____ 18. If an arrow’s mass is doubled and the speed is halved, the momentum is changed by a factor of
a. 0.25
d. 2
b. 0.5
e. 4
c. 1
____ 19. A car (of constant mass) doubles its speed while driving up a hill sloped at 45º. The factor by which its
momentum changes is
a. 0
d. 3
b. 1
e. 4
c. 2
____ 20. If the average radius of orbit for a satellite is doubled, the period will increase by a factor of
a. 0.35
d. 2.0
b. 0.50
e. 2.8
c. 1.0
____ 21. A planet being orbited by a satellite collapses to half its original radius while maintaining the same mass. The
period of orbit will increase by a factor of
a. 0.35
d. 2.0
b. 0.50
e. 2.8
c. 1.0
____ 22. Looking at the diagram above, in which part of the orbit is the planet moving the slowest?
a. A
d. D
b. B
e. none of the above
c. C
____ 23. Two charged spheres are 5.00 cm apart. One sphere has a charge of
and the other sphere has a
charge of
. Assuming k =
, the electric force between the two spheres is
a. 1.44  101 N
d. 2.88  10–4 N
b. 4.44  10–20 N
e. 4.44  10–18 N
19
c. 2.25  10 N
____ 24. The electric field 0.50 m from a small sphere with a positive charge of 7.2  10–5 C is
a. 2.6  106 N/C [inward]
d. 2.9  10–4 N/C [inward]
–4
b. 2.9  10 N/C [outward]
e. 2.6  106 N/C [outward]
c. 1.3  106 N/C [outward]
____ 25. The magnitude of the electric field between the plates of a parallel plate capacitor is 4.7  104 N/C. If the plates
were separated to a distance one-third their original separation distance, the magnitude of the electric field would
a. decrease by a factor of one-third
d. increase by a factor of three
b. increase by a factor of one-ninth
e. not be affected
c. decrease by a factor of three
____ 26. How much work must be done to carry a –4.0 C charge from negative plate A to positive plate B if the electric
potential difference between the plates is 35 V?
a. 8.8 J
d. 140 J
b. –140 J
e. –0.11 J
c. –8.8 J
____ 27. If two parallel plates are separated by a distance of 5.0 cm and the electric potential between the plates is 20.0 V,
the magnitude of would be
a. 10.0 V·m
d. 400.0 V/m
b. 0.050 m/V
e. 0.25 m/V
c. 4.0 V/m
____ 28. Which of the following statements about determining the magnetic field around a straight conductor is NOT
correct?
a. The thumb of your right hand that points is pointing in the direction of the current.
b. A compass may be used when it is orientated perpendicular to the conductor.
c. Grasp the conductor with your right hand.
d. Curl the fingers of your right hand in the direction of the magnetic field lines.
e. A compass may be used when it is orientated parallel to the conductor.
____ 29. The direction of a magnetic field
to the plane of the page.
is from right to left as shown below. A proton travels into and perpendicular
The direction of the magnetic force acting on the proton is
a. down
b. up
c. right
d. left
e. directly out of the page and perpendicular to the paper
____ 30. A magnetic force causes a positively charged particle q to undergo uniform circular motion in a uniform magnetic
field B. The radius of the circular motion is r. The magnitude of the positively charged particle’s velocity can be
described by which of the following?
a.
d.
b.
e.
c.
____ 31. A 6.0-cm straight conductor carries a current of 14 A through a uniform 0.70-T magnetic field. When the
magnitude of the force on the conductor is 0.42 N, the angle between the current and the magnetic field is
a. 0.71°
d. 36°
b. 0.25°
e. 46°
c. 0.15°
____ 32. Several experiments are performed with light. Which of the following observations is not consistent with the
wave model of light?
a. The beam of light travels in a straight line.
b. The speed of the light is less in water than in air.
c. The light can exhibit interference patterns when travelling through small openings.
d. The light can be simultaneously reflected and transmitted at certain interfaces.
e. The light can travel through a vacuum.
____ 33. An interference pattern from a monochromatic laser light is observed on a screen. If d is the distance between the
centre of the slits, w is the width of each slit, L is the distance between the slits and the screen, and x is the
distance between adjacent nodal lines in the pattern, then the wavelength of the light is
a.
d.
b.
e.
c.
____ 34. Light from a monochromatic source shines on two adjacent, narrow slits. Which of the intensity patterns shown
below best illustrates the interference pattern observed?
a.
d.
b.
e.
c.
____ 35. A student performs a double-slit experiment using two slits spaced 0.25 mm apart and located 3.0 m from a
screen. Infrared light with a wavelength of 1200 nm is used and film sensitive to infrared light is used as the
screen. What is the average distance between adjacent dark bands on the exposed film?
a. 2.8 m
d. 2.8 cm
b. 1.4 m
e. 1.4 cm
c. 14 cm
____ 36. A two-point source interference pattern is set up using blue light. What changes in the pattern would be observed
if red light was used?
a. No changes are observed.
b. The pattern disappears since red light has a very long wavelength.
c. The pattern has more nodal lines.
d. The pattern has fewer nodal lines.
e. none of the above
____ 37. The diagram above shows a single slit with lines marking the direction to a point P in the diffraction pattern
generated. X and Y are the edges of the slit. The possible value for the path difference, PY – PX, that places P at
the second intensity maximum located away from the central maximum is
a.
d.
b.
c.
e.
____ 38. Monochromatic light strikes a thin film normal to the surface. To obtain the first reflected minimum, the thickness
of the film must be
a. much less than
d.
b.
e.
c.
____ 39. Audio CDs produce better sound quality than magnetic tapes or vinyl records because
a. There is very little background distortion and noise.
b. Each note reproduced has an unvaried pitch.
c. The dynamic range of CDs is far superior than either tape or vinyl.
d. There is no physical contact between the CD surface and the reading head.
e. All of the above.
____ 40. What is the wavelength of a radio wave (travelling at 3.00  108 m/s) from a local radio station that broadcasts at
1050 kHz?
a. 286 km
d. 286 m
b. 1.05  103 m
e. 0.286 m
c. 572 m
____ 41. The rest energy, in joules, of an electron of mass 9.11  10–31 kg is
a. 5.1  105 J
d. 3.0  10–39 J
–14
b. 8.2  10 J
e. 1.0  10–47 J
–22
c. 2.7  10 J
Problem
42. An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s2 over a distance of 80.0 cm. The object
then slides with a constant speed for 4.0 s until it reaches a rough section which causes it to stop in 2.5 s.
(a) What is the speed of the object when it reaches the rough section?
(b) At what rate does the object slow down once it reaches the rough section?
(c) What total distance does the object slide throughout its entire trip?
43. A ball is thrown vertically upward from a window that is 3.6 m above the ground. Its initial speed is 2.8 m/s.
(a) With what speed does the ball hit the ground?
(b) How long after the first ball is thrown should a second ball be simply dropped from the same window so that
both balls hit the ground at the same time?
44. A car leaves point A and drives at 80.0 km/h [E] for 1.50 h. It then heads north at 60.0 km/h for 1.00 h and finally
[30.0 W of N] at 100.0 km/h for 0.50 h, arriving at point B.
(a) Determine the displacement of point B from point A.
(b) A plane flies directly from point A to point B, leaving 2.00 h after the car has departed from point A. It arrives
at point B at the same time the car arrives. There is a wind blowing at 60.0 km/h due south for the entire trip.
What is the airplane’s airspeed?
(c ) What direction must the plane head in order to arrive at point B?
(d) How long would the plane’s trip be if there was no wind?
47. A shell is fired from a cliff that is 36 m above a horizontal plane. The muzzle speed of the shell is 80.0 m/s and it
is fired at an elevation of 25 above the horizontal.
(a) Determine the horizontal range of the shell.
(b) Determine the velocity of the shell as it strikes the ground.
46.
A piece of ice (m = 500.0 g) slides down the slope of
a roof inclined at 50.0. It starts from rest and slides 8.0 m along the
roof, sliding off the edge at a height of 4.0 m above the level ground.
The coefficient of kinetic friction is 0.14.
(a) Draw a free-body diagram of the ice as it is sliding along the roof.
(b) With what speed does it leave the roof?
(c) How far away from the foot of the building does the ice land?
47.
A boy pulls a toy train
(consisting of an engine and a caboose) along a
rough floor, exerting 2.00 N of force as
indicated in the diagram. A frictional force of
0.60 N acts on the engine and a frictional force
of 0.40 N acts on the caboose.
(a) Draw free-body diagrams of both the engine
and caboose.
(b) Determine the acceleration of the entire train.
(c) Calculate the tension in the string between the engine and the caboose.
48. A 2.0  102-g mass is tied on the end of a 1.6 m long string and whirled around in a circle that describes a vertical
plane.
(a) What is the minimum frequency of rotation required to keep the mass moving in a circle?
(b) Calculate the maximum tension in the string at this frequency.
49. A flea stands on the end of a 1.0 cm long sweep second hand of a clock that rests horizontally on a table. What is
the minimum coefficient of static friction which would allow the flea to stay there without slipping? Include an
appropriate free-body diagram.
50. What force does Earth exert on a 80.0-kg astronaut at an altitude equivalent to 2.5 times Earth’s radius?
51. The gravitational field strength at the surface of a planet is 3.4 N/kg. If the planet’s mass is 7.2  1022 kg, what is
its radius?
52. A driver carelessly ignores the reduced speed limit of 40.0 km/h in a school zone and continues at 65 km/h.
Assuming a good reaction time of 0.80 s, how many more metres will it take him to stop than if he had reduced
his speed? Assume a constant emergency braking acceleration of –7.8 m/s2.
53. A child swings on a swing so that her centre of mass is located 2.4 m from the point where the rope is attached to
the tree. If she swings so that her maximum amplitude causes the rope to make an angle of 47 with the vertical,
calculate the child’s maximum speed during the swing.
54. A 1.00-kg mass and a 2.00-kg mass are set gently on a platform mounted on an ideal spring of force constant 40.0
N/m. The 2.00-kg mass is suddenly removed. How high above its starting position does the 1.00-kg mass reach?
55.
Two boxes are connected over a pulley and held at rest as shown at left.
Box A has a mass of 15 kg and box B has a mass of 12 kg. If the bottom of box A is
originally 85 cm above the floor, with what speed will it contact the floor when the
system is released? Use conservation of energy and assume that friction is negligible.
56.
A small explosive charge is placed in a rubber block resting on a
smooth surface. When the charge is detonated, the block breaks into three pieces. A
200-g piece travels at 1.4 m/s, and a 300-g piece travels at 0.90 m/s. The third piece
flies off at a speed of 1.8 m/s. If the angle between the first two pieces is 80º, calculate
the mass and direction of the third piece. Assume two significant digits for each value.
57.
During a game of billiards, the 0.30-kg cue ball, travelling at 2.1 m/s,
glances off a stationary 0.28-kg billiard ball so that the billiard ball moves off at 1.4 m/s
at an angle of 38º from the cue ball’s original path. Find the new speed of the cue ball.
58. How much work is done against gravity to fire a 7.2  102-kg weather monitor 120 km into the air? (rE = 6.38 
106 m, ME = 5.98  1024 kg)
59. Uranus orbits the Sun (MS = 1.99  1030 kg) with a mean radius of 2.87  1012 m. How long does it take Uranus to
complete one orbit? Give your answer in Earth years.
60. How fast must a satellite leave Earth’s surface to reach an orbit with an altitude of 895 km?
61.
Three identical point charges A, B, and C are located
as shown on the diagram at left. A exerts force F on B. An equal force
F is exerted by C on B (
). What is the net force on B?
62.
The electrostatic force between two point charges is
. If the distance between them is tripled, the charge of one
of the points triples and the charge of the other point halves. What
will be the force between them?
63. The electrostatic force between two point charges is
. If the distance between them is doubled, the
charge of one of the points doubles and the charge of the other point triples. What will be the force between them?
64.
Point charges W, X, Y, and Z each have the same magnitude
charge of
. Two of the charges are positive as shown and two
are negative. The corners of the square are 20.0 cm apart. Calculate the net
force on each of the point charges. Draw vector diagrams for each point
charge.
65.
Calculate the electric field 0.8 m from a small sphere with a
negative charge of 4.7  10–2 C.
66.
Calculate the electric potential a distance 0.56 m from a
point charge of –3.6  10–4 C.
67. A proton with a charge of 1.60  10–19 C is shot from plate B toward plate A with a speed of 1.5  106 m/s.
(
d = 0.0030 m)
(a) What is the electric potential between the plates?
(b) What will the speed of the proton be just before it hits plate A?
68. Calculate 1.92  10–18 C as a multiple of the elementary charge.
69. Two parallel plates labelled W and X are separated by 5.2 cm. The electric potential between the plates is 150 V.
An electron starts from rest at time tW and reaches plate X at time tX. The electron continues through the opening
and reaches point P at time tP. (Remember: e = –1.6  10–19 C and the mass of an electron is 9.1  10–31 kg.)
(a) Sketch the speed-time graph on the axes below.
(b) Determine the kinetic energy of the electron as it arrives at plate X.
70. A water wave in a ripple tank travels from a shallow to a deep region. The wavelength and speed in the shallow
region are 2.5 cm and 5.0 cm/s, respectively. If the wavelength in the deep region is 6.0 cm, find
(a) the relative index of refraction from shallow to deep water
(b) the speed of the wave in the deep water
71. A ripple tank is used to generate straight waves in region A that travel toward region B, which is separated from
region A by a straight boundary. The frequency of the generator is 2.5 Hz, and the waves travel in region A with a
speed of 15 cm/s. If the wave fronts in region A strike the boundary at 20o, and the wave fronts in region B leave
the boundary at 50o,
(a) use Snell’s law to find the relative index of refraction between the two regions
(b) find the wavelength in each region
72. A two-point source interference pattern is generated in a swimming pool. A piece of styrofoam, located on the
second nodal line, is 12.0 m from one source and 20.0 m from the other source. One wave crest takes 2.0 s to
travel the 35.0 m width of the pool. Find the speed, wavelength, and frequency of the waves.
73. A consumer wishes to test the high-end frequency response of brand new speakers as there seems to be very little
treble. To do this, a pure signal using a signal generator is fed into the two speakers, and the frequency is raised to
the highest output that is still audible. Using a microphone and the interference pattern generated, a point is found
to be located on the sixth nodal line, a distance of 2.000 m from one speaker and 2.315 m away from the other.
The speed of sound in the room is 342.8 m/s. The speakers are rated at 25 Hz–18 kHz. Should the speakers be
returned to the store?
74. A student performs Young's double-slit experiment using a slit separation of 21.6 m. A screen is placed 2.50 m
from the centre of the sources such that a point on the fifth nodal line is 37.5 cm from the centre of the
interference pattern. Find the wavelength of the light used and identify its colour.
75. Red light with a wavelength of 725 nm enters glass with index of refraction 1.52. Located within the glass block
is a single slit, and a screen is placed 1.25 m away. If the width of the slit is 18.5 m, find the width of the central
maximum (a) in degrees and (b) in centimeters.
76. Currently, CD players use a laser diode producing light with a wavelength of 780 nm. The index of refraction of
the polycarbonate plastic layer is 1.55. If blue laser light with a wavelength of 475 nm were to be used, by what
amount would the depth of the pits in the CD surface need to be changed?
77. An electron was accelerated to 0.98c in a long particle accelerator. If the proper length of the path is 15.2 km,
what length did the electron experience?
78. For an electron with speed 0.880c, calculate its kinetic energy if its total energy is 0.980 MeV.
79. With what speed would an electron be ejected from sodium that has a work function of 2.36 eV when it is
illuminated by 442-nm light?
80. Calculate the wavelength of light equivalent to an electron moving at 3.98  105 m/s.
81. An electron with energy 13.2 eV collides with an element, and the electron emerges with energy 4.5 eV. What
wavelength of light would also be emitted along with the electron?
review
Answer Section
MULTIPLE CHOICE
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
C
B
D
B
B
C
B
E
A
B
C
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
E
A
B
C
E
C
C
C
E
C
C
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
D
E
E
B
D
B
B
C
E
E
E
PROBLEM
42. ANS:
(a)
v1 = 0.0 m/s
a = 5.0 m/s2
d = 80.0 cm = 0.800 m
v2 = ?
The speed of the object upon reaching the rough section is 2.8 m/s.
(b)
v1 = 2.83 m/s
v2 = 0.0 m/s
t = 2.5 s
a=?
The object’s acceleration is 1.1 m/s2 and slowing.
(c)
During the period of acceleration:
d = 0.800 m
During the period of uniform motion:
34.
35.
36.
37.
38.
39.
40.
41.
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
E
E
D
E
B
E
D
B
v = 2.83 m/s
t = 4.0 s
d = vt = 2.83 m/s(4.0 s) = 11.32 m
During the period of deceleration:
v1 = 2.83 m/s
v2 = 0.0 m/s
t = 2.5 s
Total distance the object slides: 0.800 m + 11.32 m + 3.54 m = 16 m
The object slides a total distance of 16 m.
43. ANS:
(a) Using a sign convention of “down” as (+) and “up” as (–).
The speed of the object when it hits the ground is 8.9 m/s.
(b) Time of flight for the first ball:
Time of flight for the second ball:
The difference in flight times is 1.19 s – 0.86 s = 0.33 s.
The second ball should be dropped 0.33 s after the first one is thrown so that both hit the ground at the
same time.
44. ANS:
(a) The car drives the following displacements:
= 80.0 km/h [E](1.50 h) = 120 km [E]
= 60.0 km/h [N](1.00 h) = 60.0 km [N]
= 100.0 km/h [30W of N](0.50 h) = 50 km [30W of N]
Using the component method for the displacement of A to B:
north-south components: 60 km [N] + 50 km(cos 30) [N] = 103.3 km [N]
east-west components: 120 km [E] + 50 km(sin 30) [W] = 95 km [E]
Using Pythagoras Theorem, the magnitude of the displacement is 140 km
Using trigonometry, the direction is: [43 E of N]
The displacement of point B from point A is 1.4  102 km [43E of N].
(b) A vector diagram showing the relationship among the vectors is drawn:
vPG = velocity of plane with respect to the ground
vPA = velocity of plane with respect to the air
vAG = velocity of the air with respect to the ground
Since the car takes a total of 3.00 h to reach point B and the plane leaves 2.00 h later but arrives at the same time,
the time it takes the plane to make the flight is 1.00 h.
= 1.4  102 km/h [43 E of N]
Using cosine law:
The plane’s speed is 1.9  102 km/h.
(c) Using sine law:
,  = 13
The plane must head in a direction of [30ºE of N]. (90 – 13 – 47)
(d) If there is no wind:
The plane would take 0.74 h to fly from A to B if there was no wind.
45. ANS:
(a) Time of flight: let “up” be (–) and “down” be (+)
v1 = –80.0 m/s(sin 25) = –33.8 m/s
a = 9.8 m/s2
d = 36 m
t = ?
36 = (–33.8)t + 4.9(t)2
Solving the quadratic: t = 7.84 s
Horizontal range: d = vt = 80.0 m/s(cos 25)(7.84 s) = 5.7  102 m
The horizontal range of the shell is 5.7  102 m.
(b) Horizontal component of final velocity: 80.0 m/s(cos 25) = 72.5 m/s
Vertical component of final velocity: v2 = v1 + at = –33.8 m/s + 9.8 m/s2(7.84 s)
v2 = 43.0 m/s
Using Pythagoras:
=
The shell lands with a velocity of 84 m/s at an angle of 31 below the horizontal.
46. ANS:
(a) Free-body diagram: FN acting perpendicular to the roof (upward)
Fg acting down
FK acting up along the roof (this is the negative direction)
(b) Parallel to the roof:
ma = mg(sin ) – mg(cos )
a = 9.8 N/kg(sin 50º) – (0.14)(9.8 N/kg)(cos 50º)
a = 6.62 m/s2
The ice leaves the roof at 1.0  101 m/s.
(c) When the ice leaves the roof it becomes a projectile:
Vertically:
Solving the quadratic: t = 0.406 s
Horizontally:
d = v(cos t
= 10.3 m/s(cos 50)(0.406 s)
d = 2.7 m
The ice lands 2.7 m from the base of the building.
47. ANS:
(a)For the engine:
Free-body diagram:
For the caboose:
Free-body diagram:
FN acting up
Fg acting down
FK acting to the left
FA acting as indicated
FT acting to the left (force of caboose on engine)
FN acting up
Fg acting down
FK acting to the left
FT acting to the right (force of engine on caboose)
Let “to the right” and “upward” be (+).
(b) Considering the entire train:
The train will accelerate at 2.1 m/s2.
(c) Considering the caboose:
= 0.100 kg(2.1 m/s2) – (–0.40 N)
= 0.61 N
The tension in the string joining the engine and caboose is 0.61 N.
48. ANS:
(a) The minimum frequency occurs when the tension becomes zero.
The minimum frequency is 0.39 Hz.
(b) The maximum tension occurs at the bottom of the circle.
Let “up” be negative and “down” be positive:
The maximum tension is 3.9 N [up].
49. ANS:
(a)
The centripetal force is supplied by static friction.
The minimum coefficient of static friction is 1.1  10–5.
50. ANS:
At Earth’s surface:
Since
, then Fg(r2) is a constant.
If F1 = force at Earth’s surface
r1 = Earth’s radius
F2 = force at position in question
r2 = 2.5r1 + r1 = 3.5r1
F1(r1)2 = F2(r2)2
Earth exerts a force of 2.6  101 N on the astronaut.
51. ANS:
Fg  mg where g = 3.4 N/kg and
where M = mass of the planet and r = radius of the planet.
Both expressions represent the same force of gravity acting on mass m.
The planet’s radius is 1.2  106 m.
52. ANS:
Convert km/h to m/s:
65 km/h = 18.1 m/s
Distance travelled during reaction time:
8.1  0.80 = 14.4 m
40 km/h = 11.1 m/s
11.1  0.80 = 8.89 m
Stop time:
Total distance:
20.9 m + 14.4 m = 35.3 m
7.9 + 8.9 = 16.8 m
Difference:
35.3 m – 16.8 m = 18.5 m
It would take 19 m more to stop.
53. ANS:
Calculate the vertical displacement of the pendulum (swing):
vertical height = 2.4 – 1.637 = 0.763 m
The maximum speed at the bottom will involve zero Eg and the maximum height at the top has no EK, so
The child’s maximum speed is 3.9 m/s
54. ANS:
Calculate the original compression:
Using conservation of energy,
The 1.00-kg mass reaches a maximum height of 1.10 m above its start position.
55. ANS:
The total energy of the system will not change.
The impact speed of box A will be 1.4 m/s.
56. ANS:
The momentum of the 200-g piece, p2, is 0.20  1.4 = 0.28 kgm/s.
The momentum of the 300-g piece, p3, is 0.30  0.90 = 0.27 kgm/s.
The momentum of the unknown piece, pm, is m  1.8 = 1.8m kgm/s.
Choose the +x direction to be the direction of the 200-g piece.
 is the angle between the unknown momentum vector and opposite to the 200-g momentum vector.
Now divide Equation 1 by Equation 2:
Substitute this value into Equation 1:
The angle measured from the 200-g piece is 180º – 39º = 141º.
The mass of the third piece is 0.23 kg and it is moving 141º from the 200-g piece. (It is 139º from the 300-g
piece.)
57. ANS:
The initial momentum of the cue ball, pc, is 0.30  2.1 = 0.63 kgm/s.
The final momentum of the billiard ball, pb, is 0.28  1.4 = 0.392 kgm/s.
The final momentum of the cue ball, p'c, is 0.30  v'c = 0.30v'c kgm/s.
Choose the +x direction to be the original direction of the cue ball.
 is the angle between the original direction of the cue ball and its new direction.
Now divide Equation 1 by Equation 2:
Substitute this value into Equation 1:
The new speed of the cue ball is 1.3 m/s.
58. ANS:
The work
done against gravity is 8.3  108 J.
59. ANS:
Calculate the speed of travel:
Calculate the total distance:
Calculate the time:
Convert to years:
Uranus takes 84.0 Earth years to complete one orbit.
60. ANS:
The launch speed would need to be 8.38  103 m/s.
61. ANS:
[N]
The net force on B is
.
62. ANS:
Since
The force between the two charges will be 7.8  10 –6 N.
63. ANS:
Since
The force between the two charges will be 1.8  10–2 N.
64. ANS:
q of W, X, Y, and Z =
r = 20.0 cm = 0.200 m
Fnet = ?
We must find the distance across the diagonal to find the force between the diagonal charges:
The electric force on the charges on the side:
The electric force on the diagonal charges:
Each point charge experiences the same forces as seen in the vector diagram below.
The magnitude of the force can be found by drawing a vector diagram as shown below. (Charge W is used.)
The length of the final side of the triangle can be found as shown:
The magnitude of the net force is:
65. ANS:
q = –4.7  10–2 C
r = 0.8 m
The electric field is 6.6  108 N/C [radially inward].
66. ANS:
r = 0.56 m
q = –3.6  10–4 C
V=?
The electric potential is 5.8  106 V.
67. ANS:
(a)
d = 0.0030 m
V=?
The electric potential between the plates is 30 V.
(b)EK lost = Ep gained
The speed of the proton will be 1.5  106 m/s.
68. ANS:
Therefore, the multiple of elementary charge is 12e.
69. ANS:
(a)
(b)q = –1.6  10–19 C
V = 1.50  102 V
EK2 = ?
The kinetic energy of the electron as it arrives at plate X is +2.4  10–17 J.
70. ANS:
(a)
The relative index of refraction from shallow to deep water is 0.42.
(b)
The speed of the wave in the deep water is 12 cm/s.
71. ANS:
(a)
The index of refraction between the two regions is 0.45.
(b)
The wavelength is 13 cm.
72. ANS:
The speed is 18 m/s, the wavelength is 5.3 m, and the frequency is 3.3 Hz.
73. ANS:
Since the maximum frequency produced is 6.0 kHz, the speakers should be returned.
74. ANS:
The colour used was red with a wavelength of 7.20 x 10-7 m.
75. ANS:
(a)
The angular width of the central maximum is 2.95o.
(b)
The width of the central maximum is 6.45 cm.
76. ANS:
The CD reading head uses the properties of interference such that the pit depths are
polycarbonate plastic layer.
of the light in the
The pit depth would change by 49 nm.
77. ANS:
Ls = 15.2 km
v = 0.98c
The electron experiences a journey of 3.02 km.
78. ANS:
The electron has a kinetic energy of 0.468 MeV.
79. ANS:
v=?
The electron would be ejected from the sodium with a speed of 3.98  105 m/s.
80. ANS:
m  9.11  10–31 kg
p = mv
p  (9.11  10–31 kg)(3.98  105 m/s)
p  3.626  10–25 kg·m/s
The wavelength of a photon equivalent to the electron is 1.83 nm.
81. ANS:
The wavelength of light that would be emitted would be 140 nm.