exam review - d - blessed mother teresa...

EXAM REVIEW - D. O'Hara - SPH4U1

Multiple ChoiceIdentify the letter of 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, upb. up, south e. down, downc. south, down

____ 2. A race car completes exactly 10 laps around an oval track. Which of the following pairs of quantities concerning its motion would both have values of zero?a. displacement, average velocityb. average speed, average accelerationc. distance, average speedd. average speed, average velocitye. displacement, average speed

____ 3. A bus drives 40.0 km [E] from town A to town B, then another 30.0 km [S] to town C in a total time of 1.00 h. What are the values of its average speed and average velocity, respectively?a. 70.0 km/h, 70.0 km/h [37º S of E] d. 50.0 km/h, 70.0 km/h [37º S of E]b. 70.0 km/h, 50.0 km/h [37º S of E] e. 50.0 km/h [37º S of E], 70.0 km/hc. 50.0 km/h, 50.0 km/h [37º S of E]

____ 4. Which of the following graphs does NOT depict uniform motion?

a. A and B d. B and Db. C only e. A and Ec. D and E

____ 5. Which of the following graphs depicts uniform motion?

a. A and B d. B and Db. C and D e. E onlyc. A and C

____ 6. Which of the following statements concerning motion graphs is NOT correct?a. The slope of a position-time graph gives velocity.b. The area under a velocity-time graph gives displacement.

c. The slope of a velocity-time graph gives acceleration.d. The area under an acceleration-time graph gives velocity.e. The slope of the tangent in a position-time graph gives instantaneous velocity.

____ 7. Which of the following statements concerning motion graphs is correct?a. The slope of a position-time graph gives acceleration.b. The area under an acceleration-time graph gives instantaneous velocity.c. The slope of a velocity-time graph gives displacement.d. The area under a position-time graph gives velocity.e. The area under a velocity-time graph gives displacement.

____ 8. Which of the following descriptions best represents the acceleration-time graph of a car that pulls away from a corner when the light turns green, reaches and maintains a constant velocity, then slows down until it stops? Assume that all accelerations are uniform.a. All three sections of the graph are comprised of horizontal lines.b. Two sections of the graph are diagonal lines and one is horizontal.c. Two sections of the graph are horizontal lines and one is diagonal.d. All three sections of the graph are comprised of diagonal lines.e. All three sections of the graph are comprised of curved lines.

____ 9. 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 sb. 4.1 s e. 27 sc. 9.8 s

____ 10. Ignoring air resistance, which of the following are exhibiting “free fall”?a. an object, initially at rest, dropped out of a windowb. an object thrown vertically downward from a window c. an object projected vertically upward from a windowd. an object thrown horizontally from a windowe. all of the above

____ 11. Three identical objects are thrown from the same height through a window at the same time. Object A is thrown horizontally at 4.0 m/s, object B is thrown horizontally at 8.0 m/s, and object C is simply dropped. If air resistance is negligible, which object will reach the ground first?a. object Ab. object Bc. object Cd. objects B and C will land first and togethere. all three will land at the same time

____ 12. For a javelin thrower to maximize her throwing distance, she should release the javelin at an anglea. of 45b. of less than 45c. of greater than 45d. dependent on the javelin’s speed upon releasee. dependent on the thrower’s strength

____ 13. A boat always points directly at the opposite shore while crossing a river. The time it will take to cross will bea. less if the current is strongerb. greater if the current is strongerc. the same regardless of the currentd. dependent on the strength of the currente. impossible to predict without more information

____ 14. Three identical boats set out to cross a river that has a current. Boat A points directly across the river, boat B points 20º downstream from a point straight across the river, and boat C points 20º upstream from a point straight across the river. Which boat will arrive on the opposite shore first?a. boat Ab. boat Bc. boat Cd. it is impossible to tell with the information givene. all three boats will arrive at the same time

____ 15. The free-body diagram of a block being pushed up a rough ramp is best represented by

a. A d. Db. B e. Ec. C

____ 16. The free-body diagram of a car in a skid with its brakes locked up is best represented by


____ 17. An object sits at rest on a ramp. Which of the following free-body diagrams best represents the forces acting on the object?


____ 18. An elevator moves downward at a constant speed. What is the relationship between the gravitational force

acting on the elevator and the tension in the cable?a. d.

b. e.

c.

____ 19. According to Newton’s third law, when you walk across a floor, the force that propels you forward isa. the force applied by your feet on the floorb. the force of friction of your feet on the floorc. the force of the floor applied against your feetd. exerted upward by the floor on your feet (i.e., the normal force)e. the force acting on you working against gravity

____ 20. A 4.0-kg object, A, and a 2.0-kg object, B, are connected with a rope. A force is applied to another rope attached to the 2.0-kg object that pulls both A and B along a horizontal surface. Which of the following statements is true?a. The force that B exerts on A is greater than the force that A exerts on B.b. The force that A exerts on B is greater than the force that B exerts on A.c. The force that B exerts on A is equal to the force that A exerts on B provided that the

system slides with uniform motion.d. The force that B exerts on A is equal to the force that A exerts on B regardless of the

motion of the system.e. The sum of the applied force and the force that B exerts on A is equal to the force that A

exerts on B.____ 21. Three masses are suspended vertically as shown in the diagram below. The system is accelerating upward.

What is the relationship among the forces of tension?

a. d.

b. e.

c.

____ 22. For an object travelling with “uniform circular motion,” its acceleration isa. zero because the speed is constantb. directed tangent to the circlec. directed toward the centre of the circled. changing in magnitude depending on its position in the circlee. directed outward from the centre of the circle

____ 23. A child whirls a ball around in circles on the end of a 48 cm long string at a frequency of 2.5 Hz. What is the ball’s centripetal acceleration?a. 1.2 104 m/s2 d. 38 m/s2

b. 1.2 102 m/s2 e. 3.0 m/s2

c. 47 m/s2

____ 24. A rock is tied to the end of a 35 cm long string and whirled around in a circle that describes a vertical plane. The tension in the string becomes zero when the speed of the rock is

a. 9.8 102 cm/s d. 9.8 cm/sb. 1.9 102 cm/s e. 1.9 cm/sc. 19 cm/s

____ 25. The acceleration due to gravity on the surface of a planet having twice the Earth’s mass and twice its radius would bea. 39.2 m/s2 d. 4.9 m/s2

b. 19.6 m/s2 e. 2.45 m/s2

c. 9.8 m/s2

____ 26. Planet X has a radius 4 times that of Earth and the acceleration due to gravity at the surface of planet X is 4.9 m/s2. The mass of Planet X compared to Earth’s mass isa. 16 times d. 2 timesb. 8 times e. the samec. 4 times

____ 27. The force of gravity acting on a 10-kg object at an altitude equivalent to the Earth’s radius isa. 49 N d. 5.0 Nb. 24 N e. 2.4 Nc. 9.8 N

____ 28. The orbital speed of a satellite at an altitude equivalent to Earth’s radius (rE = 6.38 106 m) is (mE = 5.98 1024 kg, G = 6.67 10–11 N·m2/kg2)a. 9.8 103 m/s d. 4.9 103 m/sb. 7.9 103 m/s e. 2.5 103 m/sc. 5.6 103 m/s

____ 29. Astronauts on board an orbiting space station appear to be “floating” becausea. they are in the vacuum of spaceb. they are outside Earth’s gravitational influencec. the force of gravity acting on them has been reduced to an insignificant leveld. they have become truly “weightless”e. they are in free fall along with the space station itself

____ 30. Which of the following graphs best illustrates the relationship between a satellite’s orbital radius ro and its orbital speed vo?


____ 31. A ripple tank is used to generate water waves. These waves are refracted as they travel from deep to shallow water. Which of the following factors, when changed, will not affect the amount of bending observed?a. the angle between the boundary and the incident wave frontb. the difference in depth between the shallow and deep regionsc. the wavelength of the incident waved. the amplitude of the incident wavee. the frequency of the incident wave

____ 32. A two-point source interference pattern is generated in a ripple tank. Point P, on the second nodal line, is located 28 cm from one source and 37 cm from the other. The wavelength of the waves is

a. 18 cm d. 6.0 cmb. 14 cm e. 4.5 cmc. 9.0 cm

____ 33. In the diagram below (not to scale), two point sources, S1 and S2, are located 6 cm apart and are vibrating in phase. Point P is located on the first nodal line.

If the wavelength of the generated waves is cm, then PS1 – PS2 will be equal to

a. d.

b. e.

c.

____ 34. Two point sources vibrating in phase in a ripple tank are placed a fixed distance apart, creating a stationary nodal line pattern. Which of the following statements concerning the nodal lines is incorrect?a. In areas between the nodal lines, energy is transmitted away from the sources.b. When the distance from the sources is large, the nodal line separation is one-half of a

wavelength.c. The path length difference from the sources to any point on a given nodal line is a fixed

value.d. Nodal lines are a result of continuous destructive interference.e. The nodal lines are hyperbolas, becoming essentially straight lines at great distances from

the source.____ 35. In a two-point source interference pattern in a ripple tank, a point is one-half wavelength farther from one

source than the other. If the two sources are in phase, then there isa. destructive interference at this pointb. constructive interference at this pointc. both constructive and destructive interference simultaneouslyd. neither constructive nor destructive interference at this pointe. no interaction between the two waves at this point

____ 36. Two speakers producing exactly the same frequency and always in phase are located on an east–west line emitting sound toward the north. The speakers are 30 cm apart and a microphone is located 40 cm due north of one speaker. Which of the following wavelengths would produce sound with the loudest intensity at the location of the microphone?a. 50 cm d. 20 cmb. 40 cm e. 10 cmc. 30 cm

____ 37. A student counts a total of eight nodal lines on each side of a two-point source interference pattern and measures the sources to be 15 cm apart. What is the approximate wavelength of the waves?a. 0.50 cm d. 15 cmb. 2.0 cm e. none of the abovec. 7.5 cm

____ 38. A two-point source interference pattern is generated in a ripple tank by identical sources vibrating in phase and located 12.0 cm apart. There are seven nodal lines observed on each side of the centre line. If the wavelength of the sources is doubled and they remain in phasea. the number of nodal lines observed doublesb. the frequency doublesc. the speed of the wave doublesd. the number of nodal lines decreases to halfe. the average distance between nodal lines decreases

____ 39. Consider the following properties of light:I. rectilinear propagationII. reflectionIII. refractionIV. diffractionV. partial reflection-refractionWhich of these properties did Newton have trouble explaining with his theory of light?a. III only d. IV and V onlyb. IV only e. III and V onlyc. III and IV only

____ 40. A satellite receiver reflects incoming radio waves in a similar manner asa. straight water waves reflect from a straight barrierb. circular water waves reflect from a circular barrierc. straight water waves reflect from a concave barrierd. straight water waves reflect from a convex barriere. circular water waves reflect from a straight barrier

____ 41. Which of the following properties of light cannot be explained by the particle theory but can easily be explained by the wave theory?a. rectilinear propagation d. dispersionb. transmission in a vacuum e. diffractionc. refraction

____ 42. A student wishes to perform Young’s double-slit experiment to measure the wavelength of light of a given source. The student has measured the average distance between adjacent dark bands and the distance from the slits to the screen. What other measurement is required to calculate the wavelength?a. the location of the central bright lineb. the distance between the source and the slitsc. the angle between the central bright line and first nodal lined. the distance separating the two slitse. the distance between the centre of the two slits and any given dark band

____ 43. A student performs a double-slit experiment using two slits spaced 0.20 mm apart and located 1.50 m from the screen. The monochromatic light source creates an interference pattern in which the average distance between dark bands is 0.45 cm. What is the wavelength of the light being used?a. 6.0 10–9 m d. 6.0 10–5 mb. 6.0 10–8 m e. 15 mc. 6.0 10–7 m

____ 44. A student performs a double-slit experiment using a monochromatic light source, two slits spaced 0.10 mm apart, and a screen located 150 cm away. The bright fringes are located 0.30 cm apart. If the distance between the slits was changed to 0.20 mm, what would the average distance between bright fringes become?a. 0.15 cm d. 0.60 cmb. 0.30 cm e. 1.5 mc. 0.50 cm

____ 45. 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 cmb. 1.4 m e. 1.4 cmc. 14 cm

____ 46. Time isa. dependent on the observer d. simultaneous in all casesb. absolute e. never changingc. the same for different observers

____ 47. A beam of light is shone forward from a moving vehicle. The speed of light leaving the vehicle will bea. dependent on the speed of the vehicleb. the addition of the speed of light plus the speed of the vehiclec. constant at the speed of lightd. slower so that the addition of the vehicle’s speed corrects ite. none of the above

____ 48. The Lorentz contraction will apply to which of the following dimensions of an object:a. height relative to direction d. all dimensions equallyb. length relative to direction e. no contraction occursc. width perpendicular to direction

____ 49. Flying in a fast rocket ship at a speed of 0.82c, you observe both your watch and a clock outside. Which of the following statements is true?a. Time will appear to be the same for you but slower outside.b. Time will appear slower for you but normal outside.c. Neither clock runs slow.d. Both clocks run slow.e. none of the above

____ 50. Two spaceships are heading toward each other at a mutual speed of 0.999c. Both ships shine a laser beam at one another. How will each ship see the other light?a. Both ships will see the light travelling at 2c.b. Neither ship will see any light.c. Both ships will see the light travelling at c.d. One ship will see the light travelling at c, and the other will see it travelling at 2c.e. none of the above

____ 51. Mission control would like an cosmonaut to check her heart rate. She is flying at 0.50c. If a normal heart rate is 72 beats/min, what would a normal heart rate appear to be for the cosmonaut?a. 288 beats/min d. 72 beats/minb. 62 beats/min e. 18 beats/minc. 83 beats/min

____ 52. The relativistic momentum of an atomic particle of mass 1.62 10–27 kg moving at 0.92c isa. 1.14 10–18 kg·m/s d. 3.81 10–27 kg·m/sb. 1.75 10–19 kg·m/s e. 5.84 10–28 kg·m/sc. 3.29 10–19 kg·m/s

____ 53. The equation E = mc2 illustrates thata. travelling at the speed of light converts matter into energyb. rest mass and energy are equivalentc. energy can be converted into massd. matter can be converted into energye. both b and d

____ 54. The total energy of a particlea. can never increase or decreaseb. will increase due to the speed of lightc. will decrease due to the speed of lightd. will reach a limit before the speed of lighte. none of the above

____ 55. The total energy, in joules, of a 0.01-kg object moving at 0.55c isa. 7.38 1014 J d. 4.13 1015 Jb. 1.08 1015 J e. 6.58 10–3 Jc. 4.34 1014 J

____ 56. The energy, in joules, of light with a frequency of 5.6 1016 Hz isa. 5.6 1016 J d. 1.2 10–50 Jb. 8.4 1049 J e. 6.63 10–34 Jc. 3.7 10–17 J

____ 57. In the photoelectric effect, as the frequency of light shining on a piece of metal increases, the ejected electrona. gains kinetic energy d. increases in number of electronsb. loses velocity e. none of the abovec. remains unchanged

____ 58. In the photoelectric effect, the work function isa. the amount of energy required to release an electronb. the same as the threshold frequencyc. the amount of kinetic energy possessed by the electrond. dependent on the frequency of lighte. none of the above

____ 59. The momentum of a photon with a wavelength of 635 nm isa. 4.21 10–40 kg·m/s d. 1.04 10–36 kg·m/sb. 1.04 10–27 kg·m/s e. 4.21 10–31 kg·m/sc. 9.57 1026 kg·m/s

____ 60. A 200-g apple falling at 3.0 m/s has a de Broglie wavelength of a. 4.42 10–33 m d. 1.1 1–35 mb. 4.42 10–35 m e. 9.9 10–36 mc. 1.1 10–33 m

____ 61. An electron with 5.8 eV of energy strikes a mercury atom with energy levels of 4.9 eV and 6.67 eV. What energy will the electron posses after the collision?a. 0.0 eV d. 1.2 eVb. 0.9 eV e. 1.77 eVc. 0.87 eV

____ 62. The change in energy in moving from the first to the second energy level of hydrogen would be an increase of a. 2 times d.

times

b. times

e. none of the above

c. 4 times____ 63. A net force of 12 N changes the momentum of a 250-g ball by 3.7 kgm/s. The force acts for

a. 0.31 s d. 3.2 sb. 0.81 s e. 44 sc. 1.2 s

____ 64. A car with a mass of 1800 kg slows from 42 km/h [E] to 28 km/h [E]. The impulse from the brakes isa. 2.5 104 Ns [E] d. 2.1 104 Ns [W]b. 2.5 104 Ns [W] e. 7.0 103 Ns [W]c. 2.1 104 Ns [E]

____ 65. A 1.5-kg bird is flying at a velocity of 18 m/s [22º above the horizontal]. The vertical component of its momentum isa. 10 m/s [up] (2 significant digits) d. 17 m/s [up]b. 6.7 kgm/s [up] e. none of the abovec. 25 kgm/s [up]

____ 66. A bullet with a mass of 28 g is fired from a 2.8-kg gun that is stationary, but free to recoil. After the bullet is fired, the gun is observed to be moving at 1.4 m/s [left]. The velocity of the bullet isa. 140 m/s [left] d. 71 m/s [left]b. 140 m/s [right] e. 71 m/sc. 71 m/s [right]

____ 67. An arrow slows down from 43 m/s to 28 m/s as it passes through an apple. If the 493-g apple was originally at rest and sped up to 0.44 m/s, the mass of the arrow isa. 5.0 g d. 29 gb. 7.7 g e. 7.7 kgc. 14 g

____ 68. A moving curling stone, A, collides head on with a stationary stone, B. Both stones are of identical mass. If friction is negligible during this linear elastic collision,a. stone A will slow downb. after the collision, the momentum of stone B will be less than that of stone Ac. both stones will come to rest shortly after the collisiond. after the collision, the kinetic energy of the stone B will be less than that of stone Ae. after the collision, stone A will have a speed of zero

____ 69. A sabotaged curling stone explodes into three pieces as it travels across the ice. Neglecting the force of friction,a. all three pieces will travel at the same speedb. the magnitudes of the momenta for each piece will be the samec. an external net force had to act on the stone to accelerate the three piecesd. the components perpendicular to the original motion must add up to zeroe. momentum is not conserved because of the small explosive charge

____ 70. A two-dimensional collision occurs as shown below.

Which vector below most closely represents the new velocity of P?


____ 71. Two objects of equal mass with the speeds indicated by the vectors below, collide and stick together.

Which vector below best represents the velocity of the combined objects after the collision?


____ 72. A 722-kg satellite is in circular orbit 7380 km above the surface of Earth (ME = 5.98 1024 kg). The gravitational force acting on the satellite isa. 1.52 103 N d. 7.33 Nb. 5.29 109 N e. 7.08 103 Nc. 5.29 103 N

____ 73. The Sun has a mass of 1.99 1030 kg. Jupiter has a mass of 1.90 1027 kg and a mean radius of orbit around the Sun of 7.78 108 km. The speed that Jupiter travels in its orbit around the Sun isa. 1.31 104 km/s d. 4.04 102 m/sb. 4.70 104 km/h e. 1.28 104 m/s

c. 4.13 105 m/s____ 74. If the mass of Earth is 5.98 1024 kg and the radius is 6.38 106 m, the gravitational potential energy of a

1.2 103-kg satellite located in an orbit 230 km above the surface of Earth is a. –1.1 104 J d. –9.0 1012 Jb. –7.2 1010 J e. –2.1 1015 Jc. –2.1 1012 J

Short Answer75. For the same initial upward velocities, how many times higher will an object travel above the lunar surface (g

= 1.6 m/s2 [down]) than above the surface of Earth? Assume negligible air resistance on Earth.

76. With the aid of a diagram, explain why a bright band occurs on a screen at the central point of Young’s double-slit experiment.

77. A plane is dropping medical supplies to a village. Describe the path of the supplies relative to an observer on the ground watching the plane travel from left to right.

78. A plane is dropping medical supplies to a village. Describe the path of medical supplies as it appears to the person “dropping” the supplies.

79. Three observers located at A, B, and C are watching two stars located close to Observers A and C. Both stars explode simultaneously. Explain how the observers see the pattern of explosions.

A B C

80. Four observers located at A, B, and C are watching two stars located close to observers A and C. Both stars explode simultaneously. Explain how observers B and B' see the pattern of explosions if B' is moving away from B.

A B CB'

81. A car moving forward at 0.5c turns on its headlights. What would the speed of light appear to be to the driver and to a pedestrian in front of the car?

82. A person sees an UFO fly by at 0.9c and describes the ship to be the classic “flying saucer” shape. What shape of ship would you look for if the ship had landed on the ground?

83. A 57-g tennis ball travelling at 28 m/s is hit straight back with the same velocity. Determine the average force on the tennis ball if the racket is in contact with the ball for 4.9 ms.

84. A blazing spike of a 0.290-kg volleyball is blocked at the net. It is originally travelling at 18.3 m/s and bounces straight back at 14.9 m/s after being in contact with the blockers arms for a total of 18.2 ms. What average force did the blocker exert on the ball?

85. Give two observations that would enable you to conclude that the bounce of a superball is not a completely elastic collision.

86. During an elastic collision between a superball and the ground, the superball comes to rest for a brief instant. Where is the energy stored?

87. A 0.25-kg snowball moving at 15 m/s [E] collides and sticks with a 1.9-kg toy truck travelling at 2.8 m/s [W]. Neglecting friction, calculate the velocity of the snowball–truck system after the collision.

88. A 25-kg bag of cement thrown at 2.5 m/s [E] is caught by a person sliding 1.8 m/s [E] on a frictionless surface. If the velocity after the catch is 2.0 m/s, calculate the mass of the person.

89. A billiard ball collides with an identical stationary billiard ball causing the balls to travel out with speeds of 3.0 m/s and 4.0 m/s at 90º to each other as shown below. Calculate the initial speed of the moving billiard ball.

Problem90. 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?

91. An arrow is shot vertically upward with an initial speed of 25 m/s. When it’s exactly halfway to the top of its flight, a second arrow is launched vertically upward from the same spot. The second arrow reaches the first arrow just as the first arrow reaches its highest point.(a) What is the launch speed of the second arrow?(b) What maximum height does the second arrow reach?

92. A truck travels at a constant speed of 28.0 m/s in the fast lane of a two-lane highway. It approaches a stationary car stopped at the side of the road. When the truck is still 1.2 102 m behind the car, the car pulls out into the slow lane with an acceleration of 2.6 m/s2.(a) How long will it take the truck to pass the car?(b) How far will the car have travelled when the truck passes it?(c) If the car were to maintain this acceleration, how fast would it be travelling when it overtakes the truck?

93. Two canoeists, A and B, live on opposite shores of a 300.0 m wide river that flows east at 0.80 m/s. A lives on the north shore and B lives on the south shore. They both set out to visit a mutual friend X who lives on the north shore at a point 200.0 m upstream from A and 200.0 m downstream from B. Both canoeists can propel their canoes at 2.4 m/s through the water. How much time must canoeist A wait after canoeist B sets out so that they both arrive at X at the same time? Both canoeists make their respective trips by the most direct routes.

94. The graph below represents the motion of an object over a recorded time interval. Using methods of graphical analysis wherever possible, determine(a) the object’s displacement relative to its starting position at t = 6.0 s.(b) the object’s average velocity between t = 0.0 s and t = 6.0 s.(c) the object’s average speed between t = 0.0 s and t = 6.0 s.(d) Including t = 0.0 s, how many times during the entire recorded time interval is the object at its starting position?(e) During which interval is the object’s acceleration the greatest? What is the value of the acceleration during this interval?

(f) Plot the corresponding position-time graph.(g) Plot the corresponding acceleration-time graph.

95. A football quarterback attempts a pass to one of the receivers. As the ball is snapped, the receiver leaves the line of scrimmage and runs directly down field. The quarterback releases the ball 2.0 s later and from a position 3.0 m behind the line of scrimmage. He throws the ball with a speed of 26 m/s at an elevation of 60 above the horizontal. The receiver makes a diving reception, catching the ball just as it reaches the ground. See the diagram below.

(a) What is the time of flight of the football?(b) What is the average speed of the receiver?

96. A force of 3.5 N [60E of N] and a force of 2.8 N [40W of S] act on the same object. Find the net force acting on the object using (a) a trigonometric method and (b) a component method.

97. A 12.0-kg box is pushed along a horizontal surface by a 24-N force as illustrated in the diagram. The frictional force (kinetic) acting on the object is 6.0 N.

(a) What is the acceleration of the object?(b) Calculate the value of the normal force acting on the object.(c) If the 12.0-kg object then runs into a 4.0-kg object that increases the overall friction by 3.0 N, what is the new acceleration?(d) What force does the 4.0-kg object exert on the 12.0-kg object when the two are moving together?

98. A pulley device is used to hurl projectiles from a ramp (k = 0.26) as illustrated in the diagram. The 5.0-kg mass is accelerated from rest at the bottom of the 4.0 m long ramp by a falling 20.0-kg mass suspended over a frictionless pulley. Just as the 5.0-kg mass reaches the top of the ramp, it detaches from the rope (neglect the mass of the rope) and becomes projected from the ramp.

(a) Determine the acceleration of the 5.0-kg mass along the ramp. (Provide free-body diagrams for both masses.)(b) Determine the tension in the rope during the acceleration of the 5.0-kg mass along the ramp.(c) Determine the speed of projection of the 5.0-kg mass from the top of the ramp. (d) Determine the horizontal range of the 5.0-kg mass from the base of the ramp.

99. Crates of mass 50.0 kg must be hoisted onto a platform 8.0 m above the ground. A person exerts 600.0 N of force on a rope that goes up and over a pulley suspended from the ceiling. The other end of the rope is attached to the 50.0-kg mass.(a) How long will it take the person to lift a crate from the ground to the platform while exerting maximum force?(b) If a 45.0-kg person grabs the free end of the rope when the crate is next to the platform in its elevated position, what will the acceleration of the crate be as it falls?(c) What is the tension in the rope in this case?

100. 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?

101. Two blocks are connected by a “massless” string over a “frictionless” pulley as shown in the diagram.

(a) Determine the acceleration of the blocks.(b) Calculate the tension in the string .(c) If the string broke, for what minimum value of the coefficient of static friction would the 2.0-kg block not begin to slide?

102. Two masses, 4.0 kg and 6.0 kg, are connected by a “massless” rope over a “frictionless” pulley as pictured in the diagram. The ramp is inclined at 30.0º and the coefficient of kinetic friction on the ramp is 0.18.

(a) Draw free-body diagrams of both masses.(b) Determine the acceleration of the system once it begins to slide.(c) Determine the tension in the rope.(d) If the rope breaks when the 4.0-kg mass is 3.0 m from the bottom of the ramp, how long will it take for the mass to slide all the way down? Include a new free-body diagram and assume the sliding mass starts from rest.

103. 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.

104. 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.

105. A pilot of mass 75 kg takes her plane into a dive, pulling out of it along a circular arc as she nears the ground. If the plane is flying at 1.5 102 km/h along the arc, what is its radius such that the pilot feels four times heavier than normal? Provide an appropriate free-body diagram.

106. A planet has a mass of 2.5 times that of Earth and a radius 1.2 times Earth’s radius. How much would a 60.0-kg person weigh at the planet’s surface?

107. A satellite orbits Earth at an altitude of 325 km above the planet’s surface. What is its orbital period? Express your answer in minutes. (rE = 6.38 106 m, ME = 5.98 1024 kg)

108. 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

109. 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

110. 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.

111. A two-point source interference pattern is generated by sources operating in phase at 1.0 Hz. The sources are 2.0 m apart and the wavelength of the waves is 0.60 m. At what angles, measured from the centre line of the pattern, are the nodal lines produced located?

112. A student creates a two-point source interference pattern in a ripple tank with two sources operating in phase and records the following information: n = 4, = 37.5 mm, L = 1.25 m, d = 24 cm. Calculate x4.

113. A student creates a two-point source interference pattern in a ripple tank with two sources operating in phase. A point on the eighth nodal line is 1.25 m from the centre of the two sources and 48.0 cm from the perpendicular bisector of the two sources. If the source separation is 2.75 cm, find the wavelength of the waves.

114. A double-slit experiment is performed using a slit separation of 0.12 mm with a screen placed 80.0 cm away. There are 18 mm between the first and seventh nodal lines. What wavelength of light was used?

115. As you drive down the highway, you notice that the dial on your stereo is not functioning. You have the radio tuned to a station that uses two transmission towers that are 175 m apart. The towers are 25.0 km from your present location. You wish to estimate the frequency setting of your radio using the interference pattern set up by the two towers. You notice that the signal reception fluctuates between maximums as you drive a distance of 0.45 km parallel to the line joining the two towers. What is the frequency of the station to which you are listening?

116. An astronaut travelled to the next star at a speed of 0.95c, and recorded that the round trip took 11 years. How much time will have passed on Earth when he returned?

117. A pilot on a distant voyage to a star is placed in suspended animation for the journey. The ship’s clock recorded that he aged 15 years but the trip had lasted 132 years relative to Earth. How fast was the ship travelling?

118. The distance between two planets was measured by an alien spacecraft to be 4.53 light-years. If the alien pilot was travelling at 0.93c, how far away was the planet in proper distance?

119. Superman in saving Earth again, throws a 500-g bomb into space, where it will safely explode. If he throws it with a speed of 0.52c, calculate the bombs relativistic momentum.

120. If a proton moves at 0.750c, calculate the total energy of the proton in the laboratory frame of reference in MeV.

121. For an electron with speed 0.880c, calculate its kinetic energy if its total energy is 0.980 MeV.

122. Would 449-nm blue light eject electrons from silver metal with a work function of 4.74 eV?

123. 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?

124. Calculate the wavelength of light equivalent to an electron moving at 3.98 105 m/s.

125. A 112-kg satellite is given an escape velocity of 1.11 104 m/s to leave Earth’s orbit. What wavelength will the satellite experience at this speed?

126. 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?

127. What wavelength of photon could excite an electron from the third to the fifth energy state of hydrogen?

128. During a free dance program in figure skating, Victor (m = 71 kg) glides at 2.1 m/s to a stationary Shae-Lynn (52 kg) and hangs on. How far will the pair slide after the “collision” if coefficient of kinetic friction K between their skates and the ice is 0.052?

129. A spring with a force constant of 89 N/m is compressed 8.7 cm and placed between two stationary dynamics carts of mass 1.0 kg and 1.5 kg. If friction is negligible, determine the final speed of the more massive cart when the spring is released.

130. A 34-g bullet travelling at 120 m/s embeds itself in a wooden block on a smooth surface. The block then slides toward a spring and collides with it. The block compresses the spring (k = 99 N/m) a maximum of 1.2 cm. Calculate the mass of the block of wood.

131. 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.

132. A bullet with a mass of 45 g is fired into a 8.3-kg block of wood resting on a floor against a spring. This ideal spring (k = 76 N/m) has a maximum compression of 28 cm. What was the initial speed of the bullet?

133. Given that Fc = Fg for a satellite, show that the radius of orbit for an Earth satellite is .

134. 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)

135. How fast must a satellite leave Earth’s surface to reach an orbit with an altitude of 895 km?

EXAM REVIEW - D. O'Hara - SPH4U1Answer Section

MULTIPLE CHOICE1. ANS: C REF: K/U OBJ: 1.4 LOC: FM1.032. ANS: A REF: K/U OBJ: 1.1 LOC: FM1.013. ANS: B REF: K/U OBJ: 1.1 LOC: FM1.024. ANS: C REF: K/U OBJ: 1.1 LOC: FM1.025. ANS: E REF: K/U OBJ: 1.1 LOC: FM1.026. ANS: D REF: K/U OBJ: 1.1 LOC: FM1.017. ANS: E REF: K/U OBJ: 1.1 LOC: FM1.028. ANS: A REF: K/U OBJ: 1.2 LOC: FM1.019. ANS: B REF: K/U OBJ: 1.3 LOC: FM1.02

10. ANS: E REF: K/U OBJ: 1.4 LOC: FM1.0211. ANS: E REF: K/U OBJ: 1.4 LOC: FM1.0312. ANS: A REF: K/U OBJ: 1.4 LOC: FM1.0313. ANS: C REF: K/U OBJ: 1.5 LOC: FM1.0214. ANS: A REF: K/U OBJ: 1.5 LOC: FM1.0215. ANS: E REF: K/U OBJ: 2.1 LOC: FM1.0116. ANS: A REF: K/U OBJ: 2.1 LOC: FM1.0117. ANS: C REF: K/U OBJ: 2.1 LOC: FM1.0118. ANS: A REF: K/U OBJ: 2.2 LOC: FM1.0119. ANS: C REF: K/U OBJ: 2.2 LOC: FM1.0120. ANS: D REF: K/U OBJ: 2.3 LOC: FM1.0121. ANS: B REF: K/U OBJ: 2.3 LOC: FM1.0122. ANS: C REF: K/U OBJ: 3.1 LOC: FM1.0423. ANS: B REF: K/U OBJ: 3.1 LOC: FM1.0424. ANS: B REF: K/U OBJ: 3.2 LOC: FM1.0425. ANS: D REF: K/U OBJ: 3.3 LOC: FM1.0626. ANS: B REF: K/U OBJ: 3.3 LOC: FM1.0627. ANS: B REF: K/U OBJ: 3.3 LOC: FM1.0628. ANS: C REF: K/U OBJ: 3.4 LOC: FM1.0629. ANS: E REF: K/U OBJ: 3.4 LOC: FM1.0630. ANS: A REF: K/U, I OBJ: 3.4 LOC: FM1.0631. ANS: D REF: K/U OBJ: 9.1 LOC: WA1.0132. ANS: D REF: K/U OBJ: 9.3 LOC: WA1.0333. ANS: C REF: K/U, C OBJ: 9.3 LOC: WA1.0334. ANS: B REF: K/U OBJ: 9.3 LOC: WA2.0135. ANS: A REF: K/U OBJ: 9.3 LOC: WA2.0136. ANS: E REF: MC OBJ: 9.3 LOC: WA2.0437. ANS: B REF: I OBJ: 9.3 LOC: WA2.0138. ANS: D REF: I OBJ: 9.3 LOC: WA2.0139. ANS: D REF: K/U OBJ: 9.4 LOC: WA1.0540. ANS: C REF: MC OBJ: 9.4 LOC: WA3.0241. ANS: E REF: K/U OBJ: 9.4 LOC: WA1.0542. ANS: D REF: I OBJ: 9.5 LOC: WA2.03

43. ANS: C REF: K/U, C OBJ: 9.5 LOC: WA2.0244. ANS: A REF: K/U OBJ: 9.5 LOC: WA2.0245. ANS: E REF: K/U, C OBJ: 9.5 LOC: WA2.0246. ANS: A REF: K/U OBJ: 11.1 LOC: ME1.0147. ANS: C REF: K/U OBJ: 11.2 LOC: ME1.0548. ANS: B REF: K/U OBJ: 11.2 LOC: ME1.0549. ANS: A REF: K/U OBJ: 11.2 LOC: ME1.0550. ANS: C REF: MC OBJ: 11.2 LOC: ME2.0251. ANS: D REF: I OBJ: 11.2 LOC: ME2.0252. ANS: A REF: I OBJ: 11.2 LOC: ME2.0253. ANS: E REF: K/U OBJ: 11.3 LOC: ME1.0654. ANS: B REF: K/U OBJ: 11.3 LOC: ME1.0655. ANS: B REF: I OBJ: 11.3 LOC: ME1.0656. ANS: C REF: I OBJ: 12.1 LOC: ME1.0157. ANS: A REF: K/U OBJ: 12.1 LOC: ME1.0358. ANS: A REF: K/U OBJ: 12.1 LOC: ME1.0359. ANS: B REF: I OBJ: 12.1 LOC: ME1.0360. ANS: C REF: I, C OBJ: 12.2 LOC: ME1.0461. ANS: B REF: K/U OBJ: 12.4 LOC: ME1.0662. ANS: D REF: K/U OBJ: 12.5 LOC: ME1.0463. ANS: A REF: K/U OBJ: 5.1 LOC: EM1.0164. ANS: E REF: K/U OBJ: 5.1 LOC: EM1.0165. ANS: E REF: C OBJ: 5.1 LOC: EM1.0166. ANS: B REF: K/U OBJ: 5.2 LOC: EM1.0367. ANS: C REF: K/U OBJ: 5.2 LOC: EM1.0268. ANS: E REF: MC OBJ: 5.3 LOC: EM1.0469. ANS: D REF: K/U OBJ: 5.4 LOC: EM1.0370. ANS: B REF: I OBJ: 5.4 LOC: EM1.0371. ANS: D REF: I OBJ: 5.4 LOC: EM1.0372. ANS: C REF: K/U OBJ: 6.1 LOC: EM1.0673. ANS: B REF: K/U OBJ: 6.2 LOC: EM1.0674. ANS: B REF: K/U OBJ: 6.3 LOC: EM1.07

SHORT ANSWER75. ANS:

From the expression

it can be seen that the distance an object will travel is inversely proportional to the acceleration due to gravity:

Therefore, .

An object will travel 6.1 times higher on the moon than on Earth when projected vertically upward from the two surfaces with the same initial velocity.REF: C OBJ: 1.3 LOC: FM1.05

76. ANS:The central point is located equidistant from each of the two slits. Since the light through each slit originates from a single source, it is coherent (in phase) when it leaves the slits and reaches the centre of the screen in phase. This creates constructive interference, or in the case of light, a bright band.

REF: C OBJ: 9.5 LOC: WA1.0477. ANS:

The medical supplies would travel in a parabolic arch down and to the right relative to an observer on the ground.REF: MC OBJ: 11.1 LOC: ME1.01

78. ANS:It would appear that the supplies are falling straight down from the plane.REF: MC OBJ: 11.1 LOC: ME1.01

79. ANS:- Observer B will see both stars explode simultaneously.- Observer A will see the closer star at A explode, and then the farther star at C explode some time later.- Observer C will see the closer star at C explode, and then the farther star at A explode some time later.REF: K/U OBJ: 11.1 LOC: ME1.05

80. ANS:Observer B' will see the explosions as if they were simultaneous but will see them explode some time after B has seen the explosion.REF: K/U OBJ: 11.1 LOC: ME1.05

81. ANS:Both the driver and the pedestrian would see the light moving at c, the speed of light.REF: MC OBJ: 11.2 LOC: ME1.05

82. ANS:The ship would appear to be “longer” but no change in height would occur.REF: K/U OBJ: 11.2 LOC: ME2.02

83. ANS:We can neglect the force of gravity because it is so small.

The average force acting on the ball is 6.5 102 N.REF: K/U OBJ: 5.1 LOC: EM1.01

84. ANS:We can neglect the force of gravity because it is so small.

The average force acting on the ball is 529 N.REF: K/U OBJ: 5.1 LOC: EM1.01

85. ANS:- The ball does not reach its original height after the bounce. (some loss of energy)- Sound is produced. (sound energy must come from original kinetic energy)REF: K/U OBJ: 5.3 LOC: EM1.04

86. ANS:The energy is stored as elastic potential energy in the deformed shape of the ball.REF: K/U OBJ: 5.3 LOC: EM1.04

87. ANS:Choose east as the +x direction.

The final velocity is 0.73 m/s [W].REF: K/U OBJ: 5.2 LOC: EM1.02

88. ANS:Choose east as the +x direction.

The mass of the person is 62 kg.REF: K/U OBJ: 5.2 LOC: EM1.02

89. ANS:The 90º angle means we can use the Pythagorean theorem.

The initial speed of the billiard ball was 5.0 m/s.REF: K/U OBJ: 5.4 LOC: EM1.03

PROBLEM90. ANS:

(a)v1 = 0.0 m/sa = 5.0 m/s2

d = 80.0 cm = 0.800 mv2 = ?

The speed of the object upon reaching the rough section is 2.8 m/s.

(b)v1 = 2.83 m/sv2 = 0.0 m/st = 2.5 sa = ?

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:v = 2.83 m/st = 4.0 sd = vt = 2.83 m/s(4.0 s) = 11.32 m

During the period of deceleration:v1 = 2.83 m/sv2 = 0.0 m/st = 2.5 s

Total distance the object slides: 0.800 m + 11.32 m + 3.54 m = 16 mThe object slides a total distance of 16 m.REF: K/U OBJ: 1.2 LOC: FM1.02

91. ANS:

(a)Using the sign convention that “up” is (–) and “down” is (+):v1 = –25 m/sv2 = 0.0 m/sa = 9.8 m/s2

d = ?

The arrow travels 31.9 m upward to its highest point. The halfway position is 15.9 m.The time to travel the last half of its flight:d = –15.9 mv2 = 0.0 m/sa = 9.8 m/s2

t = ?

For the second arrow:d = -31.9 ma = 9.8 m/s2

t = 1.80 sv1 = ?

The speed of the second arrow at launch is 27 m/s [upward].

(b)Finding the maximum height of the second arrow:

v1 = –26.5 m/sv2 = 0.0 m/sa = 9.8 m/s2

d = ?

The second arrow reaches a maximum height of 36 m [upward].REF: K/U OBJ: 1.3 LOC: FM1.02

92. ANS:(a)Car: v1C = 0.0 m/s, aC = 2.6 m/s2

Truck: vT = 28.0 m/s

Car:

Truck: dT = vTtdT = 28.0 t

dT = dC + 1.2 102 m28.0 t = 1.3(t)2 + 1.2 102 solving the quadratic: t = 5.9 s, 16 sThe truck passes the car after 5.9 s.

(b)v1C = 0.0 m/saC = 2.6 m/s2

t = 5.9 sdC = ?

The car travels 45 m by the time the truck passes it.

(c)v1C = 0.0 m/saC = 2.6 m/s2

t = 15.6 s (the other root of the quadratic)v2C = v1C + aCt

= 2.6 m/s2(15.6 s)

v2C = 41 m/s

The car will be travelling at 41 m/s when it passes the truck if it maintains its acceleration.REF: K/U OBJ: 1.2 LOC: FM1.02

93. ANS:Canoeist B:

Using sine law: .The component of across the river is: 2.4sin(56 + 16) = 2.28 m/s.The time for B to cross to point X:

Canoeist A:

The time for A to reach point X:

Canoe A must wait 131.6 s – 125 s = 6.6 s.REF: K/U OBJ: 1.5 LOC: FM1.05

94. ANS:

(a) displacement = area under graph= 23.75 m [S] + 18.75 m [N]

displacement = 5.0 m [S]

(b)

The object’s average velocity during the first 6.0 s is 0.83 m/s [S].

(c)

The object’s average speed during the first 6.0 s is 7.1 m/s.

(d) The object is at its starting location 3 times throughout the motion.

(e) The object’s acceleration is greatest between t = 6.5 s and 7.0 s. (the greatest slope) acceleration = slope of graph = 30 m/s2 [N]

(f)

(g)

REF: K/U OBJ: 1.2 LOC: FM1.0295. ANS:

(a)Time of flight: let “up” be (–) and “down” be (+)v1 = –26 m/s(sin 60º) = –22.5 m/sa = 9.8 m/s2

d = 2.0 mt = ?

2.0 = (–22.5)t + 4.9(t)2

Solving the quadratic: t = 4.68 sThe time of flight is 4.7 s.

(b)

Horizontal range: d = vt = 26 m/s(cos 60)(4.68 s) = 60.8 mThe receiver must run: 60.8 m – 3.0 m = 57.8 m.The time the receiver has to reach the football: 4.68 s + 2.0 s = 6.68 s.

The average speed of the receiver:

The receiver must run with an average speed of 8.7 m/s.REF: K/U OBJ: 1.4 LOC: FM1.03

96. ANS:(a) Trigonometric Method

Looking at the vector triangle: = 20.Using cosine law: = 1.3 NUsing sine law: = 18.As a result: .

(b) Component MethodF1X = 3.5 N(cos 30) [E] = 3.03 N [E]F2X = 2.8 N(sin 40) [W] = 1.80 N [W]FX = 3.03 N [E] + 1.80 N [W] = 1.23 N [E]

F1Y = 3.5 N(sin 30) [N] = 1.75 N [N]F2Y = 2.8 N(cos 40) [S] = 2.14 N [S]FY = 1.75 N [N] + 2.14 N [S] = 0.39 N [S]

Using Pythagoras:

Using a trigonometric ratio: = tan–1 = 18.

As a result: .

The two methods give equivalent results.REF: C OBJ: 2.3 LOC: FM1.01

97. ANS:(a)Free-body diagram: FN acting up

Fg acting downFA acting as illustratedFK acting to the right

“Up” and “to the right” are the positive directions.

Horizontally:

The acceleration of the object is 1.0 m/s2.

(b) Vertically:

The normal force is 1.3 102 N[up].

(c)Free-body diagram: FN acting up

Fg acting downFA acting to the leftFK acting to the right


The acceleration of the two masses is 0.59 m/s2.

(d)Free-body diagram: FN acting up

Fg acting downFA acting to the leftFK acting to the rightF acting to the right (force of 4.0 kg object on 12.0 kg object)


The 4.0-kg object exerts a force of 5.3 N on the 12.0-kg object.REF: K/U OBJ: 2.3 LOC: FM1.02

98. ANS:(a)For the 5.0-kg mass:Free-body diagram: FN acting perpendicular to ramp and up

Fg acting downFT acting up along the ramp (this is the positive direction)FK acting down along the ramp (this is the negative direction)

5.0 kg(a) = FT – mg(cos ) – mg(sin )5.0 kg(a) = FT – 35.5 N

For the 20.0-kg mass:Free-body diagram: FT acting up (this is the negative direction)

Fg acting down (this is the positive direction)

20.0 kg(a) – 196 N – FT

Solving the system of equations: a = 6.4 m/s2

The acceleration of the 5.0-kg mass along the ramp is 6.4 m/s2.

(b)

The tension in the cable is 68 N.

(c)

The speed of projection of the mass off the top of the ramp is 7.2 m/s.

(d)Vertically: Let “up” be (–) and “down” be (+).a = 9.8 m/s2

d = 6.0 m

Horizontal range:

The horizontal range for the projected mass is 9.5 m.

REF: K/U OBJ: 2.3 LOC: FM1.0199. ANS:

(a)Free-body diagram of the crate: FT acting up

Fg acting down

Let “up” be (–) and “down” be (+).

= –600 N + 50.0 kg(9.8 N/kg) = –110 N

It will take 2.7 s to lift the crate.

(b)For the 50.0-kg mass:Free-body diagram: FT acting up (this is the negative direction)

Fg acting down (this is the positive direction)

50.0 kg(a) = 50.0 kg(9.8 N/kg) – FT

For the 45.0-kg mass:Free-body diagram: FT acting up (this is the positive direction)

Fg acting down (this the negative direction)

45.0 kg(a) = 45.0 kg(9.8 N/kg) – FT

Solving the system of equations:a = 0.52 m/s2

The acceleration of the crate will be 0.52 m/s2.

(c)FT = 45.0 kg(a) + 45.0 kg (9.8 N/kg)= 45.0 kg(0.52 m/s2) + 45.0 kg(9.8 N/kg)FT = 4.6 102 NThe tension in the cable would be 4.6 102 N.REF: K/U OBJ: 2.3 LOC: FM1.01

100. ANS:(a)Free-body diagram: FN acting perpendicular to the roof (upward)

Fg acting downFK 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 mThe ice lands 2.7 m from the base of the building.REF: K/U OBJ: 2.3 LOC: FM1.01

101. ANS:(a)

For the 0.80-kg mass:Free-body diagram: FN acting up

Fg acting down FT acting to the right (this is the positive direction)FK acting to the left (this is the negative direction)

0.80 kg(a) = FT – KFN0.80 kg(a) = FT – 0.14(0.80 kg)(9.8 N/kg)0.80 kg(a) = FT – 1.10 N

For the 2.0-kg mass:Free-body diagram: FN acting perpendicular to the ramp (upward)

Fg acting downFT acting up along the ramp (this is the negative direction)FK acting up along the ramp

2.0 kg(a) = 2.0 kg(9.8 N/kg)(sin 30º) – FT – 0.14(2.0 kg)(9.8 N/kg)(cos 30º)2.0 kg(a) = –FT + 7.42 N


The system will accelerate at 2.3 m/s2.

(b)FT = 0.80 kg(a) + 1.10 N= 0.80 kg(2.26 m/s2) + 1.10 NFT = 2.9 NThe tension in the string is 2.9 N.

(c)If the block remains stationary: FS = Fg sin = 2.0 kg(9.8 N/kg)(sin 30)FS = 9.8 N

The minimum coefficient of static friction required is 0.58.REF: K/U OBJ: 2.3 LOC: FM1.01

102. ANS:(a)For the 4.0-kg mass:Free-body diagram: FN acting perpendicular to the ramp (upward)

Fg acting downFT acting up along the ramp (this is the positive direction)FK acting down along the ramp (this is the negative direction)

For the 6.0-kg mass:

Free-body diagram: Fg acting down (this is the positive direction)FT acting up (this is the negative direction)

(b)For the 4.0-kg mass:

4.0 kg(a) = FT – mg(cos ) – mg(sin )4.0 kg(a) = FT – 13.5 N

For the 6.0-kg mass:

6.0 kg(a) = 58.8 N – FT


The acceleration of the 4.0-kg mass along the ramp is 4.5 m/s2.

(c)FT = 4.0 kg(a) +13.5 N= 4.0 kg(4.53 m/s2) + 13.5 NFT = 32 NThe tension in the cable is 32 N.

(d)For the block sliding down the ramp:Free-body diagram: FN acting perpendicular to the ramp (upward)

Fg acting downFK acting up along the ramp (this is the negative direction)

ma = mg(sin ) – mg(cos )a = 9.8 N/kg(sin 30º) – (0.18)(9.8 N/kg)(cos30º)a = 3.37 m/s2

It would take 1.3 s to reach the bottom of the ramp.REF: K/U OBJ: 2.3 LOC: FM1.01

103. ANS:(a)

For the engine:Free-body diagram: FN acting up

Fg acting down FK acting to the left FA acting as indicatedFT acting to the left (force of caboose on engine)

For the caboose:Free-body diagram: 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.REF: K/U OBJ: 2.3 LOC: FM1.01

104. 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].REF: K/U OBJ: 3.2 LOC: FM1.04

105. ANS:The free body diagram of the pilot at the bottom of the arc:

FN = force of seat exerted upward on the pilot (the normal force)FN = 4mg

The radius of the arc is 59 m.REF: K/U OBJ: 3.2 LOC: FM1.04

106. ANS:The weight of a 60.0-kg person at Earth’s surface:Fg = mg= 60.0 kg(9.8 N/kg)Fg = 588 N

Since , the two planets can be compared.

Earth: FE = 588 N Planet: FP = ?m1 = 60.0 kg m1 = 60.0 kgm2 = mE m2 = 2.5 mE

r = rE rP = 1.2rE

The person would weigh 1.0 103 N at the planet’s surface.REF: K/U OBJ: 3.3 LOC: FM1.06

107. ANS:The orbital radius is 6.38 106 m + 3.25 105 m = 6.705 106 mThe centripetal force acting on the satellite is supplied by gravity.FC = Fg

The orbital period is 91.0 min.REF: K/U OBJ: 3.4 LOC: FM1.06

108. 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.REF: K/U OBJ: 9.1 LOC: WA1.01

109. ANS:(a)

The index of refraction between the two regions is 0.45.

(b)

The wavelength is 13 cm.REF: K/U OBJ: 9.1 LOC: WA2.03

110. ANS:

The speed is 18 m/s, the wavelength is 5.3 m, and the frequency is 3.3 Hz.REF: K/U OBJ: 9.3 LOC: WA1.01

111. ANS:

The three nodal lines appear at angles of 8.6o, 27o, and 49o from the centre line.REF: K/U OBJ: 9.3 LOC: WA1.01

112. ANS:

The value for x4 is 68 cm.REF: K/U, C OBJ: 9.3 LOC: WA1.01

113. ANS:

The wavelength is 1.41 mm.REF: K/U, C OBJ: 9.3 LOC: WA1.01

114. ANS:

The wavelength of light used was 4.5 10–7 m.REF: K/U, C OBJ: 9.5 LOC: WA2.02

115. ANS:

The frequency of the station is 95 MHz.REF: MC OBJ: 9.5 LOC: WA3.02

116. ANS:v = 0.95c

= 11 a

The time that had passed on Earth was 35 years.REF: I OBJ: 11.2 LOC: ME2.02

117. ANS:

The ship was travelling at 0.99 the speed of light.REF: I OBJ: 11.2 LOC: ME2.02

118. ANS:v = 0.93cLm = 4.53 ly

The proper length between the planets was 12.2 ly.REF: I OBJ: 11.2 LOC: ME2.02

119. ANS:

m = 500 g (0.5 kg)v = 0.52c

The bomb's relativistic momentum is 9.1 107 kg·m/s.REF: I OBJ: 11.2 LOC: ME2.02

120. ANS:1.60 10–19 J 1 eVm 1.672 10–27 kgv 0.750c

The total energy of the proton is 142 MeV.REF: I OBJ: 11.3 LOC: ME1.06

121. ANS:

The electron has a kinetic energy of 0.468 MeV.REF: I OBJ: 11.3 LOC: ME1.06

122. ANS:

No, an electron would not be ejected because the kinetic energy is not a positive value.REF: I OBJ: 12.1 LOC: ME1.03

123. ANS:

v = ?

The electron would be ejected from the sodium with a speed of 3.98 105 m/s.REF: I, C OBJ: 12.1 LOC: ME1.03

124. ANS:m 9.11 10–31 kg

p = mvp (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.REF: I OBJ: 12.2 LOC: ME1.03

125. ANS:

The satellite will have a wavelength of 5.33 10–40 m.REF: I OBJ: 12.2 LOC: ME1.04

126. ANS:

The wavelength of light that would be emitted would be 140 nm.REF: I, C OBJ: 12.4 LOC: ME1.03

127. ANS:

The photon would have a wavelength of 1280 nm.REF: I OBJ: 12.5 LOC: ME2.01

128. ANS:Before the collision:

After the collision:

Now using the kinematics equations:

The pair coast a total of 1.4 m after the collision.REF: K/U OBJ: 5.2 LOC: EM1.02

129. ANS:Using conservation of momentum:

Now use conservation of energy:

The final speed of the more massive cart is 0.42 m/s.REF: K/U OBJ: 5.3 LOC: EM1.02

130. ANS:Momentum is conserved during the collision.

Conserved energy for the spring compression:

Now sub in to calculate the mass of the wooden block

The mass of the wooden block is 12 kg.REF: K/U OBJ: 5.3 LOC: EM1.02

131. ANS:The momentum of the 200-g piece, p2, is 0.20 1.4 = 0.28 kgm/s.The momentum of the 300-g piece, p3, is 0.30 0.90 = 0.27 kgm/s.The momentum of the unknown piece, pm, is m 1.8 = 1.8m kgm/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.)REF: K/U OBJ: 5.4 LOC: EM1.03

132. ANS:First use conservation of energy after the collision until the maximum compression:

Now use conservation of momentum for the collision:

The impact speed of the bullet was 1.6 102 m/s.REF: K/U OBJ: 5.3 LOC: EM1.02

133. ANS:Fc = Fg

REF: C OBJ: 6.2 LOC: EM1.06134. ANS:

The work

done against gravity is 8.3 108 J.REF: K/U OBJ: 6.3 LOC: EM1.07

135. ANS:

The launch speed would need to be 8.38 103 m/s.REF: K/U OBJ: 6.3 LOC: EM1.07

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