TRƯỜNG ĐẠI HỌC SƯ PHẠM KHOA VẬT LÝ BÀI TẬP VẬT LÝ CLC (CƠ & NHIỆT) DÙNG CHO SINH VIÊN KHỐI CLC ĐẠI HỌC BÁCH KHOA LƯU HÀNH NỘI BỘ Đà Nẵng, 2017 PART I: MECHANICS CHAPTER 1: PHYSICS AND MEASUREMENT No problems -CHAPTER 2: MOTION IN ONE DIMENSION 1.1 You run 100 m in 12 s, then turn around and jog 50m back toward the starting point in 30s Calculate your average speed and your average velocity for the total trip Đáp số: tốc độ trung bình: 3,57m/s ; vận tốc trung bình: 1,19m/s 1.2 Two train 75 km apart approach each other on parallel tracks, each moving at 15km/h A bird flies back and forth between the trains at 20km/h until the trains pass each other How far does the bird fly? Đáp số: t = 2,5h ; s = 50km 1.3 Upon graduation, a student throws his cap upward with in initial speed of 14,7m/s Given that its acceleration is 9,8m/s2 downward (we neglect air resistance) a) How long does it take to reach its highest point? b) What is the distance to the highest point? c) What is the total time the cap is in the air? Đáp số: a) t1 = 1,5s ; b) s = 11,0m; c) t = 3s x Highest point a = -g x=0 1.4 On a highway at night you see a stalled vehicle and brake your car to stop with an acceleration of magnitude 5m/s2 (deceleration) What is the car’s stopping distance if its initial speed is a) 15 m/s (about 54 km/h); b) 30 m/s Đáp số: a) s = 22,5m ; b) 90m 1.5 An electron in a cathode-ray tube accelerates from rest with a constant acceleration of 5,33 1012 m/s2 for 0,15 s The electron then drifts with constant velocity for 0,2 s Finally, it comes to rest with an acceleration of -0,67 1043 m/s2 How far does the electron travel ? Đáp số: s = 0,232m 1.6 While standing in an elevator, you see a screw fall from the ceiling The ceiling is 3m above the floor a) If the elevator is moving upward with a constant speed of 2,2m/s, how long does it take for the screw to hit the floor? b) How long is the screw in the air if the elevator starts from rest when the screw falls, and moves upward with a constant acceleration of a = 4m/s2 ? Đáp số: a) t = 0,78s; b) t’ = 0,66s CHAPTER 3: MOTION IN TWO DIMENSIONS 2.1 A plane is to fly due north The speed of the plane relative to the air is 200km/h and the wind is blowing from west to east at 90km/h a) In which direction should the plane head? b) How fast does the plane travel relative to the ground Đáp số: a) Máy bay phải bay theo hướng Tây-Bắc ; b) vpg = 179km/h 2.2 A helicopter drops a supply package to soldiers in a jungle clearing When the package is dropped, the helicopter is 100m above the clearing and flying at 25 m/s at an angle 0 = 36,90 above the horizontal Choose the origin to be directly below the helicopter when the package is dropped a) Where does the package land? b) If the helicopter flies at constant velocity, where is it when the package lands c) Find the time t for the package to reach its greatest height h; find h Đáp số: a) d = 126m ; b) Tọa độ x = 126m, y = 194,5m ; c) t = 1,53s ; h = 111,5m 2.3 A policeman chases a thief across city rooftops They are both running at m/s When they come to a gap between buildings that is 4m wide and has a drop of 3m The thief, having studied a little physics, leaps at m/s and at 450 and clears the gap easily The policeman did not study physics and he leaps at m/s horizontally 3m a) Does the policeman clear the gap? b) By how much does the thief clear the gap? 4m Đáp số: a) viên cảnh sát vừa chuyển động ngang vừa rơi, thời gian rơi 3m hết 0,782s, chuyển động ngang sau 0,782s đạt 3,91m Do không vượt qua b)Tên trộm nhảy lên nên có thời gian 1,22s; chuyển động ngang đạt 4,31m; dài khoảng trống 0,31m) 2.4 A hunter with a gun intends to shoot a monkey hanging from a branch The hunter aims directly at the monkey’s heart The monkey, hearing the gun discharge lets go of the branch and drops out the tree Assume that the reaction time of the monkey and air resistance are negligible Find the distance between the monkey’s heart and the point that the bullet hits the monkey Đáp số: Viên đạn chạm vào vị trí tim khỉ khỉ viên đạn rơi tự khoảng 2.5 A stone is thrown from the top of a building at an angle of 300 to the horizontal and with an initial speed of 20m/s If the height of the building is 45m, a) How long is the stone “in flight”? b) What is the speed of the stone just before it strikes the ground? Đáp số: a) t = 4,22s ; v = 35,9m/s 2.6 It has been said that in his youth George Washington threw a silver dollar across a river Assuming that the river was 75m wider, a) What minimum initial speed was necessary to get the coin across the river? b) How long was the coin in flight? Đáp số: a) vm = 27,1m/s ; b) t = 3,91s 2.7 An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 30m if her initial speed is 9m/s What is the acceleration of gravity on the planet? Đáp số: g = 2,7m/s2 2.8 The orbit of the moon about the earth is approximately circular, with mean radius of 3,84.108m It takes 27,3 days for the moon to complete one revolution about the earth Find: a) the mean orbital speed of the moon; b) its centripetal acceleration Đáp số: a) v = 1,02.103m/s ; b) a = 2,72.10-3m/s2 2.9 A worker on the roof of a house drops his hammer, which slides down the roof at a constant speed of 4m/s The roof makes an angle of 300 with the horizontal, and its lowest point is 10m from the ground What is the horizontal distance traveled by the hammer after it leaves the roof of the house and before it hits the ground? Đáp số: s = 4,26m -3 CHAPTER THE LAWS OF MOTION 4.1 The gravitational force exerted on a baseball is −𝐹𝑔 𝑗̂ A pitcher throws the ball with velocity 𝑣𝑖̂ by uniformly accelerating it along a straight horizontal line for a time interval of ∆𝑡 = 𝑡 − = 𝑡 (a) Starting from rest, through what distance does the ball move before its release? (b) What force does the pitcher exert on the ball? Answer: (a) ∆𝑥 = 𝑣𝑡 𝑣 (b) The magnitude of the force is 𝐹 = 𝑚√( ) + 𝑔2 𝑡 and its direction is 𝜃 = 𝑡𝑎𝑛−1 ( 𝑚𝑔 𝑚𝑣/𝑡 𝑔𝑡 ) = 𝑡𝑎𝑛−1 ( ) 𝑣 4.2 Two forces, ⃗⃗⃗ 𝐹1 = (−6.00𝑖̂ − 4.00𝑗̂) N and ⃗⃗⃗ 𝐹2 = (−3.00𝑖̂ + 7.00𝑗̂) N, act on a particle of mass 2.00 kg that is initially at rest at coordinates (−2.00 m, +4.00 m) (a) What are the components of the particle’s velocity at t = 10.0 s? (b) In what direction is the particle moving at t = 10.0 s? (c) What displacement does the particle undergo during the first 10.0 s? (d) What are the coordinates of the particle at t = 10.0 s? Answer: (a) 𝑣 ⃗⃗⃗𝑓 = (−45.0𝑖̂ + 15.0𝑗̂) (m/s) (b) The direction of motion makes angle θ with the x direction, where θ = −18.4°+ 180°= 162° from the + x axis (c) ∆𝑟 = −225𝑖̂ + 75.0𝑗̂ (m) (d) ⃗⃗⃗ 𝑟𝑓 = (−227𝑖̂ + 79.0𝑗̂) (m) 4.3 Three forces acting on an object are given by ⃗⃗⃗ 𝐹1 = (−2.00𝑖̂ + 2.00𝑗̂) N, ⃗⃗⃗ 𝐹2 = (5.00𝑖̂ − 3.00𝑗̂) N, and ⃗⃗⃗ 𝐹3 = (−45.0𝑖̂) N The object experiences an acceleration of magnitude 3.75 m/s2 (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s? Answer: (a) 𝑎 is at 181° counter-clockwise from the x axis (b) m = 11.2 kg (c) v = 37.5 m/s (d) 𝑣 ⃗⃗⃗𝑓 = (−37.5𝑖̂ − 0.893𝑗̂) (m/s) 4.4 Two objects are connected by a light string that passes over a frictionless pulley as shown in Fig Assume the incline is frictionless and take m1 = 2.00 kg, m2 = 6.00 kg, and θ = 55.0° (a) Draw free-body diagrams of both objects Find (b) the magnitude of the acceleration of the objects, (c) the tension in the string, and (d) the speed of each object 2.00 s after it is released from rest Fig Answer: (a) (b) a = 3.57 m/s2 (c) T = 26.7 N (d) vf = 7.14 m/s 4.5 A 3.00-kg block starts from rest at the top of a 30.0° incline and slides a distance of 2.00 m down the incline in 1.50 s Find (a) the magnitude of the acceleration of the block, (b) the friction force acting on the block, (c) the coefficient of kinetic friction between block and plane, and (d) the speed of the block after it has slid 2.00 m Answer: (a) a = 1.78 m/s2 (b) f = 9.37 N (c) µk = 0.368 (d) vf = 2.67 m/s CHAPTER CIRCULAR MOTION AND OTHER APPLICATIONS OF NEWTON’S LAWS 5.1 A car initially traveling eastward turns north by traveling in a circular path at uniform speed as shown in Fig The length of the arc ABC is 235 m, and the car completes the turn in 36.0 s (a) What is the acceleration when the car is at B located at an angle of 35.0o? Express your answer in terms of the unit vectors 𝑖̂ and 𝑗̂ Determine (b) the car’s average speed Fig and (c) its average acceleration during the 36.0-s interval Answer: (a) ⃗⃗⃗⃗ 𝑎𝑟 = (−0.233𝑖̂ + 0.163𝑗̂) (m/s2) (b) v = 6.53 m/s (c) 𝑎𝑎𝑣𝑔 = (−0.181𝑖̂ + 0.181𝑗̂) (m/s2) 5.2 Why is the following situation impossible? The object of mass m = 4.00 kg in Fig is attached to a vertical rod by two strings of length, ℓ = 2.00 m The strings are attached to the rod at points a distance d = 3.00 m apart The object rotates in a horizontal circle at a constant speed of v = 3.00 m/s, and the strings remain taut The rod rotates along with the object so that the strings not wrap onto the rod What If? Could this situation be possible on another planet? Answer: By calculation, Tb = –5.7 N, where Tb is the force Fig exerted by the lower string on the object This means that the lower string pushes rather than pulls! The situation is impossible because the speed of the object is too small, requiring that the lower string act like a rod and push rather than like a string and pull To answer the What if?, we obtain Tb = 41.2 N − 2.67g For this situation to be possible, Tb must be > 0, or g < 7.72 m/s2 This is certainly the case on the surface of the Moon and on Mars 5.3 One end of a cord is fixed and a small 0.500-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.00 m as shown in Fig When θ = 20.0o, the speed of the object is 8.00 m/s At this instant, find (a) the tension in the string, (b) the tangential and radial components of acceleration, and (c) the total acceleration (d) Is your answer changed if the object is swinging down toward its lowest point instead of swinging up? (e) Explain your answer to part (d) Answer: (a) T = 20.6 N (b) ar = 32 m/s2 inward at = 3.35 m/s2 downward tangent to the circle (c) a = 32.2 m/s2 inward and below the cord at 5.98° Fig (d) No change (e) If the object is swinging down it is gaining speed, and if the object is swinging up it is losing speed, but the forces are the same Therefore, its acceleration is regardless of the direction of swing 5.4 A person stands on a scale in an elevator As the elevator starts, the scale has a constant reading of 591 N As the elevator later stops, the scale reading is 391 N Assuming the magnitude of the acceleration is the same during starting and stopping, determine (a) the weight of the person, (b) the person’s mass, and (c) the acceleration of the elevator Answer: (a) Fg = 491 N (b) m = 50.1 kg (c) a = 2.00 m/s2 5.5 A window washer pulls a rubber squeegee down a very tall vertical window The squeegee has mass 160 g and is mounted on the end of a light rod The coefficient of kinetic friction between the squeegee and the dry glass is 0.900 The window washer presses it against the window with a force having a horizontal component of 4.00 N (a) If she pulls the squeegee down the window at constant velocity, what vertical force component must she exert? (b) The window washer increases the downward force component by 25.0%, while all other forces remain the same Find the squeegee’s acceleration in this situation (c) The squeegee is moved into a wet portion of the window, where its motion is resisted by a fluid drag force R proportional to its velocity according to R = −20.0v, where R is in newtons and v is in meters per second Find the terminal velocity that the squeegee approaches, assuming the window washer exerts the same force described in part (b) Answer: (a) Py = −2.03 N = 2.03 N down (b) ay = −3.18 m/s2 = 3.18 m/s2 down (c) vT = = 0.205 m/s down - CHAPTER 6: ENERGY AND SYSTEM 6.1 A block of mass 𝑚 = 2.50 kg is pushed a distance 𝑑 = 2.20 m along a frictionless, horizontal table by a constant applied force of magnitude 𝐹 = 16.0 N directed at an angle 𝜃 = 25.08 below the horizontal as shown in Figure P6.1 Determine the work done on the block by (a) the applied force, (b) the normal force exerted by the table, (c) the gravitational force, (d) the net force on the block Ans (a) 31.9 J; (b),(c) 0; (d) 31.9 J Figure P6.1 6.2 The force acting on a particle varies as shown in Figure P6.2 Find the work done by the force on the particle as it moves (a) from 𝑥 = to 𝑥 = 8.00 m, (b) from 𝑥 = 8.00 m to 𝑥 = 10.0 m, (c) from 𝑥 = to 𝑥 = 10.0 m Ans (a) 24.0 J; (b) −3.00 J; (d) 21.0 J Figure P6.2 6.3 A force 𝐹 = 4𝑥 î + 3𝑦ĵ, where 𝐹 in newtons and 𝑥 and 𝑦 are in meters, acts on an object as the object moves in the 𝑥 direction from the origin to 𝑥 = 5.00 m Find the work done by the force on the object Ans 50.0 J 6.4 In an electron microscope, there is an electron gun that contains two charged metallic plates 2.80 cm apart An electric force accelerates each electron in the beam from rest to 9.60% of the speed of light over this distance For an electron passing between the plates in the electron gun, determine (a) the kinetic energy of the electron as it leaves the electron gun, (b) the magnitude of the constant electric force acting on the electron, (c) the acceleration of the electron, (d) the time interval the electron spends between the plates Ans (a) 3.78 × 10−16 𝐽; (b) 1.35 × 10−14 𝑁; (c) 1.48 × 1016 𝑚/𝑠 ; (d) 1.94 × 10−9 𝑠 6.5 A single conservative force acts on a 5.00-kg particle within a system due to its interaction with the rest of the system The equation 𝐹𝑥 = 2𝑥 + describes the force, where 𝐹𝑥 is in newtons and 𝑥 is in meters As the particle moves along the 𝑥 axis from 𝑥 = 1.00 m to 𝑥 = 5.00 m, calculate (a) the work done by this force on the particle, (b) the change in the potential energy of the system, (c) the kinetic energy the particle has at 𝑥 = 5.00 m if its speed is 3.00 m/s at 𝑥 = 1.00 m Ans (a) 40.0 𝐽; (b) −40.0 𝐽; (c) 62.5 𝐽 CHAPTER 7: CONSERVATION OF ENERGY 7.1 A block of mass 0.250 kg is placed on top of a light, vertical spring of force constant 000 N/m and pushed downward so that the spring is compressed by 0.100 m After the block is released from rest, it travels upward and then leaves the spring To what maximum height above the point of release does it rise? Ans 10.2 m 7.2 A 20.0-kg cannonball is fired from a cannon with muzzle speed of 000 m/s at an angle of 37.0° with the horizontal A second ball is fired at an angle of 90.0° Use the isolated system model to find (a) the maximum height reached by each ball, (b) the total mechanical energy of the ball–Earth system at the maximum height for each ball Let 𝑦 = at the cannon Ans (a) 𝑏𝑎𝑙𝑙 1: 1.85 × 104 𝑚, 𝑏𝑎𝑙𝑙 2: 5.10 × 104 𝑚; (b) 1.00 × 107 𝐽 7.3 A bead slides without friction around a loop-theloop (Fig P7.1) The bead is released from rest at a height ℎ = 3.50𝑅 (a) What is its speed at point A? (b) How large is the normal force on the bead at point A if its mass is 5.00 g? Figure P7.1 Ans (a) √3.00𝑔𝑅; (b) 0.0980 𝑁 7.4 A block of mass m 5.00 kg is released from point A and slides on the frictionless track shown in Figure P7.2 Determine (a) the block’s speed at points B and C (b) the net work done by the gravitational force on the block as it moves from point A to point C Ans (a) 𝑣𝐵 = 5.94 𝑚/𝑠, 𝑣𝐶 = 7.67 𝑚/𝑠; Figure P7.2 (b) 147 𝐽 7.5 A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s The pulling force is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate–incline system owing to friction (c) How much work is done by the 100-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.00 m? Ans (a) −168 𝐽; (b) 184 𝐽; (c) 500 𝐽; (d) 148 𝐽; (e) 5.65 𝑚/𝑠 7.6 A 40.0-kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N The coefficient of friction between box and floor is 0.300 Find (a) the work done by the applied force, (b) the increase in internal energy in the box–floor system as a result of friction, (c) the work done by the normal force, (d) the work done by the gravitational force, (e) the change in kinetic energy of the box, (f) the final speed of the box Ans (a) 650 𝐽; (b) 588 𝐽; (c),(d) 0; (e) 62.0 𝐽; (f) 1.76 m/s 7.7 The coefficient of friction between the block of mass 𝑚1 = 3.00 kg and the surface in Figure P7.3 is 𝜇𝑘 = 0.400 The system starts from rest What is the speed of the ball of mass 𝑚2 = 5.00 kg when it has fallen a distance ℎ = 1.50 m? Ans 3.74 m/s Figure P7.3 7.8 The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21.0 ms The total mass of the train is 875 g (a) Find the minimum power delivered to the train by electrical transmission from the metal rails during the acceleration (b) Why is it the minimum power? Ans (a) 8.01 𝑊 7.9 An older-model car accelerates from to speed 𝑣 in a time interval of Δ𝑡 A newer, more powerful sports car accelerates from to 2𝑣 in the same time period Assuming the energy coming from the engine appears only as kinetic energy of the cars, compare the power of the two cars Ans The power of the sports car is four times that of the older-model car 7.10 A roller-coaster car shown in Figure P7.4 is released from rest from a height ℎ and then moves freely with negligible friction The roller-coaster track includes a circular loop of radius R in a vertical plane 10 (a) First suppose the car barely makes it around the loop; at the top of the loop, the riders are upside down and feel weightless Find the required height ℎ of the release point above the bottom of the loop in terms of 𝑅 (b) Now assume the release point is at or above the minimum required height Show that the normal force on the car at the bottom of the loop exceeds the normal force at the top of the loop by six times the car’s weight The normal force on each rider follows the same rule Such a large normal force is dangerous and very uncomfortable for the riders Roller coasters are therefore not built with circular loops in vertical planes Figure P7.5 shows an actual design Ans (a) ℎ = 5𝑅/2; (b) 6𝑚𝑔 Figure P7.4 Figure P7.5 - CHAPTER 8: LINEAR MOMENTUM AND COLLISION 8.1 In a slow-pitch softball game, a 0.200-kg softball crosses the plate at 15.0 m/s at an angle of 45.0° below the horizontal The batter hits the ball toward center field, giving it a velocity of 40.0 m/s at 30.0° above the horizontal (a) Determine the impulse delivered to the ball (b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, and then decreases linearly to zero in another 4.00 ms, what is the maximum force on the ball? Ans (a) 𝐼 = (9.05𝑖̂ + 6.12𝑗̂)Ns; (b) 𝐹𝑚 = (377𝑖̂ + 255𝑗̂)N; 8.2 As shown in Figure P8.2, a bullet of mass m and speed v passes completely through a pendulum bob of mass M The bullet emerges with a speed of v/2 The pendulum bob is suspended by a stiff rod (not a string) of length l, and negligible mass What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? 11 Figure P8.2 Ans 𝑣 = 4𝑀 𝑚 √𝑔𝑙 8.3 A 12.0-g wad of sticky clay is hurled horizontally at a 100-g wooden block initially at rest on a horizontal surface The clay sticks to the block After impact, the block slides 7.50 m before coming to rest If the coefficient of friction between the block and the surface is 0.650, what was the speed of the clay immediately before impact? Ans.: 𝑣𝑐 = 9.12 m/s 8.4 The mass of the Earth is 5.97×1024 kg, and the mass of the Moon is 7.35×1022 kg The distance of separation, measured between their centers, is 3.84×108 m Locate the center of mass of the Earth–Moon system as measured from the center of the Earth Ans.: 4.66×106 m from the Earth's center 8.5 A water molecule consists of an oxygen atom with two hydrogen atoms bound to it (Fig P8.5) The angle between the two bonds is 106° If the bonds are 0.100 nm long, where is the center of mass of the molecule? Figure P8.5 Ans.: on the dotted line in Fig P8.5, 0.00673 nm below the center of the O atom 8.6 A 7.00-g bullet, when fired from a gun into a 1.00-kg block of wood held in a vise, penetrates the block to a depth of 8.00 cm This block of wood is next placed on a frictionless horizontal surface, and a second 7.00-g bullet is fired from the gun into the block To what depth will the bullet penetrate the block in this case? Ans.: d’ = 7.94 cm - CHAPTER 9: ROTATION OF A RIGID OBJECT ABOUT A FIXED AXIS 12 9.1 A rotating wheel requires 3.00 s to rotate through 37.0 revolutions Its angular speed at the end of the 3.00-s interval is 98.0 rad/s What is the constant angular acceleration of the wheel? Ans 𝛼 = 13.7 rad /s2 9.2 A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2 During a 4.20-s time interval, the wheel rotates through 62.4 rad What is the angular speed of the wheel at the end of the 4.20-s interval? Ans 𝜔 = 3.10 rad /s 9.3 A 150-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s? Ans.: F = 177 N 9.4 A block of mass m1 = 2.00 kg and a block of mass m = 6.00 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg The fixed, wedgeshaped ramp makes an angle of 𝜃 = 30.0o as shown in Figure P9.4 The coefficient of kinetic friction is 0.360 for both blocks (a) Draw force diagrams of both blocks and of the pulley Determine (b) the acceleration of the two blocks and (c) the tensions in the string on both sides of the pulley Figure P9.4 Ans.: a = 0.309 m/s2, T = 9.22 N 9.5 Two balls with masses M and m are connected by a rigid rod of length L and negligible mass as shown in Figure P9.5 For an axis perpendicular to the rod, (a) show that the system has the minimum moment of inertia when the axis passes through the center of mass (b) Show that this moment of inertia is I = 𝜇L2, where 𝜇 = mM/(m + M) Figure P9.5 9.6 The four particles in Figure P9.6 are connected by rigid rods of negligible mass The origin is at the center of the rectangle The system rotates in the xy plane about the z axis 13 with an angular speed of 6.00 rad/s Calculate (a) the moment of inertia of the system about the z axis and (b) the rotational kinetic energy of the system Figure P9.6 Ans.: a) I = 143 kgm2, b) KR = 2.57×103 J 9.7 A cylinder of mass 10.0 kg rolls without slipping on a horizontal surface At a certain instant, its center of mass has a speed of 10.0 m/s Determine (a) the translational kinetic energy of its center of mass, (b) the rotational kinetic energy about its center of mass, and (c) its total energy Ans.: a) Ktrans = 500 J; b) Krot = 250 J; c) Ktotal = 750 J 9.8 A solid sphere is released from height h from the top of an incline making an angle u with the horizontal Calculate the speed of the sphere when it reaches the bottom of the incline (a) in the case that it rolls without slipping and (b) in the case that it slides frictionlessly without rolling (c) Compare the time intervals required to reach the bottom in cases (a) and (b) 10 Ans.: a) 𝑣𝐴 = √ 𝑔ℎ; b) 𝑣𝐵 = √2𝑔ℎ, c) 𝑡𝐴 = ( ( 2ℎ 𝑠𝑖𝑛𝜃 )√ 2ℎ ) = ( 𝑠𝑖𝑛𝜃 𝑣 2ℎ 𝑠𝑖𝑛𝜃 )√ 10𝑔ℎ , 𝑡𝐵 = 2𝑔ℎ 9.9 (a) Determine the acceleration of the center of mass of a uniform solid disk rolling down an incline making angle u with the horizontal (b) Compare the acceleration found in part (a) with that of a uniform hoop (c) What is the minimum coefficient of friction required to maintain pure rolling motion for the disk? 1 3 Ans.: a) 𝑎 = 𝑔𝑠𝑖𝑛𝜃; b) 𝑎 = 𝑔𝑠𝑖𝑛𝜃; c) 𝜇 = 𝑡𝑎𝑛𝜃 CHAPTER 10: ANGULAR MOMENTUM 10.1 A uniform solid disk of mass m = 3.00 kg and radius r = 0.200 m rotates about a fixed axis perpendicular to its face with angular frequency 6.00 rad/s Calculate the magnitude of the angular momentum of the disk when the axis of rotation (a) passes through its center of mass and (b) passes through a point midway between the center and the rim 14 Ans.: a) L1 = 0.360 kgm2/s; b) L2 = 0.540 kgm2/s; 10.2 Model the Earth as a uniform sphere (a) Calculate the angular momentum of the Earth due to its spinning motion about its axis (b) Calculate the angular momentum of the Earth due to its orbital motion about the Sun (c) Explain why the answer in part (b) is larger than that in part (a) even though it takes significantly longer for the Earth to go once around the Sun than to rotate once about its axis Ans.: a) L1 = 7.06×1033 kgm2/s; b) L2 = 2.66×1040 kgm2/s; 10.3 A uniform cylindrical turntable of radius 1.90 m and mass 30.0 kg rotates counterclockwise in a horizontal plane with an initial angular speed of 4p rad/s The fixed turntable bearing is frictionless A lump of clay of mass 2.25 kg and negligible size is dropped onto the turntable from a small distance above it and immediately sticks to the turntable at a point 1.80 m to the east of the axis (a) Find the final angular speed of the clay and turntable (b) Is the mechanical energy of the turntable–clay system constant in this process? Explain and use numerical results to verify your answer (c) Is the momentum of the system constant in this process? Explain your answer Ans.: a) 𝜔𝑓 = 11.1 rad/s; b) No, Ki = 4276 J, Kf = 3768 J; c) No; 10.4 A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 kg, traveling perpendicular to the door at just before impact Find the final angular speed of the door Does the mud make a significant contribution to the moment of inertia? Ans.: 𝜔 = 0.223 rad/s 10.5 A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length l, and of negligible mass (Fig P10.4) The rod is pivoted at the other end A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it (a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? Figure P10.4 -CHAPTER 11: no problem 15 PART II: THERMALDYNAMICS CHAPTER 12 TEMPERATURE AND THE FIRST LAW OF THERMODYNAMICS h Fig 12.1 Fig 12.2 12.1 Consider an ideal gas in a vertical cylinder of cross-sectional area A = 50.0 cm2 at 27C as shown in Fig 12.1 The piston mass is 50.0 g When the piston balances, h = 40 cm Put a 10.0-kg weight on the piston, the piston moves down a distance of Δh = cm Find the final temperature of the gas The atmosphere pressure is 1.00 atm Assume that the friction between the piston and the cylinder wall is insignificant Guide: ⃗⃗⃗𝑜 + 𝑃⃗𝑝𝑖𝑠 + 𝐹 = ⃗0  PoA + mg = PA - Nothing on the piston: 𝐹 + Initial pressure of the gas: P = Po + mg/A + Initial volume of the gas: V = Ah ⃗  PoA + (m + m’)g = P’A ⃗⃗⃗𝑜 + 𝑃⃗𝑝𝑖𝑠+𝑤 + ⃗⃗⃗ - 10-kg weight on the piston: 𝐹 𝐹′ = + Final pressure of the gas: P’ = Po + (m + m’)g/A + Initial volume of the gas: V = Ah’= A(h - Δh) - State equation: PV/T = P’V’/T’  Final temperature: T’ = P’V’T/PV 12.2 A mercury thermometer is constructed as shown in Fig 12.2 The Pyrex glass capillary tube has a diameter of 0.004 00 cm, and the bulb has a diameter of 0.250 cm Find the change in height of the mercury column that occurs with a temperature change of 30.0C Assume the dimensions of the capillary tube not change For mercury, β = 1.82 × 10-4 (°C)-1 and for Pyrex glass, α = 3.20 × 10-6 (°C)-1 Guide: The volume of the liquid increases as: ΔV = VβΔT The volume of the shell increases as: ΔV’= 3VαΔT Therefore, the overflow in the capillary is: ΔVC = ΔV - ΔV’ On the other hand: ΔVC =AΔh Δh = … 12.3 An aluminum calorimeter with a mass of 100 g contains 250 g of water The calorimeter and water are in thermal equilibrium at 10.0°C Two metallic blocks are placed into the water One is a 50.0-g piece of copper at 80.0°C The other has a mass of 16 70.0g and is originally at a temperature of 100°C The entire system stabilizes at a final temperature of 20.0°C Determine the specific heat of the unknown sample Guide: Using 𝑄𝑖𝑛 = 𝑄𝑜𝑢𝑡  (𝑚𝑤 𝑐𝑤 + 𝑚𝑐 𝐶𝑐 )(𝑇𝑓 − 𝑇𝑐 ) = 𝑚𝐶𝑢 𝑐𝐶𝑢 (𝑇𝐶𝑢 − 𝑇𝑓 ) + 𝑚𝑥 𝑐𝑥 (𝑇𝑥 − 𝑇𝑓 )  𝑐𝑥 = ⋯ 12.4 A 2.00-mol sample of helium gas initially at 300 K, and 0.400 atm is compressed isothermally to 1.20 atm Noting that the helium behaves as an ideal gas, find: (a) The final volume of the gas, (b) The work done on the gas, (c) The energy transferred by heat Guide: (a) The initial volume is: 𝑉𝑖 = 𝑛𝑅𝑇 𝑃𝑖 = 0.123 m3 For isothermal compression, 𝑃𝑖 𝑉𝑖 = 𝑃𝑓 𝑉𝑓  The final volume is: 𝑉𝑓 = 𝑉𝑖 𝑃𝑖 𝑃𝑓 = 0.0410 m3 𝑉𝑓 (b) 𝑊𝑜𝑛 = − ∫ 𝑃𝑑𝑉 = −𝑛𝑅𝑇𝑙𝑛 ( ) = 5.48 𝑘𝐽 𝑉 𝑖 (c) For isothermal compression, ∆𝐸𝑖𝑛𝑡 =  𝑄𝑖𝑛 = −𝑊𝑜𝑛 = −5.48 𝑘𝐽 12.5 A system consisting of 0.32 mol of a monatomic ideal gas, with molar heat capacity 𝑐𝑣′ = 𝑅, occupies a volume of 2.2 L at a pressure of 2.4 atm, as represented by point A in Fig 12.3 The system is carried through a cycle consisting of three processes: The gas is heated at constant pressure until its volume is 4.4 L at point B The gas is cooled at constant volume until the pressure decreases to 1.2 atm (point C) The gas undergoes an isothermal compression back to point A (a) What is the temperature at points A, B and C (b) Find Won, Qin and ΔEint for each process and for the entire cycle Fig 12.3 Guide: (a) 𝑇𝐴 = 𝑇𝐶 = 𝑃𝐴 𝑉𝐴 𝑛𝑅 = 2.0 × 102 𝐾, 𝑇𝐵 = (b) - Process A  B: 𝑊𝑜𝑛,𝐴𝐵 = −𝑃𝐴 (𝑉𝐵 − 𝑉𝐴 ) = −0.53 𝑘𝐽 17 𝑃𝐵 𝑉𝐵 𝑛𝑅 =2 𝑃𝐴 𝑉𝐴 𝑛𝑅 = 4.0 × 102 𝐾 𝑛𝑅(𝑇𝐵 − 𝑇𝐴 ) = 1.3 𝑘𝐽 = 𝑊𝑜𝑛,𝐴𝐵 + 𝑄𝑖𝑛,𝐴𝐵 = 0.80 𝑘𝐽 𝑄𝑖𝑛,𝐴𝐵 = 𝐶𝑃 ∆𝑇𝐴𝐵 = ∆𝐸𝑖𝑛𝑡,𝐴𝐵 - Process B  C: 𝑊𝑜𝑛,𝐵𝐶 = 𝐽 𝑛𝑅 (𝑇𝐶 − 𝑇𝐵 ) = −0.80 𝑘𝐽 = 𝑊𝑜𝑛,𝐵𝐶 + 𝑄𝑖𝑛,𝐵𝐶 = −0.80 𝑘𝐽 𝑄𝑖𝑛,𝐵𝐶 = 𝐶𝑉 ∆𝑇𝐵𝐶 = ∆𝐸𝑖𝑛𝑡,𝐵𝐶 - Process C  A: 𝑊𝑜𝑛,𝐶𝐴 = 𝑛𝑅𝑇𝐴 𝑙𝑛 𝑉𝐴 𝑉𝐶 = 0.37 𝑘𝐽 ∆𝐸𝑖𝑛𝑡,𝐶𝐴 = 𝐽 𝑄𝑖𝑛,𝐶𝐴 = −𝑊𝑜𝑛,𝐶𝐴 = −0.37 𝑘𝐽 - Cycle: 𝑊𝑜𝑛,𝑡𝑜𝑡𝑎𝑙 = 𝑊𝑜𝑛,𝐴𝐵 + 𝑊𝑜𝑛,𝐵𝐶 + 𝑊𝑜𝑛,𝐶𝐴 𝑄𝑖𝑛,𝑡𝑜𝑡𝑎𝑙 = 𝑄𝑖𝑛,𝐴𝐵 + 𝑄𝑖𝑛,𝐵𝐶 + 𝑄𝑖𝑛,𝐶𝐴 ∆𝐸𝑖𝑛𝑡,𝑡𝑜𝑡𝑎𝑙 = ∆𝐸𝑖𝑛𝑡,𝐴𝐵 + ∆𝐸𝑖𝑛𝑡,𝐵𝐶 + ∆𝐸𝑖𝑛𝑡,𝐶𝐴 CHAPTER 13 THE KINETIC THEORY OF GASES 13.1 (a) How many atoms of helium gas fill a spherical balloon of diameter 30.0 cm at 20.0C and 1.00 atm? (b) What is the average kinetic energy per each helium atom? (c) What is the rms speed of the helium atoms? Guide: (a) The volume of balloon: 𝑉 = 𝜋𝑅3 The quantity of gas in the balloon: 𝑛 = 𝑃𝑉 𝑅𝑇 The number of molecules: N = nNA 2 ̅ = 𝑚𝑜 ̅̅̅ (b) 𝐾 𝑣 = 𝑘𝐵 𝑇 (c) The mass of each He atom: 𝑚𝑜 = 𝑀 𝑁𝐴 , where M is the molar mass ̅ 2𝐾 The root-mean-square speed: 𝑣𝑟𝑚𝑠 = √̅̅̅ 𝑣2 = √ 𝑚𝑜 18 13.2 A cylinder contains 3.00 mol of oxygen gas at a temperature of 300 K (a) If the gas is heated at constant volume, how much energy must be transferred by heat to the gas for its temperature to increase to 500 K? (b) How much energy must be transferred by heat to the gas at constant pressure to raise the temperature to 500 K? Guide: (a) 𝑄𝑉 = 𝑛𝑐𝑉′ ∆𝑇 = 𝑛𝑅∆𝑇 (b) 𝑄𝑃 = 𝑛𝑐𝑃′ ∆𝑇 = 𝑛𝑅∆𝑇 13.3 A tank used for filling helium balloons has a volume of 0.300 m3 and contains 2.00 mol of helium gas at 20.0C Assume the helium behaves like an ideal gas (a) What is the internal energy of the gas molecules? (b) What is the average kinetic energy per molecule? Guide: (a) Because helium is monatomic, the internal energy of the gas is the total translational kinetic energy of the molecules:𝐸𝑖𝑛𝑡 = 𝐾𝑡𝑜𝑡 𝑡𝑟𝑎𝑛𝑠 = 𝑛𝑅𝑇 ̅ = 𝑚𝑜 ̅̅̅ (b) 𝐾 𝑣 = 𝑘𝐵 𝑇 2 13.4 A 1.00-mol sample of hydrogen gas is heated at constant pressure from 300 K to 420 K Calculate: (a) The energy transferred to the gas by heat, (b) The increase in its internal energy, (c) The work done on the gas Guide: (a) For a constant-pressure process, 𝑄𝑖𝑛 = 𝑛𝑐𝑃′ ∆𝑇 = 𝑛𝑅∆𝑇 (b) 𝛥𝐸𝑖𝑛𝑡 = 𝑛𝑐𝑉′ ∆𝑇 = 𝑛𝑅∆𝑇 (c) 𝛥𝐸𝑖𝑛𝑡 = 𝑄𝑖𝑛 + 𝑊𝑜𝑛  𝑊𝑜𝑛 = 𝛥𝐸𝑖𝑛𝑡 − 𝑄𝑖𝑛 13.5 A 2.00-mol sample of a diatomic ideal gas expands slowly and adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L (a) What is the final pressure of the gas? (b) What are the initial and final temperatures? (c) Find Qin, ΔEint and Won for the gas during this process Guide: 𝛾 𝛾 (a) In an adiabatic process, 𝑃𝑖 𝑉𝑖 = 𝑃𝑓 𝑉𝑓  𝑃𝑓 = ⋯ Note: for a diatomic ideal gas, γ = 1.4 19 (b) The initial temperature is: 𝑇𝑖 = The final temperature is: 𝑇𝑓 = 𝑃𝑖 𝑉𝑖 𝑛𝑅 𝑃𝑓 𝑉𝑓 𝑛𝑅 (c) The process is adiabatic, Qin = J 𝛥𝐸𝑖𝑛𝑡 = 𝑛𝑐𝑉′ ∆𝑇 = 𝑛𝑅∆𝑇 𝑊𝑜𝑛 = 𝛥𝐸𝑖𝑛𝑡 − 𝑄𝑖𝑛 -CHAPTER 14: HEAT ENGINES, ENTROPY, AND THE SECOND LAW OF THERMODYNAMICS 14.1 An engine absorbs 1.70 kJ from a hot reservoir at 277°C and expels 1.20 kJ to a cold reservoir at 27°C in each cycle (a) What is the engine’s efficiency? (b) How much work is done by the engine in each cycle? (c) What is the power output of the engine if each cycle lasts 0.300 s? Guide: |𝑄 | (a) The engine’s efficiency: 𝜀 = − |𝑄𝑐 | = − ℎ 1.20 1.70 = 0.294 (b) The work done by the engine in each cycle: 𝑊𝑏𝑦 = |𝑄ℎ | − |𝑄𝑐 |= 0.500 kJ (c) The power transferred out of the engine: 𝑃 = 𝑊𝑏𝑦 ∆𝑡 = 0.500 0.300 = 1.67 𝑘𝑊 14.2 You have half an hour before guests start arriving for your party when you suddenly realize that you forgot to buy ice for drinks You quickly put 1.00 L of water at 10.0C into your ice cube trays and pop them into the freezer Will you have ice in time for your guests? The label on your refrigerator states that the appliance has a coefficient of performance of 5.5 and a power rating P = 550 W You estimate that only 10 percent of the electrical power contributes to the cooling and freezing of the water Guide: The heat |𝑄𝑐 | absorbed from the inside of the refrigerator is the sum of the heat 𝑄𝑐𝑜𝑜𝑙 to be absorbed from the water to cool the water and the heat 𝑄𝑓𝑟𝑒𝑒𝑧𝑒 to be absorbed from the water to freeze the water: |𝑄𝑐 | = 𝑄𝑐𝑜𝑜𝑙 + 𝑄𝑓𝑟𝑒𝑒𝑧𝑒 𝑄𝑐𝑜𝑜𝑙 = 𝑚𝑐∆𝑇 = 41.8 𝑘𝐽 𝑄𝑓𝑟𝑒𝑒𝑧𝑒 = 𝑚𝐿𝐹 = 333.5 𝑘𝐽  |𝑄𝑐 | = 375 kJ 20 The work used by the refrigerator to cool and freeze water: |𝑄 | 375 𝑊𝑜𝑛 = 𝑐 = = 68 kJ (1) 𝐶𝑂𝑃 5.5 If the time the refrigerator use to freeze 1.00 L of water at 10.0C is ∆𝑡, 𝑊𝑜𝑛 = 10%𝑃 ∆𝑡 (2)  ∆𝑡 = 68/0.1 × 550 = 20.7 The guests will have ice Fig 14.2 14.3 A 1.00-mol sample of H2 gas is contained in the left side of the container shown in Fig 14.1, which has equal volumes on the left and right The right side is evacuated When the valve is opened, the gas streams into the right side (a) What is the entropy change of the gas? (b) Does the temperature of the gas change? Assume the container is so large that the hydrogen behaves as an ideal gas Fig 14.1 Guide: 𝑉𝑓 (a) For a free expansion process: ∆𝑆 = 𝑛𝑅𝑙𝑛 ( ) 𝑉𝑖 Note: 𝑉𝑓 = 2𝑉𝑖 (b) The gas is expanding into an evacuated region Therefore, no work is done on the gas, Won = It expands so fast that energy has no time to flow by heat, Qin =  𝛥𝐸𝑖𝑛𝑡 = 𝑄𝑖𝑛 + 𝑊𝑜𝑛 = For an ideal gas, the internal energy is only a function of the temperature  no change in temperature 14.4 Consider an Otto cycle, as shown in Fig 14.2, with VA/VB = 4.00 At the beginning of the compression process (state A), 0.02 moles of gas is at 1atm and 20.0°C At the beginning of the adiabatic expansion (state C), the temperature is TC = 500°C Model the working fluid as an ideal gas with γ = 1.40 Find: a State parameters (P, V, T) of the gas at each state in the Otto cycle b The work done by the gas, the heat obsorbed by the gas and the change in the internal energy of the gas in each process of the cycle c Identify the energy input, the energy exhaust and the net output work d Calculate the thermal efficiency Guide: a - State A: PA = atm; TA = 293 K; VA = nRTA/PA 21 𝛾 𝛾 - State B: VB = VA/4; 𝑃𝐴 𝑉𝐴 = 𝑃𝐵 𝑉𝐵 𝑃𝐵 = 𝛾 𝑃𝐴 𝑉𝐴 𝛾 𝑉𝐵 ; TB = PBVB/nR - State C: PB/TB = PC/TC  PC = PBTC/TB; TC = 773 K; VC = VB = 0.12 𝐿 𝛾 𝛾 𝑉 𝛾 - State D: VD = VA = 0.49 L; 𝑃𝐶 𝑉𝐶 = 𝑃𝐷 𝑉𝐷  𝑃𝐷 = 𝑃𝐶 ( 𝐶 ) ; TD = PDVD/nR 𝑉𝐷 b - Process AB: 𝑊𝑏𝑦,𝐴𝐵 = 𝑃𝐴 𝑉𝐴 −𝑃𝐵 𝑉𝐵 𝛾−1 ; Qin,AB = J  Eint,AB = 𝑊𝑜𝑛,𝐴𝐵 = −𝑊𝑏𝑦,𝐴𝐵 - Process BC: 𝑊𝑏𝑦,𝐵𝐶 = 𝐽; Qin,BC = 𝑛𝑐𝑉′ (𝑇𝐶 − 𝑇𝐵 )  Eint,BC = 𝑄𝑖𝑛,𝐵𝐶 - Process CD: 𝑊𝑏𝑦,𝐶𝐷 = 𝑃𝐶 𝑉𝐶 −𝑃𝐷 𝑉𝐷 𝛾−1 ; Qin,CD = J  Eint,CD = 𝑊𝑜𝑛,𝐶𝐷 = - 𝑊𝑏𝑦,𝐶𝐷 - Process DA: 𝑊𝑏𝑦,𝐷𝐴 = 𝐽; Qin,DA = 𝑛𝑐𝑉′ (𝑇𝐴 − 𝑇𝐷 )  Eint,DA = 𝑄𝑖𝑛,𝐷𝐴 - Cycle ABCD: Wby = 𝑊𝑏𝑦,𝐴𝐵 + 𝑊𝑏𝑦,𝐵𝐶 + 𝑊𝑏𝑦,𝐶𝐷 + 𝑊𝑏𝑦,𝐷𝐴 Qin = 𝑄𝑖𝑛,𝐴𝐵 + 𝑄𝑖𝑛,𝐵𝐶 + 𝑄𝑖𝑛,𝐶𝐷 + 𝑄𝑖𝑛,𝐷𝐴 c From B  C, the input energy is Qh = 𝑄𝑖𝑛,𝐵𝐶 From D  A, the exhaust energy is Qc = 𝑄𝑜𝑢𝑡,𝐷𝐴 = - Qin,DA d The efficiency: 𝜀 = 𝑊𝑏𝑦 𝑄ℎ 14.5 A steam engine works between a hot reservoir at 373 K and a cold reservoir at 273 K (a) What is the maximum possible efficiency of this engine? (b) If the engine is run backwards as a refrigerator, what is its maximum coefficient of performance? Guide: (a) The maximum efficiency is the Carnot efficiency: 𝜀𝑚𝑎𝑥 = − 𝑇𝑐 𝑇ℎ (b) The maximum coefficient of performance is that of Carnot refrigerator: 𝐶𝑂𝑃𝑐,𝑚𝑎𝑥 = 22 𝑇𝑐 𝑇ℎ − 𝑇𝑐 ... time interval the electron spends between the plates Ans (a) 3.78 × 10 16