Bài tập vật lí bằng tiếng anh

GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung Problem 1: (Chuyển động của vật trên mặt phẳng nghiêng) A sled is a quick push up the snow slop. The sled slides up and then comes back down; the whole trip take time t. If the coeffient of sliding friction between the sled and snow is µ , find the time tu it took the sled to reach the top point of its trajectory. The slope makes the angle θ with horizontal. Solution: Newton’s second law applied ti the sled gives accelerations upwards (positive direction up) and downwards (positive direction dow) as follows: ( ) sin u a g cos θ µ θ = − + and ( ) sin d a g cos θ µ θ = − If u and v are, respectively, the initial velocity when moving upwards and final velocity when moving downwards, then according to the (constant acceleration) kinematical formula involving squares of velocites, (try to prove this formula!). If tu and td are, respectively, the time it takes to move down, then according to the definition of acceleration and using the privious equation. Combing the latter equation with the first two, we get (noting that tan θ µ > since the sled comes back): ( ) 1 (tan ) / tan u t t θ µ θ µ = + + − New Words: • sled: xe trượt tuyết nhỏ • slope: mặt phẳng nghiêng • trip: hành trình • coefficient: hệ số • coefficient of slid friction: hệ số ma sát trượt • to take time t: mất khoảng thời gian là t • the top point: điểm cao nhất • traejectory: quỹ đạo • horizotal: (mặt phẳng) nằm ngang • to make the angle: lập một góc • law: đinh luật • acceleration: gia tốc • velocity: vận tốc • initial (final) velocity: vận tốc ban đầu (cuối cùng) • to move up (down): chuyển động đi lên ( đi xuống) • according to: theo • kinematical: (thuộc về) động học • formula: công thức (số nhiều: formulae) • square: bình phương • equation: phương trình • slide: trượt • prove: chứng minh • involve (+ing): bao gồm Problem 2: (Bài toán ném xiên) Figure (hình) 1 shows a pirate ship, moored(bỏ neo) 560m from a fort defending the harbor(bến cảng, cảng) entrance(đi vào) of an island. The harbor defense cannon, located at sea level, muzzle velocity of 82m/s. a) To what angle must the cannon be elavated (độ nâng) to hit the pirate ship? b) What are times of flight for the elevation angle calculated above? c) How far should be pirate ship be from the fort if it is to be beyond(quá, vượt xa hơn) range of the cannon balls? Solution: As known, the equation of the trajectory of the ball is: Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 1 GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung ( ) ( ) 2 2 2 0 tan 1 2 g y X X v cos θ θ = − Where θ is the elevation angle and v 0 is the muzzle velocity. To find the horizotal range of the ball, let us X = R and y = 0 in Eq (1), after a little rearrangemant, obtaining(đạt được): ( ) 2 0 .sin 2 2 v R g θ = Soving Eq (2) for θ yields(giá trị) 0 0 2 2 0 560.9,8 sin 2 0,816 27 63 82 Rg and v θ θ θ → = = = → = = . The commandent of the fort can elevate the guns to either of these two angles and hit the pirate ship. b) As known, the horizontal position of the ball is given by the equation: ( ) ( ) 0 3X v cos t θ = Soving Eq (3) for t give, for 0 27 θ = , 0 0 0 560 7,7 82. 27 X R t s v cos v cos cos θ θ = = = = Repeating the calculation for 0 63 θ = yields t = 15s. It is reasonable that the time of flight for the higher elevation angle should be larger. c) We have known that the maximun range corresponds to an elevation angle θ of 45 0 . Thus, from Eq (2). ( ) 2 2 0 0 82 sin 2 .sin 2.45 690 9,8 v R m g θ = = = As the pirate ship sails away, the two elevation angle at which the ship can bi hit draw closer togetther, eventually merging at 0 45 θ = when the ship is 690m away. Beyond that point the ship is safe. New Words: • pirate: cướp biển • (to) moor: neo • Fort: pháo đài • (to) defend: phòng thủ • Cannon: pháo, đại bác • Muzzle velocity: vận tốc đầu nòng (súng) • Elevated angle: góc tầm (góc ngắm) • (to) hit: trúng • (the) commandant (of the fort): người chỉ huy (pháo đài) • The time flight: thời gian bay • Corresponds: tương ứng, đúng với, phù hợp với Problem 3: (Bài toán tìm nhiệt dung riêng) Two identical light metal containers are filled with equal amounts(lượng, số lượng) of water and placed in a room with constant air temprature. A heavy ball is submerged(chìm, ngập dưới) into the center of one of the containers on a thin nonconducting string. The mass of ball equals to the mass of water, and the density of the ball is much greater than that of water. Both containers are heated to the boiling point of water and are then allowed to cool. The container with the ball in it takes k times longer to cool down to the room temprature than the container without the ball. Find specific heat of the material of the ball C b in terms of k and the specific heat of water C w . Solution: The equation that relates the rate of heat flow to the temprature defferece is ( ) R dQ T T dt α = − Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 2 GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung Where T is current temperature of the container’s surface, T R is room temperature, and the constant α depends on the conductivity od the materials and the geometry of the container. The container. The containers are identical so they should have the same constant α . After all, each container is made of metal and conducts heat well, minimizing the discrepancy that results from the greater surface area of the slingtly higher water level in the container with the submerged ball. ( ) ( ) ( ) ( ) . R R R R dQ T T dQ T T dt dt dT mcdT T T dt mc dt T T α α α α = − ⇒ = − = − ⇒ = − The above equation shows that the time to cool to a temprature T is proportinonal to amount of heat released by the container ( ) Q mc T= ∆ . Therefore, b w w b w w mC T mC T C C k k mC T C ∆ + ∆ + = ⇒ = ∆ And, finaly, ( ) 1 b W C k C= − New Words: • identical: đồng chất, giống hệt nhau • light: nhẹ • metal: kim loại • temprature: nhiệt độ; the current temprature: nhiệt độ hiện thời • nonconducting string: dây không dẫn (nhiệt) • density: mật độ, khối lượng riêng • boiling point: điểm sôi • specific heat: nhiệt dung riêng • to relate sth to/with sth: liên hệ cái gì với cái gì • difference: hiệu • to depend on: phụ thuộc vào • minimizing: làm giảm thiểu • discrepancy: sự khác biệt • proportional (to): tỉ lệ thuận với Problem 4: (Bài toán tìm thời gian nhỏ nhất) A child in the boat needs to cross the river, the speed of current in river is k times greater than the speed of the boat in the still water. If the child crosses the river such as to minimize the lateral displacement, it takes time t to cross. What is the minimum time required to cross the river? Solution: The two sketches below show two possible directions for the resultant velocity of the child (v 1 at the top sketche, v 2 at the bottom one). When the child chooses the direction 1 v ur (see top sketch), angle α is maximized, and she gets the minimum lateral displacement. From the top sketch: 2 2 1 1 1 sin 1 1and cos k k k k α α = = − = − For the minimum time we find: d = v.t min For the minimum displacement, the distance traveled (let us call it l ) is given by d = sinl α , and the time required to cross with minimum lateral displacement is given by: Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 3 GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung ( ) ( ) 1 .k vcos t l because v kvcos α α = = Combining these equations gives: min .sin .t k cos α α = , or: 2 min 1 . k t t k − = New Words: • speed: tốc độ • current: dòng nước (the speed of the current : tốc độ dòng nước) • lateral displacement: độ dịch chuyển theo dòng nước (nghĩa trong bài) • to minimize: làm (cho cái gì đó) đạt cực tiểu • it takes time t… : mất thời gian t để … • equation: phương trình • sketch: hình vẽ • resultant velocity: vận tốc tổng hợp Problem 5 (Bài toán tìm ứng suất) A balloom is fill with helium at the atmospheric pressure P. The volume of the balloon is V. The balloon is made of the material of mass m and density d. After being released(làm nhẹ), the balloon bursts at altitude where the atmospheric pressure is P/2. Immediatety before bursting, the balloon has a volume of 1,25V. Find the maximum stress that the balloon material can withstand. Assume(cho rằng) that the temprature of helium remains constant, the balloon remains spherical(hình cầu), and the density of the material remains virtually(gần như, hầu như) constant. Solition: Let us denote the final volume and pressure by V’ = 1,25V and P’, respectivity. Taking helium to be an ideal(lí tưởng) gas, we can write that PV = P’V’ under isothermal(đường đẳng nhiệt) conditions(điều kiện), so that P’ = 0,8P. Since the external(bên ngoài) is 0,5P, the net outward pressure foce on half of the ballon is 0,3P. Thus(như vậy, do đó ) the net(tổng cộng) pressure force on half of the balloon is 2 0,3 .P R π , where R is final radius of the balloon, given by 3 4 1,25 3 V R π = . But this must just balance(cân bằng) the elastic force A σ ∆ , where 2A Rt π ∆ = is the cross-sectional area of the ballon material at its mid-plane. Here t is the thickness(độ dày) of th ballon material, obtained(thu được) from 2 4 m d R t π = . Combining(kết hợp) the equations givens: 9 16 PVd m σ = New Words: • balloon: khí cầu • pressure: áp suất (atmospheric pressure: áp suất khí quyển) • material: vật liệu • (to) burst: nổ • density: mật độ hay khối lượng riêng • atitude: độ cao • volume: thể tích • stress: ứng suất • (to) wthstand: chịu được • (to) remain: còn là • net us denote … by…. : ta kí hiệu … là…. • the net outward pressure: áp suất tổng hợp hướng ra ngoài • pressure fore: áp lực • elastic fore: lực đàn hồi • area: diện tích (cross-sectional area: diện tích tiết diện ngang) • mid-plane: mặt phẳng giữa (đi qua tâm khí cầu) Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 4 GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung Problem 6: (Bài toán về thấu kính – gương) An eagle lands on the ground a = 5.00m behind a hiker’s back. The hiker sees two images of the eagle refleting (phản xạ) in her glasses. One image appears to be b 1 = 5.00m away, and orther is b 2 = 0.714m away. When she turns and looks at the eagle, still wearing glasses, the image of the eagle appears to be at a distance b 3 = 2.5m. Find the index of refraction n of the lenses. Solution: Let us consider(xem xét) the first situation. Since one of images is 5.00m away, it is clear that one of the lens surface is flat. For rays passing through such a lens, the lensmarker’s equation givens: 1 1 R f n = − Here R is the radius of curvature of the concave surface, n is the index of refraction of lens, ans f 1 is focal length of the lens. Also, for a concave surface with radius R, the focal length f 2 can be found as: 2 2 R f = The rays from the object are passing through the lens, reflecting from the convex surface, and passing through the lens again to emerge from the flat side to form a virtual image: ( ) 2 1 2 1 1 2 2 1 1 1 1 1 1 2 1 2 2 n n a b f f f f f R R R − − = + + = + = + = When the hiker turns, the rays from the object passing through the lens also form a virtual image: 3 1 1 1 1 1n a b f R − − = = Eliminating R from the pair of equation, ( ) 5 1 5 3 n n   − =  ÷   , so n = 1.5. New Words: • eagle: đại bàng • (to) land: đậu • round: mặt đất • hiker: lữ khách • image: ảnh (virtual image: ảnh ảo) • glasses: kính mắt (wearing glasses: đeo kính) • (to) reflect: phản xạ • index of refration: chiết suất • lens ( số nhiều: lenses): thấu kính • radius of curvarture: bán kính cong • concave surface: mặt lõm • convex surface: mặt lồi • focal length: tiêu cự • ray: tia sáng • passing through: truyền qua • (to)emerge: ló ra • (to) elimilate: khử, loại bỏ. Problem 7: (Dao động cơ học) When the system shown in the diagram is in equilibrium, the right spring is stretched by X 1 . The coefficient of the static friction the blocks is s µ ; there is no friction between the bottom block and the supporting surface. The force Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 5 GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung constant of the springs are k and 3k (see the diagram). The blocks have equal mass m. Find the maximum amplitude of the oscillatons of system shown in the diagram that does not allow the top block to slide on the bottom. Solution: The origin is at the equillibrium position and the diriection of increasing X is toward the right. If the blocks are at the origin(gốc), the net force on therm is its zero. If the blocks are a small distance X to the right of the origin, the value of the force excerted by spring on the right is less than its value with the blocks at the equillibrium position by kX. Furthermore, the value of the force exerted by spring on the left is less than (more negative than) its value with the blocks at the equillibrium position by 3kX. Therefore, with the blocks at position X the value of the net force on them is -4kX. Applying Newton’s second law to the two blocks system gives: -4kX = ma x Applying Newton’s second law to the lower block gives: K(X 1 – X) – f = ma x Where f is magnitude of the frictional force. Solving(giải) the first equation for, and substituting the result into the second equation gives: k(X 1 - X) – f = -2kX. Solving for f gives: f = k(X 1 – X). The maximum value for X is the amplitude A, and the maximum value for f is s mg µ . Thus, ( ) 1 max . s mg k X A µ = + Solving for A max gives: max 1 s mg A X k µ = − New words: • equillibrium: cân bằng (equillibrium position: vị trí cân bằng). • direction of increasing x: hướng x tăng • diagram: sơ đồ, hình vẽ • spring: lò xo (the right spring: lò xo bên phải) • (to) stretch: giãn • coeficient of the static friction: hệ số ma sát tĩnh • supporting surface: mặt đỡ • force constant: độ cứng (của lò xo) • amplitude: biến độ • oscillation: dao động • (to) slide: trượt • net force: lực tổng hợp • value: giá trị • (to) exert: tác dụng • negative: âm • law: định luật (Newton’s second law: định luật II Niuton) • magnitude: độ lớn • (to) substitute … into: thay…. Vào • Result: kết quả Problemb 8: (Chuyển động của tàu vũ trụ) A spaceship is traveling in outer space at a constant velocity v = 1000m/s. Suddenly(đột ngột), the commander observes an asteroid straight ahead. The distance to the aseroid is L = 9.0km; the diameter of the asteroid is D = 7.0km. The commander attempts to take an evasive(né tránh) maneuver by firing the emergency rocket. The rocket gives the spaceship a very quick change in Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 6 GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung velocity 300 /V m s δ = . Is it possible for the spaceship to avoid the collision ? Assume that the commander can fire the rocket in any direction. Solution: I suppose that L = 9km is the distance to the surface of the asteroid. I Fig.1 the angle that is needed to avoid a collision, is α . 0 sin 3,5 /12,5 0,28, 16,3 α α = = = In Fig.2 the maximum angle of deflection that is possible is β . 0 sin 300 /1000 0,33, 15,5 β β = = = . Clearly, : α β < there is no conllision. The only problem is that the commader experiences a change in velocity of 300m/s in a short period of time. The commander experiences an acceleration of at least 300m/s 2 . I have no ideal if he/she/it can survive this acceleration… A friendly solution is s circular trajectory with a radius R for the spaceship (that would require a continuos thrust, however). ( ) 2 2 2 3,5 12,5 20,6R R R km+ = + ⇒ = The centripetal acceleration is then: 2 2 48,6 c v m a s R = = This gives the commander a better chance of surviving this adventure. New Words: • spaceship: con tàu vũ trụ • outer space: không gian vũ trụ • commander: chỉ huy tàu • asteroid: tiểu hành tinh • (to) abserve: quan sát • straight ahead: ở phía trước ngay trên đường bay của con tàu. • (to) attempt: có ý định • maneuver: thuật bay (nghĩa trong bài) • an evasive maneuver: thuật bay né tránh • (to) fire, firing: phóng • (to) avoid: tránh • conllision: va chạm • a very quick change in velocity: một sự thay đổi rất nhanh về vận tốc • in any direction: về bất kì hướng nào • the distance to the surface of … khoảng cách tới bề mặt của … • (to) assume that: giả sử rằng • The maximum angle of deflection: góc lệch cực đại • (to) experience: cảm thấy, phải chịu • Survive: sống sót • A friendly solution: giải pháp êm dịu hơn • A circular trajectory: quỹ đạo tròn • Thrust: lực đẩy • Centripetal acceleration: gia tốc hướng tâm Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 7 GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung • Adventure: cuộc phiêu lưu Problem 9: (Dao động cơ học) On the diagram, two blocks of equal mass are connected by an ideal string. The value of m, k 1 and k 2 are given (k 1 > k 2 ). Initially, both spring are relaxed. Then the left block is slowly puleed down a dstance x and released. Find the acceleration of each block immediately after the release. Find all posible answers. Solution: Let T bi the tension in the ideal string and a be the acceleration of the blocks at instant of release. For the block on the left, the upward acceleration may be found from: 1 T k x mg ma+ − = For the block on the right, the downward acceleration may be found from: 2 k x mg T ma+ − = Adding the equations givens the acceleration of blocks as: ( ) 1 2 2 k k a m + = However, substracting the equations gives: ( ) 1 2 2 k k T mg x − = − But a negative T would indicate compression of the ideal string. So ( ) 1 2 2 k k a m + = only if 1 2 2mg k k x < + . If 1 2 2 0 mg k k T x > + ⇒ = and the blocks accelerate independently: 1 2 1 2 k x k x a g and a g m m = − = + New Words: • string: dây, lò xo (ideal string: dây lí tưởng tức dây không có khối lượng) • immediately: ngay • (to) relax: không biến dạng • (to) release: buông ra, thả ra • tension: lực căng • instant: thời điểm • adding: cộng vào • substracting: trừ • comprssion: nén • (to) vary: biến thiên Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 8 . qua tâm khí cầu) Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 4 GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung Problem 6: (Bài toán về thấu. temprature defferece is ( ) R dQ T T dt α = − Bài tập tiếng anh Vật Lý Năm học: 2008 - 2009 Trang 2 GV: Trần Danh Xuân Trường THPT Chuyên Quang Trung Where