西南交大 大学物理 英文 试题 答案No.A1-2.11348895.pdf

University Physics AI No 2 Motion in Two and Three Dimensions Class Number Name I Choose the Correct Answer 1 An object moves in the xy plane with an acceleration that has a positive x component At time 0 t the object has a velocity given by jiv 0 3 r What can be concluded about the y component of the acceleration D A The y component must be positive and constant B The y component must be negative and constant C The y component must be zero D Nothing at all can be concluded about the y component Solution According to the definition of the acceleration components t v a x x d d and t v a y y d d if we only know the initial velocity of the object nothing at all can be concluded about the y component 2 An object moves with a constant acceleration a r Which of the following expression are also constant B A t v d d r B t v d dr C t v d d 2 D t vv d d rr Solution According to the relation of the acceleration and the velocity t v a d dvv we know the velocity v v is vary with time The magnitude and the direction of the velocity may not be constant therefore A C and D are not corrected As constant d d t v a v v answer B is the correct answer 3 An object is launched into the air with an initial velocity given by m s 8 9 9 4 0 jiv r Ignore air resistance At the highest point the magnitude of the velocity is B A 0 B m s9 4 2 C m s8 9 2 D m s 8 9 9 4 22 Solution The y component of the velocity is zero at the highest point and the x component of the velocity is 4 9m s thus the velocity is m s 9 4 iv r the magnitude is 4 9m s II Filling the Blanks 1 A gypsy moth caterpillar Porthetria dispar inches along a crooked branch to a tasty oak leaf wriggling 15 cm horizontally and then 30 cm along a section of the branch inclined at 30o to the horizontal as shown in Figure 1 a The initial position vectors i 15 0 and the final position vectors ji 15 0 315 0 of the caterpillar in SI units using the crook in the branch as the origin for a set of horizontal and vertical coordinate axes b If the caterpillar traverses the distance during 1 0 min its average speed is 0 0075m s its average velocity is m s 0025 0 0068 0ji and the magnitude of its average velocity is m s1024 7 3 Solution a The coordinate system is shown in figure The initial position vector is ir 15 0 v and the final position vector is jijirf 15 0 26 0 30sin3 0 30cos3 0 oo v b The average speed is m s0075 0 60 3 015 0 t s vave The average velocity is ji iji t rr t r v if ave 0025 0 0068 0 60 15 0 15 0 26 0 vv v v m s The magnitude of the average velocity is m s 1024 70025 00068 0 0025 0 0068 0 322 jivave v 2 Use Equation 4 28 for tr r jtritrtr sin cos r and Equation 4 26 for t r ktt z r to evaluate the expression trttr rrr 0 Explain why this result is another way to see that tv r is always perpendicular to tr r for circular motion solution 2 Solution 1 ittrjttrjtritrkttrt zzz sin cos sin cos rv 15cm 30cm Branch Tree trunk Fig 1 o 30 x y 0 0 cos sin cos sin sin cos sin cos 22 tttrtttr ittrjttrjtritrtrttr zz zz rvv 2 According to tvtrt vrr and 0 trttr rrr we know 0 tvtr vr The result means that tv r is always perpendicular to tr r 3 The angle turned through by the flywheel of a generator during a time interval t is given by 43 ctbtat where a b and c are constant Its angular velocity is 32 43ctbta And its angular acceleration is 2 126ctbt Solution Its angular velocity is 3243 43 d d d d ctbtactbtat tt And its angular acceleration is 232 126 43 d d d d ctbtctbta tt 4 The flywheel of an engine I rotating at 25 2 rad s When the engine is turned off the flywheel decelerates at a constant rate and comes to rest after 19 7s The angular acceleration in rad s2 of the flywheel is 12 8 rad s2 the angle in rad through which the flywheel rotates in coming to rest is 248 1 rad and the number of revolutions made by the flywheel in coming to rest is 39 5 Solution 1 According to the equation 4 49 tt 0 the angular acceleration is 20 rad s28 1 7 19 2 25 t 2 According to the equation of motion 2 0 2 1 ttt o the change of the angle is rad1 2487 1928 1 2 1 7 192 25 2 1 22 tt o 3 The number of revolutions is 5 39 14 32 1 248 2 5 A car is traveling around a banked circular curve of radius 150 m on a test track At the instant when t 0s the car is moving north and its speed is 30 0 m s but decreasing uniformly so that after 5 0 s its angular speed will be 3 4 that it was when t 0s The angular speed of the car when t 0s is 0 2rad s the angular speed 5 0 s later is 0 15rad s the magnitude of the centripetal acceleration of the car when t 0s is 6 0m s2 the magnitude of the centripetal acceleration of the car when t 5 00s is 3 375m s2 the magnitude of the angular acceleration is 0 01rad s2 the magnitude of the tangential acceleration is 1 5m s2 Solution 1 The angular speed of the car when t 0s is rad s2 0 150 30 0 0 r v 2 The angular speed 5 0 s later is rad s15 0 4 3 0 3 The magnitude of the centripetal acceleration of the car when t 0s is 2 22 m s0 6 150 30 r v ac 4 The magnitude of the centripetal acceleration of the car when t 5 00s is 222 m s375 315015 0 rac 5 According to tt 0 the angular acceleration is 2 0 rad s01 0 5 2 015 0 t t 6 The magnitude of the tangential acceleration is 2 m s5 101 0150 ra 6 A projectile is launched at speed v0 at an angle with the horizontal from the bottom of a hill of constant slope as shown in Figure 2 The range of the projectile up the slope is 2 2 0 cos sin cos2 g v r Solution The coordinate system is shown in figure Assume the range is r the time is t we have 2 0 0 2 1 sinsin direction coscos direction gttvry tvrx Solve the two equations we have 2 2 0 cos sin cos2 g v r III Give the Solutions of the Following Problems 1 A particle leaves the origin at 0 t with an initial velocity iv m s 6 3 0 r It experiences a 0 v r Range Fig 2 y x constant acceleration jia m s4 1 m s2 1 22 r a At what time does the particle reach its maximum x coordinate b What is the velocity of the particle at this time c Where is the particle at this time Solution a According to the problem we have 2 m s2 1 x a and m s6 3 0 x v The position of the particle is 4 5 3 6 06 06 3 2 1 222 0 ttttatvx xx when max xx 0 d d t x We get t 3s the particle reach its maximum x coordinate x 5 4 b By using t v a d dvv we have j titj ti titjivv tav t tv v 4 1 2 16 3 4 1 2 1 6 3d 4 1 1 2 dd 0 0 0 0 vv v v v When t 3s the velocity of the particle is m s 2 4jv v c By using t r v d dvv we have jtitttj titr tvr t tr 7 0 6 06 3 d 4 1 1 2 3 6 dd 22 0 00 v vv v When t 3s the position vector of the particle is m 3 6 4 5 7 0 6 06 3 22 jijtittr v 2 A young basketball player is attempting to make a shot The ball leaves the hands the player at angle of 60 to the horizontal at an elevation of 2 0 m above the floor The skillful player makes the shot with the ball traveling precisely through the center of the hoop as indicated in Figure 3 To loud cheers calculate the speed at which the ball left the hands of the player 8 0m 3 0m 2 0m 60 Fig 3 y x Solution Establish the Cartesian coordinate system given in figure According to the problem we have gayxyx y m3 m8 m2 m0 00 Using the general equations 2 00 00 2 1 tatvyy tvxx yy x Then 2 2 1 60sin23 1 60cos8 2 0 0 gttv tv Solving equation 1 and 2 we have the speed at which the ball left the hands of the player is v0 9 88 m s 3 A pilot flies a plane at a speed of 700 km h with respect to the surrounding air A wind is blowing to the northeast at a speed of 100 km h In what direction must the plane be aimed so that the resulting direction of the plane is due north What is the ground speed of the aircraft Solution Establish the Cartesian coordinate system given in figure According to WGPGPW vvv rrr From the problem we have jiv jvv jiv PW PGPG WG sin700 cos700 45sin100 45cos100 r r r jijvji PG 45sin100 45cos100 sin700 cos700 oo o o 45sin100sin700 45cos100cos700 PG v y x 0 45 PG v v PW v v WG v v North East m s1 76745sin100sin700 2 84 14 2 arccos 14 2 arccos o o PG v Thus the direction which the plane must be aimed is West to North 84 2 or North to West 5 8 The speed of the plane relative to the ground is 767 1m s 4 A particle moving clockwise in a circle with a radius of 3 00 m has a total acceleration at some instant of magnitude 15 0 m s2 directed as indicated in Figure 4 a Find the magnitude of the centripetal acceleration at this instant b Find the speed of the particle at this instant c Find the angular speed of the particle at t