maximum x coordinate,)/s j.( =F cxin meters, and c a constant. At= 3.00 m, it is 11.0 J. Find3. The figure5. The angular acceleration of a wheel is 42 =6.0 4.0t t α-, with α in radians per second-squared and t in seconds. At time = 0t , the wheel has an angular velocity of+2.0 rad/s and an angular position of +1.0 rad . Write expressions for(a) the angular velocity 531.2 1.33 2.0t t ω=-+(rad/s) ; (b) the angular position 640.200.33 2.0 1.0t t t θ=-++(rad).6. An iron anchor of density 7870 kg/m 3 appears 200 N lighter in water than in air. The volume of the anchor is 232.0410 m -⨯. Its weight in the air is 31.5710 N ⨯.7. In the figure, two diverse springs of spring constant respectively 1k and 2k are in series attached to a block of mass m , the frequency of oscillation is8. A stationary motion detector sends sound waves of frequency 0.150 MHz toward a truck approaching at a speed of 45.0 m/s. The frequency of the waves reflected back to the detector is 0.195 MHz .9. The figure represents a closed cycle for a gas (the figure is not drawn to scale). The change in the internal energy of the gas as it moves from a to c along the path abc is -200 J. As it moves from c to d , 180 J must be transferred to it as heat. An additional transfer of 80 J to it as heat is needed as it moves from d to a . As it moves from c to d , the work done on the gas is 60 J .10. The figure shows the Maxwell-Boltzmann velocity distribution functions of a gas for two different temperatures 1T and 2T , then 1T < 2T (<, >, or = ).pVⅡ Calculation (Total 60 points)11．(Total 12 points, 6 points/question) A person of mass M = 32.5 kg on ice disdainfully throws a quantum text book weighing m = 2.25 kg at b v = 12 m/s. The book is thrown from zero height and the total distance between the book and the offender is 15.2 m when the book lands.(1) At what angle was this excellent book thrown? (2) How fast is the offender moving? Solution:(1) Since there are no external forces in the horizontal direction on the person-book system, thex component of the center of mass does not move while the book is in the air. We can usethis fact to find how far the book travels before it hits the ground.0() cm b b b MX x m x x M x xm M==+-⇒=+Where 15.2 m x =is the total final separation of the book and the person. Plugging in numbers,14.2 m b x =. Take the book to be thrown at an angle α with respect to the horizontal. Then thebook ’s initial velocity in the x direction is cos b v α and that in the y direction is sin b v α. In the x direction we know thatcos cos bb b b x x v t t v αα=⇒=.In they direction, the book starts and ends the same height so,2110sin sin =22b b v t gt v gt αα=-⇒Substitute t , we find2sin 2 =37.62bbgx v αα=⇒. (2) Taking the person ’s final velocity to be v , we know from conservation of momentum thatcos cos .b b mMv mv v v Mαα=-⇒=-we find 0.66/v m s =-.Answer: (1) the book is thrown at angle =37.6α; (2) the offender has a final velocity of0.66/v m s =-. (Note: because of the double angle formula, for α, there is a second solution when the angle is equal to 90α-).12. (Total 12 points, 4 points/question)(1) What is the rotational inertia CM I of a propeller with three blades (treated as rods) of massm , length L at 120o relative to each other?(2) If a torque τ acts on this propeller, how long will it take to reach an angular velocity ω? (3) How many revolutions will it have made before reaching this ω?Solution(1) We know that the rotational inertia of a single rod rotating around its end is213mL . It ’s not hard to convince oneself that if there are three of them rotating around the same axis and in thesame plane, the rotational inertia is just three times this, 2CM I mL =.(2) Sincet ωα= and CM I τα=,2CM I mL t ωωττ==.(3) From our knowledge of constant acceleration problems,222222 222CM I mL ωωωωαθθαττ=⇒===The number of revolution it made is2224mL N θωππτ==13．(Total 12 points) A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 60.0 cm, and the density of iron is 7.87 g/cm 3. Find the inner diameter.SolutionFor our estimate of submerged V we interpret “almost completely submerged ” to mean3submerged 4where 30 cm 3o o V r r π≈=Thus, equilibrium of forces (on the iron sphere) leads tog r g V g r r g m F o water submerged water i o iron iron b ⋅⋅=⋅⋅=⋅⎪⎭⎫⎝⎛-⋅=⋅=333343434πρρππρWhere i r is the inner radius (half the inner diameter). Substitute into our estimate for submerged V as well as the densities of water (1.0 g/cm 3) and iron (7.87 g/cm 3), we obtain the inner diameter:31122⎪⎪⎭⎫ ⎝⎛-=iron watero i r r ρρ=57.3cm14．(Total 12 points, 4 points/question) A progressive wave travelling along a string has maximum amplitudeA 0.0821 m =, angular frequency = 100 rad/s ωand wave number= 22.0 rad/m k . If the wave has zero amplitude at = 0t and = 0x for its starting conditions (1) State the wave function that represents the progressive wave motion for this wave travelling in the negative x -direction.(2) Find the wavelength ()λ, period ()T and the traveling speed ()v of this wave.(3) Find its amplitude at a time = 2.5 s t at a distance = 3.2 m x from its origin, for thiswave travelling in the negative x -direction.Solution(1) ()m 1000.22sin 102.82t x y -⨯=-(2) 22221000.2856 m; 0.0628 s; 4.545 m/s 2210022T v k k ππππωλω========= (3) ()[]m 10-9.85.21002.30.22sin 102.8-32⨯=⨯--⨯⨯=-y .15. (Total 12 points, 4 points/question) One mole of an ideal diatomic gas goes from a to c along the diagonal path in Figure. The scale of the vertical axis is set by = 5.0 kPa ab p and= 2.0 kPa c p , and the scale of the horizontal axis is set by 3 = 4.0 m bc V and 3 = 2.0 m a V .During the transition,(1) What is the change in internal energy of the gas? (2) How much energy is added to the gas as heat?(3) How much heat is required if the gas goes from a to c along the indirect path abc ?V a V bcVolume (m 3)SolutionTwo formulas (other than the first law of thermodynamics) will be used. It is straightforward to show, for any process that is depicted as a straight line on the pV diagram, the work isstraight2i f p p W V +⎛⎫=∆ ⎪⎝⎭Which includes, as special cases, W p V =∆ for constant-pressure process and 0W = for constant-volume processes. Furtherint 22f f E n RT pV ⎛⎫⎛⎫== ⎪ ⎪⎝⎭⎝⎭Where we have used the ideal gas law in the last step. We emphasize that, in order to obtain workand energy in joules, pressure should be in pascals (N/m 2) and volume should be in cubic meters. The degrees of freedom for a diatomic gas is 5f =.(1) The internal energy change isp abP cPressure (kPa)3333int int 355()(2.010 Pa)(4.0 m )(5.010 Pa)(2.0 m )225.010 Jc a c c a a E E p V p V -=-=⨯-⨯=-⨯(2) The work done during the process represented by the diagonal path is()333diag (3.510 Pa)(2.0 m )7.010 J 2a c ca p p W V V +⎛⎫=-=⨯=⨯ ⎪⎝⎭Consequently, the first law of thermodynamics gives()333diag int diag 5.0107.010 J 2.010 J Q E W =∆+=-⨯+⨯=⨯.(3) The fact that int E ∆ only depends on the initial and final states, and not on the details ofthe “path ” between them, means we can write 3int int int 5.010 J c a E E E ∆=-=-⨯ forthe indirect path, too. In this case, the work done consists of that done during the constant pressure part (the horizontal line in the graph) plus that done during the constant volume part (the vertical line):334indirect (5.010 Pa)(2.0 m )+0 1.010 J W =⨯=⨯Now, the first law of thermodynamics leads to343indirect int indirect ( 5.010 1.010) J 5.010 J Q E W =∆+=-⨯+⨯=⨯。