Answers to exam-style questions

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1 Answers to exam-style questions Option B 1 = 1 mark 1 a i he forces are as shown in the left diagram: F F 8.0 m m 0 W W he perpendicular distance between the axis and the line of the tension force is L sin 0. Rotational equilibrium by taking torques about an axis through the point of support gives: W L = L sin 0 Hence W = = 1 kn. ii ranslational equilibrium gives: cos 0 = F and sin 0 + F = 1 kn. Hence cos 0 = Fx = kn and Fy = 7. kn so that the magnitude of F is F = = 1 kn And the direction to the horizontal is θ = tan = b he critical case is when the worker stands all the way to the right. Rotational equilibrium in this case gives: W L + mgl = L sin 0. Solving for the tension gives: = kn. a he forces are shown in the diagram and they are the weight of the cylinder, mg. he normal reaction, N, A frictional force f. N x y f θ mg b i Newton s second law for the translational motion down the plane is Mg sinθ f = Ma. For the rotational motion by taking torques about the axis through the centre of mass is 1 1 fr = MR α MRa = 1 Mg sinθ Ma = Ma From which the result follows. physics for the IB Diploma Cambridge University Press 01 Answers to exam-style questions Option b 1

2 ii f = Mg sin θ Ma = sin sin N = c he rate of change of the angular momentum is the net torque. And this is fr = = Nm. a i When the ring makes contact with the disc and while it is sliding, it exerts a frictional force on the disc but the disc exerts equal and opposite force on the ring. Hence the net torque is zero and hence angular momentum is conserved. 1 1 ii he initial angular momentum of the disc is L = Iω = MR ω = = 7.6 Js. 1 After the ring lands the total angular momentum is L = ω ω. Hence ω ω = 7.6 Js which gives ω = 1 rad s 1. b 1 1 iii he initial kinetic energy is EK = I = 1 ω = J. 1 he final is E K = ( ) 1. 0 = J leading to a loss of J. 1.0 i α = ω = = 7.00 rad s t ii θ = αt = = 1. rad iii Which is 1..0 π = revolutions. L Γ = t Γ = = 1.6 N m.00 iv It is equal and opposite to that on the ring. Because the force on the ring is equal and opposite to that on the disc. c he change in the kinetic energy of the ring is 1 ( ) 1.0 = 9.69 J. And so the power developed is 9.69 = 1. W..00 (his can also be done through P = Γ ω = = 1. W.) 4 a i he temperature at B doubled at constant volume so the pressure also doubles at p = Pa. ii From pv = c and pv = nr we find p = c. Hence ( ) (600) = ( ) B C C B C B C leading to C = 4 K. nrb iii he volume at B is VB = = = m. pb And so V B VC V V 4 = = = = m. 00 B Answers to exam-style questions Option b physics for the IB Diploma Cambridge University Press 01

3 b i U = AB Rn = U = J AB ii his happens from C to A: W = p V = ( ) 10 = 180 J and the change in internal energy is U AB = Rn = (00 4) = 19 J. Hence Q = U + W = = J. c Any heat engine working in a cycle cannot transform all the heat into mechanical work. And this engine rejects heat into the surroundings as it should. a i A curve along which no heat is exchanged. ii An adiabatic expansion involves a piston moving outwards fast. Hence molecules bounce back from the piston with a reduced speed and hence lower temperature. b he product pressure volume is constant for an isothermal. his is the case for points A and C (product is 100 J). And the same is true for any other point on the curve, for example at p = Pa, V = m. V c i A VB V B = B = A V A B B = 70 K 0.0 = At C = 00 K since AC is isothermal. C A pv ii Using data at A: n = = R n = d i Energy is transferred out of the gas along C to A. From Q = U + W and U = 0 we find Q = 160 J. ii his happens from A to B: W = ( ) 10 = 90 J and U = (70 00) = 1 J. And so Q = = J. iii W = U (since BC is an adiabatic). BC BC And U BC = U AB = 1 (since AC is an isothermal). OR Since for the whole cycle U = 0, the net work is Q Q = 160 = 6 J. And Wnet = WAB + WBC + WCA 18 = 90 + WBC 160 WBC = 198 J. Wnet 6 iv he efficiency is e = = = Qin 6 a For an adiabatic, pv = c and since V = nr. p we find p nr = c. p Raising to the rd nr power gives p = c and so the result. p b i From constant p = we find 0 = (.0 10 ) (.0 10 ) in 6 out. physics for the IB Diploma Cambridge University Press 01 Answers to exam-style questions Option b

4 .0 10 = ii From pv.0 10 V = = K = constant we find = 0.10 m pv c he number of moles is n = = = 0. R U = Rn (804 0) = U = MJ 7 Delineate a rectangular region in the liquid whose top surface is the free surface of the liquid and has area A. Equilibrium demands that weight = net upward force. In other words that ρ Ahg = pa p 0 A. From which the result follows. b i In free fall gravity disappears and so the pressure is just the atmospheric pressure. ii When the liquid accelerates upwards there is an additional force pushing the liquid upwards and so the pressure increases. c A body immersed in a fluid experiences an upward force that is equal to the weight of the displaced liquid. d i Equilibrium demands that ρ Vg = ρ 0.7Vg. wood water Hence ρwood = ρwater 0.7 = 70 kg m. ii Equilibrium demands that ρwoodvg = ρoil 0.8Vg. ρwood 70 Hence ρoil = = = kg m a i p0 + ρ gz = p0 + ρv hence v = gz v = 9.8 (0 + 40) = m s 1 ii hat the flow is laminar, and there are no losses of energy. b he flow rate is given by Q = Av = πr v. 1 Hence Q = π (0.) 71.4 = 14 m s c i p = p0 + ρ gh = p = Pa 1 ii he pressure is given by p0 + ρ gz = p + ρv where the speed can be found from the flow rate (i.e. the continuity equation) π (0.6) v = m s 1. 4 Answers to exam-style questions Option b physics for the IB Diploma Cambridge University Press 01

5 i.e, v = ms -1. And hence 1 1 p = p0 + ρ gz ρv = = Pa d he speed at depth h is v = gh. he flow rate is Q = Av = πr gh and has to equal 0.40 m s Hence h = g πr = 9.8 π 0.0 = 7. m. 9 a he left side is connected to the holes in the tube past which the air moves fast. Hence the pressure there is low and the liquid is higher. b Call the pressure at the top of the left column p L and that on the right p R. hen 1 1 pl + ρair gz + ρvl = pr + ρair gz + ρvr and with vl = 0; vr = v, it becomes v = ( pl pr ). ρ air But pl pr = ρ gh which gives the result. ρ gh c v = =. ρ 1.0 air v = m s a Smooth streamlines. Closer together above the aerofoil b i From pl + ρ gz + ρvl = pu + ρ gz + ρvu we obtain p = pl pu = ρvu ρvl. Hence F A p A( 1 1 v v ) = = ρ U ρ L = 1.0 (8 8 ) = kn. ii hat the area above and below the foil are equal/that the flow is laminar. c he net upward force on the foil is about 16 kn and this is an estimate of the downward force on the fuselage. Ignoring effects of torque. d i he streamlines are no longer smooth but become eddy like and chaotic. ii Everywhere on the top side of the aerofoil and especially to the right. iii It will be drastically reduced. 11 a In undamped oscillations the energy is constant and so the amplitude stays the same. In damped oscillations energy is dissipated and the amplitude keeps getting smaller. b i 8.0 s ii Correct readings of amplitudes. 6 Q = π 6 Q c i Amplitude reducing more every cycle. Period staying essentially unchanged/very slightly increases. ii It will decrease. physics for the IB Diploma Cambridge University Press 01 Answers to exam-style questions Option b

6 y / cm t / s d Q = π Q 79 1 a All oscillating systems have their own natural frequency of oscillation. When a periodic external force is applied to the system the amplitude of oscillation will depnd on the relation of the external frequency to the natural frequency. he amplitude will be large when the frequency of the external force is the same as the natural frequency in which case we have resonance external. b Wider and lower curve. With peak shifted slightly to the right. See curve in blue. A f 6 Answers to exam-style questions Option b physics for the IB Diploma Cambridge University Press 01

7 c i Same intersection point. Less steep. See curve in blue. θ / rad π π 0 0 ii 10 Hz f / Hz physics for the IB Diploma Cambridge University Press 01 Answers to exam-style questions Option b 7