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1 Answers to test yourself questions Option B B Rotational dynamics ( ω + ω )t Use 0 ( +.).0 θ to get θ rad. Use ω ω0 + αθ to get ω and so ω rad s. Use ω ω0 + αθ to get.4. + α 0 π. Hence α rad s. 4 Let the forces at each support be L (for left) and R (for right). hese are vertically upwards. hen ranslational equilibrium: L + R N Rotational equilibrium: R.0 by taking torques about the support at the left. hen R 7 70 N and hence L N. Let the forces at each support be L (for left) and R (for right). hese are vertically upwards. hen just before the rod tips over we have equilibrium and the left support force tends to become zero since the rod will no longer touch the support. Hence by taking torques about the right support we find: x and x 0.60 m. 6 a he forces are as shown in the diagram below:.0 m F.0 m θ W 8 N ranslational equilibrium demands that: F x x + F 8 y y Rotational equilibrium demands that (we take torques about an axis through the point of support at the wall): 8.0 y.0 y 9 N..0 Now y sinθ and from the diagram, tanθ.. Hence θ 6.. hus.0 y N. sinθ sin 6. b Since y 9 N, F N also. Finally, Fx x 4.84 cos6. 9. N. Hence the y wall force is F Fx + Fy N. he angle it makes with the horizontal is 9 tan physics for the IB Diploma Cambridge University Press 0 ANSWERS O ES YOURSELF QUESIONS B

2 7 he torque provided by the force about the axis of rotation is Γ FR. We know that Γ I α. F 6. he moment of inertia of the cylinder is I MR and so α Γ rad s. I MR herefore ω ω + αt rad s. 0 8 Let h be the height form which the bodies are released. For the point particle will have: Mgh Mv v gh. For the others, Mgh Mv + Iω Mgh Mv + I v R I gh v + MR v So Sphere: gh I + MR v gh MR + MR gh 7 gh 7 Cylinder: v gh MR + MR gh gh Ring: v gh MR + MR gh gh Hence vring < vcylinder < vsphere < vpoint 9 a One revolution corresponds to π radians and so the disc is making 4 rev 4 rev 4 60 ω rpm. π s π 60 min π 4 b he angular acceleration has to be α. rad s. From 4.0 Γ Iα FR MR α we deduce that F MRα N. c From ω ω0 + αθ we find (.) θ and so θ 90 rad. his corresponds to π revolutions. ANSWERS O ES YOURSELF QUESIONS B physics for the IB Diploma Cambridge University Press 0

3 7 g 0 a Γ Iα MgR MR α. hus α 7 R. b It is not: the torque of the weight about the axis is decreasing because the perpendicular distance between the weight and the axis is decreasing. 7 0 g c Conservation of energy: MgR Iω. Hence MgR MR ω and so ω. 7R a I Mg L g 9.8 Γ α ML α. hus α.6. rad s. L.0 b It is not: the torque of the weight about the axis is decreasing because the perpendicular distance between the weight and the axis is decreasing. c Conservation of energy: Mg L g 9.8 Iω. Hence MgL ML ω and ω 4.9 rad s. L.0 From Newton s second law: F Ma and Fd Iα MR α. Combing these two equations gives Mad MR α MαRd MR α R d a he relevant forces are as shown. M herefore: mg ma Ma Hence, adding side by side, mg Ma + ma and so a b Using the hint we now have: mg ma and ( ) R Iα. Ma mg mg M + m. a Ia Assuming no slipping at the pulley, α and so ( ). Adding the first two equations gives R R Ia mg mg ( ) ( m + M ) a or mg ( m + M ) a i.e. mg Ma ( m + M ) a so that, finally, a R m. + M 4 a he kinetic energy initially is E K I 0 π ω J. his is also the 60 work needed to stop the disc. E.7 0 b he power developed is P t W. physics for the IB Diploma Cambridge University Press 0 ANSWERS O ES YOURSELF QUESIONS B

4 X: EK Iω ML ω J; 4 L Iω ML ω J s. Y: EK Iω ML ω.7 J; L Iω ML ω.8 J s. 6 here are no external torques so angular momentum is conserved: Iω I ω. M R MR ω ω 4 ω 0 0 leading to. 0 ω. 7 he angular momentum before and after the ring begins to move must be the same. Before it moves, the angular momentum is zero. After it begins to move it is: L Iω + mvr. mvr Hence Iω + mvr 0, i.e. ω 0.6 rad s. (he minus sign indicates that the I 0.0 ring rotates in the opposite direction to that of the car.) B hermodynamics 8 Use pv c to get 8. 0 (. 0 ) p (4.6 0 ) i.e.. 0 p p. 9 0 Pa 9 We know that pv nr c and also pv nr. From the ideal gas law we find p and V nr substituting in the first equation we find V V c or V c (another constant). Hence, 60 (.8 0 ) (4.8 0 ). Hence K 0 W p V (4..6) 0 80 J. a he process is isothermal and so U 0. Hence Q 0 + W J. b he adiabatic compression is steeper than the isotheermal and so there is more area under the graph and so more work. Isothermal is the blue curve and the red is adiabatic. he area under the isothermal curve is larger and so the work done is larger. pressure Volume / V 4 ANSWERS O ES YOURSELF QUESIONS B physics for the IB Diploma Cambridge University Press 0

5 Isothermal is the blue curve and the red is adiabatic. he area under theadiabatic curve is larger and so the work done is larger. pressure Volume / V 4 W U X positive zero zero Y zero positive positive Z positive positive positive a From Q U + W and Q 0 we find U W. he gas is compressed and so W < 0. Hence U > 0. In an ideal gas the internal energy is proportional to temperature (in kelvin) and so increasing U means increasing. b he gas is compressed by a pisotn that is rapidly moved in. he piston collides with molecules and gives kinetic energy to them increasing the average random kinetic energy of the molecules. Hence the temperature increases since temperature is proportional to the average random kinetic energy. 6 6 a Use W p V ( ).4 MJ. b From the gas law, V V and so giving 900 K 00. pv c he change in internal energy is U Rn MJ. 00 herefore Q U + W MJ 7 From Q U + W we get U Q W. So we can have Q > 0 but as long as Q W < 0, i.e. when the work done is greater than the heat supplied the temperature will actually drop. 8 a We must first determine if the gas is expanding or whether it is being compressed. Since pv nr it follows nr that p. For constant volume, the graph of pressure versus temperature would be a straight line through V the origin. his is not what we have so the volume is changing. he graph below shows two straight lines nr along each of which the volume is constant. Since p, the red line having a bigger slope corresponds to V the smaller volume. Hence the volume from P to Q decreases and so W < 0. 6 pressure Q P physics for the IB Diploma Cambridge University Press 0 ANSWERS O ES YOURSELF QUESIONS B

6 b From Q U + W we see that U < 0 since the temperature is decreasing and since the gas is being compressed, we also have that W < 0. Hence Q < 0 and thermal energy is being removed from the gas. 9 For the constant volume case we have Q U + W U + 0. For the constant pressure case we have that Q U + W. Since the heats are equal U U + W which shows that U > U since W > 0. he internal energy for an ideal gas is a function only of temperature and so the temperature increase is greater for the constant volume case. 0 Working as in the previous problem, QV U and QP U + W. he two changes in internal energy are the same because the temperature differences are the same. Since W > 0, Q P > Q V. a Steeper curve starting at. b Larger area under the adiabatic so more work done. c i Q U + W with U 0 and W kj so Q kj. Hence the change in entropy for the gas is Q 0 S 8. 8 J K 00 0 ii he surroundings receive heat kj and so their entropy increases by J K. 00 d he entropy change for the universe is zero. his is possible only for idealised reversible processes and an isothermal compression is such a process. a An adiabatic curve is a curve on a pressure-volume graph represents a process in which no heat enters or leaves a system. b BC is an isobaric process and DA is isovolumetric. c Along BC the temperature increases and so U > 0. he gas expands and so does work and hence W > 0. herefore, from the first law, Q U + W > 0 and so heat is given to the gas. Along DA the temperature drops and so U < 0; no work is done and so Q < 0. CD and AB are adiabatics and so Q 0. So heat is supplied only along BC. d i Heat is taken out along DA: the temperature at D is 6 pv nr K and that at A is K. herefore Q U + W U + 0 nr ( ) J. W Q ii he efficiency is e Q 4 in in Q Q in out Q. Hence 0.6 in.7 0 Q Qin J iii he area of the loop is the net work done which equals W Q Q J. in out in 4. his gives Let a quantity of heat Q be supplied to an ideal gas at constant volume: then Q nc V. Similarly, supply a quantity of heat Q to another ideal gas of the same mass at constant pressure so that the change in temperature will be the same. : Q nc P. Q and Q will be different temperature because in the second case some of the heat will go into doing work as the gas expands at constant pressure. From the first law of thermodynamics we have that Q nc V U + 0 and Q nc P U + p V. Subtracting these two equations gives nc P nc V p V. From the ideal gas law we find p V nr and so nc P nc V nr or c c R as required. P V 6 ANSWERS O ES YOURSELF QUESIONS B physics for the IB Diploma Cambridge University Press 0

7 4 a A and B have the same pressure and so V V i.e. hence 00 K. B and C have the same 800 volume hence P P i.e. 4 hence 600 K 00. C and D have the same pressure and so V V i.e hence 400 K 600. pv b A to B: U Rn (00 800) 80 kj 800 pv B to C: U Rn (600 00) 0 kj 00 pv C to D: U Rn ( ) 90 kj 600 pv D to A: U Rn ( ) + 0 kj 400 c From A to B: Q U + W. W + p V kj. Hence Q kj. From B to C: W 0 so Q 0 kj. From C to D: W p V kj. Hence Q kj. Finally from D to A: U 0 kj and W 0 so that Q + 0 kj. d he net work is the area of the loop which is kj. (As a check this must also be the net heat into the system i.e kj.) he heat in is 00 kj and so the efficiency is a he ice is receiving heat; the molecules are moving faster about their equilibrium positions. he disorder increases and so the entropy of the ice molecules increases. b During melting heat is provided to the ice molecules and so their entropy is again increasing. c he water must have had a temperature greater than + C. herefore as the temperature drops to + C the entropy of the water is decreasing. he overall change in entropy of the system must be positive. 6 he maximum possible efficiency for an engine working with these temperatures is, according to Carnot, e So the proposed engine is impossible a It is impossible for heat to flow from a cold to a warmer place without performing work. b Imagine the work output of the engine to be the input work in a Carnot refrigerator. hot Q Q W Carnot Q Q 4 cold Since the red engine has better efficiency than Carnot we must have that W W > i.e. that Q > Q. Notice Q Q that W Q Q Q Q4 so that Q4 Q Q Q and each of these differences is positive. Now the combined effect of the two engines is to remove transfer heat Q4 Q from the cold reservoir and deposit the (equal amount) Q Q into the hot reservoir without performing work. his violates the Clausius form of the second law. Hence an engine with an efficiency greater than Carnot (for the same temperatures) is impossible. physics for the IB Diploma Cambridge University Press 0 ANSWERS O ES YOURSELF QUESIONS B 7

8 B Fluids (HL) 8 Pressure only depends on depth and the pressure exerted on the surface so the pressures here are equal. 9 In the second case the fluid exerts an upward force on the floating wood and so the wood exerts and equal and opposite force on the fluid. Hence when put on a scale the second beaker will show a greater reading. 40 a he second beaker has a smaller mass of water and so a smaller weight by an amount equal to the weight of the displaced water. But there is an additional force pushing down on the liquid that is equal to the buoyant force. his force is equal to the weight of the displaced water and so the readings when the two beakers are put on a scale will be the same. b hey points are at the same depth so the pressure is the same. 4 Draw a horizontal line through X. Points on this line have the same pressure. Hence P P + ρ gh Pa X atm 4 Since the ice cube floats its weight is equal to the buoyant force. In turn the buoyant force is equal to the weight of the displaced water. Hence then the ice cube melts it will have a volume equal to the displaced volume of water and so the level will remain the same. W 4 From Pascal s principle, in the first case the pressure is p0 + ρ gh. In the second p0 + + ρ gh and in the third W A p0 + + ρ gh. A 44 he buoyant force on the sphere equals the weight of the sphere and also B ρ gv. Hence 4πR 4π 0. W B N. he mass is therefore m kg. he volume of the material in the sphere is 4 0. π 4π m 9.8 Hence the density is m kg m. V F F and so A A d giving π / 4 π 9.0 d 4. m. 46 Let v be the speed of the water at the faucet of area A. After falling a distance of h.0 cm the speed is imm. v v + gh and the cross sectional area is A. he equation of continuity gives Av Av A v + gh. Hence A v A v + gh ( v v ( ) A A v gha gha A A ) m s he flow rate is Q A v m s 47 he equation of continuity gives A v A v 0.60 v m s. 4. i.e. π π v and so 48 In one second a mass of water equal to m ρπr v 0 π kg. he energy of this mass of water as it leaves the pipe is E mv + mgh J. Since this is energy that is provided in s the power is also 80 W. 8 ANSWERS O ES YOURSELF QUESIONS B physics for the IB Diploma Cambridge University Press 0

9 49 We use Bernoulli s equation to find: p + ρv + ρgh p + ρv + ρgh p p kpa 0 a Imagine a streamline joining the surface to the hole and apply Bernoulli s equation to find: p + ρv + ρ gh p + ρv + ρ gh Patm Patm v + 0 v m s b he water from the upper hole has velocity v g.0 will hit the ground in a time given by h 6.0 t. It will then cover a horizontal distance of vt g m. g g g he speed from the lower hole will be v g 6.0. he lower hole water will take a time of t and cover a distance of vt g m. he ratio is then. g he water will; exit with speed given by v gd. he hole is a distance of H d from the ground and so.0 g ( H d) ( H d) will hit the ground in time t. he range is thus gd d( H d). his is clearly a g g H maximum (by graphing or differentiating) when d. 6 6 a px Patm + ρ gh (0 9) Pa; 6 p P + ρ gh Pa. Y atm b i Assuming the surface does not move appreciably, Bernoulli s equation applied to a streamline joining the surface to Y gives p + ρ v + ρ gh p + ρ v + ρgh Patm Patm v + 0 v m s ii he speed at X can be found by applying the continuity equation from X to Y: π 0.40 v π hence v m s. Now applying Bernoulli s equation to a streamline joining the surface to X gives Patm (0 9) PX PX Pa he pressure at Y is atmospheric. (Comment: the speed at X is not large enough to make much of a difference in pressure between the static case and the case of water flowing.) physics for the IB Diploma Cambridge University Press 0 ANSWERS O ES YOURSELF QUESIONS B 9

10 We use Bernoulli s equation p + ρv + ρ gh p + ρv + ρ gh. Since the heights are approximately the same this becomes p + ρ p + ρv or p p ρv ρv. From then flow rate 6 Av π(0.00) v hence v.4 m s 0.00 and v.4.8 m s. Hence π(0.00) p p.0(.8.4 ) 98 Pa. Hence ρ Hg gh 98 Pa h.99.0 mm p v m s ρ 0. he volume flow rate is Q π v and so the speed is v.0 m s. he Reynolds π 0.40 vρr number is R his is larger than 000 so the flow is turbulent. η We assume a wind speed of 0 m s a distance between buildings of 0 m an air density of kg m and a viscosity of 0 Pa s. he Reynolds number is R v ρ r 0 6 0, larger than 000 so turbulent. η 0 ρr g 4 7 he terminal speed of the droplet is equal to η and so r m. he mass of the droplet is then 4 r 4 ( ) 4 m ρv ρ π 870 π kg. When the droplet was balanced 4 mg in the electric field, mg qe and so q.4 0 C. Hence E n B4 Forced vibrations and resonance (HL) 8 Oscillations are called free if no external force acts on the system other than the restoring force that tends to bring the system back to equilibrium. In forced oscillations there is an additional force acting on the system. 9 In critically damped oscillations the system, after being displaced, returns to its equilibrium position as fast as possible without performing oscillations. In overdamped oscillations the system will again return to its equilibrium position without oscillations but will do in a ling time. 60 Damping denotes the loss of energy of an oscillating system which results in a gradual decrease of the amplitude of the oscillations. 6 When an oscillating system of natural frequency f N is exposed to an external periodic force that varies with frequency f D the system will to oscillate with the frequency f D. he amplitude will be large when f D f N and when this happens we have resonance. 0 ANSWERS O ES YOURSELF QUESIONS B physics for the IB Diploma Cambridge University Press 0

11 6 a he period is 4.0 s and so the frequency is 0. Hz. 0 b After one oscillation the amplitude drops from 0 cm to 7 cm and so Q π c he energy decreases exponentially with time: 0 7. E t d Q will decrease; Q essentially measures how many oscillations the system will perform before coming to rest. With more damping the system will come to rest with fewer oscillations. 6 a he restoring force must be proportional to and opposite to displacement. b See graph shown. y / cm t / s 0 c i and ii. See graph shown A 0 f 0 f D physics for the IB Diploma Cambridge University Press 0 ANSWERS O ES YOURSELF QUESIONS B

12 64 a 0 Hz b See graph shown. π light damping Phase difference / rad π heavier damping 0 f 0 f D c At Hz the frequency is greater than the natural frequency and so, for light damping, the phase difference between the displacement of the driver and the mass is π. his means that the pendulum mass will move to the left. ANSWERS O ES YOURSELF QUESIONS B physics for the IB Diploma Cambridge University Press 0

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