A1. An object of mass m is projected vertically from the surface of a planet of radius R p and mass M p with an initial speed v i.


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1 OBAFMI AWOLOWO UNIVRSITY, ILIF, IF, NIGRIA. FACULTY OF SCINC DPARTMNT OF PHYSICS B.Sc. (Physics) Degree xamination PHY GNRAL PHYSICS I TUTORIAL QUSTIONS IN GRAVITATION, FLUIDS AND OSCILLATIONS SCTION A: GRAVITATION A. An object of mass m is projected vertically from the surface of a planet of radius R p and mass M p with an initial speed v i. (a) Derive an expression for the maximum height h max the mass can attain in terms of the parameters given. (b) Calculate the escape velocity v esc of the mass if the planet is arth with a radius of 670 km and mass of 5.98 x kg. A. Calculate the mass of the Sun using the fact that the period of the arth s orbit around the Sun is.6x 7 s and its distance from the Sun is.496x m. A. The space shuttle releases a 470kg communication satellite while in an orbit that is 80km above the surface of the arth. A rocket engine on the satellite boosts it into a geosynchronous orbit, which is an orbit in which the satellite stays directly over a single location on the arth (a) What is the radius of the geosynchronous orbit? (b) How much energy did the engine have to provide? A4. Given that the Moon is.84 x 5 km from the arth and.50 x 8 km from the Sun, Find the net force on the Moon (m M 7.5 x kg) due to the gravitational attraction of both the arth (m 5.98 x 4 kg) and the Sun (m S.99 x 0 kg), assuming they are at right angles to each other. SCTION B: FLUIDS B. A wooden cube of side length 0. m and specific gravity 0.9 is lowered into a bowl of water. (a) Determine the mass of the cube (b) What is the buoyancy force on the cube? (c) What length of the cube is above water level? B. How much will a 700 N person weigh under water, if they have a volume of 65 L? P a g e
2 B. The diagram shows the confluence of the Niger (A) and the Benue (B) Rivers at Lokoja where they meet and flow southward in C. the channel depths and widths are respectively 40 m and.5 km at A, 0 m and.0 km at B and 50 m and.8 km at C. If the water speeds are m/s and 7 m/s at B and C respectively, determine the speed of the flow at A. B4. The barrel of a syringe has a diameter of cm. The diameter of the needle is 0.0 cm. If you apply a pressure on the plunger so that the medicine moves at cm/sec through the barrel, how fast does it move through the needle? B5. What is the pressure differential across a flat roof having an area of 40 m when the wind blows at 5 m/s? The density of air ρ is.9kg/m. B6. Cyclopropane has a density ρ of g/cm. Calculate the pressure difference (P P ) generated by a flow rate of 50 cm /second through a Venturi tube having radii of 0.50 cm and 0.00 cm? SCTION C: OSCILLATIONS C. A.0 kg block is suspended from a spring causing it to stretch by.5 cm. The block is then pulled down through another 0.50 cm. Find the block's (a) amplitude (b) period, (c) maximum speed and (d) maximum acceleration? C. A thin disk of mass M and radius R is used as a simple physical pendulum by making it oscillate about an axis perpendicular to its surface and located at its rim. (a) Derive an expression for its period of oscillation. (b) If R0.65 m, show that the system will make a good clock (c) Calculate the length of an equivalent simple pendulum. C. A small trolley of mass kg was placed on a horizontal surface between two vertical posts and attached to each post by a coil each of force constant.00 x N/m. The trolley was given a short, sharp push so that it took off from its central position with a speed of m/s. stimate the subsequent amplitude of the vibration. ND P a g e
3 OBAFMI AWOLOWO UNIVRSITY, ILIF, IF, NIGRIA. FACULTY OF SCINC DPARTMNT OF PHYSICS B.Sc. (Physics) Degree xamination PHY GNRAL PHYSICS I SOLUTIONS TO TUTORIAL QUSTIONS GRAVITATION, FLUIDS AND OSCILLATIONS SCTION A: GRAVITATION A. (a) Because the total energy is conserved the initial energy of the mass which is kinetic plus potential when r R is converted to only potential energy when the mass reaches the maximum altitude at a radius r max. Thus K i + U i K f + U f. Noting also that K f 0 at maximum altitude, Then GM m GM mvi m R r max K + U mv i GM R m GM m r max Solving the equation for the initial velocity v i results in v i GM R r max Therefore the maximum height h max becomes H max r max R vi R GM v i R (b) Let the escape velocity be v esc.. For the mass to escape, this initial velocity should be sufficient to cancel the initial potential energy at r R. That is ½mv esc GM m/r Thus v esc GM R 4. m / s.km / s P a g e
4 4 A. Using Kepler s rd Law T 4π GM s r K s r Mass of the Sun becomes: M s 4π r GT 4π kg 7 (.496 ) A. (a) Period of rocket in geosynchronous orbit is same as that of the earth: That is T day 4 hours 4 x 60 x 60 sec 8.64 x 4 s From Kepler s rd law T K r GS where K 4π GM s / m Therefore the geosynchronous radius is r GS T K 4 4 ( 8.64 ) ( 8.64 ) 7 4. m Because the initial position before the boost is 80 km.8 x 5 m and the radius of the arth is 6,70 km 6.7 x 6 m r R +.80 m Therefore i The total energy needed to boost the satellite at the geosynchronous radius is the difference of the total energy before and after the boost m GM m 6.67 s r GS r i J 4 P a g e
5 5 A4. The diagram below shows the forces on the moon due to the Sun F MS and the arth F M respectively. These are vectors whose magnitudes are: and The Resultant Force is: (.50 ) (.84 ) (4.6 ) + (9.88 ) ½ 4.77 N The angle θ which describes the direction of the Resultant Force is given by tan Therefore θ 4.6 o SCTION B: FLUIDS B. (a) Taking the density of water as ρ w 00 kg.m  then the density of the wooden cube is ρ c 0.9 x kg.m  5 P a g e
6 6 But Volume of wood V c (0.).7 x  m Therefore Mass of cube M c density x volume ρ c V c x.7 x  i.e. M c.56 kg (b) Since the density of wooden cube is less than that of water, the cube will float such that the buoyancy force balances the weight of the cube. Thus, buoyancy force F b M c g.56 kg x 9.8 m.s N (c) Let m w mass of water displaced; M c mass of wooden cube. Archimedes says: m w g M c g or ρ w V w ρ c V c Thus, the fraction of the wooden cube below water is V w / V c ρ c /ρ w. So, the portion of the cube above water is  (ρ c /ρ w )  (900 / 00) 0. Let h be the height of cube above water Therefore h x 0. x V c 0. x (0.) h 0.0 m B. The buoyant force equals the weight of fluid displaced by an object. Buoyant Force quation F b ρgv W apparent W true  F b First, convert Volume given in L to m V 65 L (m /00L) 65 m / m 6.5 x  m F b ρgv (00 kg/ m )(9.8 m/ s )(6.5 x  m ) 67 N W apparent W true  F b 700N 67N 6N B. This is an example of the application of the Continuity quation. Thus at Lokoja: A A V A + A B V B A C V C 6 P a g e
7 7 Where A i are crosssectional areas and V i are velocities Thus V A [A C V C  A B V B ]/ A A [800 x 50 x 7 00 x 0 x ]/500 x m/s B4. quation of continuity states: A v A v Assign to the Barrel. Assign to the needle (V?) V A V (π (0.5 cm) )(cm/s) cm /s A (π (0.0 cm) ).459 x 4 cm.5 x cm/s B5. From Bernoulli quation: P P ½ r(v ) ½ r(v ) Let s assume that the speed of air inside the house is 0m/s. Let s assign to the outside of the house and a to inside of the house; therefore v 5m/s and v 0m/s P P ½ ρ(v ) ½ ρ(v ) ½ (.9kg/m ) (5m/s) ½ (.9kg/m ) (0m/s) 40 Pa If we multiply this pressure times the area of the roof, we find the force of the roof from the air inside the house amounts to: (40 Pa)(40m ) 96,700 N B6. Let s call the big tube and the small tube First calculate A and A using: A tube πr and convert to SI (MKS) units A πr π(0.50cm) cm x (m/0cm) x 5 m A πr π(0.00cm).87 x  cm x (m/0cm).87 x 7 m 7 P a g e
8 8 We know the flow rate (50 cm /s) but we need the speed (V ). Flow rate A V so V flow rate/ A (Venturi tube flowmeter equation) A x 5 m P P ½ ρ(v) {(A /A ) } A.87 x 7 m V flow rate/a V 50cm /s 6.66 cm/s x m m/s cm 0cm Convert cyclopropane density to SI units ρ g x kg x (0cm/m).86 kg/m cm 00g P P ½ r(v) {(A /A ) } ½(.86 kg/m )(0.666 m/s) {(7.854 x 5 m /.87 x 7 m ) } ½(.86 kg/m )(0.666 m/s) { x 4 }.8 x 4 Pa SCTION C: OSCILLATIONS C. Mass m.0 kg xtension due to weight of mass m is x.5 x  m (a) Displacement from equilibrium position X max 5.0 x  m (b) From Hooke s law: F mg kx Therefore k mg/x.0 kg x 9.8 m/s /.5 x  m. x N/m Angular velocity ω [k/m] / [. x N/m /.0 kg] / 6 rad/s Therefore T π/ω π /6 Thus, T 0.4 s (c) Maximum speed is v max wx max 6 radian/s x 5.0 x  m 8 P a g e
9 9 0. m/s (d) Maximum Acceleration is a max w X max (6 radian/s) x 5.0 x  m.4 m/s C. (a) From table of Moments of Inertia I cm MR But we need I at the rim, so apply parallel axis theorem, hr I rim I cm + MD MR + MR Since physical pendulum frequency is MR f π MgL I T π I MgL Distance from rotation axis to cm: LR T MR π π MgR R g (b) Given that R0.65 m, then the Period T of Oscillation of the Physical Pendulum is T (0.65) π (9.8) s The calculated period of oscillation is close to sec which is the period of a good clock. This proves that the system would make a good clock. (c) For an equivalent simple pendulum, we need the simple and disk pendulums to have the same period 9 P a g e
10 T disk π R g T sp π L g R g L g R g L g L R (0.65 m) 0.48 m C. We will apply the nergy Theorem, as follows initial energy of system kinetic energy of trolley energy of system at maximum amplitude elastic potential energy of trolley Applying the principle of conservation of energy, Thus, /ka /mv Thus, A [mv /k] / [0.665 kg x (0.555 m/s) /.00 x N/m] /.0 x  m ND P a g e
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