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Physics 130 General Physics - Moustakas Fall 2012 Midterm Exam 3 November 29, 2012 Name: Instructions 1. This examination is closed book and closed notes. All your belongings except a pen or pencil and a calculator should be put away and your bookbag should be placed on the floor. 2. You will find one page of useful formulae on the last page of the exam. 3. Please show all your work in the space provided on each page. If you need more space, feel free to use the back side of each page. 4. Academic dishonesty (i.e., copying or cheating in any way) will result in a zero for the exam, and may cause you to fail the class. In order to receive maximum credit, each solution should have: 1. 2. 3. 4. 5. 6. 7. 8. 9. A labeled picture or diagram, if appropriate. A list of given variables. A list of the unknown quantities (i.e., what you are being asked to find). One or more free-body or force-interaction diagrams, as appropriate, with labeled 1D or 2D coordinate axes. Algebraic expression for the net force along each dimension, as appropriate. Algebraic expression for the conservation of energy or momentum equations, as appropriate. An algebraic solution of the unknown variables in terms of the known variables. A final numerical solution, including units, with a box around it. An answer to additional questions posed in the problem, if any. 1 Physics 130 General Physics - Moustakas Fall 2012 1. A 25 g ball of clay traveling east at 2.5 m{s collides with a 35 g ball of clay traveling 25˝ south of west at 1.5 m{s. What are the speed and the direction of the resulting 60 g ball of clay? Solution: This is an inelastic two-dimensional conservation of momentum problem. Applying conservation of momentum in the x-direction, the final momentum in the x-direction is pf x “ “ “ “ pix m1 pvix q1 ` m2 pvix q2 p0.025 kgqp2.5 m{sq ´ p0.035 kgqp1.5 m{sq cos 25˝ 0.0149 kg m{s (1) (2) (3) (4) Applying conservation of momentum in the y-direction, the final momentum in the y-direction is pf y “ “ “ “ piy m1 pviy q1 ` m2 pviy q2 p0.025 kgqp0 m{sq ´ p0.035 kgqp1.5 m{sq sin 25˝ ´0.0222 kg m{s (5) (6) (7) (8) Next, we find the magnitude of the final momentum by combining the x- and ycomponents using the Pythagorean theorem: a pf “ pf x ` pf y (9) a p0.0149 kg m{sq2 ` p´0.0222 kg m{sq2 (10) “ “ 0.0267 kg m{s (11) 2 Physics 130 General Physics - Moustakas Fall 2012 Using the final momentum, we can solve for the final speed: pf “ pm1 ` m2 qvf pf ñ vf “ m1 ` m2 0.0267 kg m{s “ 0.025 kg ` 0.035 kg “ 0.45 m{s (12) (13) (14) (15) (16) The direction is given by ˆ ˙ pf y θ “ tan p ˆ fx ˙ ´0.0222 kg m{s ´1 “ tan 0.0149 kg m{s ˝ “ ´56 ´1 3 (17) (18) (19) Physics 130 General Physics - Moustakas Fall 2012 2. Fred (mass 70 kg) is running with the football at a speed of 7.0 m{s when he is met head-on by Brutus (mass 130 kg), who is moving at 5.0 m{s. Brutus grabs Fred in a tight grip, and they fall to the ground. Which way do they slide, and how far? The coefficient of kinetic friction between football uniforms and Astroturf is 0.25. Solution: This is an inelastic collision in which Fred and Brutus are moving toward each other. Assume that Brutus is moving in the positive x-direction and Fred is moving in the negative x-direction. Because momentum is conserved, we have pix “ pf x mB pvi qB ´ mF pvi qF “ pmB ` mF qpvf q (1) (2) Solving for the final velocity of Brutus and Fred, mB pvi qB ´ mF pvi qF pmB ` mF q p130 kgqp5.0 m{sq ´ p70 kgqp7.0 m{sq “ p130 kg ` 70 kgq “ 0.80 m{s vf “ (3) (4) (5) Now that we know that Brutus and Fred moved to the right after the collision, we can find the distance that they traveled. Friction slows them down, and therefore the force of friction must act in the negative x-direction. The net force is equal to the total mass (the mass of both players) times the acceleration. ÿ Fx “ ´fk “ pmB ` mF qa (6) ´µn “ pmB ` mF qa ´µpmB ` mF qg “ pmB ` mF qa ñ a “ ´µg “ ´p0.25q ˆ p9.8 m{s2 q “ ´2.45 m{s2 4 (7) (8) (9) (10) (11) Physics 130 General Physics - Moustakas Fall 2012 Finally, we can use the kinematic equation to find the distance traveled with vf “ 0: vf2 “ vi2 ` 2ad ñd “ ´ vi2 2a ´p0.80 m{s2 q2 “ 2 ˆ p´2.45 m{s2 q “ 0.13 m “ 13 cm 5 (12) (13) (14) (15) (16) Physics 130 General Physics - Moustakas Fall 2012 3. A roller coaster car on the frictionless track shown below starts from rest at height h. The track’s valley and hill consist of circular-shaped segments of radius R. (a) What is the maximum height hmax from which the car can start so as not to fly off the track when going over the hill? Give your answer as a multiple of R. Hint First find the maximum speed for going over the hill. (b) Evaluate hmax for a roller coaster that has R “ 12 m. Solution: (a) When the car is in the valley and on the hill it can be considered to be in circular motion. We start at the top of the hill and determine the maximum speed so that the car stays on the track. We can use the equations for circular motion. The acceleration is given by v2 a“´ , (1) r where the minus sign means that the centripetal acceleration points inward, toward the center of the circular track. The free-body diagram for the car at the top of the hill illustrates the two forces acting on the car. Using Newton’s second law, we know ÿ F “ ma “ ´m FN ´ Fg “ ´m v2 R v2 R (2) (3) When the normal force is greater than zero, the car is still touching the track. Therefore, by setting the normal force equal to zero we can find the maximum speed that the car can have and still stay on the track. Fg 6 2 vmax “ m R (4) Physics 130 General Physics - Moustakas v2 mg “ m max a R ñ vmax “ gR Fall 2012 (5) (6) Now that we know the maximum velocity at the top of the hill, we can use conservation of energy to find the starting height. Kf ` Uf ` Ki ` Ui 1 2 1 2 mvmax ` mgyf “ mv ` mgyi 2 2 i 1 2 mv ` mgyf “ 0 ` mgyi 2 max (7) (8) (9) We’ll use yf “ R for the height at the top of the hill and factor the mass term, 1 2 v ` gR “ 0 ` gyi 2 max v2 yi “ max ` R 2g We can now plug in our results for vmax from above ? p gRq2 yi “ `R 2g R `R “ 2 3R “ 2 (10) (11) (12) (13) (14) (b) Plugging into the above equation, hmax for a roller coaster that has R “ 12 m we get yi “ 3R 3 ˆ 12 m “ “ 18 m 2 2 7 (15) Physics 130 General Physics - Moustakas Fall 2012 4. An 7.0 kg crate is pulled 6.0 m up a 25˝ incline by a rope angled 15˝ above the incline. The tension in the rope is 110 N, and the crate’s coefficient of kinetic friction on the incline is 0.3. (a) How much work is done by tension, by gravity, and by the normal force? (b) What is the increase in thermal energy of the crate and incline? Solution: (a) Find work done by tension, gravity and the normal WT “ T ¨ ∆r “ “ “ Wg “ Fg ¨ ∆r “ “ “ WN “ 0 J T ∆x cosp15˝ q p110 Nq ˆ p6.0 mq ˆ pcos 15˝ q 640 J mg∆x cosp90˝ ` 25˝ q p7 kgq ˆ p9.8 m{s2 q ˆ p6.0 mq ˆ pcos 115˝ q ´174 J (1) (2) (3) (4) (5) (6) (7) (b) Find the increase in thermal energy ∆Eth “ fk ¨ ∆r “ Fx ∆x “ µk Fn ∆x To find the normal force, ÿ Fy “ 0 (8) (9) (10) (11) Fn ´ Fg cos 25˝ ` T sin 15˝ “ 0 ñ Fn “ Fg cos 25˝ ´ T sin 15˝ “ p7.0 kgq ˆ p9.8 m{s2 q ˆ pcos 25˝ q ´ p110 Nq ˆ psin 15˝ q “ 33.703 N 8 (12) (13) (14) (15) Physics 130 General Physics - Moustakas Fall 2012 Plugging this result into the equation for the thermal energy, we find ∆Eth “ µk Fn ∆x “ p0.3q ˆ p33.703 Nq ˆ p6.0 mq “ 61 J 9 (16) (17) (18) (19) Physics 130 General Physics - Moustakas Fall 2012 5. A 6.0 kg cat leaps from the floor to the top of a 85 cm high table. If the cat pushes against the floor for 0.25 s to accomplish this feat, what was her average power output during the pushoff period? Solution: The average power output during the push-off period is equal to the work done by the cat divided by the time the cat applied the force. Since the force on the floor by the cat is equal in magnitude to the force on the cat by the floor, work done by the cat can be found using the workkinetic-energy theorem during the push-off period: Wnet “ Wfloor “ ∆K. We do not need to explicitly calculate Wcat , since we know that the cat’s kinetic energy is transformed into its potential energy during the leap. In other words, ∆Ug “ mgpy2 ´ y1 q “ p6.0 kgqp9.8 m{s2 qp0.85 mq “ 49.98 J. (1) (2) (3) Thus, the average power output during the push-off period is Wnet t 49.98 J “ 0.25 s “ 200 W. P “ 10 (4) (5) (6) Physics 130 General Physics - Moustakas Fall 2012 6. You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 26.0 N. If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 15.0 N. What is the statue’s density? Solution: The buoyant force, FB , on the can is given by Archimedes’ principle. The free-body diagram sketched above shows that the gravitational force of the statue, FG,statue is balanced by the spring force, Fsp , and the buoyant force, FB exerted by the water. Since the statue is in static equilibrium, we know that the net force is zero. The rest of the solution follows through a series of substitutions, recalling that the density of water is ρw “ 1000 kg m´3 : FB ` Fsp ´ FG,statue “ 0 ñ FB “ FG,statue ´ Fsp ρw Vstatue g “ FG,statue ´ Fsp FG,statue ´ Fsp Vstatue “ ρw g FG,statue ´ Fsp mstatue “ ρstatue ρw g (1) (2) (3) (4) (5) (6) ρstatue ρw g ñ “ mstatue FG,statue ´ Fsp ρw mstatue g ρstatue “ FG,statue ´ Fsp ρw FG,statue ρstatue “ FG,statue ´ Fsp p1000 kg m´3 q ˆ p26.0 Nq “ p26.0 N ´ 15.0 Nq “ 2360 kg m´3 . 11 (7) (8) (9) (10) (11) Physics 130 General Physics - Moustakas Fall 2012 7. A hurricane wind blows across a 5.0 m ˆ18.0 m flat roof at a speed of 125 km/hr. (a) Is the air pressure above the roof higher or lower than the pressure inside the house? Explain. (b) What is the pressure difference? (c) How much force is exerted on the roof? If the roof cannot withstand this much force, will it “blow in” or “blow out”? Solution: (a) The pressure above the roof is lower due to the higher velocity of the air. (b) Bernoulli’s equation with yinside « youtside is 1 pinside “ poutside ` ρair v 2 2 1 ρair v 2 ñ ∆p “ 2 ˆ ˙2 1 hr 1 125 km 1000 m 3 ˆ p1.28 kg{m q ˆ ˆ ˆ “ 2 hr km 3600 s “ 770 Pa. (1) (2) (3) (4) The pressure difference is 0.77 kPa. (c) The force on the roof is Froof “ p∆pqA “ p770 Paq ˆ p5.0 m ˆ 18.0 mq “ 6.9 ˆ 104 N. (5) (6) (7) The roof will blow up, because the pressure inside the house is greater than the pressure on the top of the roof. 12 Physics 130 General Physics - Moustakas Fall 2012 8. A 120 g ice cube at ´12˝ C is placed in an aluminum cup whose initial temperature is 65˝ C. The system comes to an equilibrium temperature of 15˝ C. What is the mass of the cup? Solution: There are two interacting systems: aluminum and ice. The system comes to thermal equilibrium in four steps: (1) the ice temperature increases from ´12 ˝ C to ´0 ˝ C; (2) the ice becomes water at 0 ˝ C; (3) the water temperature increases from 0 ˝ C to 15 ˝ C; and (4) the cup temperature decreases from 65 ˝ C to 15 ˝ C. Since the aluminum and ice form a closed system, we have Q “ Q1 ` Q2 ` Q3 ` Q4 “ 0. (1) Mice cice ∆T p0.120 kgq ˆ r2090 J{pkg Kqs ˆ p12 Kq 3010 J. Mice Lf p0.120 kgq ˆ p3.33 ˆ 105 J{kgq 40, 000 J. Mice cwater ∆T p0.120 kgq ˆ r4190 J{pkg Kqs ˆ p15 Kq 7540 J. MAl cAl ∆T MAl ˆ r900 J{pkg Kqs ˆ p´50 Kq ´p45, 000 J{kgqMAl . (2) Each term is as follows: Q1 “ “ “ Q2 “ “ “ Q3 “ “ “ Q4 “ “ “ (3) (4) (5) Inserting everything into equation (1), we get 50, 550 J ´ p45, 000 J{kgqMAl “ 0 ñ MAl “ 1.1 kg. 13 (6) Physics 130 General Physics - Moustakas Kinematics and Mechanics Energy 1 xf “ xi ` vxi t ` ax t2 2 vxf “ vxi ` ax t Kf ` Ugf “ Ki ` Ugi ∆K ` ∆U ` ∆Eth “ ∆Emech ` ∆Eth “ ∆Esys “ Wext 2 2 vxf “ vxi ` 2ax pxf ´ xi q 1 yf “ yi ` vyi t ` ay t2 2 vyf “ vyi ` ay t 1 θf “ θi ` ωi ∆t ` α∆t2 2 ωf “ ωi ` α∆t 2 ωf “ ωi ` 2α∆θ s “ rθ c “ 2πr vt “ ωr v2 “ ω2r r 2πr v“ T fk “ µk FN F~AonB “ ´F~BonA Momentum p~i “ p~f p~ “ m~v m1 ´ m2 pvix q1 m1 ` m2 2m1 pvix q1 pvfx q2 “ m1 ` m2 pvfx q1 “ P “ ∆E {∆t Fluids and Thermal Energy F P“ A m ρ“ V Q “ vA Q “ MLf for freezing/melting Q “ Mc∆T 2 v F~net “ ΣF~r “ ma “ m r Fg “ mg 0 ă fs ă“ µs FN ~ ¨ ~v P “F v1 A 1 “ v2 A 2 1 1 p1 ` ρv12 ` ρgy1 “ p2 ` ρv22 ` ρgy2 2 2 Qnet “ Q1 ` Q2 ` ... “ 0 ar “ Forces F~net “ ΣF~ “ ma Kf ` Uf ` ∆Eth “ Ki ` Ui ` Wext 1 K “ mv 2 2 U “ mgy ∆Eth “ fk ∆s ~ ¨ ~d W “F 2 2 vyf “ vyi ` 2ay pyf ´ yi q 2 Fall 2012 Constants g “ 9.8 m{s2 Mearth “ 5.98 ˆ 1024 kg rearth´sun “ 1.50 ˆ 1011 m ρwater “ 1000 kg{m 3 ρair “ 1 .28 kg{m 3 J cice “ 2090 kg K J cwater “ 4190 kg K J cAl “ 900 kg K J Lf “ 333 , 000 ice to water kg 14