Midterm Exam 3 November 29, 2012

Transcription

1 Midterm Exam 3 November 29, 202 Name: Instructions. 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:. A labeled picture or diagram, if appropriate. 2. A list of given variables. 3. A list of the unknown quantities (i.e., what you are being asked to find). 4. One or more free-body or force-interaction diagrams, as appropriate, with labeled D or 2D coordinate axes. 5. Algebraic expression for the net force along each dimension, as appropriate. 6. Algebraic expression for the conservation of energy or momentum equations, as appropriate. 7. An algebraic solution of the unknown variables in terms of the known variables. 8. A final numerical solution, including units, with a box around it. 9. An answer to additional questions posed in the problem, if any.

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 30 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 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 p ix p fx () m B pv i q B m F pv i q F pm B m F qpv f q (2) Solving for the final velocity of Brutus and Fred, v f m Bpv i q B m F pv i q F pm B m F q (3) p30 kgqp5.0 m{sq p70 kgqp7.0 m{sq p30 kg 70 kgq (4) 0.80 m{s (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. Fx f k pm B m F qa (6) µn pm B m F qa (7) µpm B m F qg pm B m F qa (8) ñ a µg (9) p0.25q p9.8 m{s 2 q (0) 2.45 m{s 2 () 2

3 Finally, we can use the kinematic equation to find the distance traveled with v f 0: v 2 f v 2 i 2ad (2) ñ d v2 i 2a (3) p0.80 m{s 2 q 2 2 p 2.45 m{s 2 q (4) 0.3 m (5) 3 cm (6) 3

4 2. A two-stage rocket is traveling at 00 m{s with respect to the earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of 25 m{s relative to the second stage. The first stage is four times as massive as the second stage. What is the speed of the second stage after the separation? Solution: The rocket is traveling at an initial speed before the explosion. The two parts separate after the explosion. To find the speed of the second stage, we ll use the conservation of momentum equation. p f p i () pm m 2 qv i m v f m 2 v 2f (2) p4m 2 m 2 qv i 4m 2 v f m 2 v 2f m 2 p4v f v 2f q (3) 5m 2 v i m 2 p4v f v 2f q (4) 5v i 4v f v 2f (5) The fact that the first stage is pushed backward at 35 m{s relative to the second can be written v f 25 m{s v 2f (6) 5v i 4p 25 m{s v 2f q v 2f (7) 5v i 00 m{s 5v 2f (8) Solving for the velocity of the second stage v 2f v 2f 5v i 00 m{s 5 (9) 4

5 5 00 m{s 00 m{s 5 (0). km{s () 5

6 3. A package of mass m is release from rest at a warehouse loading dock and slides down a 3.0 m high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. (a) Suppose the packages stick together. What is their common speed after the collision? (b) For an elastic collision, what quantities are conserved? (c) Write out the conservation of momentum equation for this problem. (d) Write out the conservation of energy equation for this problem. (e) Which quantities are not known in these equations? (f) Explain how to obtain the formulas for these unknown quantities. (g) If the collision between the packages is perfectly elastic, to what height does the package of mass m rebound? Solution: (a) From conservation of energy, we can find the speed of the package at the bottom of the chute. K i U i K f U f () 0 mgh a 2 mv2 (2) b v 2gh 2p9.8m{s 2 qp3.0mq 7.668m{s (3) If the packages stick together, the collision is inelastic and conservation of momentum is conserved. p i p f (4) m v i m 2 v 2i pm m 2 qv f (5) mv i 0 3mv f (6) 6

7 Cancelling the m s Solving for v f, we find v i 3v f (7) v f v i m{s 3 The packages move together with a speed of 2.56m{s. 2.56m{s (8) (b) In an elastic collision, both energy and momentum are conserved. (c) In this conservation of momentum equation, we ll use the velocity calculated above for the initial velocity before the collision. Also, the package at the bottom is at rest before the collision. The conservation of momentum equation gives (d) The conservation of energy equation is (9) p i p f (0) m v i m 2 v 2i m v f m 2 v 2f () m v i 0 m v f m 2 v 2f (2) mv i mv f p2mqv 2f (3) v i v f 2v 2f (4) v f v i 2v 2f (5) K i U i K f U f (6) 2 m pv i q 2 2 m 2pv 2i q 2 2 m pv f q 2 2 m 2pv 2f q 2 (7) (e) The quantities that are not known in these equations are v f and v 2f. (f) We have two unknowns and two equations. The unknowns are v f and v 2f. To find the final velocity of the second package, we can solve for v f in the momentum equation and plug this into the energy equation to eliminate v f. This will give us the equation for v 2f. Once we have an expression for v 2f, we can plug this into the momentum equation to get an expression for v f. (8) From the derivations above, the equation relating the final speed of the first mass to it s initial speed is v f m 2m m 2m v i p7.668m{sq 2.56m{s (9) 3 (20) 7

8 The package rebounds and goes back up the ramp. To determine how high the package goes, we can use the conservation of energy. K i U i K f U f (2) 2 mv2 0 0 mgh (22) h v2 2g p 2.56m{sq2 2p9.8m{s 2 q 0.33 m (23) 8

9 4. Bob can throw a 700 g rock with a speed of 32 m/s. He moves his hand forward 0.8 m while doing so. (a) How much work does Bob do on the rock? (b) How much force, assumed to be constant, does Bob apply to the rock? (c) What is Bob s maximum power output as he throws the rock? Solution: (a) Calculate work W K () W 2 mpv fq 2 2 mpv iq 2 (2) W 2 mpv fq 2 0 (3) W 2 0.7kgp32m{sq2 (4) W 358 J (5) (b) Calculate force W F r F r (6) F W r 358 J.8m (7) F 448 N (8) (c) Calculate power P F v F v (9) P 448 Np32m{sq (0) P 4, 336 W () 9

10 5. Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling resistance as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 8.0 and the coefficient of rolling friction is Use work and energy to find the length of a ramp that will stop a 6, 000 kg truck that enters the ramp at 30 m/s ( 70 mph). Solution: We ll identify the truck and the loose gravel as the system. We need the gravel inside the system because friction increases the temperature of the truck and the gravel. We will also use the model of kinetic friction and the conservation of energy equation. K f U f E th K i U i W ext () 0 U f E th K i 0 0 (2) The thermal energy created by friction is E th f k p xq (3) pµ k F N qp xq (4) pµ k mg cos θqp xq (5) pµ k mg cos θqp xq (6) Using geometry, the final gravitationl potential energy of the truck is U f mgy f (7) mgp xq sin θ (8) Finally, the initial kinetic energy is simply K i 2 mv2 i (9) 0

11 Plugging everything into the conservation of energy equation, we get mgp xq sin θ µ k mg cos θp xq 2 mv2 i (0) g xpsin θ µ k cos θq 2 v2 i () ñ x v 2 i 2gpsin θ µ k cos θq (2) p30 m{sq 2 2 p9.8 m{s 2 q psin cos 8 q (3) 82.7 m (4) km. (5)

12 6. In a hydroelectric dam, water falls 28 m and then spins a turbine to generate electricity. (a) What is U of.0 kg of water? (b) Suppose the dam is 78% efficient at converting the water s potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 60 MW of electricity? Solution: (a) The change in the potential energy of.0 kg of water falling 28 m is U g mgh () p.0 kgqp9.8 m{s 2 qp28 mq (2) 274 J. (3) (b) The dam is required to produce 60 MW W of power every second, or in other words J of energy per second. If the dam is 78% efficient at converting the water s potential energy into electrical energy, then every kilogram of water that falls produces J 24 J of electrical energy. Therefore, the total mass of water needed every second is J This is a typical value for a small hydroelectric dam..0 kg 24 J kg. (4) 2

13 7. Water flowing out of a 5 mm diameter faucet fills a.0 L bottle in 8 s. At what distance below the faucet has the water stream narrowed to 2 mm in diameter? Solution: We treat the water as an ideal fluid obeying Bernoulli s equation. The pressure at point is p and the pressure at point is p 2. Both p and p 2 are atmospheric pressure. The velocity and the area at point are v and A, and at point 2 they are v 2 and A 2. Let d be the distance of point 2 below point. First, we use the time it takes to fill a.0 L bottle to compute the flow rate, Q, the rate at which water is flowing out of the faucet, remembering that one liter equals 0 3 m 3 : Q p.0 Lq p0 3 m 3 {Lq () 8 s m 3 {s. (2) Next, we use this result to find the velocity, v, with which the water leaves the faucet through the D m diameter aperture (at point ): Q v A (3) ñ v Q A (4) Q πpd {2q 2 (5) m 3 {s π p mq 2 (6) m{s. (7) 3

14 Now, from the continuity equation we have v 2 A 2 v A (8) ñ v 2 v A A 2 (9) v πpd {2q 2 πpd 2 {2q 2 (0) v D D 2 2 () m{s m m 2 (2). m{s. (3) Finally, we turn to Bernoulli s equation to find the height, d: p 2 ρv2 ρgy p 2 2 ρv2 2 ρgy 2 (4) ρgpy y 2 q 2 ρpv2 2 v2 q (5) gd 2 pv2 2 v2 q (6) d pv2 2 v2 q 2g (7) p. m{sq2 p0.707 m{sq m{s 2 (8) m (9) 3.7 cm. (20) 4

15 8. A 20 g ice cube at 2 C is placed in an aluminum cup whose initial temperature is 65 C. The system comes to an equilibrium temperature of 5 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: () the ice temperature increases from 2 C to 0 C; (2) the ice becomes water at 0 C; (3) the water temperature increases from 0 C to 5 C; and (4) the cup temperature decreases from 65 C to 5 C. Since the aluminum and ice form a closed system, we have Q Q Q 2 Q 3 Q 4 0. () Each term is as follows: Q M ice c ice T (2) p0.20 kgq r2090 J{pkg Kqs p2 Kq 300 J. Q 2 M ice L f (3) p0.20 kgq p J{kgq 40, 000 J. Q 3 M ice c water T (4) p0.20 kgq r490 J{pkg Kqs p5 Kq 7540 J. Q 4 M Al c Al T (5) M Al r900 J{pkg Kqs p 50 Kq p45, 000 J{kgqM Al. Inserting everything into equation (), we get 50, 550 J p45, 000 J{kgqM Al 0 ñ M Al. kg. (6) 5

16 Kinematics and Mechanics x f x i v xf v xi v 2 xf v2 xi y f y i v yf v yi v 2 yf v2 yi θ f θ i ω f ω i ω f 2 ω i 2 s rθ c 2πr v t ωr v xi t a x t v yi t 2 a xt 2 2a xpx f x i q a y t 2 a yt 2 2a ypy f y i q ω i t α t a r v2 r ω2 r v 2πr T Forces 2α θ F net Σ F ma 2 α t2 F net Σ F r ma m v2 r F g mg 0 f s µ s F N f k µ k F N F AonB F BonA Momentum p i p f p m v pv fx q m m 2 m m 2 pv ix q pv fx q 2 2m m m 2 pv ix q Energy K f U gf K i U gi K U E th E mech E th E sys W ext K f U f E th K i U i W ext K 2 mv 2 U mgy E th f k s W F d P F v P E{ t Fluids and Thermal Energy P F A ρ m V Q va v A v 2 A 2 p 2 ρv 2 ρgy p 2 Q net Q Q Q ML f Q Mc T Constants g 9.8 m{s 2 for freezing/melting M earth kg r earth sun.50 0 m ρ water 000 kg{m 3 ρ air.28 kg{m 3 J c ice 2090 kg K J c water 490 kg K J c Al 900 kg K L f 333, 000 J kg ice to water 2 ρv 2 2 ρgy 2 6