Physics 201 Final Exam

Physics 201 Final Exam Information and Instructions Student ID Number: Section Number: TA Name: Please fill in all the information above. Please write and bubble your Name and Student Id number on your scantron. Also, fill in your section number under special codes. Finally question 1 asks you to fill in your test ID code which is necessary to correctly grade your exam. The exam has 25 questions (questions 2-26) and you have 2 hours and 15 to complete the exam. Please use 9.80 m/s^2 for the acceleration of gravity on the Earth. Gravitational Constant: 6.673e-11 Nm^2/kg^2. 1. Please fill in your test ID code a. A c. C b. B d. D Multiple Choice

Identify the choice that best completes the statement or answers the question. 2. A particle moving with a constant acceleration has a velocity of 20,0 cm/s when its position is x = 10.0 cm. Its position 7.0 s later is x = -30.0 cm. What is the acceleration of the particle? a. -7.3 cm/s 2 b. -8.9 cm/s 2 c. -11 cm/s 2 d. -15 cm/s 2 e. -13 cm/s 2 àa: -7.3 cm/s 2 Using: x t = 1 2 at' + v * t + x * 1 2 at' = x t v * t x * a = 2 t 2 x t v 0t x 0 = 2 49 30 20 7 10 = 7.347 = 7.3 3. A block is supported on a compressed spring, which, when released, launches the block straight up at velocity v = v 0y j. The spring and ledge it sits on then retract so they are not in the way when the block falls. You can win a prize by hitting the block with a ball. When should you throw the ball and in what direction to be sure the ball hits the block? (Assume the ball can reach the block before the block reaches the ground and that the ball is thrown from a height equal to the release position of the block.) a. At the instant when the block leaves the spring, directed at the block. b. At the instant when the block leaves the spring, directed at the spring. c. At the instant when the block is at the highest point, directed at the block. d. At the instant when the block is at the highest point, directed at the spring. e. When the block is back at the spring's original position, directed at that position. à c: At the instant when the block is at the highest point, directed at the block On the other answers a. Incorrect since the block will be moving up and the ball down b. Incorrect since the block will be moving up and the ball down and was further thrown aimed slightly lower d. Incorrect. The block and ball will both move down with the same acceleration but the ball starts off aimed further down. e. Incorrect. To get to the original springs position the block must go up and fall down past its original position. At this time it will have a large downward velocity and the thrown ball will have a small downward velocity both of which will increase with gravity. They will never meet because the ball fill be falling faster. 4. You have a machine which can accelerate pucks on frictionless ice. Starting from rest, the puck travels a distance x in time t when force F is applied. If force 3F is applied instead, the distance the puck travels in time t is a. x. b. (3/2)x. c. 3x. d. (9/2)x. e. 9x.

àc: 3x Since F=ma acceleration is increased Since x = ½ at 2 and a is three times larger x will be three times larger 5. A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide? a. 14 m b. 11 m c. 19 m d. 24 m e. 29 m àb: 11 b The box must accelerate with the truck to not slide where the force pushing on the box is the frictional force. F = ma = mu_s*m*g = 0.24*9.8*m = 2.352*m a = 2.352 then x = ½ at 2 = ½*2.353*3 2 = 10.58 = 11 m 6. An airplane moves 140 m/s as it travels around a vertical circular loop which has a 1.0-km radius. What is the magnitude of the resultant force on the 70.0-kg pilot of this plane at the bottom of this loop? a. 2.1 kn b. 1.4 kn c. 0.69 kn d. 1.5 kn e. 1.3 kn à b: 1.4 kn To move in a vertical, circle the plane must have centripetal acceleration based. The total resultant force incldes gravity so we don't have to consider it separately. is pulling downward. Fnet = Fc = mv 2 /r = 70*140^2/1000 = 1372 = 1.4kN 7. When a car goes around a circular curve on a level road without slipping, a. no frictional force is needed because the car simply follows the road. b. the frictional force of the road on the car increases when the car's speed decreases. c. the frictional force of the road on the car increases when the car's speed increases. d. the frictional force of the road on the car increases when the car moves to the outside of the curve. e. there is no net frictional force because the road and the car exert equal and opposite forces on each other. à c: the frictional force of the road on the car increases when the car's speed increases. 8. A block slides on a rough horizontal surface from point A to point B. A force (magnitude P = 2.0 N) acts on the block between A and B, as shown. Points A and B are 1.5 m apart. If the kinetic energies of the block at A and B are 5.0 J and 4.0 J, respectively, how much work is done on the block by the force of friction as the block moves from A to B?

a. -3.3 J b. +1.3 J c. +3.3 J d. -1.3 J e. +4.6 J à a: -3.3 J Using the work energy theorem W = Fnet*d = (Pcos40*d - Ff*d) = Kf Ki - Ff*d = - Pcos40*d Kf - Ki = - 2*cos40*1.5 + 4-5 = -3.298 J 9. Two clowns are launched from the same spring-loaded circus cannon with the spring compressed the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their kinetic energies at the instant of launch is a. K A = 3/2 K B. b. K A = sqrt(3/2) K B. c. K A = K B. d. K B = sqrt(3/2)k A. e. K B = 3/2 K A. à c: K A = K B. The spring potential energy is converted to kinetic energy when released and the same potential energy is stored in both cases. 10. A 12-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 0.80 kn/m). The block is initially at rest at its equilibrium position when a force (magnitude P = 80.0 N) acting parallel to the surface is applied to the block, as shown. What is the speed of the block when it is 13 cm from its equilibrium position? a. 0.78 m/s b. 0.81 m/s c. 0.71 m/s d. 0.58 m/s

e. 0.64 m/s àa: 0.78 m/s Using the work energy theorem. W = Kf-Ki Wp Ws = P*d ½ k*x 2 = ½ mv 2 80*0.13 ½ 800*0.13 2 = ½ 12v 2 v = 0.7789 = 0.78 m/s 11. A particle is acted upon by only two forces, one conservative and one nonconservative and neither being a force of friction, as it moves from point A to point B. The kinetic energies of the particle at points A and B are equal if a. the sum of the works of the two forces is zero. b. the work of the conservative force is equal to the work of the nonconservative force. c. the work of the conservative force is zero. d. the work of the nonconservative force is zero. e. None of the above. à a: the sum of the works of the two forces is zero. On the other answer b. Incorrect: If they were equal there would be work equal to twice the magnitude of either force and kinetic energy would change. c. d. The other force would still do work and change the kinetic energy 12. A 4.0-kg mass has a velocity of 4.0 m/s, east when it explodes into two 2.0-kg masses. After the explosion one of the masses has a velocity of 3.0 m/s at an angle of 60.0 north of east. What is the magnitude of the velocity of the other mass after the explosion? a. 7.9 m/s b. 8.9 m/s c. 7.0 m/s d. 6.1 m/s e. 6.7 m/s à c: 7.0 m/s Applying conservation of momentum East: pie = p1e + p2e 4.0*4.0 = 2.0*3.0 cos(60) + p2e p2e = 13 North 0 = p1n + p2n p2n = -2.0*3.0*sin(60) = 5.196 p2 = 14, v2 = 14/2 = 7 m/s 13. A wheel (radius = 0.20 m) is mounted on a frictionless, horizontal axis. A light cord wrapped around the wheel supports a 0.50-kg object, as shown in the figure. When released from rest the object falls with a downward acceleration of 5.0 m/s 2. What is the moment of inertia of the wheel?

à d: 0.019 kg m 2 For the pully a. 0.023 kg m 2 b. 0.027 kg m 2 c. 0.016 kg m 2 d. 0.019 kg m 2 e. 0.032 kg m 2 τ = TR = Iα = I a R TR = Iα = I a R I = TR' a For the mass with acceleration downward taken positive F = ma = mg-t T = mg-ma = 0.5*9.8 0.5*5.0 = 2.4 Then I = TR' a = 2,4 0.2 0.2 5 = 0.0192 = 0.019 14. A solid sphere, a solid cylinder, a spherical shell, and a hoop all have the same mass and radius. Each are rolling on a horizontal surface with the same center of mass speed, and then they roll up identical inclines. Which one goes the greatest distance up its incline? a. the hoop b. the solid sphere c. the spherical shell d. the cylinder e. They all go the same distance up their inclines. àa the hoop The hoop has the highest moment of inertial, since its mass is located at high radius, and therefore is has the highest kinetic energy due to rotation. While rolling up the hill this kinetic energy will be converted to potential energy allowing that object to go the furthest.

15. A thin rod of mass M and length L is struck at one end by a ball of clay of mass m, moving with speed v as shown in the figure. The ball sticks to the rod. After the collision, the angular momentum of the clay-rod system about A, the midpoint of the rod, is a. (m + M/3)(vL/2) b. (m + M/12)(vL/2) c. (m + M/6)(vL/2) d. mvl/2 e. mvl à d: mvl/2 mvl/2 is the original angular momentum of the ball and angular momentum is conserved in a collision. Exhibit 11-1 Two blocks of masses m 1 and m 2 are connected by a light cord that passes over a uniform density cylindrical pulley of mass M, as shown. Block m 2 slides on a frictionless horizontal surface. The blocks and pulley are initially at rest. When m 1 is released, the blocks accelerate and the pulley rotates. Use this exhibit to answer the following question(s). 16. Refer to Exhibit 11-1. The total angular momentum of the system of the two blocks and the pulley relative to the axis of rotation of the pulley is a. proportional to the radius of the pulley. b. proportional to the speed of the blocks. c. proportional to the length of the string. d. to all of the above. e. only to (a) and (b) above. à e: only to (a) and (b) above.

If you calculate the angular momentum using the method of velocity times the shortest, or perpendicular distance, you can see both objects have angular momentums of mrv where R is the radius of the pulley. Also the pulley has an angular momentum L = Iomega = Iv/R = ½ mr 2 v/r = ½ mrv. Therefore, all the elements have an angular momentum proportional to both R and V. 17. A uniform ladder 15 ft long is leaning against a frictionless wall at an angle of 53 above the horizontal. The weight of the ladder is 30 pounds. A 75-lb boy climbs 6.0-ft up the ladder. What is the magnitude of the friction force exerted on the ladder by the floor? 1 ft = 0.3048 m. 1 lb = 4.4448 N. a. 43 lb b. 34 lb c. 38 lb d. 47 lb e. 24 lb à b: 34 lb The system should be in static equilibrium. Considering the forces in x FNw Ff = 0 Considering the torques using the point on the floor as the center of rotation. (note: pounds is a weight or mass*g) -FNw*sin(53)*15 + ml*g*cos(53)*7.5 +mb*g*cos(53)*6 = 0 Ff = FNw = (30*cos(53)*7.5 +75*cos(53)*6)/(sin(53)*15) = 34 lb 18. The figure below shows a planet traveling in a counterclockwise direction on an elliptical path around a star located at one focus of the ellipse. When the planet is at point A, a. its speed is decreasing. b. its angular momentum is increasing. c. the gravitational force does no work on the planet. d. all of the above are correct. e. none of the above is correct. àe: none of the above is correct. On the other answer a. The object will accelerate as it moves toward the sun b. The gravitation force exerts no torque on the orbiting object since it is perpendicular to the velocity at any point and the angular momentum is conserved c. The gravitation force does do work and that work increases the kinetic energy of the orbiting object as it approaches the sun. 19. Find the average density of a red giant star with a mass of 20 10 30 kg (approximately 10 solar masses) and a radius of 150 10 9 m (equal to the Earth's distance from the sun). a. 1.41 10-4 kg/m 3

b. 0.007 kg/m 3 c. 1.41 kg/m 3 d. 710 kg/m 3 e. 1.41 10-3 kg/m 3 àe: 1.41 10-3 kg/m 3 ρ = ; = ; = '*AB* <?@> = = 0.00141? CD*AE > = > > 20. Some species of whales can dive to depths of one kilometer. What is the total pressure they experience at this depth? (r sea = 1020 kg/m 3 and 1.01 10 5 N/m 2 = 1.00 ATM.) a. 9.00 ATM b. 90.0 ATM c. 100 ATM d. 111 ATM e. 130 ATM à c 100 ATM P = P0 + phogh = 1.01e5 + 1020*9.8*1000 = 10097000 = 99.970 ATM = 100 ATM 21. Melanie says that when a diver enters an underwater cave of height h, the pressure on her is no greater than r W gh. Rosalind says that if the bottom of the cave is a distance H below the water surface, the pressure on the soles of the diver's feet can reach r W gh. Which one, if either, is correct? (The density of water is r W.) a. Melanie, because the roof of the cave absorbs the water pressure from above. b. Melanie, because only the fluid directly above any volume of the fluid can contribute to the pressure on that volume. c. Rosalind, because a fluid exerts equal pressure in all directions at a given depth. d. Rosalind, because the pressure also depends on the density, r c, of the material above the cave roof, so that p = r c g(h - h) + r W gh. e. Melanie, because the pressure equals p = r W gh - r c g(h - h). à c: Rosalind, because a fluid exerts equal pressure in all directions at a given depth, including up toward the feet, and the given depth is H 22. Water is flowing at 4.0 m/s in a pipe with circular cross section. If the diameter of the pipe decreases to 1/2 its former value, what is the velocity of the water downstream? a. 1.0 m/s b. 2.0 m/s c. 8.0 m/s d. 16 m/s e. 4.0 m/s à d: 16 m/s if the diameter shrinks to ½ then the area becomes ¼ and since continuity of flow requires A C v C = A ' v ' the velocity must increase by a factor of 4 23. A fountain sends water to a height of 100 meters. What must be the pressurization (above atmospheric) of the water system? 1 ATM = 1.01 10 5 N/m 2. a. 1.0 ATM b. 4.2 ATM

c. 7.2 ATM d. 9.7 ATM e. 8.2 ATM à d: 9.7 ATM Using Bernouli s law. Conceptually there are three stages. However, we only need to compare the 1 st and 3rd 1: right before the water exits, when it is under pressure 2: after it exits 3: at the height. P C + ρgy C + 1 2 ρv C ' = P ' + +ρgy ' + 1 2 ρv ' ' = P B + +ρgy B + 1 2 ρv B ' = constant P * + P C = P * + 1 2 ρv ' ' = P * + ρgy B P C = 1 2 ρv ' ' = ρgy ' P C = ρgy ' = 1000 9.8 100 = 900000 Pa = 9.7ATM 24. A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x = 5.0 cos (pt), where the amplitude, 5.0, is in meters. The magnitude of the acceleration (in m/s 2 ) of the body at t = 1.0 s is approximately a. 3.5 b. 49 c. 14 d. 43 e. 4.3 à b: 49 a = 5.0 p 2 cos (pt) = 5.0 p 2 cos (p) = 49 25. The mass in the figure slides on a frictionless surface. If m = 2 kg, k 1 = 650 N/m and k 2 = 650 N/m, the frequency of oscillation (in Hz) is approximately a. 6 b. 2 c. 4 d. 8 e. 10 à c: 4 Using the concept form lab that in a system like this you can add the spring constant k.

f = ω 2π = k m 2π = 1300 2 2π = 4 Exhibit 15-1 A graph of position versus time for an object oscillating at the free end of a horizontal spring is shown below. Use this exhibit to answer the following question(s). 26. Refer to Exhibit 15-1. The point at which the object has zero velocity and positive acceleration is a. A b. B c. C d. D e. E d: D

Physics 201 Final Exam Answer Section MULTIPLE RESPONSE 1. ANS: A, B, C, D PTS: 1 MULTIPLE CHOICE 2. ANS: A 3. ANS: C 4. ANS: C 5. ANS: B 6. ANS: B 7. ANS: C 8. ANS: A 9. ANS: C 10. ANS: A 11. ANS: A 12. ANS: C 13. ANS: D 14. ANS: A 15. ANS: D 16. ANS: E 17. ANS: B 18. ANS: E 19. ANS: E 20. ANS: C 21. ANS: C 22. ANS: D 23. ANS: D 24. ANS: B 25. ANS: C 26. ANS: D