General Physics, Final Exam Answer Key

PHYSICS 100A Final Exam NAME: _________________________________ SCORE: ___________ PART I - MULTIPLE CHOICE. EACH QUESTION IS WORTH 6 POINTS FOR A TOTAL OF 24 POINTS. 1. What is the length of a simple pendulum with a period of 2.00 s? a) b) c) d) T = 2π 0.873 m 1.62 m 0.994 m 19.6 m L g 2.00 = 2π L= 9.81 π2 ⇒ L 9.81 ⇒ = 0.99 m 2. The higher the frequency of a wave, a) b) c) d) the lower its speed. the shorter its wavelength. the greater its amplitude. the longer its period. 3. A boat at anchor is rocked by waves whose crests are 40 m apart and whose speed is 10 m/s. These waves reach the boat once every a) 400 s b) 30 s c) 4 s d) 0.25 s v=λf ⇒ 10= ( 40 m ) f f = 0.25 Hz T = 1/ f = 4.00 s 4. A cubical box 25.00 cm on each side is immersed in a fluid. The pressure at the top surface of the box is 109.4 kPa while the pressure on the bottom surface is 112.0 kPa. What is the density of the fluid? a) 1030 kg/m3 ⇒ P = ρ gh b) 1060 kg/m3 3 109.4 × 10 N/m 2 =ρ 9.8 m/s 2 h c) 1120 kg/m3 d) 1090 kg/m3 3 2 2 ( (112 ×10 ( 2.6 ×10 3 ) ( ) N/m ) = ρ ( 9.8 m/s ) ( h + 0.25 ) N/m ) = ρ ( 9.8 m/s ) ( 0.25 m ) 2 ρ = 1061 kg/m3 2 PART II - PROBLEM SOLVING FOR A TOTAL OF 36 POINTS. 1. A spring with spring constant k = 1000 N/m is attached to a mass M. It is observed that this system oscillates with period T = 0.50 seconds with an amplitude A = 1.0 cm. a) What is the mass M in kg? (4 pts) Using the relation for the period of a spring-mass system, m k T = 2π 0.50 s = 2π ⇒ M = (1000 N/m ) ( 0.50 s ) M 1000 N/m 2 4π 2 = 6.33 kg b) What is the total energy of this system? (4 pts) The total energy is E= 1 2 kA 2 (1000 N/m )( 0.01 m ) = 2 2 = 0.05 J c) What is the maximum velocity of the mass? (4 pts) The maximum kinetic energy equals the total energy. Hence, 1 2 mvmax = 0.05 J 2 vmax = 2 ( 0.05 J ) = 0.126 m/s 6.33 kg 2. A pair of speakers, one placed at the origin and the other placed 50 m away on the x-axis, emit sound that are in-phase of frequency 510 Hz. (Take speed of sound as 340 m/s.) a) What is the wavelength of the sound waves emitted? (3 pts) From the relation v=λf f = 340 m/s = ( 510 Hz ) f ⇒ 340 m/s 2 = m 510 Hz 3 b) You are standing midway between the two speakers and decide to take a step of length L to the right. If you hear no sound, what do you call this phenomenon? Explain why this phenomenon occurs. (3 pts) This is the phenomenon of destructive interference. This occurs when the wave from one speaker is such that it is opposite in phase to the wave from the other speaker so that the resultant amplitude is equal to zero. We get no wave. c) What is the smallest step L that you can take for this to happen? (6 pts) For destructive interference, the path difference must equal to an odd multiple of a half wavelength. That is, ∆L = 1 3 5 λ , λ , λ , ... 2 2 2 Since you are at the middle to begin with, your distance to each speaker is 25 m. When you take a step of length L to the right, you are now 25 + L from the speaker at the origin, and 25 – L from the speaker at 50 m. Hence the path difference is: ∆L = ( 25 + L ) − ( 25 − L ) = 2 L The smallest step you can take to get destructive interference is thus: 2L = 1 λ 2 ⇒ 1⎛2 ⎞ 1 L = ⎜ m⎟ = m 4⎝3 ⎠ 6 3. A solid iron cube has sides of length 0.32 m. The density of iron is 7860 kg/m3. a) Find the volume of the cube and hence its mass. (4 pts) The volume of the cube is V = L3 = ( 0.32 m ) = 0.033 m 3 3 Its mass is thus given by: m = ρV = ( 7860 kg/m3 )( 0.033 m3 ) = 258 kg b) This solid iron cube is fully immersed in water as shown. Find the buoyant force acting on the cube. The density of water is 1000 kg/m3 (4 pts) The buoyant force is given by F = ρ gV = (1000 kg/m3 )( 9.81 m/s 2 )( 0.033 m3 ) = 324 N c) From the results in the two parts above, find the tension in the string. (4 pts) Since the system is in equilibrium, the net force acting on the cube must equal zero. That is, T + F − mg = 0 T = mg − F ⇒ = ( 258 kg ) ( 9.81 m/s 2 ) − 324 N = 2210 N PART III – MULTIPLE CHOICE COMPREHENSIVE EXAM. EACH QUESTION IS WORTH 2 POINTS FOR A TOTAL OF 40 POINTS. 1) How long does a car with an acceleration of 2.0 m/s2 take to go from 10 m/s to 30 m/s? a. 10 s v = v0 + at ⇒ ( 30 m/s ) = (10 m/s ) + 2.0 m/s 2 t b. 20 s c. 40 s 20 m/s t= = 10 s d. 60 s 2.0 m/s 2 e. 5 s ( ) 2) A stone is thrown straight up. When it reaches its highest point, a. Both its velocity and its acceleration are zero b. Its velocity is zero and its acceleration is not zero c. Its velocity is not zero and its acceleration is zero d. Neither its velocity nor its acceleration is zero e. It cannot be determined based on the information given 3) An escalator has a velocity of 3.0 m/s at an angle of 600 above the horizontal. The vertical component of its velocity is a. 0.0 m/s b. 1.5 m/s v y = ( 3.0 m/s ) sin 600 c. 1.8 m/s = 2.6 m/s d. 2.6 m/s 600 e. 3.5 m/s G G G G G G 4) Three vectors A , B and C are related via A = B + C . Which of the following diagrams illustrates these vectors? G C G B G C G B G A a. G A b. G C G C G B G B G A G A c d. 5) A force of 40 N is needed to set a 10 kg steel box moving across a wooden floor. The coefficient of static friction is a. 0.08 f s = µ s N ⇒ ( 40 N ) = µ s (10 kg ) 9.81 m/s 2 b. 0.1 c. 0.25 40 N µs = = 0.41 d. 0.4 98.1 N e. 2.5 ( ) 6) An object of mass 2m has an acceleration a when it is acted upon by a net force of magnitude F. What is the acceleration on an object of mass m when it is acted upon by a resultant force of magnitude 2F? a. a/4 Let A = acceleration of mass m. For mass 2m, we have F = (2m)a b. a/2 Thus, for mass m, c. a d. 2a 2F = mA e. 4a A = 2F/m = 2(2ma)/m = 4a 7) An electric drill has a power rating of 600 W. How much work can this drill do in 1.0 minutes? a. 10 J Power = Work/time b. 60 J c. 600 J 600 W = Work/(60 s) d. 1000 J e. 36000 J Work = 600 x 60 J 8) Car A has a mass of 1000 kg and a speed of 60 m/s while car B has a mass of 2000 kg and a speed of 30 m/s. The kinetic energy of car A is a. a quarter that of car B KA = 0.5(1000 kg)(60 m/s)2 = 1,800,000 J KB = 0.5(2000 kg)(30 m/s)2 = 900,000 J b. half that of car B c. equal to that of car B d. twice that of car B e. four times that of car B 9) A 1 kg block slides from rest from the top of an inclined plane 2 m high and 5 m long. The potential energy is zero at the bottom of the plane. If one-half of the block’s initial mechanical energy is dissipated by friction, then the speed of the block at the bottom of the plane is most nearly a. b. c. d. 160 m/s 80 m/s The generalized work-energy theorem gives 40 m/s 20 m/s 5 m/s -0.5Ei = (0 – mghi) + (0.5mvf2 – 0) WNC = ∆U + ∆K implying that -0.5mghi = -mghi + 0.5mvf2 e. (9.81 m/s2)(2 m) = vf2 10) The average momentum of a 70 kg runner who covers 400 m in 50 s is a. 8.75 kg-m/s average speed is (400 m)/(50 s) = 8 m/s average momentum is (70 kg)(8 m/s) = 560 kg-m/s b. 57 kg-m/s c. 350 kg-m/s d. 560 kg-m/s e. 5488 kg-m/s 11) A 3.0-kg object moves to the right at 4.0 m/s. It collides head-on with a 6.0-kg object moving to the left at 2.0 m/s. Which statement is correct? a. The total momentum both before and after the collision is 24 kg · m/s. b. The total momentum before the collision is 24 kg · m/s, and the total momentum after the collision is 0 kg · m/s. c. The total momentum before the collision is 0 kg · m/s, and after the collision is 24 kg · m/s. d. The total momentum both before and after the collision is zero. e. None of the above is true. 12) A wheel is 1.0 m in diameter. When it makes 30 revolutions per minute, the linear speed of a point on its circumference is a. π/2 m/s Use v = r ω giving b. π m/s c. 30π m/s v = (0.5 m) (30 x 2π rad)/(60 s) = π/2 m/s d. 45π m/s e. 60π m/s 13) The blade of a lawn mower has a moment of inertia of 4 kg-m2. If it is rotating at 600 rpm, what it is kinetic energy? a. 4000 J b. 2 x 106 J K = 0.5 I ω2 c. 800 π2 J = 0.5 (4 kg-m2) (600 x 2π rad)2/(60 s)2 d. 40 π J = 800 π2 J e. 80 π J 14) A 1-kg rock is suspended by a massless string from one end of a 1-m measuring stick. What is mass of the measuring stick if it is balanced by a support force at the 0.25 m mark? a. 0.25 kg b. 0.5 kg c. 1 kg d. 2 kg e. 4 kg 15) A hoop is rolling without slipping along a horizontal surface with a speed of 6.50 m/s when it starts up a ramp that makes an angle of 25.00 with the horizontal. What is the speed of the hoop after it has rolled 3.00 m up the ramp? (The moment of inertia of a hoop is MR2) a. 2.06 m/s K = 0.5mv2 + 0.5Iω2 = 0.5mv2 + 0.5mr2ω2 = mv2 Energy conservation gives Uf + Kf = Ui + Ki b. 5.46 m/s Hence, (mg 3sin250) + (mv2) = 0 + m(6.5)2 c. 1.91 m/s v2 = 6.52 – 3gsin250 d. 3.79 m/s v = 5.46 m/s e. 1.27 m/s 16) A block attached to the end of an elastic rubber band hangs in equilibrium at position 2 as shown. The block is set into motion by pulling it to position 3 and releasing it from rest so that it oscillates between positions 1 and 3. Which of the following statemens are true? I. Its speed is zero at positions 1 and 3. II. The magnitude of its acceleration is greatest at position 2. III. Its kinetic energy is greatest at position 2. a. b. c. d. e. I only II only I and III only II and III only I, II and III 17) A factory siren indicating the end of a shift has a frequency of 80 Hz. What frequency is perceived by the occupant of a car traveling towards the factory at 30 m/s. Take the speed of sound in air to be 343 m/s a. 75 Hz Doppler shift for observer moving towards source is b. 87 Hz f ′ = (1 + u/v)f c. 85 Hz = (1 + 30/343)80 = 87 Hz d. 83 Hz e. 77 Hz 18) The figure represents a standing wave pattern in a stretched string of length 2.0 m. What is the wavelength of the traveling waves on this string? a. 1/3 m b. 2/3 m c. 3/4 m d. 4/3 m e. 3/2 m Distance between adjacent nodes or antinodes = 0.5λ 19) One of the harmonics of a string fixed at both ends has a frequency of 52.2 Hz and the next higher harmonic has a frequency of 60.9 Hz. What is the fundamental frequency of the string? a. 30.4 Hz For waves on a string, the frequencies are fn = n f1 where f1 is the fundamental frequency. Thus we have for some unknown n, b. 113.1 Hz c. 26.1 Hz nf1 = 52.2 Hz d. 17.4 Hz (n+1)f1 = 60.9 Hz implying f1 = 60.9 – 52.2 = 8.7 Hz e. 8.7 Hz 20) A solid object of mass m is suspended vertically from a spring balance. The spring balance reads W when the object is in air. When the object is submerged in water while still attached to the balance, the reading on the balance is a. Always exactly W b. Always greater than W c. Greater than W only if the density of the object is greater than the density of water d. Always less than W (there is always a buoyant force) e. Less than W only if the density of the object is greater than the density of water. 1 2 3