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Key for Final Exam Form P - Physics for Engineers and Scientists I | PY 205, Exams of Physics

North Carolina State University (NCSU)Physics
Prof. Thomas Pearl

Material Type: Exam; Professor: Pearl; Class: Physics for Engineers and Scientists I; Subject: Physics; University: North Carolina State University; Term: Spring 2011;

Typology: Exams

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PY205N Name (print)____________________________
Spring 2011
FINAL EXAM, version P ID# ________________________
Time limit: 3 hours
CLOSED BOOK Section#____________________
Instructions: On the blank Scantron sheet, print and bubble in your name in the upper right hand
corner where requested, similarly with your ID number where requested; your test version A or
B on line 70, and the correct answers to each of the following 40 questions on lines 1-40. Feel
free to write on the test itself, but your work will not be graded.
HONOR CODE
I have neither given nor received unauthorized aid on this test. Signed:____________________
Data: Acceleration of gravity g = 9.8 m/s2, gravitational constant G = 6.67 × 10−11 N m2/kg2,
density of water
ρ
W = 1000 kg/m3, 1 atm = 1.01 × 105 N/m2, universal gas constant R = 8.31
J/Kmol, Boltzmann constant kB = 1.38 × 10−23 J/K, moments of inertia I: hoop: mR2; cylinder or
disk: (1/2)mR2; sphere: (2/5)mR2: .
1. A sprinter runs 13 meters from rest in 4.2 seconds in a straight line. If she maintains a
constant acceleration throughout, her speed at the end of the interval is
(a) 3.1 m/s
(b) 6.2 m/s
(c) 4.6 m/s
(d) 12 m/s
(e) 1.5 m/s
2. A spring-fired toy rocket is launched vertically upward with an initial speed of 43 m/s.
Assuming free fall conditions from the time right after being fired until it returns to the
ground, the time of flight (round trip up and back) is
(a) 0.23 s
(b) 18 s
(c) 13 s
(d) 8.8 s
(e) 3.7 s
3. You hike 5 mi in a direction 37° west of north, then 2 mi west, and finally 4 mi south. Your
distance from where you started is now
(a) 5 mi
(b) 7 mi
(c) 3 mi
(d) 13 mi
(e) 9 mi
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Download Key for Final Exam Form P - Physics for Engineers and Scientists I | PY 205 and more Exams Physics in PDF only on Docsity!

PY205N Name (print)____________________________ Spring 2011 FINAL EXAM, version P ID# ________________________ Time limit: 3 hours CLOSED BOOK Section#____________________

Instructions: On the blank Scantron sheet, print and bubble in your name in the upper right hand corner where requested, similarly with your ID number where requested; your test version A or B on line 70, and the correct answers to each of the following 40 questions on lines 1-40. Feel free to write on the test itself, but your work will not be graded.

HONOR CODE I have neither given nor received unauthorized aid on this test. Signed:____________________

Data: Acceleration of gravity g = 9.8 m/s^2 , gravitational constant G = 6.67 × 10 −^11 N m^2 /kg^2 ,

density of water ρW = 1000 kg/m^3 , 1 atm = 1.01 × 105 N/m^2 , universal gas constant R = 8.

J/Kmol, Boltzmann constant k B = 1.38 × 10 −^23 J/K, moments of inertia I : hoop: mR^2 ; cylinder or disk: (1/2) mR^2 ; sphere: (2/5) mR^2 :.

  1. A sprinter runs 13 meters from rest in 4.2 seconds in a straight line. If she maintains a constant acceleration throughout, her speed at the end of the interval is (a) 3.1 m/s (b) 6.2 m/s (c) 4.6 m/s (d) 12 m/s (e) 1.5 m/s
  2. A spring-fired toy rocket is launched vertically upward with an initial speed of 43 m/s. Assuming free fall conditions from the time right after being fired until it returns to the ground, the time of flight (round trip up and back) is (a) 0.23 s (b) 18 s (c) 13 s (d) 8.8 s (e) 3.7 s
  3. You hike 5 mi in a direction 37° west of north, then 2 mi west, and finally 4 mi south. Your distance from where you started is now (a) 5 mi (b) 7 mi (c) 3 mi (d) 13 mi (e) 9 mi
  1. A soccer ball is launched at t = 0 with a velocity of 34 m/s at an angle of 40° above the horizontal. The horizontal and vertical components of the velocity v at t = 0 are (a) 22 m/s horizontal, 26 m/s vertical (b) 26 m/s horizontal, 24 m/s vertical (c) 32 m/s horizontal, 26 m/s vertical (d) 24 m/s horizontal, 26 m/s vertical (e) 26 m/s horizontal, 22 m/s vertical
  2. You throw a rock horizontally at 8 m/s off a very high cliff. Assuming free-fall conditions and neglecting air resistance, the speed of the rock 2 s after you release it is (a) 17 m/s (b) 21 m/s (c) 12 m/s (d) 31 m/s (e) 27 m/s
  3. A plane flying west is experiencing a tail wind of 60 mi/hr (tail wind = wind in same direction as the plane, pushing it along). An observer standing on the ground sees the plane moving west at apparently 570 mi/hr. The speed of the plane relative to the air is (a) 510 mi/hr (b) 700 mi/hr (c) 440 mi/hr (d) 580 mi/hr (e) 650 mi/hr
  4. A girl pushes with a force P on a crate of mass m to lower it down a frictionless inclined plane as shown below. P is directed slightly upward, parallel to the plane. The resulting acceleration a of the crate is also parallel to the plane, but directed slightly downward. Not shown in the diagram are the normal force N and the gravitational force m g. Given that the magnitudes of P , a , and N are P , a , and N , respectively, the equation describing the motion parallel to the inclined plane consistent with the diagram below is (a) mg sin(20°) + P = ma (b) − N + mg sin(20°) = − ma (c) − mg sin(20°) + P = − ma (d) mg cos(20°) + P = ma (e) − N + mg cos(20°) = − ma
  1. With a constant velocity, a girl pulls a child sitting on a sled along a horizontal sidewalk. The forces on the sled are the force exerted by the girl, gravity, the normal force exerted by the sidewalk, and friction. If the girl does 100 J of work in moving the sled from one location to another, the work done by the normal force exerted by the sidewalk, including the sign, is (a) −100 J (b) +100 J (c) cannot be determined without knowing the mass of the wagon and the coefficient of friction (d) cannot be determined without knowing the mass of the wagon and its speed (e) zero
  2. An object moving along the x axis experiences a force F whose magnitude varies with position as shown. The work done by F when the object moves from x = 0 to x = 4 m is (a) 12 J (b) 58 J (c) 36 J (d) 24 J (e) 86 J
  3. A spring has a force constant k = 800 N/m. The work required to stretch the spring by 0. m from its relaxed length is (a) 4.0 J (b) 20 J (c) 3.0 J (d) 0.30 J (e) 1.0 J
  4. A car of mass 1000 kg is traveling 40 m/s when it approaches a hill 84 m high. The driver decides to coast over the hill with the car in neutral. Neglecting friction and any kinetic energy stored in the rotation of the tires, the speed of the car at the crest of the hill is (a) 31 m/s (b) 18 m/s (c) 41 m/s (d) 57 m/s (e) The car cannot reach the crest of the hill
  1. An 8 kg crate is given a shove and released, sliding 4 m across a horizontal floor before coming to a stop. If the initial speed of the crate is 1.5 m/s, as the crate stops moving its mechanical energy (a) increases by 36 J (b) decreases by 9.0 J (c) decreases by 36 J (d) decreases by 12 J (e) increases by 12 J
  2. Four particles are attached to the corners of a rectangle with an area of 1.0 × 1.6 m^2 as shown below. Their masses re either M or 2 M as indicated. Taking the origin to be centered on the lower left mass M , the ( x , y ) coordinates of the center of mass of this 4-particle system are (a) ( x CM, y CM) = (1.0, 0.6) m (b) ( x CM, y CM) = (1.0, 0.8) m (c) ( x CM, y CM) = (0.5, 1.6) m (d) ( x CM, y CM) = (0.5, 0.8) m (e) ( x CM, y CM) = (0.8, 0.6) m
  3. Two blocks are at rest on a horizontal frictionless tabletop. The mass of one block is 0.5 kg and that of the other is 1.0 kg. They are originally pushed together, compressing a spring that is between them but not attached to either. When released from rest, the spring pushes them apart and then loses contact as the masses slide away. If the initial energy stored in the spring is 25 J, the final speed v of the smaller mass is (a) 8.2 m/s (b) 5.8 m/s (c) 7.1 m/s (d) 6.1 m/s (e) 3.3 m/s
  4. A Ping-pong ball traveling east at a speed of 4 m/s collides with a stationary bowling ball. The Ping-pong ball bounces back to the west, while the bowling ball moves very slowly to the east. Which object experienced the greater impulse during the collision? (a) Neither, the magnitude of the impulse is the same for both. (b) The Ping-pong ball. (c) The bowling ball. (d) It is impossible to decide because neither mass value is given. (e) It is impossible to decide because neither of the final velocities is given.
  1. A compound pulley is made of two pulleys joined together as shown below. The radius of the larger pulley is 3 times that of the smaller pulley, i.e., b = 3 a. Three ropes exert tangential forces of 4 N, 5 N, and F also as shown below. The value of F that produces a zero net torque about the common axis is (a) 3.0 N (b) 7.0 N (c) 17 N (d) 9.0 N (e) 3.5 N
  2. A disc of mass 4 kg and radius 0.15 m spins about an axis passing through its center at right angles to the plane of the disk. Given that its moment of inertia I = (1/2) MR^2 , its angular momentum about this axis when it is rotating at 20 rad/s is (a) 0.90 kg m^2 /s (b) 5.7 kg m^2 /s (c) 0.45 kg m^2 /s (d) 3.8 kg m^2 /s (e) 7.0 kg m^2 /s
  3. In a classroom demonstration, a student sits on a stool and holds a 2 kg mass in each hand as he rotates with the stool and weights about a vertical axis. With his arms outstretched as shown below, he is rotating at 0.25 rev/s. He now brings the weights close to the rotation axis and is observed to rotate at 1.0 rev/s. The ratio of the final moment of inertia, I fin , to the initial moment of inertia, I init , is (a) I fin/ I init = 4 (b) I fin/ I init = 2/ (c) I fin/ I init = 1/ (d) I fin/ I init = 1/ (e) I fin/ I init = 3
  4. A hoop rolls without slipping across a horizontal floor at a speed v = 2.50 m/s when it reaches a 20.0° incline. The hoop rolls up the incline again without slipping. It reaches a vertical height above the floor of (note that the question asks how high, not how far up the ramp!) (a) 0.64 m (b) 0.45 m (c) 0.32 m (d) 0.24 m (e) None of the above
  1. A 30 m long tube is placed upright in a large pond with one end a few cm underwater. The tube is then evaculated, that is, all the air is removed from it. Given an air pressure of 1.01 × 105 N/m^2 and a water density of 1000 kg/m^3 , the water rises in the tube to a height of about (a) 20 m (b) 15 m (c) 30 m (d) 5 m (e) 10 m
  2. A 30,000 N vehicle is to be lifted by a hydraulic piston in a cylinder with a diameter of 25 cm. The cylinder is connected to a smaller pump cylinder with a diameter of 5 cm. The force that needs to be applied to the piston in the smaller cylinder to raise the vehicle is (a) 520 N (b) 1,200 N (c) 13,000 N (d) 2,600 N (e) 8,300 N
  3. A mass attached to a spring moves in simple harmonic motion along the x axis between the values −0.5 m ≤ x ≤ 0.5 m. If it takes 2.0 s to move from x = −0.5 m to 0.5 m and back to −0. m again, the frequency of its motion is (a) 4.0 Hz (b) 0.50 Hz (c) 2.0 Hz (d) 0.25 Hz (e) 1.0 Hz
  4. If the amplitude of the motion of a mass in a mass-and-spring system undergoing simple harmonic motion is doubled, the mechanical energy of the system is (a) unchanged (b) increased by a factor equal to the square root of 2 (c) increased by a factor of 2 (d) increased by a factor of 3 (e) increased by a factor of 4
  5. A wave pulse on a 4 m long string with both ends fixed is shown below at time t = 0. If the wave speed is 24 m/s, the minimum time required for the string to return to exactly the same appearance that it had at t = 0 is (a) 0.50 s (b) 0.17 s (c) 0.25 s (d) 0.33 s (e) 0.82 s
  1. A container is filled with a mixture of helium atoms of atomic mass 4 and oxygen molecules of atomic mass 32. A thermometer in the container shows that the temperature is 22 °C. The gas particles that have the greater average speed are (a) neither, because the temperature is the same for both (b) helium, because helium is monotonic (c) helium, because the helium atoms are less massive than the oxygen molecules (d) oxygen, because the molecules are diatomic (e) oxygen, because the molecules are more massive than the helium atoms
  2. Using an ideal gas as its thermal substance, a hypothetical heat engine operates with the cycle ABCDA shown below. The amount of work that the engine does in one cycle is (a) P 2 V 2 − P 1 V 1 (b) ( P 2 − P 1 ) × ( V 2 − V 1 ) (c) (1/2)( P 1 + P 2 ) × (1/2)( V 1 + V 2 ) (d) P 2 V 2 + P 1 V 1 (e) P 2 V 1 − P 1 V 2
  3. An ideal gas undergoes an adiabatic process while doing 25 J of work on an external system. Its change in internal energy is (a) zero (b) 50 J (c) −50 J (d) 25 J (e) −25 J
  4. A heat engine operates between 600 °C and 100 °C with an efficiency 80 % of the maximum possible. The actual efficiency of the engine is (a) 67% (b) 57% (c) 46% (d) 36% (e) 22%

PY205N-FE practice version Spring 2011

  1. Start by finding the acceleration, ax, in order to find the final velocity, vf-x, after a certain amount of distance and time where the initial velocity, vi-x=0 m/s: ∆x=½axt 2 13 m=½ax(4.2 s)^2 ax=1.47 m/s 2 Solve for v (^) f-x: vf-x=vi-x +axt vf-x=1.47 m/s 2 *4.2 s=6.2 m/s Answer: b
  2. ∆v (^) y/∆t=-g (v (^) y-f-v (^) y-i )/ ∆t=-g (negative sign indicates the acceleration due to gravity is in the negative y direction). The rocket has an initial velocity of 43 m/s (positive y direction) and a final velocity of -43 m/s (negative y direction) for the round trip. (-43 m/s-43 m/s)/ ∆t=-9.8 m/s 2 Solve for ∆t Answer: d
  3. Find the magnitude of the net displacement vector, r net , which is the vector sum of the three displacement steps (where north=+ j , south=- j , east=+ i , west=- i ): ((sin37 o5- i +cos37o5 j )+-2 i +-4 j ) miles=-5 i miles Magnitude=5 miles Answer: a
  4. vx=34 m/scos40o, v (^) y=34 m/ssin40o Answer: e
  5. vx @t=2 s=vx-i =8 m/s v (^) y @t=2 s=-g∆t=-19.6 m/s (since v (^) y-i =0 m/s) Therefore: v @t=2 s=(8 i -19.6 j )m/s magnitude of velocity @t=2 s=21 m/s Answer: b

6.The velocity of the air with respect to the ground plus the velocity of the plane with respect to the air will give the velocity of the plane with respect to the ground: v air-ground + v plane-air= v plane-ground 60 mi/hr+ v plane-air=570 mi/hr Answer: a

  1. Use conservation of energy to determine what will happen to the car. ∆K+∆U= (Kf-Ki )+(Uf -Ui )= ½m(v (^) f)^2 -½m(vi )^2 +(mgyf -mgyi )= ½m(vf)^2 -½m(vi )^2 +(mgyf )=0 for yi =0 m Mass cancels out from the equation: ½(vf)^2 -½(v (^) i )^2 +gyf = Solve for vf: vf=(vi^2 -2gyf )½ =(1600 m 2 /s 2 -2*9.8 m/s^2 *84 m)½ =(-1646 m 2 /s 2 )½ Square root of a negative number is not physically significant for a car going up a hill, therefore, the car cannot reach the top of the hill; it does not have enough initial kinetic energy. Answer: e
  2. ∆Ecrate=∆Kcrate=Kcrate-f-Kcrate-i =-Kcrate-i (since the crate comes to a stop, Kcrate-f=0) =-½m(v (^) i )^2 =-½*8 kg(1.5 m/s)^2 Answer: b
  3. Position vector for center of mass, r CM =(m 1 r 1 +m 2 r 2 +m 3 r 3 +m 4 r 4 )/(m 1 +m 2 +m 3 +m 4 ) Written in terms of x and y coordinates for each of the four masses starting with the mass at the origin and going CCW around the rectangle: m[(0,0)+2(1,0)+(1,1.6)+2(0,1.6)] meters/6m=(3,4.8) meters/6=(0.5,0.8) meters Answer: d
  4. Use a combination of conservation of energy and conservation of linear momentum: Initial state: two blocks held against the compressed spring which has a stored energy of 25 J Final state: both blocks leave the spring System: both blocks and the spring Conservation of energy: ∆Ksystem+∆Usystem= Kf-block-1 +Kf-block-2 =-∆Uspring=25 J The spring remains stationary through the process and the initial kinetic energies of the two blocks are 0. The stored potential energy in the spring goes to 0 J when the spring is released and is transferred into the kinetic energy of the blocks. ½m(v1 )^2 +½m(v 2 )^2 =25 J =½0.5 kg(v 1 )^2 +½1.0 kg(v 2 )^2 =25 J

From conservation of linear momentum: ∆P (^) system= P (^) i =P (^) f P (^) i =0=m 1 v 1 -m 2 v 2 (blocks moved apart from each other, opposite directions) m 1 v 1 =m 2 v 2 v1 /2=v2 since m 1 =m 2 /

Substitute into conservation of energy equation for v 2 : ½0.5 kg(v 1 )^2 +½1.0 kg(v 1 /2)^2 =25 J 0.25v1^2 +0.125v 12 =25 J 0.375v1^2 =25 J v1 =8.2 m/s Answer: a

  1. The bowling ball and the ping pong ball experience the same impulse=change in momentum. This can be proven via conservation of linear momentum as well as a direct application of Newton’s 3rd^ law of motion. Answer: a
  2. Linear momentum is always conserved for any isolated system. For this case, kinetic energy is conserved and this is confirmed by calculating the kinetic energy of the system (sum of the kinetic energy of the two masses) before and after the collision: Ki =Kf. Answer: d
  3. Angular acceleration, α=∆ω/∆t=4 rev/s/30s=0.13 rev/s 2 *2π rad/rev Answer: a
  4. v=ωr=(30 rev/min1 min/60 s2π rad/rev)*1.5 m Answer: d
  5. Krot =½Iω^2 =½(2mr^2 )ω^2 +½m(2r)^2 ω^2 I=total moment of inertia of the two masses rotating about the axis shown. Krot =mr^2 ω^2 +2mr^2 ω^2 =3mr^2 ω^2 Answer: b
  6. Net torque is the sum of the torques acting on the compound pulley of two different radii where τ= F x r =Frsinθ=Fr for θ=90 o^ for all three and the sign of the torque is determined by the right-hand-rule. The problem asks for the magnitude of the unknown force in order for the net torque to be equal to zero:

τnet =5 Na+4 Nb-Fa= τnet =5 Na+4 N3a-Fa= Solve for the unknown force, F. Answer: c

  1. Time to return to same position in the same orientation is the distance to be traveled to complete one cycle divided by the wave speed: t=8 m/24 m/s Answer: d
  2. v=λf and the wave speed on a rope is determined by the tension and the linear mass density of the rope. The wave speed is unchanged if there is no change in tension or linear mass density. v=λ 1 f 1 = λ 2 f 2 Answer: c
  3. The maximum speed is the amplitude of the velocity of the traveling wave. If the position of the traveling wave is: y=0.03sin(100πx-50πt) Then: dy/dt=0.03 m50π rad/scos(100 πx-50 πt) Therefore: vmax=0.03 m*50 π rad/s Answer: e
  4. For a pipe with one end open and one end closed, for the lowest supported mode (fundamental), the length of the pipe, l, corresponds to λ/4. The frequency of this mode, f, will be equal to: v/λ=v/4l=340 m/s/4*5 m Answer: b
  5. The average or RMS speed of the gas molecules depends on the temperature therefore there is no change in the speed if the volume changes. Answer: e
  6. vRMS=(3kT/m)½ Helium atoms will have a higher RMS speed than oxygen molecules at the same temperature due to having a lower mass. Answer: c
  7. W (^) net =W (^) AB +W (^) BC +WCD+W (^) DA =P 2 (V 2 -V 1 )+0+P 1 (V 1 -V 2 )+0 (W (^) BC =W (^) DA=0 since ∆V is 0 for those steps) =P 2 (V 2 -V 1 )+-P 1 (V 2 -V 1 ) =(P 2 -P 1 ) x (V 2 -V 1 ) Answer: b
  8. ∆Eint =Q-W For an adiabatic process, Q=0, therefore ∆Eint =-W If the work done by the gas (system) is 25 J, then ∆Eint =-25 J Answer: e
  9. Ideal or maximum efficiency: e=(1-T (^) L/TH)=(1-373 K/873 K)=0. Therefore, actual efficiency=0.80*0.57=0. Answer: c