PHY 131 Final Exam, 12/2001

Remove everything from work area, except this exam, your one page of handwritten notes and your calculator. Wait for instruction to start. Use the paper provided for all work (back side is OK). Ask for empty sheets if you need more. Ask for permission if you want to leave the room, and not take this exam or any notes with you. Do not communicate to anyone while outside. When finished, copy all results to this (front) page.

Scores: Problems 1-5: 10 points for correct answer, partial credit possible. Problems 6-12: 5 points each, no partial credit. The graders will check your solution to see if the work handed in justifies your choices in the multiple choice questions. No points for guessing.

Grading/Complaints: The solutions to the exam will be posted on the course WEB page right after the exam. The grades will be posted on the WEB page by Friday.  If you think an error was made in the grading of your exam, you can complain directly to Professors Fossan and Mihaly. (Time and place will be restricted, and posted later on the course WEB page.) Except for trivial numerical errors (like incorrect addition of the scores by the grader), the returned exam book will be accepted only during this time, or by appointment. The contested exam may be compared to a xerox copy kept by the grader. Do not change your solution in any way. When partial credit is contested the whole exam will be re-graded: your grade may go up, down or remain unchanged.

1. A projectile is fired with an initial speed of 55 m/s at an angle of 35o relative to the horizontal on flat ground. Calculate the displacement from the initial position and the velocity v at the time the projectile reaches its maximum height. (You must specify values of x, y, vx, vy)

Calculate the time to reach the top: tm =v0y/g=(55m/s)(sin35o) /(9.8m/s2)= 3.22s. The coordinates are

x=(55m/s)(cos35o) tm=145m

y=(55m/s)(sin35o) tm - (1/2)g tm2=50.6m

The velocity components are

vy=0

vx=(55m/s)(cos35o)=45.1m/s

2. A block (mass m1) lying on a frictionless inclined plane is connected to a mass m2 by a massless cord passing over a pulley as shown (m1<m2). Determine a formula for the acceleration of the system in terms of m1, m2, q, and g. Write an expression (in terms of the same variables) for the friction force that acts on m1, if the coefficient of friction between the incline and m1 is m.

m2g - T = m2a

T - m1g sinq= m1a

Solve by substitution:

a = g(m2 -m1 sinq)/( m1 + m2)

Friction force: m m1g cosq (directed along the incline, downwards).

3. A piano string is L = 1.10 m long and has a mass of 9.00 g. If the string is under a tension of 680 N, what is the fundamental frequency? What is the wavelength of the sound produced by the fundamental vibration of this string (vs = 343 m/s)?

The mass density is m =0.009kg/1.1m=0.0082kg/m . The velocity of sound on the string is v = (T/m )1/2=288m/s. The fundamental frequency is v/2L=131Hz. The wavelength of the sound waves in the air is l=vs/f=2.62m.

4. The siren of a police car emitting a predominate frequency of 1600 Hz creates a sound level of 80 db at 30 m, what is the sound level for a stationary observer at 300 m? If the police car moves toward the observer at a velocity v = 25.0 m/s, what frequency will the stationary observer hear? (The sound velocity in air is vs = 343 m/s.)

The intensity at 100m distance is I' =I(30m) 2/ (100m) 2 = 0.01I. On the other hand 80dB=10log(I/I0). Therefore the intensity level at 100m is 80dB + (10) log(0.01) = 80dB-20dB = 60dB.

The frequency is Doppler shifted to f' = f/(1-vs/v) = 1600Hz / (1- 25/343) = 1726Hz

5. A neutron collides elastically with a proton (at rest initially) whose mass is about the same as the neutron's. The proton is observed to rebound at an angle of qp'=25o relative to the original direction of motion of the neutron. Determine the speed of the two particles vn' and vp', after the collision. The neutron's initial speed is vn = 5.8x105m/s.

In an elastic collision where one of the particle is initially at rest, and the particles have similar masses, the particles move in a direction perpendicular to each other. This can be easily shown by noticing that the energy conservation turns out to be a Pythagoras theorem for the velocity vectors. Therefore the x and y component of the momentum conservation yields:

mvn = m vp' cos25o + m vn' cos65o

0 = -m vp' sin25o + m vn' sin65o

From the second equation we obtain

vp' = vn' (sin65o/sin25o)= 2.15 vn'

Use this in the first equation

vn = vn' (sin65o/sin25o) cos25o + vn' cos65o = vn'(2.37)

Therefore

vn'= vn/2.37=2.45 x105m/s and vp' = 2.15 vn' = 5.26 x105m/s
 
 

6. The radius and the height of a water tube is indicated on the Figure. The pressure at point A is 10.5 atm and the water velocity is 10 m/s. What is the pressure at point D? Consider water an ideal fluid (no viscosity), of density 103kg/m3. Use 10 m/s2 for g and 105 N/m2 for 1atm.

  1. 1.5 atm
  2. 2.0 atm
  3. 4.0 atm
  4. 8.5 atm
  5. 9.5 atm
  6. 10.0 atm
According to Bernoiulli's law, p + r gy + (1/2)r v2 is constant, and the continuity equation makes Av=constant. We need the velocity: v2=v1A1/A2=40m/s. Notice that the area A is proportional to the square of the radius. The pressure is p2=p1 - rgh + (1/2)r(v12 - v22)= 2atm.
7. In the Figurethe volume at point "c" is 800 liter and the pressure is 2 atm. The volume at "a" is 200 liter and the pressure is 5 atm. What is the work done as the system goes through one complete cycle abcd?
  1. 1.6 x 105J
  2. 1.8 x 105J
  3. 2.8 x 105J
  4. 4.0 x 105J
  5. 8.0 x 105J
  6. some other value
The work is the area of the rectangle: 600liter x 3atm = 0.6m3 x 3 x 105Pa= 1.8 x 105J 8. A nuclear power plant operates at 75% of its maximum (Carnot) efficiency between the temperatures of 600oC and 350 oC. If that plant produces electrical energy at 1.30x109 W, how much heat is released in one second? A. 6.05x109J

B. 4.75x109J

C. 4.64x109J

D. 3.34x109J

E. 1.82x109J

F. something else

The Carnot efficiency is 250K/873K=0.28; the real efficiency is 0.28 x 0.75 = 0.215. By definition, e= W/QH and in one second time we get QH = 6.06 x 109J. But W = QH -QC, and therefore the amount of heat released into the cold reservoir is QC = QH -W = 4.75 x 109J.

9. In the Figure the bar is 2.0m long and its center of mass is at a distance of 0.75m from the hinge. How much is the tension in the wire, if the weight of the bar is 50N?

  1. 26.6N
  2. 35.5N
  3. 50.0N
  4. 70.9N
  5. 100N
  6. something else
The perpendicular distance between the wire and the hinge is 1.41m. The torque around the hinge is 0.75m 50N - 1.41m T = 0. This yields T=26.6N

10. One end of a cylindrical steel rod is kept in boiling water; the other and is in an ice-water mixture. In the steady state, what is the heat current in the rod if its radius is 1cm, and its length is 15cm? (Use the tables.)

  1. 2.3x10-3J/s
  2. 0.23J/s
  3. 2.67J/s
  4. 8.37J/s
  5. 267J/s
  6. 837J/s
The heat current is H=kA(TH-TC)/L = (40 W/mK) (3.14) (10-4m2) (100K )/(0.15m) = 8.37J/s

11. The density of pure oxygen gas (molecular mass: 32.0u) at pressure 1atm=101.3kPa and at temperature 20oC is closest to

  1. 32 kg/m3
  2. 20 kg/m3
  3. 1.5 kg/m3
  4. 1.4 kg/m3
  5. 1.3 kg/m3
  6. 0.75kg/m3
The density is mass/volume. Let us take n=1mol oxygen; the mass is 32g. The volume follows from the gas law: V=nRT/P, where T=293oK, and R=8.315J/molK from the Table. We get V=24 x 10-3m3. Therefore the density is 1.33 kg/m3.

12. A 0.35kg mass at the end of a spring vibrates 3.0 times per second, with an amplitude of 0.15m. Determine the total energy of the system.

  1. 0.0017 J
  2. 0.035 J
  3. 0.078 J
  4. 0.049 J
  5. 1.4 J
  6. something else
The angular frequency is w=2pf=(6.28)(3.0s-1)=18.84s-1. The maximum velocity is v= wA= (18.84s-1)(0.15m)= 2.83m/s. The total energy at that point is equal to the kinetic energy, E=1/2mv2=1.40 J.