PHY 131 Final Exam, 12/2001 (Practice)

The final will be on Wednesday, December 19, 8:00-10:30 PM, in Javits. The exam will be at two different places. If your first name (or "given name") starts with letters A to C, go to Javits 101. For all others (first names with D-Z) the exam is in Javits 100. There will be assigned seating at the exam, so you MUST go to the right place. Here are two examples: Anthony Moy should go to Javits 101, and John Black should be in Javits 100.

Exams are closed books and notes, but you will be allowed to bring your own notes on a single sheet of regular size white paper. The text on the paper should be written with your own hand (no photographic, xerox etc. copies) and it should be readable without the use of any instruments (magnifying glass, microscope, etc). You may use both sides of the paper. Numerical constants, moments of inertia, etc. needed for the exam will be printed on the exam booklet.

Bring a photo ID to the exams. Acceptable IDs: Stony Brook student ID, driver's license, green card/passport. If you do not have an ID, and you want to take the exam, we reserve the right of making a photo of you on the spot. Also bring a pen/pencil and a handheld calculator capable of doing simple arithmetic and trigonometric functions to the exam. We can not provide a replacement calculator (or allow exchange of calculators) if yours does not work or you forgot to bring one. Notebooks computers are not allowed.

No communication between students is allowed during the exams. No devices with infrared ports or any other communication options. No beepers, cell phones, buzzers, etc. Any evidence of cheating will be reported to the academic judiciary and will also result in a stiff grade penalty. Leaving the exam room is by permission only. Use the exam book for all calculations, ask for more paper if you need it, and hand in all work. When finished, copy all results to the front page.

Grading/complaints: Problems 1-5: 10 points for correct answer, partial credit possible. Problems 6-12: 5 points each, no partial credit. Show your work for all problems. The graders may check your solution to see if the work handed in justifies your choices in the multiple choice questions. No points will be given for guessing. 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, and the graded exams will be placed in front of the PHY 131 laboratory. If you think an error was made in the grading of your exam, you can complain directly to Professors Fossan and Mihaly. Except for trivial numerical errors (like incorrect addition of the scores by the grader), the exam book will be accepted for re-grading only at a time and place that is going to be posted on the course WEB page. 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.

Moments of inertia:

Solid sphere, mass M, radius R: 2/5 MR2
Solid cylinder mass M, radius R: 1/2 MR2
Thin walled hollow cylinder mass M, radius R: MR2
Thin walled sphere mass M, radius R: 2/3 MR2
Rod of length l, rotating around an axis in center 1/12 Ml2
Thin solid rectangle, length l, width w, rotating around an axis along an edge parallel to its length: 1/3 Mw2

1. Mr. Smith, weighting 800N, is standing on the top of an "A" shaped ladder that has 5.00m long legs. The distance between the legs on the floor is 6.00m. The (horizontal) chain connecting the two legs is 1.50m above the floor. What is the tension in the chain? (Neglect friction, neglect the weight of the ladder.)

Make a free body diagram for one of the legs of the ladder. There is no friction, therefore the force at the floor is vertical. We know that the floor is supporting this leg with half of the weight of Mr. Smith (the other half is supported at the other leg; this is a symmetric ladder). We have no idea about the forces at the top of the leg: there is certainly a vertical component (due to Mr. Smith) and a horizontal one (due to the other leg). We can avoid all trouble with these forces, if we pick the axis of rotation at this point. The torque equation becomes: 2.5m T - 3.0m 400N = 0. The tension is 480N.

2. Two crates, of mass 80kg and 210kg, are in contact, and slide on a horizontal surface, due to a 750N force, pushing the 80kg crate. The coefficient of kinetic friction between the 80kg crate and the supporting surface is 0.12; the other crate slides without friction. Calculate the magnitude of the force that the 80kg crate exerts on the 210kg crate.

Draw two free body diagrams, and take the usual xy frame of reference. The acceleration in the y direction is zero. In the x direction both crates has the same acceleration, a. Looking at the y components tells us that the normal forces are equal to the weight for each crate. Accordingly, the friction forces are: F1=m m1g = 94N (172N) and F2=0 (no friction there). Newton's law for the motion in the x direction:

750N- F1-Fp = m1a and Fp=m2a.

Add the two equations (to eliminate Fp), and solve for the acceleration:
750N- F1 = (m1 + m2 )a
a = (750N-94N)/290kg =2.26m/s2
a = (750N-172N)/290kg =1.99m/s2
Use the second equation to get Fp = m2a = 475N.

3. In the Figure the tension in the diagonal string is 60.0N. Find the magnitude of the horizontal forces F1 and F2 that must be applied to hold the strings in the position shown.

We can draw two free body diagrams, one for each of the two knots connecting three ropes. We take the usual xy system of reference. Let us denote the tension in the diagonal rope by T3. Newton’s law for the x components yields:

-F1 + T3 cos 45o = 0
F2 - T3 cos 45o = 0
There is no need to write down the equations for the y components. The magnitude of the forces are
F1 = T3 cos 45o = 28N (42N)
F2 = T3 cos 45o = 28N (42N)

4. A 0.35kg mass at the end of a spring vibrates 3.0 times per second, with an amplitude of 0.15m. Determine the velocity when it passes through the equilibrium position.

The angular frequency is w=2pf=(6.28)(3.0s-1)=18.84s-1. The velocity is v=wA=(18.84s-1)(0.15m)=2.83m/s

5. The typical game of squash is played by two men, hitting a soft rubber ball to a wall until they are about to die due to general dehydration and exhaustion. Assume that the ball hits the wall at a velocity of 20m/s, and bounces back with a velocity of 10m/s. The kinetic energy lost in the process heats the ball (neglect energy loss to the environment). What will be the temperature increase of the ball after one bounce? (The specific heat capacity of rubber is about 1200 J/kgK.)

The kinetic energy loss is1/2m(v12 - v22). This must be equal to mcDT, the heat taken up by the ball. We get 1/2(v12 - v22)= cDT, and DT= (v12 - v22)/2c = 0.12K. Indeed, if play squash, you can feel the ball getting quite warm during the game.

6. Refer to the figure, and use 10 m/s2 for g and 100000 N/m2 for 1atm. The pressure at A is 9.5 atm and the water velocity is 10 m/s. Water can be considered an ideal fluid (no viscosity). What is the water velocity at point B?

  1. 2.5 m/s
  2. 5 m/s
  3. 10 m/s
  4. 20 m/s
  5. 40 m/s
  6. some other value
Use the continuity equation, v1A1=v2A2. Since the area A is proportional to the square of the radius, we get v2=v1(r1/r2)2. Therefore vB=(10m/s)(4cm/8cm)2=2.5m/s

7. How much work is done by the gas if its state changes from a to c? The 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.

  1. 1.2 x105 J
  2. 1.8 x105 J
  3. 3 x105 J
  4. 4 x105 J
  5. Something else
The work is the area under the curve: W=(5atm)(800L-200L)= 3 x105 J

8. A bowling ball weighting 71.2 N is attached to the ceiling on a 4.20 m long rope. The ball is pulled to one side and released; it will swing back and forth as a pendulum. As the rope swings through the vertical, the speed of the ball is 5.2 m/s. What is the tension in the rope at this instant?

  1. 1156 N
  2. 118 N
  3. 72.4 N
  4. 46.8 N
  5. 24.2 N
  6. 1.24 N
The tension works against the weight of the ball, and still keeps it on circular orbit. Accordingly:
T-W = mv2/r. We get T = mg + mv2/r =118N

9. On a very clear dry dark night little radiation is received from the sky. How much energy, in one hour, will a perfectly black object with 2.0 m2 of surface area exposed only to the sky radiate if its temperature is 50 oC? (The Stefan-Boltzman constant is s=5.67 x 10-8 W/m2K4.)

  1. 7.4x104 J
  2. 2.2x106 J
  3. 1226 J
  4. 4.4x106 J
  5. some other value
The heat current is H=AesT4, where A=2.0m2, e=1, s=5.67 x 10-8 W/m2K4, and T=323K. We get 1226J/s. Multiply this with the time (3600s) to obtain Q=4.4x106J.

10. How much (what mass) of steam at 100oC must be added to 1.00kg water at 10oC to yield liquid water at 80oC? (The heat of vaporization of water is 2256 x 103J/kg. The specific heat capacity of water is 4190 J/kgK.)

  1. 3.50kg
  2. 0.130kg
  3. 0.125kg
  4. 0.050kg
  5. something else
The steam will condense, releasing Q = m(2256 x 103J/kg) heat. It will then cool from 100oC to 80oC, releasing another Q' = m(20K) (4190 J/kgK) heat. This is all taken up by the cold water, warming in the process by 70K: (1.00kg) (70K) (4190 J/kgK) = m (2256 x 103J/kg + 20K x 4190 J/kgK) . We get m =0.125kg.

11. A freight car of mass 24,000kg is coasting without friction on a level track with a velocity of 3.0m/s. A conveyor belt loads the moving car with 4000kg sand. The sand is falling vertically downward with a velocity of 5.0m/s. What will be the velocity of the car when it is loaded?

  1. -2.0m/s
  2. 2.0m/s
  3. 2.6m/s
  4. 4.3m/s
  5. 5.0m/s
  6. something else
Apply conservation of momentum, horizontal component only. This component of the momentum of the sand is initially zero. The car has an initial momentum of 24,000kg x 3 m/s. At the end the total mass is 28,000kg, and the momentum must be the same. The velocity is 2.6m/s.

12. A body is rotated in a horizontal circle of radius 6.3m. What is the period of rotation at which the centripetal (radial) acceleration has a magnitude of 5g? (Note: g = 9.81m/s2)

  1. 0.36s
  2. 0.81s
  3. 2.25s
  4. 5.04s
  5. 7.05s
  6. something else
We will use a = 4p2R/T2. Express the period: T = 2p[R/a]1/2 . With the values given we get
T = 2p [6.3m/(5x9.81m/s2)]1/2 =2.25 s