The second midterm will be on Monday, November 12, 8:30-10:00 PM, in Javits. The exam will be at two different places. If your first name (or "given name") starts with letters A to D, go to Javits 103. With first names E-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 103, 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 1,2,3,4: 10 points for correct answer, partial credit possible. Problems 5-9: 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 will be posted on the course WEB page. The graded exam will be handed back to you by your recitation instructor. If you feel that a mistake was made, return the exam and a brief written statement of your grievance to your recitation instructor. Except for trivial mistakes (like incorrect addition of the scores by the grader), the returned exam will be accepted only on the same day that you received the graded exam in recitation. Preferably, you should ask for the re-grading at the end of the recitation session. Do not change your solution in any way. The contested exams may be compared to a xerox copy made by the grader. If you make notes on the exam sheet, use a writing instrument that is very different from the pen/pencil used during the exam. If 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. 1.5m long, 80.0kg board is propped against a wall. The board makes a 60^o angle to the horizontal ground. How great a force does the wall exert on the board? Assume that there is no friction between the board and the wall, but the board is firmly set against the ground.



Draw the free body diagram for the board. The force on the top is horizontal, since the friction between the board and the door is neglected. The torque equation , taking the bottom of the board as the center of rotation, is:
F L sin 60^o -mg (L/2) cos 60^o = 0, where L is the length of the board. This yields
F = mg / (2 tan 60^o ) = 226N
 

2. A ball (solid sphere) rolls without sliding on a horizontal surface with velocity v. It reaches a loop curving upwards, and its center of mass climbs to the maximum height of 0.50m before the it rolls back. What was the velocity of the ball?


Use the conservation of mechanical energy. The kinetic energy of the ball at the beginning is 1/2 mv²+ 1/2 Iw², with w = v/r and I = 2/5 mr². The potential energy is zero. The total energy must be equal to the total energy at the end, when the kinetic energy is zero and the potential energy is mgh = m (9.81m/s² ) (0.5m). Therefore (1/2) (1+2/5) mv² = mgh yielding 0.7 v² = 4.91 m²/s², and v = 2.65 m/s.
 
3. A 5.00g bullet is fired horizontally into a 1.5-kg block resting on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.2 . The bullet remains embedded into the block, which is observed to slide 0.250m along the surface before stopping. What was the initial speed of the bullet?


This is very similar to the problem of "ballistic pendulum". First there is an inelastic collision, and we can use the conservation of momentum: mv = (m + M) v', where v' is the velocity of the block and bullet after the collision. Next we use the work - energy principle. The potential energy is zero all the time. The kinetic energy of the block/bullet is 1/2 (m + M) v'² at the beginning. It becomes zero due to the work done by the friction force: 1/2 (m + M) v'² - s m N = 0, where s = 0.25m is that total distance of motion. The normal force is equal to the weight N = (m + M) g, and we get v' = [2 smg]^1/2 =0.99m/s. Now we can use the first equation to get : v = v' (m + M)/ m = 0.99m/s 1505g/5g = 298m/s. The velocity of the bullet is 298m/s.
 

4. A string is wrapped around a solid cylinder of radius 0.080m and mass 1.20kg. The free end of the string is held in place and the cylinder is released from rest. What is the tension in the string?


Make a free body diagram, indicate positive directions for the rotation, and linear motion. Newton's law for the vertical motion is mg - T = ma. Pick the axis of rotation at the center axis of the cylinder. The torque equation is: I a = T r, where I is the moment of inertia, 1/2 mr² for a cylinder. We also know that a = a/r . Using these, from the second equation we obtain 1/2 ma = T. Plug in this to the first equation: mg - 1/2 ma = ma. This yields a = 2/3g, and than the tension is T = 1/3 mg = 3.92N. (Note: There was no need for r)
 
5. To compress a spring 4.00cm from its unstretched length 12.0 J work must be done. How much work must be done to compress the same spring 3.00cm from its unstretched length?

1. 6.75 J
2. 9.00 J
3. 10.39 J
4. 12 J
5. 16.0 J
6. other

1. 1.92 x 10^-13 increasing
2. 1.63 x 10^-13 increasing
3. 1.34 x 10^-13 increasing
4. 1.34 x 10^-13 decreasing
5. 1.63 x 10^-13 decreasing
6. 1.92 x 10^-13 decreasing

7. A 3.00m long diving board is supported by a vertical force at a point 1.00m from the end, and a diver weighting 580N stands at the other end. The board is of uniform cross section, and its weight is 420N. What is the magnitude of the force at the support point P?

1. 3000N
2. 2370N
3. 1170N
4. 1000N
5. 580N
6. other

8.  A solid wood door 1.00m wide and 2.00m high is hinged around one side and has a total mass of 50kg. Someone keeps pushing the door at its center with a force of 10N, perpendicular to the door. How long will it take for the door to rotate by 90^o, if it starts at rest? (Neglect friction.)

1. 5.23s
2. 10.5s
3. 3.23s
4. 1.7s
5. 2.4s
6. other

9.  A 140kg astronaut (including space suit) acquires a speed of 2.50m/s by pushing off with his legs from a 1800kg spacecraft. What is the kinetic energy of the spacecraft after the push?