Exam III

Exam III will take place during the lab period on Thursday, April 17. As before, you will have the full three hours. The exam will consist of 6 to 8 problems and cover from Chapter 12 through Chapter 16. Below are a few sample exam problems.

You are responsible for all material covered in lecture and homework assignments. DO NOT ASSUME that a subject not covered in the sample exam below will not show up on Tuesday's exam! During the exam, you will be provided with an equation sheet. You may download the Equation sheet. You will be allowed to use only the equations on the sheet, or others that you derive from those on the sheet. I will email that sheet to you. Use only that when trying the problems below. Resist the temptation to look at the answers, as that will better prepare you for the exam.

  1. A fully suited 100 kg astronaut stands on a spherical asteroid of diameter 1.0 km and mass 2.5 x 1012 kg.
    a) What is the astronaut's "weight" on this asteroid?
    b) With what speed must she jump to escape the asteroid?

    F = Gm1m2/r2 where r = 500 m yields F = 0.0667 N. Vesc = sqrt(2GM/r) = 0.82 m/s ... easy to do!

  2. Five 80 kg students load into a 2000 kg car. Another student measures that the car lowered by 1.00 cm when the students got in.
    a) What is the effective spring constant for the car suspension system?
    b) Assuming really bad shock absorbers, what would be the frequency of oscillation of the loaded car after hitting a bump in the road?

    F =-kx, where the added weight that caused the 1.0 cm drop is w = 5 x (80kg) x 9.8 m/s2 = 3920 N = k(.01m) -> k = 3.92 x 105N/m.

    The frequency of oscillation is f = 1/2p sqrt[3.92 x 105/(2000+5x80)] = 2.04 Hz.

  3. The inside of an underwater research station is maintained at atmospheric pressure. What net force acts on the 10m by 5m roof which is 8m below the sea surface?

    F = PA = rghA = 3.92 x 106N.

  4. A giant piece of styrofoam has dimensions 1.0.m x 2.0 m x 0.5 m and weighs 200N. How many 70 kg students can balance on top as it floats in the sea?

    FB = rVg = (1030)(1.0x2.0x0.5)(9.8) = weight of students + 200N. Weight of students = 9,900 N, m = 1000 kg, divde by 70 kg to get 14.4 -> 14 students.

  5. The maximum water flow from your faucet should fill a 1.0 L bottle in 10 sec.
    a) If the faucet has radius 1.0 cm, what flow velocity is required?
    b) What must be the minimum gauge pressure in the mainpipe which has radius 4.0 cm and is located 2.0 m below the faucet during maximum flow at the faucet?

    Av = flowrate -> pi(0.01)2v = 0.001m3/10 sec -- > v = 0.32 m/s.

    Use Bernoulli's for part b ... Need v2 from A1v1 = A2v2 -> v2 = 1/16(v1) = 0.02m/s, so p1 + 1/2r0.322 + rg(0) = p2 + 1/2r0.022 + rg(-2.0m) -- > P2 - P1 = 19700 Pa.

  6. An ocean wave moves by as you are floating. You move through a total vertical distance of 2 m from the crest to the adjacent trough of the wave in just 1.2 seconds. You note that two adjacent crests are 7 m apart. What is the amplitude, frequency, wavelength and velocity of the wave?

    Amplitude = 1/2 of total vertical motion = 1 m. The period is 2.4 seconds, so f = .42 Hz. Wavelength is 7 m and so v = lf = 2.9 m/s

  7. A trombone player plays a note that has an intensity level of 50 dB at a distance of 10 m from the player.
    a) What would the decibel level be 100 m away from the player?
    b) How many identical trombones players would be needed to increase the level to 60 dB at 10 m?

    a) Since P ~ 1/r2, the intensity will drop by 102 = 100. That is a drop of 20 dB. (A factor of 10 for every 10 dB.) So the dB level at 100m is 30 dB.
    b) Increase of 10 dB requires a tenfold increase in intensity ... so 10 trombones! You should also do this problem the long way using b = 10 log(I/Io). That will be necessary for those cases without easy ratios of 10.

  8. A mast wire is held under tension 1000N. If the 10 m length of wire has mass 0.8 kg what is the frequency of the fundamental (lowest) mode of vibration in the wire?

    v = sqrt(T/m) = sqrt(1000N/(0.8 kg/10m)) = 112 m/s. The lowest mode of vibration is 1/2 wavelength between the ends, so l = 20 m and f = v/l = 5.6 Hz (Note: You should be able to draw the first few harmonic vibration modes for strings and both open-end and closed-end pipes. You can deduce the wavelengths from your drawings.)

  9. You are walking at 1 m/s away from a stationary trumpet player playing a 512 Hz note. At the same instant an identical trumpeter is walking at 1 m/s towards you. What frequency do you hear from each and what other effect do you hear hear?

    f' = [(343-1)/343] 512 = 511 Hz for the first trumpet.
    f' = [(343 +1)/(343-1)]512 = 515 Hz for the second trumpet.
    You will hear these two tones and the beat between them of 4 Hz.