AP Physics - 250: Tests, Fall 2005

John. A. Shaw

 

The tests are closed book and will cover the material from Jones and Childers and the lectures. If you have a question about a test or the date of a test, you can email me at jshaw@lsmsa.edu .

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  • Test 3 sample solutions are here (400k pdf file)
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  • Test#1: Spring ’03 -- Chapters 15-17  (Friday February 14) 
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    • Sample test solutions below:
    • 1. a) simplest proof drops a vertical from H to line connecting O to P.  This gives two right triangles: the side of the one with P at a vertex has a length 1.2-0.958cos(52.5) and a height of  0.958sin52.5.  Squaring and adding gives 0.98^2 for the hypotenuse and the tan of the angle is 0.958sin52.5/(1.2-0.958cos52.5) or 50.1 degrees, QED.
    •     b) The field of each hydrogen nuclei is 1.5x10^11 Volts/meter and the field of the oxygen nucleus is 8x10^11 Volts/meter.  These are vectors, but symmetry saves us from too much work.  The x-components of the hydrogen's E-field add to the x-component of the oxygen's E-field.  The y-components cancel.
    • ==> 2cos(50.1) (1.5x10^11) + 8x10^11 = 9.9x10^11 Volts/meter.
    •     c) F=qE, so F=-1.6x10^-7 Newtons along the x-axis.  The minus means the electron is pulled to the left at P.
    •    d) The voltage has no direction so you just add the voltages due to each charge. The voltage at P due to each hydrogen is 24.8 Volts and the voltage at P due to the oxygen nucleus is 96 Volts: Vnet = 2 V_H +V_O=145.7Volts.
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    • 2. a) E is the same as a point charge with charge +Q when a<r<b since Q_enclosed is just +Q.  The net flux is E A where A is 4*pi*r^2.  Gauss's Law just gives Coulomb's law: E=kQ/r^2 since k=1/(4*pi*eps_o).
    •     b) When r>b then the charge enclosed is +Q on surface a and -Q on surface b.  The sum vanishes, so  E=0 is the only solution for E consistent with the symmetry.
    •     c) from a, E = kQ/r^2 so set r=a and solve for Q: Q=2.08x10^-7 Coulombs.
    • 3. a) see lab writeup for sketch of E-field
    •     b) see lab writeup for sketch of equipotential curves.
    •     c) Work =0, equipotential
    • 4. a) 2x10^-4 Farads
    •     b) 0.22 mm 
    •     c) 9x10^8 Volts/meter
    •     d) 400 Joules
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·         Sample test solutions below:

·         1.  a) a = -g sin q - mu_k g cos q = -9.8 sin (40) -0.45 9.8 cos(40) = -9.68 m/s^2

·              b) v^2=vo^2 + 2 a d, and v=0 at top so d = -vo^2/(2a) = +3^2/(2*9.68)=0.465 meters

·              c) if mu_k=0, then a = -gsinq so d=3^2/(2*6.3)= 0.71 meters

·         2. a) tanq=T_1/T_2, so find T_1 and T_2 first.  T_1 =m_a g =20*9.8=196N, 

·                 T_2 = mu_s m_b g =0.25*50*9.8=122.5N, q=32 degrees.

·             b) T_3 = m_a g /cosq = 231 N

·             c) T_1=196, and T_2 = 123 N from above

·         3. a) T^2 = 4pi^2/(GM) r^3, so M=4pi^2/(GT^2) r^3 = 4(3.14159)^2/(6.67x10^-11*16^2*86400^2) (1.22x10^9)^3 = 5.6x10^26 kg.

·             b) T = 2 pi /(G M)^0.5 r^1.5 = 1.65x10^5 seconds = 1.9 days

·         4. a) diagram will give mu_s  g =  v^2/r, for v maximum 

·              b) v=(mu_s g r )^0.5 = (0.32*32 ft/s^2 68 ft)^0.5 = 27.6 ft/s 

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·         Sample Test solutions below:

·         1.a) -127 Joules, friction opposite displacement

·            b) +254 Joules, gravity has component along displacement

·            c) +118 Joules

·             d) 6.9 m/s

·         2. a) v1f=0.7 m/s, v2f=2.2 m/s, both positive

·             b) pf-pi=-3.6 kgm/s for m1, +3.6 for m2, equal and opposite

·             c) -3600 N for m1, +3600 N for m2, equal and opposite (N3L!)

·         3. a) vf/v1=0.4 or v1/vf=2.5 if inverted

·             b) vf=1.62 m/s after collision

·             c) v1=4.05 m/s before collision

·         4.  a) 10.3 km/s

·              b) 11.2 km/s

·         Solutions to last year's final (large non-text file, ca. 200kB)

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Final Exam Solutions (from ’02, sorry, you still have to take the ’03 version!):

I. Short answer section

            1.E=7200 V/m

            2. V=0 halfway between equal and opposite charges

            3. E=V/d = 90,000 V/m

            4. Q=CV = 7.97x10^-8 C

            5. Ctotal = 6.67 pF or 6.67x10^-12 F

            6. Rtotal=13.2 Ohms

            7. i = 0.45 Amps through 10 Ohm resistor.  At node i splits into 0.36 Amps through the 4 Ohm and 0.09 Amps             through the 16 Ohm resistors respectively.  Note the sum is 0.45 as it must be.

            8. r =1.6 cm or 1.6x10^-2 meters

            9. B=6.28x10^-5 Tesla

            10. Tau = RC = 5x10^-7 seconds

            11. Lambda’ = Lambda/n  so n=5/4=1.25 (dimensionless ratio)

            12. n = 1.59 (dimensionless ratio).

            13. theta_2 = 14.9 degrees from the normal

            14. i = 12 cm

            15. M = +2/3, upright and smaller

            16. angular separation = 6.6x10^-3 radians

II. Longer responses:

            1. a) Read book for a detailed discussion of Gauss’ law. You need a closed surface surrounding the charge.  The surface should be perpendicular to the electric field lines so the normal and the electric field are parallel.  This is easy for a point charge if you surround the charge by a sphere with radius r.  E is a constant so summing over the surface just gives E (A) = q/e_o since q is inside the sphere.  A = 4pi r^2 for a sphere so solving for E gives Coulomb’s law.

            b) Gauss just says that the flux of the field through a closed surface is proportional the amount of charge inside the surface.  No charge inside, no flux.  Charge inside, then positive or negative flux depending on the sign of the net charge enclosed.

            2. a) 6-i_2 10 – 3 – i_1 10 = 0 and 3 +i_2 10 – i_3 20 – i_3 30 = 0

            b) i_1 – i_2 –i_3 =0 at A (  i_1 flows in from left and i_2 and i_3 go down and to right respectively).

            c) haven’t worked this out, but you can do it on a calculator TI-84 or higher…

 

            3. a) read book for detailed discussion of Ampere’s Law.  You need a closed curve going around the wire.  The curve should be parallel to the magnetic field so B and the tangent to the curve are parallel.  This is easy for a wire if you surround the wire by a circle.  B has a constant magnitude on the curve so summing B dS around the curve just gives B (L) = mu_o i since i in the wire is enclosed by the curve.  L = 2 pi r for a circle so solving for B gives B = mu_o i/(2 pi r).

            b) The magnetic field is proportional to the enclosed current.  No current enclosed, no B.  The sum of B dS around the curve (technically the line integral of B along the curve C) doesn’t have a neat name like “flux.”

            4. See book for details on ray tracing. 

            a) If o > f but o< 2f then the image is enlarged, inverted, and real.  The rays intersect to form an image on the far side of the lens:  e.g. o=1.5f gives i =+3 f  so M =-3/(3/2) = -2 => real, enlarged, and inverted as in ray trace..

            b) If o =f then there is no image formed.  The rays are parallel: e.g. o=f gives 1/i = 0 so i is infinitely large => no intersection, no image. Ray trace should give almost exactly parallel lines.

            c) If o < f then the image is enlarged, upright, and virtual.  The rays intersect to form an image on the same side of the lens as the object: e. g. o = 0.5f gives i= -f so M = +2 => virtual, enlarged, and upright as in ray trace.

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