( ) ( ) 9.8 m / s 2 0.35 kg F = kx k = = 28.58 N / m 0.12 m m - - PDF document

1 1) When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displaced from its equilibrium position and released, and undergoes simple harmonic oscillations. What is the period of the oscillations? A) 0.695 s B) 0.483 s C) 0.286 s D) 0.0769 s E) 1.44 s Answer: A

F = kx k = 0.35kg

( ) 9.8m / s2

( )

0.12m = 28.58N / m T = 2 m k = 6.28 0.35kg 28.58N / m = 0.695s

2) One crying baby emits sound with an intensity of 8.0 10-8 W/m2. What is the intensity level for a set of quintuplets (5 babies)? The lowest detectable intensity is 1.0 10-12 W/m2. A) 79 dB B) 69 dB C) 56 dB D) 49 dB E) 36 dB Answer: C

dB]

[ ] = 10log10

I I0

• I0 = 1012 W/m2 = 0 dB

[ ]

5BABIES dB]

[ ] = 10log10

5I1BABY I0

• = 10log10

40.0 108 1012

• = 56dB

2 3) Two strings with their ends fixed are otherwise identical (same mass per unit length, stretched with the same tension), but one string is 0.33 cm longer than the other. Waves on these strings propagate at 34.0 m/s. The fundamental frequency of the shorter string is 258 Hz. What is the beat frequency when each string is vibrating at its fundamental frequency? A) 12.3 Hz B) 9.00 Hz C) 12.0 Hz D) 12.7 Hz E) 11.3 Hz Answer: A waves on a string:

S1 = 2LS1

• fS1 = v

S1 = v 2LS1 258Hz = 34m / s 2LS1 LS1 = 0.0659m S2 = 2LS2 =, fS2 = v 2LS2 = v 2(LS1 + 0.0033m) = 34m / s 0.138m = 245.7Hz fBEAT = fS1 fS2 = 258 254.7 = 12.3Hz

4) A Doppler weather radar operates at a frequency of 3.40 GHz. The waves from this radar system reflect from an approaching weather system moving with a speed of 39.0 m/s. What is the difference in frequency between the outgoing and returning waves? A) 442 Hz B) 351 Hz C) 419 Hz D) 322 Hz E) 670 Hz Answer: A

Doppler effect for EM waves : f ' = f 1± u c

• f = f ' f = f

u c

• (u = relative speed, use + for approaching, - for moving apart)

f = 3.4 109 Hz 39m / s 3108m / s

• = 442Hz

3 5) An object is located 3.8 m in front of a plane mirror. The image formed by the mirror appears to be A) 1.9 m in front of the mirror. B) on the mirror's surface. C) 1.9 m behind the mirror's surface. D) 3.8 m in front of the mirror. E) 3.8 m behind the mirror's surface. Answer E For plane mirror, image distance = object distance. 6) A 4.0-cm-tall object is placed 50.0 cm from a diverging lens of focal length 25.0 cm. What is the nature and location of the image? A) A real image, 4.0 cm tall, 20 cm other side of the object B) A virtual image, 4.0 cm tall, 20 cm other side of the object C) A virtual image, 2.0 cm tall, 10 cm other side of the object D) A virtual image, 1.3 cm tall, 16.7 cm same side as the object E) A real image, 1.3 cm tall, 16.7 cm same side as the object Answer: D

1 dO + 1 dI = 1 f 1 dI = 1 f 1 dO = 1 25cm 1 50cm = 3 50cm , dI = 16.67cm Minus sign means image is on same side as object, and therefore must be virtual. m = dI dO = 16.67 50 = +0.333, + means erect, so image is upright and 1.33cm tall.

7) In Young's two-slit experiment, the distance between the slits and the screen is 1.10 m and the distance between the slits is 0.0400 mm. If the second order bright fringe is measured to be 4.20 cm from the centerline, what is the wavelength of light? A) 200 nm B) 381 nm C) 401 nm D) 620 nm E) 763 nm Answer: E For m=2:

2-slit pattern, slit spacing d, mth bright fringe: sinm = m d Fringe position on screen at distance L: ym = L tanm L m d

( )

ym = 1.10m 2 0.000040m

( ) = 0.042m ( )0.000040m / 2 1.10m ( )

• = 7.63107 m

4 8) What is the limiting angle of resolution for a telescope with mirror diameter 4 m, at wavelength of light of 600 nm? A) 1.8 10-4 rad B) 2.1 10-5 rad C) 1.8 10-7 rad D) 1.5 10-8 rad E) 2.6 10-4 rad Answer: C

Rayleigh criterion: min = 1.22 D = 1.22 6 107 m 4.0m

• = 1.8 107rad

9) You are moving at a speed (2/3)c relative to Randy, and Randy shines a light toward you. At what speed do you see the light passing you by? A) (1/3)c B) (2/3)c C) (4/3)c D) c E) It depends on whether you are moving towards or away from Randy. Same question as in a previous exam…if you come away from this class remembering only

• ne fact, let it be this: the speed of light is the same to observers in any inertial frame.

10) In the Bohr model, if a hydrogen atom’s electron is originally in a state with principal quantum number n and is excited to state n' = 2n, then A) its radius and binding energy will double. B) its radius will quadruple and the binding energy will double. C) its radius will double and the binding energy will quadruple. D) its radius will quadruple and the binding energy will be reduced by a factor of four. E) its radius and binding energy will quadruple. Answer: D

rn = h2 4 2mkZe2

• n2,

En = 13.6 eV

( ) Z 2

n2 n 2n r 4r, En En 4

5 11) Which one of the following statements is true? A) All nuclei of a given element have the same number of neutrons. B) Different isotopes of a given element have different numbers of protons. C) The strong nuclear force acts only on neutral particles. D) Protons are not fundamental particles but have constituents called quarks. Answer D All nuclei of a given element have the same number of protons. The strong nuclear force acts only on particles that are hadrons, which are made of 3 quarks (or 3 anti-quarks). 12) The symbol for a certain isotope of strontium is . How many protons are in the nuclei

• f this isotope?

A) 38 B) 52 C) 88 D) 90 E) 128 Answer: A 13) An electron and a positron annihilate each other, emitting two identical photons in the

• process. What is the wavelength of these photons? The mass of an electron or positron is 9.11

10-31 kg and h = 6.63 10-34 Js. A) 1.21 10-12 m B) 1.73 10-12 m C) 2.43 10-12 m D) 3.07 10-12 m E) 3.46 10-12 m Answer: C Photons share the energy released equally, so each gets E = rest energy of one electron. ETOTAL = 2 me c2; E = me c2 = hc = hc me c2 = h me c = 6.6310-34 J-s 9.1110-31 kg

( )3108m / s

= 2.42 10-12m

6 14) Carbon-14 has a half-life of 5730 years. A sample of wood has been recovered by an

• archaeologist. The sample is sent to a laboratory, where it is determined that the activity of the

sample is 0.144 Bq/g. By comparing this activity with the activity of living organic matter, 0.230 Bq/g, the scientist determines how old the wood sample is, or more precisely, when the tree that the sample came from died. How old is the sample of wood? A) 3870 years B) 4250 years C) 4590 years D) 2630 years E) 2940 years Answer: A = ln 2 T

1/2

= 0.693 5730y

( )(365d / y) 86,400s / d ( )

= 3.84 1012s1 R = N N = 0.144 Bq/g 3.84 1012s1 = 3.751010 C14 nuclei per gram of wood. N0 = 0.23 Bq/g 3.84 1012s1 = 5.99 1010 C14 nuclei per gram of wood, assuming

• riginal wood was same as contemporary wood when tree was alive.

N(t) = N0et ln N / N0

( ) = t

t = ln N / N0

( )

3.84 1012s1 = 0.46 3.84 1012s1 = 2.19 1015s = 3.87 103 y 15) Two deuterium nuclei, , fuse to produce a helium-3 nucleus, , and a neutron. A neutral deuterium atom has a mass of 2.014102 u; a neutral helium-3 atom has a mass of 3.016030 u; a neutral hydrogen atom has a mass of 1.007825 u; a neutron has a mass of 1.008665 u; and a proton has a mass of 1.007277 u. How much energy is released in the process? 1 u = 931.494 MeV/c2. A) 3.27 MeV B) 3.57 MeV C) 4.00 MeV D) 4.50 MeV E) 6.58 MeV Answer: A Atomic mass includes electrons. Deuterium nucleus =(2.014102 u) (931.494 MeV/c2 / u) - (electron mass) Initial mass =2((1876.124 MeV/c2) - (0.511MeV/c2) )= 3751.226 MeV/c2 He-3 nucleus =(3.016030 u) (931.494 MeV/c2 / u) - 2(electron mass) Final mass =(2809.414 MeV/c2 - 2(0.511MeV/c2)) + ((1.008665 u) (931.494 MeV/c2 )) = 3747.957 MeV/c2 Difference = mass that was converted into energy = 3.27 MeV/c2 E = mc2 = (mass loss) c2 = (3.27 MeV/c2 )c2 = 3.27 MeV