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AP Physics Physics B Exam - 2004 Solutions to Multiple Choice

BASIC IDEA SOLUTION ANSWER

#1. F= 0 a = 0 None of the situations described would necessarily meet Ethe conditions of Newton's First Law of Motion.

#2 F= ma 50N 30N = 30N9.8

m

s2

a , taking 9.8 m/s2 as 10m/s2 yields 6.7m/s2 B

#3. components The spring would have given the ball a radial velocity and the Erotation would have provided a velocity toward the top of thepage.

#4 v = 0 a = g so it is not in equilibrium ruling out choices A) and B) Ev = 0 which it was not before or after so kinetic energy andthe magnitude of the momentum are 0 therefore less then theywere before and after this moment so they are clearly not at theirmaxima.

#5.

F= 0

a = 0 Archimedes' Principle. Weight =0.4kg(9.8m/s2)=4N. The Bbuoyant force is then 4N-3N = 1N.

#6. Ueqyuk= min Both are going at their maximum speed at the equilibrium point. A

therefore K = max

#7. T = 2m

k m

1

k=

g k=

m1g

E

T = 2

g

#8. The amplitude is not one of the independent variables. E

#9. momentum cons. No horizontal force external to the system -sled, m, 2m- therefore Cthere will be no change in the horizontal component of themomentum. The order in which the objects are dropped does not

change the horizontal component of momentum or the final velocity.#10. Doubling M would of itself double the persons weight, but making B

the radius twice as great gives us a factor of 4 in the denominatorsince (2r)2 = 4r2. The total effect then is2/4 or one half the force.

#11. D

#12. Zeroth Law Thermal equilibrium is the definition of "same temperature." C

#13. State Variable Temperature for an ideal gas is directly proportional to the internal Akinetic energy. For an ideal gas this is the only internal energy. Theinternal energy depends on the state of the gas not the path taken.The Second Law of Thermodynamics tells us that U = Q+W andTherefore a vast variety of combinations of Q and W will result in

Same overall change in internal energy in going from one state toAnother.

#14. F= IBsin The effect on one wire on the other can be found by combining Cthese two equations. (here sin = 1)

#15. conductors & The electric field due to static charges is zero inside a conductor. Acharge distributions The charge resides on the outer surface of the conductor so here

there is no way for the field to be "transmitted" to points C and P.

F= 0 a = 0

F= ma

F= GMm

r2

vav=

x

tvav=

8m 2m

2s 1s= 6m / s

T = 2m

k

B =oi

2r F=oiIB

2r

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#16. The total charge that the capacitor can hold is the simple sum of Athe charge on each capacitor.

#17. B

#18. P = F v EP=IV

e =W

out

Win

#19. by convention the Dfield is in the directionof the force on a + charge

#20. Because the charges are of the same magnitude this applies to Deither leg of the triangle depicted in #19, therefore the triangle isa 45-90 triangle and the hypotenuse =

#21. Because both particle have the same magnitude of charge the force will be Ethe same on both.

#22. uniform expansion Because each portion of the plate expands by 1% the dimensions of the hole Dmust expand by the same amount. Simply picture the plate expanding withoutthe material in the hole removed.

#23. PV=nRT If the average speed of the gas molecules increases the temperature, which is Ca manifestation of the kinetic energy of the molecules also increases. If thevolume is held constant the pressure must increase.

#24. "b" is at the estimated half-way point between the center of curvature "c" Band the mirror.

#25. harmonics Harmonics are integer multiple of the fundamental. Here the first harmonic Bis f, which must be twice the fundamental. Then, of course the second harmonicis 1.5f or three times the fundamental,

#26. That tells us the image is virtual therefore A

upright. The second equation tells us the image is twice the size of the object.(This is one of those problems that would go quickly from experience withlenses in the laboratory. There would be no need for a mathematical solution.)

#27. radio is an electro- Cmagnetic emission.f=v

#28. general nature of waves Sound: longitudinal therefore cannot be polarized, mechanical relatively low speed ELight: transverse therefore can be polarized, electromagnetic so the speed is c.

#29. A

#30. Ft=p Here the quick way to get p is to find the area under the graph: a rectangle Cand a triangle give 5N(4s) +0.5(4s)(25N-5N) = 60N.s

#31. Ft=p The change in momentum is the same, but by increasing the time the force Ecan be reduced.

#32. F= 0

Fy = 0

FL + 250 -125 - 500 = 0 therefore FL = 375N B

e =Wout

Win=

Fvt

IVt=

(9kg)(9.8m / s2)(0.5m / s)

(0.5A)(120V) 0.75

E= kQ

r2

2kQ

d2

R =

AR

1=

L

r2 R

2=

L2

r2( )

2= 2

L

r2= 2R

1

E=

F

q

f =R

2

1

si+

1

0.5 f=

1

fsolving for si = f

(100 106)= 310

8m / s

m =y

x=

F

t

F

t=

25N 5N

4s 0s= 5N / s

1

p+1

q=1

f

hi

ho=

q

p

C=Q

1+Q

2+ ...+Q

n

V

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#33. for rotational Taking torques about the left end of the bar: Bequilibrium 125N(2m) +500N(x) -250N(4m) = 0 and x = 1.5 mwhere t=rFsin

#34. E= hf - W A linear function with a negative y-intercept. A

#35. Since we are dealing with photons of the same frequency the energy DE=hf energy arriving each second is proportional to the number of photons/s which

intern is proportional to the number of electrons emitted each second. A linearrelationship, which passes through the origin since no photons no emitted

electrons.

#36. binding energy As the nuclei become larger stability requires a ratio of neutrons to Cprotons that is greater than 1 this rules out B). A) and D) can be ruled outby the stable nucleus of hydrogen, , and the stable helium nucleusrules out E).

#37. P = Po + gh Pascal's principle says the pressure is independent of the shape of the vessel. EAlso, since all of the factors which determine pressure are the same all of the

pressures must be the same.The force on the bottom is PA therefore with

equal pressure and equal area the Force is the same on the bottom of all of the

beakers.

#38. Bernoulli's Principle The velocity will be greatest in the smaller diameter tube so the pressure A

will be less and the liquid will be forced up the tube.

#39. Archimedes' Principle Apparent loss in weight is equal to the weight of the fluid displaced. Because E

at the same elevation the weight is proportional to the mass, we could restate

the principle that the apparent loss of mass is equal to the mass of the fluid

displaced. Here the loss is 0.45 - 0.36 = 0.09 kg. That mass corresponds to a

volume of water of 0.09kg/1000 kg/m3 = 9x10-5 m3. The density is then

0.45kg/9 x10-5 m3= 5x103kg/m3.

#40. momentum mA(5m/s) + mB(-2m/s) = 0 yields B

conservation

#41. 10N - 30Ncos60 = 10a therefore a = -0.5 m/s2 A

#42. B

#43. Notice that only the period of the mass on the string is dependent on g. EThe larger mass means more gravitational attraction and larger g thereforea shorter period. On the other hand the period of the mass on the spring isnot affected by gravitation and remains the same.

#44. Because the object is moving in a horizontal circle there is a centripetal Eacceleration caused by a centripetal force.

#45. uniform field 10V-2V=E (0.04m) so E = 200 V/m. Then from the bottom plate we have DV = Ed VP - 2V = 200(.01) and VP = 4V.

#46. uniform field 10V-2V=E (0.04m) so E = 200 V/m. D

#47. Faraday's Law of Loop 1 has a flux into the paper, but the flux is decreasing as the loop moves CInduction: away so the induced emf will be in such a direction as to try and maintain the

Flux. The right rule (current in direction of curled fingers - thumb in thedirection of the field) gives us a clockwise current. Loop 2 has no change in

Lenz's Law & the flux and therefore no emf or induced current.Right Hand Rule

= 0

intensity energy/s

1

1H 2

4He

P =F

A

p = 0 mAmB

=

2

5

Fx = maa =

v

2

r

vT = 2r

a =

2RT( )

2

R=

42R

T2

T = 2

g

T = 2m

k

F= ma

= d

dt

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#48. Resistors in For the parallel portion: andRP = 1500 . That DSeries and Parallel in series with 2500 gives us a total resistance of 4000 . ThenRs=R1+R2++Rn using Ohm's Law we get: 12V = I (4000 ) and I = 3.0mA

V=IR

#49. cons. of charge Since I1 is the total current it must be the larger of the three. Because AV=IR R2 and R3 have the same potential drop across them, the smaller resistance, R2,

has the larger current.

#50. n1sin1=n2sin2 Going from air into glass (n2>n1) the light will bend toward the normal to the Esurface and when it passes back into the air it will bend away from the normal.

#51. n1sin1=n2sin2 Going from the glass into the air (n2n1 surface changes phase and gets one-half wave ahead so the path difference need

only be an additional half wave length giving us a thickness of/4.

#54. then q = 2.67cm E

#55. image formation Rays from all parts of the object strike each portion of the lens and Dform the image. (Otherwise people with bifocals or "granny" glasseswould only see a part of the image.)

#56. Here x = d the thickness of the slab. Decreasing A and T and holding Dd constant would reduce the rate. The other choices all involve at leastone change that would increase the rate.

#57. U=Q + Won U = -16J-32J = -48J A

#58. Conservation of The original kinetic energy is "supplemented" by energy from the reduced Emass+energy over all mass. (Even without a knowledge of the particular nuclear reactionthe loss can be inferred from the choices.)

#59. Rutherford -Soddy The atomic number goes up but the mass number remains the same. ELaw (n p + e- + )

#60. f =c divide this energy/photon CE=hf into the energy per second in order to get the photons/s. (keep in mind no

Calculators are allowed on this part.)

#61. Same force mean same size acceleration in both directions. Same acceleration C

vf=vo + at from restfor the same time produces the same final velocity in both directions.

#62. For the first second the magnitude of the velocity is proportional to the B

time, therefore the kinetic energy will be proportional to t2. Then at one second

vf=vo + at the speed in the y is constant but the speed in the x increases from zero also in

a manner proportional to t2 for the second second. After that there is no

acceleration so the kinetic energy remains constant.

#63. 3Mvf- Mv+2Mv=0 therefore vf =v/3. D

Inelastic collision

1

RP

=1

R1

+1

R2

++1

Rn

1

RP

=1

2000+

1

6000

1

p+1

q=1

f

1

8cm+1

q=

1

2cm

Q

tA

T

x

E= hc

= 6.6310

34J s

3.0 108 m

s

4m 5 10

26J

50,000W

5 1026J= 110

30photons / s

F= ma

K =1

2mv

2

p = 0Kf Ki =

1

23M

v

3

2

Mv2 2Mv2

=

4

3Mv2

K =1

2mv

2

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#64. Lenz's Law The flux goes from increasing to decreasing each time the plane of the loop D

becomes perpendicular to the field, that is once each half rotation.

#65. K+U = 0 In the first case the resulting K = QV. In the second case it is 2QV, in other D

U =qV words twice as much.

#66. A negative charge will experience a force in the direction opposite to the field. D

#67. solving for C

#68. P=900N(4m/s)cos0 = 3 600 W E

#69. F=qvB qvB=qE and then C

F=qE

F= 0 a = 0#70. A

F= q EF=G

Mm

r2

F= maa =

v2

r

GMm

r2= ma = m

v2

rr =

Gm

v2

P = Fvcos

v =E

B=

6N /C

2T= 3m / s

E= kq

r2

E= 9 109N m

2/C

2( )4 10

6C

2m( )2

= 9 103N /C