55) If a particle travels at a constant speed along the track shown, what is the correct ordering of the magnitudes of the acceleration at the labeled points? Let aP be the acceleration at point P, etc.

A) aR < aS < aP < aQB) aP < aQ < aR < aSC) aR = aS = aP = aQD) aP < aQ < aS < aRE) aR < aS < aQ < aP

 

Solution Since the particle is moving at a constant speed, for all points the tangential acceleration is zero and the total acceleration is equal to the radial acceleration. The radial acceleration is inversely proportional to the radius of curvature. From looking at the figure, we see that the correct order is aR < aS < aQ < aP i.e. the tighter the curve, the larger the radial acceleration. 56) Which of the following is an accurate statement about circular motion? A) The vector sum of the tangential acceleration and the centripetal acceleration can be zero for a point on the rim of a rotating disk. B) All points on a rotating disk have the same angular velocity. C) All points on a rotating disk have the same linear speed. D) All points on a car tire have zero acceleration if the car is moving with constant linear velocity. E) All points on a rotating disk experience the same radial acceleration. Solution Tangential and centripetal acceleration are perpendicular to each other. Their vector sum can only be zero if both are zero. A point on a rotating disk has a radial or centripetal acceleration of ω2R = v2/R ie. any point on a rotating disk has centripetal acceleration except the center. Linear speed for a point on a disk with angular velocity ω is given by v = ωR. Thus, linear speed varies with radius while angular velocity is the same for all points on a rotating disk.

57) A long distance swimmer is able to swim through still water at 4 km/h. She

wishes to cross an ocean current that runs from west to east at 3 km/h. In what direction should she swim to cross the current along a straight line from the south shore to the north shore?

A) 37 o east of north B) 41 o east of north C) 37 o west of north D) 49 o west of north E) 41 o west of north Solution From the sketch we see that

θ = arcsinvwater ,shorevswimmer ,water

⎛

⎝⎜⎞

⎠⎟= arcsin(0.75) = 49o

58) In the figure, a constant external force P =

170 N is applied to a 20 kg box, which is on a rough horizontal surface. The angle between the force and the horizontal is 30o.The force pushes the box a distance of 8.0 m, in a time interval of 6.0 s, and the speed changes from

= 0.3 m/s to = 2.5 m/s. The work done by the external force P is closest to:

A) 1180 J B) 810 J C) 1060 J D) 940 J E) 680 J Solution:

W =F ⋅ Δr = FΔxcosθ = (170 N)(8.0 m)cos30° = 1178 J

59) In the system shown in the figure, a 1kg

block rests on a horizontal table. The coefficient of static friction between the block and the table is µS = 0.6. The coefficient of kinetic friction between the block and the table is µK = 0.3. The 1kg block is attached via an ideal massless string over an ideal massless pulley to a 3kg block, which can move vertically without resistance. Initially, the 1kg block is held fixed and is released a t = 0. What is the magnitude of the acceleration of the 1kg block after it is released?

A) 0 m/s2 B) 5.9 m/s2 C) 6.6 m/s2 D) 23.5 m/s2 E) 26.5 m/s2 Solution Let m = 1kg and M = 3 kg. First check if the system is static. For this to be the case the force of static friction must balance the tension in the string, which, in this case, is the weight of the 3kg block. Thus fS = gM = (9.81 m

s2)(3kg) = 29.4N .

The static friction limit, however, is fs,max = µSmg = 5.9N . The required static friction exceeds the limit son indeed the block slides. Now that we know, the block is sliding, we need to draw the free body diagram of each block. If a is the acceleration of the blocks, Newton’s 2nd law gives: 1kg block :T − fk = ma3kg block :Mg − T = Ma

Adding these equations together, we get Mg − fk = (m + M )a ⇒ Mg − µkmg = (m + M )a

⇒ a = g M − µkmM + m

= 6.6 ms2

60) A 10kg weight hangs from two ideal massless strings as shown in the figure. Give the tension in the left rope.

A) 33 N B) 49 N C) 85 N D) 98 N E) 113 N Solution The angle between the two ropes is 90o. Let us take a coordinate system where the x-axis is parallel to the right rope and the y-axis is parallel to the left rope. In this coordinate system the tension in the left rope must cancel the y component of the weight. This tension is therefore TL = (10kg)(9.8m/s2) cos30o = 84.9 N. 61) In the following, K = kinetic energy, U = potential energy, and W = work. In using the expression K2 +U2 = K1 +U1 +Wother we recognize that A) we are assuming no friction is present. B) gravitational and elastic potential energies would be included in and . C) K and U can be either positive or negative, depending on the reference point chosen. D) is always positive. E) gravitational and elastic potential energies would be included in . Solution Wother represents the work done by non-conservative forces (e.g. work done by friction). The Ui represent potential energies and include gravitational and elastic potential energies. Kinetic energies K can never be negative. If index 1 represents the initial state and index 2 the final state, energy lost due to friction would be negative in the equation.

 

62) The potential energy function of a particle moving in one dimension is

U = kx2e− x2 a2

where a = 7.80 nm and k = 20 eV. At what value of x is a point of stable equilibrium located? A) 12.62 nm B) 7.80 nm C) 11.03 nm D) 6.81 nm E) 0 nm Solution:

dUdx

= 0→ 2kxe− x2 a2

+ x2 (−2kx a2 )e− x2 a2

= 0

→ 2kxe− x2 a2

(1− x2 a2 ) = 0→ x2 a2 = 1 or x = 0→ x = ±a or 0k > 0 Sketch curve → x = maximum at ± a and minimum at 0

63) The figure to the right shows a graph of potential energy U versus position for a particle moving in a straight line. From this curve, for the region shown, we deduce that A) there are three positions of stable equilibrium. B) this could not represent an actual physical situation, since the drawing shows the potential energy going negative, which is not physically realizable. C) for a given value of x, the particle can have a total energy that lies either above or below the value given by the curve at that point. D) the force on the particle would be greatest when the particle is near point D. E) the force on the particle would be greatest when the particle is near the origin. Solution The figure indicates two stable equilibriums, which are the local minima of the potential energy function. We can always add an arbitrary integration constant to the potential energy function and thus U can be negative. Total energy E is the sum of potential energy U and kinetic energy K. Since kinetic energy is never negative E ≥ K. The slope of U is zero at D. Since F = −dU/dx, F is zero at D. The slope of U is largest near the origin and thus the force is strongest there.

 

 

64) In the figure on the right, a bullet of mass 0.01 kg moving horizontally strikes a block of wood of mass 1.5 kg which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upward a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block? The length of the string is 2 meters. A) 366 m/s B) 423 m/s C) 646 m/s D) 66.7 m/s E) 34.4 m/s Solution:

In a pendulum, mechanical energy is conserved, we can deduce the velocity of

the bullet and the block right after collision v = 2gh = 2.80 m/sUse conservation of momentum for the completely inelastic collision to deducethe initial velocity of the bullet (0.01 kg)vi = (1.51 kg)v → vi = 423 m/s

65) The radius of a 3.0 kg wheel is 6.0 cm. The wheel is released from rest at point A on a 30 o incline. The wheel rolls without slipping and moves 2.4 m to point B in 1.20s. The angular acceleration of the wheel is closest to: A) 56 rad/s2 B) 82 rad/s2 C) 65 rad/s2 D) 73 rad/s2 E) 48 rad/s2 Solution: The wheel moves a distance d = 2.4 m in t = 1.2 second, while undergoing constant acceleration a. From d = at2/2 we see that a = 2d/t2. For rolling without slipping the angular acceleration α is related the linear acceleration a by α = a/R, where R is the radius of the wheel. Thus,

α = a / R =2dRt 2 =

2 × 2.4m0.06m × (1.2s)2 = 55.6 rad/s2

66) If both the mass of a simple pendulum and its length are doubled, the period will A) increase by a factor of 4. B) be unchanged. C) increase by a factor of 0.71. D) increase by a factor of 2. E) increase by a factor of 1.4 Solution

Since T =2πω

= 2π Lg

, doubling the length L increases T by 2 = 1.4.

67) In the figure on the right, which of the curves best represents the variation of wave speed as a function of tension for transverse waves on a stretched string? A) A B) B C) C D) D E) E Solution

The speed of a transverse wave on a stretched string is given by v = Fµ

,

where F is the tension in the string and µ is the linear mass density. Onlyfunction B has the shape of a square root function. The wave speed has togo to zero for zero tension. Function C is linear and function D is a parabola.

68) Two loudspeakers placed 6.0 m apart are driven in phase by an audio oscillator, whose frequency range is 1032 Hz to 1396 Hz. A point P is located 5.3 m from one loudspeaker and 3.6 m from the other. The speed of sound is 343 m/s. The frequency produced by the oscillator, for which constructive interference of sound occurs at point P is closest to: A) 1265 Hz B) 1211 Hz C) 1315 Hz D) 1366 Hz E) 1164 Hz Solution: For constructive interference the distances between P and either speaker have to differ by an integer multiple of the wave length

path difference = n × λ = n × (v f ) [n =1,2,3,...]

f = n × (v λ) = n × (343m/s)/(1.7m) = 1211 Hz when n = 6 69) A train is approaching a signal tower at a speed of 40 m/s. The train engineer sounds the 1020 Hz whistle and a switchman in the tower responds by sounding the 1215 Hz siren. The air is still and the speed of sound is 343 m/s. The wavelength of the train whistle tone reaching the switchman is given by: A) 0.30 m B) 0.36 m C) 0.32 m D) 0.34 m E) 0.38 m Solution

The frequency fL that the switchman perceives is given by fL = fSv

v + vS. The

corresponding wavelength is then λ =vfL

=v + vSfS

=(343− 40)m/s1020Hz

= 0.30m .

70) A 23.4-kg solid aluminum cylindrical wheel of radius 0.41 m is rotating about its axle on frictionless bearings with angular velocity ω = 32.8 rad s. If its temperature is now raised from 20.0°C to 75.0°C, what is the fractional change in (The coefficient of linear expansion of aluminum is

2.5 ⋅10−5 C ( )−1 .)

A) −2.8 × 10−3 B) +2.1 × 10−3 C) −4.8 × 10−3 D) −1.2 × 10−3 E) −5.9 × 10−3

Solution The change in radius with heating does not cause a torque on the rotating wheel, and so the wheel’s angular momentum does not change. Also recall that for a cylindrical wheel rotating about its axis, the moment of inertia is .

L0= L

final → I

0ω

0= I

finalω

final → ω

final=

I0ω

0

Ifinal

=12

mr02ω

0

12

mr 2=

r02ω

0

r 2

Δωω

=ω

final−ω

0

ω0

=

r02ω

0

r 2−ω

0

ω0

=r

02

r 2− 1 =

r02

r0+ Δr( )2

− 1 =r

02

r0+αr

0ΔT( )2

− 1 =r

02

r0+αr

0ΔT( )2

− 1

=1

1+αΔT( )2− 1 =

1− 1+ 2αΔT + αΔT( )2( )1+αΔT( )2

=−2αΔT − αΔT( )2

1+αΔT( )2= −αΔT

2 +αΔT

1+αΔT( )2

Now assume that αΔT 1 , and so

Δωω

= −αΔT2 +αΔT

1+αΔT( )2≈ −2αΔT . Evaluate at the

given values. −2αΔT = −2 25 × 10−6 Co( ) 55Co( ) = −2.8 × 10−3

71) Approximately how long should it take 11.0 kg of ice at 0°C to melt when it is

placed in a carefully sealed Styrofoam ice chest of dimensions 25 cm × 35 cm × 55 cm whose walls are 1.5 cm thick? Assume that the heat conductivity of Styrofoam is 0.023J/ s ⋅m ⋅C( ) , latent heat of fusion of ice is

3.33 ⋅105J/kg , and that the outside temperature is 32°C. A) 10h B) 12h C) 16h D) 18h E) 25h Solution This is an example of heat conduction. The heat conducted is the heat released by the melting ice, Q = m

iceL

fusion . The area through which the heat is conducted is the total area of the six surfaces of the box, and the length of the conducting material is the thickness of the Styrofoam. We assume that all of the heat conducted into the box goes into melting the ice, and none into raising the temperature inside the box. The time can then be calculated by equation Qt= kAT1 − T2

l⇒ t = miceLl

kAΔT

t =11.0kg( ) 3.33 ⋅105 J / kg( ) 1.5 ⋅10−2m( )

0.023J / s ⋅m ⋅C ( )( ) 2 0.25m( ) 0.35m( ) + 2 0.25m( ) 0.55m( ) + 2 0.35m( ) 0.55m( )⎡⎣ ⎤⎦ 32C( )

t = 9.0 ⋅104 s ≈ 24.6h

72) A sample of some material is contained in a closed, well-insulated container. Heat is added at a constant rate and the sample temperature is recorded. The resulting data is sketched in the figure. Which of the following conclusions is justified from the data given? A) The heat of fusion is greater than the heat of vaporization. B) At t = 23 minutes the sample was all liquid. C) The heat capacity of the solid phase is greater than that of the liquid phase. D) The sample never boiled. E) At t = 7 minutes the sample was a mixture of solid and liquid. Solution Between 5 and 10 min the material underwent a solid-liquid phase transition. Thus at t = 7 min the sample was a mixture of solid and liquid. At t = 19 min the samples started to boil and thus at t = 23 min was mixture of liquid and gas. More heat is supplied during the liquid-gas phase transition than during solid-liquid phase transition. Therefore, the heat of vaporization is larger than the heat of fusion. Since dT/dQ = 1/(c m) and Q ∝ t, the slope is inversely proportional to the heat capacity c, a smaller slope means a larger heat capacity. The slope dT/dQ is larger for the solid phase than for the liquid phase. Thus, the heat capacity of the liquid phase is greater than the heat capacity of the solid phase. 73) During an isobaric volume expansion of an ideal gas, the average kinetic energy of the gas molecules A) increases. B) does not change. C) decreases. D) may either increase or decrease, depending on whether or not the gas is monatomic. E) may or may not change, but insufficient information is given at make such a determination. Solution The average kinetic energy of an ideal gas is given by KE = nCVT . From the ideal gas law pV = nRT, we see that an increase in volume at constant pressure corresponds to an increase in temperature: dT = p dV / (nR) and thus an increase in the average kinetic energy.

 

 

74) In the figure on the right, a diatomic ideal gas is going clockwise through a cyclic process. Which of the following is an accurate statement? A) The work done in the process is equal to the area under the curve abc. B) The work done in the process is equal to the area enclosed by the cyclic process. C) The work done in the process is equal to the area under the curve adc. D) The work done in the process is zero. E) The work done in the process is equal to the area under ab minus the area under dc. Solution The work done is equal to the area under the curve abc minus the area under the curve cda. This is equal to the area enclosed by the cyclic process. 75) A heat engine takes 9.0 moles of a diatomic ideal gas through the reversible cycle abca, on the p-V diagram, as shown. The path bc is an isothermal process. The temperature at c is 640 K, and the volumes at a and c are 0.03 m3 and 0.22 m 3, respectively. In the figure on the right, for the path ab, the heat absorbed by the gas, in kJ, is closest to: A) 145 B) −100 C) zero D) −145 E) 100 Solution:

From the path ab, ΔV = 0→W = 0→ Q = ΔU

ΔU = nCvΔT = (9 mol)(20.78 J/mol ⋅K)(640 K −0.030.22

× 640 K) = 103 kJ

76) One and one-half moles of an ideal monatomic gas expand adiabatically,

performing 7500 J of work in the process. What is the change in temperature of the gas during this expansion?

A) 200K B) −200K C) 400K D) −400K E) 0 Solution Since the expansion is adiabatic, there is no heat flow into or out of the gas. Use the first law of thermodynamics to calculate the temperature change.

ΔU = Q −W → 32

nRΔT = 0 −W →

ΔT = − 23

WnR

= −2 7500 J( )

3 1.5 mol( ) 8.315J moliK( ) = −401K = −4.0 × 102 K

77) A Carnot engine whose high-temperature reservoir is at 620 K takes in

550 J of heat at this temperature in each cycle and gives up 335 J to the low-temperature reservoir. What is the temperature of the low-temperature reservoir?

A) 224 K B) 310 K C) 340 K D) 378 K E) 412 K Solution TCTH

=QC

QH

⇒ TC = THQC

QH

= 620K 335J550J

= 378K

78) A force probe is constructed in such a way that the force F applied results in

a potential difference V that can be read in a voltmeter. The relation between the force and the potential difference is linear. To calibrate the probe, we observe that when the probe is hanging ready to use but with nothing attached to the hook, the reading is 10 volts. When 5.0 kg are hung from the hook, the reading is 60 volts. When we hang a mass M from the hook, the reading is 120 volts. What is the value of M?

A) 9.0 kg B) 10 kg C) 11 kg D) 12 kg E) 13 kg Solution Let a and b that describe the linear relationship between V and F: V = a F + b. For F = 0, the scale reads V = 10 volts. Thus, b = 10 volts. For F = (5kg)(9.8m/s2) = 49 N, the scale reads 60 volts. Thus, 60 volts = 49N x a + 10N and a = 1.0 volts/N. Therefore, 120 volts = Mg x a + b or M = (120 volts – b) / (a g) = (120 – 10) volts / (1 volts/N 9.8 m/s2) = 11 kg

79) Two scales 1 and 2 are used to measure the weight of an object. The results are shown below, with the maximum errors (the real value is within the quoted range with 100% probability). Scale 1: 50 N ± 4% Scale 2: 45 N ± E What is the minimum value of the error E in case 2 that would allow us to conclude that these measurements are consistent with one another? A) 2 N B) 3 N C) 4 N D) 5 N E) 6 N Solution The smallest the weight can be according to scale 1 is 50N – 50N x 0.04 = 48N. The error in the measurement of scale 2 needs to be at least 45N + E = 48N to be consistent with the measurement of scale 1. Thus, E = 3N. 80) In the Rotational Motion experiment, you observed the rotation of a disk about its axis. Shown below are three graphs of angular velocity versus time. In which case(s) is the disk always moving in the same direction?

A) Case 1 alone. B) Case 2 alone. C) Case 3 alone. D) Cases 1 and 2. E) All of them. Solution The two directions of rotations correspond to positive and negative angular velocities. In cases 1 and 2 the disk is rotating with positive angular velocity only.

t  

ω  

Case  1  

t  

ω  

Case  2  

t  

ω  

Case  3  

81) A glass flask full of air sealed with a well-fitting cap is sitting in a cold-water bath. Hot water is slowly added to the bath, in small amounts and stirred well in between additions. The flask is submerged during the whole process.

Which of the following graphs best describes the contents of the flask?

Solution The air in the flask can be described by the ideal gas law pV = nRT. The flask is sealed so n is constant. The volume of the flask V is constant, too. While the temperature of the bath is increased, the temperature T of the air in the flask, which is in thermal contact with the bath, increases, too. The pressure of the air in the flask p = nRT/V increases linearly with temperature T as shown in figure B.

V  

p  

A  T  

p  

B  T  

V  

C  

V  

p  

D  T  

p  

E  

Cold  water  

Air