Physics 157 Final Exam Review Package - Solutions - UBC ...

Physics 157 Final Exam Review Package - Solutions

UBC Engineering Undergraduate Society

Attempt questions to the best of your ability. This review package consists of 28 pages, including 1 coverpage and 18 questions. Problems are ranked in difficulty as (∗) for easy, (∗∗) for medium, and (∗ ∗ ∗) fordifficult. Difficulty is subjective, so don’t be discouraged if you are stuck on a (∗) problem.

Solutions posted at: https://ubcengineers.ca/tutoring/

If you believe that there is an error in these solutions, or have any questions, comments, or suggestionsregarding EUS Tutoring sessions, please e-mail us at: [email protected]. If you are interested inhelping with EUS tutoring sessions in the future or other academic events run by the EUS, please [email protected].

Want a warm up? Short on study time? Want a challenge?These are the easier problems These cover most of the material These are some tougher questions

1, 2, 4, 5 1, 8, 10, 11, 17, 18 13, 14, 15, 16, 17, 18

Some of the problems in this package were not created by the EUS. Those problems originated from oneof the following sources:

• Fundamentals of Physics / David Halliday, Robert Resnick, Jearl Walker. – 9th ed.

• Exercises for the Feynman Lectures on Physics / Matthew Sands, Richard Feynman, Robert Leighton.

All solutions prepared by the EUS.

EUS Health and Wellness Study Tips

• Eat Healthy—Your body needs fuel to get through all of your long hours studying. You should eata variety of food (not just a variety of ramen) and get all of your food groups in.

• Take Breaks—Your brain needs a chance to rest: take a fifteen minute study break every couple ofhours. Staring at the same physics problem until your eyes go numb won’t help you understand thematerial.

• Sleep—We have all been told we need 8 hours of sleep a night, university shouldn’t change this. Getto know how much sleep you need and set up a regular sleep schedule.

Good Luck!

1

https://ubcengineers.ca/tutoring/

mailto:[email protected]

mailto:[email protected]

Physics 157 Final Exam Review Package Page 2 of 28

1.(∗) A 103 kg non-uniform log hangs by two steel wires, A and B, both of radius 1.20 mm. Initially, wire Awas 2.5 m long, and wire B 2 mm longer than that. The log is now horizontal. If the Young’s modulusfor steel is Y = 200 GPa, what are the magnitudes of the forces on the log from,

(a) wire A?

(b) wire B?

(c) what is the ratio dA/dB?

wire A

wire B

centre of mass

Solution:

• Let A = 0.00122π be the cross sectional area of the wires.

• Let lA = 2.5 m, lB = 2.502 m, and l′A = l′B be the lengths of the wires after stretching.

From the stress/strain relations, we have

FAA

= Yl′A − lAlA

(1.1)

FBA

= Yl′B − lBlB

(1.2)

From the force equilibrium, we have

FA + FB = 103 · g (1.3)

Multiplying both (1.1) and (1.2) through by A, then substituting into (1.3), we then have

FA + FB = AYl′A − lAlA

+AYl′B − lBlB

= 103 · g (1.4)

Substituting in some known numerical values, as well as l′A = l′B ,

l′A − 2.5

2.5+l′A − 2.502

2.502=

103g

AY= 0.00111 (1.5)

(a) From this, we obtainl′A = l′B = 2.502395 m

Substituting this back in to (1.1) and (1.2), we find that

FA = 866 N


(b)FB = 143 N

(c) From the torque equilibrium, we have

dAFA = dBFB (1.6)

sodAdB

=FBFA

= 0.165 (1.7)


2.(∗) Three equal length straight rods, of aluminum, Invar, and steel, all at 20.0 ◦C, form an equilateraltriangle with hinge pins at the vertices. At what temperature will the angle opposite the Invar be 59.95?The coefficients of thermal expansion for aluminum, Invar, and steel, are 23 · 10−6 /◦C , 0.7 · 10−6 /◦C,and 11 · 10−6 /◦C respectively.

Solution:

• Let the length of each bar at 20◦C be L.

• Let L′A be the length of the aluminum bar after changing temperature

• Let L′I be the length of the Invar bar after changing temperature

• Let L′S be the length of the steel bar after changing temperature.

Then, using the thermal expansion formula,

L′A = L(1 + 23 · 10−6∆T ) (2.1)

L′I = L(1 + 0.7 · 10−6∆T ) (2.2)

L′S = L(1 + 11 · 10−6∆T ) (2.3)

Plugging these expanded lengths (all of which depend on ∆T ) into the cosine law,

L′2I = L′2A + L′2S − 2L′AL′S cos(59.95◦) (2.4)

Any second order terms will be ignored because thermal expansion coefficients are already small,and their squares are even smaller. Thus, plugging in (2.1), (2.2), (2.3) into (2.4), we solve for ∆T .The result is

∆T = 48◦C

so then we conclude that the temperature needs to be raised to 68 ◦C.


3.(∗) Given the following stress vs strain curve for a material under tension,find the following values/intervals:

2

Problem 1 A metal specimen was subjected to a tensile test. From the load-deflection data, the stress-strain curve shown in Figure 1 was computed. Note: The bottom curve gives the details of the initial part of the original curve where the strain axis is zoomed in by a factor of 10.

100

200

300

400

Stress σ (MPa)

0.02 (0.002)

0.04 (0.004)

0.06 (0.006)

0.08 (0.008)

0.10 (0.010)

Strain ε 0

(×10 Zoomed on Strain Scale)

Original Scale

Figure 1 Stress versus strain curve of the tensile specimen.

Using the given curves determine the following parameters and indicate the proper units for them: (a) the proportional limit (5 points) (b) the young’s modulus (10 points) (c) the yield strength (by 0.2% offset method) (10 points) (d) the ultimate strength (5 points) (e) the fracture stress (5 points) (f) the % Elongation (at fracture) (5 points) (g) the modulus of resilience (5 points) (h) the modules of toughness (5 points)

(a) Proportional limit

(b) region of elastic behaviour

(c) region of plastic deformation

(d) Young’s modulus of the material

(e) Fracture point

Solution: This question only tests one’s knowledge of the definitions of the terms described in thetextbook about stress-strain curves.

(a) Proportional limit is at strain = 0.004

(b) region of elastic behaviour is for strain between 0 and 0.004

(c) region of plastic deformation is for strain between 0.004 and 0.1 (this is where the materialwould not return to regular length if released from tension)

(d) Young’s modulus is the slope of the curve in the elastic behaviour region, so

Y =225 MPa

0.004= 56.3 GPa

(e) Fracture point is at strain = 0.1 (this is where the material breaks)


4.(∗) A rigid beam ABC of length 2L is hinged to a fixed rigid wall at end A and horizontally supported usinga vertical hanger rod hinged at the other end C. The hanger rod DC is uniform with cross sectional areaA = 100mm2, and length l = 3 m. The end D of the hanger rod is hinged to a fixed rigid ceiling asshown. A vertical downward load P = 70 kN is applied at the mid-point B of the beam (AB = L =BC). The material of the hanger rod has obeys the stress-strain curve shown. Determine the followingvalues:

(a) Normal (tensile) stress in the hanger rod

(b) Downward movement at the point of load application (downward movement at point B)

2

Problem 1 A rigid beam ABC of length 2L is hinged to a fixed rigid wall at end A and horizontally supported using a vertical hanger rod hinged at the other end C. The hanger rod DC is uniform with area of X-section 2100mmA , and length l=3m. The end D of the hanger rod is hinged to a fixed rigid ceiling (see Figure 1(a)). A vertical downward load P=70kN is applied at the mid-point B of the beam (AB = L = BC). The material of the hanger rod obeys the stress-strain curve shown in Figure 1(b). Determine: (a) Normal (tensile) stress in the hanger rod. (20 points)

(b) Downward movement at the point of load application (B). (20 points)

Note 1: Assume that the beam is light (or its weight is included in P). Note 2: Assume that the movements due to P are small compared to the

dimensions of the beam and the rod (Hence, the circular arc of movement of the beam at C may be assumed to correspond to the extension of the rod DC. Similarly, the circular arc of movement of the beam at B may be assumed to be approximately vertical).

Note: 3: Under the present loading condition, the hanger rod may not be in the

linear region of the stress-strain curve. (a)

BL L

C

A

2100mmA

Hanger Rod

l=3m

Beam70kNP

D

3

(b)

0

100

200

300

400

0 0.001 0.002 0.003 0.004Normal Strain

(m/m)

Normal Stress V

(MPa)

Figure 1: (a) A rigid beam supported by a hanger rod; (b) Stress versus strain curve

of the hanger rod. Problem 2 A shaft ABC of circular X-section is made by welding a steel segment AB of length 4L=4m and radius 50mmsa to a brass segment BC of length L=1m and radius

25mmba . The two ends A and C are rigidly fixed (Figure 2). A torque BT is applied to the shaft at B. Find: (a) Maximum torque maxBT that can be applied at B without exceeding the allowable

levels of shear stress. (20 points)

(b) Corresponding reaction torques AT and TC at the two ends of the shaft. (10 points)

(c) Corresponding maximum shear stresses in the two segments of the shaft. (15 points)

(d) Corresponding angular movement BI at joint B. (15 points)

Note: The allowable shear stresses for the steel segment and the brass segment are, respectively: 50MPasallwW , 25MPaballwW . The shear moduli of the two segments are:

80GPasG , 40GPabG .

Solution:

(a) Let Ay be the upward force on the horizontal rod at A, and Cy be the upward force on thehorizontal rod at C. Then

Ay + Cy = P

and by symmetry, Ay = Cy = P/2.The stress in the hanger rod will be 35 · 103/100 · 10−6 = 350 MPa.

(b) Note that the value of 350 MPa falls outside the elastic region. Thus we have to use the graphvalue to find the strain at this value of stress. We can see that we have a strain of 0.002, so then

δ = 0.002l = 0.002(3) = 0.006 m

Thus point C moves down by 6 mm. This means that point B moves down by 3 mm, since itis halfway along the rod. (Later, in Solid Mechanics, you will learn that this isn’t actually thecase, but for now, this approximation will do)


5.(∗) In the figure, string 1 has a linear density of 3.00 g/m, and string 2 has a linear density of 5.00 g/m.They are under tension due to the hanging block of mass M = 500 g. Calculate the wave speed on

(a) string 1

(b) string 2

(c) Find M1,M2 such that the wave speeds in the two strings are equal.

Solution: Both strings will have tension T = 0.25g in them.

(a)

v1 =√T/µ1 =

√0.25g/0.003 = 28.6 m/s

(b)

v2 =√T/µ2 =

√0.25g/0.005 = 22.1 m/s

(c) We set the two speeds equal, so v1 = v2. Thus√T1µ1

=

√T2µ2

Since the block is divided, we have a system of equations: M1 +M2 = 0.5T1 = M1gT2 = M2g

Solving the system of equations,M1 = 188 grams

M2 = 312 grams


6.(∗∗) A heat engine is proposed that operates with heat reservoirs at temperatures T1 and T2, using amonatomic ideal gas as the working fluid. The engine cycle consists of the following three steps:

(i) Expansion at constant pressure, P , from V1 and T1 to V2 and T2.

(ii) Cooling at constant volume from T2 to T1

(iii) Reversible isothermal compression at T1 from V2 to V1

(a) Draw the cycle on a P–V diagram.

(b) Calculate the heat and work for each step in the cycle.

(c) Derive an expression for the efficiency of the engine in terms of T2 and T1.

(d) For T2 = 600 K and T1 = 300 K, what is the efficiency of this engine? How does this compare withthe efficiency of a Carnot cycle engine operating between reservoirs at the same two temperatures?

Solution:

(a)V

P

P

V₁ V₂

T₂

T₁

(b) (i) This process is isobaric, so the equations for W and Q will be:

W = P (V2 − V1) (6.1)

Q = U +W = nCP (T2 − T1) (6.2)

(ii) This process is isochoric, so there is no work done.

W = 0 (6.3)

Q = nCV (T1 − T2) (6.4)

(iii) This process is isothermal, so then there is no change in internal energy.

W = nRT1 ln

(V1V2

)(6.5)

Q = nRT1 ln

(V1V2

)(6.6)


(c) Only some of these Q’s are positive. Only the ones that are positive contribute to the heatadded. The positive one is the one from Step (i).

e =

∑W∑QH

=P (V2 − V1) + nRT1 ln(V1/V2)

nCP (T2 − T1)(6.7)

Now we need to perform some substitutions so that the answer is only in terms of T1 and T2.Since CP = 5R/2 (monatomic gas), and we have the ideal gas law

PV1 = nRT1

PV2 = nRT2

Which means that we have the ratioV1V2

=T1T2

we can do some substitutions and cancellations to find:

e =T2 − T1 + T1 ln(T1/T2)

(5/2)(T2 − T1)(6.8)

(d) Plugging in those values, we obtain

e =2(1 + ln(0.5))

5= 12.3%

The Carnot efficiency is 50%, which is significantly higher than what was obtained for thisparticular engine.


7.(∗∗) A sinusoidal wave of angular frequency 1200 rad/s and amplitude 3.00 mm is sent along a cord withlinear mass density 2.00 g/m and tension 1200 N.

(a) What is the average rate at which energy is transported by the wave to the opposite end of thecord?

(b) If, simultaneously, an identical wave travels along a separate, identical cord, what is the totalaverage rate at which energy is transported to the opposite ends of the two cords by the waves?

(c) If, instead, those two waves are sent along the same cord simultaneously, what is the total averagerate at which they transport energy when their phase difference is

(i) 0

(ii) 0.4π rad

(iii) π rad

Solution:

(a) We use the formula for power transfer along a string, so

P =1

2µω2A2v = 10 W

(b) The total power transferred will just be double of the previous one, so 20 W. This is becausethere is no superposition involved, since the waves travel on distinct strings.

(c) (i) The amplitudes will add together, so we will have double the amplitude and thus fourtimes the power of part (a). So it becomes 40 W.

(ii) If we add two waves, we will have to use the trigonometric identity

sin θ1 + sin θ2 = 2 cos

(θ1 − θ2

2

)sin

(θ1 + θ2

2

)Since θ1 = ωt, and θ2 = ωt+0.4π, we have 2A cos(0.2π) as the new amplitude, so pluggingthis amplitude into the formula yields

P = 26.3 W

(iii) There will be destructive interference all along the wire, so then no power is transferredand

P = 0 W


8.(∗∗) Consider a planet in a binary star system. There are two stars, A and B. The planet is rA away fromstar A, which emits total power PA. The planet is rB away from star B, which emits total power PB .The planet has albedo α and emissivity e. What is the temperature T of the planet? Express youranswer in terms of the given variables.

Solution: The intensities of radiation of each star at the planet are

IA =PA

4πr2A(8.1)

IB =PB

4πr2B(8.2)

Let S be the total surface area of the planet. The power absorbed by the planet is given by

Pin =S

4(1− α)(IA + IB) (8.3)

=S(1− α)

16π

[PAr2A

+PBr2B

](8.4)

Note that we use S/4 because the cross sectional area of a sphere is a quarter of the sphere’s totalsurface area. The power emitted by the planet is given by

Pout = eσST 4 (8.5)

Because the planet is in thermal steady state, we equate these two quantities.

Pin = Pout (8.6)

S(1− α)

16π

[PAr2A

+PBr2B

]= eσST 4 (8.7)

Solving for T yields

T = 4

√(1− α)

16πeσ

[PAr2A

+PBr2B

](8.8)


9.(∗∗) A girl is sitting near the open window of a train that is moving at a velocity of 10.00 m/s to the East.The girl’s uncle stands near the tracks and watches the train move away. The locomotive whistle emitssound at frequency 500.0 Hz. The air is still.

(a) What frequency does the uncle hear?

(b) What frequency does the girl hear?

A wind begins to blow from the east at 10.00 m/s.

(c) What frequency does the uncle now hear?

(d) What frequency does the girl now hear?

Solution: The shifted frequency due to Doppler shift is given by

f ′ = f ·(v ± vDv ∓ vS

)(9.1)

where f is the unshifted frequency, v is the speed of the wave (speed of sound), vD is the speed ofthe detector/observer, and vS is the speed of the source. We will take velocities with respect to theuncle/ground.

(a) The uncle (detector/observer) is at rest with respect to the ground, so vD = 0. The speed ofthe source is vS = 10 m/s. The train is moving away, choose the plus sign for the denominator.Thus

f ′ = 500

(343

343 + 10

)= 485.8 Hz

(b) The girl is now the detector/observer. She is in the same reference frame as the locomotive’swhistle, and thus has no relative velocity to the whistle. Thus (v + vD) = (v + vS) and

f ′ = f = 500.0 Hz

(c) For parts (c) and (d) we will have to take velocities relative to the air. This is because the waveis moving with its wave speed 343 m/s with respect to the air, and we can then use the familiarformula. Relative to the air the locomotive is moving at vS = 20.00 m/s away from the uncle.Use the plus sign in the denominator. Relative to the air the uncle is moving at vD = 10.00 m/stoward the locomotive. Use the plus sign in the numerator. Thus

f ′ = f ·(v ± vDv ∓ vS

)= 500

(343 + 10

343 + 20

)= 486.2 Hz

(d) Relative to the air the locomotive is moving at vS = 20.00 m/s away from the girl and the girlis moving at vD = 20.00 m/s toward the locomotive. Use the plus signs in both the numeratorand the denominator. Thus (v + vD) = (v + vS) and

f ′ = f = 500.0 Hz


10.(∗∗) In the figure, an aluminum wire, of length L1 = 60.0cm, cross-sectional area 1.00 · 10−2 cm2, and density2.60 g/cm3, is joined to a steel wire, of density 7.80 g/cm3 and the same cross-sectional area. Thecompound wire, loaded with a block of mass m = 10.0 kg, is arranged so that the distance L2 from thejoint to the supporting pulley is 86.6 cm. Transverse waves are set up on the wire by an external sourceof variable frequency; a node is located at the pulley.

(a) Find the lowest frequency that generates a standing wave having the joint as one of the nodes.

(b) How many nodes are observed at this frequency? (including the nodes at the endpoints)

Solution: Note: In this solution, m is not the mass of the block featured in the figure, it is aninteger indexing the frequencies of standing modes of the string.

(a) • The frequency of the wave is the same for both sections of the wire

• The wave speed and wavelength can be different in the different sections

• We have the equation fm = mv/2L for the frequency of the mth standard mode, and wewant it to be satisfied in both sections, where m is a positive integer.

Since we want these frequencies to be the same in both sections (otherwise the wire would bediscontinuous!), we have

mv12L1

=nv22L2

(10.1)

where m is the integer corresponding to wire L1 and n is the integer corresponding to wire L2.In order to compute the velocity of the wave in each medium, we must first compute the linearmass densities µ in each medium. We have:

µ1 = Aρ1 = 0.0026 kg/m

µ2 = Aρ2 = 0.0078 kg/m

Now we have (for a tension T = mg)

v1 =√T/µ1 = 194.1 m/s

v2 =√T/µ2 = 112.1 m/s

Substituting these in to (7.1), we have m = 0.4n, or

5m = 2n

The smallest integer solution to this equation is m = 2, n = 5. Thus the frequency will be

f =mv12L1

= 324 Hz

(b) Since m = 2 corresponds to 3 nodes, and n = 5 corresponds to 6 nodes, but there is one nodeshared by the two sections (where L1 meets L2), there are 8 nodes in total.


11.(∗∗) Two bulbs of volumes 200 cm3 and 100 cm3, as shown, are connected by a short tube containing aninsulating porous plug that permits equalization of pressure but not of temperature between the bulbs.The system is sealed at 27◦C when it contains oxygen under atmospheric pressure. The small bulb isimmersed in an ice bath at 0◦C and the large bulb is placed in a steam bath at 100◦ C. What is thefinal pressure P inside the system? Neglect thermal expansion of the bulbs.

Solution: We will call the smaller volume (1), and the larger volume (2).Then we can write the ideal gas law equations:

P1V1 = n1RT1 (11.1)

P2V2 = n2RT2 (11.2)

Some known variables are

• P1 = P2 = 101.3 kPa

• T1 = T2 = 300 K

Let the primed variables be the values at equilibrium. Then we can write the ideal gas law for eachbulb at equilibrium:

PV1 = n′1RT′1 (11.3)

PV2 = n′2RT′2 (11.4)

Since the temperatures at equilibrium will be the same as the baths, we know that T ′1 = 273 K, andT ′2 = 373 K. Let n = n1 +n2 = n′1 +n′2 be the total number of moles in the system. Using the idealgas law, we can calculate the number of moles initially in each bulb. To do that, plug

• V1 = 0.0001 m3

• V2 = 0.0002 m3

• P1 = P2 = 101.3 kPa

• T1 = T2 = 300 K

n1 = 0.00406 mol, and n2 = 0.00813 mol, so n = 0.0122 mol. Thus we three have equations in threeunknowns, n′1 + n′2 = 0.0122

PV1 = n′1RT′1

PV2 = n′2RT′2

Solving the system, we obtainP = 112.2 kPa


12.(∗∗) Leidenfrost effect : A water drop that is slung onto a skillet with a temperature between 100◦C and about200◦C will last about 1 s. However, if the skillet is much hotter, the drop can last several minutes, aneffect named after an early investigator. The longer lifetime is due to the support of a thin layer of airand water vapour that separates the drop from the metal (by distance L as shown). Let L = 0.100 mm,and assume that the drop is flat with height h =1.50 mm and bottom face area A = 4.00 ·10−6 m2. Alsoassume that the skillet has a constant temperature Ts = 300◦C and the drop has a temperature of 100◦C.Water has density ρ = 1000 kg/m3, latent heat of vaporization Lv = 2265 J/g, and the supporting layerhas thermal conductivity k = 0.026 W/m ·K.

(a) At what rate is energy conducted from the skillet to the drop through the drop’s bottom surface?

(b) If conduction is the primary way energy moves from the skillet to the drop, how long will the droplast?

Solution:

(a) The rate of energy flow through the surface is given by

H =kA∆T

L(12.1)

Plugging in the given numbers,

H =(0.026)(4.00 · 10−6)(300− 100)

0.0001= 0.21 W

(b) First we find mass of the droplet, which is given by:

m = ρAh = 6 · 10−6 kg

Then we observe that the heat transferred can be related to the latent heat of vapourization

Q = Ht = Lvm = (6 · 10−6)(2265 · 103)

so thent = 65 s


13.(∗ ∗ ∗) A massless rod with mass M on the end of it, length L swings as a pendulum with two horizontal springsof negligible mass and constants k1 and k2 acting at the bottom end, as shown. Both springs are relaxedwhen the rod is vertical. What is the period P of small oscillations? Express your answer in terms ofM , L, k1, and k2.

Solution: There will be a tension T in the rod, and thus the gravitational force on the mass Mg isbalanced by T cos θ, where θ is the angle between the rod and the vertical. Since θ is small, T ≈Mg.Thus the horizontal force on the mass due to the rod will be T sin θ ≈ Mgθ = Mgx/L. Thus thetotal force on the mass will be

F = − [Mgx/L+ (k1 + k2)x] (13.1)

Setting up Newton’s second law F = ma = md2x

dt2,

Md2x

dt2+Mgx

L+ (k1 + k2)x = 0 (13.2)

d2x

dt2+gx

L+

(k1 + k2)x

M= 0 (13.3)

d2x

dt2+ x

(g

L+k1 + k2M

)= 0 (13.4)

Thus

ω =

√g

L+k1 + k2M

=

√Mg + L(k1 + k2)

ML(13.5)

By the definition of angular frequency,

P =2π

ω(13.6)

this means that

P = 2π

√ML

Mg + L(k1 + k2)(13.7)


14.(∗ ∗ ∗) Suppose we expand 1.00 mol of a monatomic gas initially at 5.00 kPa and 600 K from initial volumeVi = 1.00 m3 to final volume Vf = 2.00 m3. Suppose that at any instant during the expansion, thepressure P and volume V of the gas are related by P = 5e(Vi−V )/a, with P in kilopascals, Vi and V incubic metres, and a = 1.00 m3. What are

(a) The final pressure

(b) Final temperature

(c) Find the work done by the expansion

(d) What is ∆S for the expansion?

Solution: Let n = 1, Pi = 5, Ti = 600, and both Pf , Tf unknown.

(a) The final pressure is

P (2) = 5e1−2 =5

ekPa = 1839.4 Pa

(b) To find the temperature, plug the final pressure from part (a) into the ideal gas law.

PfVf = nRTf (14.1)

This produces

Tf =10000

eR= 441 K

(c) We use the knowledge of antiderivatives learned in Math 100.The work is given by

W =

∫PdV (14.2)

= 5

∫eVi−V dV (14.3)

= −5eVi−V + C (14.4)

To solve for the constant C, we note that no work has been done before the gas has beenexpanded, so W (Vi) = 0. This means that C = 5. The work is then

W = −5eVi−V + 5 (14.5)

We evaluate at Vf , so

W (Vf ) =−5

e+ 5 kJ

Then we haveW = 5000(1− 1/e) J = 3160.6 J

(d) To find ∆S, we will have to recall how the work formula for a isothermal expansion is derived,and use that information to find the formula for ∆S in this case.


Remark 1. Recall the process of finding the work for an isothermal process (in the textbook).For isothermal process,

W =

∫PdV (14.6)

=

∫nRTdV

V(14.7)

= nRT ln

(VfVi

)(14.8)

We can use a similar procedure to find the formula for part (d). From the first law of thermo-dynamics,

dQ = dW + dU (14.9)

the second law of thermodynamics

dS =dQ

T(14.10)

We can combine (3.9) and (3.10) to make

dS =dW + dU

T(14.11)

Thus

∆S =

∫dW

T+dU

T(14.12)

=

∫PdV

T+nCV dT

T(14.13)

=

∫nRdV

V+nCV dT

T(14.14)

= nR ln

(VfVi

)+ nCV ln

(TfTi

)(14.15)

Note that this is the same form as the work formula that we’ve been given, so we can logicallyargue that a result of the same form is obtained here.Therefore

∆S = R ln

(2

1

)+

3R

2ln

(10000

600eR

)= 1.94 J/K


15.(∗ ∗ ∗) A crude musical instrument is constructed by stretching a wire of negligible mass under tension Tbetween two points and firmly attaching a mass m to the wire at a distance x from one end, as shownin figure. The mass is displaced from equilibrium by a small distance A (i.e. A � x,A � l − x), andthen allowed to vibrate. Neglect the force of gravity.

(a) Find the frequency ν of the sound.

(b) Write the equation for the displacement of the mass from equilibrium as a function of time y(t).

(c) As x is varied, what are the minimum and maximum frequencies available?

Express your answers in terms of T , m, l, x, and A.

Solution:

(a) Let θ1 be the angle that the string on the left makes with the horizontal, and θ2 be the angle thatthe string on the right makes with the horizontal. Since the tension T is uniform throughoutthe wire, the force Fy on the mass m is given by

Fy = −(T sin θ1 + T sin θ2) (15.1)

= −T(y

x+

y

l − x

)(15.2)

= −y T l

x(l − x)(15.3)

Setting up Newton’s second law, Fy = ma = md2y

dt2

md2y

dt2+ y

T l

x(l − x)= 0 (15.4)

d2y

dt2+ y

T l

mx(l − x)= 0 (15.5)

As discussed in class, the coefficient of y is ω2. Thus we find ω to be:

ω =

√T l

mx(l − x)= 2πν (15.6)

sTherefore

ν =1

2π

√T l

mx(l − x)(15.7)


(b) The general equation for oscillation will then be:

y(t) = A cos(ωt) (15.8)

= A cos

(t

√T l

mx(l − x)

)(15.9)

Note that you could have used sine instead of cosine and it would also be completely correct.

(c) When x is taken close to 0 or l, the frequency becomes unbounded and races off towards infinity,so

νmax →∞

When x is l/2 (mass is in the middle) we have

νmin =1

π

√T

ml


16.(∗ ∗ ∗) A cylinder, filled with argon, is equipped with a spring loaded piston of mass m and area A. Atequilibrium the argon is at total pressure P0, the piston is distance L0 from either end of the system asshown, and the spring (constant K) is compressed by x0 (its free length being L0+x0). Find the angularfrequency ω of small oscillations (x0 � L0) of the piston if the gas compresses isothermally. Hint: Youmay use 1/(1 + x) ≈ 1 for x� 1. Express your answer in terms of K, P0, A, L0, and m.

Solution: At equilibrium, the forces on the piston are balanced, so then P0A = Kx0. For conve-nience, set x = 0 at the equilibrium point. Since the process is isothermal, we know that (for thepressure and volume inside the argon filled cylinder)

P0V0 = P (x) · V (x) (16.1)

P0AL0 = P (x)A · (L0 + x) (16.2)

To find the contribution to the force that this pressure (caused by a small displacement x) will exerton the piston, we take

P0 − P (x) = P0 −P0L0

L0 + x(16.3)

=P0x

L0 + x(16.4)

Thus the total force on the piston will be the sum of the spring force Kx and the force due to thepressure change (11.5)

Kx+AP0x

L0 + x= x

(K +

P0A

L0 + x

)(16.5)

Since we can approximate 1/(1 + x) ≈ 1 for x� 1, we can perform the following manipulation:

F = x

(K +

P0A

L0(1 + x/L0)

)(16.6)

≈ x

(K +

P0A

L0

)(16.7)

Then using Newton’s second law, we have F = ma = md2x

dt2, which, for this physical scenario, is

md2x

dt2+ x

(K +

P0A

L0

)= 0 (16.8)

d2x

dt2+ x

(K + P0A

L0

)m

= 0 (16.9)


The coefficient of the x will be ω2, so we have

ω2 =

(K + P0A

L0

)m

(16.10)

and the final answer becomes

ω =

√√√√(K + P0AL0

)m

(16.11)


17.(∗ ∗ ∗) An object of mass 5.0 kg is found to oscillate with neglible damping when suspended from a spring whichcauses it to perform 10 complete cycles in 10.0 seconds. Thereafter, a certain small magnetic dampingis applied, proportional to the velocity of the motion and the amplitude decreases from an initial 0.2 mto 0.1 m in 10 cycles.

(a) Write the differential equation associated with this motion, with the coefficients of d2x/dt2, dx/dtand x represented by numbers in SI units.

(b) What is now the period T ′ of the motion? Find T ′ to six decimal places.

(c) In how many cycles, N (starting from the 0.2 m amplitude) will the amplitude reach

(i) 0.05 m?

(ii) 0.02 m?

(d) What is average rate of energy dissipation P by damping during the first cycle?

Solution:

(a) The general form of the differential equation for the motion is

md2x

dt2+ b

dx

dt+ kx = 0 (17.1)

It is clear that the mass m = 5. Since T = 1 because the oscillation does 10 cycles in 10 seconds,we have that since

ω =2π

T(17.2)

plugging in T = 1 for the undamped motion, we find

ω0 = 2π (17.3)

which is the angular frequency for the undamped motion. Since

ω20 =

k

m(17.4)

we havek = 20π2

To find b is more complicated. When the damping force is applied, the period will change, so10 cycles no longer means 10 seconds. For damped motion we will have a new period T ′, withcorresponding angular frequency ω. This new ω is given by

ω =

√ω20 −

(b

2m

)2

(17.5)

=

√(2π)2 −

(b

10

)2

(17.6)

Since (12.2) still holds, but this time for different period and angular frequency, we have

T ′ =2π

ω(17.7)

Plugging (12.6) into (12.7), we have

T ′ =2π√

(2π)2 −(b10

)2 (17.8)


The amplitude of the motion as a function of time is

A(t) = A0e−bt/2m (17.9)

and since the amplitude of the motion drops from 0.2 m to 0.1 m after 10 cycles (10 periods),we have so we will have

0.1 = 0.2e−b10T′/10 (17.10)

Rearranging,

ln

(1

2

)= −bT ′ (17.11)

Plugging (12.8) into this, we have

ln(2) =b(20π)

10

√(2π)2 −

(b10

)2 (17.12)

Solving for b, we find thatb = 0.693 kg/s

Thus the differential equation is

5d2x

dt2+ 0.693

dx

dt+ 20π2x = 0 (17.13)

(b) Now that we have b, we can plug it back in to the (12.6) to find ω.

ω =

√(2π)2 −

(0.693

10

)2

= 6.282803 rad/s

and thus, by (12.7),T ′ = 1.000061 s

(c) (i) Setting the amplitude to 0.05, we obtain

0.05/0.2 = e−bNT′/10

and since we already know b and T ′, we solve for N to find

N = 20 cycles

(ii) A similar calculation to (i) will yield

N = 33 cycles

(d) We must find the initial energy E0. This can be calculated with the formula for spring potentialenergy.

E0 =kA2

0

2=

(20π2)(0.2)2

2=

2π2

5Since

E(t) = E0e−bt/m (17.14)

we take the difference quotient

E(T ′)− E0

T ′= −0.2644 J/s (17.15)

Thus the average rate of energy dissipation over the first period will be 0.2644 J/s.


18.(∗ ∗ ∗) Energy can be removed from water as heat at and even below the normal freezing point (0.0 ◦C atatmospheric pressure) without causing the water to freeze; the water is then said to be supercooled.Suppose a 1.00 g water drop is supercooled until its temperature is that of the surrounding air, whichis at −5.00 ◦C. The drop then suddenly and irreversibly freezes, transferring energy to the air as heat.What is the entropy change for the drop? The specific heat of ice is 2220 J/K·kg, the specific heat ofwater is 4186 J/K·kg, and the latent heat of fusion of ice is 334 kJ/kg.

Solution: Since entropy is a state function, this means that the change in entropy is independentof the path taken between the initial state and the final state. For this problem, we consider a threestep process:

(i) In the first process, consider the water drop heating up to 0 ◦C

(ii) In the second process, consider the water drop freezing at constant temperature of 0◦C

(iii) In the third process, consider the water drop (now as a solid ice) cooling down to its originaltemperature

First, we need to find out how to compute the change in entropy for a non-constant temperatureprocess. This can be done in the following way: Since

dS =dQ

T(18.1)

anddQ = mcdT (18.2)

we can plug (17.2) into (17.1):

dS =mcdT

T(18.3)

Taking the antiderivative of both sides, we have

∆S = mc ln

(TfTi

)(18.4)

Now we are ready to calculate the appropriate entropies:

(i)

∆S1 = mc ln

(TfTi

)= (0.001)(4168) ln

(273

268

)= 0.077 J/K

(ii)

∆S2 = −mLT

= −0.001(334000)/273

= −1.22 J/K


(iii)

∆S3 = mc ln

(TfTi

)= (0.001)(2220) ln

(268

273

)= −0.041 J/K

Summing these together,

∆S = ∆S1 + ∆S2 + ∆S3

= 0.077− 1.22− 0.041

= −1.18 J/K


Useful Constants and Conversion Ratios:R = Ideal Gas constant = 8.31451 J/molK, 1 atm = 1.013× 105 Pa, 1 atm · litre = 101.3 Jσ = Stefan-Boltzmann constant = 5.6704× 10−8 W/m2K4, γair = 1.4, CVair

= 20.8 J/molKρwater = Density of water = 1 gram/cm3 = 1000 kg/m3

Mechanics:Linear Motion: x = x0 + 1

2 (v0 + v)t, x = x0 + v0t+ 12at

2, v = v0 + at, v2 = v20 + 2a(x− x0)

Circular Motion: ac =v2

r

Forces: F = ma =d

dtp, Friction: |F| = µ|N|, Spring: F = −kx, Damping: F = −bv

Buoyant |F| = ρV g

W = Work =

∫ rf

ri

F · dr = F ·∆r, K = 12mv

2, ∆Ugravity = mg∆h, ∆Uspring = 12kx

2

P =dW

dt= F · v

Thermodynamics:

Thermal Expansion: ∆L = αL0∆T , Stress and Strain:|F|A

= Y∆L

L, Ideal Gas Law: PV = nRT

Kav = 32kT

Thermal Conductivity: I =∆Q

∆t= kA

∆T

∆xBlack Body Radiation: P = eσAT 4, λmaxT = 2.8977685× 10−3m ·KInternal Energy: U = nCV TFirst Law of Thermodynamics: dQ = dU + dW For an ideal gas, dW = PdVWork for an isothermal process W = nRT ln(Vf/Vi)Work for an adiabatic expansion TV γ−1 = constant, if the number of moles is constant PV γ = Cwhere C is a constant and γ = CP /CV

Work for adiabatic process: W =

∫ V2

V1

PdV = C

∫ V2

V1

dV

V γ=

C

1− γ(V 1−γ

2 − V 1−γ1 )

Heat Transfer: Q = mc∆T , Q = mL, CP = CV +R, CV =f

2R, where f = degrees of freedom.

f = 3 for monatomic and f = 5 for diatomic.

dS =dQ

T

e = W/QH , COPCooling =|QC ||W |

, COPHeating =|QH ||W |

, eCarnot = 1− TCTH

Integrals:∫xndx =

xn+1

n+ 1+ C, n 6= 1

∫x−1dx = lnx+ C

Trigonometry:

sin θ1 + sin θ2 = 2 cos

(θ1 − θ2

2

)sin

(θ1 + θ2

2

)Area and Volume:Surface Area of a sphere: A = 4πr2. Lateral surface area of a cylinder: A = 2πrl.Area of a circle: A = πr2. Volume of a cylinder: V = lπr2 Volume of a sphere: V = 4

3πr3

Oscillations:

ω = 2πf , T =1

f, x = A cos(ωt+ φ), ω2 =

k

m

Damped Oscillations: x = A0e− bt

2m cos(ωt+ φ), where ω =

√w2

0 −(

b

2m

)2

, Q = 2πE

∆E

Energy for damped E = E0e− bt

m


Waves:

v =

√T

µ, k =

2π

λ, P = 1

2µω2A2v, po = ρωvs0

v =

√γRT

M, I =

Pav

4πr2, β = 10dB log10

(II0

), Doppler Effect f ′ = f0

(v ± vLv ∓ vS

)Beats: ∆f = f2 − f1, y = A cos(kx∓ ωt+ φ)Interference: k∆x+ ∆φ = 2πn or π(2n+ 1), n = 0,±1,±2,±3,±4, . . .

Standing Waves fm =mv

2L, m = 1, 2, 3, . . . , fm =

mv

4L, m = 1, 3, 5, . . .

Constants:

k =1

4πε0≈ 9× 109 Nm2/C2, ε0 = 8.84× 10−12 C2/Nm2, e = 1.6× 10−19 C

µ0 = 4π × 10−7 Tm/A, c =1

√ε0µ0

= 299, 792, 458 m/s

Point Charge:

|F| = k|q1q2|r2

, |E| = k|q|r2

, V =kq

r+ Constant

Electric potential and potential energy ∆V = Va − Vb =

∫ b

a

E · dl = −∫ a

b

E · dl

Ex = −dVdx

, E = −∇V , ∆U = Ua − Ub = q(Va − Vb)Maxwell’s Equations: ∫

S

E · dA =Qenc

ε0= 4πkQenc

∫S

B · dA = 0

∫C

B · dl = µ0 (Ienclosed) + ε0µ0dΦEdt

∫C

E · dl = −dΦBdt

Where S is a closed surface and C is a closed curve. ΦE =

∫E · dA and ΦB =

∫B · dA

Energy Density:

uE =1

2ε0E

2 and uB =1

2µ0B2 (energy per volume)

Forces:F = qE + qv×B, F = IL×BCapacitors:

q = CV , UC =1

2· q

2

C, For parallel plate capacitor with vacuum (air): C =

ε0A

d, Cdielectric = KCvacuum

Inductors:EL = −LdIdt , UL = 1

2LI2, where L = NΦB/I and N is the number of turns.

For a solenoid B = µ0nI where n is the number of turns per unit length.DC Circuits: VR = IR, P = V I, P = I2R(For RC circuits) q = ae−t/τ + b, τ = RC, a and b are constants(For LR circuits) I = ae−t/τ + b, τ = L/R, a and b are constantsAC circuits: XL = ωL, XC = 1/(ωC), VC = XCI, VL = XLI

V = ZI, Z =√

(XL −XC)2 +R2, Paverage = I2rmsR, Irms =Imax√

2

If V = V0 cos(ωt), then I = Imax cos(ωt− φ), where tanφ =XL −XC

R,Pav = VrmsIrms cosφ

Additional Equations: dB =µ0

4π· Idl× r

r3

LRC Oscillations: q = A0e−Rt

2L cos(ωt+ φ), where ω =

√ω20 −

(R2L

)2and ω2

0 = 1LC

Physics 157 Final Exam Review Package - Solutions - UBC ...

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