Physics 211 Practice Final NAME: SECTION:

Physics 211 Practice FinalAugust 31, 2011

useful constants: cw = 4186J/kgK, Lf = 3.33 × 105J/kg, Lv = 22.6 × 105J/kg, cice =2100J/kgK, kB = 1.38 × 10−23J/K, σ = 5.67 × 10−8W/(m2K4), ρsteel = 7.8 × 103kg/m3,Ysteel = 20 × 1010N/m2, αsteel = 11 × 10−6C−1, csteel = 460J/kgK, ksteel = 46W/m◦C,Yal = 7.0 × 1010N/m2, αal = 24 × 10−6C−1, cal = 920J/kgK, kal = 240W/m◦C, Yglass =5.7 × 1010N/m2, αglass = 9.0 × 10−6C−1, cglass = 840J/kgK, kglass = 0.84W/m◦C, STP :T = 0◦C and P = 1atm = 101.3kPa.

1. (10 points) Micheal Ballack is kicking a soccer ball that is headed by Lucas Podolskia horizontal distance of 18.0m away, Podolski heads the ball while it is 2.00m abovethe ground. If the ball is headed 0.800s after being kicked what is the ball’s initialvelocity? ~v0 =SOLUTION

given:x0 = 0m xf = 20.0m y0 = 0.0myf = 2.00m t = 0.800s ay = −g = −9.8m/s2

v0 =?find inital velocities in the x and y directions before adding them together.

d(t) = d0 + v0t+1

2at2 (1)

v0 =d(t) − d0

t− 1

2at (2)

v0x =xft

=20.0m

0.500s= 10.3m/s(9.21m/s)(8.92m/s) (3)

v0y =yf − y0

t+

1

2gt (4)

= 22.5m/s(25.0m/s)(25.0m/s) (5)

v =√v20x + v20y = 24.8m/s(26.6m/s)(26.5m/s) (6)

θ = tan−1(6.42m/s

25.0m/s) = 24.7◦ (7)

Answer: ~v = 40.9m/s at θ = 11.9◦

2. (10 points) A uniform solid cylinder (mass 100.kg and radius 50.0cm) is mounted soit is free to rotate about a fixed horizontal axis that passes through the centers ofthe cylinders ends. A 10.0kg block is hung from a massless cord wrapped around thecylinder’s circumference. When the block is released the cord unwinds and the blockaccelerates downward. What is the block’s acceleration? a =

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Physics 211 Practice Final NAME: SECTION:

Solution:Given: M = 100.kg R = 0.500m m = 10.0kg g = 9.8m/s2∑

τ = rFT = Iα (8)∑F = FT −mg = −ma (9)

a = rα (10)

I =1

2MR2 (11)

R(mg −ma) =1

2MR2 a

R(12)

mg =1

2Ma−ma (13)

a =2m

M + 2mg = 1.63m/s2 (14)

3. (10 points) A 5.00kg mass sliding up a frictionless incline strikes a spring with a stiff-ness of 10.0N/m attached to the top of the incline. If the spring compresses 5.0cm andthe incline makes an angle of 30◦ what is the speed of the block just before it strikesthe spring? v0 =SolutionGiven: m = 5.0kg l = 0.05m k = 10N/m θ = 30◦

∆E = ∆K + ∆Ug + ∆Us = 0 (15)

1

2mv2f −

1

2mv20 +mg∆y +

1

2k∆l2 = 0 (16)

1

2mv20 = mg∆y +

1

2k∆l2 = mgl cos θ +

1

2k∆l2 (17)

v =

√2gl cos θ +

k

ml2 (18)

= 0.924m/s (19)

4. (10 points) A container of ideal gas at STP undergoes an isothermal expansion and itsentropy changes by 2.20J/K. How much work does it do? W =SOLUTIONGiven: T = 0◦C = 273K ∆S = 2.20J/K W =?

∆S =Q

T(20)

∆U =3

2nR∆T = 0 (21)

Q = ∆U +W (22)

W = Q = ∆ST = 600J(1000J)(800J) (23)

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Physics 211 Practice Final NAME: SECTION:

5. (10 points) What is the ratio of power levels of two sounds with intensity levels of35.dB and 45.dB (Assume the same distance from the source)? PB : PA =SOLUTION:given: β1 = 35dB β2 = 45dBWe know that the ratio of powers is equal to the ratio intensities, and that the log ofthe ratios of intensities is proportional to the difference in sound intensity levels.

I =P

4πR2

I1I2

=P1

P2

(24)

β = 10 log(I1I0

β1 − β2 = 10 log(I1I2

) (25)

10(β1−β2

10) =

I1I2

=P1

P2

(26)

P1 : P2 = 10 : 1 (27)

6. (10 points) A block rests on a frictionless plane inclined at θ = 60◦. If it is releasedfrom rest, what is the speed of the block after it has slid 30cm down the incline? v =?Solution

given:∆d = 30cm θ = 60◦ ay = −g = −9.8m/s2

a‖ = ay sin θ v =? v0 = 0m/s

1

2mv2 = mgl cos θ (28)

v =√

2gl cos θ = 1.7m/s (29)

7. (10 points) The speed of sound in steel is about 4.50km/s. A steel rail is struck witha hammer, and an observer 0.400km away has one ear to the rail. How much timewill elapse from the time the sound is heard through the rail until the time it is heardthrough the air? Assume that the temperature of the is 20◦C and that no wind isblowing. ∆t =Solutiongiven: vs = 4.50km/s l = 0.400km(.500km)(.600km) T = 20◦C ∆t =?

∆t = ts − ta (30)

t =l

v(31)

∆t = ts − ta =l

vs− l

va=

l

vs− l

331 + 06T= 1.08s (32)

8. (10 points) A 10g bullet moving horizontally at 400m/s penetrates a 3.0kg wood blockresting on a horizontal surface. If the bullet slows down to 300m/s after emerging fromthe block, what is the speed of the block immediately after the bullet emerges?v =?Solution

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Physics 211 Practice Final NAME: SECTION:

given: m1 = 10g m2 = 3.0kg v1 = 400m/s v1f = 300m/s

p1 = p2 (33)

m1v1 = m1v1f +m2v2f (34)

v2f =m1v1 −m1v1f

m2

= 0.33m/s (35)

9. (10 points) If 0.050kg of ice at 0◦C is added to 0.550kg(0.650kg)(0.750kg) of waterat 25◦C in a 0.100kg aluminum calorimeter cup, What is the final temperature of thewater? TF =Solution

given:mi = 0.050kg Ti = 0◦C Tf =? mw = 0.550kgTw = 25◦C mc = 0.100kg

miLf +micw∆T1 +mccc∆T2 +mwcw∆T2 = 0 (36)

miLf +micw(Tf − Ti) + (mccc +mwcw)(Tf − Tw) = 0 (37)

miLf +micwTf −micwTi + (mccc +mwcw)(Tf ) − (mccc +mwcw)(Tw) = 0 (38)

(micw +mccc +mwcw)Tf = micwTi + (mccc +mwcw)(Tw) −miLf (39)

Tf =(micwTi + (mccc +mwcw)(Tw) −miLf )

(micw +mccc +mwcw)= 16.6◦C (40)

10. (10 points) A flat cylindrical grinding wheel is spinning at 20.0rpm (clockwise whenviewed head-on) when its power is suddenly turned off. Normally, if left alone, it takes45.0s to coast to rest. Assume the grinder has a moment of inertia of 2.00kgm2. Howmuch work was done by friction on the axle. W =Solutiongiven: ω = 20rpmt = 45.0s I = 2.43kgm2 d = 0.075m

W = τθ = ∆K =1

2Iω2 = 4.39J (41)

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