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PHYS%2013.%%General%Physics%1.%
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Exam%3%study%guide%and%useful%information%
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Instructor:!Dr.!Marc!S.!Seigar!
Phone:!(218)!726<6704!
Email:[email protected]!
Office:!MWAH!374!
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The!material!you!will!need!to!know!for!Exam!3!is!covered!in!Chapters!8,!9,!and!10!of!
Young!&!Freedman.!!This!document!lists!some!important!information!you!will!need!
to!know!to!do!well!in!the!exam.!!There!is!an!information!sheet!given!to!you!as!part!of!
the!exam.!!
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This!is!what!you!will!need!to!know:!
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You!should!understand!the!definition!of!displacement,!velocity,!and!acceleration,!
and!the!derivatives!that!relate!them.!!You!should!also!understand!the!definition!of!
average!velocity!(or!speed)!and!average!acceleration.!
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You!should!be!able!to!determine!velocity!from!a!graph!of!x!vs!t.!
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You!should!be!able!to!determine!acceleration!from!a!graph!of!x!vs!t,!and!from!a!
graph!of!v!vs!t.!
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You!should!know!the!kinetic!equations:!
!
v f = vi + at !
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!
1
x f = xi + vi t + at 2 !
2
2
2
v f = vi + 2a ( Δx ) !
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x f = xi +
!
1
(v f + vi ) t !
2
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Newton’s!laws!of!motion.!
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Friction!
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Masses!on!frictionless!slopes!
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Masses!on!slopes!with!friction!
!
1
Definition!of!kinetic!energy! k = mv 2 !
2
Definition!of!potential!energy! U = mgy !
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Definition!of!work! W = Fd !(work!=!force!x!distance!moved)!or!in!integral!form!!
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 
W = ∫ F• dr !
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The!work!energy!theorem!(i.e.,!work!done!is!change!in!kinetic!energy!of!work!done!
is!change!in!potential!energy).!
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For!conservative!forces,!total!energy!(kinetic!energy!+!potential!energy)!is!
conserved.!
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Friction!is!a!non<conservative!force.!!Gravity!is!a!conservative!force.!
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Definition!of!momentum,! p = mv .!
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Conservation!of!(linear)!momentum.!
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Elastic!and!(perfectly)!inelastic!collisions.!!!Momentum!is!always!conserved.!!Energy!
is!conserved!for!elastic!collisions.!
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Impulse,! I = Δp = FΔt ,!and!its!uses.!
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Rotational!kinematics,!angular!velocity,!and!angular!acceleration!definitions.!
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The!kinetic!equations!for!rotational!kinematics:!
!
ω f = ωi + α t !
!
!
1
θ f = θi + ωit + α t 2 !
2
2
2
ω f = ωi + 2α ( Δθ ) !
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θ f = θi +
!
1
(ω f + ωi ) t !
2
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  
Definition!of!torque,! τ = rF sin θ ,!or!more!generally,! τ = r × F .!!Torque!has!the!same!
units!as!work.!
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A!body!in!equilibrium!has! Στ = 0 !and! ΣF = 0 !(static!equilibrium).!
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If!not!in!equilibrium!then,! Στ = Iα !(recognize!similarity!to! ΣF = ma ).!
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Work!and!energy!in!a!rotational!system.!
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  
Definition!of!angular!momentum,! L = r × p = rpsin φ = rmvsin φ .!
!
dL
dL
dp
And! τ =
!or! Στ = tot !!(which!is!analogous!to! ΣF = tot ).!
dt
dt
dt
!
For!a!rotating!body,! L = Iω !(analogous!to! p = mv ).!
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Conservation!of!angular!momentum.!
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Energy!of!rotating!systems.!
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Example!questions!
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1. A 1220 kg car traveling initially with a speed of 25 m/s in an
easterly direction crashes into the back of a 8300 kg truck
moving in the same direction at 20 m/s. The velocity of the
car right after the collision is 18 m/s to the east.
(a) What is the velocity of the truck right after the collision?
(Give your answer to five significant figures.)
(b) What is the change in mechanical energy of the car–truck
system in the collision?
(c) Account for this change in mechanical energy.
2. A 10.1-g bullet is fired into a stationary block of wood having
mass m = 4.93 kg. The bullet imbeds into the block. The
speed of the bullet-plus-wood combination immediately after
the collision is 0.607 m/s. What was the original speed of the
bullet? (Express your answer with four significant figures.)
3. A car of mass m moving at a speed v1 collides and couples
with the back of a truck of mass 4m moving initially in the
same direction as the car at a lower speed v2. (Use any
variable or symbol stated above as necessary.)
(a) What is the speed vf of the two vehicles immediately after the
collision?
(b) What is the change in kinetic energy of the car–truck system
in the collision?
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4. A 90.2-kg fullback running east with a speed of 5.10 m/s is
tackled by a 95.8-kg opponent running north with a speed of
2.90 m/s.
(a) Explain why the successful tackle constitutes a perfectly
inelastic collision.
(b) Calculate the velocity of the players immediately after the
tackle.
(c) Determine the mechanical energy that disappears as a result
of the collision.
(d) Account for the missing energy.
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5. A racing car travels on a circular track of radius 290 m.
Suppose the car moves with a constant linear speed of 52.0 m/s.
(a) Find its angular speed.
(b) Find the magnitude and direction of its acceleration.
6. A discus thrower accelerates a discus from rest to a speed of
25.8 m/s by whirling it through 1.29 rev. Assume the discus
moves on the arc of a circle 1.01 m in radius.
(a) Calculate the final angular speed of the discus.
(b) Determine the magnitude of the angular acceleration of the
discus, assuming it to be constant.
(c) Calculate the time interval required for the discus to
accelerate from rest to 25.8 m/s.
7. Find the net torque on the wheel in the figure below about the
axle through O, taking a = 7.00 cm and b = 21.0 cm. (Assume
that the positive direction is counterclockwise.)
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8. Calculate the net torque (magnitude and direction) on the
beam in the figure below about the following axes.
(a) an axis through O, perpendicular to the page
(b) an axis through C, perpendicular to the page
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9. A cylinder with moment of inertia I1 rotates about a vertical,
frictionless axle with angular velocity ωi. A second cylinder;
this one having a moment of inertia of I2 and initially not
rotating, drops onto the first cylinder. Because of friction
between the surfaces, the two eventually reach the same
angular speed ωf.
(a) Calculate ωf. (Use any variable or symbol stated above as
necessary.)
(b) Show that the kinetic energy of the system decreases in this
interaction by calculating the ratio of the final to initial
rotational energy. Express your answer in terms of ωi. (Use
any variable or symbol stated above as necessary.)
10. A playground merry-go-round of radius R = 1.80 m has a
moment of inertia I = 255 kg · m2 and is rotating at 12.0
rev/min about a frictionless vertical axle. Facing the axle, a
26.0-kg child hops onto the merry-go-round and manages to
sit down on the edge. What is the new angular speed of the
merry-go-round?
Solutions
1. (a) conservation of momentum, mC vCi + mT vTi = mC vCf + mT vTf
i.e.,! 30500 +166000 = 21960 + 8300vTf !
Solve!for! vTf :!
!
!
vTf =
!
!
174540
= 21.029 !m/s!
8300
"1
% "1
%
1
1
(b)!Change!in!KE,! Δk = $ mC vCi2 + mT vTi2 ' − $ mC vCf2 + mT vTf2 ' !
#2
& #2
&
2
2
! !
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Δk = 381250 +1660000 −197640 −1835208 = 8402 !J!
!
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! (c)!The!energy!has!been!transformed!into!internal!energy!and!sound.!
!
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2. Conservation!of!momentum,! mbullet vbullet = (mblock + mbullet )v f !
i.e.,! 0.0101vbullet = (4.94 + 0.0101)0.607 !
Therefore,! vbullet =
2.9986
= 296.9 !m/s!
0.0101
!
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3. (a)!From!conservation!of!momentum,! mv1 + 4mv2 = 5mv f !
so,! v f =
!
v1 + 4v2
!
5
2
# v + 4v2 &
(b)! Δk = 0.5mv + 2mv − 2.5mv = 0.5mv + 2mv − 2.5m % 1
( !
$ 5 '
2
! Δk = m ( v12 + v22 − 2v1v2 ) !
5
2
1
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2
2
2
f
2
1
2
2
4. (a)!!The!opponent!grabs!the!fullback!and!does!not!let!go,!so!the!two!players!
move!together!at!the!end!of!their!interaction.!!Thus!the!collision!is!perfectly!
inelastic.!
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(b)!!Conservation!of!momentum:!
! !x<direction,! 90.2(5.1) = 186v fx ,!and!so! v fx = 2.47 !m/s!
!
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!y<direction,! 95.8(2.9) = 186v fy ,!and!so! v fy = 1.49 !m/s!
! !
and!therefore,! v = 2.472 +1.49 2 = 2.9 !m/s!
!
(c)! Δk = 0.5(90.2)5.12 + 0.5(95.8)2.9 2 − 0.5(186)2.9 2 = 794 !J!
!
(d)!The!kinetic!energy!is!lost!to!internal!energy!
5.! (a)! ω =
!
!
(b)! ac =
!
!
6.! (a)! ω =
!
!
!
!
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(b)! ac =
v 52
=
= 0.18 !rad/s!
r 290
v 2 52 2
=
= 9.32 !m/s2!!toward!the!center!of!the!circle!
r 290
v 25.8
=
= 25.5 !rad/s!
r 1.01
v 2 25.82
=
= 659 !m/s!
r
1.01
(c)! v f = vi + at !
! 25.8 = 659t !
25.8
! t=
= 0.04 !s!
659
!
!
7.! Στ = 12(0.07) −10(0.21) − 9(0.21) = −3.15 !N!m!
!
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8.! (a)! Στ = 25(2)sin 60 −10(4)sin 20 = 43.3−13.7 = 29.6 !N!m!clockwise!
! (b)! Στ = 30(2)sin 45 −10(2)sin 20 = 42.4 − 6.8 = 35.6 !N!m!counterclockwise!
!
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9.! (a)!Angular!momentum!is!conserved:!! I1ω i = (I1 + I 2 )ω f !
!
!
!
Therefore,! ω f =
I1ω i
!
I1 + I 2
!
!
(b)!Initial!KE,! ki =
!
!
!
1
I1ω i2 !
2
2 2
1
1 ( I1 + I 2 ) I1 ω1
I12ω i2
Ik
2
=
= 1 i < ki !
Final!KE,! k f = (I1 + I 2 )ω f =
2
2
2 ( I1 + I 2 )
2 ( I1 + I 2 ) ( I1 + I 2 )
!
!
10.!!
Conservation!of!angular!momentum:!
!
!
!
Iω i = ( I + mr 2 )ω f !
!
!
!
255(1.256) = (255 + 26(1.82 ))ω f !
!
Solve!for! ω f =
!
and! ω = 12 rev/min =
!
!
!
320.28
0.94(60)
= 0.94 rad/s =
= 9 rev/min !
339.24
2π
12(2π )
= 1.256 rad/s !
60
Information Page
g = standard gravity at earth’s surface = 9.80 m/s2;
G = gravitational constant = 6.67 x 10-11 m3/(kg s2)
1 day = 8.640 x 104 s;
Re = equatorial radius of earth = 6.374 x 106 m;
Me = mass of earth = 5.976 x 1024 kg;
Mm= mass of moon = 7.350 x 1022 kg
Mean radius of moon’s orbit around earth = 3.844 x 108 m
Ms = mass of sun = 1.989 x 1030 kg
Moon’s orbital period = 27.3 days
Avg Earth to Sun distance = 1.5x1011 m
kB = 1.38 x 10-23 J/K; R = 8.31 J/(mol*K); 1 cal = 4.185 J
1 atmosphere (atm) = 1.013 x 105 Pa
Density of water = 1 x 103 kg/m
Density of Ice = 0.917 x 103 kg/m3
Density of air (STP) = 1.29 kg/m3
Specific Heat of Water =1.0kcal/(kg K)
Latent Heat of fusion for water =80 kcal/kg
Specific Heat of Aluminum = 0.215 cal/g C
Specific Heat of Copper = 0.0924 cal/g C
Volsphere = (4/3)πr3
NA = 6.023 x 1023
Weight = mg
xf = xi + vxit + 0.5axt2
θ(t) = θ o + ωot + (1/2) αt2
PV = nRT = N kB T
vxf = vxi + axt
ω(t) = ω o + αt
Q = C T = m cv T
vxf2 = vxi2 + 2ax(xf-xi)
ω2 - ω o2 = 2 α (θ - θ o)
Eint = Q + W
vave = Δx/Δt; aave = Δv/ Δt
ω
ΣFi = Fnet = m a
ave
= Δθ /t; α ave = Δω / t
W = - p dV
KEave = (3/2) kB T = (1/2) m vrms2
Στ i = I α
v =ω r
acentripetal = v2/r
acentripetal = ω 2 r
p=mv
L = Iω
Kinetic Energy = (1/2) m v2
KErot = (1/2) Iω 2
ΔUg = PE = mgh
PE (spring) = (1/2) k x2
W = Work = F o r = F Δx cos (θ )
Work = τ Δθ
Ug = -G m1 m2 / r
friction, f = µN
FG = (G M1 M2 )/ ( R2 ) is Newton’s Law of Universal Gravitation
v= λf
ω= 2 π f
Isphere = (2/5)MR2;
vstring = (T/µ)1/2
f = (1/T)
Tpendulum = 2 π (l/g)1/2 Tspring = (2π ) (m/k)1/2
SHM: x(t) = A sin (ωt);
Ithin cylinder = MR2
Wave: y(x) = A sin (2 π x/λ )