midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
Mechanics - Basic Physical Concepts
3
Math: Circle: 2 π r, π r2 ; Sphere: 4 π r2 , (4/3)
√ πr
2 −4 a c
−b±
b
2
Quadratic Eq.: a x + b x + c = 0, x =
2a
Cartesian and polar coordinates:
y
x = r cos θ, y = r sin θ, r2 = x2 + y 2 , tan θ = x
Trigonometry: cos α cos β + sin α sin β = cos(α − β)
α−β
sin α + sin β = 2 sin α+β
2 cos 2
α−β
cos α + cos β = 2 cos α+β
2 cos 2
sin 2 θ = 2 sin θ cos θ, cos 2 θ = cos2 θ − sin2 θ
1 − cos θ = 2 sin2 2θ , 1 + cos θ = 2 cos2 2θ
~ = (Ax , Ay ) = Ax ı̂ + Ay ̂
Vector algebra: A
~ =A
~+B
~ = (Ax + Bx , Ay + By )
Resultant:
R
~·B
~ = A B cos θ = Ax Bx + Ay By + Az Bz
Dot: A
Cross product: ı̂ × ̂ = k̂, ̂ × k̂ = ı̂, k̂ × ı̂ = ̂
¯
¯
¯ ı̂
̂
k̂ ¯
¯
¯
~
~
~
¯
C = A × B = ¯ Ax Ay Az ¯¯
¯
¯ B
x By Bz
C = A B sin θ = A⊥ B = A B⊥ , use right hand rule
d xn = n xn−1 ,
d
1
Calculus: dx
dx ln x = x ,
d sin θ = cos θ,
d cos θ = − sin θ,
d
dθ
dθ
dx const = 0
Measurements
Dimensional analysis: e.g.,
2
F = m a → [M ][L][T ]−2 , or F = m vr → [M ][L][T ]−2
PN
PN
Summation:
i=1 (a xi + b) = a i=1 xi + b N
Motion
One dimensional motion: v = ddts , a = ddtv
s −s
v −v
Average values: v̄ = tff −tii , ā = tff −tii
One dimensional motion (constant acceleration):
v(t) : v = v0 + a t
s(t) : s = v̄ t = v0 t + 12 a t2 , v̄ = v02+v
v(s) : v 2 = v02 + 2 a s
Nonuniform acceleration: x = x0 + v0 t + 12 a t2 +
1 j t3 + 1 s t4 + 1 k t5 + 1 p t6 + . . ., (jerk, snap,. . .)
6
24
120
720
ttrip
v0y
2 = g
1
2
h = 2 g tf all , R = vox ttrip
2
Circular: ac = vr , v = 2 Tπ r , f = T1 (Hertz=s−1 )
q
Curvilinear motion: a = a2t + a2r
Relative velocity: ~v = ~v ′ + ~u
Projectile motion: trise = tf all =
Law of Motion and applications
Force: F~ = m ~a, Fg = m g, F~12 = −F~21
2
Circular motion: ac = vr , v = 2 Tπ r = 2 π r f
Friction: Fstatic ≤ µs N
Fkinetic = µk N
P
Equilibrium (concurrent forces):
~
i Fi = 0
Energy
Work (for all F): ∆W = WAB = WB − WA
1
RB
F~ · d~s (in Joules)
Fk s = F s cos θ = F~ · ~s → A
Effects due to work done: F~ext = m ~a − F~c − f~nc
Wext |A→B = KB − KA + UB − UA + Wdiss |A→B
RB
Kinetic energy: KB −KA = A
m ~a ·d~s, K = 12 m v 2
R
K (conservative F~ ): U − U = − B F~ · d~s
B
A
A
Ugravity = m g y,
Uspring = 12 k x2
∂
U
From U to F~ : Fx = − ∂x , Fy = − ∂∂yU , Fz = − ∂∂zU
Fgravity = − ∂∂yU = −m g,
U = −k x
Fspring = − ∂∂x
2
∂
U
∂
U
Equilibrium: ∂x = 0, ∂x2 > 0 stable, < 0 unstable
W = F v = F v cos θ = F
~ · ~v (Watts)
Power: P = ddt
k
Collision
Rt
Impulse: I~ = ∆~
p = p~f − p~i → tif F~ dt
Momentum: p~ = m ~v
x1 +m2 x2
Two-body: xcm = m1m
1 +m2
pcm ≡ M vcm = p1 + p2 = m1 v1 + m2 v2
Fcm ≡ F1 + F2 = m1 a1 + m2 a2 = M acm
K1 + K2 = K1∗ + K2∗ + Kcm
Two-body collision: p~i = p~f = (m1 + m2 ) ~vcm
vi′ = vi∗′ + vcm
vi∗ = vi − vcm ,
Elastic: v1 − v2 = −(v1′ − v2′ ),
vi∗′ = −vi∗ , vi′ = 2 vcm − vi
R
P
~r dm
m ~r
Many body center of mass: ~rcm = P i i = R
mi
mi
P
p
Force on cm: F~ext = d~
=
M~
a
,
p
~
=
p
~
cm
i
dt
Rotation of Rigid-Body
Kinematics: θ = rs , ω = vr , α = art
R
P
Moment of inertia: I =
mi ri2 = r2 dm
1
1
2
Idisk = 2 M R , Iring = 2 M (R12 + R22 )
1 M ℓ2 , I
1
2
2
Irod = 12
rectangle = 12 M (a + b )
Isphere = 25 M R2 , Ispherical shell = 23 M R2
I = M (Radius of gyration)2 , I = Icm + M D2
Kinetic energies: Krot = 12 I ω 2 , K = Krot + Kcm
Angular momentum: L = r m v = r m ω r = I ω
Torque: τ = ddtL = m ddtv r = F r = I ddtω = I α
Wext = ∆K +∆U +Wf ,
K = Krot + 12 m v 2 ,
P =τω
Rolling, angular
momentum
and
´
³
´
³ torque
Ic + M v 2
Rolling: K = 12 Ic + M R2 ω 2 = 12 R
2
~
Angular momentum: L = ~r × p~, L = r⊥ p = I ω
~
Torque: ~τ = d L = ~r × d~p = ~r × F~ , τ = r F = I α
dt
dt
1 dL = τ = mgh
Gyroscope: ωp = ddtφ = L
L
Iω
dt
Static equilibrium
P
P~
~τi = 0
Fi = 0, about any point
+mB ~rBcm
Subdivisions: ~rcm = mA ~rAcm
mA +mB
Elastic modulus = stress/strain
stress: F/A
strain: ∆L/L, θ ≈ ∆x/h, −∆V /V
⊥
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
2
.
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midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
Question 1, chap 5, sect 5.
part 1 of 1
10 points
Two students sit in identical office chairs
facing each other. Bob has a mass of 95 kg,
while Jim has a mass of 77 kg. Bob places
his bare feet on Jim’s knees, as shown to the
right. Bob then suddenly pushes outward
with his feet, causing both chairs to move.
Explanation:
North
vre
vbe
East
Taking North and East be positive, let
vbe = velocity of boat relative to earth
vbr = velocity of boat relative to river
vre = velocity of river relative to earth
vre = 2.77 m/s
vbr = 7.3 m/s .
Jim
In this situation, while Bob’s feet are in contact with Jim’s knees,
1. Each student exerts a force on the other,
but Jim exerts a larger force.
2. None of these answers is correct.
3. Bob exerts a force on Jim, but Jim doesn’t
exert a force on Bob.
4. Each student exerts the same amount of
force on the other. correct
5. Each student exerts a force on the other,
but Bob exerts a larger force.
6. Neither student exerts a force on the
other.
Explanation:
By Newton’s third law, action and reaction
are of the same magnitude and in the opposite
direction.
Question 2, chap 4, sect 6.
part 1 of 3
10 points
A river flows due North at 2.77 m/s. A boat
crosses the river from the West shore to the
East shore by maintaining a constant velocity
of 7.3 m/s due East relative to the water.
What is the magnitude of the velocity of
the boat relative to the shore?
Correct answer: 7.80787.
θ
vbr
Given :
Bob
3
and
~vbe = ~vbr + ~vre .
The velocities are perpendicular, so
q
2 + v2
vbe = vbr
re
q
= (7.3 m/s)2 + (2.77 m/s)2
= 7.80787 m/s .
Question 3, chap 4, sect 6.
part 2 of 3
10 points
By how many degrees off course is the boat
forced by the current?
Correct answer: 20.7794.
Explanation:
tan θ =
vre
vbr
vre
θ = tan
v
br
−1 2.77 m/s
= tan
7.3 m/s
−1
= 20.7794 ◦ .
Question 4, chap 4, sect 6.
part 3 of 3
10 points
If the river is 255 m wide, how far downstream is the boat when it reaches the East
shore?
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
Correct answer: 96.7603.
Explanation:
Given : ∆y = 255 m .
∆x = vre ∆t and
∆y = vbr ∆t
The times are the same, so
∆t =
Explanation:
• Check y = A cos(k x − ω t):
Since k is an inverse length, k x is dimensionless, and similarly, ωt is also dimensionless,
so k x − ω t is dimensionless, as an argument
of a trigonometry function must always be.
The value of a trigonometry function (e.g.
cos(k x − ω t)) is also dimensionless. So the
right hand side has the dimension of length.
∆y
∆x
=
vbr
vre
[y] = L
[A cos(k x)] = L .
and
vre ∆y
vbr
(2.77 m/s) (255 m)
=
7.3 m/s
∆x =
This equation is dimensionally correct.
• Check F = m ω 2 r:
ML
The dimension of force is [F ] =
. The
T2
dimension of right hand side is
= 96.7603 m .
Question 5, chap 1, sect 6.
part 1 of 1
10 points
Identify the equation below which is dimensionally incorrect.
A, x, y and r have units of length. k here
has units of inverse length. v0 and v have units
of velocity. a and g have units of acceleration.
ω has units of inverse time. t is time, m is
mass, V is volume, ρ is density, and F is force.
1. g =
ρ
F
+ gV
m m
2. y = A cos(k x − ω t)
q
−v0 + v02 + 2 a x
v2
3. t =
+ 2
a
a t
v
correct
4. F = m g 1 +
rg
p
5. v = ±ω A2 − x2
p
6. v0 = v 2 − 2 a x
r
p
rF
7. v = 2 g y +
m
8. F = m ω 2 r
4
2
ML
1
L= 2 .
mω r = M
T
T
2
So it is also dimensionally correct.
r
p
rF
:
• Check v = 2 g y +
m
L
F
= 2 = [g], and [y] = [r] = L,
Since
m
T
#
"r
hp
i
rF
2gy =
m
r
L2
L
= .
=
2
T
T
L
The left hand side of the equation is [v] = .
T
So this equation is also dimensionally correct.
v
:
• Check F = m g 1 +
rg
v
F
= [g], so 1 +
As mentioned earlier
m
rg
should be dimensionless. However,
v
=
rg
L
T =T.
L
L
L 2
T
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
and
2
L
2 v
T
= 2 = T .
2
a t
L
T
T2
So this equation is dimensionally correct.
p
• Check v0 = v 2 − 2 a x:
As mentioned earlier, this one is correct.
p
• Check v = ±ω A2 − x2 :
i
h p
1√ 2 L
L = = [v] .
ω A 2 − x2 =
T
T
So this equation is dimensionally correct.
F
ρ
+ gV:
m m
F
= [g] , and
As mentioned earlier
m
• Check g =
hρ
m
i
gV =
M
L3 [g] L3 = [g] .
M
So this equation is also dimensionally correct.
Question 6, chap 4, sect 4.
part 1 of 1
10 points
A cannon fires a 0.242 kg shell with initial
velocity vi = 10 m/s in the direction θ = 50 ◦
above the horizontal.
The acceleration of gravity is 9.8 m/s2 .
m
/s
∆h
10
Therefore the right hand side of this equation
is not dimensionally correct because quantities of different dimensions can not be added
up together.
q
−v0 + v02 + 2 a x
v2
• Check t =
+ 2 :
a q
a t L2
2
v0 + 2 a x =
Since [a x] = 2 , we have
T
L
= [v0 ], therefore
T
q


L
−v0 + v02 + 2 a x
= T =T

L
a
T2
5
∆y
◦
50
y
∆x
The shell’s trajectory curves downward because of gravity, so at the time t = 0.404 s
the shell is below the straight line by some
vertical distance.
Your task is to calculate this distance ∆h
in the absence of air resistance.
Correct answer: 0.799758.
Explanation:
In the absence of gravity, the shell would fly
along the straight line at constant velocity:
x̂ = t vi cos θ ,
ŷ = t vi sin θ .
The gravity does not affect the x coordinate
of the shell, but it does pull its y coordinate
at constant downward acceleration ay = −g,
hence
x = t vi cos θ,
g t2
.
2
1
Thus, x = x̂ but y = ŷ − g t2 , or in other
2
words, the shell deviates from the straight-line
path by the vertical distance
y = t vi sin θ −
∆h = ŷ − y =
g t2
.
2
Note: This result is completely independent on the initial velocity vi or angle θ of
the shell. It is a simple function of the flight
time t and nothing else (besides the constant
g = 9.8 m/s2 ).
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
Explanation:
v = ωR,
1
I = M R2 , and
2
1
1
Kdisk = I ω 2 = M v 2 .
2
4
Consider the free body diagrams
a
m3
m3 g
m2
T1
R
m1
m1 g
a
T1
Given: The pulley is massless and frictionless. A massless inextensible string is
attached to these masses: m1 = m, m2 = m,
and m3 = 4 m.
T3
Let :
T2
Question 7, chap 5, sect 6.
part 1 of 2
10 points
9. T1 = m g
m2 g
g t2
∆h =
2
(9.8 m/s2 ) (0.404 s)2
=
2
= 0.799758 m .
6
ω
T2
T3
m2
T1
ℓ
m3
m1
What is the tension T1 in the string connecting the two objects of equal mass?
1. T1 =
2. T1 =
3. T1 =
4. T1 =
3
mg
2
1
mg
2
2
mg
3
4
m g correct
3
5. T1 = 3 m g
6. T1 = 4 m g
1
7. T1 = m g
4
8. T1 = 2 m g
Basic Concept : For each mass in the
system
Fnet = m a .
Solution : The acceleration a of the system will be down to the right, and each mass
in the system accelerates at the same rate.
For the lower left-hand mass, the acceleration
is up and
m a = T1 − m g .
Multiplying this Eq. by (−5), we have
−5 m a = −5 T1 + 5 m g .
(1)
For the upper left-hand mass, the acceleration is up and
m a = T2 − m g − T1 .
(2)
For the right-hand mass, the acceleration is
down and
(4 m) a = (4 m) g − T2 .
(3)
Adding Eq. 1, 2, and 3 together,
0 = 8 m g − 6 T1 .
8
mg
6
4
= mg
3
T1 =
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
Alternate Solution : Considering that a
pulley only redirects the motion, and doesn’t
itself act as a force, then we can consider this
problem a little bit differently. Consider the
all of the masses to be a system and consider
over the pulley to the right to be positive.
Then, the sum of the forces is the sum of the
masses times the acceleration, or
4ma+ 2ma = 4mg − 2mg
= 2mg.
So
g
2g
= .
6
3
Then we examine the free body diagram of
the bottom left mass, m1 ,
a=
T1 − m g = m a
mg
.
=
3
7
9. Fnet = (m3 − m1 ) a
10. Fnet = m2 a
Explanation:
The net force on m1 is Fnet = m1 a.
Question 9, chap 5, sect 2.
part 1 of 2
10 points
The following 2 questions refer to a coin
which is tossed straight up into the air. After it is released it moves upward, reaches its
highest point and falls back down again. Indicate the force acting on the coin for each of
the cases described below.
The coin is moving upward after it is released.
1. The force is down and decreasing.
Then, finally,
4
T1 = m g .
3
Question 8, chap 5, sect 6.
part 2 of 2
10 points
Let: m3 > m1 + m2 and let m1 > m2 .
The magnitude of the acceleration of the
masses is just a.
What is the magnitude of the net force on
mass m1 ?
1. Fnet = m3 a
2. Fnet = m1 g
3. Fnet = m1 a correct
4. Fnet = m2 g
2. The force is up and constant.
3. The force is up and decreasing.
4. The force is zero.
5. The force is up and increasing.
6. The force is down and increasing.
7. The force is down and constant. correct
Explanation:
The force acting on the coin is the gravity.
Question 10, chap 5, sect 2.
part 2 of 2
10 points
The coin is at its highest point.
5. Fnet = (m1 − m2 ) g
1. The force is down and increasing.
6. Fnet = (m3 − m1 ) g
2. The force is up and decreasing.
7. Fnet = (m1 − m2 ) a
3. The force is zero.
8. Fnet = m3 g
4. The force is up and increasing.
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
8
and
5. The force is down and constant. correct
B y = R y − Ay
= R sin θR − A sin θA
= (191 cm)(sin 52.8◦ )
− (225 cm)(sin 157◦ )
= 64.223 cm ,
6. The force is down and decreasing.
7. The force is up and constant.
Explanation:
The same reason as Part 1.
Question 11, chap 4, sect 3.
part 1 of 2
10 points
A man pushing a mop across a floor causes
the mop to undergo two displacements. The
first has a magnitude of 225 cm and makes an
angle of 157◦ with the positive x-axis.
The resultant displacement has a magnitude of 191 cm and is directed at an angle of
52.8◦ to the positive x-axis.
Find the magnitude of the second displacement.
Correct answer: 328.923.
Explanation:
Let :
A = 225 cm ,
θA = 157◦ ,
R = 191 cm ,
θR = 52.8◦ .
which act at right angles to each other.
From the Pythagorean Theorem,
q
Bx2 + By2
q
= (322.592 cm)2 + (64.223 cm)2
B=
= 328.923 cm .
Question 12, chap 4, sect 3.
part 2 of 2
10 points
Find the direction of the second displacement with respect to positive x axis. (Consider counterclockwise to be positive and answer within the limits of −180◦ to 180◦ .)
Correct answer: 11.2595.
Explanation:
and
tan θ =
1c
m
19
Since
cm
28.
B=3
923 cm
~ +B
~ =R
~,
A
~ =R
~ −A
~.
B
~ has components
Thus, the vector B
B x = R x − Ax
= R cos θR − A cos θA
= (191 cm)(cos 52.8◦ )
− (225 cm)(cos 157◦ )
= 322.592 cm
By
θ = tan
B
x
64.223 cm
−1
= tan
322.592 cm
−1
=
225
R
A=
By
Bx
= 11.2595◦ ,
which is directed above the x axis.
Question 13, chap 4, sect 5.
part 1 of 1
10 points
In the spin cycle of a washing machine, the
tub of radius 0.422 m rotates at a constant
rate of 939 rev/min.
What is the maximum linear speed of the
water inside the machine?
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
Correct answer: 41.4962.
Explanation:
First we have to convert the rate of rotation
n (expressed in revolutions per minute) into
angular speed ω (in radians per second or,
equivalently, in s−1 ):
2 π rad
min
ω = (939 rev/min )
rev
60 s
= 98.3321 s−1 .
Then we use the formula connecting linear
and angular velocities and obtain
v = rω
= (0.422 m) (98.3321 s−1 )
= 41.4962 m/s .
Question 14, chap 5, sect 4.
part 1 of 1
10 points
You weigh 198 lb on Earth (4.448 N = 1
lb). When you reach Alderon, you discover
that its gravity is 1.29 that of earth.
What is your weight on Alderon?
Correct answer: 1136.11.
Explanation:
Assume: An astronaut drops a ball off the
rim of the crater and that the free fall acceleration of the ball remains constant throughout
the ball’s 34 km fall at a value of 4 m/s2 . (We
assume that the crater is as deep as the volcano is tall, which is not usually the case in
nature.)
a) Find the time for the ball to reach the
crater floor.
Correct answer: 130.384.
Explanation:
Let : h = 34 km and
a = 4 m/s2 .
Basic Concepts: Free fall
1
s = so + v0 t + a t2
2
2
2
v = v0 + 2 a h .
Solution: The distance an object falls
from rest under an acceleration a is
1
h = a t2 , so
2
r
2h
t=
s a
WA = m gA
= m (1.29 gE )
= (1.29) m gE
= (1.29) W E
= (1.29) (198 lb)
= 1136.11 N .
2 (34 km)
(4 m/s2 )
=
=
given : WE = 198 lb , and
gA = 1.29 gE .
9
q
(272000 m2 /s2 )
= 130.384 s .
Question 16, chap 2, sect 6.
part 2 of 2
10 points
b) Find the magnitude of the velocity with
which the ball hits the crater floor.
Correct answer: 521.536.
4.448 N
1 lb
Question 15, chap 2, sect 6.
part 1 of 2
10 points
The tallest volcano in the solar system is the
34 km tall Martian volcano, Olympus Mons.
Explanation:
Since the object falls from rest,
v 2 = 2 a h so
√
v = q2 a h
= 2 (4 m/s2 ) (34 km)
q
= (272000 m2 /s2 )
= 521.536 m/s .
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
A subway train starting from rest leaves a
station with a constant acceleration. At the
end of 6.11 s, it is moving at 13.1976 m/s.
What is the train’s displacement in the first
4.68026 s of motion?
Correct answer: 23.6572.
Explanation:
Basic Concepts: Uniform Acceleration of
Motion
a=
v
t
Displacement of Motion from Rest
s=
a t2
2
Solution: The acceleration of the train is
v
t1
(13.1976 m/s)
=
(6.11 s)
= 2.16 m/s2 .
a=
So, the displacement in the first 4.68026 s is
Distance (meters)
Question 17, chap 2, sect 5.
part 1 of 1
10 points
10
Distance vs Time
80
75
70
65
60
55
50
45
40
35
30
25
20
15
10
5
0
0 2 4 6 8 10 12 14 16 18 20 22 24
Time (minutes)
Jill has been riding for 16 min ± 2.24 min.
She has gone a distance of 55 m ± 7.15 m.
Estimate her speed.
Correct answer: 3.4375.
Explanation:
Distance equals velocity times time (d =
v t), so we have
d
t
(55 m)
=
(16 min)
v=
(1)
= 3.4375 m/min .
1
2
1
=
2
1
=
2
s=
a t22
v 2
t
t1 2
(13.1976 m/s)
(4.68026 s)2
(6.11 s)
= 23.6572 m .
Question 18, chap -1, sect -1.
part 1 of 2
10 points
The graph below shows the measurement
and accuracy of Jill riding a bicycle at a constant speed on a country path.
Question 19, chap -1, sect -1.
part 2 of 2
10 points
What is the uncertainty in her speed, as
calculated in Part 1?
Correct answer: 0.928125.
Explanation:
Fractional expansion of the calculation used
in Part 1, Eq. 1, defines error propagation.
Let
∆t
2.24 min
=
= 0.14
t
16 min
7.15 m
∆d
=
= 0.13 .
d
55 m
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
Substituting
∆v = v
∆d
d≡d 1±
,
(2)
d
∆v
, and
v ≡v 1±
v
∆t
t≡t 1∓
,
t
d
, we have
into Eq. 1 v =
t


∆d 

 1±

d
∆v
d 
=
v 1±
∆t 
v
t


 1∓

t


∆d 




 1±
∆v
d
=
1±

∆t 
v


 1∓

t
∆d
∆t
∆v
= 1±
1±
1±
v
d
t
∆v
∆t ∆d ∆d ∆t
±
(3)
=±
+
+
v
t
d
d t
∆v
= (0.14) + (0.13) + (0.13) (0.14)
v
= 0.2882 , so
∆v
v
∆v =
v
= (0.2882) (3.4375 m/min)
= 0.990688 m/min ,
1
≈ 1 − χ, where χ ≪ 1.
since
1+χ
∆d ∆t
However, the second order term
d t
in the calculation usually can be neglected, so
we have
∆v
∆t ∆d
=
+
v
t
d
= (0.14) + (0.13)
= 0.27 , so
∆v
∆v = v
v
= (3.4375 m/min) (0.27)
∆v
v
d ∆t ∆d
=
+
t
t
d
d
1
= 2 ∆t + ∆d
t
t
(55 m)
(2.24 min)
=
(16 min)2
1
(7.15 m)
+
(16 min)
11
(4)
= 0.928125 m/min , or using Eq. 4
(5)
= 0.928125 m/min .
The full answer is
v = v ± ∆v
= 3.4375 m/min ± 0.928125 m/min .
Note: Usually, only the first couple of digits
are useful. Books and journals require the
removal of unneeded digits for the purpose of
saving paper.
Alternate Solution: This approximation
(keeping only first order terms) is equivalent
to
∂ v ∂ v
∆v = ∆t+ ∆d
∂t
∂d
1
d
(5)
= 2 ∆t + ∆d ,
t
t
which the same as the previous Eq. 5, since
∂v
∂ d
d
=
=− 2,
∂t
∂t t
t
∂ d
1
∂v
=
=+ .
∂d
∂d t
t
and
The use of algebraic parsing programs (e.g.,
Mathematica, Maple, etc.) make error analysis just a “click”.
Question 20, chap 1, sect 5.
part 1 of 1
10 points
A piece of wire has a density of 7.6 g/cm.
What is the mass of 12.8 cm of the wire?
Correct answer: 97.28.
Explanation:
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
Let : ρ = 7.6 g/cm
ℓ = 12.8 cm .
and
g
Since
· cm = g, we must multiply the dencm
sity by the length of the wire to obtain the
mass. This can also be determined by the
definition of linear mass density:
m
, so that
ρ=
ℓ
m = ρℓ
= (7.6 g/cm) (12.8 cm)
12
A cat chases a mouse across a 1.6 m high
table. The mouse steps out of the way, and
the cat slides off the table and strikes the floor
3.0 m from the edge of the table.
The acceleration of gravity is 9.81 m/s2 .
What was the cat’s speed when it slid off
the table?
Correct answer: 5.25268.
Explanation:
Basic Concept:
∆x = vx ∆t
1
∆y = − g (∆t)2
2
= 97.28 g .
Given:
∆y = −1.6 m
∆x = 3.0 m
g = 9.81 m/s2 .
Question 21, chap 1, sect 99.
part 1 of 1
10 points
Two airplanes leave an airport at the same
time. The velocity of the first airplane is
740 m/h at a heading of 59.4◦ . The velocity
of the second is 560 m/h at a heading of 180◦ .
How far apart are they after 1.5 h?
Correct answer: 1699.11.
Explanation:
Under constant velocity, the displacement
for each plane in the time t is
d = v t.
These displacements form two sides of a triangle. The angle between the displacements
is
α = θ2r − θ1r .
Given two sides, d1 and d2 of the triangle
and the angle α between them, by the law of
cosines, the distance between the planes is
q
d = d21 + d22 − 2 d1 d2 cos(α)
= (1110 m)2 + (840 m)2
1/2
−(−949262 m2 )
= 1699.11 m .
Question 22, chap 4, sect 4.
part 1 of 1
10 points
Solution:
∆t =
vx =
=
r
s
s
2 ∆y
∆x
=
−g
vx
−g
∆x
2∆y
−(9.81 m/s2 )
(3 m)
2 (−1.6 m)
= 5.25268 m/s .
Question 23, chap 1, sect 5.
part 1 of 1
10 points
A newly discovered Jupiter-like planet has
an average radius 11 times that of the Earth
and a mass 313 times that of the Earth.
Calculate the ratio of new planet’s mass
density to the mass density of the Earth.
Correct answer: 0.235162.
Explanation:
Let : Rnp = 11 RE and
Mnp = 313 ME .
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
The volume of a sphere of radius R is V =
4
π R3 . The (average) density ρ of a body is
3
the ratio of the body’s mass to its volume, ρ =
M
. Planets are spherical, so the (average)
V
density of a planet of a given mass M and a
given radius R is
ρ=
M
4
π R3
3
ρnp
ρE
no initial velocity in the vertical direction, so
the time spent in the air may be deduced from
the kinematic equation,
y=
Mnp
4
3
π Rnp
3
=
ME
4
3
π RE
3
Mnp
M
= E 3
Rnp
RE
313
=
(11)3
= 0.235162 .
Question 24, chap 4, sect 5.
part 1 of 1
10 points
A ball on the end of a string is whirled
around in a horizontal circle of radius 0.316 m.
The plane of the circle is 1.11 m above the
ground. The string breaks and the ball lands
2.7 m away from the point on the ground
directly beneath the ball’s location when the
string breaks.
The acceleration of gravity is 9.8 m/s2 .
Find the centripetal acceleration of the ball
during its circular motion.
Correct answer: 101.839.
Explanation:
In order to find the centripetal acceleration
of the ball, we need to find the initial velocity
of the ball. Let y be the distance above the
ground. After the string breaks, the ball has
1 2
gt .
2
Solving for t,
.
Comparing the newly discovered planet to the
Earth, we have
13
⇒t=
r
2y
.
g
Let d be the distance traveled by the ball.
Then
d
d
.
vx = = r
t
2y
g
Hence, the centripetal acceleration of the ball
during its circular motion is
vx2
r
d2 g
=
2yr
ac =
= 101.839 m/s2 .
Question 25, chap 2, sect 1.
part 1 of 2
10 points
You drive a car 3 h at 45 km/h, then 3 h at
68 km/h.
What is your average velocity?
Correct answer: 56.5.
Explanation:
Basic Concept
Average velocity is defined by
vav =
∆d
∆t
Solution
If you drive for t hours at a speed of v1 and
for the same time at a speed of v2 , then
∆d = d1 + d2 = v1 t + v2 t
in a time of
∆t = t1 + t2 = 2t
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
14
so that the average velocity is
v1 t + v2 t
2t
v1 + v2
=
2
(45 km/h) + (68 km/h)
=
2
= 56.5 km/h .
710
vav =
Question 26, chap 2, sect 1.
part 2 of 2
10 points
What is your average velocity if you drive
a distance of 169.5 km at a speed of 45 km/h,
then the same distance at a speed of 68 km/h?
Correct answer: 54.1593.
N θ1
and
d
d
+
v1 v2
so that the average velocity will be
N
θ
685 N
a) What is the angle θ1 which the righthand cable makes with respect to the ceiling?
Correct answer: 29.8661.
Explanation:
Observe the free-body diagram below.
θ2
F2
θ1
F1
Explanation:
If you drive the distance d at v1 and the
same distance at v2 , then
∆d = d1 + d2 = 2 d
730
Wg
Note: The sum of the x- and
y-components of F1 , F2 , and
Wg are equal to zero.
∆t = t1 + t2 =
vav =
2d
·
v1 v2
v1 v2
d
d
+
v1 v2
2 d v1 v2
=
d v2 + d v1
2 v1 v2
=
v1 + v2
2 (45 km/h) (68 km/h)
=
(45 km/h) + (68 km/h)
= 54.1593 km/h .
Question 27, chap 5, sect 5.
part 1 of 2
10 points
Consider the 685 N weight held by two
cables shown below. The left-hand cable has
tension 710 N and makes an angle of θ with
the wall. The right-hand cable has tension
730 N and makes an angle of θ1 with the
ceiling.
Given : Wg
F1
θ1
θ2
= 685 N ,
= 730 N ,
= 29.8661◦ ,
= 90◦ − θ .
and
Basic Concepts:
X
Fx = 0
F1x
F1 cos θ1
F12 cos2 θ1
X
and
Fy
= F2x
= F2 cos θ2
= F22 cos2 θ2
(1)
(2)
=0
F1y + F2y + F3y = 0
F1 sin θ1 + F2 sin θ2 − F3 = 0
F1 sin θ1 = −F2 sin θ2 + F3
F12 sin2 θ1 = F22 sin2 θ2
−2 F2 F3 sin θ2 + F32 , since (3)
F3 sin θ3 = F3 sin 270◦ = −F3 , and
F3 cos θ3 = F3 cos 270◦ = 0 .
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
= 29.8661◦ .
Question 28, chap 5, sect 5.
part 2 of 2
10 points
b) What is the angle θ which the left-hand
cable makes with respect to the wall?
Correct answer: 63.1528.
m/
.8
10
Building
F22 = F12 − 2 F1 F3 sin θ1 + F32
F 2 + F32 − F22
sin θ1 = 1
2 F1 F3
2
F3 + F12 − F22
θ1 = arcsin
2 F1 F3
(685 N)2 + (730 N)2
= arcsin
2 (730 N) (685 N)
(710 N)2
−
2 (730 N) (685 N)
s
Solution: Since sin2 θ + cos2 θ = +1 and
adding Eqs. 2 and 3, we have
◦
59
h
15.3 m
How tall, h, is the building?
Correct answer: 11.6091.
Explanation:
Basic Concepts:
vx = vx0 = constant
vy = vy0 − g t
Explanation:
Using Eq. 1, we have
F1
cos θ2 =
cos θ1
F2
F1
cos θ1
θ2 = arccos
F2
(730 N)
◦
= arccos
cos 29.8661
(710 N)
= 26.8472◦
θ = 90◦ − θ2
= 90◦ − 26.8472◦
= 63.1528◦ .
x = vx0 t
y = vy0 t −
1 2
gt .
2
Let : θ = 59 ◦ ,
v0 = 10.8 m/s ,
∆x = 15.3 m , and
m = 0.84 kg .
Solution: The flying time can be determined
by:
Question 29, chap 4, sect 4.
part 1 of 1
10 points
A 0.84 kg rock is projected from the edge of
the top of a building with an initial velocity of
10.8 m/s at an angle 59 ◦ above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 15.3 m from the base
of the building.
Assume: The ground is level and that the
side of the building is vertical. The acceleration of gravity is 9.8 m/s2 .
15
x = v0x t , or
x
t=
v0x
x
=
.
v0 cos θ
From the point where the rock was projected
(set to be the origin O), the y-coordinate of
the point where the rock struck the ground is
y = v0y t −
1 2
gt
2
midterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 14 2007, 10:00 pm
2
x
1
x
= v0 sin θ
− g
v0 cos θ
2
v0 cos θ
2
gx
= x tan θ −
, so
2 (v0 cos θ)2
h = |y|
g x2
− x tan θ
=
2 (v0 cos θ)2
(9.8 m/s2 ) (15.3 m)2
=
2 [(10.8 m/s) cos 59 ◦ ]2
− (15.3 m) tan 59 ◦
= 11.6091 m .
16