oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
1
= (39 m) + (180 m)
Question 1, chap 2, sect 4.
part 1 of 2
10 points
= 219 m .
The velocity v(t) of some particle is plotted
as a function of time on the graph below.
The scale on the horizontal axis is 2 s per
grid square and on the vertical axis 9 m/s per
grid square.
Initially, at t = 0 the particle is at x0 =
39 m.
What is the particle’s acceleration?
Correct answer: −3.375 m/s2 (tolerance ± 1
%).
Explanation:
The average acceleration of the particle is
6
velocity × (9 m/s)
Question 2, chap 2, sect 4.
part 2 of 2
10 points
5
∆v
∆t
vf − v0
=
t
(9 m/s) − (36 m/s)
=
(8 s)
= −3.375 m/s2 .
4
ā =
v(t)
3
2
1
0
0
1
2
3 4 5 6
time × (2 s)
7
8
9
What is the position x of the particle at
time t = 8 s?
Correct answer: 219 m (tolerance ± 1 %).
Explanation:
Looking at the v(t) plot we see that over
time t = 4 × 2 s = 8 s, the particle’s velocity
decreases from the initial v0 = 4 × 9 m/s =
36 m/s to final vf = 1 × 9 m/s = 9 m/s. The
v(t) line is straight, which indicates constant
deceleration rate, hence the average velocity
is given by
v0 + vf
v̄ =
2
(36 m/s) + (9 m/s)
=
2
= 22.5 m/s .
Consequently, the particle’s displacement
during this time is simply
∆x = t v̄
= (8 s) (22.5 m/s)
= 180 m ,
and its final position
x = x0 + ∆x
The acceleration is the slope of the velocity
vs time graph. Since the v(t) line is straight,
the acceleration is constant, hence a = ā =
−3.375 m/s2 .
Question 3, chap 2, sect 5.
part 1 of 1
10 points
A ranger in a national park is driving at
37.4 mi/h when a deer jumps into the road
193 ft ahead of the vehicle. After a reaction
time of t the ranger applies the brakes to
produce and acceleration of −9.04 ft/s2 .
What is the maximum reaction time allowed if she is to avoid hitting the deer?
Correct answer: 0.48455 s (tolerance ± 1 %).
Explanation:
Given :
vi = 37.4 mi/h = 54.8533 ft/s ,
vf = 0 , and
a = −9.04 ft/s2 .
The total distance covered is defined by
vi t + ∆xstop = ∆x ,
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
where t is the reaction time.
vf2 = vi2 + 2 a ∆x = 0
since vf = 0 mi/h, so
∆xstop =
and it starts its free fall motion from this
point. Its initial velocity is equal to the velocity of the helicopter at that time
v0 =
−vi2
.
2a
dh
= 3 c t2
dt
= 3 (1.2 m/s3 ) (4 s)2 = 57.6 m/s .
Thus the equation of motion governing the
mailbag is
Thus
vi t = ∆x − ∆xstop
∆x ∆xstop
−
t=
vi
vi
vi2
∆x
=
+ 2a
vi
vi
vi
∆x
+
=
vi
2a
193 ft
54.8533 ft/s
=
+
54.8533 ft/s 2 (−9.04 ft/s2 )
= 0.48455 s .
Question 4, chap 2, sect 6.
part 1 of 1
10 points
The height of a helicopter above the ground
is given by h = c t3 , where c = 1.2 m/s3 , h is
in meters, and t is in seconds. The helicopter
takes off at t = 0 s. After 4 s it releases a
small mailbag.
The acceleration of gravity is 9.8 m/s2 .
How long after its release does the mailbag
reach the ground?
Correct answer: 12.9641 s (tolerance ± 1 %).
Explanation:
Given : t = 4 s .
y(t) = 0 = h0 + v0 t −
1 2
gt .
2
In quadratic form,
1 2
g t − v0 t − h0 = 0 .
2
From the quadratic formula,
q
v0 ± v02 + 2 g h0
.
t=
g
Since
D = v02 + 2 g h
= (57.6 m/s)2 + 2 9.8 m/s2 (76.8 m)
= 4823.04 m2 /s2 ,
the time for the mailbag to reach the ground
is
√
v0 ± D
t=
g
p
57.6 m/s ± 4823.04 m2 /s2
=
.
9.8 m/s2
The helicopter takes off at t = 0 s.
negative solution is rejected, so
p
57.6 m/s + 4823.04 m2 /s2
t=
9.8 m/s2
The
= 12.9641 s .
Under free fall,
h(t) = y(t) = y0 + v0 t +
2
1 2
at .
2
The initial height of the mailbag is the
height of the helicopter 4 s after takeoff
h0 = h(t) = (1.2 m/s3 ) (4 s)3 = 76.8 m ,
Question 5, chap 4, sect 4.
part 1 of 1
10 points
A 0.92 kg rock is projected from the edge of
the top of a building with an initial velocity of
12 m/s at an angle 35 ◦ above the horizontal.
Due to gravity, the rock strikes the ground at
Building
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
3
2
a horizontal distance of 20.2 m from the base
x
1
x
= v0 sin θ
− g
of the building.
v0 cos θ
2
v0 cos θ
Assume: The ground is level and that the
2
gx
side of the building is vertical. The accelera= x tan θ −
, so
2 (v0 cos θ)2
tion of gravity is 9.8 m/s2 .
h = |y|
/s
m
g x2
12 ◦
− x tan θ
=
2 (v0 cos θ)2
35
(9.8 m/s2 ) (20.2 m)2
=
2 [(12 m/s) cos 35 ◦ ]2
h
− (20.2 m) tan 35 ◦
20.2 m
= 6.54804 m .
How tall, h, is the building?
Correct answer: 6.54804 m (tolerance ± 1
%).
Explanation:
Basic Concepts:
vx = vx0 = constant
vy = vy0 − g t
x = vx0 t
y = vy0 t −
Let :
1 2
gt .
2
θ = 35 ◦ ,
v0 = 12 m/s ,
∆x = 20.2 m , and
m = 0.92 kg .
Solution: The flying time can be determined
by:
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
Question 6, chap 4, sect 4.
part 1 of 3
10 points
A boy is standing on a cliff. The boy’s
hands are a height b above the ground level at
the base of the cliff.. A monkey is in a tree.
The monkey is at a height h above the boy’s
hands. At t = 0 the boy throws a coconut
upward at a speed v0 , and at the same time
the monkey releases his grip, falling downward
to catch the coconut.
Assume the initial speed of the monkey is 0
m/s, and the cliff is high enough so that the
monkey is able to catch the coconut before
hitting the ground.
The acceleration of gravity is 10 m/s2 .
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
monkey height
C
h
v0
B
hand height
Let yB = 0. The y coordinate of the coconut as a function of time is
1
ycoconut (t) = v0 t − g t2 .
2
The y coordinate of the monkey as a function
of time is
1
ymonkey (t) = h − g t2 .
2
The monkey is able to catch the coconut when
ycoconut = ymonkey .
v0 t −
b
1 2
1
g t = h − g t2
2
2
v0 t = h , so
t=
A
v0
2g
v0
2. t =
g
2h
3. t =
v0
h
4. t =
correct
v0
h
5. t =
2v
s0
2 h v0
+
6. t =
g
g
v0
h
7. t =
+
2 g 2 v0
s
2h
h
8. t =
+
g
2 v0
2 v0
9. t =
g
s
2h
10. t =
g
Explanation:
h
.
v0
ground level
Figure: Consider only vertical motion.
How long t does it take for the monkey to
reach the coconut?
1. t =
4
Question 7, chap 4, sect 4.
part 2 of 3
10 points
Suppose it takes 2 s for the monkey to catch
the coconut and the initial upward speed of
the coconut is 5 m/s . Let the y coordinate
of the ground, the boy, and the monkey be,
respectively, yA = −b , yB = 0 , and yC = h .
What is the y coordinate of the location
where the monkey catches the coconut?
1. y = +5 m
2. y = −5 m
3. y = 0 m
4. y = −20 m
5. y = +15 m
6. y = −10 m correct
7. y = +2 m
8. y = +9 m
9. y = +8 m
10. y = +10 m
Explanation:
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
ycoconut = v0 t −
1 2
gt
2
1
(10 m/s2 ) (2 s)2
2
= (10 m) − (20 m)
= (5 m/s) (2 s) −
= −10 m .
Question 8, chap 4, sect 4.
part 3 of 3
10 points
Caution: Let upward motion be associated
with positive velocity.
Using the same numerical values as part
2, what is the velocity v of the coconut just
before the monkey catches the coconut?
1. v = −5 m/s
2. v = −15 m/s correct
3. v = +10 m/s
4. v = +0 m/s
5. v = +20 m/s
6. v = +5 m/s
7. v = −20 m/s
8. v = +15 m/s
9. v = +25 m/s
10. v = −10 m/s
Explanation:
v = v0 − g t
= (5 m/s) − (10 m/s2 ) (2 s)
= −15 m/s .
Question 9, chap 2, sect 6.
part 1 of 1
10 points
A stone is thrown straight up from the
ground with an initial speed of 23.6 m/s .
5
At the same instant, a stone is dropped from
a height of h meters above ground level. The
two stones strike the ground simultaneously.
The acceleration of gravity is 9.8 m/s2 .
Find the height h.
Correct answer: 113.665 m (tolerance ± 1
%).
Explanation:
Let : v0 = 23.6 m/s .
Consider the thrown stone
1
h0 = h0 + v0 T − g T 2
2
1
0 = T v0 − g T
2
2v
T =
g
2 (23.6 m/s)
=
(9.8 m/s2 )
= 4.81633 s .
Applying the time to the free fall of the second
stone
1
g T2
2
2
1
2v
= g
2
g
2
v
=2
g
(23.6 m/s)2
=2
(9.8 m/s2 )
h=
= 113.665 m .
Question 10, chap 3, sect 3.
part 1 of 2
10 points
A particle undergoes two displacements.
The first has a magnitude of 11 m and makes
an angle of 63 ◦ with the positive x axis.
The result after the two displacements is
8.7 m directed at an angle of 123 ◦ to the
positive x axis using counter-clockwise as the
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
positive angular direction (see the figure below).
= (7.29643 m) − (9.80107 m)
= −2.50464 m .
m
~ is
The magnitude of B
11
8. 7
m
123 ◦
6
63 ◦
~ =
kBk
Find the magnitude of the second displacement.
Correct answer: 10.0494 m (tolerance ± 1
%).
=
q
q
Bx2 + By2
(−9.73226 m)2 + (−2.50464 m)2
= 10.0494 m .
Explanation:
and
R
θR
Find the angle of the second displacement
(measured from the positive x axis, with
counter-clockwise positive within the limits
of −180◦ to +180◦ ).
Correct answer: −165.568 ◦ (tolerance ± 1
%).
Explanation:
A
~ = 11 m ,
Given : kAk
θa = 63 ◦ ,
~ = 8.7 m ,
kRk
θR = 123 ◦ .
Question 11, chap 3, sect 3.
part 2 of 2
10 points
θA
tan θB =
B
θB
~ =A
~ +B
~ , we have B
~ =R
~ − A.
~
Since R
The components of the second displace~ are
ment B
B x = R x − Ax
= R cos θR − A cos θa
= (8.7 m) cos 123 ◦ − (11 m) cos 63 ◦
= (−4.73836 m) − (4.9939 m)
= −9.73226 m and
B y = R y − Ax
= R sin θR − A sin θa
= (8.7 m) sin 123 ◦ − (11 m) sin 63 ◦
By
Bx
By
θB = arctan
B
x
(−2.50464 m)
= arctan
(−9.73226 m)
= −165.568 ◦ .
Question 12, chap 6, sect 99.
part 1 of 3
10 points
A ball tied to the end of a string swings in a
vertical arc under the influence of gravity, as
in the figure.
The acceleration of gravity is 9.8 m/s2 .
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
7
Question 13, chap 6, sect 99.
part 2 of 3
10 points
23. 6
◦
9.8 m/s2
1. 1
9m
m
. 43
/s
1
Find the magnitude of the radial acceleration.
Correct answer: 1.7184 m/s2 (tolerance ± 1
%).
When the ball is at an angle 23.6 ◦ to the
vertical, it has a tangential acceleration of
magnitude g sin θ (produced by the tangential
component of the force m~g ). Therefore, at
θ = 23.6◦ , using the formula aθ = g sin θ , we
obtain aθ = 3.92342 m/s2 .
Find the magnitude of the total acceleration at θ = 23.6◦ .
Correct answer: 4.28324 m/s2 (tolerance ± 1
%).
Explanation:
Since ~a = ~ar + ~at , and the vectors ~ar and
~at are perpendicular, the magnitude of k~ak at
θ = 23.6◦ is
Explanation:
Let : v
r
θ
g
a = k~ak
q
= a2r + a2θ
q
= (1.7184 m/s2 )2 + (3.92342 m/s2 )2
= 1.43 m/s ,
= 1.19 m ,
= 23.6◦ , and
= 9.8 m/s2 .
= 4.28324 m/s2 .
Question 14, chap 6, sect 99.
part 3 of 3
10 points
θ
r
g
v 6= 0
v = 1.43 m/s
Explanation:
From the figure we see that
aθ
= tan φ ,
ar
ar
φ
Calculate the angle between the total acceleration ~a and the string at θ = 23.6◦ .
Correct answer: 66.3473 ◦ (tolerance ± 1 %).
a
aθ
Since v = 1.43 m/s and r = 1.19 m , we find
that the magnitude of the radial acceleration
at this instant is
v2
ar =
r
(1.43 m/s)2
=
(1.19 m)
= 1.7184 m/s2 .
so
aθ
φ = arctan
ar
3.92342 m/s2
= arctan
1.7184 m/s2
= (1.15798 rad) (57.2958 deg/rad)
= 66.3473◦ .
Question 15, chap 4, sect 6.
part 1 of 3
10 points
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
A river flows due North at 2.1 m/s. A boat
crosses the river from the West shore to the
East shore by maintaining a constant velocity
of 9.8 m/s due East relative to the water.
What is the magnitude of the velocity of
the boat relative to the shore?
Correct answer: 10.0225 m/s (tolerance ± 1
%).
Explanation:
North
8
Question 17, chap 4, sect 6.
part 3 of 3
10 points
If the river is 211 m wide, how far downstream is the boat when it reaches the East
shore?
Correct answer: 45.2143 m (tolerance ± 1
%).
Explanation:
vbe
vre
θ
vbr
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
Given : ∆y = 211 m .
∆x = vre ∆t
∆y = vbr ∆t
and
The times are the same, so
∆t =
Given : vre = 2.1 m/s and
vbr = 9.8 m/s .
∆y
∆x
=
vbr
vre
and
~vbe = ~vbr + ~vre .
The velocities are perpendicular, so
q
2 + v2
vbe = vbr
re
q
= (9.8 m/s)2 + (2.1 m/s)2
vre ∆y
vbr
(2.1 m/s) (211 m)
=
9.8 m/s
∆x =
= 45.2143 m .
= 10.0225 m/s .
Question 16, 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: 12.0948 ◦ (tolerance ± 1 %).
Explanation:
Question 18, chap 5, sect 99.
part 1 of 2
10 points
A spherical mass rests upon two wedges, as
seen in the figure below. The sphere and the
wedges are at rest and stay at rest. There is no
friction between the sphere and the wedges.
M
tan θ =
vre
vbr
vre
θ = tan
v
br
−1 2.1 m/s
= tan
9.8 m/s
−1
= 12.0948 ◦ .
The following figures show several attempts
at drawing free-body diagrams for the sphere.
Which figure has the correct directions for
each force?
Note: The magnitude of the forces are not
necessarily drawn to scale.
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
normal
normal
normal
normal
weight
correct
weight
3. Since the sphere is not moving, no forces
act on it.
normal
normal
friction
friction
weight
5.
weight
weight
Fsphere weight
weight
normal
normal
n
normal
1. Since the sphere is not moving, no forces
act on it.
2.
normal
weight
8.
normal
weight
io
ct
fri
7.
normal
weight
The wedges themselves lie on a horizontal
table. There is friction between the table and
the wedges. The following figures show several
attempts at drawing free-body diagrams for
the left wedge.
Which figure has the correct directions for
each force?
Note: The magnitude of the forces are not
necessarily drawn to scale.
normal
3.
weight
e
6.
Question 19, chap 5, sect 99.
part 2 of 2
10 points
he
r
4.
Explanation:
Weight – the force of gravity – pulls the
sphere down. The normal force of the left
wedge upon the sphere acts perpendicular to
(normal to) their surfaces at the point of contact; i.e., diagonally upward and rightward.
Likewise, the normal force of the right wedge
upon the sphere acts diagonally upward and
leftward.
sp
2.
normal
9.
F
1.
weight
9
weight
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
Explanation:
The wedge’s weight pulls it straight down.
The force of the sphere upon the wedge is
the reaction to the force of the wedge upon
~ sphere pushes diagonally
the sphere, hence F
downward and leftward. The normal force
of the table upon the sphere acts straight
up, perpendicular to the table’s horizontal
surface. Finally, the friction force pushes the
wedge rightward in order to prevent it from
~ sphere .
sliding to the left due to F
normal
4.
F
sp
he
r
e
friction
weight
normal
5.
correct
friction
Question 20, chap 5, sect 6.
part 1 of 1
10 points
e
n
normal
weight
fri
ct
io
he
r
sp
F
6.
10
A 7.5 kg object hangs at one end of a rope
that is attached to a support on a railroad
boxcar. When the car accelerates to the right,
the rope makes an angle of 22◦ with the vertical.
The acceleration of gravity is 9.8 m/s2 .
22 ◦
Fsphere weight
7.5 kg
sp
he
r
e
normal
F
7.
a
friction
Find the acceleration of the car.
Correct answer: 3.95946 m/s2 (tolerance ± 1
%).
weight
Explanation:
he
r
e
normal
sp
Given : m = 7.5 kg ,
θ = 22◦ , and
F
8.
friction
g = 9.8 m/s2 .
weight
T cos θ θ
9.
normal
Fs
weight
er
ph
T
T sin θ
e
mg
Vertically
X
Fy = T cos θ − m g = 0
T cos θ = m g .
(1)
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
Horizontally,
X
11
so
Fx = T sin θ = m a .
(2)
Dividing Eqs 1 and 2, we have
T sin θ
a
=
T cos θ
g
a
tan θ =
g
a = g tan θ
= 9.8 m/s2 tan 22◦
= 3.95946 m/s2 .
Question 21, chap 5, sect 2.
part 1 of 1
10 points
Newton’s law of universal gravitation is
F =G
Mm
.
r2
Here, M and m are masses and r is the separation distance. The dimension of force is
specified by the equation F = ma.
What are the SI units of the constant G?
The common expression is [G] = m3 /kg/s2 .
Question 22, chap 1, sect 99.
part 1 of 1
10 points
Given: An acre is an area equivalent to
that of a rectangle 60.5 yd wide and 80 yd
long. There are 36 inches in one yard. There
are 2.54 cm in one inch.
In May 1998, forest fires in southern Mexico
and Guatemala spread smoke all the way to
Austin. Those fires consumed forest land at a
rate of 25000 acres/week.
On the average, how many square meters
of forest are burned down every minute?
Correct answer: 10036.8 m2 /min (tolerance
± 1 %).
Explanation:
week
A conversion factor for
can be easily
min
derived
1. [G] = m/kg/s2
2. [G] = m2 /kg2 /s2
3. [G] = m3 /kg/s2 correct
1 wk 1 day
1 hr
wk
·
·
≡
.
7 days 24 hr 60 min
min
m2
can also be deA conversion factor for
acre
rived
2
yd2
1m
m2
36 in 2.54 cm
·
·
≡
,
1 yd
1 in
100 cm
acre
acre
2
4. [G] = m /kg
5. [G] = N m
6. [G] = kg/m2 /s2
7. [G] = J s/kg
8. [G] = m3 /kg2 /s2
9. [G] = W/m3
10. [G] = N m/s2
Explanation:
F =G
r2
.
Mm
Dimensional analysis of G:
kg m
m2
m3
=
.
s2
(kg)2
(kg s2 )
G=F
Mm
r2
yd2
where
is given in the problem.
acre
Therefore the rate R in square meters per
minutes is
2
m /acre
R=
(25000 acres/week)
week/min
(0.836127 m2 /yd2 )(4840 yd2 /acre)
=
(10080 min/week)
× (25000 acres/week)
= 10036.8 m2 /min .
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
Question 23, chap 2, sect 2.
part 1 of 1
10 points
7.
The graph below shows the velocity v as a
function of time t for an object moving in a
straight line.
v
12
x
t
tQ
tR
tS
tP
0
8. None of these graphs are correct.
x
9.
t
tQ
tR
tS
tP
0
Which of the following graphs shows the
corresponding displacement x as a function of
time t for the same time interval?
x
t
1.
tQ
0
tR
tS
t
0
tQ
tR
tS
tP
Explanation:
The displacement is the integral of the velocity with respect to time:
tP
x
2.
t
tQ
0
3.
tR
tS
tP
x
~x =
t
tQ
0
tR
tS
Z
~v dt .
tP
x
4.
t
0
correct
tQ
tR
tS
tP
x
5.
Because the velocity increases linearly from
zero at first, then remains constant, then decreases linearly to zero, the displacement will
increase at first proportional to time squared,
then increase linearly, and then increase proportional to negative time squared.
From these facts, we can obtain the correct
answer.
x
t
0
tQ
tR
tS
tP
t
x
0
6.
t
0
tQ
tR
tS
tP
tQ
tR
tS
tP
Question 24, chap 2, sect 5.
part 1 of 3
10 points
Consider the plot below describing motion
along a straight line with an initial position of
x0 = 10 m.
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
b
7
b
calculated:
6
v = vi + a (tf − ti )
5
= (0 m/s) + (3.5 m/s2 ) [(2 s) − (0 s)]
= 7 m/s .
4
3
2
Question 26, chap 2, sect 5.
part 3 of 3
10 points
1
0 b
b
−1
−2
1
2
3
b
9
4 5 6 7 8
time (s)
What is the acceleration at 1 second?
Correct answer: 3.5 m/s2 (tolerance ± 1 %).
Explanation:
Basic Concepts: The plot shows a curve
of velocity versus time.
Theacceleration is the
slope of the velocity
dv
∆v
curve a =
.
=
dt
∆t
The position is the area under the velocity
curve Z
1 2
x = x0 + v dt = x0 + v0 t + a t .
2
The area of a triangle is one-half the base
× the height.
The area of a trapezoid is one-half the
height × the sum of the bases.
Solution: The slope of the velocity curve
from 0 seconds to 2 seconds is
vf − vi
a=
tf − ti
(7 m/s) − (0 m/s)
=
(2 s) − (0 s)
= 3.5 m/s2 .
Question 25, chap 2, sect 5.
part 2 of 3
10 points
What is the velocity at 2 seconds?
Correct answer: 7 m/s (tolerance ± 1 %).
Explanation:
The velocity at 2 seconds can be read off
the plot (7 m/s); however, it can also be
What is the position at 2 seconds?
Correct answer: 17 m (tolerance ± 1 %).
Explanation:
The initial position given in the problem is
10 m.
bb
7
b
6
5
4
velocity (m/s)
velocity (m/s)
13
3
2
1
0 bb
−1
b
1
2
3
4 5
time (s)
b
6 7
8
−2
9
b
The position at 2 seconds is 10 meters plus
the area of the triangle (shown in gray in the
above plot)
1
[(2 s) − (0 s)]
2
× [(7 m/s) − (0 m/s)]
= 17 m ;
x = (10 m) +
however, it can also be calculated:
1
(tf − ti )2
2
= (10 m) + (0 m/s) [(2 s) − (0 s)]
1
+ (3.5 m/s2 ) [(2 s) − (0 s)]2
2
= 17 m .
x = xi + vi (tf − ti ) +
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
Question 27, chap 2, sect 5.
part 1 of 1
10 points
An engineer in a locomotive sees a car stuck
on the track at a railroad crossing in front of
the train. When the engineer first sees the
car, the locomotive is 330 m from the crossing
and its speed is 14 m/s.
If the engineer’s reaction time is 0.72 s,
what should be the magnitude of the minimum deceleration to avoid an accident?
Correct answer: 0.306327 m/s2 (tolerance ±
1 %).
Explanation:
While the engineer takes 0.72 s to react, the
train moves
∆d = v t = (14 m/s) (0.72 s) = 10.08 m
forward, so it now has to decelerate to rest
within a displacement of
∆x = (330 m) − (10.08 m) = 319.92 m .
From the equation
vf2 = vi2 + 2 a ∆x
let vf = 0 and solve for a:
−vi2
2 ∆x
−(14 m/s)2
=
2 (319.92 m)
= −0.306327 m/s2
a=
which has a magnitude of 0.306327 m/s2 .
Question 28, chap 4, sect 4.
part 1 of 2
10 points
In a local bar, a customer slides an empty
beer mug on the counter for a refill. The
bartender is momentarily distracted and does
not see the mug, which slides off the counter
and strikes the floor d = 4.72 m from the base
of the counter. The height of the counter is
h = 0.927 m.
The acceleration of gravity is 9.8 m/s2 .
14
How long after sliding off the edge of the
counter does the mug hit the floor?
s
2d
1. t =
g
2d
2. t = √
hg
s
2h
3. t =
correct
g
p
4. t = h g
r
g
5. t =
d
s
2 (d + h)
6. t =
g
r
g
7. t =
h
d
8. t =
2g
9. t = d g
2h
10. t = √
dg
Explanation:
Basic Concept: Two-dimensional motion
with constant acceleration
Solution: Set up the coordinate axes with
the y-axis vertical. The y-component of the
mug’s position vector during free fall is
1
y = y0 + v0y + ay t2
2
Take the ground to be y = 0. Start the stopwatch (t = 0) at the instant the mug leaves
the counter top; so y0 = h. The mug’s velocity is in the horizontal direction as it slides
along the counter, so the instant it leaves
the counter the y-component of its velocity is
zero; v0y = 0. During free fall (ignoring air
resistance) ay = −g, so now we have
1
y = h − g t2 .
2
To find out how long the mug is in free fall
set y = 0 and solve for t
1
0 = h − g t2
2
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
1 2
gt = h
2
s
2h
.
t=
g
Question 29, chap 4, sect 4.
part 2 of 2
10 points
With what speed did the mug leave the
counter?
Correct answer: 10.8518 m/s (tolerance ± 1
%).
Explanation:
Since the mug’s velocity is in the horizontal direction as it slides along the counter, its
speed at the instant it leaves the counter is
equal to the magnitude of the horizontal component of its velocity at that instant. Neglecting air resistance, there is no acceleration in
the horizontal direction so the horizontal component of its velocity remains constant until it
hits the floor. We know it travels a horizontal
distance d in time t, so
15
Correct answer: 6.43877 m/s (tolerance ± 1
%).
Explanation:
vbw = 6.13 m/s is the velocity of the boat
relative to the water, and is a vector directed
Northward. vws = 1.97 m/s, the velocity of
the water relative to the shore, and is directed
East. vbs = the velocity of the boat relative to
the shore, and is a vector directed at an angle
of θ, relative to the Northward direction.
~vbs = ~vbw + ~vws .
The Northward component of vbs is
vbs cos θ = 6.13 m/s .
The Eastward component of vbs is
vbs sin θ = 1.97 m/s .
Then we have
tan θ =
1.97 m/s
,
6.13 m/s
and
θ = 17.8159◦ .
So
d
|v0 | =
t
d
=r
2h
g
r
g
=d
2h
= (4.72 m)
vbs =
6.13 m/s
= 6.43877 m/s .
cos(17.8159◦ )
Question 31, chap 4, sect 6.
part 2 of 2
10 points
s
9.8 m/s2
2 (0.927 m)
= 10.8518 m/s .
Question 30, chap 4, sect 6.
part 1 of 2
10 points
A river flows due East at 1.97 m/s. A
boat crosses the river from the South shore
to the North shore by maintaining a constant
velocity of 6.13 m/s due North relative to the
water. The river is 432 m wide.
What is the velocity of the boat relative to
shore?
How far downstream has the boat moved
by the time it reaches the North shore?
Correct answer: 138.832 m (tolerance ± 1
%).
Explanation:
The time to cross the river is
t=
y
432 m
= 70.4731 s .
=
vbw
6.13 m/s
The Eastward drift is
x = vws t
= 138.832 m .
Question 32, chap 4, sect 6.
part 1 of 3
10 points
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
L
vsg
1
=L
vsg
t1 t2
=
t2 + t1
(30 s) (36.9 s)
=
(36.9 s) + (30 s)
= 16.5471 s .
A shopper in a department store can walk
up a stationary (stalled) escalator in 30 s.
The normally functioning “up” escalator can
carry the standing shopper to the next floor
in 36.9 s. Assume the same walking effort for
the shopper whether the escalator is stalled
or moving.
How long will it take the shopper to walk
up the moving escalator?
Correct answer: 16.5471 s (tolerance ± 1 %).
Explanation:
vse =the velocity of the shopper relative to
the escalator,
veg =the velocity of the escalator relative to
the ground, and
vsg =the velocity of the shopper relative to
the ground.
~vsg = ~vse + ~veg .
t=
Question 33, chap 4, sect 6.
part 2 of 3
10 points
How long will it take the shopper to walk
down this upward moving escalator?
Correct answer: 160.435 s (tolerance ± 1 %).
Explanation:
Assume: The length of the escalator is L.
When veg =0 (the escalator is stalled) and
vsg = vse and
t1 =
L
L
=
= 30 s,
vsg
vse
L
vse =
.
30 s
If vse =0 (the shopper rides up without walking), vsg = veg and
t2 =
~vsg = ~vse − ~veg .
So,
L
t
= vsg − veg
L
L
= −
t1 t2
t2 − t1
=L
.
t1 t2
vsg =
L
L
=
= 36.9 s,
vsg
veg
L
.
36.9 s
If the escalator is running and the shopper
walks up, the time required is
veg =
t=
L
,
vsg
or
So,
L
t
= vsg + veg
L
L
= +
t1 t2
t2 + t1
.
=L
t1 t2
vsg =
16
L
vsg
1
=L
veg
t1 t2
=
t2 − t1
(30 s) (36.9 s)
=
(36.9 s) − (30 s)
= 160.435 s .
t=
Question 34, chap 4, sect 6.
part 3 of 3
10 points
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
The shopper has gone up and down the
upward moving escalator.
However, if the escalator is stalled, the
shopper will make the trip up and down the
escalator
1. quicker. correct
Which figure has the correct directions for
each force?
Note: The magnitude of the forces are not
necessarily drawn to scale.
1.
normal
friction
2. slower.
force
weight
3. None of these.
4. in the same time.
Explanation:
If the escalator is stalled, the travel time is
2.
friction
force
t = 2 t1
= 2 (30 s)
= 60 s .
The times from Part 1 and Part 2 added
together for the round-trip is
normal
weight
3.
t = (16.5471 s) + (160.435 s)
= 176.982 s .
friction
normal
force
weight
Consequently, the shopper’s round-trip is
quicker if the escalator is stalled.
Question 35, chap 5, sect 2.
part 1 of 2
10 points
A book is at rest on an incline as shown
above. A hand, in contact with the top of
the book, produces a constant force Fhand
vertically downward.
4.
force
friction
normal
5.
Fhand
weight
weight
friction
force
B
oo
k
normal
6.
force
weight
friction
normal
The following figures show several attempts
at drawing free-body diagrams for the book.
17
correct
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
7.
normal
force
friction
18
Explanation:
The force that completes the third law pair
with the normal force of the wedge on the
book is the normal force of the book on the
wedge.
weight
Question 37, chap 5, sect 2.
part 1 of 2
10 points
8.
force
normal
friction
weight
Three objects can only move along a
straight, level path. The graphs below show
the position d of each of the objects plotted
as a function of time t.
I
Explanation:
The normal force points perpendicular to
the surface of the inclined plane. The weight
force points down. The Fhand also points
down. The friction force keeps the book from
sliding and consequently points up the incline.
d
II
d
t
III
d
t
t
The magnitude of the velocity k~v k of the
object increases in which of the cases?
1. I and III only
Question 36, chap 5, sect 2.
part 2 of 2
10 points
For the normal force exerted on the book
by the wedge in the diagram, which force(s)
complete(s) the force pair for Newton’s third
law (action-reaction)?
1. the pull of the earth on the book
2. the normal force exerted on the wedge by
the book correct
3. the component of gravity pointing parallel
to the surface of the incline
4. the component of gravity pointing perpendicular to the surface of the incline
5. the component of Fhand pointing perpendicular to the surface of the incline
6. the pull of the book on the earth
7. the sum of the component of gravity perpendicular to the surface of the incline and
the component of Fhand perpendicular to the
surface of the incline
2. II only
3. I and II only
4. III only correct
5. I only
6. II and III only
7. I, II, and III
Explanation:
Case I: The object moves at constant speed.
Case II: The object remains at rest.
Case III: The speed of the object increases
with time; i.e., constant acceleration.
Thus, the magnitude of the velocity of the
object increases only in case III.
Question 38, chap 5, sect 2.
part 2 of 2
10 points
X
The sum of the forces
Fi = 0 on the
object is zero in which of the cases?
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
19
1. I, II, and III
on the car than the car exerts on the truck.
2. II and III only
3. Not enough information is given to pick
one of the answers.
3. I and III only
4. Neither exerts a force on the other; the
car gets smashed simply because it is in the
way of the truck.
4. II only
5. III only
5. None of the answers describes the situation correctly.
6. I only
7. I and II only correct
Explanation:
When the sum of forces on the object is
zero, the acceleration of object is zero by
Newton’s 2nd law.
In cases I and II, the velocity of the object
doesn’t change with the time, so the sum of
the forces on the object is zero.
In case III, the object is accelerating, so the
sum of forces on the object is not zero.
Question 39, chap 5, sect 5.
part 1 of 2
10 points
The following 2 questions refer to the collisions between a car and a light truck whose
weight is equal to that of the car (M = m).
For each description of a collision below,
choose the one answer from the possibilities
that best describes the size (or magnitude) of
the forces between the car and the truck.
Assume: Friction is so small that it can be
ignored.
M
m
v
v
They are both moving at the same speed
when they collide.
1. The truck exerts the same amount of
force on the car as the car exerts on the truck.
correct
2. The truck exerts a greater amount of force
6. The car exerts a greater amount of force
on the truck than the truck exerts on the
car.
Explanation:
By Newton’s third law, action and reaction
are of the same magnitude and in the opposite
direction.
Question 40, chap 5, sect 5.
part 2 of 2
10 points
The light truck is standing still when the
car hits it.
1. Not enough information is given to pick
one of the answers.
2. None of the answers describes the situation correctly.
3. The truck exerts a greater amount of force
on the car than the car exerts on the truck.
4. The truck exerts the same amount of
force on the car as the car exerts on the truck.
correct
5. Neither exerts a force on the other; the
car gets smashed simply because it is in the
way of the truck.
6. The car exerts a greater amount of force
on the truck than the truck exerts on the
car.
Explanation:
The same reason as Part 1.
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
Question 41, chap 5, sect 5.
part 1 of 2
10 points
Consider the 646 N weight held by two
cables shown below. The left-hand cable had
tension T2 and makes an angle of θ2 with
the ceiling. The right-hand cable had tension
440 N and makes an angle of 43 ◦ with the
ceiling.
The right-hand cable makes an angle of 43◦
with the ceiling and has a tension of 440 N .
θ2
T2
0
44
N
a) What is the tension T2 in the left-hand
cable slanted at an angle of θ2 with respect to
the wall?
Correct answer: 472.455 N (tolerance ± 1
%).
Explanation:
Observe the free-body diagram below.
F2
F1
θ1
Wg
Note: The sum of the x- and
y-components of F1 , F2 , and
Wg are equal to zero.
Given : Wg
F1
θ1
θ2
Basic Concept: Vertically and Horizontally,
we have
x
Fnet
= F1x − F2x = 0
= F1 cos θ1 − F2 cos θ2 = 0
(1)
y
y
y
Fnet = F1 + F2 − Wg = 0
= F1 sin θ1 + F2 sin θ2 − Wg = 0 (2)
Solution: Using Eqs. 1 and 2, we have
F2x = F1 cos θ1
= (440 N) cos 43◦
= 321.796 N , and
(1)
◦
43
646 N
θ2
20
= 646 N ,
= 440 N ,
= 43◦ , and
= 90◦ − θ .
F2y = F3 − F1 sin θ1
(2)
◦
= 646 N − (440 N) sin 43
= 646 N − 300.079 N
= 345.921 N , so
q
F2 = (F2x )2 + (F2y )2
q
= (321.796 N)2 + (345.921 N)2
= 472.455 N .
Question 42, chap 5, sect 5.
part 2 of 2
10 points
b) What is the angle θ2 which the left-hand
cable makes with respect to the ceiling?
Correct answer: 47.0692 ◦ (tolerance ± 1 %).
Explanation:
Using Eq. 2, we have
y
F1
θ2 = arctan
Fx
1
300.079 N
= arctan
321.796 N
◦
= 47.0692 .
Question 43, 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 0.215 m.
The plane of the circle is 1.91 m above the
ground. The string breaks and the ball lands
oldmidterm 01 – JYOTHINDRAN, VISHNU – Due: Feb 12 2007, 4:00 am
2.41 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: 69.304 m/s2 (tolerance ± 1
%).
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
no initial velocity in the vertical direction, so
the time spent in the air may be deduced from
the kinematic equation,
y=
Basic Concepts:
⇒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 =
= 69.304 m/s2 .
θ
A
~ has a magnitude of 42.9 and
Vector A
~
points in the positive x-direction. Vector B
has a magnitude of 67.3 and makes an angle
of 39.7◦ with the positive x-axis.
~ − B?
~
What is the magnitude of A
Correct answer: 43.8968 (tolerance ± 1 %).
Explanation:
A-B
Solution:
~ along the x-axis
The component of vector B
is
Bx = B cos θ
= 67.3 cos 39.7◦
= 51.7806
and the component along the y-axis is
By = B sin θ
= 67.3 sin 39.7◦
= 42.9891
~ points in the x-direction, so it has
Vector A
no component along y-axis, and the two com~ −B
~ thus are
ponents for vector A
~ − B)
~ x = Ax − B x
(A
= 42.9 − 51.7806
= −8.88058
and
~ − B)
~ y = Ay − B y
(A
= 0 − 42.9891
= −42.9891
~ −B
~ is
So, magnitude of vector A
q
~
~
|A − B| = (−8.88058)2 + (−42.9891)2
= 43.8968
Question 44, chap 3, sect 3.
part 1 of 1
10 points
Magnitude of Vector
B
1 2
gt .
2
Solving for t,
21