Download Version 055 – Midterm 1

Version 055 – Midterm 1 – OConnor (05141)
This print-out should have 36 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering. V1:1, V2:1, V3:3, V4:2,
V5:1.
001 (part 1 of 1) 10 points
The graph below shows the velocity v as a
function of time t for an object moving in a
straight line.
1
x
t
7.
0
8.
tQ
tR
tS
tP
x
0
t
tQ
tR
tS
tP
x
9.
t
v
0
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
1.
t
0
correct
tQ
tR
tS
tP
x
2.
tQ
tR
tS
tP
Explanation:
The displacement is the integral of the velocity with respect to time
Z
~x = ~v dt .
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
3.
tQ
tR
tS
tP
x
0
0
t
tQ
tR
tS
tP
x
4.
t
0
tQ
tR
tS
t
tP
tQ
5.
t
tQ
tR
tS
1. 2.304 m
3. 3.36 m
tP
0
6. None of these graphs are correct.
tS
tP
002 (part 1 of 1) 10 points
A man jogs at a speed of 1.5 m/s. His dog
waits 1.6 s and then takes off running at a
speed of 4 m/s to catch the man. How far will
they have each traveled when the dog catches
up with the man?
2. 2.94458 m
x
tR
4. 3.705 m
5. 3.84 m correct
Version 055 – Midterm 1 – OConnor (05141)
6. 4.03636 m
2
y
7. 5.032 m
t
3.
8. 6.5025 m
9. 7.26042 m
y
10. 10.472 m
Explanation:
We calculate first the distance that separates the man and the dog when the dog takes
off. It is d0 = vman ∗ twait = 2.4 m. Once the
dog takes off, the position of the dog and the
man at any time t is
t
4.
y
t
5.
xman = d0 + vman t
xdog = vdog t.
They meet when xman = xdog , so d0 +vman t =
vdog t and the time is
t=
d0
= 0.96 s.
vdog − vman
y
t
6.
The distance is
d = vdog ∗ t = 3.84 m.
y
003 (part 1 of 1) 10 points
An object was suspended in a fixed place
(y = 0) and then allowed to drop in a free fall.
Taking up as the positive vertical direction,
which of the following graphs correctly represents its vertical motion as displacement vs
time?
t
7.
y
t
8.
y
t
1.
y
9.
correct
y
2.
t
t
Version 055 – Midterm 1 – OConnor (05141)
3
y
hmax
v0
Explanation:
The object is undergoing a constant downward gravitational acceleration g. The slope
of a position vs time curve represents the
velocity. Given
g 6= 0 m/s2
a < 0 m/s2
a = −g m/s2
v0 = 0 m/s
y0 = 0 m .
Since
y = y0 + v 0 t +
1 2
at ,
2
therefore
1
y = − g t2 .
2
This is a parabolic shaped curve starting a
y = 0 with a continuously decreasing slope
as time increases. Only the figure below is
correct.
y
t
004 (part 1 of 1) 10 points
A ball is thrown upward. Its initial vertical speed v0 , acceleration of gravity g, and
maximum height hmax are shown in the figure
below.
Neglect: Air resistance. The acceleration
of gravity is 9.8 m/s2 .
9.8 m/s2
t
10.
What is its maximum height, hmax (in
terms of the initial speed v0 )?
√ 2
3v
1. hmax = √ 0
2 2g
5 v02
2. hmax =
8g
v2
3. hmax = 0
g
3 v02
4. hmax =
4g
√ 2
5v
5. hmax = √ 0
2 2g
√ 2
3 v0
6. hmax =
2g
v2
7. hmax = 0
4g
v2
8. hmax = 0 correct
2g
v2
9. hmax = √ 0
2g
Explanation:
Basic Concept: For constant acceleration, we have
(1)
v 2 = v02 + 2 a (y − y0 ) .
Solution: The velocity at the top is zero.
Since we know velocities and acceleration, Eq.
1 containing v, a, y0 , and y (no time) is
the easiest one to use. Choose the positive
direction to be up; then a = −g and
0 = v02 + 2 (−g) (hmax − 0)
Version 055 – Midterm 1 – OConnor (05141)
or
Explanation:
Basic Concept:
For constant acceleration, we have
v2
= 0 .
2g
hmax
005 (part 1 of 1) 10 points
A ball is thrown upward. Its initial vertical speed v0 , acceleration of gravity g, and
maximum height hmax are shown in the figure
below.
Given: g = 9.8 m/s2 .
Neglect: Air resistance.
v = v0 + a t .
0 = v0 + (−g) tup
v0
or
v0
.
g
tup =
9.8 m/s2
hmax
(1)
Solution: The velocity at the top is zero.
Since we know velocities and acceleration, Eq.
1 containing v, a, and t . Choose the positive
direction to be up; then a = −g and
4
006 (part 1 of 1) 10 points
A ball is thrown upward. Its initial vertical speed v0 , acceleration of gravity g, and
maximum height hmax are shown in the figure
below.
Neglect: Air resistance. The acceleration
of gravity is 9.8 m/s2 .
2. tup =
3. tup =
4. tup =
5. tup =
6. tup =
7. tup =
8. tup =
9. tup =
v0
2g
4 v0
g
√
2 v0
g
√
3 v0
g
2 v0
g
v
√0
2g
√
2v
√ 0
3g
v0
correct
g
v0
4g
v0
1. tup =
9.8 m/s2
Whatis its time interval, tup (in terms of
the initial speed v0 ), between the release of
the ball and the time it reaches its maximum
height?
hmax
What is its time interval, tup (in terms of
the maximum height hmax ), between the release of the ball and the time it reaches its
maximum height?
√
2g
1. tup =
hmax
√
4g
2. tup =
hmax
√
2 g
3. tup =
hmax
Version 055 – Midterm 1 – OConnor (05141)
5. tup
6. tup
7. tup
8. tup
9. tup
10. tup
while it is in Michael’s hand and after he lets
it go, assuming it has not yet hit the ground?
s
4 hmax
g
√
g
=
2h
s max
2 hmax
=
correct
g
s
hmax
=
2g
g
=
hmax
hmax
=
g
s
hmax
=2
g
4. tup =
a
1 2
at .
2
t
1.
correct
a
t
2.
a
3.
Explanation:
Basic Concept:
For constant acceleration, we have
y = y0 + v 0 t +
5
a
(1)
Solution: The velocity at the top is zero.
Since we know velocities and acceleration, Eq.
1 containing hmax = y − y0 , a, and t . Choose
the positive direction to be up, then a = −g
and
1 2
gt
2
1
= g t 2 − g t2
2
1 2
= gt ,
2
t
t
4.
a
5.
t
hmax = v0 t −
where v0 = g t . Therefore
r
2 ymax
tup =
.
g
007 (part 1 of 1) 10 points
Michael stands motionless holding a baseball in his hand. After a while he tosses it
upwards, and it travels up for a while before turning about and heading towards the
ground. Define upwards to be positive.
Which of the following diagrams can describe the vertical acceleration of the ball,
a
6.
t
Explanation:
The ball first experiences a period of zero
acceleration when Michael is just holding the
ball. Then, as the ball is thrown upward, it
feels an upward acceleration. Quickly the ball
is let loose. Once the ball leaves Michael’s
hand it is in free-fall. In free-fall all objects feel a downward (here negative) constant
gravitational acceleration. This scenario is
shown below.
Version 055 – Midterm 1 – OConnor (05141)
6
Basic Concepts:
a
~a =
t
∆~v
∆t
For constant acceleration
008 (part 1 of 3) 10 points
Given: The acceleration of gravity on Earth
is 9.8 m/s2 .
Consider a ball thrown up from the ground
(the point O). It passes a window (the segment AB) in the time interval 0.314 s (see the
figure). The points in the figure represent the
sequential order, and are not drawn to scale.
The distance AB = 1.42 m.
C
B
y
vi + v f
∆s
=
∆t
2
v = v0 + a t .
v̄ =
For the acceleration of gravity (a = −g)
v = v0 − g t .
Solution: The average velocity is given by
|AB|
∆s
=
∆t
∆t
1.42 m
=
0.314 s
= 4.52229 m/s .
v̄ =
1.42 m
Alternative Part 1:
x
A
d = yB − yA = v A t −
1 2
gt
2
O
Find the average speed as the ball passes
the window.
1. 2.59936 m/s
d+
vA =
1 2
gt
2
t
1
(9.8 m/s2 ) (0.314 s)2
2
=
(0.314 s)
= 6.06089 m/s
1.42 m +
2. 2.67491 m/s
3. 2.80435 m/s
4. 3.20225 m/s
5. 3.47692 m/s
6. 3.68421 m/s
7. 4.52229 m/s correct
8. 5.05576 m/s
9. 5.2459 m/s
10. 5.45455 m/s
Explanation:
vB = v A − g t
= 6.06089 m/s − (9.8 m/s2 ) (0.314 s)
= 2.98369 m/s
vA + v B
2
6.06089 m/s + 2.98369 m/s
=
2
= 4.52229 m/s .
v=
009 (part 2 of 3) 10 points
What is the magnitude of the decrease of the
velocity from A to B?
1. 2.156 m/s
Version 055 – Midterm 1 – OConnor (05141)
2. 2.3912 m/s
5. 0.21444 s
3. 2.6362 m/s
6. 0.304458 s correct
4. 2.7048 m/s
7. 0.313938 s
5. 2.7734 m/s
8. 0.381394 s
6. 2.9498 m/s
9. 0.413296 s
7. 2.989 m/s
8. 3.0772 m/s correct
10. 0.446586 s
Explanation:
The velocity change is
vA − vB = ∆v
9. 3.1654 m/s
10. 3.4888 m/s
Explanation:
For a constant acceleration
~a =
=⇒ vA = vB + ∆v
and the average velocity is
∆~v
,
∆t
=⇒| ∆~v |=| ~a ∆t |= g ∆t .
v̄ =
2 vB + ∆v
vA + v B
=
.
2
2
Thus
Thus the decrease in velocity is
∆v = g ∆t
= (9.8 m/s2 )(0.314 s)
= 3.0772 m/s .
Alternative Part 2:
∆v = vA − vB
= 6.06089 m/s − 2.98369 m/s
= 3.0772 m/s .
010 (part 3 of 3) 10 points
If the ball continues its path upward without
obstruction, find the travel time between B
and C, where point C is at the ball’s maximum
height.
1. 0.109241 s
2. 0.13145 s
3. 0.148158 s
4. 0.192288 s
7
vB = v̄ −
∆v
2
= 4.52229 m/s −
3.0772 m/s
2
= 2.98369 m/s .
The ball reaches its maximum height when its
velocity is zero. Based on v = v0 + a t
vC = 0 = v B − g t
vB
g
2.98369 m/s
=
9.8 m/s2
t=
= 0.304458 s .
011 (part 1 of 3) 10 points
The position of a softball tossed vertically
upward is described by the equation
y = c 1 t − c 2 t2 ,
Version 055 – Midterm 1 – OConnor (05141)
where y is in meters, t in seconds, c1 =
8.59 m/s, and c2 = 7.4 m/s2 .
Find the ball’s initial speed v0 at t0 = 0 s.
1. 3.15 m/s
5. −5.7912 m/s
6. −3.0588 m/s
2. 5.69 m/s
7. −1.63 m/s
3. 5.89 m/s
8. −0.5417 m/s
4. 6.04 m/s
9. 3.64864 m/s
5. 6.61 m/s
8
10. 6.75688 m/s
Explanation:
Substituting t = 1.08 s into the above formula for v, we obtain
6. 6.93 m/s
7. 8.4 m/s
8. 8.59 m/s correct
v = 8.59 m/s − 2(7.4 m/s2 )(1.08 s)
= −7.394 m/s .
9. 10.3 m/s
013 (part 3 of 3) 10 points
Find its acceleration at t = 1.08 s.
1. −15.4 m/s2
10. 10.8 m/s
Explanation:
Basic Concepts:
dx
v=
dt
a=
dv
d2 x
= 2
dt
dt
Solution:
The velocity is simply the derivative of y
with respect to t:
v=
dy
= 8.59 m/s − 2(7.4 m/s2 )t,
dt
which at t = 0 is
v0 = 8.59 m/s .
012 (part 2 of 3) 10 points
Find its velocity at t = 1.08 s.
1. −20.4 m/s
2. −12.989 m/s
3. −7.394 m/s correct
4. −6.6046 m/s
2. −14.8 m/s2 correct
3. −13.9 m/s2
4. −13.14 m/s2
5. −12.98 m/s2
6. −11.3 m/s2
7. −10.1 m/s2
8. −6.46 m/s2
9. −4.56 m/s2
10. −3.96 m/s2
Explanation:
The acceleration is the derivative of velocity
with respect to time:
a=
dv
= −2(7.4 m/s2 ) = −14.8 m/s2 .
dt
014 (part 1 of 2) 10 points
Version 055 – Midterm 1 – OConnor (05141)
A commuter airplane starts from an airport
and takes the route shown in the figure. It
first flies to city A located at 198 km in a
direction 29 ◦ north of east. Next, it flies
135 km 16◦ west of north to city B. Finally,
it flies 180 km due west to city C.
y (km)
N
250
C
200
180 km
km
135
150
R100
50
BW
198
29◦
50
km
16◦
E
S
A
x (km)
100 150 200
How far away from the starting point is city
C?
1. 212.275 km
2. 219.693 km
3. 230.017 km correct
4. 234.482 km
β
b
a
α
The x-component of the resultant is
rx = a x + bx + c x
= a cos α − b sin β − c
= (198 km) cos 29◦ − (135 km) sin 16◦
− 180 km
= −44.0363 km .
The y-component of the resultant is
ry = a y + by + c y
= a sin α + b cos β + 0
= (198 km) sin 29◦ + (135 km) cos 16◦
= 225.763 km .
and the resultant is
q
R = rx 2 + r y 2
q
= (−44.0363 km)2 + (225.763 km)2
= 230.017 km .
5. 239.753 km
6. 246.565 km
015 (part 2 of 2) 10 points
What is the direction of the final position
vector r, measured from North? Use counterclockwise as the positive angular direction,
between the limits of −180◦ and +180◦ .
1. 9.37141 ◦
7. 249.497 km
8. 258.343 km
9. 269.981 km
2. 11.0373 ◦ correct
10. 283.602 km
Explanation:
3. 13.8464 ◦
4. 15.0734 ◦
5. 16.0841 ◦
Given :
9
a = 198 km ,
α = 29◦ ,
b = 135 km ,
β = 16◦ , and
c = 180 km .
6. 16.5333 ◦
7. 19.6103 ◦
8. 21.2758 ◦
Version 055 – Midterm 1 – OConnor (05141)
9. 21.8767 ◦
6. tmax
rx
7. tmax
8. tmax
r
ry
γ
9. tmax
10. tmax
Since γ is the angle between r and the yaxis, for r is in the second quadrant
tan γ =
Thus
|rx |
.
|ry |
Explanation:
The cannonball’s time of flight is
t=
¶
|rx |
γ = tan
|ry |
¶
µ
−1 44.0363 km
= tan
225.763 km
−1
µ
1. hmax
016 (part 1 of 3) 10 points
Denote the initial speed of a cannon ball fired
from a battleship as v0 . When the initial
projectile angle is 45◦ with respect to the
horizontal, it gives a maximum range R.
2. hmax
3. hmax
v0
4. hmax
◦
5. hmax
45
x
R
The time of flight tmax of the cannonball for
this maximum range R is given by
1 v0
4 g
√ v0
= 3
g
√ v0
= 2
correct
g
v0
=4
g
6. hmax
7. hmax
1. tmax =
8. hmax
2. tmax
9. hmax
4. tmax
2 v0y
2 v0 sin 45◦ √ v0
=
= 2
.
g
g
g
017 (part 2 of 3) 10 points
The maximum height hmax of the cannonball
is given by
= 11.0373◦ .
3. tmax
2 v0
3 g
1 v0
=
2 g
√ v0
= 3
g
v0
=2
g
v0
=
g
1 v0
=√
3 g
5. tmax =
10. 22.687 ◦
Explanation:
y
10
10. hmax
v02
=4
g
1 v2
=√ 0
2 g
2 v02
=
3 g
1 v02
=
2 g
√ v02
= 2
g
√ v02
= 3
g
2
1 v0
correct
=
4 g
v02
=
g
v2
=2 0
g
1 v2
=√ 0
3 g
Explanation:
Version 055 – Midterm 1 – OConnor (05141)
oo
k
Use the equation
11
B
2
vy2 = vy0
− 2gh.
At the top of its trajectory vy = 0. Solving
for h yields
h=
2
vy0
2g
v 2 sin2 45◦
= 0
2g
2
1 v0
=
.
4 g
018 (part 3 of 3) 10 points
The speed vhmax of the cannonball at its maximum height is given by
1. vhmax
Which free body diagram best describes
this system?
normal
1.
friction
gravitational
1
= v0
2
2. vhmax = v0
√
3. vhmax = 3 v0
2.
normal
gravitational
4. vhmax = 2 v0
5. vhmax = 4 v0
√
6. vhmax = 2 v0
1
7. vhmax = √ v0 correct
2
2
8. vhmax = v0
3
1
9. vhmax = √ v0
3
1
10. vhmax = v0
4
Explanation:
At the top of the cannonball’s trajectory,
vy = 0. Hence the speed is equal to vx .
1
|v| = vx = v0 cos 45◦ = √ v0 .
2
019 (part 1 of 3) 10 points
Consider a book that remains at rest on an
incline.
friction
3.
normal
gravitational
normal
4.
friction
gravitational
5.
normal
gravitational
correct
Version 055 – Midterm 1 – OConnor (05141)
12
021 (part 3 of 3) 10 points
Are they opposite in direction?
6.
friction
1. Yes, the normal force acts opposite the
weight force because the book is stationary.
gravitational
2. Yes, the normal force always acts opposite
the weight force. Otherwise, the book would
fall through the inclined plane.
7.
friction
normal
3. No, the normal force acts perpendicular
to the surface of the inclined plane. correct
gravitational
4. No, the normal force acts up the incline
so that the book does not slide down.
8.
friction
gravitational
Explanation:
The frictional force keeps the book from
sliding down. The normal force acts perpendicular to the inclined surface. The gravitational force acts down.
020 (part 2 of 3) 10 points
Compare the normal force exerted on the
book by the inclined plane and the weight
force exerted on the book by the earth.
Are they equal in magnitude?
Explanation:
The normal force acts perpendicular to the
surface of the inclined plane. Because the
plane is inclined, this is not opposite in direction to the weight force.
022 (part 1 of 1) 10 points
A man stands in an elevator in the university’s
administration building and is accelerating
upwards. (During peak hours, this does not
happen very often.)
Elevator
Cable
1. Yes
2. No correct
3. Their magnitudes cannot be determined
since the forces are not in the same direction.
Explanation:
They are not equal in magnitude. The
normal force given by N = m g cos θ, with θ
the angle the incline makes with the ground.
Since | cos θ| is less than 1 as long as the incline
is not horizontal, the magnitude of the normal
force, N , will be less than the magnitude of
the weight, m g.
Choose the correct free body diagram for
the man, where Fi,j is the force on the object
i, from the object j.
1.
Fman, elevator Felevator, cable
Version 055 – Midterm 1 – OConnor (05141)
2.
Felevator, cable
13
shown below. The left-hand cable had tension T2 and makes an angle of 44 ◦ with the
ceiling. The right-hand cable had tension T1
and makes an angle of 48◦ with the ceiling.
Fman, earth
T
3.
44 ◦ 48◦
T
1
2
Fman, cable
679 N
a) What is the tension in the cable labeled
T1 slanted at an angle of 48◦ ?
1. 381.046 N
4.
Felevator, cable Fman, floor
2. 385.203 N
3. 392.64 N
Fman, earth
4. 417.991 N
5.
Fman, floor
5. 429.023 N
6. 438.515 N
Fman, earth
7. 452.597 N
8. 457.621 N
correct
6.
Fman, acceleration
9. 463.535 N
Explanation:
Only the forces acting directly on the man
are to be in the free body diagram. Therefore,
the force from the cable should be omitted,
while those from gravity and from the floor’s
normal force should be included.
023 (part 1 of 2) 10 points
Consider the 679 N weight held by two cables
F
2
F
θ2
1
10. 488.729 N correct
Explanation:
Observe the free-body diagram below.
Wg
θ1
Version 055 – Midterm 1 – OConnor (05141)
Note: The sum of the x- and
y-components of F1 , F2 , and
Wg are equal to zero.
Given : Wg = 679 N ,
θ1 = 48◦ , and
θ2 = 44◦ .
Basic Concept: Vertically and Horizontally,
we have
x
Fnet
y
Fnet
F1x
− F2x
=
=0
= F1 cos θ1 − F2 cos θ2 = 0
(1)
y
y
= F1 + F2 − Wg = 0
= F1 sin θ1 + F2 sin θ2 − Wg = 0 (2)
Solution: Using Eq. 1, we have
F2 = F 1
cos θ1
.
cos θ2
14
7. 468.891 N
8. 488.842 N
9. 502.789 N
10. 512.395 N
Explanation:
Using Eq. 1, we have
F2 = F 1
cos θ1
cos θ2
= (488.729 N)
cos 48◦
cos 44◦
= 454.617 N .
025 (part 1 of 3) 10 points
Consider a man standing on a scale which is
placed in an elevator. When the elevator is
stationary, the scale reading is Ss .
Substituting F2 from Eq. 2 into Eq. 1, we have
F1 sin θ1 + F2 sin θ2 = Wg
cos θ1
sin θ2 = Wg
F1 sin θ1 + F1
cos θ2
F1 sin θ1 + F1 cos θ1 tan θ2 = Wg
Wg
F1 =
sin θ1 + cos θ1 tan θ2
679 N
F1 =
◦
sin 48 + cos 48◦ tan 44◦
= 488.729 N
024 (part 2 of 2) 10 points
a) What is the tension in the cable labeled T2
slanted at an angle of 44◦ ?
1. 397.788 N
2. 414.851 N
3. 421.94 N
4. 428.768 N
5. 447.713 N
6. 454.617 N correct
Scale
Find Sup , the scale reading when the elevator is moving upward with acceleration
1
~a = g ̂, in terms of Ss .
5
7
1. Sup = Ss
6
2
2. Sup = Ss
3
4
3. Sup = Ss
5
4. Sup = 0 m/s2
Version 055 – Midterm 1 – OConnor (05141)
3
5. Sup = Ss
5
6
6. Sup = Ss correct
5
3
7. Sup = Ss
2
5
8. Sup = Ss
4
7
9. Sup = Ss
5
5
10. Sup = Ss
6
Explanation:
Basic Concepts: Newton’s 2nd law
X
~ = m ~a .
F
Equation 1 in the upward direction reads
Sup − m g = m a ,
and since a =
1
mg
5
Sup = m g +
6
mg
5
6
= Ss .
5
1
mg
5
=
Thus the scale gives a higher reading when
the elevator is accelerating upwards.
(1)
026 (part 2 of 3) 10 points
Find Sdown , the scale reading when the elevator is moving downward with constant velocity, in terms of Ss .
1. Sdown =
2. Sdown =
3. Sdown =
4. Sdown =
5. Sdown =
Ss − m g = 0 ,
or Ss = m g as we would expect.
Call the scale reading for this part Sup (in
units of Newtons). Consider the free body
diagram for each the case where the elevator
is accelerating down (left) and up (right). The
man is represented as a sphere and the scale
reading is represented as S.
Sup
a
Sdown
a
mg
1
g at this particular instant,
5
Sup − m g =
Solution
We consider the forces acting on the man.
Taking up (̂) as positive, we know that m g
acts on the man in the downward (−̂) direction. The only other force acting on the
~ s from the scale. By
man is the normal force S
the law of action and reaction, the force on
the scale exerted by the man (i.e., the scale
reading) is equal in magnitude but opposite
~ s vector. Initially, the elin direction to the S
evator is moving upward with constant speed
(no acceleration) so
mg
15
6. Sdown =
7. Sdown =
7
Ss
5
5
Ss
6
4
Ss
5
5
Ss
4
3
Ss
2
3
Ss
5
6
Ss
5
8. Sdown = Ss correct
2
Ss
3
7
10. Sdown = Ss
6
Explanation:
At constant velocity the acceleration is
zero.
Equation (1) in the upward direction reads
9. Sdown =
Sdown − m g = 0 ,
Version 055 – Midterm 1 – OConnor (05141)
16
since a = 0 at this particular instant
027 (part 3 of 3) 10 points
Suppose all of a sudden the rope breaks when
the elevator is moving upward.
What is the reading of the scale Sb after the
rope is broken?
1. Sb =
2. Sb =
3. Sb =
4. Sb =
5. Sb =
6. Sb =
7. Sb =
8. Sb =
9. Sb =
2
Ss
3
3
Ss
2
7
Ss
6
4
Ss
5
5
Ss
6
7
Ss
5
3
Ss
5
5
Ss
4
6
Ss
5
θ2
42
0
θ1
0N
Thus the scale gives the same reading as when
stationary if the elevator is moving downward
at constant velocity.
028 (part 1 of 2) 10 points
Hint: sin2 θ + cos2 θ = 1 .
Consider the 674 N weight held by two
cables shown below. The left-hand cable had
tension 420 N and makes an angle of θ2 with
the ceiling. The right-hand cable had tension
500 N and makes an angle of θ1 with the
ceiling.
N
50
Sdown − m g = 0
Sdown = m g − 0
= mg
= Ss .
674 N
a) What is the angle θ1 which the righthand cable makes with respect to the ceiling?
1. 35.9706 ◦
2. 36.5949 ◦
3. 37.0824 ◦
4. 37.9641 ◦
5. 39.627 ◦
6. 41.9197 ◦
7. 48.4217 ◦
8. 49.1665 ◦
10. Sb = 0 N correct
Explanation:
When the rope breaks, the whole system
moves under the influence of gravity with the
acceleration ~a = −g ̂, so
Sb − m g = −m g
Sb = 0 N .
Intuitively this implies that there is no force
between the person and the scale.
Note: There are four different presentations
of this problem.
9. 50.9586 ◦
10. 51.5544 ◦ correct
Explanation:
Observe the free-body diagram below.
Version 055 – Midterm 1 – OConnor (05141)
2
θ1
1
F
F
θ2
(420 N)2
−
2 (500 N) (674 N)
= 51.5544◦ .
¸
029 (part 2 of 2) 10 points
b) What is the angle θ2 which the left-hand
cable makes with respect to the ceiling?
1. 41.1969 ◦
Wg
Note: The sum of the x- and
y-components of F1 , F2 , and
Wg are equal to zero.
Given : Wg = 674 N ,
F1 = 500 N ,
F2 = 420 N .
Basic Concepts:
X
17
2. 42.211 ◦ correct
3. 43.6346 ◦
4. 44.1286 ◦
and
5. 46.0622 ◦
6. 46.9763 ◦
7. 47.6435 ◦
x
F =0
F1x
F1 cos θ1
F12 cos2 θ1
X
and
Fy
= F2x
= F2 cos θ2
= F22 cos2 θ2
8. 51.5378 ◦
(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 .
Solution: Since sin2 θ + cos2 θ = 1 and
adding Eqs. 2 and 3, we have
F22 = F12 − 2 F1 F3 sin θ1 + F32
F 2 + F32 − F22
sin θ1 = 1
2 F 1 F3
µ 2
¶
F3 + F12 − F22
θ1 = arcsin
2 F 1 F3
·
(674 N)2 + (500 N)2
= arcsin
2 (500 N) (674 N)
9. 53.4745 ◦
10. 54.7926 ◦
Explanation:
Using Eq. 1, we have
F1
cos θ1
F2
µ
¶
F1
θ2 = arccos
cos θ1
F2
¶
µ
500 N
◦
cos 51.5544
= arccos
420 N
= 42.211◦ .
cos θ2 =
030 (part 1 of 3) 10 points
A 3.14 kg block slides down a smooth, frictionless plane having an inclination of 32◦ .
The acceleration of gravity is 9.8 m/s2 .
Version 055 – Midterm 1 – OConnor (05141)
18
W|| = m g sin θ = m a
5m
4
3.1
3. 8
Motion has a constant acceleration. Recall
the kinematics of motion with constant acceleration.
Solution: Because the block slides down
along the plane of the ramp, it seems logical
to choose the x-axis in this direction. Then
the y-axis must emerge perpendicular to the
ramp, as shown.
Let us now examine the forces in the xdirection. Only the weight has a component
along that axis. So, by Newton’s second law,
kg
µ=
0
32◦
Find the acceleration of the block.
1. 3.3518 m/s2
2. 3.67115 m/s2
X
3. 3.98602 m/s2
Fx = m a = m g sin θ
Thus
4. 4.14166 m/s2
a = g sin θ
5. 4.29604 m/s2
With our particular value of θ,
6. 4.60082 m/s2
a = (9.8 m/s2 ) sin 32◦ = 5.19321 m/s2
7. 5.04737 m/s2
8. 5.19321 m/s2 correct
031 (part 2 of 3) 10 points
What is the block’s speed when, starting from
rest, it has traveled a distance of 3.85 m along
the incline.
1. 4.03606 m/s
9. 5.7603 m/s2
10. 5.89779 m/s2
Explanation:
2. 4.15573 m/s
Given :
m = 3.14 kg ,
θ = 32◦ , and
µs = 0 .
3. 4.93036 m/s
4. 5.11916 m/s
Consider the free body diagram for the
block
N
m
in
gs
θ
N
os
µN g c
m
=
W = mg
Basic Concepts:
Fx,net = F cos θ − W|| = 0
θ
5. 5.3329 m/s
6. 5.8349 m/s
7. 6.0893 m/s
8. 6.2011 m/s
9. 6.32358 m/s correct
10. 6.74584 m/s
Explanation:
Version 055 – Midterm 1 – OConnor (05141)
Since v0 = 0, a = 5.19321 m/s2 and L =
3.85 m,
vf2 = v02 + 2 a (x − x0 )
= 2 (5.19321 m/s2 ) (3.85 m)
vf = 6.32358 m/s .
It is very important to note at this point that
neither of these values depended on the mass
of the block. This may seem odd at first, but
recall what Galileo discovered 300 years ago –
objects of differing mass fall at the same rate.
032 (part 3 of 3) 10 points
What is the magnitude of the perpendicular
force that the block exerts on the surface of
the plane at a distance of 3.85 m down the
incline?
1. 18.6733 N
2. 19.6422 N
3. 20.138 N
4. 24.5779 N
5. 24.9669 N
6. 26.0961 N correct
19
Two blocks on a frictionless horizontal surface
are connected by a light string.
The acceleration of gravity is 9.8 m/s2 .
12.4 kg
T
20.7 kg
49.8 N
µ
Find the acceleration of the system.
1. 1.45455 m/s2
2. 1.50453 m/s2 correct
3. 1.63548 m/s2
4. 1.67869 m/s2
5. 1.93723 m/s2
6. 1.97674 m/s2
7. 2.04829 m/s2
8. 2.14503 m/s2
9. 2.17932 m/s2
10. 2.25905 m/s2
Explanation:
7. 32.7944 N
Let :
8. 36.7439 N
9. 38.4731 N
10. 39.3247 N
Explanation:
By examining the free body diagram again,
we see that the force in the y direction is given
by
X
F = −m g cos θ = FN .
=⇒
| FN | = m g cos θ
= 3.14 kg (9.8 m/s2 ) cos 32◦
= 26.0961 N.
033 (part 1 of 4) 10 points
m1 = 12.4 kg ,
m2 = 20.7 kg ,
F = 49.8 N , and
µ = 0 , Parts 1 and 2
µ = 0.109 , Parts 3 and 4.
Apply Newton’s second law to the each
block
F1net = m1 a = T ,
(1)
F2net = m2 a = F − T .
Adding these equations gives us
(m1 + m2 ) a = F
a=
F
.
m1 + m 2
(2)
Version 055 – Midterm 1 – OConnor (05141)
034 (part 2 of 4) 10 points
What is the tension in the string between the
blocks?
1. 7.21356 N
20
10. 1.31433 m/s2
Explanation:
The friction force on each block is
Ff r = µ m g .
2. 8.64446 N
3. 9.12074 N
4. 9.31809 N
5. 9.78168 N
6. 11.6628 N
7. 12.9232 N
8. 18.6562 N correct
so apply Newton’s second law to each block
F1net = m1 a1 = T1 − µ m1 g
(3)
F2net = m2 a1 = F − T1 − µ m2 g
(4)
Adding these equations gives
(m1 + m2 ) a1 = F − µ m1 g − µ m2 g
a1 =
a1 =
9. 19.0545 N
10. 19.893 N
Explanation:
From equation 1,
T = m1 a .
F − µ (m1 + m2 ) g
m1 + m 2
F
− µg.
m1 + m 2
036 (part 4 of 4) 10 points
What would be the new tension in the string
between the blocks?
1. 7.21356 N
2. 8.64446 N
035 (part 3 of 4) 10 points
If the surface were frictional, and the coefficient of kinetic friction between each block
and the surface is 0.109, what would be the
new acceleration?
1. 0.436332 m/s2 correct
3. 9.12074 N
4. 9.31809 N
5. 9.78168 N
2. 0.552573 m/s2
6. 11.6628 N
3. 0.714369 m/s2
7. 12.9232 N
4. 0.750544 m/s2
8. 18.6562 N correct
5. 0.870439 m/s2
9. 19.0545 N
6. 0.898744 m/s2
7. 0.927834 m/s2
8. 1.09643 m/s2
9. 1.18841 m/s2
10. 19.893 N
Explanation:
From equation 3,
T1 = µ m 1 g + m 1 a 1 .