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Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
This print-out should have 72 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Displacement Curve 02
001 (part 1 of 2) 10.0 points
Consider a moving object whose position x
is plotted as a function of the time t. The
object moved in different ways during the
time intervals denoted I, II and III on the
figure.
x
4
2. During each of the three intervals correct
3. During none of the three intervals
4. During interval III only
5. During interval I only
Explanation:
For each of the three intervals I, II or III, the
x(t) curve is linear, so its slope (the velocity
v) is constant. Between the intervals, the
velocity changed in an abrupt manner, but it
did remain constant during each interval.
Distance Time Graph 01
003 (part 1 of 6) 10.0 points
Consider the following graph of motion.
2
I
II
2
III
4
50
t
Distance (m)
6
1
6
During these three intervals, when was the
object’s speed highest? Do not confuse the
speed with the velocity.
40
30
20
10
0
1. Same speed during each of the three intervals.
2. During interval I
3. Same speed during intervals II and III
4. During interval II
5. During interval III correct
Explanation:
The velocity v is the slope of the x(t) curve;
the magnitude v = |v| of this slope is the
speed. The curve is steepest (in absolute
magnitude) during the interval III and that is
when the object had the highest speed.
002 (part 2 of 2) 10.0 points
During which interval(s) did the object’s velocity remain constant?
0
1
2 3 4
Time (sec)
5
How far did the object travel between 2 s
and 4 s?
1. 30 m
2. 10 m
3. 40 m
4. 50 m
5. 20 m correct
Explanation:
The particle moved from 40 m to 20 m, so
∆d = 40 m − 20 m = 20 m .
004 (part 2 of 6) 10.0 points
The graph indicates
1. During interval II only
1. constant position.
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
2. decreasing velocity.
3. 400 m
3. constant velocity. correct
4. 250 m
4. no motion.
5. 350 m correct
5. increasing velocity.
6. 150 m
Explanation:
The slope of the graph is the same everywhere, so the graph indicates constant positive velocity.
005 (part 3 of 6) 10.0 points
What is the speed from 2 s to 4 s?
2. 10 m/s correct
8. 500 m
9. 50 m
10. 100 m
∆d = 400 m − 50 m = 350 m .
3. 5 m/s
4. 20 m/s
007 (part 5 of 6) 10.0 points
The graph indicates
5. 0 m/s
Explanation:
1. no motion.
40 m − 20 m
∆d
=
= 10 m/s .
∆t
2s
006 (part 4 of 6) 10.0 points
Consider the following graph of motion.
400
2. increasing velocity. correct
3. constant velocity.
4. constant position.
5. decreasing velocity.
350
Explanation:
The slopes are steeper as time goes on, so
the velocities are increasing.
300
Distance (m)
7. 450 m
Explanation:
The particle moved from 50 m to 400 m, so
the distance
1. 15 m/s
v=
2
250
200
150
008 (part 6 of 6) 10.0 points
What is the average speed from 3 s to 9 s?
100
50
0
0
1
2
3
4 5 6
Time (sec)
7
8
9
How far did the object travel between 3 s
and 9 s?
1. 20 m/s
2. 60 m/s
1. 200 m
3. 30 m/s
2. 300 m
4. 58 m/s correct
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
3
a
5. 47 m/s
t
4.
6. 40 m/s
7. 36 m/s
a
8. 50 m/s
t
5.
9. 25 m/s
Explanation:
a
v=
∆d
400 m − 50 m
=
= 58 m/s .
∆t
6s
t
6.
a
Acceleration Time Graph 01
009 (part 1 of 5) 10.0 points
t
7.
Consider a toy car which can move to the
right (positive direction) or left on a horizontal surface along a straight line.
a
car
t
8.
v
+
O
What is the acceleration-time graph if the
car moves toward the right (away from the
origin), speeding up at a steady rate?
a
t
1.
Explanation:
Since the car speeds up at a steady rate,
the acceleration is a constant.
010 (part 2 of 5) 10.0 points
What is the acceleration-time graph if the car
moves toward the right, slowing down at a
steady rate?
a
2.
correct
t
a
1.
3. None of these graphs is correct.
t
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
a
4
a
t
2.
t
1.
correct
a
a
t
3.
t
2.
a
a
t
4.
t
3.
a
a
t
5.
t
4.
6. None of these graphs is correct.
a
a
a
a
8.
t
5.
t
7.
t
correct
Explanation:
Since the car slows down, the acceleration
is in the opposite direction.
011 (part 3 of 5) 10.0 points
What is the acceleration-time graph if the car
moves towards the left (toward the origin) at
a constant velocity?
t
6.
7. None of these graphs is correct.
a
8.
t
Explanation:
Since the car moves at a constant velocity,
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
the acceleration is zero.
5
a
012 (part 4 of 5) 10.0 points
What is the acceleration-time graph if the car
moves toward the left, speeding up at a steady
rate?
t
8.
a
t
1.
Explanation:
The same reason as Part 1.
a
013 (part 5 of 5) 10.0 points
What is the acceleration-time graph if the car
moves toward the right at a constant velocity?
t
2.
a
3. None of these graphs is correct.
t
1.
a
t
4.
a
t
2.
a
t
5.
a
t
3.
a
t
6.
a
7.
t
4.
a
t
a
correct
5.
t
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
6
a
t
vx
a
correct
a
vf
y
θ
6.
vy
b
t
7.
8. None of these graphs is correct.
Explanation:
The same reason as Part 3.
Accelerating Between Walls 02
014 (part 1 of 2) 10.0 points
A particle travels horizontally between two
parallel walls separated by 18.4 m. It moves
toward the opposing wall at a constant rate
of 8.5 m/s. Also, it has an acceleration in the
direction parallel to the walls of 3.2 m/s2 .
d
The horizontal motion will carry the particle to the opposite wall, so
d = vx tf
d
18.4 m
tf =
=
= 2.16471 s
vx
8.5 m/s
is the time for the particle to reach the opposite wall.
Horizontally, the particle reaches the maximum parallel distance when it hits the oppod
site wall at the time of t =
, so the final
vx
parallel velocity vy is
(3.2 m/s2 ) (18.4 m)
ad
=
vy = a t =
vx
8.5 m/s
= 6.92706 m/s .
3.2 m/s2
8.5 m/s
18.4 m
What will be its speed when it hits the
opposing wall?
Correct answer: 10.9651 m/s.
Explanation:
The velocities act at right angles to each
other, so the resultant velocity is
q
vf = vx2 + vy2
q
= (8.5 m/s)2 + (6.92706 m/s)2
= 10.9651 m/s .
015 (part 2 of 2) 10.0 points
At what angle with the wall will the particle
strike?
Correct answer: 50.8218◦ .
Let :
d = 18.4 m ,
vx = 8.5 m/s ,
a = 3.2 m/s2 .
Explanation:
When the particle strikes the wall, the vertical component is the side adjacent and the
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
horizontal component is the side opposite the
angle, so
tan θ =
7
vertical velocity and its cosine component is
the initial horizontal velocity. Thus
voy = v sin θ
vx
vy
θ = arctan
= arctan
vx
vy
8.5 m/s
6.92706 m/s
and
vox = v cos θ
= 50.8218◦ .
Frustrated With It All
016 (part 1 of 2) 10.0 points
A disgruntled physics student, frustrated with
finals, releases his tensions by bombarding the
adjacent building, 12 m away, with water balloons. He fires one at 40◦ from the horizontal
with an initial speed of 24.4 m/s.
The acceleration of gravity is 9.8 m/s2 .
For how long is the balloon in the air?
1. 0.76511 s
2. 1.10657 s
3. 0.642004 s correct
4. 4.97959 s
5. 2.21313 s
6. 0.491803 s
7. 1.56492 s
The horizontal motion carries the projectile
to its destination. Since there is no horizontal
acceleration, the horizontal velocity remains
constant and
x = vox t
so that
t=
x
vox
017 (part 2 of 2) 10.0 points
How far above the initial launch height does
the balloon hit the opposing building?
1. 8.04957 m correct
2. 17.6845 m
3. 14.0196 m
4. 9.98037 m
5. 10.0692 m
6. 2.01963 m
7. 13.6453 m
8. 12.0888 m
8. 2.4898 s
Explanation:
Basic Concepts
1
s = so + vo t + at2
2
v = vo + at
The only thing the horizontal and vertical
motions have in common is the time the projectile was in flight.
Solution
Since the initial velocity is inclined at θ to
the horizontal, its sine component is the initial
Explanation:
Vertically, the initial velocity acts upward
and gravity supplies the (downward) acceleration, so
1
y = voy t − gt2
2
Troglodyte War
018 (part 1 of 2) 10.0 points
A colony of troglodytes has been in a lengthy
feud with its neighbors on the adjacent cliff.
Colony A finally develops an important military breakthrough: it rolls bombs off its cliff
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
at known rates of speed, thus gaining pinpoint
accuracy in its attacks.
COLONY
A
The vertical motion carries the bomb to its
target, so
1
h = − g t21
s2
−2 h
t1 =
g
COLONY
B
h
x
If the cliffs are separated by 47.7 m and a
bomb is rolled at 6.6 m/s, how far down the
opposite cliff will it land? The acceleration
due to gravity is 9.8 m/s2 .
8
and the required initial horizontal speed
would be
s
r
−g
−9.8 m/s2
x
=x
= (47.7 m)
v=
t1
2h
2 (−140 m)
= 8.92385 m/s .
Correct answer: 255.944 m.
Explanation:
Let :
x = 47.7 m ,
vo = 6.6 m/s , and
g = 9.8 m/s2 .
Horizontally, there is no acceleration, so
x = vo t ; vertically, there is no initial velocity,
1
so y = − g t2 .
2
The horizontal motion carries the bomb to
x
the opposite cliff, so t =
and the distance
vo
it drops is
9.8 m/s2 (47.7 m)2
1 2 g x2
=
y = gt =
2
2 vo2
2 (6.6 m/s)2
AP M 1998 MC 11
020 10.0 points
A satellite of mass M moves in a circular orbit
of radius R with constant speed v.
True statements about this satellite include
which of the following?
v
I) Its angular speed is .
R
II) Its tangential acceleration is zero.
III) The magnitude of its centripetal acceleration is constant.
1. I only
2. I, II, and III correct
3. II only
4. II and III only
= 255.944 m .
5. I and III only
019 (part 2 of 2) 10.0 points
The troglodyte war continues, and a particularly offensive member of colony B is located
140 m below the top.
At what speed must a bomb be rolled to get
him?
Correct answer: 8.92385 m/s.
Explanation:
Let :
h = −140 m .
Explanation:
Here we will only list some facts of the
circular orbital movement:
v
1. The angular speed is ω = .
R
2. The acceleration points to the center of
v2
the circular orbit with magnitude
= ω 2 R.
R
3. The centripetal force is therefore the
mass M times the acceleration.
2π
2 πR
=
.
4. The period is
v
ω
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
Ball of a Rope
021 (part 1 of 3) 10.0 points
An athlete swings a 3.54 kg ball horizontally
on the end of a rope. The ball moves in a
circle of radius 0.67 m at an angular speed of
0.43 rev/s.
What is the tangential speed of the ball?
9
m v2
rr
Tr
=
m
(152.6 N) (0.67 m)
=
3.54 kg
T =
vmax
= 5.37419 m/s .
Correct answer: 1.81019 m/s.
Moon Orbit
024 (part 1 of 2) 10.0 points
The orbit of a Moon about its planet is approximately circular, with a mean radius of
4.87 × 108 m. It takes 26.3 days for the Moon
to complete one revolution about the planet.
Find the mean orbital speed of the Moon.
Explanation:
Let
r = 0.67 m and
ω = 0.43 rev/s .
Correct answer: 1346.6 m/s.
vt = r ω
= (0.67 m) (0.43 rev/s)
Explanation:
Dividing the length C = 2πr of the trajectory
of the Moon by the time
2π
1 rev
= 1.81019 m/s .
T = 26.3 days = 2.27232 × 106 s
of one revolution (in seconds!), we obtain that
the mean orbital speed of the Moon is
022 (part 2 of 3) 10.0 points
What is its centripetal acceleration?
2πr
C
=
T
T
2 π (4.87 × 108 m )
=
2.27232 × 106 s
= 1346.6 m/s .
Correct answer: 4.8907 m/s2 .
v=
Explanation:
ac = r ω 2
= (0.67 m) (0.43 rev/s)
= 4.8907 m/s2 .
2
2π
1 rev
2
025 (part 2 of 2) 10.0 points
Find the Moon’s centripetal acceleration.
Correct answer: 0.00372349 m/s2 .
023 (part 3 of 3) 10.0 points
If the maximum tension the rope can withstand before breaking is 152.6 N, what is the
maximum tangential speed the ball can have?
Correct answer: 5.37419 m/s.
Explanation:
Let :
T = 152.6 N .
Explanation:
Since the magnitude of the velocity is constant, the tangential acceleration of the Moon
is zero. The centripetal acceleration is
v2
r
(1346.6 m/s )2
=
4.87 × 108 m
= 0.00372349 m/s2 .
ac =
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
Conceptual 04 05
026 (part 1 of 3) 10.0 points
Suzie (of mass 54 kg) is roller-blading down
the sidewalk going 24 miles per hour. She
notices a group of workers down the walkway
who have unexpectedly blocked her path, and
she makes a quick stop in 0.2 seconds.
What is Suzie’s average acceleration?
Correct answer: −53.6333 m/s2 .
10
1. the friction between the air and Suzie
2. the gravity
3. All of these
4. the friction between the ground and the
skates correct
5. the upward force exerted by the ground
Explanation:
Let :
vi = 24 mi/h ,
t = 0.2 s , and
vf = 0 m/s .
First we need to convert Suzie’s initial
speed into meters per second:
24 mi/h ×
Explanation:
The upward force exerted by the ground
balances the gravity. The friction between
the air and Suzie is so small that it can be
ignored. It is the friction between the ground
and the skates that stopped Suzie.
1 hour 1609 m
×
= 10.7267 m/s
3600 s
1 mile
vf − vi
∆v
a=
=
t
t
0 m/s − 10.7267 m/s
=
0.2 s
Force and Motion 04
029 (part 1 of 8) 10.0 points
Consider a toy car which can move to the
right or left along a horizontal line (the positive part of the distance axis) and a force
applied to that car.
= −53.6333 m/s2 .
027 (part 2 of 3) 10.0 points
What force was exerted to stop Suzie?
Correct answer: −2896.2 N.
Explanation:
Let : m = 54 kg .
F = ma
= (54 kg) (−53.6333 m/s2 )
car
v
+
O
Identify the force that would allow the car
to move toward the right (away from the origin) with a steady rate (constant velocity).
Assume friction is so small that it can be ignored.
F
1.
= −2896.2 N .
028 (part 3 of 3) 10.0 points
Where did this force come from?
2. None of these
t
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
F
11
F
t
3.
t
10.
F
t
4.
Explanation:
Constant velocity (in either direction)
means zero acceleration: F = m a = 0 .
F
t
5.
cor-
030 (part 2 of 8) 10.0 points
Identify the force that would allow the car to
remain at rest.
F
rect
t
1.
F
t
6.
F
t
2.
F
t
7.
F
t
3.
F
t
8.
F
t
4.
F
9.
t
rect
F
5.
t
cor-
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
F
F
t
6.
F
t
t
4.
5. None of these
F
t
8.
t
3.
F
7.
12
F
t
6.
F
t
9.
F
t
7.
cor-
10. None of these
rect
Explanation:
Zero velocity means zero acceleration: F =
ma = 0.
F
t
8.
031 (part 3 of 8) 10.0 points
Identify the force that would allow the car
to move toward the right and speed up at a
steady rate (constant acceleration).
F
t
9.
F
t
1.
F
10.
t
F
2.
t
Explanation:
Positive velocity speeding up at a constant
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
rate means constant positive acceleration and
a constant positive force: F = m a > 0 .
032 (part 4 of 8) 10.0 points
Identify the force that would allow the car to
move toward the left at a steady rate (constant velocity).
F
t
8.
F
t
9.
F
13
t
1.
F
2. None of these
t
10.
F
t
3.
Explanation:
Constant velocity (in either direction)
means zero acceleration: F = m a = 0 .
F
t
4.
033 (part 5 of 8) 10.0 points
Identify the force that would allow the car to
move toward the right and slow down at a
steady rate (constant acceleration).
F
t
5.
F
t
1.
F
t
6.
F
t
2.
F
7.
t
corF
3.
rect
t
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
F
F
t
4.
t
1.
F
F
t
5.
t
2.
F
t
6.
14
F
cor-
t
3.
rect
F
F
t
7.
F
F
t
8.
t
4.
t
5.
F
F
9.
t
t
6.
rect
10. None of these
Explanation:
Positive velocity slowing down at a steady
rate means constant negative acceleration and
constant negative force: F = m a < 0 .
034 (part 6 of 8) 10.0 points
Identify the force that would allow the car to
move toward the left and speed up at a steady
rate (constant acceleration).
7. None of these
F
8.
t
cor-
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
F
F
t
9.
t
6.
F
F
t
10.
15
t
7.
Explanation:
Negative velocity speeding up at a steady
rate means constant negative acceleration and
a constant negative force: F = m a < 0 .
F
t
8.
035 (part 7 of 8) 10.0 points
Identify the force that would allow the car
to move toward the right; first speeding up,
then slowing back down.
F
t
9.
F
t
1.
F
10.
t
F
t
2.
F
Explanation:
When the car speeds up, the force is in the
same direction as the velocity: F = m a > 0;
when it slows down, the force is in the opposite
direction: F = m a < 0. The car has to
transition from the positive to the negative
force.
t
3.
F
4.
rect
5. None of these
t
cor-
036 (part 8 of 8) 10.0 points
The car is pushed toward the right and then
released.
Which graph describes the force after the
car is released?
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
F
16
F
t
1.
t
9.
F
F
t
2.
t
10.
Explanation:
After the car is released, no force acts on it.
3. None of these
F
t
4.
rect
cor-
Force on a Bullet 02
037 10.0 points
A 5.5 g bullet leaves the muzzle of a rifle with
a speed of 635.7 m/s.
What constant force is exerted on the bullet
while it is traveling down the 0.9 m length of
the barrel of the rifle?
F
t
5.
Correct answer: 1234.79 N.
Explanation:
Average acceleration can be found from
vf2 = vo2 + 2 a ℓ
F
Since vo = 0, we have
t
6.
a=
v2
2ℓ
Thus
F
t
7.
v2
2ℓ
(5.5 g)(635.7 m/s)2 1 kg
·
=
2 (0.9 m)
1000 g
F = ma = m
= 1234.79 N .
F
8.
t
Net Forces 02
038 (part 1 of 3) 10.0 points
A 54.7 N object is in free fall.
What is the magnitude of the net force
which acts on the object?
Correct answer: 54.7 N.
Explanation:
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
During free fall, the net acceleration is
down, so
X
X
~ net =
~ down −
~ up .
F
F
F
Correct answer: 4.14634 cm.
Explanation:
During free fall with no air resistance, the only
force acting is the weight, so the net force is
its weight.
Let : k1 = 3 N/cm ,
k2 = 5.2 N/cm ,
W = 34 N ,
039 (part 2 of 3) 10.0 points
What is the magnitude of the net force when
the object encounters 14.3 N of air resistance?
Correct answer: 40.4 N.
Explanation:
The weight acts down and the air resistance acts up with the net acceleration acting
downward, so
~ net = W
~ −F
~ = 54.7 N − 14.3 N = 40.4 N .
F
17
and
The springs stretch the same amount x because of the way they were positioned.
Then F = k x , so
X
Fup =
X
Fdown
k1 x + k2 x = W
x=
040 (part 3 of 3) 10.0 points
What is the magnitude of the net force when it
falls fast enough to encounter an air resistance
of 54.7 N?
=
W
k1 + k2
34 N
3 N/cm + 5.2 N/cm
= 4.14634 cm .
Correct answer: 0 N.
Explanation:
The net force is now zero, since the air
resistance is the same as the weight.
5.2 N/cm
3 N/cm
Parallel and Series Springs
041 (part 1 of 5) 10.0 points
In the parallel spring system, the springs are
positioned so that the 34 N weight stretches
each spring equally. The spring constant
for the left-hand spring is 3 N/cm and the
spring constant for the right-hand spring is
5.2 N/cm .
34 N
How far down will the 34 N weight stretch
the springs?
042 (part 2 of 5) 10.0 points
In this same parallel spring system, what is
the effective combined spring constant kpar of
the two springs?
Correct answer: 8.2 N/cm.
Explanation:
Considering this as a one-spring system, it
would react with a force F = −k x = −W due
the law of action and reaction, so
kpar =
−W
W
34 N
=
=
= 8.2 N/cm ,
−x
x
4.14634 cm
which is the sum k1 + k2 of the individual
constants.
043 (part 3 of 5) 10.0 points
Now consider the same two springs in series.
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
5.2 N/cm
3 N/cm
kseries =
34 N
What distance will the spring of constant
3 N/cm stretch?
Correct answer: 11.3333 cm.
W
34 N
=
x1 + x2
11.3333 cm + 6.53846 cm
= 1.90244 N/cm .
AP M 1998 MC 14 15
046 (part 1 of 2) 10.0 points
A spring has a force constant of 977 N/m and
an unstretched length of 6 cm. One end is
attached to a post that is free to rotate in the
center of a smooth table, as shown in the top
view below. The other end is attached to a
3 kg disk moving in uniform circular motion
on the table, which stretches the spring by
3 cm.
Note: Friction is negligible.
Explanation:
In the series system, the springs stretch
a different amount, but each carries the full
weight W = 34 N.
977 N/m
3 kg
9 cm
W = k 1 x1
W
34 N
x1 =
=
= 11.3333 cm .
k1
3 N/cm
044 (part 4 of 5) 10.0 points
In this same series spring system, what distance will the spring of constant 5.2 N/cm
stretch?
18
What is the centripetal force Fc on the disk?
Correct answer: 29.31 N.
Explanation:
Correct answer: 6.53846 cm.
Explanation:
W = k 2 x2
W
34 N
x2 =
=
= 6.53846 cm .
k2
5.2 N/cm
045 (part 5 of 5) 10.0 points
In this same series spring system, what is the
effective combined spring constant kseries of
the two springs?
Correct answer: 1.90244 N/cm.
Explanation:
xtotal = x1 + x2 , so
W = kseries xtotal
Let : r = 6 cm = 0.06 m ,
∆r = 3 cm = 0.03 m ,
m = 3 kg , and
k = 977 N/m .
The centripetal force is supplied only by
the spring. Given the force constant and the
extension of the spring, we can calculate the
force as
Fc = k ∆r
= (977 N/m) (0.03 m)
= 29.31 N .
047 (part 2 of 2) 10.0 points
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
What is the work done on the disk by the
spring during one full circle?
19
m
r
1. W = 11.0496 J
v
2. W = 0 J correct
3. W = 5.5248 J
Mg
Since the suspended mass is in equilibrium,
the tension is
T = M g = (0.88 kg) 9.8 m/s2
4. W = 0.00633096 J
5. W = 1.7586 J
= 8.624 N .
Explanation:
Since the force is always perpendicular to
the movement of the disk, the work done by
the spring is zero .
Serway CP 07 25
048 (part 1 of 3) 10.0 points
An air puck of mass 0.27 kg is tied to a string
and allowed to revolve in a circle of radius
0.91 m on a horizontal, frictionless table. The
other end of the string passes through a hole
in the center of the table and a mass of 0.88 kg
is tied to it. The suspended mass remains in
equilibrium while the puck revolves.
0.27 kg
0.91 m
049 (part 2 of 3) 10.0 points
What is the horizontal force acting on the
puck?
Correct answer: 8.624 N.
Explanation:
The horizontal force acting on the puck is
the tension in the string, so
Fc = T = 8.624 N .
050 (part 3 of 3) 10.0 points
What is the speed of the puck?
Correct answer: 5.3913 m/s.
Explanation:
v
0.88 kg
What is the tension in the string? The
acceleration due to gravity is 9.8 m/s2 .
m v2
Fc =
r
s
r
Fc r
(8.624 N) (0.91 m)
v=
=
m
0.27 kg
= 5.3913 m/s .
Correct answer: 8.624 N.
Work Done on a Block
051 10.0 points
Lee pushes horizontally with a force of 51 N
on a 26 kg mass for 27 m across a floor.
Calculate the amount of work Lee did.
Explanation:
Let :
M = 0.88 kg and
g = 9.8 m/s2 .
Correct answer: 1377 J.
Explanation:
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
Let :
F = 51 N and
d = 27 m .
The mass of the object is ignored since there
is no friction present, so
W = F d = (51 N) (27 m) = 1377 J .
Sliding Into a Loop
052 10.0 points
A 5.26 kg block is released from A at height
5.3 m on a frictionless track shown. The
radius of the track is 2.97 m.
The acceleration of gravity is 9.8 m/s2 .
A
20
So, the total acceleration at P is
q
a = a2t + a2r
q
= (9.8 m/s2 )2 + (15.3764 m/s2 )2
= 18.2339 m/s2 .
Dropping a Baseball
053 10.0 points
You drop a 150 g baseball from a window 15 m
above the ground.
What is the kinetic energy of the baseball
when it hits the ground? The acceleration
due to gravity is 10 m/s2 .
1. 22.5 J correct
2. 0 J
3. 2.25 × 104 J
h
P
R
4. 1.125 × 105 J
5. 1.125 J
Determine the magnitude of the acceleration for the block at P.
Correct answer: 18.2339 m/s2 .
Explanation:
From conservation of energy
(K + U )A = (K + U )P
we obtain
mgh =
m v2
2
Let :
m = 150 g ,
h = 15 m , and
g = 10 m/s2
Energy is conserved:
K i + Ui = K f + Uf
The initial kinetic energy and final potential energy are zero,
+ mgR.
K i = Uf = 0 ,
and the initial potential energy is the gravitational potential energy,
Therefore
v=
Explanation:
p
2 g (h − R) = 6.75781 m/s .
Then the radial acceleration at P is
ar =
v2
= 15.3764 m/s2
R
and the tangential acceleration is
at = g = 9.8 m/s2 .
Ui = m g h ,
so the kinetic energy of the baseball when it
hits the ground is
K f = Ui = m g h
kg
= 150 g × 3
(10 m/s2 )(15 m)
10 g
= 22.5 J .
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
21
q
= 2 (9.8 m/s2 )(2.8 m − 0.5 m)
Girl on Swing 02
054 (part 1 of 2) 10.0 points
= 6.71416 m/s .
A girl swings on a playground swing in such
a way that at her highest point she is 2.8 m
from the ground, while at her lowest point she
is 0.5 m from the ground.
055 (part 2 of 2) 10.0 points
At what height above the ground will the
girl be moving at a speed half of her maximum
speed?
Correct answer: 2.225 m.
7.4 m
2.8 m
0.5 m
Explanation:
Let h1/2 = the height, where v = v1/2 =
1
vmax .
2
What is her maximum speed?
The acceleration of gravity is 9.8 m/s2 .
r
v1
Correct answer: 6.71416 m/s.
/2
Explanation:
Let : r = 7.4 m ,
htop = 2.8 m ,
hbot = 0.5 m .
h1/2
v htop
hbottom
From conservation of energy,
and
We can solve this by using the principle of
conservation of energy. We need to know the
kinetic and potential energies at two points
in time. The girl will be moving the fastest
when her kinetic energy is largest which occurs when her potential energy is smallest.
This means that she will be moving fastest at
the bottom of the swing. By conservation of
energy,
Etop = Ebot
Ktop + Utop = Kbot + Ubot
1
1
2
2
m vtop
+ m g htop = m vbot
+ m g hbot .
2
2
Since vtop = 0,
1
2
m vbot
= m g (htop − hbot ) so
2
vmax = vbot
q
= 2 g (htop − hbot )
K1/2 + U1/2 = Ktop + Utop
1
2
m v1/2
+ m g h1/2 = 0 + m g htop
2
1 1 2
m g h1/2 = m g htop −
.
v
2 4 max
From Part 1,
1
2
m vmax
= m g (htop − hbot ) .
2
Substitution yields
m g h1/2 = m g htop −
1
m g (htop − hbot )
4
1
(3 htop + hbot )
4
1
= [3 (2.8 m) + 0.5 m]
4
= 2.225 m .
h1/2 =
Decaying Uranium Nucleus
056 10.0 points
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
A uranium nucleus 238 U may stay in one piece
for billions of years, but sooner or later it decays into an α particle of mass 6.64 × 10−27 kg
and 234 Th nucleus of mass 3.88 × 10−25 kg,
and the decay process itself is extremely fast
(it takes about 10−20 s). Suppose the uranium
nucleus was at rest just before the decay.
If the α particle is emitted at a speed of
7.51×106 m/s, what would be the recoil speed
of the thorium nucleus?
Correct answer: 1.28522 × 105 m/s.
Explanation:
22
Correct answer: 1.10414 m/s.
Explanation:
Let : m = 19.3 kg ,
M = 2150 kg , and
v = 123 m/s .
The cannon’s velocity immediately after it
was fired is found by using conservation of
momentum along the horizontal direction:
M V + mv = 0
Let : vα = 7.51 × 106 m/s ,
Mα = 6.64 × 10−27 kg ,
MTh = 3.88 × 10−25 kg .
and
1
Since the speeds involved are less than
10
of the speed of light, we can ignore the effects of special relativity. Use momentum
conservation: Before the decay, the Uranium
nucleus had zero momentum (it was at rest),
and hence the net momentum vector of the
decay products should total to zero:
~ tot = Mα ~vα + MTh ~vTh = 0 .
P
This means that the Thorium nucleus recoils
in the direction exactly opposite to that of the
α particle with speed
k~vα k Mα
MTh
(7.51 × 106 m/s) (6.64 × 10−27 kg)
=
3.88 × 10−25 kg
k~vTh k =
= 1.28522 × 105 m/s .
Cannon Recoil
057 (part 1 of 2) 10.0 points
A revolutionary war cannon, with a mass of
2150 kg, fires a 19.3 kg ball horizontally. The
cannonball has a speed of 123 m/s after it has
left the barrel. The cannon carriage is on a
flat platform and is free to roll horizontally.
What is the speed of the cannon immediately after it was fired?
m
v
M
where M is the mass of the cannon, V is the
velocity of the cannon, m is the mass of the
cannon ball and v is the velocity of the cannon
ball. Thus, the cannon’s speed is
⇒ −V =
m
|v|
M
19.3 kg
(123 m/s)
=
2150 kg
|V | =
= 1.10414 m/s .
058 (part 2 of 2) 10.0 points
The same explosive charge is used, so the total
energy of the cannon plus cannonball system
remains the same.
Disregarding friction, how much faster
would the ball travel if the cannon were
mounted rigidly and all other parameters remained the same?
Correct answer: 0.550836 m/s.
Explanation:
By knowing the speeds of the cannon and
the cannon ball, we can find out the total
kinetic energy available to the system
Knet =
1
1
m v2 + M V 2 .
2
2
This is the same amount of energy available
as when the cannon is fixed. Let v ′ be the
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
speed of the cannon ball when the cannon is
held fixed. Then,
1
1
m v ′2 = (m v 2 + M V 2 ) .
2
2
r
M 2
V
v2 +
m
r
m
=v 1+
M
s
⇒ v′ =
= (123 m/s) 1 +
23
060 (part 1 of 2) 10.0 points
An impulse of 149 N s is required to stop a
person’s head in a car collision.
If the face is in contact with the steering
wheel for 0.0279 s, what is the average force
on the cheekbone?
Correct answer: 5340.5 N.
Explanation:
19.3 kg
2150 kg
= 123.551 m/s .
∆p = 149 N s and
∆t = 0.0279 s .
Force is related to impulse by
Thus, the velocity difference is
I ≡ ∆p = F ∆t ,
v ′ − v = 123.551 m/s − 123 m/s
so
= 0.550836 m/s .
F =
∆p
149 N s
=
= 5340.5 N .
∆t
0.0279 s
Boxing Match
059 10.0 points
The linear impulse delivered by the hit of a
boxer is 320 N · s during the 0.422 s of contact.
What is the magnitude of the average force
exerted on the glove by the other boxer?
061 (part 2 of 2) 10.0 points
If an average force of 902 N fractures the
cheekbone, how long must it be in contact
with the steering wheel in order to fracture?
Correct answer: 758.294 N.
Correct answer: 0.165188 s.
Explanation:
Explanation:
Let :
Let : ∆t = 0.422 s and
∆p = 320 N · s .
The average force exerted is
F=
∆p
,
∆t
and its magnitude is
∆p
∆t
320 N · s
=
0.422 s
= 758.294 N .
F =
∆t =
F1 = 902 N .
∆p
149 N s
=
= 0.165188 s .
F1
902 N
Bullet Moves a Block
062 10.0 points
A(n) 23.3 g bullet is shot into a(n) 4976 g
wooden block standing on a frictionless surface. The block, with the bullet in it, acquires
a speed of 1.55 m/s.
Calculate the speed of the bullet before
striking the block.
Correct answer: 332.571 m/s.
Your Body in a Collision
Explanation:
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
Basic concepts: Momentum of any object
is
24
of mass, as shown below. Each star has mass
M and speed v.
p = mv.
Solution: The collision is inelastic, and by
conservation of momentum
pbef ore = paf ter
v
M
v
M
mb vb + 0 = (mb + mw ) vf
vb =
(mb + mw ) vf
.
mb
AP B 1993 MC 6
063 10.0 points
If Spacecraft X has twice the mass of Spacecraft Y , then what is true about X and Y ?
I) On Earth, X experiences twice the gravitational force that Y experiences;
II) On the Moon, X has twice the weight of
Y;
III) When both are in the same circular orbit,
X has twice the centripetal acceleration
of Y .
1. I only
2. I, II, and III
3. III only
4. I and II only correct
5. II and III only
Explanation:
I) gravitational force ∝ mass.
II) weight ∝ mass.
III) The centripetal acceleration is determined by
v2
ac =
,
r
so X and Y should have the same centripetal
acceleration when they are in the same circular orbit.
AP M 1998 MC 20
064 10.0 points
Two identical stars, a fixed distance D apart,
revolve in a circle about their mutual center
D
Which of the following is a correct relationship among these quantities? G is the
universal gravitational constant.
1. v 2 = M G D
4 G M2
D
4GM
3. v 2 =
D
2 G M2
4. v 2 =
D
2GM
5. v 2 =
D
GM
6. v 2 =
correct
2D
GM
7. v 2 =
D2
GM
8. v 2 =
D
Explanation:
From circular orbital movement, the cenv2
2 v2
tripetal acceleration is a =
=
.
D
D
2
Using Newton’s second law of motion, the
acceleration is
2. v 2 =
a=
1 G M2
GM
F
=
·
=
2
M
M
D
D2
2 v2
F
GM
=a=
=
D
M
D2
GM
.
v2 =
2D
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
Circular Moon Orbit
065 (part 1 of 3) 10.0 points
A satellite is in a circular orbit just above the
surface of the Moon.
What is the satellite’s acceleration?
The value of gravitational constant is
6.67259 × 10−11 Ncdotm2 /kg2 and the mass
of the moon is 7.36 × 1022 kg and its radius is
2155.2 km .
25
Correct answer: 2.49185 h.
Explanation:
2πR
v
2 π (2.1552 × 106 m)
1h
=
×
1509.53 m/s
3600 s
T =
= 2.49185 h .
2
Correct answer: 1.0573 m/s .
Explanation:
Let : G = 6.67259 × 10−11 Ncdotm2 /kg2 ,
Mmoon = 7.36 × 1022 kg , and
rmoon = 2155.2 km = 2.1552 × 106 m .
Fc = m ac = FG , so
G m Mmoon
r2
G Mmoon
ac =
r2
= (6.67259 × 10−11 Ncdotm2 /kg2 )
7.36 × 1022 kg
×
(2.1552 × 106 m)2
m ac =
= 1.0573 m/s2 .
066 (part 2 of 3) 10.0 points
What is the satellite’s speed?
Correct answer: 1509.53 m/s.
Explanation:
v2
r
√
v = ac r
q
= (1.0573 m/s2 )(2.1552 × 106 m)
ac =
= 1509.53 m/s .
Tipler PSE5 11 58
068 10.0 points
The gravitational field at some point is given
by ~g = (2.8 × 10−6 N/kg) ̂.
What is the gravitational force on a mass
of 7.12 g at that point?
Correct answer: 1.9936 × 10−8 N.
Explanation:
Let : ~g = (2.8 × 10−6 N/kg) ̂ and
m = 7.12 g = 0.00712 kg .
The gravitational force is
~
F
m
~ = m~g
F
= (0.00712 kg) (2.8 × 10−6 N/kg) ̂
~g =
= (1.9936 × 10−8 N) ̂ .
Launch Energy
069 10.0 points
What is the minimum kinetic energy needed
to launch a payload of mass m to an altitude
that is one Earth radius, RE , above the surface of the Earth (the payload will then fall
back to Earth)? (Note that ME is the mass of
the Earth.)
G m ME
RE
G m ME
2. 0.25
RE
1. 0.1
067 (part 3 of 3) 10.0 points
What is the period of the satellite’s orbit?
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
G m ME
RE
G m ME
correct
4. 0.5
RE
G m ME
5.
RE
Explanation:
Energy is conserved:
3. 2
K i + Ui = K f + Uf
The kinetic energy after the payload has
been launched to an altitude RE above the
Earth’s surface is
Kf = 0 .
The potential energy at Earth’s surface before
the launch is
Ui = −
G ME m
,
RE
and the potential energy at altitude RE above
the Earth’s surface after the launch is
Ui = −
Ki =
G m ME
G ME m G ME m
−
=
RE
2RE
2 RE
New Satellite Orbit 02
070 (part 1 of 3) 10.0 points
What is the kinetic energy of a satellite of
mass m which is in a circular orbit of radius
3 Re about the earth?
m v2
3 Re
m v2
2. K =
6 Re
1. K =
3. K = m g Re
4. K =
G Me m
Re2
G Me m
6. K =
3 Re
5. K =
7. K = 3 m g Re
8. K = −
G Me m
correct
6 Re
G Me m
3 Re
9. K = 3 G Me m
10. K = −
G Me m
6 Re2
Explanation:
The acceleration of the satellite in circular
v2
orbit of radius 3 Re is ac =
, so the force
3 Re
on the satellite is
F = m ac =
m v2
G Me m
=
3 Re
(3 Re)2
G Me m
m v2 =
e Re
and the kinetic energy is
G ME m
.
2RE
So, the minimum energy needed to launch
the payload to RE above Earth’s surface is
26
K=
1
G Me m
m v2 =
.
2
6 Re
071 (part 2 of 3) 10.0 points
What is the total energy of the satellite?
G Me m
6 Re2
G Me m
2. E =
3 Re
Gm
3. E =
Re
1. E = −
4. E = 3 G Me m
5. E = m g Re +
6. E =
1
m v2
2
G Me m
Re
7. E = 3 m g Re +
8. E = −
G Me m
3 Re
G Me m
3 Re
Version 001 – Review 1: Mechanics – tubman – (IBII20142015)
G Me m
correct
6 Re
Explanation:
The potential energy of the satellite is
G Me m
U =−
3 Re
so the total energy is
G Me m G Me m
−
E =K +U =
6 Re
3 Re
G Me m
=−
.
6 Re
9. E = −
072 (part 3 of 3) 10.0 points
How much work must an external force
do on the satellite to move it from a circular orbit of radius 2 Re to 3 Re , if its mass
is 2000 kg?
The universal gravitational
constant 6.67 × 10−11 N · m2 /kg2 , the mass
of the Earth 5.98 × 1024 kg and its radius
6.37 × 106 m.
Correct answer: 1.04361 × 1010 J.
Explanation:
Let :
G = 6.67 × 10−11 N · m2 /kg2 ,
Me = 5.98 × 1024 kg , and
Re = 6.37 × 106 m .
The work done by an external force to move
the satellite from the closer orbit to the further orbit will be the work against gravity (a
positive number which yields the change in
potential energy) plus the change in kinetic
energy (a negative number since the kinetic
energy is smaller in the orbit with the greatest
radius):
G Me m
G Me m
W = Ef − Ei = −
− −
6 Re
4 Re
G Me m
=
12 Re
1
=
(6.67 × 10−11 N · m2 /kg2 )
12
(5.98 × 1024 kg)(2000 kg)
×
6.37 × 106 m
= 1.04361 × 1010 J .
27