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Physics 7B - AB
Lecture 9
May 29
Detailed Relation of Force to Motion
Recap Newtonian Model, Circular Motion
Simple Harmonic Motion
1
Quiz 3 Re-evaluation Request Due
TODAY
Quiz 4 Due June 5 (next Thursday)
Quiz 5 & 6
Due June 9 at the time of Final
Quiz 5 Rubrics on the website
TODAY Quiz 6 (Last Quiz!!!)
2
11 days till…
3
11 days till…
7B Final June 9 Mon 1- 3pm
• Practice Final as well as Quiz problems from Fri
lecture sections are on the course website (solutions
will be posted on Tuesday, June 3)
• Next Week, June 5 is Last lecture
will focus on Final Review = Practice Final Problems
Come prepared!
• Review session schedule (June 5 - 8) will be on the
course web site next week.
4
Final format
6 ~ 8 questions (most likely…)
Quantitative and qualitative questions
Questions are on any material throughout the quarter.
Chapter 5 Fluids, Circuits, Transport, Capacitor/Exponential
Chapter 6 Vectors/Force (Galilean Space-Time Model)
Chapter 7 Momentum/Force, Angular Momentum/Torque
Chapter 8 Newtonian Model, SHM
To do science, one must practise!
But make sure your practice is useful......
available resources : Quiz problems from this quarter, Quiz problems from lecture
section C/D , Practice Final Problems.
5
Recap Detailed Relation of Force to Motion
Which takes longer to hit the ground: a bullet shot
horizontally or a bullet dropped from the same height?
A) The dropped bullet hits the ground first
B) The fired bullet hits the ground first
C) It depends on the mass of the bullet
D) They both hit the ground at the same time
6
Recap Detailed Relation of Force to Motion
Some relevant questions to ask:
What is the vertical component of the initial velocity in
two cases? Are they different?
How is the force diagram look like in two cases?
What is the vertical component of acceleration (while
the bullet is moving toward the ground)?
A) The dropped bullet hits the ground first
B) The fired bullet hits the ground first
C) It depends on the mass of the bullet
D) They both hit the ground at the same time
7
Recap Detailed Relation of Force to Motion
Some relevant questions to ask:
What is the vertical component of the initial velocity in
two cases? Are they different?
How is the force diagram look like in two cases?
What is the vertical component of acceleration (while
the bullet is moving toward the ground)?
A) The dropped bullet hits the ground first
B) The fired bullet hits the ground first
C) It depends on the mass of the bullet
D) They both hit the ground at the same time
8
Recap Detailed Relation of Force to Motion
A rider in a “barrel of fun” is shown to the right. The
18. Afinds
rider
in astuck
“barrel
of fun”
isthe
shown
rider
herself
with her
back to
wall.to the
right. The rider finds herself stuck with her
Which
diagram below correctly shows the forces acting
back to the wall. Which diagram below
on her?
correctly shows the forces acting on her?
Rotating at constant speed
Rotation
direction
other
A)
B)
C)
D)
E)
9
Recap Detailed Relation of Force to Motion
A rider in a “barrel of fun” is shown to the right. The
18. Afinds
rider
in astuck
“barrel
of fun”
isthe
shown
rider
herself
with her
back to
wall.to the
right. The rider finds herself stuck with her
Which
diagram below correctly shows the forces acting
back to the wall. Which diagram below
on her?
correctly shows the forces acting on her?
Rotating at constant speed
Rotation
direction
other
A)
B)
C)
D)
E)
10
Recap Detailed Relation of Force to Motion
19. Consider two carts of masses m and 2m, at rest on a
Consider
two carts of masses M
and 2M, at rest on a frictionless track. If you push
frictionless track.
If you push one cart for 3s and then the
oneother
cart forfor
3s and
thelength
other for
same exerting
length of time,
exerting
force
thethen
same
ofthe
time,
equal
force equal
on each,
the momentum
of the
light
is
on each,
the momentum
of the
lightcart
cart is:
A) four times.
A)
Four times
B) twice
equal
B)C) Twice
to
D) one-half
one-quarter
C)E) Equal
to
the momentum of the heavy cart.
M
2M
D) One-half
E) One quarter
The momentum of the heavy cart
11
Recap Detailed Relation of Force to Motion
19. Consider two carts of masses m and 2m, at rest on a
Consider
two carts of masses M
and 2M, at rest on a frictionless track. If you push
frictionless track.
If you push one cart for 3s and then the
oneother
cart forfor
3s and
thelength
other for
same exerting
length of time,
exerting
force
thethen
same
ofthe
time,
equal
force equal
on each,
the momentum
of the
light
is
on each,
the momentum
of the
lightcart
cart is:
A) four times.
A)
Four times
B) twice
equal
B)C) Twice
to
D) one-half
one-quarter
C)E) Equal
to
the momentum of the heavy cart.
M
2M
D) One-half
E) One quarter
The momentum of the heavy cart
Impulseext = ∆ p = F ave.ext x ∆ t
12
Recap Detailed Relation of Force to Motion
19. Consider two carts of masses m and 2m, at rest on a
Consider
two carts of masses M
and 2M, at rest on a frictionless track. If you push
frictionless track.
If you push one cart for 3s and then the
oneother
cart forfor
3s and
thelength
other for
same exerting
length of time,
exerting
force
thethen
same
ofthe
time,
equal
force equal
on each,
the momentum
of the
light
is
on each,
the momentum
of the
lightcart
cart is:
A) four times.
A)
Four times
B) twice
equal
B)C) Twice
to
D) one-half
one-quarter
C)E) Equal
to
the momentum of the heavy cart.
M
2M
D) One-half
E) One quarter
The momentum of the heavy cart
Impulseext = ∆ p = F ave.ext x ∆ t
13
Recap Detailed Relation of Force to Motion
20. A person spins a tennis ball on a
string in a horizontal circle (so
A personthat
spinsthe
a tennis
a string in
axisball
of on
rotation
isa
horizontal
circle. At theAtpoint
in the
vertical).
theindicated
point indicated
in
the
figure,
ballblow
is in
given
a sharp
figure, the
ball
is giventhe
a sharp
the forward
blow in the forward direction. This
direction.causes
This causes
a change
in the
angular
a change
in the
angular
momentum
L in the
momentum, 

L
z
y
x
F
v
, in the
A) x-direction.
B) y-direction.
C) z-direction.
A) x direction
B) y direction
C) z direction
14
Recap Detailed Relation of Force to Motion
20. A person spins a tennis ball on a
string in a horizontal circle (so
A personthat
spinsthe
a tennis
a string in
axisball
of on
rotation
isa
horizontal
circle. At theAtpoint
in the
vertical).
theindicated
point indicated
in
the
figure,
ballblow
is in
given
a sharp
figure, the
ball
is giventhe
a sharp
the forward
blow in the forward direction. This
direction.causes
This causes
a change
in the
angular
a change
in the
angular
momentum
L in the
momentum, 

L
A) x-direction.
B) y-direction.
C) z-direction.
, in the
z
y
x
∆L
F
v
 (torque exerted
by the blow)
Net Angular Impulseext = ∆ L =  ave.ext x ∆ t
A) x direction
B) y direction
C) z direction
15
Recap Detailed Relation of Force to Motion
20. A person spins a tennis ball on a
string in a horizontal circle (so
A personthat
spinsthe
a tennis
a string in
axisball
of on
rotation
isa
horizontal
circle. At theAtpoint
in the
vertical).
theindicated
point indicated
in
the
figure,
ballblow
is in
given
a sharp
figure, the
ball
is giventhe
a sharp
the forward
blow in the forward direction. This
direction.causes
This causes
a change
in the
angular
a change
in the
angular
momentum
L in the
momentum, 

L
A) x-direction.
B) y-direction.
C) z-direction.
, in the
z
y
x
∆L
F
v
 (torque exerted
by the blow)
Net Angular Impulseext = ∆ L =  ave.ext x ∆ t
A) x direction
B) y direction
C) z direction
16
Recap Detailed Relation of Force to Motion
23. An asteroid is traveling to the
An asteroid
travelingdeep
to thespace
right through
right isthrough
at a
constant
velocity.
The The
path
of of
deep space
at a constant
velocity.
path
the asteroid
shown
the
the asteroid
is shown is
to the
right.toSuddenly,
right. Suddenly it is hit fairly
it is hithard
fairly
by a comet
comes
byhard
a comet
that that
comes
flying
from above
aboveand
andthen
then
bounces
flyingin
in from
bounces
off.
So,
thea asteroid feels a
So theoff.
asteroid
feels
downward force, which acts only
downward
only for
a very
short
for aforce,
verywhich
shortacts
time.
Which
path
intime.
the
Asteroid
is hit
here.
A
B
E
D
C
picture is the most
reasonable for the asteroid to follow after the impact?
Which path in the picture is the most reasonable for the asteroid to follow after
the impact?
17
Recap Detailed Relation of Force to Motion
23. An asteroid is traveling to the
An asteroid
travelingdeep
to thespace
right through
right isthrough
at a
constant
velocity.
The The
path
of of
deep space
at a constant
velocity.
path
the asteroid
shown
the
the asteroid
is shown is
to the
right.toSuddenly,
right. Suddenly it is hit fairly
it is hithard
fairly
by a comet
comes
byhard
a comet
that that
comes
flying
from above
aboveand
andthen
then
bounces
flyingin
in from
bounces
off.
So,
thea asteroid feels a
So theoff.
asteroid
feels
downward force, which acts only
downward
only for
a very
short
for aforce,
verywhich
shortacts
time.
Which
path
intime.
the
Asteroid
is hit
here.
A
B
E
D
C
C
picture is the most
reasonable for the asteroid to follow after the impact?
Which path in the picture is the most reasonable for the asteroid to follow after
the impact?
18
Recap Detailed Relation of Force to Motion
25. An asteroid is traveling to the
Rocket
An asteroid
is traveling
to space
the right
Rocket
engine
right through
deep
at a
starts
startshere.
here
through
deep space
at a constant
constant
velocity
as shown.
Suddenly
a giant
rocket
velocity.
Suddenly,
a giant
rocketengine
that is attached to the asteroid
engine
attached
the asteroid
is which
fired is
upward
so tothat
there is
a constant
is fired
upward sodownward
that thereforce
is a on the
asteroid.
Which
path
in the
E
constant
downward
force
on the
D C
picture is the most reasonable
asteroid.
for the asteroid to follow after
the path
impact?
Which
in the picture is the most reasonable for the asteroid to follow
after the impact?
A
B
19
Recap Detailed Relation of Force to Motion
25. An asteroid is traveling to the
Rocket
An asteroid
is traveling
to space
the right
Rocket
engine
right through
deep
at a
starts
startshere.
here
through
deep space
at a constant
constant
velocity
as shown.
Suddenly
a giant
rocket
velocity.
Suddenly,
a giant
rocketengine
that is attached to the asteroid
engine
attached
the asteroid
is which
fired is
upward
so tothat
there is
a constant
is fired
upward sodownward
that thereforce
is a on the
asteroid.
Which
path
in the
E
constant
downward
force
on the
D C
picture is the most reasonable
asteroid.
for the asteroid to follow after
the path
impact?
Which
in the picture is the most reasonable for the asteroid to follow
after the impact?
A
BB
20
Recap
Detailed Relation of Force to Motion
33. The moon does not crash into the earth because:
The moon
notnot
crashaccelerating
into the Earth because:
A) does
it is
B)
C)
D)
E)
moon
it is not accelerating too much
it is not accelerating toward the earth
it is accelerating away from the earth
more than one of the above
A) It is not accelerating too much
B) It is not accelerating toward the
Earth
Earth
C) It is accelerating away from the
Earth
D) More than one of the above
21
Recap
Detailed Relation of Force to Motion
33. The moon does not crash into the earth because:
The moon
notnot
crashaccelerating
into the Earth because:
A) does
it is
B)
C)
D)
E)
moon
it is not accelerating too much
it is not accelerating toward the earth
it is accelerating away from the earth
more than one of the above
A) It is not accelerating too much
B) It is not accelerating toward the
Earth
Earth
C) It is accelerating away from the
Earth
D) More than one of the above
22
Detailed Relation of Force to Motion
A lot of things oscillate (periodically)
Tuning fork
Atoms in Liquids
and Solids
23
Simple harmonic motion:
is simply a type of motion which follows a repetitive
pattern caused by a restoring force
∑F = – k x
Force is zero at equilibrium. For many systems, the net force
takes this form near equilibrium, provided equilibrium is stable
Particle in a bowl
equilibrium
equilibrium
“Stable” means the net force
pushes back to equilibrium
24
equilibrium
Not all systems are “stable”
We don’t find many unstable systems, as any
small “bump” has already disrupted them
equilibrium
SHM not applicable
tipping point
tipping point
equilibrium
Most realistic systems have SHM like behaviour
close to equilibrium, but behave in very different
ways if they get a large push.
new equilibrium
SHM applicable for small oscillations
near (stable) equilibrium.
The environment
The stock market
etc.
25
Simple harmonic motion:
SHM means that:
∑F = – k x
The nice thing about SHM is we can solve it!
From Newton’s Second Law, ∑F = – k x = ma
From the definition of a, ∑F = – k x = ma = m d2x/dt2
This means, a(t) = d2x(t)/dt2 = – (k/m) x(t)
Math Question
What kind of function x(t) is a function whose second derivcative is proportional to
the negative of the original function?
26
Simple harmonic motion:
SHM means that:
∑F = – k x
The nice thing about SHM is we can solve it!
From Newton’s Second Law, ∑F = – k x = ma
From the definition of a, ∑F = – k x = ma = m d2x/dt2
This means, a(t) = d2x(t)/dt2 = – (k/m) x(t) = – (constant) x(t)
Math Question
What kind of function x(t) is a function whose second derivcative is proportional to
the negative of the original function?
Answer: Sine function!
Where T = 2√m/k, A and  depend on the initial condition,e.g.
27
how far you pull the spring before letting it go.
Simple harmonic motion:
∑F = – k x
Position of the object with above
restoring force exerted on it is SHM,
i.e.
A is the amplitude
x(t)
is the phase constant
responsible for the offset at t = 0
T is the period: time it takes for one
cycle (crest to crest, or trough to trough)
T
A
time
A
T
The motion is identical one period later at any point.
28
Explaining the parameters in SHM:
A is the amplitude
is the phase constant
responsible for the offset
at t = 0
k: spring constant
m: mass
T: is the period
f : frequency
Set by what you do
to the system
Set by what the system
is made of.
 A may change, but T must remain the same.
 The same setup with a different starting push always have the same periods
29
Shown to the right are two systems
undergoing SHM. The vertical
position represents displacement
and the horizontal axis represents
time. How are the two systems
different?
A) They have different periods
B) They have different amplitudes
C) They have different phase constants
D) Only two of the above
E) a, b and c correct
30
Shown to the right are two systems
undergoing SHM. The vertical
position represents displacement
and the horizontal axis represents
time. How are the two systems
different?
A) They have different periods
B) They have different amplitudes
C) They have different phase constants
D) Only two of the above
E) a, b and c correct
31
Shown to the right are two systems
undergoing SHM. The vertical
position represents displacement
and the horizontal axis represents
time. How are the two systems
different?
A) They have different periods
B) They have different amplitudes
C) They have different phase constants
D) Only two of the above
E) a, b and c correct
32
Shown to the right are two systems
undergoing SHM. The vertical
position represents displacement
and the horizontal axis represents
time. How are the two systems
different?
A) They have different periods
B) They have different amplitudes
C) They have different phase constants
D) Only two of the above
E) a, b and c correct
33
Shown to the right are two systems
undergoing SHM. The vertical
position represents displacement
and the horizontal axis represents
time. How are the two systems
different?
A) They have different periods
B) They have different amplitudes
C) They have different phase constants
D) Only two of the above
E) a, b and c correct
34
Shown to the right are two systems
undergoing SHM. The vertical
position represents displacement
and the horizontal axis represents
time. How are the two systems
different?
A) They have different periods
B) They have different amplitudes
C) They have different phase constants
D) Only two of the above
E) a, b and c correct
35
Be sure to write your name, ID number & DL section!!!!!
1
MR 10:30-12:50
Dan Phillips
2
TR 2:10-4:30
Abby Shockley
3
TR 4:40-7:00
John Mahoney
4
TR 7:10-9:30
Ryan James
5
TF 8:00-10:20
Ryan James
6
TF 10:30-12:50
John Mahoney
7
W 10:30-12:50
Brandon Bozek
7
F 2:10-4:30
Brandon Bozek
8
MW 8:00-10:20
Brandon Bozek
9
MW 2:10-4:30
Chris Miller
10 MW 4:40-7:00
Marshall Van Zijll
11 MW 7:10-9:30
Marshall Van Zijll
36