Physics Exam Preparation - Power, Energy, Momentum
Exam#1 (
chapter 1-6
)
time:
Tuesday 02/19 8:00 pm- 10:00 pm
Location:
physics building room 112
If you can not make it, please let me know by Friday 02/15 so that I can arrange
a make-up exam.
If you have special needs, e.g. exam time extension, and has not contact me
before, please bring me the letter from the Office of the Dean of Students before
Friday 02/15.
AOB
•
Prepare your own scratch paper, pencils, erasers, etc.
•
Use only
pencil
for the answer sheet
•
Bring your own calculators
•
No cell phones, no text messaging, no computers
•
No crib sheet of any kind is allowed. Equation sheet will be provided.
1
2
2/16/2025
2
Power
•
It is not only important how much
work is done but also how quick, i.e.
the rate, at which work is done
•
So the quantity
•
Power = P = W/t (unit is a watt)
•
is very important.
•
Generally energy supplies, motors, etc are rated by power and one can
determine how much work can be done by multiplying by time.
•
W = Pt (joules).
3
Examples of Watt and Joules
T
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–
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.
A
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–
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1
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0
0
w
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3
6
0
0
s
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c
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n
d
s
1
k
W
H
r
=
1
0
0
0
*
3
6
0
0
=
3
.
6
m
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f
1
0
0
0
k
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1
0
0
m
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q
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s
1
0
0
0
*
9
.
8
*
1
0
0
j
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b
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2
0
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c
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d
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.
T
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p
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w
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d
i
s
1
0
0
0
*
9
.
8
*
1
0
0
/
2
0
=
4
9
0
0
0
W
a
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s
o
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>
4
9
0
0
0
w
a
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t
s
Quiz
: A sled and rider with a total mass of 40 kg are perched at the
top of the hill shown. Suppose that 2000 J of work is done against
friction as the sled travels from the top (at 40 m) to the second
hump (at 30 m). Will the sled make it to the top of the second hump
if no kinetic energy is given to the sled at the start of its motion?
a)
yes
b)
no
c)
It depends.
E
n
e
r
g
y
c
o
n
s
e
r
v
a
t
i
o
n
:
m
g
H
=
m
g
h
+
K
E
+
h
e
a
t
40kg X 9.8N/s
2
X 40m = 40kg X
9.8N/s
2
X 30m + KE + 2000J
KE = 1920 J > 0
, i.e. yes, he
can reach the 2
nd
hump.
4
Momentum and Impulse
Momentum and Impulse
How can we describe the change in velocities of
colliding football players, or balls colliding with bats?
How does a strong force applied for a very short time
affect the motion?
Can we apply Newton’s Laws to collisions?
What exactly is
momentum
? How is it different from
force or energy?
What does “
Conservation of Momentum
” mean?
5
What happens when a ball bounces?
When it reaches the floor, its
velocity quickly changes direction.
There must be a strong force
exerted on the ball by the floor during
the short time they are in contact.
This force provides the upward
acceleration necessary to change the
direction of the ball’s velocity.
6
Momentum and Impulse
Multiply both sides of Newton’s second
law by the time interval over which the
force acts:
T
h
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l
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f
t
s
i
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t
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e
q
u
a
t
i
o
n
i
s
i
m
p
u
l
s
e
,
the (average) force acting on an object
multiplied by the time interval over which
the force acts.
How a force changes the motion of an object depends on both the size of the
force and how long the force acts.
T
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t
.
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b
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a
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b
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c
t
t
i
m
e
s
i
t
s
v
e
l
o
c
i
t
y
.
7
Impulse-Momentum Principle
The impulse acting on an object produces a change
in momentum of the object that is equal in both
magnitude and direction to the impulse.
8
When
F = 0,
p = 0
, i.e. total
momentum has no change as a
function of time, i.e.
conserved
9
Conservation of Momentum
When
F = 0,
p = 0
, i.e. total
momentum has no change as a
function of time, i.e.
conserved
p = F
t
This is the impulse equation
Does the momentum of the
fullback conserve?
A). Yes, because the
external force is zero.
B). No, because the
external force is NOT zero
10
Does the sum of the momentum of
the two players conserve?
A). Yes, because the
external force is zero.
B). No, because the
external force is NOT zero
11
Conservation of Momentum
•
Newton’s third law
–
For every action, there is an equal but
opposite reaction.
–
The defensive back exerts a force on the
fullback, and the fullback exerts an equal
but opposite force on the defensive back.
–
The impulses on both are equal and
opposite.
–
The changes in momentum for each are
equal
and
opposite
.
–
The total change of the momentum for the
two players is zero.
12
Conservation of Momentum
For any system, if the
net external force acting
on a system of objects is
zero, the total
momentum of the system
is
conserved
.
p = F
t
F = 0,
p = 0
13
A 100-kg fullback moving straight downfield collides with a
75-kg defensive back. The defensive back hangs on to the
fullback, and the two players move together after the
collision. What is the initial momentum of each player?
A). Full back: 500 kg·m/s
2,
defensive back: -300 kg·m/s
2
B). Full back: 500 kg·m/s
2,
defensive back: 300 kg·m/s
2
C). Full back: 200 kg·m/s
2,
defensive back: -300 kg·m/s
2
D). Full back: 250 kg·m/s
2,
defensive back: 300 kg·m/s
2
p = mv = (100 kg) (5
m/s) = 500 kg·m/s
2
p = mv = (75 kg)(-4
m/s)= -300 kg·m/s
2
14
What is the total momentum of the system?
Total momentum:
p
total
= p
fullback
+ p
defensive back
= 500 kg·m/s - 300 kg·m/s
= 200 kg·m/s
A). 150 kg·m/s
2
B). 500 kg·m/s
2
C). 100 kg·m/s
2
D). 200 kg·m/s
2
15
What is the velocity of the two players
immediately after the collision?
A.
4.5m/s
B.
1.14m/s
C.
153 m/s
D.
0 m/s
E.
0.25m/s
Total mass:
m = 100 kg +
75 kg = 175
kg
Velocity of
both:
v = p
total
/ m = (200
kg·m/s) / 175
kg = 1.14 m/s
16
Two skaters of different masses prepare to push
off against one another. Which one will gain the
larger velocity?
a)
The more massive one
b)
The less massive one
c)
They will each have equal but
opposite velocities.
The net external force acting on the system is zero,
so conservation of momentum applies.
Before the push-off, the total initial momentum is zero.
The total momentum after the push-off should also be
zero.
B
o
t
h
m
u
s
t
m
o
v
e
w
i
t
h
m
o
m
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n
t
u
m
v
a
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q
u
a
l
i
n
m
a
g
n
i
t
u
d
e
b
u
t
o
p
p
o
s
i
t
e
i
n
d
i
r
e
c
t
i
o
n
:
p
2
=
p
1
When added together, the total final momentum of the
system is then zero.
S
i
n
c
e
m
o
m
e
n
t
u
m
i
s
m
a
s
s
t
i
m
e
s
v
e
l
o
c
i
t
y
p
=
m
v
,
t
h
e
s
k
a
t
e
r
w
i
t
h
t
h
e
s
m
a
l
l
e
r
m
a
s
s
m
u
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t
h
a
v
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l
a
r
g
e
r
v
e
l
o
c
i
t
y
:
m
1
v
1
=
m
2
v
2
17
Recoil
•
The explosion of the powder causes the shot to move
very rapidly forward.
•
If the gun is free to move, it will recoil backward with a
momentum equal in magnitude to the momentum of
the shot.
18
•
Even though the mass of the shot is small, its
momentum is large due to its large velocity.
•
The shotgun recoils with a momentum equal in
magnitude to the momentum of the shot.
•
The recoil velocity of the shotgun will be smaller than
the shot’s velocity because the shotgun has more mass,
but it can still be sizeable.
19
How can you avoid a bruised shoulder?
•
If the shotgun is held firmly against your shoulder, it
doesn’t hurt as much.
WHY?
20
Elastic and Inelastic Collisions
•
Different kinds of collisions
produce different results.
–
Sometimes the objects stick
together.
–
Sometimes the objects bounce
apart.
What is the difference between
these types of collisions?
Is energy conserved as well as
momentum?
21
•
A ball bouncing off a floor or wall with no decrease in the
magnitude of its velocity is an
elastic collision
.
–
The kinetic energy does not decrease.
–
No energy has been lost.
•
A ball sticking to the wall is a
perfectly inelastic collision
.
–
The velocity of the ball after the collision is zero.
–
Its kinetic energy is then zero.
–
All of the kinetic energy has been lost.
–
Does the total momentum become zero?
•
Most collisions involve some energy loss, even if the
objects do not stick, because the collisions are not perfectly
elastic.
–
Heat is generated, the objects may be deformed, and sound
waves are created.
–
These would be
partially inelastic collisions
.
22
Four railroad cars, all with the same mass of
20,000 kg, sit on a track. A fifth car of identical
mass approaches them with a velocity of 15 m/s.
This car collides and couples with the other four
cars. What is the initial momentum of the system?
a)
200,000 kg·m/s
b)
300,000 kg·m/s
c)
600,000 kg·m/s
d)
1,200,000 kg·m/s
m
5
= 20,000 kg
v
5
= 15 m/s
p
initial
=
m
5
v
5
= (20,000 kg)(15 m/s)
=
300,000 kg·m/s
23
What is the velocity of the five coupled cars
after the collision?
a)
1 m/s
b)
3 m/s
c)
5 m/s
d)
10 m/s
m
total
= 100,000 kg
p
final
=
p
initial
v
final
=
p
final
/
m
total
= (300,000 kg·m/s)/(100,000 kg)
=
3 m/s
24
Is the kinetic energy after the railroad cars collide
equal to the original kinetic energy of car 5?
a)
yes
b)
no
c)
It depends.
KE
initial
= 1/2
m
5
v
5
2
= 1/2 (20,000 kg)(15 m/s)
2
=
2250 kJ
KE
final
= 1/2
m
total
v
final
2
= 1/2 (100,000 kg)(3 m/s)
2
=
450 kJ
KE
final
≠
KE
initial
25
Q
u
i
z
:
W
h
e
n
a
b
a
l
l
b
o
u
n
c
e
s
b
a
c
k
w
i
t
h
t
h
e
s
a
m
e
s
p
e
e
d
,
t
h
e
m
o
m
e
n
t
u
m
c
h
a
n
g
e
s
f
r
o
m
m
v
t
o
-
m
v
,
s
o
t
h
e
i
m
p
u
l
s
e
a
n
d
t
h
e
c
h
a
n
g
e
i
n
m
o
m
e
n
t
u
m
a
r
e
A.
2mv, 2mv
B.
mv, mv
C.
No change in
momentum.
Impulse is 0.
D.
No change in
momentum.
Impulse is mv
26
BACKUPS
27
28
Review Chapters 1 - 6
•
Units----Length, mass, time SI units m, kg, second
•
Coordinate systems
•
Average speed = distance/time = d/t
•
Instantaneous speed = d/
Δ
t
•
Vector quantities---magnitude and direction
•
Magnitude is always positive
•
Velocity----magnitude is speed
•
Acceleration = change in velocity/time =
Δ
v/
Δ
t
•
Force = ma Newtons
28
29
Speed, velocity and acceleration
•
v
=
Δ
d
/
Δ
t
a
=
Δ
v
/
Δ
t
T
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m
a
g
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f
a
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l
a
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d
t
o
t
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e
m
a
g
n
i
t
u
d
e
o
f
v
t
h
e
d
i
r
e
c
t
i
o
n
o
f
a
i
s
n
o
t
r
e
l
a
t
e
d
t
o
t
h
e
d
i
r
e
c
t
i
o
n
o
f
v
v = v
0
+ at constant acceleration
d = v
0
t + 1/2at
2
d = 1/2(v + v
0
) t
v
2
= v
0
2
+ 2ad
d
,
v
0
v
,
a
c
a
n
b
e
+
o
r
–
i
n
d
e
p
e
n
d
e
n
t
l
y
30
2/16/2025
One dimensional motion and gravity
•
v = v
0
+ at d = v
0
t + 1/2at
2
d = ½(v + v
0
)t
•
v
2
= v
0
2
+ 2ad
+
+
g
=
-
9
.
8
m
/
s
2
A
t
t
h
e
t
o
p
v
=
0
a
n
d
t
=
v
0
/
9
.
8
A
t
t
h
e
b
o
t
t
o
m
t
=
2
v
0
/
9
.
8
31
Equations
•
v = v
0
+ at d = v
0
t + 1/2at
2
•
Sometimes you have to use two equations
.
•
v
0
= 15m/s v = 50m/s What is h?
•
v = v
0
+ at
•
50 = 15 + 9.8t t = 3.57 s
•
h = v
0
t + 1/2at
2
•
h = 15 x 3.57 + 1/2x9.8x3.57
2
•
= 116m
•
v
0
v
g
h
31
32
Projectile Motion
•
axis 1 v
1
= constant and d
1
= v
1
t
•
axis 2 v
v
= v
0v
+ at and d = v
0v
t + 1/2at
2
9
.
8
m
/
s
2
U
s
e
+
d
o
w
n
s
o
g
i
s
+
a
n
d
h
i
s
+
h
=
v
0
v
t
+
1
/
2
a
t
2
v
0
v
=
0
,
t
2
=
2
h
/
a
R
=
v
1
t
v
=
v
0
v
+
a
t
33
Complete Projectile
•
highest point the vertical velocity is zero
•
•
v
v
= v
0v
+ at so t = v
0v
/9.8 h = v
0v
t + 1/2at
2
•
end t = 2v
0v
/9.8 and R = v
1
x 2v
0v
/9.8
•
and the vertical velocity is minus v
0v
v
0
v
33
34
Newton’s Second and First Law
•
Second Law
F = ma unit is a Newton (or pound)
•
First Law
F = 0 a = 0 so v = constant
•
Third law
For every force there is an equal and opposite reaction
force
W
e
i
g
h
t
=
m
g
F
=
m
a
v
=
v
0
+
a
t
d
=
v
0
t
+
½
a
t
2
34
35
Examples
•
30 – 8 – T = 4a
•
T – 6 = 2a
•
30 – 8 – 6 = 6a
g
+
N – mg = ma
a + N > mg
a – N < mg
35
gravitation
36
37
Examples of circular motion
•
m
g
–
N
=
m
v
2
/
r
N
-
m
g
=
m
v
2
/
r
v
L
o
o
k
i
n
g
d
o
w
n
N
N
=
m
v
2
/
r
S
i
d
e
m
g
F
f
m
g
=
F
f
V
e
r
t
i
c
a
l
m
o
t
i
o
n
v
T
m
g
m
g
+
T
=
m
v
2
/
r
t
o
p
T
-
m
g
=
m
v
2
/
r
b
o
t
t
o
m
38
Work energy and Power
•
Kinetic energy = 1/2mv
2
•
W = Fd and can be + or –
•
F is net force parallel to d.
•
Units are joules
•
Power = W/t watts
•
Potential energy = mgh
•
Spring = 1/2kx
2
•
Oscillations
•
Transfer of KE
PE
•
Conservative force
•
Transfer of KE
PE
•
v
Explore topics in power, energy, and momentum in preparation for an upcoming physics exam. Understand concepts such as work, power, joules, and momentum through practical examples and quiz questions. Be ready for your exam with a solid grasp of key principles in physics.
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Presentation Transcript
Exam#1 (chapter 1-6) time: Tuesday 02/19 8:00 pm- 10:00 pm Location: physics building room 112 If you can not make it, please let me know by Friday 02/15 so that I can arrange a make-up exam. If you have special needs, e.g. exam time extension, and has not contact me before, please bring me the letter from the Office of the Dean of Students before Friday 02/15. AOB Prepare your own scratch paper, pencils, erasers, etc. Use only pencil for the answer sheet Bring your own calculators No cell phones, no text messaging, no computers No crib sheet of any kind is allowed. Equation sheet will be provided. 1
Power It is not only important how much work is done but also how quick, i.e. the rate, at which work is done So the quantity Power = P = W/t (unit is a watt) is very important. Generally energy supplies, motors, etc are rated by power and one can determine how much work can be done by multiplying by time. W = Pt (joules). 2/16/2025 2 2
Examples of Watt and Joules The unit for electrical usage is the kilowatt hour. A kilowatt hour is the energy used by a 1000 watt device for 3600 seconds 1kWHr = 1000*3600 = 3.6 million joules In order to lift an elevator with a mass of 1000kg to 100 meters requires 1000*9.8*100 joules but we need to do it in 20 seconds. The power we need is 1000*9.8*100/20 = 49000 Watts so we need to install a motor rated at > 49000 watts 3
Quiz: A sled and rider with a total mass of 40 kg are perched at the top of the hill shown. Suppose that 2000 J of work is done against friction as the sled travels from the top (at 40 m) to the second hump (at 30 m). Will the sled make it to the top of the second hump if no kinetic energy is given to the sled at the start of its motion? a) b) c) yes no It depends. Energy conservation: mgH = mgh + KE + heat 40kg X 9.8N/s2 X 40m = 40kg X 9.8N/s2 X 30m + KE + 2000J KE = 1920 J > 0, i.e. yes, he can reach the 2nd hump. 4
Momentum and Impulse How can we describe the change in velocities of colliding football players, or balls colliding with bats? How does a strong force applied for a very short time affect the motion? Can we apply Newton s Laws to collisions? What exactly is momentum? How is it different from force or energy? What does Conservation of Momentum mean? 5
What happens when a ball bounces? When it reaches the floor, its velocity quickly changes direction. There must be a strong force exerted on the ball by the floor during the short time they are in contact. This force provides the upward acceleration necessary to change the direction of the ball s velocity. Fnet= ma = m v t 6
Momentum and Impulse Fnet= ma = m v Multiply both sides of Newton s second law by the time interval over which the force acts: t The left side of the equation is impulse, the (average) force acting on an object multiplied by the time interval over which the force acts. Fnet t = m v How a force changes the motion of an object depends on both the size of the force and how long the force acts. The right side of the equation is the change in the momentum of the object. The momentum of the object is the mass of the object times its velocity. p = mv 7
Impulse-Momentum Principle The impulse acting on an object produces a change in momentum of the object that is equal in both magnitude and direction to the impulse. impulse = change in momentum v F = m t net = p When F = 0, p = 0, i.e. total momentum has no change as a function of time, i.e. conserved 8
Conservation of Momentum p = F t This is the impulse equation When F = 0, p = 0, i.e. total momentum has no change as a function of time, i.e. conserved 9
Does the momentum of the fullback conserve? A). Yes, because the external force is zero. B). No, because the external force is NOT zero 10
Does the sum of the momentum of the two players conserve? A). Yes, because the external force is zero. B). No, because the external force is NOT zero 11
Conservation of Momentum Newton s third law For every action, there is an equal but opposite reaction. The defensive back exerts a force on the fullback, and the fullback exerts an equal but opposite force on the defensive back. The impulses on both are equal and opposite. The changes in momentum for each are equal and opposite. The total change of the momentum for the two players is zero. 12
Conservation of Momentum For any system, if the net external force acting on a system of objects is zero, the total momentum of the system is conserved. p = F t F = 0, p = 0 13
A 100-kg fullback moving straight downfield collides with a 75-kg defensive back. The defensive back hangs on to the fullback, and the two players move together after the collision. What is the initial momentum of each player? A). Full back: 500 kg m/s2, defensive back: -300 kg m/s2 B). Full back: 500 kg m/s2, defensive back: 300 kg m/s2 C). Full back: 200 kg m/s2, defensive back: -300 kg m/s2 D). Full back: 250 kg m/s2, defensive back: 300 kg m/s2 p = mv = (100 kg) (5 m/s) = 500 kg m/s2 p = mv = (75 kg)(-4 m/s)= -300 kg m/s2 Fullback: 14 Defensive back:
What is the total momentum of the system? A). 150 kg m/s2 B). 500 kg m/s2 C). 100 kg m/s2 D). 200 kg m/s2 Total momentum: ptotal = pfullback + pdefensive back = 500 kg m/s - 300 kg m/s = 200 kg m/s 15
What is the velocity of the two players immediately after the collision? Total mass: m = 100 kg + 75 kg = 175 kg A. 4.5m/s B. 1.14m/s C. 153 m/s Velocity of both: v = ptotal / m = (200 kg m/s) / 175 kg = 1.14 m/s D. 0 m/s E. 0.25m/s 16
Two skaters of different masses prepare to push off against one another. Which one will gain the larger velocity? a) b) c) The more massive one The less massive one They will each have equal but opposite velocities. The net external force acting on the system is zero, so conservation of momentum applies. Before the push-off, the total initial momentum is zero. The total momentum after the push-off should also be zero. Both must move with momentum values equal in magnitude but opposite in direction: p2 = p1 When added together, the total final momentum of the system is then zero. Since momentum is mass times velocity p = mv, the skater with the smaller mass must have the larger velocity: m1v1 = m2v2 17
Recoil The explosion of the powder causes the shot to move very rapidly forward. If the gun is free to move, it will recoil backward with a momentum equal in magnitude to the momentum of the shot. 18
Even though the mass of the shot is small, its momentum is large due to its large velocity. The shotgun recoils with a momentum equal in magnitude to the momentum of the shot. The recoil velocity of the shotgun will be smaller than the shot s velocity because the shotgun has more mass, but it can still be sizeable. 19
How can you avoid a bruised shoulder? If the shotgun is held firmly against your shoulder, it doesn t hurt as much. WHY? 20
Elastic and Inelastic Collisions Different kinds of collisions produce different results. Sometimes the objects stick together. Sometimes the objects bounce apart. What is the difference between these types of collisions? Is energy conserved as well as momentum? 21
A ball bouncing off a floor or wall with no decrease in the magnitude of its velocity is an elastic collision. The kinetic energy does not decrease. No energy has been lost. A ball sticking to the wall is a perfectly inelastic collision. The velocity of the ball after the collision is zero. Its kinetic energy is then zero. All of the kinetic energy has been lost. Does the total momentum become zero? Most collisions involve some energy loss, even if the objects do not stick, because the collisions are not perfectly elastic. Heat is generated, the objects may be deformed, and sound waves are created. These would be partially inelastic collisions. 22
Four railroad cars, all with the same mass of 20,000 kg, sit on a track. A fifth car of identical mass approaches them with a velocity of 15 m/s. This car collides and couples with the other four cars. What is the initial momentum of the system? m5 v5 pinitial = 20,000 kg = 15 m/s = m5v5 = (20,000 kg)(15 m/s) = 300,000 kg m/s 200,000 kg m/s 300,000 kg m/s 600,000 kg m/s 1,200,000 kg m/s a) b) c) d) 23
What is the velocity of the five coupled cars after the collision? 1 m/s 3 m/s 5 m/s 10 m/s mtotal pfinal vfinal = 100,000 kg = pinitial = pfinal / mtotal = (300,000 kg m/s)/(100,000 kg) = 3 m/s a) b) c) d) 24
Is the kinetic energy after the railroad cars collide equal to the original kinetic energy of car 5? KEinitial= 1/2 m5 v52 = 1/2 (20,000 kg)(15 m/s)2 = 2250 kJ KEfinal = 1/2 mtotal vfinal2 = 1/2 (100,000 kg)(3 m/s)2 = 450 kJ KEfinal KEinitial yes no It depends. a) b) c) 25
Quiz: When a ball bounces back with the same speed, the momentum changes from mv to -mv, so the impulse and the change in momentum are A. 2mv, 2mv B. mv, mv C. No change in momentum. Impulse is 0. D. No change in momentum. Impulse is mv 26
BACKUPS 27
Review Chapters 1 - 6 - +x d Units----Length, mass, time SI units m, kg, second Coordinate systems Average speed = distance/time = d/t Instantaneous speed = d/ t Vector quantities---magnitude and direction Magnitude is always positive Velocity----magnitude is speed Acceleration = change in velocity/time = v/ t Force = ma Newtons 28 28
Speed, velocity and acceleration v = d/ t a = v/ t 4 23 The magnitude of a is not related to the magnitude of v 1 the direction of a is not related to the direction of v v = v0 + at constant acceleration d = v0t + 1/2at2 d = 1/2(v + v0) t v2 = v02 + 2ad d,v0 v,a can be + or independently 29
One dimensional motion and gravity v = v0 + at d = v0t + 1/2at2 d = (v + v0)t v2 = v02 + 2ad + g = -9.8m/s2 + At the top v = 0 and t = v0/9.8 At the bottom t = 2v0/9.8 2/16/2025 30
Equations v = v0 + at d = v0t + 1/2at2 Sometimes you have to use two equations. v0 = 15m/s v = 50m/s What is h? v = v0 + at 50 = 15 + 9.8t t = 3.57 s h = v0t + 1/2at2 h = 15 x 3.57 + 1/2x9.8x3.572 = 116m v0 g h v 31 31
Projectile Motion axis 1 v1 = constant and d1 = v1t axis 2 vv = v0v + at and d = v0vt + 1/2at2 v1 g 9.8m/s2 h v R Use + down so g is + and h is + h = v0vt + 1/2at2 v0v = 0, t2 = 2h/a R = v1t v = v0v + at 32
Complete Projectile v0v 9.8m/s2 v1 v1 v1 v0v highest point the vertical velocity is zero vv = v0v + at so t = v0v/9.8 h = v0vt + 1/2at2 end t = 2v0v/9.8 and R = v1 x 2v0v/9.8 and the vertical velocity is minus v0v 33 33
Newtons Second and First Law Second Law F = ma unit is a Newton (or pound) First Law F = 0 a = 0 so v = constant Third law For every force there is an equal and opposite reaction force N Weight = mg mg F Ff F Ff F = ma v = v0 + at d = v0t + at2 34 34
Examples + T N g 30 8 T = 4a T 6 = 2a 30 8 6 = 6a mg N mg = ma a + N > mg a N < mg 35 35
gravitation = m1g ? = ??2 ?2 36
Examples of circular motion Looking down Vertical motion N N v v W = mg N = mv2/r v mg N = mv2/r Side T mg N Ff mg mg mg + T = mv2/r top mg = Ff N - mg = mv2/r T - mg = mv2/r bottom 37
Work energy and Power Kinetic energy = 1/2mv2 W = Fd and can be + or F is net force parallel to d. Units are joules F v d Power = W/t watts g F = mg Potential energy = mgh Spring = 1/2kx2 h Oscillations Transfer of KE PE Conservative force Transfer of KE PE 38
