Transcript Motion

Motion II
Momentum and Energy
Momentum
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Obviously there is a big difference
between a truck moving 100 mi/hr and a
baseball moving 100 mi/hr.
We want a way to quantify this.
Newton called it Quantity of Motion. We
call it Momentum.
Definition: Momentum = (mass)x(velocity)
p = mv
Momentum is a vector.
 Combination of mass and velocity
 Large momentum for massive objects
moving slowly or small objects moving
quickly
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Newton’s 2nd Law
p
F
t
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Note: Sometimes we write t instead of just t.
They both represent the time interval for
whatever change occurs in the numerator.
Newton’s 2nd Law
p mv
F

 ma
t
t
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Since p = mv, we have p = (mv)
If m is constant (not true for rockets and
ballons), p = mv and v/t = a
Rearrange Newton’s 2nd Law
p  Ft
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This is the impulse
Impulse is the total change in momentum
Example
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Baseball
p = pf – pi
= mvf – (-mvi)
= m(vf + vi)
F = p/t
Where t is the time the
bat is in contact with
the ball
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Realistic numbers
M = 0.15 kg
Vi = -40 m/s
Vf = 50 m/s
t = 0.0007 s
F = 19286 N
a = 128571 m/s2
slow motion
photography
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Ion Propulsion
Engine: very weak
force (0.1 N, about
the weight of a paper
clip) but for very long
times (months)
Large momentum
change
Very efficient
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Crumple zones on
cars increase the time
for a car to stop. This
greatly reduces the
average force during
accident.
Air bags do the same
thing
A 70 kg person traveling at +15 m/s stops.
What is the impulse?
1.
2.
3.
4.
-500 kgm/s
-1000 kgm/s
-1500 kgm/s
-2000 kgm/s
25%
1
25%
2
25%
3
25%
4
Example
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70 kg person traveling at 15 m/s stops.
p = pf – pi = 0 kgm/s – 1050 kg·m/s = 1050 kg·m/s
Example
70 kg person traveling at 15 m/s stops.
 p = pf – pi = 0 kgm/s – 1050 kg·m/s = 1050 kg·m/s
 If the person is stopped by the windshield
t = 0.01 s
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What is the force?
Example
70 kg person traveling at 15 m/s stops.
 p = pf – pi = 0 kgm/s – 1050 kg·m/s = 1050 kg·m/s
 If the person is stopped by the windshield
t = 0.01 s
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F = -1050 kgm/s/0.01 s = -105000 N
Example
70 kg person traveling at 15 m/s stops.
 p = pf – pi = 0 kgm/s – 1050 kg·m/s = 1050 kg·m/s
 If the person is stopped by an airbag
t = 0.40 s
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What is the force?
Example
70 kg person traveling at 15 m/s stops.
 p = pf – pi = 0 kgm/s – 1050 kg·m/s = 1050 kg·m/s
 If the person is stopped by an airbag
t = 0.40 s
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F = -1050 kgm/s/0.40 s = -2625 N
A woman has survived after falling from the 23rd
floor of a hotel in the Argentine capital, Buenos
Aires.
Her fall was broken by a taxi, whose driver got out moments before the impact
crushed the roof and shattered the windscreen. Eyewitness said the woman
had climbed over a safety barrier and leapt from a restaurant at the top of the
Hotel Crowne Plaza Panamericano. She was taken to intensive care for
treatment for multiple injuries. The woman, who has not been named, is
reported to be an Argentine in her 30s. The taxi driver, named by local media
as Miguel, said he got out of his vehicle just before the impact after noticing a
policeman looking up. "I got out of the car a second before. If I had not got
out, I would have been killed," he told Radio 10. "I was only 10 metres from
the impact. It made a terrible noise," he added.
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Estimate m = 50kg
H= 23 x 3.5m = 80.5m
1 2
x  at
2
2x
2
t 
a
2x
(2)(80.5m)
2
t


16
.
4
s
 4.05s
2
a
9.8m / s
v  at
v  (9.8m / s 2 )( 4.05s )  39.7 m / s  89mph
Estimate Acceleration
v v x
a

t x t
BUT
x
1
v
 vavg  v  0 
t
2
2
v  v  0  v
Thus
a
v v 2 
x

v2
2x 
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Case 1: stops in deformation distance of
body ~ 0.30 m (3 cm).

39.7m / s 
a

2x  20.03m 
v2
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2
 26268m / s
2
Case 2: stops in height of roof, ~ 0.50 m
(50 cm).
2
v2

39.7m / s 
a

 1576m / s 2
2x  20.50m 
Conservation of Momentum
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Newton’s 3rd Law requires F1 = -F2
p1/t = -p2/t
Or
p1/t + p2/t = 0
Or
p1 + p2 = 0
NO NET CHANGE IN MOMENTUM
Example: Cannon
1000 kg cannon
 5 kg cannon ball
Before it fires
Pc + Pb = 0
Must also be true after it fires.
If Vb = 400 m/s
McVc + MbVb = 0
Vc = -MbVb/Mc
= -(5kg)(400m/s)/(1000kg)
= -2m/s
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Conservation of Momentum
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Head on collision
Football
Train crash
Wagon rocket
Cannon Fail
ENERGY & WORK
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In the cannon example, both the cannon
and the cannon ball have the same
momentum (equal but opposite).
Question: Why does a cannon ball do so
much more damage than the cannon?
ENERGY & WORK
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In the cannon example, both the cannon
and the cannon ball have the same
momentum (equal but opposite).
Question: Why does a cannon ball do so
much more damage than the cannon?
Answer: ENERGY
ENERGY & WORK
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Essentially, energy is the ability to
make something move.
What transfers or puts energy into
objects?
A push
WORK
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We will begin by considering the “Physics”
definition of work.
Work = (Force) x (displacement traveled in
direction of force)
W=Fd
1) if object does not move, NO WORK is done.
2) if direction of motion is perpendicular to
force, NO WORK is done
Example
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Push a block for 5 m using a 10 N force.
Work = (10 N) x (5 m)
= 50 Nm
= 50 kgm2/s2
= 50 J (Joules)
I have not done any work if I
push against a wall all day long
but is does not move.
1.
2.
True
False
0%
1
0%
2
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Doing WORK on a body changes its
ENERGY
ENERGY is the capacity to do work.
The release of energy does work, and
doing work on something increases its
energy
E=W=Fd
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Two basics type of energy: Kinetic and
Potential.
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Kinetic energy: Energy of motion
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Potential energy: Energy of situation
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Position
Chemical structure
Electrical charge
Kinetic Energy: Energy of Motion
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Derivation: Push a ball of mass m with a
constant force.
KE = W = Fd = (ma)d
But d = ½ at2
KE = (ma)(½ at2) = ½ m(at)2
But v = at
KE = ½ mv2
A 6000 kg truck travels at 10 m/s.
What is its kinetic energy?
1.
2.
3.
4.
3000 J
30000 J
300000 J
3000000 J
25%
1
25%
2
25%
3
25%
4
If the speed of the truck doubles, what happens to
the energy?
1.
2.
3.
Increases by ½x
Increases by 2x
Increases by 4x
33%
1
33%
2
33%
3
Potential Energy
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Lift an object through a height, h.
PE = W = Fd= (mg)h
PE = mgh
Note: We need to decide where to set h=0.
Units of energy and work
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W = Fd Nm  kg  m m  kg  m  Joule  J
s2
s2
2
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2
m
kg

m
KE = mv2/2 kg  
 Joule  J
2
s
s
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m
kg  m
kg 2 m 
 Joule  J
2
s
s
2
PE = mgh
2
Other types of energy
Electrical
 Chemical
 Thermal
 Nuclear
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Conservation of Energy
We may convert energy from one type to
another, but we can never destroy it.
 We will look at some examples using only
PE and KE.
KE + PE = const.
OR
KEi + PEi = KEf + PEf
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Drop a rock of 100m cliff
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Call bottom of cliff h = 0
Initial KE = 0; Initial PE = mgh
Final PE = 0
Use conservation of energy to find final
velocity
mgh = ½mv2
v2 = 2gh
v  2 gh
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For our 100 m high cliff
v  (2(9.8m / s )(100m)
2
 44.27 s
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Note that this is the same as we got from
using kinematics with constant acceleration
A 5 kg ball is dropped from the top of the Sears
tower in downtown Chicago. Assume that the
building is 300 m high and that the ball starts from
rest at the top.
1.
2.
3.
4.
5880 m/s
2940 m/s
76.7 m/s
54.2 m/s
A 5 kg ball is dropped from rest from the top (300
m high) of the Sears Tower.
From conservation of energy, what is its speed at
the bottom?
1.
2.
3.
4.
5880 m/s
2940 m/s
76.7 m/s
54.2 m/s
25%
1
25%
2
25%
3
25%
4
Pendulum
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At bottom, all energy
is kinetic.
At top, all energy is
potential.
In between, there is a
mix of both potential
and kinetic energy.
Bowling Ball Video
Energy Content of a Big Mac
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540 Cal = 540000 cal
1 Cal = 4186 Joules
540Cal
41861CalJoule   2,260,440 Joules
How high can we lift a 1kg
object with the energy content
of 1 Big Mac?
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PE = mgh
h = PE/(mg)
PE = 2,260,440 J
m = 1 kg
g = 9.8 m/s2
h
2, 260, 440 J
(1kg )(9.8m / s 2 )
 230,657m  230.7km  143miles
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2 Big Macs have enough energy content to
lift a 1 kg object up to the orbital height of
the International Space Station.
150 Big Macs could lift a person to the
same height.
It would take about 4,000,000 Big Macs to
get the space shuttle to the ISS height.
Note: McDonald’s sells approximately
550,000,000 Big Macs per year.
Revisit: Cannon Momentum
1000kg cannon
 5kg cannon ball
Before it fires
Pc + Pb = 0
Must also be true after it fires.
Assume Vb = 400 m/s
McVc + MbVb =0
Vc= -MbVb/Mc
= -(5kg)(400m/s)/(1000kg)
= -2m/s
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Cannon: Energy
KEb= ½ mv2 = 0.5 (5kg)(400m/s)2
= 400,000 kgm2/s2
= 400,000 J
KEc= ½ mv2 = 0.5 (1000kg)(2m/s)2
= 2,000 kgm2/s2
= 2,000 J
Note: Cannon Ball has 200 times more
ENERGY
When I drop a rock, it start off with PE which gets
converted in to KE as it falls. At the end, it is not
moving, so it has neither PE or KE. What
happened to the energy?
1.
2.
3.
Energy is not conserved
in this case.
The energy is converted
into Thermal energy
The energy is transferred
to whatever it is dropped
on.
33%
1
33%
2
33%
3
Important Equations
d
v t

v
a t
2
1
d  2 at
p  mv
I  p
p
F  ma  t
2
1
KE  2 mv
PE  mgh