Transcript Momentum

Momentum
Chapter-6
Momentum
• Symbol is lower case p
• Momentum is a product of mass and
velocity
• p = mv
Units
• p = mv,
so the units are:
• p = (kg)(m/s)
kgm
s
Or, N-s
because velocity is a vector
momentum is a vector quantity
82,288 kg x 0.02m/s = 1600 kgm/s
100kg astronaut in
a 60 kg suit moving
At 10 m/s
1600 kg m/s
Conservation of momentum
• Momentum of a system is conserved:
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•
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Before and after objects join
Before and after an explosion
Newton’s Third Law applies
FB on A = - FA on B
1600 kgm/s astronaut grabs a 100
kg Sputnik, how fast do both go?
10 m/s
m1v1 = m2v2
1600 kgm/s = 260 kg v2
v2 = 6.2 m/s
In a collision, the total momentum before doesn’t change:
2500 N-s
+ 800 N-s = 3300 N-s
1500 N-s
+ 1800 N-s = 3300 N-s
Explosions between vehicles
Recoil
Let’s say it’s a 5 kg rifle and
a 10 g bullet fired at 1000m/s
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•
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•
m1v1 = m2v2
(0.01 kg)(1000m/s) = 5kg v2
10 kgm/s = 5kg v2
2m/s = v2
of course you usually put your
shoulder there
Converting to Newton-S
• A Newton is a kgm/s2
• A Newton-second is a (kgm/s2)(s) = kgm/s
• So Newton-seconds are equivalent to units
of momentum
Deriving Impulse Theorem from
Newton’s 2nd Law
• F = ma = m Dv/Dt
• F Dt = m Dv
• So, the average force over the time interval is equal to the
change in momentum
• F Dt = p2 - p1
So it’s the same rifle on your
shoulder, & the recoil lasts 0.01s
• mDv = FDt
• 10 kgm/s = F(0.01 s)
• F = 1000N
That kind of hurts
Hitting the wall at the Indianapolis Speedway
The average force you feel depends on rate of change in momentum
Lets say you’re going 100km/hr, How fast do you decelerate to
0 km/hr as the front of your car crumples on the concrete?
What force is felt by the 100kg driver?
(100 km/h)(1000 m/km)(1 h/3600 s) = 27.8 m/s
Let’s say the hood is one meter long d = 1/2vt t = 0.072 s
The impulse is F Dt = m Dv
F (0.0726 s) = (100 kg) (27.8 m/s - 0 m/s)
F = 38,600 N
This is fatal
So we use straw bales around race tracks.
… and some soft squishy spectators.
(100 km/h)(1000 m/km)(1 h/3600 s) = 27.8 m/s
Let’s say the straw makes the car stop in 10m, d = 1/2vt t = 0.72 s
The impulse is F Dt = m Dv
F (0.72 s) = (100 kg) (27.8 m/s - 0 m/s)
F = 3,860 N
3860 N / 980 N = about 4g (4x gravity)
this is likely survivable
Billiards: the break
1-D conservation of momentum
Very few shots line up perfectly
Momentum of the constituents before and after contact is equal
A stationary ball
(red 3-ball) hit with
a cue-ball.
They share
momentum
after collision.
Forward or back spin
can affect the trajectory
of the cue ball.
Little to no spin is
transferred to the target
ball.
Conservation of Momentum in
2-D
Cueball-1
o
60
Greenball
Cueball-2
0.1 kg ball
Rolls at 0.5 m/s
Conservation of Momentum in
2-D
Cueball-1
o
60
Greenball
Cueball-2
Cue ball starts with a momentum p=(0.1 kg)(0.5 m/s)= 0.05 kgm/s
After collision (time 2):
pcb2 = (sinq)(pcb1)
pcb2 = (sin 60o))(0.05kgm/s)
pcb2 = 0.0433 kgm/s
pgb = (cosq)(pcb1)
pgb = (cos 60o))(0.05kgm/s)
pgb = 0.025 kgm/s
vcb2 = 0. 433 m/s
vgb = 0. 25 m/s
Conservation of Momentum in
2-D
Cueball-1
o
60
Greenball
Cueball-2
0.433 m/s
0.1 kg ball
Rolls at 0.5 m/s
0.25 m/s
So, I looked this over…
• No this is right…it seems counterintuitive but it
checks
• Momentum is conserved here because these are in
different directions, their vector sum is the same as
the original input momentum
• And, if you check the sum total kinetic energy of
the system KE = 1/2 mv2, that too is conserved.
Impulse in Rocket Flight
• A rocket engine
provides thrust force
over a short time
interval
• Solid fuel rocket
engines are rated in
letters
So let’s say I use an A8-3 engine
on a 45 gram rocket
• F∆t = m∆v
• 2.5 N-s = (.045kg) ∆v
• ∆v = 55.6 m/s
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How much acceleration?
F=ma
8N = (.045 kg) a
a = 177.8 m/s2 -9.81 m/s2
a = 168 m/s2
How far up does it get before it
coasts and then in total before *?
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There are 2 phases to the flight
v2 = vo2 + 2a(d-do)
(55.6 m/s)2 = 02 + 2(168m/s 2)(d-do)
d = 9.2 m (under acceleration)
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Then its simple ballistic flight for 3 s
d = do + vot + 1/2at2
d = 9.2m + (55.6m/s)(3s) + 1/2(-9.8m/s2)(3s)2
d = 131.9 m total height attained
Multiple stage rockets drop
excess mass during the flight.
What about calculating the
Height attained by a rocket
that has multiple stages?
These have three phases,
Initial acceleration at mr+b
With the the thrust of the
initial engine C-6-0,
no coasting time
Then an acceleration and
coast phase for the rocket
mass without booster mr
Release of booster from second stage
So, currently under construction
in my laboratory is this kit:
We will mass and launch this on Friday,
weather permitting
We will calculate the booster, upper stage,
and parachute deployment heights of this
Two-stage rocket.
We must mass the rocket parts and record the
Specs of the Engines used
We must have several students sight and
record angles and horizontal observation
Distances. Then share the data Monday.
Angular Momentum
The spinning “beacon” of a neutron star can make it a “pulsar”
as observed from Earth
Radiotelescopes
http://www.jb.man.ac.uk/~pulsar/Education/Sounds/sounds.html
So, lets say I have a
massive star, 10x the mass
of our sun. It runs out of
hydrogen fuel collapses and
explodes most of its mass
into space in a supernova.
The core of the star has a
Mass 1.5 x that of our sun.
3 x 1030kg. Its initial radius
is 7x108 m it collapses to
2x104 m as the matter is
compressed into neutrons.
What is the velocity of the surface
of the neutron star?
Rotation speed of a star
v = 2πr
T
Sunspots rotate in 30 d
v = 2π( 7 x 108m)
30d x 24h/d x 3600s/h
v = 1700 m/s
w = mvr
Conservation of angular
momentum (w)
m1v1r1 = m2v2r2
(3 x 1030 kg) (1700m/s) (7 x 108m) = (3 x 1030 kg) v2 (2 x 104m)
6 x 107 m/s = v2
If the speed of light is 3 x 108 m/s its about 20% of light speed
v = 2πr
T
6 x 107 m/s = 2π (2 x 104m)
T
http://www.jb.man.ac.uk/~pulsar/Education/Sounds/crab.au
T = 0.002 sec….