Download Momentum

Momentum
• What is it?
• It plays a huge part in sports, both technique,
and equipment.
• If you have ever been in an auto accident
physics may have saved your life or at least
prevented more serious injuries.
• Conservation of Momentum as a forensic tool.
What do we know already?
Some things to chew on
• Describe mass
• What is inertia?
• Define Newton’s Laws of Motion, particularly
the first one.
• If an object has mass and velocity what does it
have?
• Momentum or inertia in motion!
Momentum
Inertia in motion
• We have a sense that it is harder to stop a
truck than a car if both are moving at the
same speed.
• It is more difficult to stop a car moving quickly
than one that is moving slowly.
Momentum
• Momentum = mass x velocity
•
=mv
• Momentum is a vector
• We often write momentum as p,
p =mv
• The unit of momentum is kg m/s; there is no
derived unit for momentum.
Momentum
• An object can have a large momentum if
• m is large,
mv , or if
v
• It has a large speed, m , or
• Both
• Can a skateboard have more
momentum than a cement
truck?
A) Yes
B) No
Impulse
• To change momentum we must change either
mass or velocity (or both).
• If m is constant to change p we must change
v (acceleration)
•
Δp = mΔv
• What causes a? A force F.
• But there is another variable when changing
momentum, time!
Impulse
• If you are pushing a car without gas the final
velocity depends on both how hard and how
long you push.
• The final momentum thus depends on how large a
force and on how long you apply the force.
• Definition of impulse
p  Ft
• Can also define an average impulse when force is variable
Impulse-examples
p  Ft
Increase momentum (increase Δt)
• Follow through- golf, baseball
• Barrel of rifle
Decrease momentum
Increase Δt to decrease F
Impulse
• The concept of impulse leads to a more general
form of Newton’s 2nd law.
•
FΔt = Δp
•
F = Δp/Δt
• In this form we can handle
problems where the mass changes
• If m is constant Δp = mΔv
•
F = m Δv/Δt = ma
• Look familiar?
Example: An object of mass 3.0 kg is allowed to fall
from rest under the force of gravity for 3.4 seconds.
What is the change in momentum? Ignore air
resistance.
Want p = mv.
v  at
v   gt  33.3 m/sec
p  mv  100 kg m/s (downward)
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Example : A force of 30 N is applied for 5 sec to each of
two bodies of different masses.
30 N
Take m1 < m2
m1 or m2
(a) Which mass has the greater momentum change?
p  Ft
Since the same force is applied to each
mass for the same interval, p is the
same for both masses.
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Example continued:
(b) Which mass has the greatest velocity change?
p
v 
m
(c) Which
Since both masses have the same p, the
smaller mass (mass 1) will have the larger
change in velocity.
mass has the greatest acceleration?
v
a
t
Since av the mass with the greater
velocity change will have the greatest
acceleration (mass 1).
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Example (text problem 7.10): What average force is
necessary to bring a 50.0-kg sled from rest to 3.0 m/s in
a period of 20.0 seconds? Assume frictionless ice.
p  Favt
p mv
Fav 

t
t

50.0 kg 3.0 m/s 
Fav 
 7.5 N
20.0 s
The force will be in
the direction of
motion.
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Question….Whaaattt?
As a linebacker on the football team, which
player would be the easiest for you to stop
when you tackle him?
A) The 350-lb lineman who can cover 10
yards in 2.5 seconds.
B) The 160-lb running back, who can make
it down the field (100 M) in ten seconds.
C) The 240-lb halfback, who can make it
down the field in twenty seconds.
D) They are all equally hard to stop.
Question…So many
As a kid playing on the playground, you would bend
your knees when you landed after jumping off
the monkey bars to reduce the "sting" in your
feet. This worked because
A) bending your knees gave you upward
momentum which partly canceled the
downward momentum.
B) bending your knees lowered your center of
gravity reducing the force of your fall.
C) bending your knees increased the time of
contact for the ground to bring you to rest.
D) you didn't do it on purpose, your knees just
buckled.
Momentum
Consider two interacting bodies with m2 > m1:
F21
F12
m1
m2
If we know the net force on each body then
Fnet
v  at 
t
m
The velocity change for each mass will be different if
the masses are different.
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Rewrite the previous result for each body as:
m1v1  F21t
m2v2  F12t  F21t
Newton’s 3rd
Law
Combine the two results:
m1v1  m2 v 2
m1 v1 f  v1i   m2 v 2 f  v 2i 
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m1v1  m2 v 2
p1  p 2
The change in momentum of the two bodies is “equal
and opposite”. Total momentum is conserved during
the interaction; the momentum lost by one body is
gained by the other.
Big hairy discourse to illustrate the conservation of
momentum, the total momentum (p) before has to
equal the total momentum after in an isolated system
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Conservation of Momentum
• If the net external force acting on a system is
zero, then the momentum of the system does
not change.
• When a quantity in physics does not change
we say it is conserved. Momentum is
conserved.
• We have to be careful to define the system.
• It is only forces external to the system that can
change momentum
Conservation of Momentum
• Internal forces do not change the momentum
of the system.
Momentum is a vector. Each component is
conserved separately.
Vectors can cancel each other out.
0
Is anything truly ever conserved?
• The answer is no, nothing is truly conserved
nor is it truly isolated but it makes the math
manageable and in most cases the external
forces are negligible for the immediate
reaction being considered
-M v + m V = 0
Conservation of Momentum
Crash! What happens in collisions
Energy and Momentum in collisions.
Conservation of Momentum
• If the net external force acting on a system is
zero, then the momentum of the system does
not change.
• We have to be careful to define the system.
• It is only forces external to the system that can
change momentum.
• If there are no external forces to the system
the internal changes in momentum cancel
Question
You peddle frantically to get your bicycle up to a
speed of 15 m/s. On level ground, you relax
and start to coast. Your speed
A) stays the same, because momentum is
conserved.
B) increases, now that you are not expending
energy turning the pedals.
C) decreases, as there are still forces acting on
the bicycle.
Conservation of Momentum
v1i
m1>m2
v2i
m1
m2
A short time later the
masses collide.
m1
m2
What happens?
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During the interaction:
N1
N2
y
F21
F12
x
w1
F
F
y
 N1  w1  0
x
  F21  m1a1
w2
F
F
y
 N 2  w2  0
x
 F12  m2 a2
There is no net
external force on
either mass.
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The forces F12 and F21 are internal forces. This means
that:
p1  p 2
p1 f  p1i  p 2 f  p 2i 
p1i  p 2i  p1 f  p 2 f
In other words, pi = pf. That is, momentum is
conserved. This statement is valid during the
interaction only.
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To be clear
• The momentum has to be conserved if we
consider the system isolated.
• Both masses have an initial momentum pi or
mvi and a final momentum pf aka mvf
• pi + pi = pf + pf
Be careful of signs!!!!! Everything one way must
be positive and everything the other must be
negative.
Example: A rifle has a mass of 4.5 kg and it fires a bullet
of 10.0 grams at a muzzle speed of 820 m/s. What is the
recoil speed of the rifle as the bullet leaves the barrel?
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Question
You paddle your own canoe forward by
pushing back on the water. If you can
change the velocity of 7.3 kg of water by
3.0 m/s with each stroke, your speed
changes by (myou + mcanoe = 93 kg)
A) 226 m/s
B) 38 m/s
C) 4.2 m/s
D) 0.24 m/s
Collisions
When there are no external forces present, the
momentum of a system will remain unchanged.
(pi = pf)
If the kinetic energy before and after an interaction is the
same, the “collision” is said to be perfectly elastic. If the
kinetic energy changes, the collision is inelastic.
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Elastic collisions
Inelastic collisions
If, after a collision, the bodies remain stuck together, the loss of
kinetic energy is a maximum, but not necessarily a 100% loss of
kinetic energy. This type of collision is called perfectly inelastic.
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Example: In a railroad freight yard, an empty freight car of mass m
rolls along a straight level track at 1.0 m/s and collides with an
initially stationary, fully loaded, boxcar of mass 4.0m. The two cars
couple together upon collision.
(a)
What is the speed of the two cars after the collision?
pi  p f
p1i  p2i  p1 f  p2 f
m1v1  0  m1v  m2 v  m1  m2 v
 m1 
v1  0.2 m/s
v  
 m1  m2 
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(b) Suppose instead that both cars are at rest after the collision.
With what speed was the loaded boxcar moving before the
collision if the empty one had v1i = 1.0 m/s.
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Example: A projectile of 1.0 kg mass approaches a stationary body of
5.0 kg mass at 10.0 m/s and, after colliding, rebounds in the reverse
direction along the same line with a speed of 5.0 m/s. What is the
speed of the 5.0 kg mass after the collision?
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The New York Times, Jan.13, 1920, p.
12
…its flight would be neither accelerated nor maintained
by the explosion of the charges… To claim that it would
be, is to deny a fundamental law of dynamics… That
Professor Goddard, with his ‘chair’ in Clark College and
the countenancing of the Smithsonian Institution, does
not know the relation of action to reaction, and of the
need to have something better than a vacuum against
which to react – to say that would be absurd.
The New York Times, July 17, 1969,
p. 34
…further investigation and experimentation have
confirmed the findings of Isaac Newton in the 17th
century, and it is now definitely established that a rocket
can function in a vacuum as well as in an atmosphere.
The Times regrets the error.
…an editorial feature of the New York Times dismissed
the notion that a rocket could function in a vacuum and
commented on the ideas of Robert H. Goddard.
Question
A boxcar with mass m moving to the right with
speed v collides with a second boxcar of the
same mass which is at rest. After the
collision, both boxcars move off to the right
with speed v/2. In this collision
A) momentum is conserved.
B) energy is conserved.
C) momentum and energy are conserved.
D) neither momentum nor energy is
conserved.
Center of Mass
• Why is it important?
• Is the center of the object always the center of
mass?
• Can I have multiple centers on a body?
• Does the center always have to be on the
body?
There is something special about this point
Gravitational Torque
• Remember Rotational motion? Rotating
objects also have momentum, but first lets
look again at what makes them spin.
• Center of mass. Gravity acts on an objects
center of mass. Anytime that center is not
directly over an object gravitational force
provides torque
• T = rw or (xcg –x1) mg
• Previously we determined cm by hanging from
a string
More torque stuff
• Many objects have more than one center of
mass, and sometimes the center of mass may
not even be located on the body.
• The center of mass (CM) is the point
representing the mean (average) position of
the matter in a body.
• An example of a body with two center of
masses: T1 = rw and T2 = -rw so from the
previous equation Tnet = 0 = T1 + T2 so……
• (xcg –x1) mg - (xcg –x1) mg = 0
Page 250
The center of mass (of a two body system) is found
from:
xcm
m1 x1  m2 x2

m1  m2
This is a “weighted” average of the positions of the
particles that compose a body.
A larger mass is more important.
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What about 3 particles in a coordinate plane (big boy stuff here)?
Example The positions of three particles are (4.0 m, 0.0 m), (2.0 m,
4.0 m), and (1.0 m, 2.0 m). The masses are 4.0 kg, 6.0 kg, and
3.0 kg respectively. What is the location of the center of mass?
y
2
1
3
x
M
r
4 (4,0)
6 (2,4)
3 (-1,-2)
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Example continued:
xcm
m1 x1  m2 x2  m3 x3

m1  m2  m3

4 kg 4 m   6 kg 2 m   3 kg  1 m 

4  6  3 kg
 1.92 m
ycm
m1 y1  m2 y2  m3 y3

m1  m2  m3

4 kg 0 m   6 kg 4 m   3 kg  2 m 

4  6  3 kg
 1.38 m
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Where is the CM
A
of a Donut?
B
C
D) It could be A or C or anywhere on the donut.
Motion of the Center of Mass
For an extended body, it can be shown
that p = mvcm.
From this it follows that Fext = macm.
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Fig. 07.11
Rotational/Angular Momentum
•
•
•
•
•
•
Angular acceleration = T/I = w/t
Torque is the rotational equivalent to Force
Moment of inertia I is the equivalent to mass
Angular velocity w is equivalent to velocity
So… if linearly p = mv, rotationally L = Iw
Linear momentum is conserved so rotational
momentum is conserved as well.
More stuff you should already
know
• If pi = pf then Li = Lf
• Remember rotationally the moment of inertia
(I) is the equivalent to mass (m)
• I = mr2
• 1 Rev = 2(pi)r
• Frequency and period are the inverse of each
other