LINEAR

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LINEAR MOMENTUM IMPULSE
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(Linear) Momentum, p

• ? is mass times velocity
• p m ? v vector!
• ? (p) kg m/s
• ? a 1 kg object moving at 1000 m/s has the same
  momentum as a 1000 kg object moving at 1 m/s (p
  1000 kg m/s)
• ? a roller skate rolling has more momentum than
• stationary truck.

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How can the momentum of an object be changed?
By changing its mass, or, more usually, by
exerting a force causing an acceleration that
changes its velocity.
Lets go back to Newtons second law F ma.
Actually, Newton formulated his second law
as Force time rate of change of momentum

?p is the change in momentum produced by the
force F in time ?t
If the mass doesnt change, then
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is in fact the form in which you should remember
the second law of motion since the law in the
form F ma is actually, as we have seen, a
special case it can not be applied to
situations in which mass can change.
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? we can get a very useful form of Newtons 2.
law
F?t ?p Dp mv - mu

F?t is called the impulse of the force.
impulse (action of a force F over time Dt ) will
produce change in momentum Dp
units (F?t) Ns Ns kg m/s
REMEMBER Although we write F for simplicity, we
actually mean Fnet , because only Fnet and not
individual forces can change momentum (by
producing an acceleration)
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? Achieving the same change in momentum over a
long time requires smaller force and over a short
time greater force.
Lets think about the time it takes to slow the
truck to zero.
EXAMPLES
?p F?t
You could stop it with your own force just if
you exert it over a long, long period of time.
or, you could exert a huge force over very short
period of time.
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For better understanding well do another example
EXAMPLES
h 2 m m 30 g 0.03 kg
both eggs fall the same distance, so the velocity
of both eggs just before impact is
h
h
Impact before impact u 6 m/s just after
impact is v 0
?p mv mu 0.18 kg m/s
In both cases momentum is reduced to zero during
impact/interaction with the floor. But the time
of interaction is different. In the case of
concrete, time is small while in the case of
pillow, the stopping time is greatly
increased. If you look at the impulse-momentum
relation F? t ?p, you see that for the same
change in momentum ( 0.18 kg m/s in this case),
if the time is smaller the ground must have
exerted greater force on the egg. And vice versa.
The pillow will exert smaller force over greater
period of time.
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? Often you want to reduce the momentum of an
object to zero but with minimal impact force (or
injury). How to do it? Try to maximize the time
of interaction this way stopping force is
decreased.
Getting smart and smarter by knowing physics
EXAMPLES
? Car crash on a highway, where theres either a
concrete wall or a barbed-wire fence to
crash into. Which to choose? Naturally, the
wire fence your momentum will be decreased by
the same amount, so the impulse to stop you
is the same, but with the wire fence, you
extend the time of impact, so decrease the force.
? Bend your knees when you jump down from high!
Try keeping your knees stiff while landing
it hurts! (only try for a small jump, otherwise
you could get injured) Bending the knees
extends the time for momentum to go to zero,
by about 10-20 times, so forces are 10-20 times
less.
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? Safety net used by acrobats, increases impact
time, decreases the forces. ? Catching a
ball let your hand move backward with the ball
after contact ? Bungee jumping ? Riding
with the punch, when boxing, rather than moving
into
EXAMPLES
By moving away, the time of contact is extended,
so force is less than if he hadnt moved.
By moving into the glove, he is lessening the
time of contact, leading to a greater force, a
bigger ouch!
? Wearing the gloves when boxing versus boxing
with bare fists.
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? Sometimes you want to increase the force over
a short time
This is how in karate (tae kwon do), an expert
can break a stack of bricks with a blow of a
hand Bring in arm with tremendous speed (large
momentum), that is quickly reduced on impact with
the bricks. The shorter the time, the
larger the force on the bricks.
EXAMPLES
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Determine the change in impulse due to a time
varying force
Formulas we had are for the constant force. What
if the force changes over time ?t ?
The graph shows the variation with time of the
force on the football of mass 0.5 kg.
ball was given an impulse of approximately
100x0.01 1Ns during this 0.01s.
area under graph is the total impulse given to
the ball 2x(100x0.05)/2 5 Ns
F?t ?p ? ?p 5 kg m/s
?p m?v ? ?v 10 m/s
v u ?v
Change in momentum, ?p, in time ?t is the area
under the graph force vs. time.
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In actuality one is much more likely to use
the measurement of the speed of the football to
estimate the average force that is exerted by the
foot on the football. The time that the foot is
in contact with the ball can be measured
electronically.
Favg m ?v/ ?t
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Till now we were concentrated on ONE object. Now
we move to the system of (usually) two
objects exerting strong forces over a short time
intervals on each other like collisions,
explosions, ejections
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Collisions - conservation of momentum

• ? collisions can be very complicated
• ? two objects bang into each other and exert
  strong forces over short time intervals which are
  very hard to measure
• ? fortunately, we can predict the future without
  going into pesky details of force.
• ? What will help us is the law of conservation
  of linear momentum

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Law of Conservation of Momentum
? consider system particle 1 and particle
2 collide with one another.
velocities just before interaction (collision)
velocities just after interaction (collision)
(p1 p2 p)
v1
pafter pbefore
v2
m1v1 m2v2 m1u1 m2u2
The total linear momentum of a system of
interacting particles is conserved - remains
constant, provided there is no resultant external
force. Such a system is called an isolated
system.
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Certain situations (collisions, explosions,
ejections) do not allow detailed knowledge of
forces (strength, direction, duration) or
acceleration. Of course that these situations
must follow Newtons laws. The only problem is
that it is difficult to see exactly how to apply
them. One cannot easily measure neither forces
involved in the collision nor acceleration
(velocity appears to be instantaneously
acquired).
The law of conservation of momentum gives us an
easy and elegant way to predict the outcome
without knowing forces involved in process. It is
much easier to measure velocities and masses
before and after interaction.
WE CAN APPLY THE LAW OF CONSERVATION OF MOMENTUM
TO COLLISIONS AND EXPLOSIONS (EJECTIONS) IF
DURING INTERACTION THE NET EXTERNAL FORCE IS ZERO
OR IT CAN BE NEGLECTED.
Example baseball is struck with a bat duration
of the collision is about 0.01 s, and the average
force the bat exerts on the ball is several
thousand Newtons what is much greater than the
force of gravity, so you can ignore it. And as we
consider velocities just before and just after
interaction, there is no much change due to
gravity. The system can be considered isolated
and momentum is conserved.
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beauty of the law of conservation of momentum

• ? if we know what the objects were doing before
  they collided, we can figure out what can happen
  after they collide.
• ? We can work backward sometimes to figure out
  from the collision scene what was going on before
  the collision.

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Momentum is conserved in every isolated system.
Internal forces can never change momentum of the
system.
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Example how to use law of conservation of
momentum in the case of ejections or explosions.
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A 60.0-kg astronaut is on a space walk when her
tether line breaks. She throws her 10.0-kg oxygen
tank away from the shuttle with a speed of 12.0
m/s to propel herself back to the shuttle. What
is her velocity?
after
before
12.0 m/s
60
70
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v1 ?
u 0
pbefore pafter
0 m1 v1 m2 v2 0
60.0 v1 10.0 (12.0) v1 - 2.0 m/s

moving in the negative direction means toward
shuttle
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Very similar case is spaceship propulsion which
is actually example of conservation of momentum.
Since no outside forces act on the system
(spaceship its fuel) or it is very small
compared to the explosion, the momentum gained by
fuel ejected in the backward direction must be
balanced by forward momentum gained by the
spaceship.
hot gas ejected at very high speed
pbefore pafter 0 m1 v1 m2 v2 m1 v1 - m2
v2
? the same as untied balloon.
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Similar examples are recoil of the firing gun,
recoil of the firing cannon, ice-skaters recoil,
throwing of the package from the boat etc.
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? Two stationary ice skaters push off ? both
skaters exert equal forces on each other ?
however, the smaller skater acquires a larger
speed (due to larger acc.) than the larger
skater. ? momentum is conserved!
pbefore pafter 0 m1 v1 m2 v2 m1 v1 - m2
v2
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If you consider momentum before 0, so after
must be zero too, therefore the speeds gained
(while the force of interaction acted) are pretty
different.
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Example how to use law of conservation of
momentum in the case of collisions.
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There are two fish in the sea. A 6 kg fish and a
2 kg fish. The big fish swallows the small one.
What is its velocity immediately after lunch?

• the big fish swims at 1 m/s toward and
• swallows the small fish that is at rest.

before lunch
after lunch
Net external force is zero. Momentum is
conserved.
v ?
p before lunch p after lunch
momentum is vector, direction matters choose
positive direction in the direction of big
fish.
Mu1 mu2 (M m)v
(6 kg)(1 m/s) (2 kg)(0 m/s) (6kg 2 kg) v
6 kg m/s (8 kg) v
v 0.75 m/s
in the direction of the large fish before lunch
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b. Suppose the small fish is not at rest but is
swimming toward the large fish at 2 m/s.
before lunch
after lunch
- 2 m/s
1 m/s
v ?
p before lunch p after lunch
Mu1 mu2 (Mm)v
(6 ) (1 ) (2 ) (2 ) (6 2 ) v 6 4
8 v
v 0.25 m/s
in the direction of the large fish before lunch
The negative momentum of the small fish is very
effective in slowing the large fish.
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c. Small fish swims toward the large fish at 3
m/s.
before lunch
after lunch
- 3 m/s
1 m/s
v ?
p before lunch p after lunch
Mu1 mu2 (Mm)v
(6 ) (1 ) (2 ) (3 ) (6 2 ) v 6 6
(8 ) v
v 0 m/s
fish have equal and opposite momenta. Zero
momentum before lunch is equal to zero momentum
after lunch, and both fish come to a halt.
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d. Small fish swims toward the large fish at 4
m/s.
before lunch
after lunch
- 4 m/s
1 m/s
v ?
p before lunch p after lunch
Mu1 mu2 (Mm)v
(6 ) (1 ) (2 ) (4) (6 2 ) v 6 8 8 v

v 0.25 m/s
The minus sign tells us that after lunch the
two-fish system moves in a direction opposite to
the large fishs direction before lunch.
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A red ball traveling with a speed of 2 m/s along
the x-axis hits the eight ball. After the
collision, the red ball travels with a speed of
1.6 m/s in a direction 37o below the positive
x-axis. The two balls have equal mass. At what
angle will the eight ball fall in the side
pocket? What is the speed of the blue (8th) ball
after collision.
v2
before collision after
collision
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u2 0
u1
?2
8
370
the point of collision
pbefore pafter
v1
in x direction m u1 0 m v1 cos
370 m v2 cos q2
v2 cos q2 u1 - v1 cos 370 0.72 m/s
(1)
in y direction 0 - m v1 sin 370
m v2 sin ?2
v2 sin ?2 v1
sin 370 0.96 m/s (2)
direction of v2 (2)/(1)
tan ?2 1.33 ?2 530
(2) ? v2 0.96 / sin 530
v2 1.2 m/s
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Derivation of the Law of Conservation of Momentum
? consider system particle 1 and particle 2
collide with one another with no net
external force acting on neither of them.
velocities just before interaction (collision)
velocities just after interaction (collision)
forces during collision
v1
v2
? During the time interval the collision takes
place, ?t, impulse F1?t given to particle 1
will cause its momentum change ?p1. During
the same time interval impulse F2 ?t will
change particles 2 momentum by ?p2.
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(p1 p2 p)
particle 1 F1?t ?p1
particle 2 F2 ?t ?p2
F1 F2 (N3.L) ? ?p2 ?p1
What one object loses in the collision the other
one gains.
???????????????? ?p1 ?p2 0 ? ?(p1 p2 )
0 ? ?p 0 ? pafter pbefore Total
momentum of a system before and after collision
is the same.
Conservation of Momentum if no external force
act on a system, the total momentum of the system
is conserved it will not change. Such a system
is called an isolated system.
This argument can be extended up to any number of
interacting particles so long as the system of
particles is still isolated.