Transcript Document 7390519

(Linear) Momentum, p
● is mass times velocity
p=mv
vector!
p = mv
● (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.
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.
Let’s go back to Newton’s second law: F = ma.
Actually, Newton formulated his second law as:
Δp
Force = time rate of change of momentum F =
Δt
Δp is the change in momentum produced by the force F in time Δt
Δ(mv)
Δv
=m
= ma
If the mass doesn’t change, then F =
Δt
Δt
Δp
F=
Δt
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.
→ we can get a very useful form of Newton’s 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)
● Achieving the same change in momentum over a long time
requires smaller force and over a short time greater force.
Let’s think about the time it
takes to slow the truck to zero.
∆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.
For better understanding we’ll do another example:
h=2m
h
h
m = 30 g = 0.03 kg
both eggs fall the same distance,
so the velocity of both eggs just
before impact is:
v = u2 +2gh = 2gh = 6m/s
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.
● 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:
►
Car crash on a highway, where there’s 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.
►
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
By moving away, the time of
contact is extended, so force is
less than if he hadn’t 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.
● 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.
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
Change in momentum, Δp, in time Δt is
the area under the graph force vs. time.
v = u + ∆v
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
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
● 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:
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)
m1
(p1 + p2 = p)
v1
u1
pafter = pbefore
u2
m2
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”.
Certain situations (collisions, explosions, ejections) do not allow detailed
knowledge of forces (strength, direction, duration) or acceleration. Of course
that these situations must follow Newton’s 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.
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.
Momentum is
conserved in every
isolated system.
Internal forces can
never change
momentum of the
system.
Example how to use law of conservation of
momentum in the case of ejections or explosions.
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
70
60
u=0
v1 = ?
10
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
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.
Similar examples are: recoil of the firing gun, recoil of the
firing cannon, ice-skater’s recoil, throwing of the package
from the boat etc.
● 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
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.
Example how to use law of conservation of
momentum in the case of collisions.
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?
a. the big fish swims at 1 m/s toward and
swallows the small fish that is at rest.
before lunch
after lunch
1 m/s
6
Net external force is zero
Momentum is conserved.
8
2
v=?
p before lunch = p after lunch
Mu1 + mu2 = (M + m)v
momentum is vector, direction matters;
choose positive direction in the
direction of big fish.
+
(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
+
b. Suppose the small fish is not at rest but is
swimming toward the large fish at 2 m/s.
before lunch
1 m/s
6
- 2 m/s
after lunch
2
8
v=?
p before lunch = p after lunch
Mu1 + mu2 = (M+m)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.
+
c. Small fish swims toward the large fish at 3 m/s.
before lunch
1 m/s
6
- 3 m/s
after lunch
2
8
v=?
p before lunch = p after lunch
Mu1 + mu2 = (M+m)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.
+
d. Small fish swims toward the large fish at 4 m/s.
before lunch
1 m/s
6
- 4 m/s
after lunch
2
8
v=?
p before lunch = p after lunch
Mu1 + mu2 = (M+m)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
fish’s direction before lunch.
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:
u1
8
in y – direction
θ2
370
the point of
v1
collision
m u1 + 0 = m v1 cos 370 + m v2 cos q2
v2 cos q2 = u1 - v1 cos 370 = 0.72 m/s
(1)
0 = - m v1 sin 370 + m v2 sin θ2
v2 sin θ2 = v1 sin 370 = 0.96 m/s
(2)
direction of v2 ; (2)/(1)
(2) →
8
u2 = 0
pbefore = pafter
in x – direction
after collision:
v2 = 0.96 / sin 530
tan θ2 = 1.33
θ2 = 530
v2 = 1.2 m/s
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)
m1
forces during
collision
F1
u1
v1
F2
u2
m2
velocities just after
interaction (collision)
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 particle’s 2
momentum by ∆p2.
particle 1 : F1∆t = ∆p1
(p1 + p2 = p)
particle 2 : F2 ∆t = ∆p2
F1 = – F2 (N3.L)
∆p = 0 →
pafter = pbefore
→ ∆p2 = – ∆p1
What one object loses in the
collision the other one gains.
∆p1 + ∆p2 = 0 → ∆(p1 + p2 ) = 0 →
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.