Transcript Momentum and Collisions

Momentum and
Collisions
Linear Momentum

The linear momentum of a particle or an
object that can be modeled as a particle of
mass m moving with a velocity v is defined to
be the product of the mass and velocity:

p=mv



We will usually refer to this as “momentum”, omitting the
“linear”.
Momentum is a VECTOR.
It has components. Don’t forget that.
Linear Momentum



The dimensions of momentum are ML/T
The SI units of momentum are kg · m / s
Momentum can be expressed in component
form:

px = m v x
py = m v y
pz = m v z
A 3.00-kg particle has a velocity of . (a) Find its x and y
components of momentum. (b) Find the magnitude and
direction of its momentum.
Newton and Momentum


Newton called the product mv the quantity of
motion of the particle
Newton’s Second Law can be used to relate
the momentum of a particle to the resultant
force acting on it
dv d  mv  dp
F  ma  m


dt
dt
dt
with constant mass.
Newton’s Second Law

The time rate of change of the linear momentum of
a particle is equal to the net force acting on the
particle




This is the form in which Newton presented the Second
Law
It is a more general form than the one we used previously
This form also allows for mass changes
Applications to systems of particles are particularly
powerful
Two Particles (1 dimension for now)
dptotal dp1 dp2


dt
dt
dt
IF there are NO external forces then
the only forces acting on this system are
those that are one particle producing a
force on the other.
Continuing
dptotal dp1 dp2


dt
dt
dt
dptotal
Fexternal  0 
  Fint ernal  0
dt
NO EXTERNAL FORCES: ptotal is
constant

Force on P1 is from particle 2 and is F21
Force on P2 is from particle 1 and is F12

F21=-F12

Summary
The momentum of a SYSTEM of particles that are
isolated from external forces remains a constant of
the motion.
Conservation of Linear Momentum
Textbook Statement

Whenever two or more particles in an
isolated system interact, the total
momentum of the system remains
constant


The momentum of the system is conserved,
not necessarily the momentum of an individual
particle
This also tells us that the total momentum of
an isolated system equals its initial momentum
Two blocks of masses M and
3M are placed on a horizontal,
frictionless surface. A light
spring is attached to one of
them, and the blocks are
pushed together with the
spring between them). A cord
initially holding the blocks
together is burned; after this,
the block of mass 3M moves
to the right with a speed of
2.00 m/s. (a) What is the speed
of the block of mass M? (b)
Find the original elastic
potential energy in the spring
if M = 0.350 kg.
An Easy One…
A particle of mass m moves with
momentum p. Show that the kinetic
energy of the particle is K = p2/2m. (b)
Express the magnitude of the particle’s
momentum in terms of its kinetic energy
and mass.
Consider a particle roaming around that is
suddenly subjected to some kind of FORCE
that looks something like the last slide’s graph.
NEWTON :
p  mv
dp
dv
m
 ma  F
dt
dt
p  constant if F  0
If F  0 then p will change
Let’s drop the vector notation and stick to
one dimension.
dp
F
dt
dp  Fdt
f
tf
i
ti
 dp   Fdt
tf
p f  pi   Fdt
ti
Change of Momentum
Impulse
NEW LAW

IMPULSE = CHANGE IN MOMENTUM
A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at
an angle of 60.0° with the surface. It bounces off with the
same speed and angle (Fig. P9.9). If the ball is in contact
with the wall for 0.200 s, what is the average force exerted
on the ball by the wall?
LETS
TALK
ABOUT
Consider two particles:
v2
v1
m1
m2
1
1
V1
2
2
V2
What goes on during the collision? N3
Force on m2
=F(12)
Force on m1
=F(21)
The Forces
Equal and Opposite
More About Impulse: F-t The Graph




Impulse is a vector quantity
The magnitude of the
impulse is equal to the area
under the force-time curve
Dimensions of impulse are
ML/T
Impulse is not a property of
the particle, but a measure
of the change in momentum
of the particle
Impulse



The impulse can also
be found by using the
time averaged force
I = F Dt
This would give the
same impulse as the
time-varying force does
SKATEBOARD DEMO
Conservation of Momentum, Archer
Example


The archer is standing on
a frictionless surface (ice)
Approaches:



Newton’s Second Law –
no, no information about F
or a
Energy approach – no, no
information about work or
energy
Momentum – yes
Archer Example, 2




Let the system be the archer with bow (particle 1)
and the arrow (particle 2)
There are no external forces in the x-direction, so it
is isolated in terms of momentum in the x-direction
Total momentum before releasing the arrow is 0
The total momentum after releasing the arrow is p1f
+ p2f = 0
Archer Example, final

The archer will move in the opposite direction
of the arrow after the release


Agrees with Newton’s Third Law
Because the archer is much more massive
than the arrow, his acceleration and velocity
will be much smaller than those of the arrow
An estimated force-time curve for a baseball struck by a bat
is shown in Figure P9.7. From this curve, determine (a) the
impulse delivered to the ball, (b) the average force exerted
on the ball, and (c) the peak force exerted on the ball.
Overview: Collisions – Characteristics



We use the term collision to represent an event
during which two particles come close to each other
and interact by means of forces
The time interval during which the velocity changes
from its initial to final values is assumed to be short
The interaction force is assumed to be much greater
than any external forces present

This means the impulse approximation can be used
Collisions – Example 1


Collisions may be the
result of direct contact
The impulsive forces
may vary in time in
complicated ways


This force is internal to
the system
Momentum is
conserved
Collisions – Example 2



The collision need not
include physical contact
between the objects
There are still forces
between the particles
This type of collision can
be analyzed in the same
way as those that include
physical contact
Types of Collisions

In an elastic collision, momentum and kinetic
energy are conserved



Perfectly elastic collisions occur on a microscopic level
In macroscopic collisions, only approximately elastic
collisions actually occur
In an inelastic collision, kinetic energy is not
conserved although momentum is still conserved

If the objects stick together after the collision, it is a
perfectly inelastic collision
Collisions, cont



In an inelastic collision, some kinetic energy
is lost, but the objects do not stick together
Elastic and perfectly inelastic collisions are
limiting cases, most actual collisions fall in
between these two types
Momentum is conserved in all collisions
Perfectly Inelastic Collisions


Since the objects stick
together, they share the
same velocity after the
collision
m1v1i + m2v2i =
(m1 + m2) vf
A 10.0-g bullet is fired into a stationary block of wood (m =
5.00 kg). The relative motion of the bullet stops inside the
block. The speed of the bullet-plus-wood combination
immediately after the collision is 0.600 m/s. What was the
original speed of the bullet?
. Velcro couplers make the carts stick together after colliding.
Find the final velocity of the train of three carts. (b) What If ?
Does your answer require that all the carts collide and stick
together at the same time? What if they collide in a different
order?
Most of us know intuitively that in a head-on collision between a
large dump truck and a subcompact car, you are better off being
in the truck than in the car. Why is this? Many people imagine
that the collision force exerted on the car is much greater than
that experienced by the truck. To substantiate this view, they
point out that the car is crushed, whereas the truck is only
dented. This idea of unequal forces, of course, is false.
Newton’s third law tells us that both objects experience forces
of the same magnitude. The truck suffers less damage because
it is made of stronger metal. But what about the two drivers?
Do they experience the same forces? To answer this question,
suppose that each vehicle is initially moving at 8.00 m/s and that
they undergo a perfectly inelastic head-on collision. Each driver
has mass 80.0 kg. Including the drivers, the total vehicle masses
are 800 kg for the car and 4 000 kg for the truck. If the collision
time is 0.120 s, what force does the seatbelt exert on each driver?
Elastic Collisions

Both momentum and
kinetic energy are
conserved
m1v1i  m2 v 2 i 
m1v1 f  m2 v 2 f
1
1
2
m1v1i  m2 v 22i 
2
2
1
1
2
m1v1 f  m2 v 22 f
2
2
Elastic Collisions, cont






Typically, there are two unknowns to solve for and so you need two
equations
The kinetic energy equation can be difficult to use
With some algebraic manipulation, a different equation can be used
v1i – v2i = v1f + v2f
This equation, along with conservation of momentum, can be used
to solve for the two unknowns
 It can only be used with a one-dimensional, elastic collision
between two objects
The solution is shown on pages 262-3 in the textbook. (Lots
of algebra but nothing all that difficult.
We will look at a special case.
Let’s look at the case of equal masses.
Second Particle initially at rest
Explains the demo!
Elastic Collisions, final

Example of some special cases



m1 = m2 – the particles exchange velocities
When a very heavy particle collides head-on with a very
light one initially at rest, the heavy particle continues in
motion unaltered and the light particle rebounds with a
speed of about twice the initial speed of the heavy particle
When a very light particle collides head-on with a very
heavy particle initially at rest, the light particle has its
velocity reversed and the heavy particle remains
approximately at rest
Collision Example – Ballistic
Pendulum




Perfectly inelastic collision –
the bullet is embedded in
the block of wood
Momentum equation will
have two unknowns
Use conservation of energy
from the pendulum to find
the velocity just after the
collision
Then you can find the
speed of the bullet