Download Linear Momentum

Center of Mass
& Linear
Momentum
Physics
Montwood High School
R. Casao
Center of Mass

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We have discussed parabolic trajectories using a
“particle” as the model for objects.
But clearly objects are not particles – they are
extended and may have complicated shapes & mass
distributions.
So if we toss something like a baseball bat into the air
(spinning and rotating in a complicated way), what can
we really say about it’s trajectory?
There is one special location in every object that
provides us with the basis for our earlier model of a
point particle.
That special location is called the center of mass.
The center of mass will follow a parabolic trajectory –
even if the rest of the bat’s motion is very complicated
Center of Mass
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To start, let’s suppose that we have two masses m1
and m2, separated by some distance d.
We have also arbitrarily aligned the origin of our
coordinate system to be the center of mass m1.
We define the center of mass for these two particles
to be:
xcom
m2

d
m1  m2
Center of Mass

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

From this we can see that
if m2 = 0, then xcom = 0.
Similarly, if m1 = 0, then
xcom = d.
Finally, if m1 = m2, then
xcom = ½·d.
So we can see that the
center of mass in this case
is constrained to be
somewhere between x = 0
and x = d.
Center of Mass



Now lets shift the origin of
the coordinate system a
little.
We now need a more
general definition of the
center of mass.
The more general
definition (for two
particles) is:
xcom
m1  x1  m2  x2

m1  m2
Center of Mass


Now let’s suppose that we have lots of particles –
all lined up nicely for us on the x axis
The equation would now be:
xcom

m1  x1  m2  x2  ...  mn  xn

M
where M = m1 + m2 + … + mn
The collection of terms in the numerator can be
rewritten as a sum resulting in:
1 n
xcom 
M
  mi  xi
i 1
Center of Mass

This result is only for
one dimension
however, so the more
generalized result for
3 dimensions is
shown here:
n
xcom
1

  mi  xi
M i 1
n
ycom
1

  mi  yi
M i 1
n
zcom
1

  mi  zi
M i 1
Center of Mass


You know by intuition that the center of mass of
a sphere is at the center of the sphere; for a rod
it lies along the central axis of the rod; for a flat
plate it lies in the plane of the plate.
Note however that the center of mass doesn’t
necessarily have to lie within the object or have
any mass at that point:
 The COM of a horseshoe is somewhere in the
middle along the axis of symmetry.
 The COM of a doughnut is at it’s ‘geographic’
center, but there is no mass there either.
Center of Mass


Going back to our
baseball bat, the COM
will lie along the central
axis (the axis of
symmetry).
And it is the COM that
faithfully follows the line
of a parabola.
Newton’s 2nd Law for a System of Particles
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We know from experience that if you roll the cue ball into
another billiard ball that is at rest, that the two ball system
will continue on away from you after the impact.
You would be very surprised if one or the other of the balls
came back to you (without putting ‘English’ on the ball).
What continued to move away from you was the COM of
the two ball system!
Remember that the COM is a point that acts as though all
of the mass in the system were located there.
So even though we may have a large number of particles –
possibly of different masses, we can treat the assembly as
having all of it’s mass at the point of it’s COM.
Newton’s 2nd Law for a System of Particles
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
So we can assign that point a position, a
velocity and an acceleration.
And it turns out that Newton’s 2nd law holds
for that point:


Fnet  M  acom

where Fnet is the sum of all the external forces
acting on the mass; M is the total mass; and acom
is the acceleration of the center of mass.
Newton’s 2nd Law for a System of Particles

Circling back around to where we started from,
we can break the one vector equation down into
an equation for each dimension:
Fnet, x  M  acom, x
Fnet, y  M  acom, y
Fnet, z  M  acom, z
Newton’s 2nd Law for a System of Particles

Now let’s re-examine what is going on when we have
the collision between the two billiard balls:



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Once the cue ball has been set in motion, there are no external
forces acting on the system.
Thus, Fnet = 0 and as a consequence, acom = 0.
This means that the velocity of the COM of the system must be
constant.
So when the two balls collide, the forces must be
internal to the system (the balls exert forces on each
other; no external forces act on the balls) – which
doesn’t affect Fnet.
Thus the COM of the system continues to move forward
unchanged by the collision.
Newton’s 2nd Law for a System of Particles

When a fireworks rocket
explodes, the COM of the
system does not change;
while the fragments all
fan out, their COM
continues to move along
the original path of the
rocket.

This is also how a
ballerina seems to defy
the laws of physics and
float across the stage
when doing a grand jeté.

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Suppose a cannon shell traveling in a parabolic
trajectory (neglecting air friction) explodes in flight,
splitting into two fragments of equal mass.
The fragments follow new parabolic paths, but the
center of mass continues on the original parabolic path
as if all the mass were still concentrated at that point.
Linear Momentum
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Linear momentum of an object is the mass of the
object multiplied by its velocity.
Momentum: p = m·vcom
Unit: kg·m/s or N·s
Newton expressed his 2nd law of motion in terms of
momentum:
 The time rate of change of the momentum of a
particle is equal to the net force acting on the particle

and is in the direction of that force.

Fnet

dp

dt
Both momentum and kinetic energy describe the
motion of an object and any change in mass
and/or velocity will change both the momentum
and kinetic energy of the object.
Linear Momentum
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Momentum refers to inertia in motion.
Momentum is a measure of how difficult it is to
stop an object; a measure of “how much motion”
an object has.
More force is needed to stop a baseball thrown at
95 mph than to stop a baseball thrown at 45
mph, even though they both have the same mass.
More force is needed to stop a train moving at 45
mph than to stop a car moving at 45 mph, even
though they both have the same speed.
Both mass and velocity are important factors
when considering the force needed to change the
motion of an object.
Impulse
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Impulse (J) = force·time
Equation: J = F·t
Unit: N·s
The impulse of a force is equal to the change in
momentum of the body to which the force is
applied. This usually means a change in
velocity.
F·t = m·v where v = vf - vi
The same change in momentum can be
accomplished by a small force acting for a long
time or by a large force acting for a short time.
Impulse

The area under the curve in a force
vs. time graph represents the
change in momentum (m·v).

If your car runs into a
brick wall and you come
to rest along with the
car, there is a significant
change in momentum.
If you are wearing a seat
belt or if the car has an
air bag, your change in
momentum occurs over
a relatively long time
interval. If you stop
because you hit the
dashboard, your change
in momentum occurs
over a very short time
interval.
Impulse
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If a seat belt or air bag brings you to a stop over a time
interval that is five times as long as required to stop when
you strike the dashboard, then the forces involved are
reduced to one-fifth of the dashboard values. That is the
purpose of seat belts, air bags, and padded dashboards.
By extending the time during which you come to rest,
these safety devices help reduce the forces exerted on you.
If you want to increase the momentum of an object as
much as possible, you apply the greatest force you can for
as long a time as possible.
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A 1000 kg car moving at 30 m/s
(p = 30,000 kg m/s) can be stopped by 30,000 N of force
acting for 1.0 s (a crash!)
or by 3000 N of force acting for 10.0 s (normal stop)
Impulse and Bouncing
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Impulses are greater
when bouncing takes
place.
The impulse required
to bring an object to a
stop and then throw it
back again is greater
than the impulse
required to bring an
object to a stop.
Conservation of Linear
Momentum

In a closed system of objects, linear
momentum is conserved as the objects
interact or collide. The total vector
momentum of the system remains
constant.
p before interaction = p after interaction
Collisions
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A collision is an isolated event in which
two or more bodies (the colliding bodies)
exert relatively strong forces on each
other for a relatively short time.
An interesting point to note is that the
definition doesn’t necessarily require the
bodies to actually make contact – a near
miss (near enough so that there are
“relatively strong forces” involved) will
suffice…
Collisions
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To analyze a collision we
have to pay attention to
the 3 parts of a collision:
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Before
During
After
Momentum & Kinetic Energy in Collisions
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If we have a system where two bodies collide,
then there must be some kinetic energy (and
therefore linear momentum) present.
During the collision, the kinetic energy and linear
momentum of each body is changed by the
impulse from the other body.
We define two different kinds of collisions:

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An elastic collision is one where the total kinetic energy
of the system is conserved.
An inelastic collision is one where the total kinetic energy
is not conserved.
A collision is not necessarily completely elastic or
completely inelastic – most are in fact somewhere
in between…
Momentum & Kinetic Energy in Collisions
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For any kind of collision however, linear
momentum must be preserved.
This is because in a closed, isolated system,
the total linear moment cannot change
without the presence of an external force (and
the forces involved in a collision are internal to
the system – not external).
This does not mean that the linear momentum
of the various colliding bodies cannot change.
The linear momentum of each body involved in
the collision may indeed change, but the total
linear momentum of the system cannot
change.
Perfectly Inelastic Collisions
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Perfectly inelastic
collisions are those in
which the colliding objects
stick together and move
with the same velocity.
Kinetic energy is lost to
other forms of energy in
an inelastic collision.
Inelastic Collision Example
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Cart 1 and cart 2 collide and stick together
Momentum equation:
m1  v1  m2  v 2  (m1  m2 )  v'

Kinetic energy equation:
2
1
K i  0.5  m1  v
 0.5  m2  v 2
K f  0.5  (m1  m2 )  v'
2
K lost  K f  K i
v1 and v2 = velocities before collision
v  = velocity after collision
2
Directions for Velocity
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Momentum is a vector, so direction is
important.
Velocities are positive or negative to indicate
direction.
Example: bounce a ball off a wall
Inelastic Collisions
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Kinetic energy is lost when the objects
are deformed during the collision.
Momentum is conserved.
Elastic Collisions
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Momentum and
kinetic energy are
conserved in an
elastic collision.
The colliding objects
rebound from each
other with NO loss
of kinetic energy.
Elastic Collision Example
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Example: mass 1 and mass 2 collide and
bounce off of each other
Momentum equation:
m1  v1  m2  v 2  m1  v1'm2  v 2 '

Kinetic energy equation:
0.5  m1  v12  0.5  m2  v 22  0.5  m1  v1' 2 0.5  m2  v 2 ' 2
v1 and v2 = velocities before collision
v1 and v2 = velocities after collision
 Velocities are + or – to indicate directions.
Elastic Collisions Involving an Angle
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Momentum is conserved in both the xdirection and in the y-direction.
Before:
v1x  v1  cosθ1
positive
v1y  v1  sinθ1
negative
v 2 x  v 2  cosθ2
positive
v 2 y  v 2  sinθ2
positive
Elastic Collisions Involving an Angle

v1x '  v1' cos θ 3
positive
After: v1y '  v1 ' sin θ 3
positive
v 2 x '  v 2 ' cos θ 4
positive
v 2 y '  v 2 ' sin θ 4
negative
Elastic Collisions Involving an Angle
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Directions for the velocities before and
after the collision must include the
positive or negative sign.
The direction of the x-components for v1
and v2 do not change and therefore remain
positive.
The directions of the y-components for v1
and v2 do change and therefore one
velocity is positive and the other velocity is
negative.
Elastic Collisions Involving an Angle

px before = px after
m1  v1x  m2  v 2 x  m1  v1x 'm2  v 2 x '

py before = py after
m1  v1y  m2  v2 y  m1  v1y 'm2  v2 y '

Velocity after collision:
2
2
v1'  v1x '  v1y '
2
2
v 2 '  v 2x '  v 2 y '
Elastic Collisions
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Perfectly elastic collisions do not
have to be head-on.
Particles can divide or break
apart.
Example: nuclear decay
(nucleus of an element emits an
alpha particle and becomes a
different element with less mass)
Elastic Collisions


mn  v n  mn  mp  v n 'mp  v p

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mn = mass of nucleus
mp = mass of alpha particle
vn = velocity of nucleus before event
vn’ = velocity of nucleus after event
vp = velocity of particle after event
Recoil

Recoil is the term that describes the backward
movement of an object that has propelled
another object forward. In the nuclear decay
example, the vn’ would be the recoil velocity.
(mgun  mbullet )  v  mgun  v gun  mbullet  v bullet
Head-on and Glancing
Collisions
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Head-on collisions occur when all of the
motion, before and after the collision, is
along one straight line.
Glancing collisions involve an angle.
A vector diagram can be used to
represent the momentum for a glancing
collision.
Vector Diagrams

Use the three vectors and construct a
triangle.
Vector Diagrams
mB  v B
mR  v R '

sin 115 sin 30
mB  v B
mB  v B '

sin 115 sin 35
mR  v R ' mB  v B '

sin 30 sin 35

Use the
appropriate
expression to
determine the
unknown
variable.
Vector Diagrams
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Total vector momentum is conserved. You
could break each momentum vector into an x
and y component.
px before = px after
py before = py after
You would use the x and y components to
determine the resultant momentum for the
object in question
Resultant momentum = p 2  p 2
x
y
Vector Diagrams

Right triangle trigonometry can be used
to solve this type of problem:
Vector Diagrams

Pythagorean theorem:
ma  va 
2

 mb  v b   ma  mb   v T 
2
2
If the angle  for the direction in which the cars
go in after the collision is known, you can use
sin, cos, or tan to determine the unknown
quantity. Example: determine final velocity vT
if the angle is 25°.
ma  v a
sin 25 
ma  mb   v T
mb  v b
cos 25 
ma  mb   v T
Vector Diagrams

To determine the angle at which the cars go off
together after the impact:
1  ma
 va
θ  tan 
 mb  v b




Special Condition



When a moving ball strikes a stationary
ball of equal mass in a glancing collision,
the two balls move away from each other
at right angles.
ma = mb
va = 0 m/s
Special Condition

Use the three vectors to construct a
triangle.
Special Condition
m B  v B m A  v A '

sin 90
sin 50
mB  v B mB  v B '

sin 90
sin 40
m A  v A ' mB  v B '

sin 50
sin 40
Use the
appropriate
expression to
determine the
unknown
variable.
Ballistic Pendulum



In the ballistic pendulum lab, a ball of known
mass is shot into a pendulum arm. The arm
swings upward and stops when its kinetic
energy is exhausted.
From the measurement of the height of the
swing, one can determine the initial speed of
the ball.
This is an inelastic collision. As always, linear
momentum is conserved.
Ballistic Pendulum
Ballistic Pendulum

Potential energy of
ball in gun:

Ball embeds in
pendulum:
Ballistic Pendulum

Pendulum rises to
a maximum
height:

Solving for the initial
speed of the projectile
we get:
vb 
mb  mp
mb
 2g h
Series of Collisions



What happens when there are multiple
collisions to consider?
Consider an object that is securely bolted
down so it can’t move and is continuously
pelted with a steady stream of projectiles.
Each projectile has a mass of m and is
moving at velocity v along the x axis.
Series of Collisions



The linear momentum of each projectile is
m·v.
Suppose that n projectiles arrive in an interval
of Δt.
As each projectile hits (and is absorbed by)
the mass, the change in the projectile’s linear
momentum is Δp; thus the total change in
linear momentum during the interval Δt is
n·Δp.
Series of Collisions

The total impulse on the target is the same
magnitude but opposite in direction to the
change in linear momentum – thus:
J  n  p

But we also know that:

This leads us to:
Favg
Favg
J

t
J
n
n


 p 
 m  v
t t
t
Series of Collisions



So now we have the average force in terms of
the rate at which projectiles collide with the
target (n/Δt) and each projectile’s change in
velocity (Δv).
In the time interval Δt, we also know that an
amount of mass Δm = n·m collides with the
target.
Thus our equation for the average force
finally turns out to be:
Favg
m

 v
t
Series of Collisions

Assuming that each projectile stops upon
impact, we know that:
v  v f  vi  0  v  v

In this case the average force is:
Favg
m

v
t
Series of Collisions

Suppose instead that each projectile rebounds
(bounces directly back) upon impact – in this
case we have:
v  v f  vi   v   v  2  v

In this case the average force is:
Favg
m
 2 
v
t
Elastic Collision Example


Example: mass 1 and mass 2 collide and
bounce off of each other
Momentum equation:
m1  v1  m2  v 2  m1  v1'm2  v 2 '

Kinetic energy equation:
0.5  m1  v12  0.5  m2  v 22  0.5  m1  v1' 2 0.5  m2  v 2 ' 2

v1 and v2 = velocities before collision
v1 and v2 = velocities after collision
Velocities are + or – to indicate directions.
Elastic Collision Example

Working with kinetic energy:
0.5  m1  v12  0.5  m2  v 22  0.5  m1  v1'2 0.5  m2  v 2 '2

0.5 cancels out.
2
m1  v1
2
m1  v1

2
 m2  v 2  m1  v1'2 m2  v 2 '2
2
2
 m1  v1'  m2  v 2 ' m2  v 2


m1  v12  v1 '2  m2  v 2 '2 v 22
v 2 '2  v 2 2
m1
 2
m2 v1  v1 '2

2
Elastic Collision Example

The velocity terms are perfect squares and can
be factored:
a2-b2 = (a – b)·(a + b)

v 2 'v 2   v 2 ' v 2 
m1

m2 v1  v1'  v1  v1'

We will use this equation later.
Elastic Collision Example

Momentum equation:
m1  v1  m2  v 2  m1  v1'm2  v 2 '
m1  v1  m1  v1'  m2  v 2 'm2  v 2
m1  v1  v1'  m2  v 2 'v 2 
v 2 'v 2
m1

m2 v1  v1'
Elastic Collision Example


Both the kinetic energy and momentum
equations have been solved for the ratio of
m1/m2.
Set m1/m2 for kinetic energy equal to m1/m2
for momentum:
v 2 'v 2   v 2 'v 2   v 2 'v 2
v1  v1'  v1  v1' v1  v1'
Elastic Collision Example

Get all the v1 terms together and all the v2
terms together:
v 2 ' v 2   v 2 ' v 2   v1  v1 '  v1  v1 '
v 2 ' v 2

v1  v1 '
Cancel the like terms:
v 2 'v 2  v1  v1'
Elastic Collision Example

Rearrange to get the initial and final
velocities back together on the same side of
the equation:
v 2  v1  v1'v 2 '

This equation can be solved for one of the
two unknowns (v1´ or v2´), then substituted
back into the conservation of momentum
equation.
Change in Momentum Example

A 0.5 kg rubber ball is thrown towards a wall with a
velocity of 14 m/s. It hits the wall, causing it to
deflect in the opposite direction with a speed of
9
m/s. What is the change in the ball’s momentum?
∆p
∆p
∆p
∆p
∆p

=
=
=
=
=
F·∆t
m·vf – m·vi
(0.5 kg)·(-9 m/s) – (0.5 kg)·(14 m/s)
(-4.5 kg·m/s) – (7 kg·m/s)
-11.5 kg·m/s
The change in the ball’s momentum is 11.5 kg·m/s.