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
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The terms momentum and linear momentum will be
used interchangeably in the text
Linear momentum is a vector quantity
 Its direction is the same as the direction of the
velocity
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 vx
py = m vy
pz = m vz
Newton and Momentum
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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
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
Conservation of Linear Momentum

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
 Conservation of momentum can be expressed mathematically in
various ways
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ptotal = p1 + p2 = constant
p1i + p2i = p1f + p2f
In component form, the total momenta in each direction are
independently conserved
 pix = pfx
piy = pfy piz = pfz
Conservation of momentum can be applied to systems with any
number of particles
This law is the mathematical representation of the momentum
version of the isolated system model
Conservation of momentum is required by Newton’s 2nd and 3rd laws:
Newton’s 2nd law shows the net force on a particle equals the rate of change of momentum:
dv d(mv) dp


(1)
dt
dt
dt
Consider the gravitational attraction between 2 objects:
F  ma  m
F21 
m1

But F12  F21
(2)
Gm1m 2
r2
i.e.
F12 
Gm1m2
r2
m2
r

dp1 dp2 d( p1  p2 )
F12  F21  0 


dt
dt  dt
(3)
That is, considering the case of a gravitational interactions, the rate of change of total
momentum is zero, i.e. total momentum is conserved.
More generally, Newton’s 3rd law requires all forces to come in pairs – action and reaction
are equal and opposite – i.e. for any 2 interacting bodies, equation (2) applies, so equation
(3) applies generally, not just for gravity, so conservation of linear momentum is thus
required by Newton’s 3rd law.
Archer Example
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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
mAvAi + mavai = mAvAf + mavaf
0+ 0 = mAvAf + mavaf  mAvAf = -mavaf
So if mA >> ma then vAf << vaf

The final velocity of the archer (A) is
negative
 Indicates he moves in a direction
opposite the arrow (a)
 Archer has much higher mass than
arrow, so velocity is much lower
Impulse and Momentum
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From Newton’s Second Law, F  dp

F =ma = mdv/dt = dp/dt
dt
Solving for dp gives dp 
Fdt
Integrating to find the change in momentum over some time interval
tf
Dp  pf  pi   Fdt  I
ti
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The integral is called the impulse, I , of the force acting on an object
over Dt
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This equation expresses the impulse-momentum theorem: The
impulse of the force acting on a particle equals the change in the
momentum of the particle
 This is equivalent to Newton’s Second Law
Dp  I
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Impulse is a vector quantity
The magnitude of the
impulse is equal to the area
under the force-time curve
tf
Dp  pf  pi   Fdt  I
ti
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The force may vary with
time
Dimensions of impulse are
ML/T
i.e. kg ms-1, same as momentum
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Impulse is not a property of
the particle, but a measure
of the change in momentum
of the particle
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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
In many cases, one force acting on a
particle acts for a short time, but is much
greater than any other force present
dp = Fdt  Dp = FDt  mDv = FDt
When using the Impulse Approximation, we
will assume this is true
Especially useful in analyzing
collisions
The force will be called the impulsive force
The particle is assumed to move very little
during the collision
pi and pf represent the momenta
immediately before and after the collision
Crash Test Example
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Assume force exerted by wall is large compared with other forces
Gravitational and normal forces are perpendicular and so do not
effect the horizontal momentum
Can apply impulse approximation
If wall doesn’t move
Initial momentum, mcvci + mwvwi = mcvci = -15.0mc kg ms-1
Final, mcvcf + mwvwf = mcvcf = 2.60mc kg ms-1
Inelastic
|Final momentum| < |Initial momentum| collision
 Impulse imparted to wall
– kinetic
energy
Impulse = Dp = pf – pi = 2.60 –(-15.0) = 17.6 kg ms-1
lost
Average force, FDt = mDv  F = 17.6m/Dt
If car “weighs” 1500 kg and collision lasts for 0.15 s
 F = 17.6x1500/0.15 = 176,000 N
Note, if collision lasted longer, e.g. 1 hour  F = 17.6x1500/3600 =
7.3 N, which is a lot more gentle. This is why landing on a mattress
isn’t as sore as landing on the road and the idea behind crumple zones
in cars (also why it’s nicer to be slapped with a sponge than a brick).
Collisions – Characteristics
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We use the term collision to represent an event
during which two particles come close to each other
and interact by means of forces
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May involve physical contact, but must be generalized to
include cases with interaction without physical contact
The time interval during which the velocity changes
from its initial to final values is assumed to be short
The interaction forces are assumed to be much
greater than any external forces present
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This means the impulse approximation can be used
Collisions – Example 1
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Collisions may be the
result of direct contact
The impulsive forces
may vary in time in
complicated ways
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This force is internal to
the system
Observe the variations in
the active figure
Momentum is
conserved
Collisions – Example 2
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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, kinetic energy is conserved (momentum
is always conserved)
 Perfectly elastic collisions occur on a microscopic level
 In macroscopic collisions, only approximately elastic
collisions actually occur
 Generally some energy is lost to deformation, sound, etc.
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
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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
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Since the objects stick
together, they share the
same velocity after the
collision
m1v1i  m2 v2i   m1  m2  vf
Elastic Collisions
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Both momentum and
kinetic energy are
conserved
m1v1i  m2 v 2 i 
m1v1f  m2 v 2f
1
1
2
m1v1i  m2 v 22 i 
2
2
1
1
2
m1v1f  m2 v 22f
2
2
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
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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
Ballistic Pendulum
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Conceptualize
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Observe diagram
Categorize
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Isolated system of projectile and
block
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
Some energy was transferred
during the perfectly inelastic
collision
Conservation of momentum
m1v1A + 0 = (m1+m2)vB
vB = m1v1A/(m1+m2)
Conservation of energy KEi + PEi = KEf + PEf
Just after the collision KE = ½ (m1+m2)vB2 , all of which goes into
lifting the block to height h
i.e. ½ (m1+m2)vB2 = (m1+m2)gh
In ballistic tests want to measure speed of bullet from known
masses and measuring height travelled by block -> get rid of vB
½ (m1+m2) [m1v1A/(m1+m2)]2 = (m1+m2)gh
m12v1A2 = 2(m1+m2)2gh  v1A = (2gh)½(m1+m2)/m1
Two-Dimensional Collisions
The momentum is conserved in all directions
Use subscripts for identifying the object indicating
initial or final values of the velocity components
If the collision is elastic, use conservation of kinetic
energy as a second equation
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Particle 1 is moving at
velocity v1i and particle 2 is
at rest
In the x-direction, the initial
momentum is m1v1i
In the y-direction, the initial
momentum is 0
For example
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After the collision, the
momentum in the xdirection is m1v1f cos q 
m2v2f cos f
After the collision, the
momentum in the ydirection is m1v1f sin q 
m2v2f sin f
If the collision is elastic,
apply the kinetic energy
equation
This is an example of a
glancing collision
Two-Dimensional Collisions - tips
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If the collision is inelastic, kinetic energy of the system is
not conserved, and additional information is probably
needed
If the collision is perfectly inelastic, the final velocities of the
two objects are equal. Solve the momentum equations for
the unknowns.
If the collision is elastic, the kinetic energy of the system is
conserved
 Equate the total kinetic energy before the collision to the
total kinetic energy after the collision to obtain more
information on the relationship between the velocities
Check to see if your answers are consistent with the
mental and pictorial representations
Check to be sure your results are realistic
Example
A 1500 kg car travelling east with a speed of 25 m/s collides at an
intersection with a 2500 kg van travelling north at a speed of 20
m/s. Find the direction and magnitude of the wreckage after the
collision, assuming the vehicles undergo a perfectly inelastic
collision (i.e. they stick together).
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Ignore friction
Model the cars as particles
The collision is perfectly
inelastic
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The cars stick together
FIRST! Draw a diagram and write
down all the information given
Examine momentum in each component
x: 25mc + 0mv = (mc+mv)vfcosq
Van has no velocity
component in x-direction
Likewise for y: 0mc + 20mv = (mc+mv)vfsinq
25(1500) = (1500+2500)vfcosq & 20(2500) = (1500+2500)vfsinq
37500 = 4000vfcosq & 50000 = 4000vfsinq
2 unknowns, but 2 equations – can solve
Divide Eq. 2 by Eq1. sinq/cosq = tan q = 50000/37500
q = 53.1o and plugging this into either Eq. 1 or 2 gives vf = 15.6 m/s
Could also be 233.1o, but wouldn’t make sense