Linear Momentum
Physics
Montwood High School
R. Casao
Linear Momentum
• Linear momentum of an object is the
mass of the object multiplied by its
velocity.
• Momentum: p = m·v
• Unit: kg·m/s or N·s
• 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
• 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.
Linear Momentum
• 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
• 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
• 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.
• The area under the curve in a force vs.
time graph represents the change in
momentum (m·v).
Impulse
• 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.
• 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
• 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
Perfectly Inelastic Collisions
• 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
• 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
• 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
• Kinetic energy is lost when the objects
are deformed during the collision.
• Momentum is conserved.
Elastic Collisions
• 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
• 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
• 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
• After:
v1x '  v1 ' cos θ 3
positive
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
• 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
• 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
•
•
•
•
•
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
• 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
m R  v R ' • Use the

appropriate
sin 115 sin 30
expression
mB  v B
mB  v B '

sin 115 sin 35
mR  v R ' mB  v B '

sin 30 sin 35
to determine
the
unknown
variable.
Vector Diagrams
• 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 =
2
2
px  py
Vector Diagrams
• Right triangle trigonometry can be used to
solve this type of problem:
Vector Diagrams
• Pythagorean theorem:
ma  v a 
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
mB  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.
Rocket Propulsion (Jet Propulsion)
• As fuel burns and exhaust gases leave the
rocket or jet engine, momentum is carried
with them.
• To conserve momentum, the rocket or jet
must gain the same amount of momentum
in the opposite direction.
Rocket Propulsion (Jet Propulsion)
• Thrust = the magnitude (size) of the force
exerted by an engine or rocket.
Fthrust
Δm  v

Δt
• Mass flow rate for fuel = Δm
Δt
• The direction of the thrust is opposite to the
direction of the exhaust gases coming from
the rocket or jet engine.
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
Helpful Websites
• Physics Classroom – Momentum and
Collisions
• ExploreScience – Two Dimensional Collisions
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.
m1  v12  m2  v 22  m1  v1'2 m2  v 2 '2
m1  v12  m1  v1 '2  m2  v 2 '2 m2  v 22
m1 

2
v1
 v1'
2
  m  v ' v 
v 2 '2  v 2 2
m1
 2
m2 v1  v1'2
2
2
2
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, then substituted back into
the conservation of momentum equation.