Impulse and Momentum and
Collisions and Stuff
We will now quantify exactly how it
hits the fan.
Momentum
• Momentum = product of mass X velocity
• Symbol for momentum is “p” (don’t ask me why)
– So, p = mv
• Momentum is also a vector and points in the
direction of velocity
• Units for mass = kg, units for velocity = m/s
• So, units for momentum = kg•m/s
– Sadly, there’s no other, more convenient unit for it
• Plural is ‘momenta’
A closer look at momentum
• Think of momentum as being how easy/hard
it is to get something to stop.
• Remember, p = mv
• So if mass or velocity is small, it’s likely that
the object has a small momentum, unless the
other of the two is very large
• What would be the momentum of an object at
rest?
5 Scenarios
• Let’s look at the momenta of….
• A car at rest
• Mass = 1000 kg, velocity = 0 m/s
• P = mv = (1000kg)(0m/s) = 0 kg m/s
• A bumblebee in flight (low mass, low velocity)
• Mass = 0.05 kg, velocity = 2 m/s
• P = mv = (0.05kg)(2m/s) = 0.1 kg m/s
• Projectile from BB gun (low mass, high velocity)
• Mass = 0.01 kg, velocity = 150 m/s
• P = (0.01 kg)(150 m/s) = 1.5 kg m/s
5 Scenarios continued
• A giant land tortoise (large mass, low velocity)
• Mass = 300 kg (I looked it up!), velocity = 0.5 m/s
• P = (300kg)(0.5m/s) = 150 kg m/s
• Tony Stewart at the Brickyard (large mass, high
velocity)
– Mass = 1000 kg, velocity = 90 m/s
– P = (1000kg)(90 m/s) = 90,000 kg m/s
Understanding momentum
• Which is easier to stop…
• A slow moving baseball or a fast moving
baseball?
• A chihuahua racing down the hall towards you
or Mr. Barbini racing down the hall towards
you?
• A speeding Volkswagen beetle or a speeding
dump truck?
I need a volunteer
• Rank these momenta from lowest to highest
– A fast flying bee
– The earth in orbit around the sun
– A parked garbage truck
– A slowly flying bee
– Your grandmother driving down the street in her
1959 Edsel
– An oil tanker sailing the seven seas.
– Tony Hawk grinding on a rail
Impulse
• Impulse = Product of Force and how long that
force acts.
– Impulse = F∆t
• Impulse is a vector. It acts in the same
direction as the force
• Units of force = Newtons, units of time =
seconds. So, units for impulse =
Newton•Seconds
Okay, but what does that really mean?
• Let’s say you have a force of 10 N. You apply
this force to a hockey puck on (frictionless) ice
for 2 seconds.
• Impulse = (10N)(2 sec) = 20 Newton seconds
• Now, let’s say you apply this same force for 4
seconds
• So the impulse now = 40 Newton seconds
Impulse-momentum theorem
• Remember the work-energy theorem? This is
kind of like that.
• Impulse = change in momentum
• In general, the mass of the object(s) in
questions stays constant
• ∆p = pF – pI = mvF – mvI
• So F∆t = mvF – mvI
Well, Duh…
• Let’s think about 2 situations:
• Situation 1: pushing someone in a rolling chair
– If you push on them with a certain force for two
different times, the longer time will result in a
greater change in momentum
– So if the chair was at rest to begin with, it will end
up going faster at the end for the longer time the
force was applied
Less Duh
• Situation 2: crashing your 1,000 kg car
• Let’s say you are driving along at 45 m/s
(about 96 mph) and you crash into one of two
objects: a solid wall or a series of water-filled
plastic bins (like they have on the highway)
• If your car goes from 45 m/s to 0 m/s, you
have a change in momentum of:
• P = mv, so momentum = (1000kg)(45m/s) =
45,000 kg•m/s
The car crash continued
• Recall that impulse = change in momentum,
so Ft = mv = 45,000 kg•m/s
• So, Force X time = 45,000 kg•m/s
• Now, would you rather have that force spread
out over a long time, or over a short time?
Compare stopping times
• Change in momentum is 45,000 kg•m/s
• For a stopping time of 5 sec (slamming on your
brakes): (F)(5 sec) = 45,000 kg•m/s
– So F = 9,000 N
• For a stopping time of 1.5 sec (smashing into the
plastic water bins): (F)(1.5 sec) = 45,000 kg•m/s
– So F = 30,000 N
• For a stopping time of 0.2 sec (smashing into the
concrete wall): (F)(0.2 sec) = 45,000 kg•m/s
– So F = 225,000 N
Let’s look at those numbers more
closely…
• Remember we have a 1,000 kg car and F = ma
• If F = 9000N, then accel = 9 m/s (just under
acceleration due to gravity)
• If F = 30,000N, then accel = 30 m/s (just over
3X acceleration due to gravity)
• If F = 225,000N, then accel = 225 m/s (about
23X the acceleration due to gravity)
• So, which is the most survivable?
Let’s look at some examples
• Hitting a baseball (mass = 0.14 kg)
• Let’s say that a baseball is travelling to the left
(negative velocity) at -38m/s. After being hit, it
moves to the right at 58 m/s. What is the impulse
applied to the ball? If the contact time was
1.6X10-3 sec, what was the force?
• Impulse = ∆p = pF – pI = mvF – mvI
• Impulse = (0.14kg)(58m/s) – (0.14kg)(-38m/s)
• Impulse = 13.4 kg m/s
• Impulse = F∆t, so F = impulse/∆t
• Force = (13.4 kg m/s)/(1.6X10-3 sec) = 8400 N
Conservation of Momentum
• In general, momentum is conserved
• This means that the momentum at the
beginning is the same as the momentum at
the end
• I.e. momentum is not created or destroyed
• Remember, symbol for momentum is ‘p’
• So PI = ‘initial momentum’ and PF = ‘final
momentum’
• Let’s look at some examples
Conservation of Momentum: Example 1
• Cannon and cannonball
• What is momentum
before cannon is fired (PI)?
• Zero
• So what does final
momentum (PF) have to
be if momentum is
conserved?
• Zero
http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section3.rhtml
Conservation of Momentum: Example 1
• Let’s say that the cannon
has a mass of 1000 kg
and Cannonball has a
mass of 10 kg
• So total PI = PI of ball + PI
of cannon
• PI of ball = (10kg)(0m/s)
= 0 kg m/s
• PI of cannon =
(1000kg)(0m/s) = 0 kg
m/s
Conservation of Momentum: Example 1
• Now let’s say that the
cannonball moves to the
right at 75 m/s. How fast
does the cannon move to
the left to conserve
momentum?
• PI = PF
• So PF must equal zero
Conservation of Momentum: Example 1
• PF must equal zero
• Total PF = PF of ball + PF
of cannon
• Remember, moving to
the right = positive
velocity
• Moving to the left =
negative velocity
Conservation of Momentum: Example 1
• PF = 0, so
• (10kg)(75 m/s) +
(1000kg)(Vcannon) = 0
• So -Vcannon =
(10kg)(75m/s) / (1000kg)
• Vcannon = -0.75 m/s
• If ball moves to the right,
cannon moves to the
left, so Vcannon should be
negative, which it is
Conservation of Momentum: Example 2
• Newton’s cradle (or
“Executive ball clicker”
or, more crassly,
“Newton’s Balls”)
• How does it work?
• If one ball is used, how
many come up on the
other side?
• If two are used…?
Now, Dr. Mason, remind us again when
conservation of momentum holds…
• Momentum is conserved when there are NO
EXTERNAL FORCES ACTING ON THE SYSTEM.
• If any net external force acts, momentum will
NOT be conserved.
• Let’s look at a contrived example involving
pool balls.
Billiards
• Imagine the scenario where one pool ball is
stationary on a pool table. A second pool ball
smacks into it.
– The ‘system’ is the two pool balls.
– Are there any net external forces?
• Now, imagine that just before they hit, a hole
opens up underneath them.
– The ‘system’ is the two pool balls.
– Now are there any net external forces?
• Okay, this time, no sudden trapdoor. Same two
pool balls collide.
– The ‘system’ this time is just one pool ball.
– Now are there any net external forces?
Freight trains example
• A freight train is being assembled out of two
boxcars.
• Car 1 has a mass of 65,000 kg and is moving to
the right at V01 = 0.8m/s.
• Car 2 has a mass of 92,000 kg and is also moving
to the right at V02 = 1.3 m/s.
• Car 2 collides with car 1 and couples with it. What
is the speed of the train after the coupling?
• Momentum is conserved.
• (m1 + m2)Vf = m1V01 + m2V02
• Vf = 1.1 m/s
Ballistic Pendulum
• A ballistic pendulum is a device that is used to
determine the muzzle velocity of a gun.
http://session.masteringphysics.com/problemAsset/1010989/26/1010989A.jpg
Collisions in 2-D
• Conservation of momentum holds in 2
dimensions as well
• As one may expect, you can treat the X- and Ycomponents independently
– So, Pf = Pi
– And PfX = PiX
– And PfY = PiY