### Momentum & Impulse

```Momentum & Impulse
Momentum (p)
•
•
•
•
“inertia of motion”
p = mv
Units for momentum  Kg*m/s
Vector Quantity
• One way of looking at it…How much an object in
motion… wants to stay in motion
• Lot of momentum  hard to stop
How can you change an object’s
momentum??
• Newton’s 2nd Law states a net force causes an
acc.
• An acc. Changes the velocity
• Changing the velocity, changes the momentum
Impulse Momentum Theorem
• Applying a force over a time interval changes the
momentum
▫ F changes v, therefore (mv) changes
• Never looked at a relationship between ‘F’ and ‘t’
• F x t = Impulse
▫ Since an impulse changes ‘v’, this changes
momentum
• Ft = Δ(mv)  Impulse-Momentum Theorem
Newton’s 2nd Law reworked…
• F = ma
and a = (Δv/t)
• F= m(Δv/t) then multiply both sides by ‘t’
• Ft = mΔv which is the same thing as
• Ft = Δ mv
• Impulse- Momentum Theorem is just Newton’s
2nd Law written a different way
Examples
• Boxing gloves vs. MMA gloves
• http://www.yourdiscovery.com/vi
deo/future-cars-nido/
• Features on a car??
• Pillow punch vs. brick punch
• Bungee jump w/ elastic cord vs.
rigid cord
• Egg toss competition
• “rolling with a punch”
Bouncing?
• Greater Δ(mv) than just stopping an
object??
• Why??  ….greater Δv
▫ Going from -5 m/s to 5 m/s is a greater
velocity change than going from -5 m/s to 0
m/s, therefore greater Δmv and impulse
Pelton Wheel Example
• Paddles are cups instead of just flat planks
• Allows water to change directions
• Greater Δmv of water which means more impulse
and wheel is turned much more effectively
Example Problem
Other Examples
• Karate chop
eature=related
w
• Mr. Schober gets assaulted by strangers… Story
Conservation of Momentum
• If no outside force is applied, then the total
amount of momentum in a closed system will
remain constant.
▫ Only external forces can change momentum.
• Σpi= Σpf
• m1v1i +m2v2i …= m1v1f + m2v2f…
Conservation of Momentum
pai =
m(v)
pbi = m(0)
paf = m(0) pbf = m(v)
Conservation of Momentum
• Momentum is
conserved for all
objects in the
interaction, even if
one doesn't stop
pai + pbi = paf + pbf
Is momentum conserved here?
Yes, due to the vector nature of momentum.
Is momentum conserved?
A
B
• Initial
velocities of
both objects is
0.
• pai = ma(0)
• pbi = mb(0)
• Σpi = 0
Is momentum conserved?
• paf = ma(-va)
• pbf = mb(vb)
• pf = 0
• Σpi = Σpf, so
momentum is
conserved!!
A
B
pf = ma(-va) + mb(vb)
Why do internal forces result in
momentum being conserved?
• When Girl A pushes on Girl B, according to
Newton’s 3rd Law, Girl B pushes on Girl A
▫ How much?
• These forces are equal in magnitude and
opposite in direction
• The time over which these forces act is exactly
the same
▫ Only while the girls are in contact, in this case
How does a gun work?
How does the gun work?
• Only forces are internal (no net external
forces are adding impulse to the system)
• The momentum of both will add up
to zero (bullet is +, gun is -)
Why do internal forces result in
momentum being conserved?
• Impulse is equal in magnitude but opposite in
direction
▫ I = (ΣF)(Δt)
▫ Forces are equal and opposite, times are equal
• Δp is equal in magnitude, opposite in direction,
resulting in Σp = 0!!
Collisions
• Inelastic
▫ Any collision in which momentum is conserved but kinetic energy is not
▫ Most ‘real’ collisions are of this kind
▫ KE is not conserved because some is lost to the deformation
▫ m1v1i+ m2v2i= m1v1f + m2v2f
• Perfectly Inelastic
▫ Objects collide and stick together
▫ KE not conserved
▫ m1v1i + m2v2i = (m1 + m2) vf
• Elastic
http://www.flixxy.com/golfball-slow-motion.htm Golf Ball during a surprising
inelastic collision
h?v=pQ9NiazPYI8 --baseball
▫ Both momentum and KE are conserved
▫ “perfectly “elastic collisions only occur in real life at the subatomic level, but will treat any
collision labeled as “elastic” as being ‘perfectly’ elastic
▫ Collisions between billiard balls or between air molecules and the surface of a container are
both highly elastic
▫ No Energy lost to deformation
▫ m1v1i+ m2v2i= m1v1f + m2v2f
And
▫ ½m1v1i2 + ½m2v2i2 = ½m1v1f2 + ½m2v2f2
▫
When combining these two and reducing we get….
V1i – v2i =-(v1f – v2f)
Example problem
Problem Solving #1
• A 6 kg fish swimming at 1 m/sec swallows a 2 kg fish
that is at rest. Find the velocity of the fish
immediately after “lunch”.
• System is both fish, and collision is perfectly
inelastic so …..
• Σpi = Σpf
• (m1v1i) + (m2v2i) = (m 1+ m2)vf
• 6(1) + (2)(0) = (6+2) vf
• Vf =6/8 = .75 m/s
Problem Solving #2
• Now the 6 kg fish swimming at 1 m/sec swallows a 2
kg fish that is swimming towards it at 2 m/sec. Find
the velocity of the fish immediately after “lunch”.
• System is both fish, so….
• Σpi = Σpf
•
(Σ(mv))i = (Σ(mv))f
• (m1v1i) + (m2v2i) = (m 1+ m2)vf
• (6 kg)(-1 m/s) + (2 kg)(2 m/s) = (6 kg + 2 kg)(vf)
• -6 kg.m/s + 4 kg.m/s = (8 kg)(vf)
•
vf = -2 kg.m/s / 8 kg
•
vf = -.25 m/s
•
Collisions in 2-D (more to be posted
later)
• Σpxi = Σpxf
• Σpyi = Σpyf
Momentum is a vector, so momentum must be
conserved in the x-direction, and in the ydirection
Inelastic in 2-D
??
??
1 kg
2.2 m/s
.5 kg
33°
1.5 m/s
Perfectly Inelastic in 2-D
1 kg
1.5 kg
2.5 m/s
1.3 m/s
.5 kg
```