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Chapter 10
Collisions
Review
Momentum:


p  mv
If Fext = 0, then momentum does not change


pi  p f
For continuous momentum transfer (Rockets):
 Mi
v f =vi+vex ln 
M
 f
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



2
Rockets:
Continuous Momentum Transfer
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3
Momentum in a Collision
In a collision, objects only
exert forces on each other, so
Fext=0.
Total momentum is conserved
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Impulse
During a collision, the momentum on an object
changes
This change in momentum is called “Impulse”

 

J  p  p f  pi
When objects A and B collide


J A  JB
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Impulse
Recall:
 p
F
t

 
J  p  Ft
(constant force)
In the limit of small t:

pf 
tf 
J   dp   F dt
pi
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ti
(changing force)
6
Impulse in a Collision
 p
F
t
Blue
p/t Large
Red
p/t Small
Large F:
p changes rapidly
Small F:
p changes slowly
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Momentum
Different collisions with the
same total impulse:
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Example: The Impulsive Spiderman
Spiderman, who has a mass of 70 kg, jumps
from a train 5 meters high moving at 20 m/s
(about 40 mph).
He lands standing up, taking t = 0.1 s to stop
himself after making contact with the ground.
How much force did his knees feel?
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Example
Treat as collision between
Spiderman and the ground
Get force from the impulse:


J  p  Ft
Initial: p = mvtotal
Final: p = 0
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
F 

p
mvtotal

t
t
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Example

 p mvtotal
F 

t
t
Need to find vy:
U initial  K final
1
2
vtotal  vx  v y
2
2
mvy  mgh
 (20 m / s) 2  2(9.8 m / s 2 )(5 m)
v y  2 gh
 22.3 m / s
2
2
 (70 kg)( 22.3 m / s)
F 
 16,000 N
0.1 s
If he wasn’t a superhero, he’d break his legs!
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Example
What if he rolls on landing for
t = 2 sec?
 (70 kg)( 22.3 m / s )
F 
 780 N
2s
Much easier on the knees!
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Cannon Recoil
Cannon: mc=1134 kg
Ball: mb=13.6 kg
Ball shot at ~ speed of sound
 vb = 340 m/s
The cannon and ball are initially at rest:
pi  p f  0
pball  pcannon  0
pcannon   pball
pball = mballvball = (13.6kg)(340 m/s) = 4620 kg m/s
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So, pcannon= -4620 kg m/s
12
Cannon Recoil
Cannon recoil stopped in
~2 s by ropes. What is the
tension in the ropes?
pc
T
T
p
Fnet  2T 
t
1 p 1 4620 kg m / s
T

 1160 N ~ 260 lbs
2 t 2
2s
A rope can easily handle this much force without
breaking
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Momentum Conservation in
Different Frames
Simple 1D problem
m
v
-v
m
PTOT = mv - mv =0
Stick
together
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2m
v=0
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Momentum Conservation in
Different Frames
Same 1D problem viewed from right hand block, or
with right hand block at rest
m
2v
m
PTOT = 2mv + 0 = 2mv
2m
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v
15
Changes in Momentum
Independent of Frame
Case 1
Case 2
i
f
i
f
Left
mv
0
2mv
mv
Right
-mv
0
0
mv
PTf – PTi = 0 – 0 = 0
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PTf – PTi = 2mv – 2mv = 0
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Center of Momentum Frame
There is always a frame of reference where
PTOT=0.
‘Center of mass’ frame
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A Limitation of Momentum
V=30 MPH
V=0
Before
vT
vc
BOOM!
After
pcar,i  (1500 lbs )(0 MPH))
 (681 kg)(0 m/s ) = 0
ptruck,i  (3500 lbs )(30 MPH )
 (1,589 kg)(13.4 m/s )
 21,292 kg  m/s
ptotal,i  pcar,i  ptruck,i
 21,292 kg  m/s
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ptotal,f  pcar,f  ptruck,f
How do we determine
the velocities?
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A Limitation of Momentum
ptotal,f  pcar,f  ptruck,f  Constant
pcar,f  ptotal,f  ptruck,f
pcar
There are many
possibilities
Conservation of
Momentum can’t
tell them apart
ptruck
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Elastic Collisions
Momentum and kinetic energy are conserved
Two equations:




m1v1,i  m2 v2,i  m1v1, f  m2 v2, f
2 1 2 1 2 1 2
1
2 m1v1,i  2 m2 v2 ,i  2 m1v1, f  2 m2 v2 , f
Good approximation for a lot of collisions, and
exact for some
Examples: Billiard Balls, superball on floor…
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Elastic Collisions in One Dimension
Before
After
m1
m2
m1
V2,i
V1,i
m2
V1,f
V2,f
Two conservation laws
Momentum
m1v1,i  m2 v2,i  m1v1, f  m2 v2, f
(Always)
Energy
1
2
m1v12,i  12 m2v22,i  12 m1v12, f  12 m2v22, f
(Elastic only - Mechanical Energy is conserved)
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A Unique Solution
We now have two equations and two unknowns:
m1v1,i  m2 v2,i  m1v1, f  m2 v2, f
1
2
m1v12,i  12 m2v22,i  12 m1v12, f  12 m2v22, f
Lots of Algebra
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v1,f
m1  m2
2m2

v1,i 
v2 ,i
m1  m2
m1  m2
v2 ,f
2m1
m2  m1

v1,i 
v2 ,i
m1  m2
m1  m2
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Limiting Cases
v1,f
m1  m2
2m2

v1,i 
v2 ,i
m1  m2
m1  m2
v2 ,f
2m1
m2  m1

v1,i 
v2 ,i
m1  m2
m1  m2
Consider limiting case:
m1 = m2
How do we understand
what types of motion
these predict?
v1,f  v2 ,i
v2 ,f  v1,i
The two objects simply trade values of velocity!
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Limiting Cases
What if m1 >> m2?
v1,f  v1,i
v1,f
m1  m2
2m2

v1,i 
v2 ,i
m1  m2
m1  m2
v2 ,f
2m1
m2  m1

v1,i 
v2 ,i
m1  m2
m1  m2
v2 ,f  2v1,i  v2 ,i
Semi truck hits a parked VW bug: Truck keeps going
Bug bounces off with twice truck’s speed!
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Demonstration
m1>>m2
A Question:
v1,f  v1,i
v2 ,f  2v1,i  v2 ,i
Before:
vbasket,i  v0
vtennis,i  v0
After:
vbasket,f  v0
vtennis,f  3v0
What
Happens?
The Slingshot Effect
9.6 km/s
v probe,f  2v jupiter  v probe,i
 2(9.6 km/s)  (10 km/s)
 29.2 km/s
-10 km/s
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Car-Truck Crash
A 2000 kg car has a head-on collision with a
10,000 kg truck. They each are travelling at 10 m/s
and they collide elastically (solid bumpers!).
What are their final velocities?
m1
v1i
Choose positive x direction
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v2i
m2
+x
27
Car-Truck Crash (continued)
m1
v1i
v2i
v1i = 10 m/s
m1 = 2,000 kg
v2i = -10 m/s
m2 = 10,000 kg
v2f = -3.33 m/s
Truck slows down
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m2
v1,f
m1  m2
2m2

v1,i 
v2 ,i
m1  m2
m1  m2
v2 ,f
2m1
m2  m1

v1,i 
v2 ,i
m1  m2
m1  m2
v1f = -23.3 m/s
Car goes flying backwards!
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Car-Truck Crash (continued)
If the two vehicles are being driven by 60 kg PSU
students, what are the impulses they feel?
In truck: J = p = mv
= m(v2f - v2i)
= 60(-3.33 – (-10))
= 400 kg m/s
In car:
J = p = mv
= m(v1f – v1i)
= 60(-23.3 – (10))
= -2000 kg m/s
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Car-Truck (question)
Which would you rather be driving?
Say collision lasts Δt = 0.2 seconds
Force on student is given by F = Δp/Δt
Student in truck feels 2,000 N (survivable)
Student in car feels 10,000 N (not good)
What if instead of a 2000 kg car, she was on a
500 kg motorcycle!
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Example: 2-D Elastic Collision
v1,i=(1 m/s)i+(2 m/s)j
Two billiard balls collide elastically on a table. The
initial velocity of the first ball is v1,i=(1 m/s)i+(2 m/s)j.
The second ball is initially at rest. Both balls have
the same mass. Determine the final velocity of both
after the collision.
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Inelastic Collisions
Momentum is conserved (NOT Kinetic Energy)
Completely Inelastic:
Two objects stick together



m1v1,i  m2 v2,i  (m1  m2 )v f
Examples: Spit wads, football player being
tackled,…
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Inelastic Collisions…
http://www.baylortv.com/streaming/000026/300kbps_ref.mov
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Car Crash
m1=750 kg
v1=20 m/s
m2=1000 kg
v2=30 m/s
Two cars collide and stick together after the
collision. What is the final velocity of the
system?
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Car Crash
m1=750 kg
m2=1000 kg
v1=20 m/s
v2=30 m/s
Using conservation of momentum:
pi  p f
m1v1,i  m2 v2,i  (m1  m2 )v f
vf 
m1v1,i  m2 v2,i
m1  m2
(750 kg)( 20 m / s )  (1000 kg)( 30 m / s )

(750 kg)  (1000 kg)
v f  8.6 m / s
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Basketball Cannon
A ball projected from a cannon hits the trash can such that:
1) It sticks into the trash can.
2) It hits the trash can and bounces back.
Will the velocity of the trash can be bigger
for case 1, case 2, or exactly the same?
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Basketball Cannon
M
m
v
vtrash=0
Consider an elastic collision:
vtrash,f
M m
2m

vtrash,i 
v
M m
M m
vtrash,f
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2mv

M m
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Basketball Cannon
M
m
v
vtrash=0
Consider a perfectly inelastic collision:
mv  ( M  m)v f
mv
vf 
M m
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Basketball Cannon
Elastic:
Inelastic:
2mv
vf 
M m
mv
vf 
M m
Elastic collision results in twice the velocity!
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