Wed.
Lab
Fri.
10.1-.4 Introducing Collisions Quiz 9
L9 Multi-particle Systems
10.6-.8 Scattering
RE 10.a
RE 10.b
Mon.,
Lab
Wed.,
Fri.,
10.9-.10 Collision Complications
L10 Collisions 1
10.5, .11 Different Reference Frames
1.1 Translational Angular Momentum Quiz 10
RE 10.c
EP9
RE 11.a; HW10: Pr’s 13*, 21, 30, “39”
Collisions! Short, Sharp Shocks
Collision: Operational definition: quick enough/strong enough interaction that all
others are negligible
Short, Sharp Example
System = ball
Can ignore Earth’s pull during “smack”.
Say an 0.5 kg ball is in contact is in contact with the
bat for 5 ms. During that time, it switches from going
forward at 100mph = 44.7 m/s to going 80 mph
backward = 35.8 m/s.
EF
Smack!
BF
EF
N050,8t
pFnet
s
smkgsmkg
005.0
/7.445.0/8.355.0
Of that, the Earth’s pull accounts for only
NsmkgmgF Eb 9.4/8.95.0 2
1-D Collision
A B
Pow! A B
Example
t
System = carts A & B
initially
finally
frictionAF
frictionBF
frictionAF
frictionAF
Often know initial motion, want to predict final motion
Two unknowns: and fAv .
fBv .
Generally need two equations to solve for them
0.. tFpFdt
pdextnetsystemextnet
system
A B
iAp .
iBp .
Pow! A B
fAp .
fBp .
Example
t
System = carts A & B
initially
finally
frictionAF
frictionBF
frictionAF
frictionAF
0
...)(
.
BA
frictionBfrictionABA
extnetsystemsystem
pp
tFFpp
tFp
If we look right before and right after collision
One of the Equations: Momentum Principle
1-D Collision
A B
iAp .
iBp .
Pow! A B
fAp .
fBp .
t
System = carts A & B
initially
finally
frictionAF
frictionBF
frictionAF
frictionAF
0 BA pp
0.... iBBfBBiAAfAA vmvmvmvm
csv '
One of the Equations: Momentum Principle
1-D Collision
A space satellite of mass 500 kg
has velocity < 12, 0, –8 > m/s just
before being struck by a rock of
mass 3 kg with velocity
< –3000, 0, 900 > m/s.
After the collision the rock’s
velocity is < 700, 0, –300 > m/s.
Now what is the velocity of the
space satellite?
a. < –5100, 0, –400 > m/s
b. < –10.2, 0, –0.8 > m/s
c. < 10.2, 0, 0.8 > m/s
d. < –3688, 0, 1191 > m/s
e. < 3688, 0, –1192 > m/s
0.... iBBfBBiAAfAA vmvmvmvm
A B
iAp .
iBp .
Pow! A B
fAp .
fBp .
t
System = carts A & B
initially
finally
frictionAF
frictionBF
frictionAF
frictionAF
0 BA pp
0.... iBBfBBiAAfAA vmvmvmvm
csv '
One of the Equations: Momentum Principle
1-D Collision
Special Case: “Maximally Inelastic” – hit & stick
Other Equations: for “Maximally Inelastic”
fBfAf vvv
A squishy clay ball collides in
midair with a baseball, and
sticks to the baseball, which
keeps going.
Initial momenta:
and
Final momentum of clay+ball :
Which equation correctly describes
this collision?
1)
2)
3) CLAYp _1
BALLp _1
2p
BALLCLAY ppp _1_12
BALLCLAY ppp _1_12
BALLCLAY ppp _1_12
A bullet of mass m traveling
horizontally at a very high speed
v embeds itself in a block of
mass M that is sitting at rest on
a nearly frictionless surface.
What is the speed of the block
just after the bullet embeds
itself in the block?
(1) (4)
(2) (5)
(3)
v
vM
m
vmM
m
vmM
m
vm
mM
m v
Initial
final
?
M
Maximally Inelastic
1-D Collision Special Case: “Maximally Inelastic” – hit & stick
Multi-step Example: A 60kg kid’s on a swing, say she’s gotten herself going so that at
her highest her center of mass is 4m above the ground and when she swoops down her
center of mass is just 0.5 m above the ground. On her down-sweep, she reaches down
and picks up her 3kg backpack that’s sitting on the ground (maybe her cell phone
started ringing).
How high will she get on her upswing with the pack in her lap?
A B C D
YA = 4m
YB= YC = 0.5m
YD= ?
mk = 60kg
mb= 3kg
A B
iAp .
iBp .
Pow! A B
fAp .
fBp .
t
System = carts A & B
initially
finally
Second Equations: Energy Principle
1-D Collision 0& BAE
BABBAABA UEKEKE &int.int.&
(2)
(3)
(4)
(5)
(1)
22
2
1
2
1if mvv)mM(
22
2
1
2
1if mvMv
2
internal
2
2
1)(
2
1if mvEvmM
2
internal2
1imvE
internal
22
2
1)(
2
1EmvvmM if
Which is an accurate energy equation for this collision
for the system of bullet + block?
m v
Initial final
? M
Maximally Inelastic
A squishy clay ball collides in midair
with a baseball, and sticks to the
baseball, which keeps going.
Initial kinetic energies:
K1clay, K1baseball
Final kinetic energy of clay+ball :
Which equation correctly describes
this collision?
1) Kclay+ball= K1clay + K1baseball
2) Kclay+ball> K1clay + K1baseball
3) Kclay+ball< K1clay + K1baseball
final initial
A B
iAp .
iBp .
Pow! A B
fAp .
fBp .
t
System = carts A & B
initially
finally
02
.212
.212
.212
.21 iBBfBBiAAfAA vmvmvmvm
csv '
Second Equations: Energy Principle
1-D Collision 0& BAE
BABBAABA UEKEKE &int.int.&
Special Case: Perfectly Elastic (all internal changes ‘bounce back’)
BABBAABA UEKEKE &int.int.&
02222
2
.
2
.2
.
2
.
B
iB
B
fB
A
iA
A
fA
m
p
m
p
m
p
m
p
Which of the following is a property of all “elastic” collisions?
(1) The colliding objects interact through springs.
(2) The kinetic energy of one of the objects doesn’t change.
(3) The total kinetic energy is constant at all times -- before, during, and after the
collision.
(4) The total kinetic energy after the collision is equal to the total kinetic energy
before the collision.
(5) The elastic spring energy after the collision is greater than the elastic spring
energy before the collision.
1-D Collision
A B
Pow! A B
Example
System = carts A & B
initially
finally
frictionAF
frictionBF
frictionAF
frictionAF
Often know initial motion, want to predict final motion
Two unknowns: and fAv .
fBv .
Generally need two equations to solve for them
0& BABA ppp
True for all collisions
0&& BABABA UEEE
BB EK .intAA EK .int
A B
iAp .
iBp .
Pow! A B
fAp .
fBp .
System = carts A & B
initially
finally
02
.212
.212
.212
.21 iBBfBBiAAfAA vmvmvmvm csv '
1-D Collision
BABBAABA UEKEKE &int.int.&
Special Case: Perfectly Elastic (all internal changes ‘bounce back’)
02222
2
.
2
.2
.
2
.
B
iB
B
fB
A
iA
A
fA
m
p
m
p
m
p
m
p
iBiAfBfA pppp ....
Equation 1
Equation 2
0
11
2
11
2
1 ..
2
....
2
.
A
iBiA
BA
iB
A
iAiBfB
BA
fBm
pp
mm
p
m
ppp
mmp
Deriving relation for Final Speed
1-D Collision
02222
2
.
2
.2
.
2
.
B
iB
B
fB
A
iA
A
fA
m
p
m
p
m
p
m
p
fBiBiAfA pppp ....
2...
2
. fBiBiAfA pppp
Special Case: Perfectly Elastic (all internal changes ‘bounce back’)
Equation 1
Equation 2
fBiAfBiBiBiAfBiBiAfA pppppppppp ......
2
.
2
.
2
.
2
. 222
Form of 0.
2
. cbpap fBfB Solved by a
acbbp fB
2
42
.
Ack!
0
11
2
11
2
1 ..
2
....
2
.
A
iBiA
BA
iB
A
iAiBfB
BA
fBm
pp
mm
p
m
ppp
mmp
Extra special case: Say B is initially stationary
1-D Collision Special Case: Perfectly Elastic (all internal changes ‘bounce back’)
A B
Pow! A B
fAp .
fBp .
System = carts A & B
initially
finally
011
2
1 ..
2
.
A
iAfB
BA
fBm
pp
mmp
BA
A
iAfB
mmm
pp
112 .
.
AB
iAB
mm
pm
.2
1
Extra special case: Say B is initially stationary
1-D Collision Special Case: Perfectly Elastic (all internal changes ‘bounce back’)
A B
Pow! A B
fAp .
fBp .
System = carts A & B
initially
finally
AB
iABfB
mm
pmp
.
. 2
fBiAfA ppp ...
AB
iABiAfA
mm
pmpp
.
.. 2
iA
AB
BAfA p
mm
mmp ..
iA
AB
AfB v
mm
mv ..
2
iA
AB
BAfA v
mm
mmv ..
Extra special case: Say B is initially stationary
1-D Collision Special Case: Perfectly Elastic (all internal changes ‘bounce back’)
A B
Pow! A B
fAp .
fBp .
System = carts A & B
initially
finally
iA
AB
A vmm
m.
2
iA
AB
BA vmm
mm.
iAv .2
iAv .
1 iAv .
0
𝑣 𝐵.𝑓 𝑣 𝐴.𝑓
𝑚𝐵
𝑚𝐴
Initially Stationary
𝑣 𝐴.𝑓 𝑣 𝐵.𝑓
Wed.
Lab
Fri.
10.1-.4 Introducing Collisions Quiz 9
L9 Multi-particle Systems
10.6-.8 Scattering
RE 10.a
RE 10.b
Mon.,
Lab
Wed.,
Fri.,
10.9-.10 Collision Complications
L10 Collisions 1
10.5, .11 Different Reference Frames
1.1 Translational Angular Momentum Quiz 10
RE 10.c
EP9
RE 11.a; HW10: Pr’s 13*, 21, 30, “39”
Collisions! Short, Sharp Shocks
P
mp
ipp .
T
mT
0. iTp
T
P
fpp .
fTp .
Fast (v~c) Collision
P
mp
0,0,.. ipip pp
T
mT
0. iTp
T
P
0,sin,cos ... pfppfpfp ppp
p
0,sin,cos ... TfTTfTfT ppp T
Conservation of Momentum
Fast (v~c) Collision
00sinsin:ˆ
0coscos:ˆ
0
..
...
...
TfTpfp
ipTfTpfp
ipfTfp
ppy
pppx
ppp
P
mp
0,0,.. ipip pp
T
mT
0. iTp
T
P
0,sin,cos ... pfppfpfp ppp
p
0,sin,cos ... TfTTfTfT ppp T
Conservation of Momentum
where
21cv
vmp
Fast (v~c) Collision
00sinsin:ˆ
0coscos:ˆ
0
..
...
...
TfTpfp
ipTfTpfp
ipfTfp
ppy
pppx
ppp
P
mp
0,0,.. ipip pp
T
mT
0. iTp
T
P
0,sin,cos ... pfppfpfp ppp
p
0,sin,cos ... TfTTfTfT ppp T
Conservation of Momentum
Conservation of Energy
0.... iTipfTfp EEEE
where
21cv
vmp
Fast (v~c) Collision
00sinsin:ˆ
0coscos:ˆ
0
..
...
...
TfTpfp
ipTfTpfp
ipfTfp
ppy
pppx
ppp
where
2
2
1cv
mcE
P
mp
0,0,.. ipip pp
T
mT
0. iTp
T
P
0,sin,cos ... pfppfpfp ppp
p
0,sin,cos ... TfTTfTfT ppp T
Conservation of Momentum
Conservation of Energy
0.... iTipfTfp EEEE
where
21cv
vmp
Fast (v~c) Collision
00sinsin:ˆ
0coscos:ˆ
0
..
...
...
TfTpfp
ipTfTpfp
ipfTfp
ppy
pppx
ppp
where
2
2
1cv
mcE
CW: show
By plugging that into it and
recovering that.
222mcpcE