Physics 211: Lecture 16 Todays Agenda

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Physics 211 Lecture 16Todays Agenda

• Elastic collisions in two dimensions
• Examples (nuclear scattering, billiards)
• Impulse and average force

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Elastic Collision of 2 objects in 2-D
Ice table
Precollision
Postcollision
m1
v1,f
m1
v1,i
CM
CM
VCM
VCM
v2,i
m2
m2
v2,f
VCM is constant since P is conserved!!
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Energy in Elastic Collisions
Recall from the previous lecture In 1
dimension this means v1,f -v 1,i
v2,f -v2,i In 2 or more
dimensions
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Elastic Collisions

• So we see that

CM Frame
backscattering is 180 degrees
? scattering angle
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Lecture 16, Act 1Elastic Collisions
Golf bowling

• Consider the two elastic collisions shown below.
  In 1, a golf ball moving with speed V hits a
  stationary bowling ball head on. In 2, a bowling
  ball moving with the same speed V hits a
  stationary golf ball.
• In which case does the golf ball have the greater
  speed after the collision?

(a) 1 (b) 2 (c) same

V
V
1
2
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Lecture 16, Act 1Solution

• The speed of approach of two objects before an
  elastic collision is the same as the speed of
  recession after colliding.

• Since the bowling ball is much heavier than the
  golf ball, its speed will be changed very little
  in either collision.

V
V
1
2
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Lecture 16, Act 1Solution
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2-D Elastic Collision of 2 objects

• Suppose we know what the pre-collision
  velocities are.
• We want to find out about the motion of both
  objects after the collision.
• We want v1x,f , v1y,f , v2x,f , v2y,f (the
  final velocities of the 2 particles)
• What else do we know
• In an elastic collision, kinetic energy is
  conserved as well as momentum. This leads to 3
  equations
• Ef Ei
• Px,f Px,i (where Px p1x p2x m1v1x
  m2v2x etc)
• Py,f Py,i
• We have 3 equations and 4 unknowns
• We need more information (scattering angle,
  masses).
• Many collisions satisfy these equations and have
  the same initial velocities
• How do we know what happens??
• Whats missing????

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Impact parameter

• Lets say you know the initial momenta of two
  particles
• And you know that the collision between them is
  elastic
• Go to the CM intertial reference frame
• And have the particles approach each other along
  trajectories parallel to the x-axis.

Head on (is 1-D)
just like 1-D
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Impact parameter

• Now offset the trajectory of one particle with
  respect to the other, but keep the CM fixed at
  the origin
• The particles scatter at an angle q that is
  determined by the particle sizes and the impact
  parameter

offset
This distance is impact parameter
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Impact parameter

• Now offset the trajectory of one particle with
  respect to the other, but keep the CM fixed at
  the origin
• The particles scatter at an angle q that is
  determined by the particle sizes and the impact
  parameter

offset
This distance is impact parameter
q
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Impact parameter

• So this is why we have 4 unknowns and three
  equations involving velocities
• Thats not enough to uniquely fix the solution
• One more thing is needed to pin everything down
• Thats the impact parameter

offset
This distance is impact parameter
q
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2-D Elastic CollisionNuclear Scattering

• A particle of unknown mass M is initially at
  rest. A particle of known mass m is shot at it
  with initial momentum pi . After the particles
  collide, the new momentum of the shot particle pf
  is measured.
• Figure out what M is in terms of pi and pf and m.

at rest
P
pi
pf
initial
final
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2-D Elastic CollisionNuclear Scattering
We know pi, pf, m We want to find Px, Py,
M We have 3 equations 1) Momentum
conservation in the x direction 2) Momentum
conservation in the y direction 3) Energy
conservation
P
pf
So we can solve the problem!
final
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Aside Kinetic Energy

• We know that K 1/2mv2

Kinetic energy can also be expressed in terms of
momentum
K 1/2mv2
2
2
v
m

m
2
v
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2-D Elastic CollisionNuclear Scattering
pf
P

• Using momentum conservation pi pf P
• So P2 (pi -pf )2

pi

• Using kinetic energy conservation

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2-D Elastic CollisionNuclear Scattering
pf
P

• So we find that
• If we measure pi and pf and we know m we can
  measure M.
• We can learn about something we cant see!
• This is the basic idea behind a large body of
  work done in atomic, nuclear and particle
  physics.

pi
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Rutherford Backscattering

• Shoot a beam of ? particles (helium nuclei)
  having known energy Ei into a sample of unknown
  composition. Measure the energy Ef of the ?
  particles that bounce back out at 180o with
  respect to the incoming beam.

unknown stuff
particle detector (measures energy)
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Rutherford Backscattering
vectors
P
magnitudes
pf
pi

• In the 180o case, this simplifies significantly

)
(
)
(
)
(
é
ù
2



p
p
v
v
v
v
i
f
i
f
i
f


M
m
ê
ú
m
(
)
(
)
2
2

-
-
v
v
v
v
p
p
ê
ú
ë
û
i
f
i
f
i
f
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Rutherford Backscattering

• Shoot a beam of ? particles (helium nuclei)
  having known energy Ei into a sample of unknown
  composition. Measure the energy Ef of the ?
  particles that bounce back out at 180o with
  respect to the incoming beam.

Ei
unknown stuff
Ef
particle detector (measures energy)
So we learn about the mass of the nuclei in the
unknown stuff. (We learn what the stuff is).
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Rutherford Backscattering

• For example Suppose we are shooting ? particles
  that have an initial energy of Ei 2 MeV at a
  target made of an unknown material. The ?
  particles return with final energy Ef 1.1 MeV.
  What is the weight of the unknown material?
• m(?) 4 (2 protons, 2 neutrons)
• So
  M 27

Aluminum!! (13 protons, 14 neutrons)
22

University of Illinois
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RBS example
How many La, Sr and Mn atoms did we deposit?
2.5 MeV a-particles
BS detector

• Analyze content of thin film of LaSrMn2O6 grown
  on top of MgO area of peaks proportional to
  film content of that element

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Another example of 2-D elasticcollisions
Billiards.elastic collisions of equally massive
balls

• If all we know is the initial velocity of the cue
  ball, we dont have enough information to solve
  for the exact paths after the collision. But we
  can learn some useful things...

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Billiards.

• Consider the case where both balls have the same
  mass and one ball is initially at rest.
• Note that all these have the same mass

pf
pi
vcm
Pf
F
initial
final
The final direction of the red ball will depend
on where the balls hit.
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Billiards

• We know momentum is conserved pi pf Pf
• We also know that kinetic energy is conserved
• Comparing these two equations tells us that
  

pi2 (pf Pf )2 pf2 Pf2 2 pf ? Pf
2
2
2
P
p
p


f
f
i
pf ? Pf 0
Pf
pf
Therefore, Pf and pf must be orthogonal!
pi
Example of equations providing qualitative
information
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Billiards.

• The final directions are separated by 90o.

pf
pi
vcm
Pf
F
initial
final
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Billiards.

• So, we can sink the red ball without sinking the
  white ball.

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Billiards.

• So, we can sink the red ball without sinking the
  white ball.
• However, we can also scratch. All we know is
  that the angle between the balls is 90o.

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Billiards.

• Tip If you shoot a ball spotted on the dot,
  you will sink both balls !

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Lecture 16, Act 2Elastic Collisions in 2-D

• A moving ball initially traveling in the
  direction shown hits an identical but stationary
  ball. The collision is elastic.
• Describe one possible direction of both balls
  just after the collision.

(a) (b) (c)

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Lecture 16, Act 2Solution

• In the first solution, the angle between the
  balls is not 90o.
• In the second solution, there are no downward y
  components to balance out the upward y components.

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Lecture 16, Act 2Solution

• The third choice both balances the y components
  and has 90o between the final direction vectors
  of the two balls.
• As a result, the third choice is the only one of
  the three that fits all necessary criteria.

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Collision timescales

• Collisions typically involve interactions that
  happen quickly.

vf
vi
Vf
F
initial
final
The balls are in contact for a very short time.
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Collision timescales

• During this brief time, the forces involved can
  be quite large

?t
t1
t5
t2
t4
t3
p1
p2
p3 0
p5
p4
F2
F4
F3
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Force and Impulse

• The diagram shows the force vs. time for a
  typical collision. The impulse, I, of the force
  is a vector defined as the integral of the force
  during the collision.

F
Impulse I area under this curve !
t
?t
ti
tf
Impulse has units of Ns.
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Force and Impulse

• Using

the impulse becomes
F
t
?t
impulse change in momentum!
ti
tf
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Force and Impulse
Egg

• Two different collisions can have the same
  impulse since I dependsonly on the change in
  momentum,not the nature of the collision.

same area
F
t
?t
?t
ti
tf
ti
tf
?t big, F small
?t small, F big
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Force and Impulse
soft spring
F
stiff spring
t
?t
?t
ti
tf
ti
tf
?t big, F small
?t small, F big
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Lecture 16, Act 3Force Impulse

• Two boxes, one heavier than the other, are
  initially at rest on a horizontal frictionless
  surface. The same constant force F acts on each
  box for exactly 1 second.
• Which box has the most momentum after the force
  acts?

(a) heavier (b) lighter
(c) same
F
F
heavy
light
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Lecture 16, Act 3Solution
In this problem F and Dt are the same for both
boxes!
F
F
heavy
light
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Force and Impulse

• We can use the notion of impulse to define
  average force, which is a useful concept, but
  not very fundamental

F
The time average of a force for the time interval
?t tf - ti is
Fav
t
?t
or
ti
tf
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Force and Impulse
Ball-Block Collisions
soft spring
Fav
F
stiff spring
Fav
t
?t
?t
ti
tf
ti
tf
?t big, Fav small
?t small, Fav big
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Force and ImpulseBaseball Example

• A pitcher pitches the ball (m .7 kg) at 145
  km/hr (about 90 mph).
• The batter makes contact with the ball for .001 s
  causing the ball to leave the bat going 190 km/hr
  (about 120 mph).
• Find the average force on the ball, disregarding
  gravity.

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Baseball Example
Finally find the average force
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Recap of todays lecture

• Two-dimensional elastic collisions. (Text
  8-6)
• Examples (nuclear scattering, billiards). (Text
  8-6)
• Impulse and average force. (Text 8-6)
• Look at textbook problems Chapter 8 59, 61,
  63, 98, 128