Physics 211: Lecture 16, Pg 1
Physics 211: Lecture 16
Physics 211: Lecture 16
Today’s Agenda
Today’s Agenda
zElastic collisions in two dimensions
zExamples (nuclear scattering, billiards)
zImpulse and average force
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Two-Dimensional Elastic Collisions: Impulse, Force, and Conservation Laws, Study notes of Physics
The concepts of two-dimensional elastic collisions, including impulse and force, conservation of momentum and energy, and examples of nuclear scattering and billiards. Students will learn how to apply these principles to solve problems involving elastic collisions.
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Physics 211: Lecture 16, Pg 1
Physics 211: Lecture 16
Physics 211: Lecture 16
Today’s Agenda
Today’s Agenda
z
Elastic collisions in two dimensions
z
Examples (nuclear scattering, billiards)
z
Impulse and average force
Physics 211: Lecture 16, Pg 2
Elastic Collision of 2 objects in 2
Elastic Collision of 2 objects in 2
D
D
Ice table
m
1
m
2
v v
2,i
v v
1,i
CM
V
V
CM
Precollision
CM
V
V
CM
v v
2,f
v v
1,f
m
1
m
2
Postcollision
V
V
CM
is constant since
PP
is conserved!!
Physics 211: Lecture 16, Pg 4
Elastic Collisions:
Elastic Collisions:
z
So we see that:
f
,
2
i
,
2
f
,
1
i
,
1
- v * v * v * v
v*
1,f
v*
2,f
θ
v*
1,i
v*
2,i
CM
CM Frame:
“backscattering”
is 180 degrees
θ
= “scattering angle”
Physics 211: Lecture 16, Pg 5
Lecture 16,
Lecture 16,
Act 1
Act 1
Elastic Collisions
Elastic Collisions
Golf & bowling
z
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.
Î
In which case does the golf ball have the greater speed after the collision?
(a) (a)
(b) (b)
(c) (c)
same
V
V
Physics 211: Lecture 16, Pg 7
Lecture 16,
Lecture 16,
Act 1
Act 1
Solution
Solution
V
z
In case
the bowling ball will almost remain at rest, and the
golf ball will bounce back with speed close to
V
V
2V
z
In case
the bowling ball will keep going with speed close to
V
, hence the golf ball will rebound with speed close to 2
V
Physics 211: Lecture 16, Pg 8
D Elastic Collision of 2 objects
D Elastic Collision of 2 objects
z
Suppose we know what the “pre-collision” velocities are.
z
We want to find out about the motion of both objects after the collision.
Î
We want
v
1x,
f
, v
1y,
f
, v
2x,
f
, v
2y,
f
(the final velocities of the 2 particles)
z
What else do we know :
Î
In an elastic collision, kinetic energy is conserved as well asmomentum. This leads to 3 equations:
Î
E
f
= E
i
Î
P
x,f
= P
x,i
(where
P
x
= p
1x
- p
2x
= m
1
v
1x
- m
2
v
2x
etc)
Î
P
y,f
= P
y,i
z
We have 3 equations and 4 unknowns:
Î
We need more information (scattering angle, masses).
Î
Many collisions satisfy these equations and have the same initialvelocities
Î
How do we know what happens??
Î
What’s missing????
Physics 211: Lecture 16, Pg 10
Impact parameter
Impact parameter
z
Now offset the trajectory of one particle with respect to theother, but keep the CM fixed at the origin
z
The particles scatter at an angle
θ
that is determined by
the particle sizes and the impact parameter
offset
This distance is
impact parameter
Physics 211: Lecture 16, Pg 11
Impact parameter
Impact parameter
z
Now offset the trajectory of one particle with respect to theother, but keep the CM fixed at the origin
z
The particles scatter at an angle
θ
that is determined by
the particle sizes and the impact parameter
offset
This distance is
impact parameter
θ
Physics 211: Lecture 16, Pg 13
D Elastic Collision:
D Elastic Collision:
Nuclear Scattering
Nuclear Scattering
z
A particle of unknown mass
M
is initially at rest. A particle of known
mass
m
is “shot” at it with initial momentum
pp
i
. After the particles
collide, the new momentum of the shot particle
pp
f
is measured.
Î
Figure out what
M
is in terms of
pp
i
and
p p
f
and
m
.
M
p p
i
at rest
p p
f
P
P
m
m
M
initial
final
Physics 211: Lecture 16, Pg 14
D Elastic Collision:
D Elastic Collision:
Nuclear Scattering
Nuclear Scattering
p p
i
at rest
initial
m
M
y
x
We know:
p p
i
p p
f
m
We want to find:
P
P
x
PP
y
M
We have 3 equations:
- Momentum conservation in the x direction 2) Momentum conservation in the y direction 3) Energy conservation
p p
f
P
P
final
So we can solve the problem!
Physics 211: Lecture 16, Pg 16
D Elastic Collision:
D Elastic Collision:
Nuclear Scattering
Nuclear Scattering
p p
i
P
P
p p
f
z
Using momentum conservation:
pp
i
=
p p
f
P P
Î
So
P
2
= (
pp
i
pp
f
)
2
p
m
p
m
P
M
i
f
2
2
2
P
M
p
m
p
m
i
f
2
2
2
z
Using kinetic energy conservation:
(
)
P
i
f
2
2
=
−
p
p
and using
(
)
M
m
p
p
i
f
i
f
p
p
2
2
2
Physics 211: Lecture 16, Pg 17
D Elastic Collision:
D Elastic Collision:
Nuclear Scattering
Nuclear Scattering
z
So we find that
z
If we measure
pp
i
and
p p
f
and we know
m
we can measure
M
.
Î
We can learn about something we can’t see!
z
This is the basic idea behind a large body of work done in atomic, nuclear and particle physics.
p p
i
P
P
p p
f
(
)
M
m
p
p
i
f
i
f
p
p
2
2
2
Physics 211: Lecture 16, Pg 19
Rutherford Backscattering
Rutherford Backscattering
z
In the 180
o
case, this simplifies significantly:
p p
i
P P
pp
f
(
)
M
m
p
p
i
f
i
f
p
p
2
2
2
vectors
)
(
)(
)
(
)(
)
M
m
p
p
p
p
v
v
v
v
v
v
v
v
i
f
i
f
i
f
i
f
i
f
i
f
2
2
2
m
(
magnitudes
(
)
(
)
M
m
v
v
v
v
i
f
i
f
v
v
M
m
M
m
f
i
E
E
M
m
M
m
f
i
2
2
i
2
f
mv
mv
Solving for M:
i
f
i
f
E
E
E
E
m
M
Physics 211: Lecture 16, Pg 20
Rutherford Backscattering
Rutherford Backscattering
z
Shoot a beam of
α
particles (helium nuclei) having known
energy
E
i
into a sample of unknown composition. Measure
the energy
E
f
of the
α
particles that bounce back out at
o
with respect to the incoming beam.
particle detector(measures energy)
E
i
E
f
So we learn about the mass of the nucleiin the unknown stuff. (We learn what thestuff is).
unknownstuff
i
f
i
f
E
E
E
E
m