Download Lecture 28: More on Collisions

Lecture 28: More on Collisions
• Last time we started discussing the fraction of the kinetic
energy retained by m1 after an elastic collision in the LAB
frame is:
T1 v12
= 2
T0 u1
• To relate this to the LAB scattering angle, we use
conservation of energy and momentum:
p x = m1u1 = m1v1 cosψ + m2 v2 cos ζ
p y = 0 = m1v1 sinψ − m2 v2 sin ζ
1
1
1
2
2
T = m1u1 = m1v1 + m2 v22
2
2
2
3 equations
Treat u1/v1, v2, ψ and
ζ as variables
Can solve for u1/v1 as a
function of ψ
• From the energy equation, we find:
m1 2
v2 =
( u1 − v12 )
m2
• Plugging this into the py equation gives:
m1v1 sinψ = m2
sin ζ =
m1 2
( u1 − v12 ) sin ζ
m2
m1v1 sinψ
m1m2 ( u12 − v12 )
cos ζ = 1 −
= 1 − cos2 ζ
m v sin ψ
=
2
2
m1m2 ( u1 − v1 )
2 2
1 1
2
m2 ( u12 − v12 ) − m1 v12 sin 2 ψ
m2 ( u12 − v12 )
• This means that:
m2 v2 cos ζ = m2
2
2
2
2
m
u
v
m
v
−
−
sin
ψ
(
)
m1 2
2
1
1
1 1
2
( u1 − v1 )
m2
m2 ( u12 − v12 )
= m1 m2 ( u12 − v12 ) − m1 v12 sin 2 ψ
• We can now insert this into the px equation to find:
m1u1 = m1v1 cosψ + m1 m2 ( u12 − v12 ) − m1 v12 sin 2 ψ
m1 ( u1 − v1 cosψ ) = m1 m2 ( u12 − v12 ) − m1 v12 sin 2 ψ
m12 u12 − 2m12 v1 cosψ + m12 v12 cosψ = m1 m2 ( u12 − v12 ) − m1 v12 sin 2 ψ
• Or:
v12 ( m1 + m2 ) − 2m1u1v1 cosψ + u12 ( m1 − m2 ) = 0
v12
v1
m + m2 ) − 2m1 cosψ + m1 − m2 = 0
2 ( 1
u1
u1
from which we can find v1/u1 with the quadratic formula:
2
2
2
2
v1 2m1 cosψ ± 4m1 cos ψ − 4 ( m1 − m2 )
=
u1
2 ( m1 + m2 )
2
m1
m
=
cosψ ± cos2 ψ − 1 − 22
m1 + m2
m1
m1
m22
2
=
cosψ ±
−
sin
ψ
2
m1 + m2
m1
Inelastic Collisions
• All collisions conserve energy (just like elastic collisions
do)
• But, if systems of particles collide, the internal energy of
the systems may change
– Therefore, the kinetic energy associated with the motion of
the center of mass is not conserved
• Such collisions are called inelastic
– The extreme case is a collision between two objects that
stick together after they collide (two blobs of clay or silly
putty might behave this way)
– These collisions are called totally inelastic
• Most collisions lie somewhere between elastic and totally
inelastic
Quantifying Inelasticity
• There are two ways to measure how inelastic a collision is
1. Measure the kinetic energy before and after the collision,
and call the difference Q:
T f − Ti = Q
1 2 2 1 2 2 1 2 2 1 2 2
Q + m1 u1 + m2 u2 = m1 v1 + m2 v2
2
2
2
2
The larger |Q| is, the more inelastic the collision
–
–
If Q > 0, the final kinetic energy is greater than the initial.
Such collisions are called exoergic – an explosion is one
example
If Q < 0, the final kinetic energy is less than the initial.
These collisions are endoergic – a collision between blobs
of clay falls into this category
Coefficient of Restitution
2. Another way to think about inelasticity is to consider the
relative velocity in a head-on collision. In the CM frame:
Elastic collision
m2
m1 u
u2
1
v1= -u1
v2 = -u2
v1 − v2
=1
u1 − u2
•
•
Totally inelastic collision
m1 u
m2
u
1
2
m1 + m2
v1 = v2 = 0
v1 − v2
=0
u1 − u2
In general we define v1 − v2 = ε as the coefficient of
u1 − u2
restitution
For head-on collisions in non-CM reference frames, the
velocity components normal to the collision plane enter
the formula
Impulse
• Even though we may not know what forces act during a
collision, we can determine something about those forces
from Newton’s Second Law
• During the collision:
d ( mv )
F=p=
dt
Fdt = d ( mv )
t2
Fdt = ∆ ( mv ) = m∆v if m is constant
t1 ∆(mv) is called the impulse, and givcn the
• The quantity
symbol P
• The left-hand side is related to the time-average of the
force acting during the collision, so we have:
Favg ∆t = P
Scattering Cross Sections
• We’ve seen that if we know the initial velocities, and know
the forces involved, we can exactly predict the results of
the collision
• In particle physics we want to reverse the process
– i.e., we want to learn about the forces acting between
particles by observing the results of a collision
• There’s a catch, though: we can’t specify the initial state
exactly – there’s a limit to how well we can aim the
incoming particles
– We can typically direct a beam of particles (say, electrons or
protons) into a block of material, or at each other
– In each case, we don’t know which two particles are going to
collide, or the initial distance between them
• In the CM frame, the collision looks like:
θ
T1
b
T2
• The quantity b is called the “impact parameter”, and we
usually don’t know its value for a collision
– We typically do control the initial energies accurately, and
can measure the scattering angle and final energies
– We also know the intensity (number of particles per unit
area) of the incoming beams
• So we define a quantity in terms of the things we know or
can measure:
Collisions per particle that cause
scattering into a solid angle element
d Ω′ at angle θ
σ (θ ) =
Beam intensity
• Note that the numerator has no dimension, while the
denominator has dimensions of 1/area
• So, σ(θ) has dimensions of area, which is why we give it
the name “cross section”
• As defined, σ(θ) relates to scattering into a differential
element of solid angle, so it’s called the “differential cross
section”
• The equation for differential cross section is
dN / d Ω′
(
)
σ θ =
I
dN
(
)
σ θ d Ω′ =
I
• If the force between the colliding particles is symmetric
about the collision axis (as all central forces are), we can
integrate over φ to find:
d Ω′ = 2π sin θ dθ
• For particles of a given energy, the impact parameter
determines the scattering angle
• If collisions with impact parameter between b and b + db
result in scattering into d Ω,′ we have:
dN = IdA = I ⋅ 2π bdb
• Relating the two expressions for dN gives:
I ⋅ 2π bdb = I σ (θ ) d Ω′ = I σ (θ ) 2π sin θ dθ
σ (θ ) =
b db
sin θ dθ
• Actually, for most collision forces db/dθ is negative
– the closer the objects come, the greater the scattering angle
• Since the cross section should be a positive quantity, we
write:
b db
σ (θ ) =
sin θ dθ