Classical Mechanics
LECTURE 9:
INELASTIC COLLISIONS
Prof. N. Harnew
University of Oxford
MT 2016
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OUTLINE : 9. INELASTIC COLLISIONS
9.1 Examples of 2D elastic collisions
9.1.1 Example 1: Equal masses, target at rest
9.1.2 Example 2: Elastic collision, m2 = 2m1 , θ1 = 30◦
9.2 Inelastic collisions in the Lab frame in 1D (u2 = 0)
9.2.1 Coefficient of restitution
9.3 Inelastic collisions viewed in the CM frame
9.3.1 Kinetic energy in the CM : alternative treatment
9.3.2 Coefficient of restitution in the CM
9.3.3 Example of inelastic process
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9.1.1 Example 1: Equal masses, target at rest
Before
After
Magnitude of velocities:
I
I
I
I
I
3
vCM =
m1 u1 +m2 u2
m1 +m2
=
u10 = u0 − vCM = u20
u20 = −vCM = − u20
|v10 | = |u10 | = u20
|v20 | = |u20 | = u20
u0
2
Relationships between angles and speeds
Angles:
I
Cosine rule:
( u20 )2 = ( u20 )2 + v12 − 2v1 u20 cos θ1
I
v1 u0 cos θ1 = v12
I
cos θ1 =
I
cos θ2 =
v1
u0
v2
u0
as before
Opening angle:
I
Cosine rule:
u02 = v12 + v22 − 2v1 v2 cos(θ1 + θ2 )
I
I
But u02 = v12 + v22 (conservation of
energy)
cos(θ1 + θ2 ) = 0 → θ1 + θ2 =
π
2
NB: Lines joining opposite corners of rhombus cross at 90◦
4
9.1.2 Example 2: Elastic collision, m2 = 2m1 , θ1 = 30◦
Find the velocities v1 and v2 and the angle θ2
Magnitude of velocities:
m1 u1 +m2 u2
m1 +m2
I
vCM =
I
u10 = u0 − vCM =
I
u20 = −vCM = − u30
I
|v10 | = |u10 | =
I
|v20 | = |u20 | =
5
2u0
3
u0
3
=
u0
3
2u0
3
Relationships between angles and speeds
I
Sine rule:
(sin 30/ 2u30 ) = (sin α/ u30 )
→ sin α =
1
4
→ α = 14.5◦
I
β = 30 + α = 44.5◦
I
sin 30/ 2u30 = sin(180 − 44.5)/v1
→ v1 = 0.93u0
I
Cosine rule:
v22 = ( u30 )2 + ( u30 )2 − 2( u30 )2 cos β
→ v2 = 0.25u0
I
Sine rule:
(sin 44.5/v2 ) = (sin θ2 / u30 )
→ θ2 = 68.0◦
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9.2 Inelastic collisions in the Lab frame in 1D (u2 = 0)
An inelastic collision is where
energy is lost (or there is
internal excitation).
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Take m2 at rest & in 1D. Momentum : m1 u1 = m1 v1 + m2 v2 (1)
I
Energy :
I
Square Equ.(1) and subtract 2m1 × Equ.(2)
→
I
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= 12 m1 v12 + 21 m2 v22 + ∆E
(2)
m2 (m2 − m1 )v22 + 2m1 m2 v1 v2 − 2m1 ∆E = 0
Substitute for m1 v1 from Equ.1 to get quadratic in v2
→
I
1
2
2 m1 u1
m2 (m2 + m1 )v22 − 2m1 m2 u1 v2 + 2m1 ∆E = 0
Solve, taking consistent solutions with elastic case (∆E = 0)
√
2m m u + 4m12 m22 u12 −8m1 m2 (m1 +m2 )∆E
→ v2 = 1 2 1
(3)
2m2 (m1 +m2 )
√
2m2 u − 4m12 m22 u12 −8m1 m2 (m1 +m2 )∆E
→ v1 = 1 1
(4)
2m1 (m1 +m2 )
1D inelastic collisions viewed in the Lab frame (u2 = 0)
I
We see from Equ. (3) & (4) there is a limiting case:
4m12 m22 u12 − 8m1 m2 (m1 + m2 )∆E ≥ 0
I
i.e. ∆E ≤
I
This corresponds to the two bodies sticking together in a
single object of mass (m1 + m2 ) → v1 = v2
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From momentum cons. m1 u1 = m1 v1 + m2 v2
m1 m2 u12
2(m1 +m2 )
if v1 = v2 = v , then v =
For equal mass m1 = m2
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m1 u1
(m1 +m2 )
(the CM velocity)
v2 , v1 =
u1
2
1±
q
1−
4∆E
mu12
9.2.1 Coefficient of restitution
e=
General definition :
=
Speed of relative separation
Speed of relative approach
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From Equ.(3) & (4) previously
√
2m m u + 4m12 m22 u12 −8m1 m2 (m1 +m2 )∆E
v2 − v1 = 1 2 1
2m (m +m )
√ 2 22 2 1 2
2
2m u − 4m1 m2 u1 −8m1 m2 (m1 +m2 )∆E
− 1 1
2m1 (m1 +m2 )
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Factorizing, then simplifying, then dividing by u1 gives
r
q
2 )∆E
e = 1 − 2(mm1 +m
=
1 − ∆E
T0
m u2
1
where
I
I
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|v2 −v1 |
|u1 −u2 |
T0
=
2 1
1
2
2 µu1
with µ =
T0
m1 m2
m1 +m2
(the reduced mass)
We see later that
is the initial energy in the CM frame,
hence e is related to the fractional energy loss in this frame
e = 1 completely elastic; e = 0 perfectly inelastic,
in general 0 < e < 1
9.3 Inelastic collisions viewed in the CM frame
Case of perfectly inelastic collision (e = 0)
After collision, total mass (m1 + m2 ) is at rest in CM:
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2
KE in CM: TCM = TLAB − 12 (m1 + m2 )vCM
Differentiate: Loss in KE ∆TCM = ∆TLAB (obvious)
Max. energy that can be lost = TCM =
2
= 12 m1 u12 + 21 m2 u22 − 12 (m1 + m2 )vCM
9.3.1 Kinetic energy in the CM : alternative treatment
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Revisit kinetic energy in the CM frame: TLab = TCM + 12 MvCM
I
TCM = 12 m1 u102 + 12 m2 u202
I
2
x10 = − m1m+m
x = − mM2 x , x20 =
2
I
u10 = − mM2 ẋ , u20 =
I
TCM =
1
2
I
TCM =
1 m1 m2
2 M2
I
Also
m1
M x
m1
M ẋ
m1 (− mM2 )2 + m2 ( mM1 )2 ẋ 2
(m2 + m1 ) ẋ 2 =
1 m1 m2 2
2 M ẋ
ẋ = ẋ2 − ẋ1 = u20 − u10
TCM =
1 m1 m2 2
2 M ẋ
=
1
2
2 µẋ
=
1
0
0 2
2 µ(u1 −u2 )
=
1
2
2 µ(u1 −u2 )
These expressions give the CM kinetic energy in terms of the
relative velocities in the CM & Lab and the reduced mass µ
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9.3.2 Coefficient of restitution in the CM
I
i
Initial KE in the CM : TCM
= 21 µ(u10 − u20 )2
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f
Final KE in the CM : TCM
= 12 µ(v10 − v20 )2
I
i
f
Conservation of energy : TCM
= TCM
+ ∆E
→
→
→
1
0
2 µ(u1
− u20 )2 = 21 µ(v10 − v20 )2 + ∆E
0 0 2
v1 −v2
= 1 − T∆E
i
u10 −u20
CM
0 0
q
v1 −v2
= ± 1 − T∆E
i
u 0 −u 0
1
2
CM
i
Same expression as before with T 0 = TCM
e=
Coefficient of restitution
q
|v −v |
= |u2 −u1 |
= 1−
|v02 −v01 |
|u01 −u02 | CM
1
2
LAB
∆E
i
TCM
ONLY in CM frame can ALL the KE be used to create ∆E
→
For e = 0 the two particles coalesce and are at rest in CM
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9.3.3 Example of inelastic process
A calcium nucleus (A=20), mass m, travels with velocity u0 in the Lab.
It decays into a sulphur nucleus (A=16), mass 45 m, and an α-particle
(A=4), mass 15 m. Energy ∆T is released as KE in the calcium rest
frame (CM). A counter in the Lab detects the sulphur nucleus at 90◦
to the line of travel. What is the speed and angle of the α-particle in
the Lab?
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I
Energy ∆T is released as
KE in the CM. vCM = u0
I
Momentum in CM:
4
1
0
0
5 mv1 − 5 mv2 = 0
→ v20 = 4v10
I
Energy: ∆T = 12 ( 45 m)v10 2 +
1 1
02
02
2 ( 5 m)16v1 = 2mv1
1
2
→ v10 = [ ∆T
2m ]
1
2
→ v20 = [ 8∆T
m ]
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Transform to Lab by
boosting by vCM (= u0 )
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cos α =
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2mu02 1
2
∆T ]
Cosine rule: v22 = v20 2 +
α
Sine rule: sinv 0θ2 = sin
v2
2
u0
v10
=[
u0 2 + 2v20 u0 cos α
Solve for
v2 , θ2