Download Document

Chapter 7 – Lecture II
Collisions
The physics used to analyze collisions is as follows:
r
r
dP1
F12 =
dt
r
r
dP
F21 = 2
dt
r
r
2, and F21 = force on particle 2 because of particle
Where F12 is force on particle 1 because of particle
1. Adding these together we have:
r
r
d r r
F12 + F21 = (P1 + P2 )
dt
1
r
r
But from Newton’s 3 Law F12 + F21 = 0 , so we have
r v
P1 + P2 = constant.
rd
In other words, if we consider our system to be two
particles then the forces on one because of the
other is an internal force and as a system, doesn’t
affect the center of mass. The physics used to
analyze collisions is two equations
r r
P1 + P2 = constant
(E1 + E2 )before collison =
(E ′ + E ′ )
1
2 after collison
+Q
2
Page 2 of 9
11/13/2002
Where Q is energy added (or absorbed) during the
collision.
The coordinate system most often used to analyze
a collision problem is one where the origin moves
with the center of mass.
In this system, we have
x = xcm + x′
y = y′
and
vx = vcm + vx′
v =v ′
y
y
Look at a collision in each coordinate system:
m1v1a
m1v10 →
φ1
φ2
m2v2a
Laboratory
System
m 1 u 10
θ
Center of
Mass
System
m1u1a
θ
m2u 2a
m2u20
3
Page 3 of 9
11/13/2002
From the transformation equations we have:
v1a cos φ1 = vcm + u1a cos θ
v1a sin φ1 = u1a sin θ
Divide these:
sin φ1
u1a sin θ
=
cos φ1 vcm + u1a cos θ
tan φ1 =
γ≡
sin θ
cos θ + γ
4
vcm
u1a
Likewise for φ2 :
v2 a cos φ2 = vcm − u 2a cos θ
v2 a sin φ2 = u 2a sin θ
tan φ2 =
sin θ
vcm
− cos θ
u2 a
=
sin θ
α − cos θ
5
Page 4 of 9
11/13/2002
α≡
vcm
u2 a
α=
m2
γ since m1u1a = m2u 2a
m1
In the Laboratory System our conservation of
momentum and energy equations become:
m1v10 = m1v1 cos φ1 + m2v2 cos φ2
0 = m1v1 sin φ1 − m2v2 sin φ2
1
1
1
2
m1v10
+ 0 = m1v12 + m2v22 + Q
2
2
2
These are three equations with 8 algebraic
quantities.
In the Center of Mass System, our conservation of
momentum and energy equations become:
m1u10 + m2u20 = 0
m1u1 + m2u 2 = 0
1
1
1
1
2
2
m1u10
+ m2u20
= m1u12 + m2u22 + Q′
2
2
2
2
You can see the Center of Mass System yields
simpler equations.
Page 5 of 9
11/13/2002
Example 1
Textbook Problem 7.10
A moving particle of mass m1 collides elastically
with a target particle of mass m2 which initially is
at rest. If the collision is head-on, show the
incident particle lose a fraction
4µ
m of its original
kinetic energy, where µ =reduced mass and
m=m 1+m2.
Our momentum equation gives us:
m1v10 = m1v1 + m2v2
1
Our energy equation gives us:
1
1
1
2
m1v10
= m1v12 + m2v22
2
2
2
Let
m2
2
m1 ≡ α , then these become
2
v10
− v12 = αv22
3
v10 − v1 = αv2
4
Divide 3 by 4:
v10 + v1 = v2
5
Page 6 of 9
11/13/2002
Substitute this into 4:
v10 − v1 = α(v10 + v1 )
v10 (1 − α) = v1 (1 + α)
v1 =
1− α
v10
1+ α
The problem asks for the functional loss in
kinetic energy.
T10 − T1
v12
= 1− 2
T10
v10
1 − α 
= 1− 

1
+
α


2
(
1 + α )2 − (1 − α )2
=
(1 + α )2
=
=
4α
(1 + α )2
4m1m2
=
(m1 + m2 )2
4
m2
m1
1 + m2 

m1 

=4
2
m1m2
1
4µ
=
m1 + m2 m1 + m2 m
Page 7 of 9
11/13/2002
Example 2
Textbook Problem 7.14
A proton of mass mp with initial velocity v 0 collides
with a helium atom, mass 4mp, that is initially at
rest. If the proton leaves the point of impact at an
angle of 450 with its original line of motion, find
the final velocities of each particle. Assume that
the collision is perfectly elastic.
Momentum mv0 = mv1 cos 45 + 4 mv2 cos θ
0 = mv1 sin 45 − 4mv2 sin θ
Energy
1 2 1 2 1
mv0 = mv1 + 4mv22
2
2
2
or rewriting:
v0 = v1 cos 45 + 4v2 cos θ
0 = v1 sin 45 − 4v2 sin θ
v02 = v12 + 4v22
From 2 we have v1 =
1
2
3
4v2 sin θ
sin 45
4
Page 8 of 9
11/13/2002
Put this into 1:
v0 = 4v2 sin θ
cos 45
+ 4v2 cos θ
sin 45
v0 = 4v2 (sin θ + cos θ)
5
Now put 4 and 5 into 3:
16v22
(sin θ + cos θ)
2
16v22 sin 2 θ
2
=
+
4
v
2
sin 2 45
Divide out the common factors:
4(sin θ + cos θ)2 = 8 sin 2 θ + 1
Simplify:
(
)
4 sin 2 θ + cos 2 θ + 2 sin θ cos θ = 8 sin 2 + 1
8 sin θ cos θ = 8 sin 2 θ − 3
4 sin 2θ = 4(1 − cos 2θ) − 3
4(sin 2θ + cos 2θ) = 1
sin 2θ + cos 2θ = 1 4
Page 9 of 9
11/13/2002
2
2
21
sin 2θ +
cos 2θ =
2
2
2 4
2
sin 2θ + 450 =
8
(
)
 2
 − 450
2θ = sin −1 
 8 
1  −1 2 
1
0


(169.82 − 45 )
θ = sin 
−
45
=


2
 8 
 2
θ = 62.410
Then from 5
v2 = v0 [4(sin θ + cos θ)]
v2 = .185 v0
and from 4
v1 = 4v2 sin θ sin 45
v1 = .929 v0