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Chapter 15 Collision Theory
- 15.1 Introduction
- 15.2 Reference Frames Relative and Velocities..........................................................
- 15.2.1 Center of Mass Reference Frame
- 15.2.2 Relative Velocities
- 15.3 Characterizing Collisions
- 15.4 One-Dimensional Elastic Collision Between Two Objects
- Reference Frame 15.4.1 One-Dimensional Collision Between Two Objects – Center of Mass
- 15.5 Worked Examples
- Example 15.1 Elastic One-Dimensional Collision Between Two Objects
- Collision Between Two Objects Example 15.2 The Dissipation of Kinetic Energy in a Completely Inelastic
- Example 15.3 Elastic Two-Dimensional Collision................................................
- Emerge at Right Angles Example 15.4 Equal Mass Particles in a Two-Dimensional Elastic Collision
- Example 15.5 Bouncing Superballs
Chapter 15 Collision Theory
Despite my resistance to hyperbole, the LHC [Large Hadron Collider]
belongs to a world that can only be described with superlatives. It is not
merely large: the LHC is the biggest machine ever built. It is not merely
cold: the 1.9 kelvin (1.9 degrees Celsius above absolute zero) temperature
necessary for the LHC’s supercomputing magnets to operate is the coldest
extended region that we know of in the universe—even colder than outer
space. The magnetic field is not merely big: the superconducting dipole
magnets generating a magnetic field more than 100,000 times stronger than
the Earth’s are the strongest magnets in industrial production ever made.
And the extremes don’t end there. The vacuum inside the proton-containing
tubes, a 10 trillionth of an atmosphere, is the most complete vacuum over
the largest region ever produced. The energy of the collisions are the highest
ever generated on Earth, allowing us to study the interactions that occurred
in the early universe the furthest back in time.
1
Lisa Randall
15.1 Introduction
When discussing conservation of momentum, we considered examples in which two
objects collide and stick together, and either there are no external forces acting in some
direction (or the collision was nearly instantaneous) so the component of the momentum
of the system along that direction is constant. We shall now study collisions between
objects in more detail. In particular we shall consider cases in which the objects do not
stick together. The momentum along a certain direction may still be constant but the
mechanical energy of the system may change. We will begin our analysis by considering
two-particle collision. We introduce the concept of the relative velocity between two
particles and show that it is independent of the choice of reference frame. We then show
that the change in kinetic energy only depends on the change of the square of the relative
velocity and therefore is also independent of the choice of reference frame. We will then
study one- and two-dimensional collisions with zero change in potential energy. In
particular we will characterize the types of collisions by the change in kinetic energy and
analyze the possible outcomes of the collisions.
15.2 Reference Frames Relative and Velocities
We shall recall our definition of relative inertial reference frames. Let R
be the
vector from the origin of frame S^ to the origin of reference frame S^ ′. Denote the
position vector of particle i with respect to the origin of reference frame S by r i
and
1 Randall, Lisa, Knocking on Heaven's Door: How Physics and Scientific Thinking Illuminate the Universe
and the Modern World , Ecco, 2011.
respect to origin of reference frame S by
r i and similarly, denote the position vector of
particle i with respect to origin of reference frame S cm by
r cm , i (Figure 15.2).
Figure 15.2 Position vector of i
th particle in the center of mass reference frame.
The position vector of particle i in the center of mass frame is then given by
r cm , i
r i
R
cm
The velocity of particle i^ in the center of mass reference frame is then given by
v cm , i
v i
V
cm
There are many collision problems in which the center of mass reference frame is the
most convenient reference frame to analyze the collision.
15.2.2 Relative Velocities
Consider two particles of masses m 1 and m 2 interacting via some force (Figure 15.3).
Figure 15.3 Two interacting particles
Choose a coordinate system (Figure 15.4) in which the position vector of body 1 is given
by r 1
and the position vector of body 2 is given by r 2
. The relative position of body 1
with respect to body 2 is given by r 1 (^) , 2 = r 1 (^) − r 2
Figure 15.4 Coordinate system for two bodies.
During the course of the interaction, body 1 is displaced by d r 1
and body 2 is displaced
by d r 2
, so the relative displacement of the two bodies during the interaction is given by
d r 1 (^) , 2 = d r 1 (^) − d r 2
. The relative velocity between the particles is
(^1 2 1 ) 1 2 1 2
, ,
d (^) d d
dt dt dt
r (^) r r v v v
We shall now show that the relative velocity between the two particles is independent of
the choice of reference frame providing that the reference frames are relatively inertial.
The relative velocity v 1 ′ 2
in reference frame S ′ can be determined from using Eq. ( 0. 3. 4 )
to express Eq. ( 0. 3. 7 ) in terms of the velocities in the reference frame S ′,
v 1 , 2
v 1
v 2
v ′ 1 +
V ) − (
v ′ 2 +
V ) =
v 1 ′ −
v ′ 2 =
v ′ 1 , 2 ( 0. 3. 8 )
and is equal to the relative velocity in frame S.
For a two-particle interaction, the relative velocity between the two
vectors is independent of the choice of relatively inertial reference frames.
We showed in Appendix 13 A that when two particles of masses m 1 and m 2 interact, the
change of kinetic energy between the final state B and the initial state A due to the
interaction force is equal to
Δ K =
μ( v B
2 − v A
2 ) , ( 0. 3. 9 )
where v B
v 1 , 2
B
v 1
B
v 2
B denotes the relative speed in the state B and
v A
v 1 , 2
A
v 1
A
v 2
A denotes the relative speed in state A , and
μ = m 1 m 2 / ( m 1
- m 2 ) is the reduced mass. (If the relative reference frames were non-
inertial then Eq. ( 0. 3. 9 ) would not be valid in all frames.
collides with an object of mass m 2 and initial x - component of the velocity v 2 x , i
. The
scalar components v 1 x , i and v 1 x , i can be positive, negative or zero. No forces other than
the interaction force between the objects act during the collision. After the collision, the
final x - component of the velocities are v 1 x , f and v 2 x , f
. We call this reference frame the
“laboratory reference frame”.
Figure 15. 5 One-dimensional elastic collision, laboratory reference frame
For the collision depicted in Figure 15.5, v 1 x , i
0 , v 2 x , i < 0 , v 1 x , f < 0 , and v 2 x , f
Because there are no external forces in the x - direction, momentum is constant in the x -
direction. Equating the momentum components before and after the collision gives the
relation
m 1 v 1 x , i
- m 2 v 2 x , 0 = m 1 v 1 x , f
- m 2 v 2 x , f
Because the collision is elastic, kinetic energy is constant. Equating the kinetic energy
before and after the collision gives the relation
m 1 v 1 x , i
2
m 2 v 2 x , i
2
m 1 v 1 x , f
2
m 2 v 2 x , f
2 ( 0. 4. 7 )
Rewrite these Eqs. ( 0. 4. 6 ) and ( 0. 4. 7 ) as
m 1 ( v 1 x , i − v 1 x , f ) = m 2 ( v 2 x , f − v 2 x , i
m 1 ( v 1 x , i
2 − v 1 x , f
2 ) = m 2 ( v 2 x , f
2 − v 2 x , i
2 ). ( 0. 4. 9 )
Eq. ( 0. 4. 9 ) can be written as
m 1 ( v 1 x , i − v 1 x , f )( v 1 x , i
- v 1 x , f ) = m 2 ( v 2 x , f − v 2 x , i )( v 2 x , f
- v 2 x , i
Divide Eq. ( 0. 4. 9 ) by Eq. ( 0. 4. 8 ), yielding
v 1 x , i
- v 1 x , f = v 2 x , i
- v 2 x , f.^ (^0.^4.^11 )
Eq. ( 0. 4. 11 ) may be rewritten as
v 1 x , i − v 2 x , i = v 2 x , f − v 1 x , f
Recall that the relative velocity between the two objects in state A is defined to be
v A
rel
v 1 , A
v 2 , A.^ (^0.^4.^13 )
where we used the superscript “rel” to remind ourselves that the velocity is a relative
velocity. Thus v x , i
rel = v 1 x , i − v 2 x , i is the initial x - component of the relative velocity, and
v x , f
rel = v 1 x , f − v 2 x , f is the final x - component of the relative velocity. Therefore Eq. ( 0. 4. 12 )
states that during the interaction the initial x - component of the relative velocity is equal
to the negative of the final x - component of the relative velocity
v x , i
rel = − v x , f
rel
. ( 0. 4. 14 )
Consequently the initial and final relative speeds are equal. We can now solve for the
final x - component of the velocities, v 1 x , f and v 2 x , f , as follows. Eq. ( 0. 4. 12 ) may be
rewritten as
v 2 x , f = v 1 x , f
Now substitute Eq. ( 0. 4. 15 ) into Eq. ( 0. 4. 6 ) yielding
m 1 v 1 x , i
- m 2 v 2 x , i = m 1 v 1 x , f
- m 2 ( v 1 x , f
- v 1 x , i − v 2 x , i
Solving Eq. ( 0. 4. 16 ) for v 1 x , f involves some algebra and yields
v 1 x , f
m 1 − m 2
m 1
v 1 x , i
2 m 2
m 1
v 2 x , i
To find v 2 x , f , rewrite Eq. ( 0. 4. 12 ) as
v 1 x , f = v 2 x , f − v 1 x , i
Now substitute Eq. ( 0. 4. 18 ) into Eq. ( 0. 4. 6 ) yielding
m 1 v 1 x , i
- m 2 v 2 x , i = m 1 ( v 2 x , f − v 1 x , i
- v 2 x , i ) v 1 x , f
- m 2 v 2 x , f
We can solve Eq. ( 0. 4. 19 ) for v 2 x , f and determine that
v ′ 1 x , i = v 1 x , i − v x ,cm = ( v 1 x , i − v 2 x , i
m 2
m 1
v ′ 2 x , i = v 2 x , i − v x ,cm = ( v 2 x , i − v 1 x , i
m 1
m 1
In the CM frame the momentum of the system is zero before the collision and hence the
momentum of the system is zero after the collision. For an elastic collision, the only way
for both momentum and kinetic energy to be the same before and after the collision is
either the objects have the same velocity (a miss) or to reverse the direction of the
velocities as shown in Figure 15.6.
Figure 15.6 One-dimensional elastic collision in center of mass reference frame
In the CM frame, the final x - components of the velocities are
v 1 ′ x , f = − v 1 ′ x , i = ( v 2 x , i − v 1 x , i )
m 2
m 1 + m 2
v ′ 2 x , f = − v ′ 2 x , i = ( v 2 x , i − v 1 x , i
m 1
m 1
The final x - components of the velocities in the “laboratory frame” are then given by
v 1 x , f = v ′ 1 x , f
= ( v 2 x , i − v 1 x , i
m 2
m 1
m 1 v 1 x , i
m 1
= v 1 x , i
m 1 − m 2
m 1 + m 2
2 m 2
m 1 + m 2
as in Eq. ( 0. 4. 17 ) and a similar calculation reproduces Eq. ( 0. 4. 20 ).
15.5 Worked Examples
Example 15.1 Elastic One-Dimensional Collision Between Two Objects
Consider the elastic collision of two carts along a track; the incident cart 1 has mass m 1
and moves with initial speed v 1 , i
. The target cart has mass m 2 = 2 m 1 and is initially at
rest, v 2 , i = 0. Immediately after the collision, the incident cart has final speed v 1 , f and the
target cart has final speed v 2 , f
. Calculate the final x - component of the velocities of the
carts as a function of the initial speed v 1 , i
Solution Draw a “momentum flow” diagram for the objects before (initial state) and after
(final state) the collision (Figure 15.7).
Figure 15.7 Momentum flow diagram for elastic one-dimensional collision
We can immediately use our results above with m 2 = 2 m 1 and v 2 , i = 0. The final x -
component of velocity of cart 1 is given by Eq. ( 0. 4. 17 ), where we use v 1 x , i = v 1 , i
v 1 x , f = −
v 1 , i. ( 0. 4. 29 )
The final x - component of velocity of cart 2 is given by Eq. ( 0. 4. 20 )
v 2 x , f
v 1 , i
Example 1 5.2 The Dissipation of Kinetic Energy in a Completely Inelastic Collision
Between Two Objects
An incident object of mass m 1 and initial speed v 1 , i collides completely inelastically with
an object of mass m 2 that is initially at rest. There are no external forces acting on the
objects in the direction of the collision. Find Δ K / K initial = ( K final − K initial ) / K initial.
unknown angle θ 2 , f (as shown in the Figure 15.8). Find the final speeds of each of the
objects and the angle θ 2 , f
Figure 15.8 Momentum flow diagram for two-dimensional elastic collision
Solution: The components of the total momentum
p i
sys = m 1
v 1 , i
v 2 , i in the initial state
are given by
p x , i
sys = m 1 v 1 , i
p y , i
sys = 0.
The components of the momentum
p f
sys = m 1
v 1 , f
v 2 , f in the final state are given by
p x , f
sys = m 1 v 1 , f cos θ 1 , f
p y , f
sys = m 1 v 1 , f sin θ 1 , f − m 1 v 2 , f sin θ 2 , f
There are no any external forces acting on the system, so each component of the total
momentum remains constant during the collision,
px ,i
sys = px , f
sys ( 0. 4. 37 )
py , i
sys = py , f
sys
. ( 0. 4. 38 )
Eqs. ( 0. 4. 37 ) and ( 0. 4. 37 ) become
m 1 v 1 , i = m 1 v 1 , f cos θ 1 , f
0 = m 1 v 1 , f sin θ 1 , f − m 1 v 2 , f sin θ 2 , f
The collision is elastic and therefore the system kinetic energy of is constant
K
i
sys = K f
sys
. ( 0. 4. 40 )
Using the given information, Eq. ( 0. 4. 40 ) becomes
m 1 v 1 , i
2
m 1 v 1 , f
2
m 1 v 2 , f
2
. ( 0. 4. 41 )
We have three equations: two momentum equations and one energy equation, with three
unknown quantities, v 1 , f , v 2 , f and^ θ 2 , f because we are given v 1 , i =^ 3.0^ m^ ⋅s
− 1 and
θ 1 , f
o
. We first rewrite the expressions in Eq. ( 0. 4. 39 ), canceling the factor of m 1 , as
v 2 , f cos θ 2 , f = v 1 , i − v 1 , f cos θ 1 , f
v 2 , f sin θ 2 , f = v 1 , f sin θ 1 , f
We square each expressions in Eq. ( 0. 4. 42 ), yielding
v 2 , f
2 (cos θ 2 , f
2 = v 1 , f
2 − 2 v 1 , i v 1 , f cos θ 1 , f
2 (cos θ 1 , f
2 ( 0. 4. 43 )
v 2 , f
2 (sin θ 2 , f
2 = v 1 , f
2 (sin θ 1 , f
2 ( 0. 4. 44 )
We now add together Eqs. ( 0. 4. 43 ) and ( 0. 4. 44 ) yielding
v 2 , f
2 (cos
2 θ 2 , f
2 θ 2 , f ) = v 1 , f
2 − 2 v 1 , i v 1 , f cos θ 1 , f
2 (cos θ 1 , f
2 θ 1 , f
We can use identity cos
2 θ + sin
2 θ = 1 to simplify Eq. ( 0. 4. 45 ), yielding
v 2 , f
2 = v 1 , i
2 − 2 v 1 , i v 1 , f cos θ 1 , f
2
. ( 0. 4. 46 )
Substituting Eq. ( 0. 4. 46 ) into Eq. ( 0. 4. 41 ) yields
m 1 v 1 , i
2
m 1 v 1 , f
2
m 1 ( v 1 , i
2 − 2 v 1 , i v 1 , f cos θ 1 , f
2 ). ( 0. 4. 47 )
Eq. ( 0. 4. 47 ) simplifies to
0 = 2 v 1 , f
2 − 2 v 1 , i v 1 , f cos θ 1 , f
which may be solved for the final speed of object 1,
p i
sys
p f
sys , ( 0. 4. 54 )
which becomes
m 1
v 1 , i = m 1
v 1 , f + m 1
v 2 , f. ( 0. 4. 55 )
Eq. ( 0. 4. 55 ) simplifies to v 1 , i
v 1 , f
v 2 , f
Recall the vector identity that the square of the speed is given by the dot product
2 v ⋅ v = v
. With this identity in mind, we take the dot product of each side of Eq. ( 0. 4. 56 )
with itself,
v 1 , i
v 1 , i
v 1 , f
v 2 , f
v 1 , f
v 2 , f
v 1 , f
v 1 , f
v 1 , f
v 2 , f
v 2 , f
v 2 , f
This becomes
v 1 , i
2 = v 1 , f
2
v 1 , f
v 2 , f
2
. ( 0. 4. 58 )
Recall that kinetic energy is the same before and after an elastic collision, and the masses
of the two objects are equal, so Eq. ( 0. 4. 41 ) simplifies to
v 1 , i
2 = v 1 , f
2
2
. ( 0. 4. 59 )
Comparing Eq. ( 0. 4. 58 ) with Eq. ( 0. 4. 59 ), we see that
v 1 (^) , f ⋅ v 2 , f = 0
The dot product of two nonzero vectors is zero when the two vectors are at right angles to
each other justifying our claim that the collision particles emerge at right angles to each
other.
Example 15.5 Bouncing Superballs
Two superballs are dropped from a height h i above the ground, one on top of the other.
Ball 1 is on top and has mass m 1 , and ball 2 is underneath and has mass m 2 with
m 2
m 1
. Assume that there is no loss of kinetic energy during all collisions. Ball 2 first
collides with the ground and rebounds. Then, as ball 2 starts to move upward, it collides
with the ball 1 which is still moving downwards. How high will ball 1 rebound in the air?
Hint: consider this collision as seen by an observer moving upward with the same speed
as the ball 2 has after it collides with ground (Figure 15.9). What speed does ball 1 have
in this reference frame after it collides with the ball 2?
Solution: The system consists of the two balls and the earth. There are five special states
for this motion shown in the figure above.
Initial state: the balls are released from rest at a height h i above the ground.
State A: the balls just reach the ground with speed v 1 , A = v 2 , A = v A = 2 gh i
State B: ball 2 has collided with the ground and reversed direction with the same speed,
v 2 , B = v A , and ball 1 is still moving down with speed v 1 , B = v A
State C: because we are assuming that m 2
m 1 , ball 2 does not change speed as a result
of the collision so it is still moving upward with speed v 2 , C = vA. As a result of the
collision, ball 1 moves upward with speed v 1 , C.
Final State: ball 1 reaches a maximum height h f = v b
2 / 2 g above the ground.
Figure 15.9 States in superball collisions
Choice of Reference Frame: For states B and C, the collision is best analyzed from the
reference frame of an observer moving upward with speed v A , the speed of ball 2 just
after it rebounded from the ground. In this frame ball 1 is moving downward with a speed
v ′ 1 , B that is twice the speed seen by an observer at rest on the ground (lab reference
frame).
v ′ 1 , B = 2 v A
