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Chapter 15 Collision Theory
Example 15.6 Two-dimensional elastic collision between particles of equal mass
- 15.1 Introduction...........................................................................................................
- 15.2 Reference Frames and Relative Velocities..........................................................
- 15.2.1 Relative Velocities
- 15.2.2 Center-of-mass Reference Frame
- 15.2.3 Kinetic Energy in the Center-of-Mass Reference Frame
- 15.2.4 Change of Kinetic Energy and Relatively Inertial Reference Frames......
- 15.3 Characterizing Collisions
- 15.4 One-Dimensional Collisions Between Two Objects...........................................
- 15.4.1 One Dimensional Elastic Collision in Laboratory Reference Frame
- Reference Frame 15.4.2 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 Bouncing Superballs
- 15.6 Two Dimensional Elastic Collisions
- 15.6.1 Two-dimensional Elastic Collision in Laboratory Reference Frame......
- Example 15.5 Elastic Two-dimensional collision of identical particles..............
- Example 15.7 Two dimensional collision between particles of unequal mass
- 15.7 Two-Dimensional Collisions in Center-of-Mass Reference Frame
- 15.7.1 Two-Dimensional Collision in Center-of-Mass Reference Frame...........
- 15.7.2 Scattering in the Center-of-Mass Reference Frame
- Example 15.8 Scattering in the Lab and CM Frames
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 and Relative 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
(^1) Randall, Lisa, Knocking on Heaven's Door: How Physics and Scientific Thinking Illuminate the Universe
and the Modern World , Ecco, 2011.
15.2. 1 Relative Velocities
Consider two particles of masses m 1 and m 2 interacting via some force (Figure 15.2).
Figure 15. 2 Two interacting particles
Choose a coordinate system (Figure 15.3) 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. 3 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
d r 1 2 , d r 1 d r 2 v 1 2 , = = − = v 1 − v 2. (15.2.5) dt dt dt
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 12 ′^ in reference frame S ′ can be determined from using Eq.
(15.2.4) to express Eq. (15.2.5) in terms of the velocities in the reference frame S ′ ,
v 1 ′ v 1, 2 ′
v 1, 2 = v 1 −
v 2 = ( v
1 ′^ +^ V )^ −^ ( v
V ) = −
v ′ 2 = (15.2.6)
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.
15.2. 2 Center-of-mass Reference Frame
Let
r cm be the vector from the origin of frame S to the center-of-mass of the
system of particles, a point that we will choose as the origin of reference frame S (^) cm ,
called the center-of-mass reference frame. Denote the position vector of the (^) j th^ particle
with respect to origin of reference frame S by
r j and similarly, denote the position
vector of the j th^ particle with respect to origin of reference frame S (^) cm by
r ′ j (Figure
15.4).
S (^) cm
r cm
r j
j th^ particle r j
S!
Figure 15.4 Position vector of j th^ particle in the center-of-mass reference frame.
The position vector of the j^ th particle in the center - of - mass frame is then given by
r ′ j = r j r cm. (15.2.7)
The velocity of the j th^ particle in the center-of-mass reference frame is then given by
−
v ′ j = v (^) j v (^) cm. (15.2.8)
There are many collision problems in which the center-of-mass reference frame is the most convenient reference frame to analyze the collision.
Consider a system consisting of two particles, which we shall refer to as particle 1 and particle 2. We can use Eq. (15.2.8) to determine the velocities of particles 1 and 2 in the center-of-mass,
μ v 1 ′ −
v (^) cm = v 1 −
m 1 v 1 + m 2 v 2 m 2
m 1 + m 2 m 1 + m 2
v 1, −
v 2 ) = m 1
v 1, 2.
= v 1 (15.2.9)
KS = m 1 ( v 1 ′ + v (^) cm ) ⋅( v 1 ′ + v (^) cm ) + m 2 ( v ′ 2 + v (^) cm ) ⋅( v ′ 2 + v (^) cm ) (^2 2). (15.2.16) 1 1 1 = m 1 v 1 ′ 2 + m 2 v 2 ′ 2 + ( m 1 + m 2 ) v (^) cm^2^ + ( m 1 v 1 ′ + m 2 v ′ 2 ) ⋅ v cm 2 2 2
The last term is zero due to the fact that the momentum of the system in the center of mass reference frame is zero (Eq. (15.2.11)). Therefore Eq. (15.2.16) becomes
KS = m 1 v 1 ′ 2 + m 2 v 2 ′ 2 + ( m 1 + m 2 ) v (^) cm. (15.2.17) 2 2 2
The first two terms correspond to the kinetic energy in the center of mass frame, thus the kinetic energies in the two reference frames are related by
KS = K +
cm ( m 1 +^ m 2 ) v^ cm^2.^ (15.2.18) 2
We now use Eq. (15.2.13) to rewrite Eq. (15.2.18) as
KS = μ v 1, 2 2 + ( m 1 + m 2 ) v (^) cm (15.2.19) 2 2
Even though kinetic energy is a reference frame dependent quantity, because the second term in Eq. (15.2.19) is a constant, the change in kinetic energy in either reference frame is equal to
(^1 2 )
Δ K = μ (( v 1, 2 ) − ( v 1, 2) ). (15.2.20)
2 f^ i
This generalizes to any two relatively inertial reference frames because the relative velocity is a reference frame independent quantity,
the change in kinetic energy is independent of the choice of relatively inertial reference frames.
We showed in Appendix 13A that when two particles of masses m 1 and m 2 interact, the
work done by the interaction force is equal to
W = μ (( v 1, 2 ) − ( v 1, 2 ) ). (15.2.21)
2 f^ i
Hence we explicitly verified that for our two-particle system
W = Δ Ksys. (15.2.22)
1 5.3 Characterizing Collisions
In a collision, the ratio of the magnitudes of the initial and final relative velocities is called the coefficient of restitution and denoted by the symbol e ,
v (^) B e =. (15.2.23) v (^) A
If the magnitude of the relative velocity does not change during a collision, e = 1 , then the change in kinetic energy is zero, (Eq. (15.2.21)). Collisions in which there is no change in kinetic energy are called elastic collisions ,
Δ K = 0, elastic collision. (15.2.24)
If the magnitude of the final relative velocity is less than the magnitude of the initial relative velocity, e < 1 , then the change in kinetic energy is negative. Collisions in which the kinetic energy decreases are called inelastic collisions ,
Δ K < 0, inelastic collision. (15.2.25)
If the two objects stick together after the collision, then the relative final velocity is zero, e = 0. Such collisions are called totally inelastic. The change in kinetic energy can be found from Eq. (15.2.21),
(^1) 2 1 m 1 m 2 2 Δ K = − μ vA = − vA , totally inelastic collision. (15.2.26) 2 2 m 1 + m 2
If the magnitude of the final relative velocity is greater than the magnitude of the initial relative velocity, e > 1 , then the change in kinetic energy is positive. Collisions in which the kinetic energy increases are called superelastic collisions ,
Δ K > 0, superelastic collision. (15.2.27)
15.4 One-Dimensional Collisions Between Two Objects
15.4.1 One Dimensional Elastic Collision in Laboratory Reference Frame
Consider a one-dimensional elastic collision between two objects moving in the x -
direction. One object, with mass m 1 and initial x -component of the velocity v 1 x , i ,
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
where we used the superscript “rel” to remind ourselves that the velocity is a relative
velocity (and to simplify our notation). Thus vx^ rel , i =^ v 1 x , i −^ v 2 x , i is the initial x^ -component
of the relative velocity, and vx^ rel , f =^ v 1 x , f −^ v 2 x , f is the final x -component of the relative
velocity. Therefore Eq. (15.3.7) states that during the interaction the initial relative velocity is equal to the negative of the final relative velocity
(^) rel rel v i = − v
f ,^ (1−^ dimensional energy-momentum prinicple)^.^ (15.3.9)
Consequently the initial and final relative speeds are equal. We shall call this relationship between the relative initial and final velocities the one-dimensional energy-momentum principle because we have combined these two principles to realize this result. The energy-momentum principle is independent of the masses of the colliding particles.
Although we derived this result explicitly, we have already shown that the change in kinetic energy for a two-particle interaction (Eq. (15.2.20)), in our simplified notation is given by
(^1) rel Δ K = μ(( v ) (^2) f^ − ( v^ rel^ ) i^2 ) (15.3.10) 2
Therefore for an elastic collision where Δ K = 0 , the square of the relative speed remains constant rel (^) ) 2 f
rel (^) ) 2 ( v = ( v (^) i. (15.3.11)
For a one-dimensional collision, the magnitude of the relative speed remains constant but
the direction changes by 180 ^.
We can now solve for the final x -component of the velocities, v 1 x , f and v 2 x , f , as
follows. Eq. (15.3.7) may be rewritten as
v 2 x , f = v 1 x , f + v 1 x , i − v 2 x , i. (15.3.12)
Now substitute Eq. (15.3.12) into Eq. (15.3.1) 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 ) (^). (15.3.13)
Solving Eq. (15.3.13) for v 1 x , f involves some algebra and yields
m 1 − m 2 2 m 2 v 1 x , f = v 1 x , i + v 2 x , i. (15.3.14) m 1 + m 2 m 1 + m 2
To find v 2 x , f , rewrite Eq. (15.3.7) as
v 1 x , f = v 2 x , f − v 1 x , i + v 2 x , i. (15.3.15)
Now substitute Eq. (15.3.15) into Eq. (15.3.1) 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. (15.3.16)
We can solve Eq. (15.3.16) for v 2 x , f and determine that
m 2 − m 1 2 m 1 v 2 x , f = v 2 x , i + v 1 x , i. (15.3.17) m 2 + m 1 m 2 + m 1
Consider what happens in the limits m 1 >> m 2 in Eq. (15.3.14). Then
v 1 x , f → v 1 x , i +
m 2 v 2 x , i ; (15.3.18) m 1
the more massive object’s velocity component is only slightly changed by an amount proportional to the less massive object’s x -component of momentum. Similarly, the less massive object’s final velocity approaches
v 2 x , f → − v 2 x , i + 2 v 1 x , i = v 1 x , i + v 1 x , i − v 2 x , i. (15.3.19)
We can rewrite this as
v 2 x , f − v 1 x , i = v 1 x , i − v 2 x , i = v rel x , i^. (15.3.20)
i.e. the less massive object “rebounds” with the same speed relative to the more massive object which barely changed its speed.
If the objects are identical, or have the same mass, Eqs. (15.3.14) and (15.3.17) become
v 1 x , f = v 2 x , i , v 2 x , f = v 1 x , i ; (15.3.21)
the objects have exchanged x -components of velocities, and unless we could somehow distinguish the objects, we might not be able to tell if there was a collision at all.
v 1 x , f = v 1 ′ x , f + vx ,cm
m 2 m 1 v 1 x , i +^ m 2 v 2 x , i = ( v 2 x , i − v 1 x , i ) + (15.3.25) m 1 + m 2 m 1 + m 2 m 1 − m 2 2 m 2 = v 1 x , i + v 2 x , i m 1 + m 2 m 1 + m 2
as in Eq. (15.3.14) and a similar calculation reproduces Eq. (15.3.17).
15.5 Worked Examples
Example 15.1 Elastic One-Dimensional Collision Between Two Objects
ˆ i
1 2
ˆ i v 1, i = v 1, x , i v 2, i = (^0) initial state m 2 = 2 m 1
v ˆ i^ ˆ i final state ˆ 1,^ f^ =^ v 1, x ,^ f^ v 2,^ f^ =^ v 2, x ,^ f i m 2 = 2 m 1
1 2
Figure 15.7 Elastic collision between two non-identical carts
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 , (Figure 15.7). 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 The momentum flow diagram for the objects before (initial state) and after (final state) the collision are shown in Figure 15.7. 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. (15.3.14), where we use v 1 x , i = v 1, i
v 1 x , f = −
v 1, i. ( 15. 4. 1 )
The final x - component of velocity of cart 2 is given by Eq. ( 15. 3. 17 )
v 2 x , f =
v 1, i. ( 15. 4. 2 )
Example 1 5.2 The Dissipation of Kinetic Energy in a Completely Inelastic Collision Between Two Objects
ˆ i v 1, i v 2, i = 0 initial state 1 2
ˆ i final state v^ f
1 2
Figure 15.7b Inelastic collision between two non-identical carts
An incident cart of mass m 1 and initial speed v 1, i collides completely inelastically with a
cart of mass m 2 that is initially at rest (Figure 15.7b). 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.
Solution: In the absence of any net force on the system consisting of the two carts, the momentum after the collision will be the same as before the collision. After the collision the carts will move in the direction of the initial velocity of the incident cart with a common speed vf found from applying the momentum condition
m 1 v 1, i = ( m 1 + m 2 ) vf ⇒ m 1 (^15.^4.^3 ) vf = v 1, i. m 1 + m 2
The initial relative speed is vi^ rel^ = v 1, i. The final relative velocity is zero because the carts
stick together so using Eq. ( 15. 2. 26 ), the change in kinetic energy is
Δ K = −
μ( vi^ rel^ )^2 = − 1 m 1 m 2 v 1,^2 i. ( 15. 4. 4 ) 2 m 1 + m 2
Solution
The system consists of the two balls and the earth. There are five special states for this motion shown in the figure below.
part a)
Initial State: the balls are released from rest at a height hi above the ground.
State A: the balls just reach the ground with speed va = 2 ghi. This follows from
Δ Emech = 0 ⇒ Δ K = −Δ U. Thus (1 / 2) mv^2 a^ − 0 = − mg Δ h = mghi ⇒ v (^) a = 2 ghi.
State B: immediately before the collision of the balls. Ball 2 has collided with the ground and reversed direction with the same speed, va , but ball 1 is still moving downward with
speed va.
State C: immediately after the collision of the balls. Because we are assuming that
m 2 >> m 1 , ball 2 does not change its speed as a result of the collision so it is still moving
upward with speed va. As a result of the collision, ball 1 moves upward with speed vb.
Final State: ball 1 reaches a maximum height hf = vb^2 / 2 g above the ground. This again
follows from Δ K = −Δ U ⇒ 0 − (1 / 2) mvb^2^ = − mg Δ h = − mgh (^) f ⇒ h (^) f = vb^2 / 2 g.
Choice of Reference Frame:
As indicated in the hint above, this collision is best analyzed from the reference frame of
an observer moving upward with speed va , the speed of ball 2 just after it rebounded with
the ground. In this frame immediately, before the collision, ball 1 is moving downward
with a speed vb ′ that is twice the speed seen by an observer at rest on the ground (lab
reference frame).
va ′ = 2 va (15.4.7)
The mass of ball 2 is much larger than the mass of ball 1, m 2 >> m 1. This enables us to
consider the collision (between States B and C) to be equivalent to ball 1 bouncing off a hard wall, while ball 2 experiences virtually no recoil. Hence ball 2 remains at rest in the
reference frame moving upwards with speed va with respect to observer at rest on
ground. Before the collision, ball 1 has speed va ′ = 2 va. Since there is no loss of kinetic
energy during the collision, the result of the collision is that ball 1 changes direction but maintains the same speed,
vb ′ = 2 v (^) a. (15.4.8)
However, according to an observer at rest on the ground, after the collision ball 1 is moving upwards with speed
vb = 2 v (^) a + v (^) a = 3 v (^) a. (15.4.9)
While rebounding, the mechanical energy of the smaller superball is constant (we consider the smaller superball and the Earth as a system) hence between State C and the Final State,
Δ K + Δ U = 0. (15.4.10)
The change in kinetic energy is
1 Δ K = − m 1 (3 va )^2. (15.4.11) 2
The change in potential energy is Δ U = m 1 g hf. (15.4.12)
So the condition that mechanical energy is constant (Equation (15.4.10)) is now
m 1 (3 v 1 a )^2 + m 1 g hf = 0. (15.4.13) 2
We can rewrite Equation (15.4.13) as 1 m 1 g hf = 9 m 1 ( v (^) a )^2. (15.4.14) 2
of each of these vectors, v 1, f , v 2, f , θ1, f , and θ2, f. These quantities are related by the two
equations describing the constancy of momentum, and the one equation describing constancy of the kinetic energy. Therefore there is one degree of freedom that we must specify in order to determine the outcome of the collision. In what follows we shall express our results for v 1, f , v 2, f , and θ2, f in terms of v 1, i and θ1, f.
The components of the total momentum
p sys i = m 1
v 1, i + m 2
v 2, i in the initial state are given
by
p sys x , i = m 1 v 1, i sys^ (15.5.1) py , i = 0.
The components of the momentum
p sys f = m 1
v 1, f + m 2
v (^) 2, f in the final state are given by
p^ sys v x , f =^ m 1 1, f cos^ θ1, f +^ m 2 v 2, f cos^ θ2, f (15.5.2) p sys y , f^ = m 1 v 1, f sin θ1, f − m 2 v 2, f sin θ2, f.
There are no any external forces acting on the system, so each component of the total momentum remains constant during the collision,
p^ sys^ = sys (15.5.3) x ,i px , f p^ sys^ = p sys. (15.5.4) y , i y , f Eqs. (15.5.3) and (15.5.4) become
m 1 v 1, i = m 1 v 1, f cos θ1, f + m 2 v 2, f cos θ2, f , (15.5.5) 0 = m 1 v 1, f sin θ1, f − m 2 v 2, f sin θ2, f. (15.5.6)
The collision is elastic and therefore the system kinetic energy of is constant
K^ sys^ sys i =^ K^ f.^ (15.5.7)
Using the given information, Eq. (15.5.7) becomes
m 1 v 1, i = m 1 v 1, f + m 2 v 2, f. (15.5.8) 2 2 2
Rewrite the expressions in Eqs. (15.5.5) and (15.5.6) as
m 2 v 2, f cos θ2, f = m 1 ( v 1, i − v 1, f cos θ1, f ), (15.5.9)
m 2 v 2, f sin θ2, f = m 1 v 1, f sin θ1, f. (15.5.10)
Square each of the expressions in Eqs. (15.5.9) and (15.5.10), add them together and use
the identity cos^2 θ + sin^2 θ = 1 yielding
2 m 1
2 = 2 − 2 v^2 v 2, f (^) m 2 ( v 1, i 1, iv 1, f cos θ1, f + v 1, f ). (15.5.11) 2
Substituting Eq. (15.5.11) into Eq. (15.5.8) yields
1 2 1 2 1 m 12 2 2 m 1 v 1, i = m 1 v 1, f + ( v 1, i − 2 v 1, i v 1, f cos θ1, f + v 1, f ). (15.5.12) 2 2 2 m 2
Eq. (15.5.12) simplifies to
⎛ (^) m 1 ⎞ (^2) m 1 ⎛ (^) m 1 ⎞ 2 0 = 1 + ⎠^ ⎟^
v 1, f − 2 v 1, i v 1, f cos θ1, f − 1 − ⎠^ ⎟^
v 1, i , (15.5.13) ⎝^ ⎜^ m 2 m 2 ⎝^ ⎜^ m 2
Let α = m 1 / m 2 then Eq. (15.5.13) can be written as
0 = (1+ α ) v 1,^2 f − 2 α v 1, i v 1, f cos θ1, f − (1−α ) v 1,^2 i , (15.5.14)
The solution to this quadratic equation is given by
2 2 1/ α v 1, i cos θ1, f ± ( α 2 v 1, i cos^2 θ1, f + (1−α ) v 1, i ) v 1, f =. (15.5.15) (1+ α )
Divide the expressions in Eq. (15.5.9), yielding
v 2, f sin θ2, f v 2, f cos θ2, f
v 1, f sin θ1, f v 1, i − v 1, f cos θ1, f
Eq. (15.5.16) simplifies to
tan θ2, f =
v 1, f sin θ1, f v 1, i − v 1, f cos θ1, f
The relationship between the scattering angles in Eq. (15.5.17) is independent of the masses of the colliding particles. Thus the scattering angle for particle 2 is
