MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
There are many problems involving conservation of momentum in which the analysis is easier depending on your choice of reference frame. In the notes that follow we describe how velocities of objects are related in different reference frames and in particular describe the center of mass reference frame.
Let R be the vector from the origin of a reference frame S to the origin of a second reference frame S

. Denote the position vector of particle i with respect to the origin of reference frame S by r and similarly, denote the position vector of particle i i with respect to the origin of reference frame S

by r
 i
(Figure 1).
Figure 1 Position vector of i th
particle in two reference frames.
The position vectors are related by r i
  i

R .
The relative velocity between the two reference frames is given by
(1)
V
 d R dt
. (2)
When the relative velocity between the two reference frames is constant then the relative acceleration between the two reference frames is zero,
A
 d V dt

0 , (3)

When Eq. (3) is satisfied, the reference frames S and S
 reference frames.
are called relatively inertial
Suppose the i th
particle in Figure 1 is moving; then observers in different reference frames will measure different velocities. Denote the velocity of frame
r v i

S by
 d r r

/ dt
r v i
 d r r i
/ dt
. Since the derivative of the position is velocity, the velocities of the particles in two different reference frames are related according to i th
particle in
, and the velocity of the same particle in frame S
 by v i v
 i
V . (4)
This relation is called the law of addition of velocities and is the primary tool that we will use when we describe velocities in different reference frames.
Example: velocity
Suppose a cart has velocity reference frame fixed to the ground and let
V

V i
ˆ
r
V

V
ˆ i with respect to the ground. Let S denote a
S

denote the reference frame moving with
with respect to the ground. In thrown from the cart with velocity
r u
   u ball relative to the ground reference frame
S
ˆ i relative to the cart. Then the velocity of the
is
S

the cart is at rest. Suppose aball is
r u
 r u
  r
V
  u
ˆ i

V
ˆ i

( V
 u )ˆ i . (5)
Center of Mass Reference Frame
Consider a system of N particles. Let
r
R cm
be the vector from the origin of a r reference frame S to the center of mass of the system of particles. Let
r j
denote the position of the
j th particle. Recall the definition of the position of the center of mass
r
R cm
 j

N  j

1 j j

 N

1 m j r r j m j

1 m
T j j

N  m j

1 r r j
(6) where
 j

 N j

1 m j
is the mass of the system. The velocity of the center of mass is given m
T by
r
V cm

1 m
T j j

N 

1 m j r v j
(7)

where
r v j
is the velocity of the
j th particle with respect to reference frame S . Choose the center of mass to be the origin of reference frame S cm
, called the center of mass reference frame . Denote the position vector of particle i with respect to origin of reference frame S by
r r i
and similarly, denote the position vector of particle i with respect to origin of reference frame S cm
by
r r cm , i
(Figure 2).
Figure 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 r cm , i
 r r i
 r
R cm
.
The velocity of particle i in the center of mass reference frame is then given by
r v cm , i
 r v i
 r
V cm
.
(8)
(9)

IC_W08D3-1 Group Problem: Railroad Gun
A railroad gun of mass the horizontal as measured in a reference frame moving with the final velocity of the gun r
v
1 f m
1
fires a shell of mass m
2
at an angle of
after the shell has been fired.
After the firing is complete, the final speed of the projectile relative to the gun (muzzle velocity) is r v
2

 with respect to
(with speed
u
 r v

2
).
The gun recoils with speed
v f
 r v
1 f
and the instant the projectile leaves the gun, it makes an angle  with respect to the ground.
Solution:
Solution: a) What is the velocity, r v
2
, of the projectile with respect to the ground?
The projectile is shot at an angle  with respect to the horizontal as measured relative to the gun. Therefore the velocity of the projectile relative to the gun is given by
 ˆ i
 u sin
 ˆ j .
After the firing is complete, the gun is moving relative to the ground with velocity
r v
1 f
  v f i
ˆ
. The velocity , r v
2
, of the projectile with respect to the ground is given by
r v
2
 r v
2
  r v
1 f
 u cos
r v

2
 ˆ
 u cos i
 u sin
 ˆ j
 v f
ˆ i

( u cos
  v f
)ˆ i
 u sin
 ˆ j .
(10)
(11) b) Find v f
and  in terms of m
1
, m
2
,
u
, and  .
The final momentum of the projectile according to the observer on the ground is
r p
2
 m
2 r v
2
 m
2
(( u cos
  v f
)ˆ i
 u sin
 ˆ j ) . (12)

There are no external forces in the
x -direction, so the system is constant. Therefore we have the
x
-component of the momentum of
0
 p x ,1
 p x ,2
  m
1 v f
 m
2
( u cos
  v f
) . (13)
We can solve Equation
(13)
for the final speed of the gun as measured by the observer on the ground, v f
 m
2 u cos
 m
1
 m
2
. (14)
Therefore we can find the final velocity of the ball relative to the ground by substituting
Eq. into Eq. (11) yielding
r v
2
 m
1 u cos m
1
 m
2

ˆ i
 u sin
 ˆ j . (15)
Note that the
y
-component of the momentum is not constant because as the gun fires the projectile the normal force that the ground exerts on the gun exceeds the gravitational force, and so the net external force in the y
-direction is non-zero.
The angle
 that the projectile makes with ground as seen by the observer on the ground is given by tan
  tan

1
( v y ,2
/ v x ,2
)
 tan

1


( m
1
 m
2
) u sin m
1 u cos

 

. (16)
From Equation (16),
 
0 ; the observer on the ground sees the projectile moving closer to the vertical. Notice that this result is independent of the muzzle speed u .

IC_W08D3-2 Group Problem Helium-Lithium Collision:
This problem is on pset 7 and is a difficult problem so we will get you started today.
A thin target of lithium is bombarded by helium nuclei of energy E
0
. The lithium nuclei are initially at rest in the target but are essentially unbound. When a helium nucleus enters a lithium nucleus, a nuclear reaction can occur in which the compound nucleus splits apart into a boron nucleus and a neutron. The collision is inelastic, and the final kinetic energy is less than E
0
by 2.8 MeV . (
6  -13
1 MeV=10 eV=1.6 10 J ). The relative masses of the particles are: helium, mass 4 ; lithium, mass 7 ; boron mass 10 ; neutron, mass 1 . The reaction can be symbolized
7 4
Li + He
 10
B
 1 n

2.8 MeV .
Let’s consider only the outgoing particles, boron and neutrons that are moving along the same line as the initial helium nucleus, (call this the x -direction). a) Draw momentum diagrams for the initial and final states in a reference frame in which the target lithium nucleus is initially at rest. b) What is the center of mass velocity of the system of lithium and helium? Does this velocity change due to the collision?
Answer:
r
V cm

( m
He r v m
He
He
 m
Li r v
Li
 m
Li
)
 m
He m
He r v )
He
 m
Li

4
4 mv m

He
ˆ
7 i m

4
11 v
He
ˆ i
There are no external forces so the velocity of the center of mass remains constant. e) Draw momentum diagrams for the initial and final states in a reference frame moving with the velocity of center of mass. What are the velocities of the helium nucleus and the lithium in the center of mass reference frame? What is the momentum of the system in the center of mass reference frame?
Answer: The velocity of the helium in the center of mass reference frame is given by

r v cm , He
 r v
He
 r
V cm
 r v
He

4
11 r v
He

7
11 r v
He
.
The velocity of the lithium in the center of mass reference frame is given by
r v cm , Li
 r v
Li
 r
V cm
 
4
11 r v
He
.
The momentum of the system in the center of mass reference frame is zero.
r p cm , He
 r p cm , Li

4 m
7
11 r v
He

7 m
4
11 r v
He
 r
0 .
Below threshold energy necessary to create the boron and neutron the helium and lithium just bounce off. The momentum diagrams for the initial and final states in a reference frame moving with the velocity of center of mass for that case are shown below. c) The minimum initial kinetic energy necessary for the reaction to take place is called the threshold energy, E
0,threshold
. Draw a momentum diagram for the initial and final states in the center of mass reference frame when the initial kinetic energy of the incident particles is at the threshold energy (keep in mind the lithium target is also moving in the center of mass reference frame). Hint: What are the velocities of the outgoing particles at threshold?
Answer. At the threshold the energy is just enough to create the boron and the neutron so they are at rest in the center of mass reference frame.

d) Write down equations for conservation of energy and momentum in the center of mass frame when the initial kinetic energy of the incident particles is at the threshold energy. What is the threshold energy in the center of mass reference frame?
Answer: Momentum is zero.
r
0
 r p cm , He
 r p cm , Li
 r p cm , B
 r p cm , n
 r
0 .
The boron and neutron are at rest at threshold energy. The incident kinetic energy is just equal to the energy necessary to create the boron and neutron which is 2.8 MeV .
1
2
2
(4 m ) v cm , He

1
2
2
(7 m ) v cm , Li
E
0, Threshold

2.8 MeV . e) Draw momentum diagrams for the initial and final states in the center of mass reference frame when the initial kinetic energy of the incident particles lies in the range E
0,threshold

E
0

E
0,threshold

0.27 MeV . How many possible final states are there? (Hint: Which way is the neutron moving?)
Answer: The momentum diagrams for the initial and final states in the center of mass reference frame when the boron and neutron have non-zero kinetic energy above threshold are shown below. There are two possible final states corresponding to the two possible directions of the neutron and boron. f) Draw momentum diagrams for the initial and final states in which the target is initially at rest for the case of part e). How many possible final states are there?
(Hint: Which way is the neutron moving?)

Answer: There are two final states: state (a) corresponding to a slow neutron moving in the positive x -direction; and state (b) corresponding to a slow neutron moving in the positive x -direction. g) What is the initial kinetic energy of the helium in this reference frame at threshold? How many possible final states are there? (Hint: Which way is the neutron moving?)
Answer: We have already determined that v cm , He

(7 / 11) v
He
and v cm , Li

(4 / 11) v
He
Therefore the initial kinetic energy in the center of mass reference frame at threshold is
.
K cm , i

1
2

1
2
2
(4 m ) v cm , He
2
(4 m ) v
He


49
121


28
1
2

121

2
(7 m ) v cm , Li

1
2

(4 m ) v
2
He
1
2
(4
7
11 m )
49
121
2 v
He

Note that the incident kinetic energy in lab frame is
1
2
16
(7 m )
121
2 v
He
.
K lab , i

1
2
(4 m ) v 2
He
.
Therefore
K cm , i

1
2
2
(4 m ) v
He
7
11

7
11
K lab .
i

2.8 MeV
We set this equal to the threshold energy of 2.8 MeV in order to determine the incident kinetic energy at threshold in the lab reference frame

Thus
7
11
K lab .
i

2.8 MeV .
K lab .
i

4.4 MeV; (at threshold)