Physics I
Class 9
Conservation of Momentum
in Two and Three Dimensions
Rev. 11-Aug-03 GB
09-1
Class #8
Take-Away Concepts (Review)
1.
2.
Systems; internal/external forces in systems.
Momentum defined for a system:
 N 
P   pi
i 1
3.
Newton’s
 Second Law for a system:

dP
  Fext
dt
4. Conservation
of momentum when


dP
  Fext  0
dt
Pafter  Pbefore
09-2
Conservation of Momentum in
Multiple Dimensions
 Each direction of motion is independent.
 Conservation of momentum occurs (or not) separately in each direction.
d Px
  Fext , x
dt
d Py
dt
  Fext , y
d Pz
  Fext ,z
dt
09-3
Collisions in Multiple Dimensions
Y
X
Before:
After:
Px ,before  m1v1, x ,before  m 2 v 2, x ,before
Px ,after  m1v1, x ,after  m 2 v 2, x ,after
Py ,before  m1v1, y ,before  m 2 v 2, y ,before
Py ,after  m1v1, y ,after  m 2 v 2, y ,after
09-4
Solving Multi-Dimensional
Two-Body Collision Problems
That sounds complicated. Even with only two bodies in two dimensions,
there are 8 components of velocity to consider.
However:
 Momentum is conserved (or not) in each direction separately.
 Conserved directions of momentum do not mix with other directions.
So all we really need to do is to solve two one-dimensional problems.
If conserved:
m1v1, x ,before  m 2 v 2, x ,before  m1v1, x ,after  m 2 v 2, x ,after
m1v1, y ,before  m 2 v 2 , y ,before  m1v1, y ,after  m 2 v 2 , y ,after
09-5
Center of Mass
Center of mass defined for a system:
N
M   mi
i 1
x cm
y cm
1 N
  mi x i
M i 1
1 N
  mi yi
M i 1
09-6
Center of Mass
Example
Y
6
x cm
m = 4 kg
center of mass
2
y cm
m = 2 kg
1

4  2  2  8   4
42
1

4  6  2  2   4 2
42
3
X
2
8
09-7
Velocity of the Center of Mass
System Momentum
Velocity of center of mass:

v cm

v cm


N
d x cm 1
d xi 1 N
1 N 


  mi
  mi vi   pi
dt
M i 1
d t M i 1
M i 1

N


1
P

P

M
v
  pi 
or
cm
M i 1
M
If the momentum of a system is conserved (constant),
so is the velocity of the center of mass.
09-8
Class #9
Take-Away Concepts
1.
2.
3.
Momentum is conserved (or not) separately for each direction
if Fext for that direction is negligible (or not).
Conserved components of momentum do not mix with each
other.
Center of mass defined (x equation for example):
x cm
4.
1 N
  mi x i
M i 1
Velocity of the center of mass and system momentum:


P  M v cm
09-9
Class #9
Problems of the Day
___ 1. In the figure below, penguin A is at the right edge of a uniform sled which
lies on the frictionless ice of a frozen lake, initially at rest. Penguin A sees
penguin B on the shore of the lake, and waddles to the left edge of the sled to
get closer, as indicated by the arrow.
While penguin A is waddling from the right edge to the left edge of the sled,
the center of mass of the system made up of penguin A and the sled
A. is moving toward the shore.
B. is moving away from the shore.
C. remains the same distance from shore.
D. could go either way, depending on the masses of the penguin and the sled.
E. who cares, they are only penguins.
09-10
Answer to Problem 1 for Class #9
The answer is C.
The momentum of the system (penguin plus sled) starts at zero and
since the only forces are internal, remains at zero. The momentum
is the total mass times the velocity of the center of mass, and so the
center of mass has zero velocity, remaining at the same distance
from shore. The penguin gets closer and the sled gets further away
so that the center of mass stays in the same place.
09-11
Class #9
Problems of the Day
2. A hockey puck sliding on frictionless ice (the best kind) collides
with a second puck that is initially at rest. Both pucks have the same
mass. Define the +X direction as the direction of the initial velocity
of the first puck and +Y as 90º counter-clockwise from +X. After
the collision, the first puck has a velocity of 2.0 m/s in a direction
30º counter-clockwise from +X. What is the Y component of the
velocity of the second puck? (include + or – sign)
09-12
Answer to Problem 2 for Class #9
X components are irrelevant.
Since initial Py=0 by the definition of the Y axis, and the two
pucks have equal mass, they must have opposite Y velocities
after the collision.
Vy1  V sin(30)  1 m/s
Vy 2  Vy1  1 m/s
09-13
Activity #9 - Conservation of
Momentum in Two Dimensions
Objectives of the Activity:
1.
2.
3.
Use VideoPoint to study conservation of momentum for a
two-object system in two dimensions.
Practice calculating and understanding the center of mass.
Practice solving two-dimensional conservation of momentum
problems.
09-14