Transcript Systems of Particles - University of Central Florida

Linear momentum and
Collisions
Chapter 9
I. Linear Momentum and its Conservation
II. Impulse and Momentum
III. Collisions in One Dimension
IV. Collisions in Two Dimensions
V. The Center of Mass
VI. Motion of a System of Particles
VII. Deformable Systems
VIII. Rocket Propulsion
Linear Momentum
The linear momentum of a particle, or an
object that can be modeled as a particle of
mass m moving with a velocity v, is defined
to be the product of the mass and velocity:
p  mv
The terms momentum and linear momentum
will be used interchangeably in the text
I. Linear momentum
The linear momentum of a particle is a vector p defined as:


p  mv
Momentum is a vector with magnitude equal mv and has
direction of v .
The dimensions of momentum are ML/T
SI unit of the momentum is kg-meter/second
Momentum can be expressed in component form:
px = m vx py = m vy
pz = m vz
Newton and Momentum
Newton’s Second Law can be used to
relate the momentum of a particle to the
resultant force acting on it
dv d  mv  dp
F  ma  m


dt
dt
dt
with constant mass
Newton called the product mv the quantity of motion of the
particle
Newton II law
in terms of momentum:
The time rate of change of the momentum of a particle is
equal to the net force acting on the particle and is in the
direction of the force.





dp d (mv )
dv
Fnet 

m
 ma
dt
dt
dt
System of particles:
The total linear moment P is the vector sum of the
individual particle’s linear momentum.
   





P  p1  p2  p3  .... pn  m1v1  m2v2  m3v3  ... mnvn


P  Mvcom
The linear momentum of a system of particles is
equal to the product of the total mass M of the
system and the velocity of the center of mass.





dvcom
dP
dP
M
 Macom  Fnet 
dt
dt
dt
Net external force acting on the system.
Conservation:
If no external force acts on a closed, isolated system of
particles, the total linear momentum P of the system cannot
change.

P  const

Fnet
(Closed, isolated system)



dP
0
 Pf  Pi
dt
Closed: no matter passes through the system
boundary in any direction.
Conservation of Linear Momentum
If no net external force acts on the system of particles the total
linear momentum P of the system cannot change.
Each component of the linear momentum is conserved
separately if the corresponding component of the net external
force is zero.
If the component of the net external force on a closed
system is zero along an axis  component of the linear
momentum along that axis cannot change.
The momentum is constant if no external forces act on a
closed particle system. Internal forces can change the
linear momentum of portions of the system, but they cannot
change the total linear momentum of the entire system.
Conservation of Linear Momentum
• Whenever two or more particles in an
isolated system interact, the total
momentum of the system remains
constant
– The momentum of the system is conserved,
not necessarily the momentum of an
individual particle
– This also tells us that the total momentum of
an isolated system equals its initial
momentum
Conservation of Momentum
• Conservation of momentum can be expressed
mathematically in various ways
p total = p 1 + p 2 = constant
p1i + p 2i = p 1f + p 2f
• In component form, the total momenta in each direction
are independently conserved
pix = pfx
piy = pfy
piz = pfz
• Conservation of momentum can be applied to systems
with any number of particles
• This law is the mathematical representation of the
momentum version of the isolated system model
Conservation of Momentum,
Archer Example
• The archer is standing on a
frictionless surface (ice)
• Approaches:
– Newton’s Second Law – no,
no information about F or a
– Energy approach – no,
no information about work or
energy
– Momentum – yes
Archer Example
• Conceptualize
– The arrow is fired one way and the archer recoils in
the opposite direction
• Categorize
– Momentum
Let the system be the archer with bow (particle 1)
and the arrow (particle 2)
There are no external forces in the x-direction, so it
is isolated in terms of momentum in the x-direction
• Analyze
– Total momentum before releasing the arrow is 0
Archer Example
• Analyze.
– The total momentum after releasing the arrow is
• Finalize
p 1f  p 2 f  0
– The final velocity of the archer is negative
• Indicates he moves in a direction opposite the arrow
• Archer has much higher mass than arrow, so velocity
is much lower
Impulse and Momentum
• From Newton’s Second Law,
• Solving for dp gives d p 
dp
F
dt
 Fdt
• Integrating to find the change in momentum
over some time interval
p  p f  p i 

tf
ti
F dt  I
• The integral is called the impulse, I, of the
force acting on an object over Δt
IV. Collision and impulse
Collision: isolated event in which two or more bodies exert
relatively strong forces on each other for a relatively short time.
Impulse: Measures the strength and duration of the collision
force
Third law force pair
FR = - FL
Single collision

pf

 dp
tf 
 

F
 dp  F (t )dt   dp   F (t )dt
ti

dt
pi
 tf 



I   F (t )dt  p f  pi  p
ti
Impulse-linear momentum theorem
The change in the linear momentum of a body in a
collision is equal to the impulse that acts on that body.
 
 
p  p f  pi  I
Units: kg m/s
p fx  pix  p x  I x
p fy  piy  p y  I y
p fz  piz  p z  I z
Favg such that:
Area under F(t) vs Δt curve
= Area under Favg vs t
I  Favg t
An estimated force-time curve for a baseball struck by a bat
is shown in Figure. From this curve, determine (a) the
impulse delivered to the ball, (b) the average force exerted
on the ball, and (c) the peak force exerted on the ball.
An estimated force-time curve for a baseball struck by a bat is shown in
Figure. From this curve, determine (a) the impulse delivered to the ball,
(b) the average force exerted on the ball, and (c) the peak force exerted
on the ball.
(a)
I  Fdt area under curve
I
(b)
(c)


1
1.50  103 s 18 000 N   13.5 N  s
2
13.5 N  s
F
 9.00 kN
3
1.50  10 s
Fm ax  18.0 kN
More About Impulse
• Impulse is a vector quantity
• The magnitude of the
impulse is equal to the area
under the force-time curve
– The force may vary with
time
• Dimensions of impulse are
ML/T
• Impulse is not a property of
the particle, but a measure
of the change in momentum
of the particle
Impulse
• The impulse can also
be found by using the
time averaged force
I
 Ft
• This would give the
same impulse as the
time-varying force does
Impulse Approximation
• In many cases, one force acting on a particle acts
for a short time, but is much greater than any
other force present
• When using the Impulse Approximation, we will
assume this is true
– Especially useful in analyzing collisions
• The force will be called the impulsive force
• The particle is assumed to move very little during
the collision
• p i and p f represent the momenta immediately
before and after the collision
Impulse-Momentum:
Crash Test Example
• Categorize
– Assume force exerted by
wall is large compared
with other forces
– Gravitational and normal
forces are perpendicular
and so do not effect the
horizontal momentum
– Can apply impulse
approximation
Crash Test Example
• Analyze
– The momenta before and after the collision
between the car and the wall can be determined
– Find
• Initial momentum
• Final momentum
• Impulse
• Average force
• Finalize
– Check signs on velocities to be sure they are
reasonable
Collisions – Characteristics
• We use the term collision to represent an event
during which two particles come close to each
other and interact by means of forces
– May involve physical contact, but must be generalized
to include cases with interaction without physical
contact
• The time interval during which the velocity
changes from its initial to final values is assumed
to be short
• The interaction forces are assumed to be much
greater than any external forces present
– This means the impulse approximation can be used
Collisions – Example 1
• Collisions may be the
result of direct contact
• The impulsive forces
may vary in time in
complicated ways
– This force is internal to
the system
• Momentum is
conserved
Collisions – Example 2
• The collision need not
include physical
contact between the
objects
• There are still forces
between the particles
• This type of collision
can be analyzed in the
same way as those
that include physical
contact
Types of Collisions
• In an elastic collision, momentum and kinetic
energy are conserved
– Perfectly elastic collisions occur on a microscopic
level
– In macroscopic collisions, only approximately
elastic collisions actually occur
• Generally some energy is lost to deformation,
sound, etc.
• In an inelastic collision, kinetic energy is not
conserved, although momentum is still
conserved
– If the objects stick together after the collision, it is
a perfectly inelastic collision
Collisions
• In an inelastic collision, some kinetic energy
is lost, but the objects do not stick together
• Elastic and perfectly inelastic collisions are
limiting cases, most actual collisions fall in
between these two types
• Momentum is conserved in all collisions
Perfectly Inelastic Collisions
• Since the objects stick
together, they share
the same velocity after
the collision
m1v 1i  m 2 v 2 i   m1  m 2  v f
Elastic Collisions
• Both momentum and
kinetic energy are
conserved
m1v1i  m2 v 2i 
m1v1f  m2 v 2f
1
1
2
2
m1v1i  m2 v 2i 
2
2
1
1
2
m1v1f  m2 v 22f
2
2
Elastic Collisions
• Typically, there are two unknowns to solve for and so
you need two equations
• The kinetic energy equation can be difficult to use
• With some algebraic manipulation, a different
equation can be used
v1i – v2i = v1f + v2f
• This equation, along with conservation of momentum,
can be used to solve for the two unknowns
– It can only be used with a one-dimensional, elastic
collision between two objects
Elastic Collisions
• Example of some special cases
m1 = m2 – the particles exchange velocities
– When a very heavy particle collides head-on with a very
light one initially at rest, the heavy particle continues in
motion unaltered and the light particle rebounds with a
speed of about twice the initial speed of the heavy
particle
– When a very light particle collides head-on with a very
heavy particle initially at rest, the light particle has its
velocity reversed and the heavy particle remains
approximately at rest
Problem-Solving Strategy:
One-Dimensional Collisions
• Conceptualize
– Image the collision occurring in your mind
– Draw simple diagrams of the particles before
and after the collision
– Include appropriate velocity vectors
• Categorize
– Is the system of particles isolated?
– Is the collision elastic, inelastic or perfectly
inelastic?
Problem-Solving Strategy: OneDimensional Collisions
• Analyze
– Set up the mathematical representation of
the problem
– Solve for the unknown(s)
• Finalize
– Check to see if the answers are consistent
with the mental and pictorial
representations
– Check to be sure your results are realistic
Example: Stress Reliever
• Conceptualize
– Imagine one ball coming
in from the left and two
balls exiting from the right
– Is this possible?
• Categorize
– Due to shortness of time,
the impulse approximation
can be used
– Isolated system
– Elastic collisions
Example: Stress Reliever
• Analyze
– Check to see if momentum is conserved
• It is
– Check to see if kinetic energy is conserved
• It is not
• Therefore, the collision couldn’t be elastic
• Finalize
– Having two balls exit was not possible if
only one ball is released
Example: Stress Reliever
• What collision is
possible
• Need to conserve both
momentum and kinetic
energy
– Only way to do so is
with equal numbers of
balls released and
exiting
Collision Example – Ballistic Pendulum
• Conceptualize
– Observe diagram
• Categorize
– Isolated system of projectile
and block
– Perfectly inelastic collision –
the bullet is embedded in the
block of wood
– Momentum equation will
have two unknowns
– Use conservation of energy
from the pendulum to find the
velocity just after the collision
– Then you can find the speed
of the bullet
Ballistic Pendulum
• A multi-flash
photograph of a ballistic
pendulum
• Analyze
– Solve resulting
system of equations
• Finalize
– Note different
systems involved
– Some energy was
transferred during
the perfectly inelastic
collision
V. Momentum and kinetic energy in collisions
Assumptions: Closed systems (no mass enters or leaves
them)
Isolated systems (no external forces act on
the bodies within the system)
Elastic collision:
If the total kinetic energy of the system of
two
colliding
bodies
is
unchanged
(conserved) by the collision.
Example: Ball into hard floor.
Inelastic collision: The kinetic energy of the system is not
conserved  some goes into thermal
energy, sound, etc.
Completely inelastic collision: After the collision the bodies lose
energy and stick together.
Example: Ball of wet putty into floor
VII. Elastic collisions in 1D
(Total kinetic energy before collision)  (Total kinetic energy after collision)
In an elastic collision, the
kinetic
energy
of
each
colliding body may change,
but the total kinetic energy of
the system does not change.
Stationary target:
Closed, isolated
system 
m1v1i  m1v1 f  m2v2 f
1
1
1
2
2
m1vi1  m1v1 f  m2 v22 f
2
2
2
m1 (v1i  v1 f )  m2v2 f
Linear momentum
Kinetic energy
(1)
m1 (v12i  v12f )  m2v22 f  m1 (v1i  v1 f )(v1i  v1 f )
(2)
Stationary target:
Dividing (2) /(1)  v2 f  v1i  v1 f
From (1)  v2 f
m1

(v1i  v1 f )
m2
(1) in (3) v1 f  v2 f
m1  m2
v1 f 
v1i
m1  m2
(3)
m1
 v1i 
(v1i  v1 f )  v1i 
m2
v2 f
2m1

v1i
m1  m2
v2f >0 always
v1f >0 if m1>m2  forward mov.
v1f <0 if m1<m2  rebounds
Equal masses: m1=m2  v1f=0 and v2f = v1i  In head-on
collisions bodies of equal masses simply exchange
velocities.
Massive target: m2>>m1  v1f ≈ -v1i
and
v2f ≈ (2m1/m2)v1i  Body 1 bounces back with
approximately same speed. Body 2 moves forward
at low speed.
Massive projectile: m1>>m2  v1f ≈ v1i
and
v2f ≈ 2v1i  Body 1 keeps on going scarcely lowed by
the collision. Body 2 charges ahead at twice the
initial speed of the projectile.
VII. Elastic collisions in 1D
Moving target:
Closed, isolated system 
m1v1i  m2v2i  m1v1 f  m2v2 f
Linear momentum
1
1
1
1
2
2
2
m1vi1  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
m1 (v1i  v1 f )  m2 (v2i  v2 f )
Kinetic energy
(1)
m1 (v1i  v1 f )(v1i  v1 f )  m2 (v2i  v2 f )(v2i  v2 f )
m1  m2
2m2
Dividing (2) /(1)  v1 f 
v1i 
v2 i
m1  m2
m1  m2
v2 f
2m1
m2  m1

v1i 
v2 i
m1  m2
m1  m2
(2)
VIII. Collisions in 2D
Closed, isolated system 
 


P1i  P2i  P1 f  P2 f
Linear momentum conserved
Elastic collision 
K1i  K2i  K1 f  K2 f
Example:
Kinetic energy conserved
x  axis  m1v1i  m1v1 f cos1  m2v2 f cos2
y  axis  0  m1v1 f sin 1  m2v2 f sin 2
If the collision is elastic 
1
1
1
2
2
m1vi1  m1v1 f  m2 v22 f
2
2
2
Two blocks of masses M and 3M are placed on a horizontal,
frictionless surface. A light spring is attached to one of them,
and the blocks are pushed together with the spring between
them. A cord initially holding the blocks together is burned; after
this, the block of mass 3M moves to the right with a speed of
2.00 m/s. (a) What is the speed of the block of mass M? (b)
Find the original elastic potential energy in the spring if M =
0.350 kg.
Two blocks of masses M and 3M are placed on a horizontal,
frictionless surface. A light spring is attached to one of them,
and the blocks are pushed together with the spring between
them. A cord initially holding the blocks together is burned; after
this, the block of mass 3M moves to the right with a speed of
2.00 m/s. (a) What is the speed of the block of mass M? (b)
Find the original elastic potential energy in the spring if M =
0.350 kg.
(a)
p  0
0  M vm   3M
vm  6.00 m s
pi  pf
  2.00 m s
(motion toward
the left).
Two blocks of masses M and 3M are placed on a horizontal,
frictionless surface. A light spring is attached to one of them,
and the blocks are pushed together with the spring between
them. A cord initially holding the blocks together is burned; after
this, the block of mass 3M moves to the right with a speed of
2.00 m/s. (a) What is the speed of the block of mass M? (b)
Find the original elastic potential energy in the spring if M =
0.350 kg.
(b)
1 2 1
1
2
kx  M vM   3M  v32M  8.40 J
2
2
2
A tennis player receives a shot with the ball (0.0600 kg)
traveling horizontally at 50.0 m/s and returns the shot with
the ball traveling horizontally at 40.0 m/s in the opposite
direction. (a) What is the impulse delivered to the ball by
the racquet? (b) What work does the racquet do on the
ball?
A tennis player receives a shot with the ball (0.0600 kg)
traveling horizontally at 50.0 m/s and returns the shot with
the ball traveling horizontally at 40.0 m/s in the opposite
direction. (a) What is the impulse delivered to the ball by
the racquet? (b) What work does the racquet do on the
ball?
Assume the initial direction of the ball in the –x direction.
(a)
 
ˆ  0.060 0  50.0  i
ˆ  5.40i
ˆN  s
I p  p f  pi   0.060 0  40.0 i
(b)
1
2
2

W  K f  K i   0.060 0  40.0   50.0   27.0 J
2
Two blocks are free to slide along the frictionless wooden track
ABC shown in Figure. A block of mass m1 = 5.00 kg is
released from A. Protruding from its front end is the north pole
of a strong magnet, repelling the north pole of an identical
magnet embedded in the back end of the block of mass
m2 = 10.0 kg, initially at rest. The two blocks never touch.
Calculate the maximum height to which m1 rises after the
elastic collision.
Two blocks are free to slide along the frictionless wooden track ABC shown
in Figure. A block of mass m1 = 5.00 kg is released from A. Protruding from
its front end is the north pole of a strong magnet, repelling the north pole of
an identical magnet embedded in the back end of the block of mass
m2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the
maximum height to which m1 rises after the elastic collision.
v1 - speed of m 1 at B before collision.
1
m 1v12  m 1gh
2
v1  2 9.80 5.00  9.90 m s
Two blocks are free to slide along the frictionless wooden track ABC
shown in Figure. A block of mass m1 = 5.00 kg is released from A.
Protruding from its front end is the north pole of a strong magnet, repelling
the north pole of an identical magnet embedded in the back end of the
block of mass m2 = 10.0 kg, initially at rest. The two blocks never touch.
Calculate the maximum height to which m1 rises after the elastic collision.
v1f , speed of m 1 at B just after collision
.
m1v1i  m1v1 f  m2v2 f
1
1
1
2
2
m1vi1  m1v1 f  m2 v22 f
2
2
2
m1 (v1i  v1 f )  m2v2 f
(1)
m1 (v12i  v12f )  m2v22 f  m1 (v1i  v1 f )(v1i  v1 f )
(2)
Two blocks are free to slide along the frictionless wooden track ABC shown
in Figure. A block of mass m1 = 5.00 kg is released from A. Protruding from
its front end is the north pole of a strong magnet, repelling the north pole of
an identical magnet embedded in the back end of the block of mass m2 =
10.0 kg, initially at rest. The two blocks never touch. Calculate the
maximum height to which m1 rises after the elastic collision.
m1 (v1i  v1 f )  m2v2 f
(1)
m1 (v12i  v12f )  m2v22 f  m1 (v1i  v1 f )(v1i  v1 f )
Dividing (2) /(1)  v2 f  v1i  v1 f
From (1)  v2 f
m1

(v1i  v1 f )
m2
(1) in (3) v1 f  v2 f
m1  m2
v1 f 
v1i
m1  m2
(3)
m1
 v1i 
(v1i  v1 f )  v1i 
m2
v2 f
2m1

v1i
m1  m2
(2)
Two blocks are free to slide along the frictionless wooden track ABC shown
in Figure. A block of mass m1 = 5.00 kg is released from A. Protruding from
its front end is the north pole of a strong magnet, repelling the north pole of
an identical magnet embedded in the back end of the block of mass
m2
= 10.0 kg, initially at rest. The two blocks never touch. Calculate the
maximum height to which m1 rises after the elastic collision.
v1f , speed of m 1 at B just after collision
.
m1  m 2
1
v1f 
v1    9.90 m s  3.30 m s
m1  m 2
3
m 1ghm ax
hm ax 
1
2
 m 1  3.
30
2
 3.30 m s

2
2
2 9.80 m s

 0.556 m
As shown in Figure, a bullet of mass m and speed v passes
completely through a pendulum bob of mass M. The bullet
emerges with a speed of v/2. The pendulum bob is suspended
by a stiff rod of length ℓ and negligible mass. What is the
minimum value of v such that the pendulum bob will barely
swing through a complete vertical circle?
As shown in Figure, a bullet of mass m and speed v passes
completely through a pendulum bob of mass M. The bullet
emerges with a speed of v/2. The pendulum bob is suspended
by a stiff rod of length ℓ and negligible mass. What is the
minimum value of v such that the pendulum bob will barely
swing through a complete vertical circle?
Energy is conserved for the bob-Earth
system between bottom and top of swing.
At the top the stiff rod is in compression
and the bob nearly at rest.
Ki U i  K f  U
f
1
2
M vb  0  0  M g2
2
2
vb
 g4
vb  2 g
As shown in Figure, a bullet of mass m and speed v passes
completely through a pendulum bob of mass M. The bullet
emerges with a speed of v/2. The pendulum bob is suspended
by a stiff rod of length ℓ and negligible mass. What is the
minimum value of v such that the pendulum bob will barely
swing through a complete vertical circle?
Momentum of the bob-bullet system
is conserved in the collision:

v
mv m  M 2 g
2
4M
v
m
g

A small block of mass m1 = 0.500 kg is released from rest at
the top of a curve-shaped frictionless wedge of mass
m2 = 3.00 kg, which sits on a frictionless horizontal surface as
in Figure (a). When the block leaves the wedge, its velocity is
measured to be 4.00 m/s to the right, as in Figure (b).
(a) What is the velocity of the wedge after the block reaches
the horizontal surface? (b) What is the height h of the wedge?
A small block of mass m1 = 0.500 kg is released from rest at
the top of a curve-shaped frictionless wedge of mass
m2 = 3.00 kg, which sits on a frictionless horizontal surface as
in Figure (a). When the block leaves the wedge, its velocity is
measured to be 4.00 m/s to the right, as in Figure (b). (a)
What is the velocity of the wedge after the block reaches the
horizontal surface? (b) What is the height h of the wedge?
(a) The initial momentum of the system is zero,
which remains constant throughout the motion.
Therefore, when m1 leaves the wedge, we must
have:
m 2vw edge  m 1vblock  0
v wedge
v block = 4.00 m/s
 3.00 kg vw edge  0.500 kg4.00 m s  0
vw edge  0.667 m s
+x
A small block of mass m1 = 0.500 kg is released from rest at the top of a
curve-shaped frictionless wedge of mass m2 = 3.00 kg, which sits on a
frictionless horizontal surface as in Figure (a). When the block leaves the
wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure (b).
(a) What is the velocity of the wedge after the block reaches the horizontal
surface? (b) What is the height h of the wedge?
(b)
Using conservation of energy for the
block-wedge-Earth system as the block
slides down the smooth (frictionless) wedge,
we have:
v wedge
v block = 4.00 m/s
+x
 K block  U system    K w edge    K block  U system    K w edge 

i 
i 
f 
f
1
 1
2
2
0  m 1gh  0   m 1  4.00  0  m 2  0.667
2
 2
h 0.952 m
Two objects have equal kinetic energies. How do
the magnitudes of their momenta compare?
25%
1.
2.
3.
4.
25%
25%
2
3
25%
p1 < p2
p1 = p2
p1 > p2
not enough
information to tell
1
4
In a perfectly inelastic one-dimensional collision between two
moving objects, what condition alone is necessary so that the
final kinetic energy of the system is zero after the collision?
1.
2.
3.
4.
The objects must have
momenta with the same
magnitude but opposite
directions.
The objects must have the
same mass.
The objects must have the
same velocity.
The objects must have the
same speed, with velocity
vectors in opposite
directions.
25%
1
25%
25%
2
3
25%
4
A table-tennis ball is thrown at a stationary bowling ball. The tabletennis ball makes a one-dimensional elastic collision and bounces back
along the same line. Compared with the bowling ball after the collision,
table-tennis ball will have:
1.
2.
3.
4.
5.
a larger magnitude of
momentum and more kinetic
energy.
a smaller magnitude of
momentum and more kinetic
energy.
a larger magnitude of
momentum and less kinetic
energy.
a smaller magnitude of
momentum and less kinetic
energy.
the same magnitude of
momentum and the same
kinetic energy.
20%
1
20%
2
20%
20%
3
4
20%
5
Two-Dimensional Collisions
• The momentum is conserved in all directions
• Use subscripts for
– Identifying the object
– Indicating initial or final values
– The velocity components
• If the collision is elastic, use conservation of
kinetic energy as a second equation
– Remember, the simpler equation can only be used for
one-dimensional situations
Two-Dimensional Collision, example
• Particle 1 is moving at
velocity v1i and
particle 2 is at rest
• In the x-direction, the
initial momentum is
m1v1i
• In the y-direction, the
initial momentum is 0
Two-Dimensional Collision,
example
• After the collision, the
momentum in the x-direction
is m1v1f cos θ + m2v2f cos φ
• After the collision, the
momentum in the y-direction
is m1v1f sin θ+ m2v2f sin f
• If the collision is elastic,
apply the kinetic energy
equation
• This is an example of a
glancing collision
Problem-Solving Strategies – TwoDimensional Collisions
• Conceptualize
– Imagine the collision
– Predict approximate directions the particles will
move after the collision
– Set up a coordinate system and define your
velocities with respect to that system
• It is usually convenient to have the x-axis coincide
with one of the initial velocities
– In your sketch of the coordinate system, draw
and label all velocity vectors and include all the
given information
Problem-Solving Strategies – TwoDimensional Collisions
• Categorize
– Is the system isolated?
– If so, categorize the collision as elastic, inelastic or
perfectly inelastic
• Analyze
– Write expressions for the x- and y-components of the
momentum of each object before and after the collision
• Remember to include the appropriate signs for the
components of the velocity vectors
– Write expressions for the total momentum of the system
in the x-direction before and after the collision and
equate the two. Repeat for the total momentum in the
y-direction.
Problem-Solving Strategies – TwoDimensional Collisions
– If the collision is inelastic, kinetic energy of the system
is not conserved, and additional information is probably
needed
– If the collision is perfectly inelastic, the final velocities of
the two objects are equal. Solve the momentum
equations for the unknowns.
– If the collision is elastic, the kinetic energy of the
system is conserved
• Equate the total kinetic energy before the collision to the
total kinetic energy after the collision to obtain more
information on the relationship between the velocities
Problem-Solving Strategies – TwoDimensional Collisions
• Finalize
– Check to see if your answers are
consistent with the mental and pictorial
representations
– Check to be sure your results are realistic
Two-Dimensional Collision Example
• Conceptualize
– See picture
– Choose East to be the
positive x-direction and
North to be the positive ydirection
• Categorize
– Ignore friction
– Model the cars as
particles
– The collision is perfectly
inelastic
• The cars stick together
Two dimensional collision
m1 = 1500.0kg
m2 = 2500.0 kg
Find vf .
Two dimensional collision
m1 = 800.0kg
m2 = 1400.0 kg
Find vf .
m1 v1 + m2 v2 = (m1 + m2) vf
(800kg) (25m/s) + 0 = (2200kg) vf cosθ – x-component
(1400kg) (20m/s) + 0 = (2200kg) vf sinθ - y-component
1400
0.8
 tan
800
  54.50
(1400)(20)  (2200)v f sin 54.50

v f  (9.07iˆ)m / s  (12.72 ˆj )m / s
v f  15.63m / s
An unstable atomic nucleus of mass 17.0  10–27 kg initially
at rest disintegrates into three particles. One of the
particles, of mass 5.00  10–27 kg, moves along the y-axis
with a speed of 6.00  106 m/s. Another particle, of mass
8.40  10–27 kg, moves along the x-axis with a speed of
4.00  106 m/s. Find (a) the velocity of the third particle and
(b) the total kinetic energy increase in the process.
An unstable atomic nucleus of mass 17.0  10–27 kg initially at rest disintegrates into three
particles. One of the particles, of mass 5.00  10–27 kg, moves along the y-axis with a speed of
6.00  106 m/s. Another particle, of mass 8.40  10–27 kg, moves along the x-axis with a speed
of 4.00  106 m/s. Find (a) the velocity of the third particle and (b) the total kinetic energy
increase in the process.
vi  0 (the parent nucleus)
m 0  17.0  1027 kg
v1  6.00  106 ˆ
jm s
ˆm s
m 2  8.40  1027 kg v2  4.00  106 i
(a) m 1v1  m 2v2  m 3v3  0
m 3  m 0  m 1  m 2  3.60  1027 kg
m 1  5.00  1027 kg
 5.00  10   6.00  10 ˆj  8.40  10   4.00  10 iˆ   3.60  10  v
27

v3 
27
6
27
6
3
0

ˆ 8.33  106 ˆ
9.33  106 i
j m s
(b) E  1 m v2  1 m v2  1 m v2
1 1
2 2
3 3

2

2
1
E   5.00  1027 6.00  106
2 
E  4.39  1013 J
 
2
2

 8.40  1027 4.00  106
 
2


2
 3.60  1027 12.5  106 

The Center of Mass
• There is a special point in a system or
object, called the center of mass, that
moves as if all of the mass of the system
is concentrated at that point
• The system will move as if an external
force were applied to a single particle of
mass M located at the center of mass.
M is the total mass of the system
II. Newton’s second law for a system of particles
Motion of the center of mass:
Center of the mass of the system moves as a particle
whose mass is equal to the total mass of the system.


Fnet  Macom
Fnet is the net of all external forces that act on the
system. Internal forces (from one part of the system to
another are not included).
The system is closed: no mass enters or leaves the
system during the movement. (M=total system mass).
acom is the acceleration of the system’s center of mass.
I. Center of mass
The center of mass of a body or a system of
bodies is the point that moves as though all the
mass were concentrated there and all external
forces were applied there.
System of particles:
Two particles of masses m1 and m2 separated by a distance d
Origin of reference
system
coincides
with m1
m2
xcom 
d
m1  m2
System of particles:
Choice of the reference
origin is arbitrary  Shift
of the coordinate system
but center of mass is still
at the same distance
from each particle.
The center of mass lies somewhere between the two particles.
General:
m1 x1  m2 x2 m1 x1  m2 x2
xcom 

m1  m2
M
M = total mass of the system
System of particles:
We can extend this equation to a general situation for n particles
that strung along x-axis. The total mass of the system
M=m1+m2+m3+……+mn The location of center of the mass:
xcom
m1 x1  m2 x2  m3 x3  ........ mn xn
1


M
M
n
m x
i 1
i i
3D:
1 n
xcom 
 mi xi
M i 1
1 n
ycom 
 mi yi
M i 1
1 n
zcom 
 mi zi
M i 1
System of particles:
3D: The vector form
Position of the
particle:
Position COM:

ri  xiiˆ  yi ˆj  zi kˆ

rcom  xcomiˆ  ycom ˆj  zcomkˆ

1 n 
rcom 
 mi ri
M i 1
M = total mass
of the object
Center of Mass, Extended Object
• Similar analysis can be
done for an extended
object
• Consider the extended
object as a system
containing a large
number of particles
• Since particle
separation is very small,
it can be considered to
have a constant mass
distribution
Center of Mass, position
• The center of mass in three dimensions can be
located by its position vector, rCM
– For a system of particles,
rCM
1

M
m r
ri is the position of the i
i i
i
th
particle, defined by
ri  x i ˆi  y i ˆj  z i kˆ
– For an extended object,
rCM
1
  r dm
M
Solid bodies:
Continuous distribution of matter. Particles = dm
(differential mass elements).
3D: xcom
1

x dm

M
ycom
1

y dm

M
zcom
1

z dm

M
M = mass of the object
Assumption: Uniform objects  uniform density
1
xcom   x dV
V
1
ycom   y dV
V
dm M


dV V
1
zcom   z dV
V
Motion of a System of Particles
• Assume the total mass, M, of the system
remains constant
• We can describe the motion of the system in
terms of the velocity and acceleration of the
center of mass of the system
• We can also describe the momentum of the
system and Newton’s Second Law for the
system
Velocity and Momentum of a
System of Particles
• The velocity of the center of mass of a system of
particles is
d rCM
1
v CM 
  mi v i
dt
M i
• The total momentum of the system can be
expressed as
M v CM   m i v i   p i  p tot
i
i
• The total linear momentum of the system equals
the total mass multiplied by the velocity of the
center of mass
Acceleration of the Center of Mass
• The acceleration of the center of mass can be
found by differentiating the velocity with
respect to time
a CM
d v CM
1


dt
M
m a
i
i
i
Forces In a System of Particles
• The acceleration can be related to a force
M a CM   Fi
i
• If we sum over all the internal forces, they
cancel in pairs and the net force on the system
is caused only by the external forces
Newton’s Second Law for a System of
Particles
• Since the only forces are external, the net
external force equals the total mass of the
system multiplied by the acceleration of the
center of mass:
F
ext
 M a CM
• The center of mass of a system of particles of
combined mass M moves like an equivalent
particle of mass M would move under the
influence of the net external force on the system
Impulse and Momentum of a System of
Particles
• The impulse imparted to the system by
external forces is
I
F
ext
dt  M  d v CM   p tot
• The total linear momentum of a system of
particles is conserved if no net external force
is acting on the system
M v CM  p tot  constant
when
F
ext
0
The center of mass of an object with a point, line or
plane of symmetry lies on that point, line or plane.
The center of mass of an object does not need to lie
within the object (Examples: doughnut, horseshoe )
Finding Center of Mass, Irregularly
Shaped Object
• Suspend the object
from one point
• Then suspend from
another point
• The intersection of
the resulting lines is
the center of mass
Center of Gravity
• Each small mass element of an extended object
is acted upon by the gravitational force
• The net effect of all these forces is equivalent to
the effect of a single force Mg acting through a
point called the center of gravity
– If g is constant over the mass distribution, the center
of gravity coincides with the center of mass
Center of Mass, Rod
Show that the center of mass
of a rod of mass M and length
L lies midway between its
ends, assuming the rod has a
uniform mass per unit length.
Center of Mass, Rod
Show that the center of mass
of a rod of mass M and length
L lies midway between its
ends, assuming the rod has a
uniform mass per unit length.
The rod is aligned along the x-axis, so ycm = zcm = 0. Lets call the
mass per unit length λ (the linear mass density) λ = M/L for the
uniform rod. If we divide the rod into elements of length dx, then
the mass of each element is dm = λ dx, and for xcm we have:
xcm
1

M
1
 xdm  M
Because λ = M/L, this
reduces to:
 x
 x dx  M
L2
xcm 
2M
2 L
2
0

 L2
2M
M  L
 
 L 2
Problem solving tactics:
(1) Use object’s symmetry.
(2) If possible, divide object in several parts. Treat each of
these parts as a particle located at its own center of mass.
(3) Chose your axes wisely. Use one particle of the system as
origin of your reference system or let the symmetry lines be
your axis.
A water molecule consists of an oxygen atom with two
hydrogen atoms bound to it. The angle between the two
bonds is 106. If the bonds are 0.100 nm long, where is
the center of mass of the molecule?
A water molecule consists of an oxygen atom with two
hydrogen atoms bound to it. The angle between the two
bonds is 106. If the bonds are 0.100 nm long, where is
the center of mass of the molecule?
y
yCM  0
xCM
xCM
m ixi



mi
0  1.008 u  0.100 nm  cos53.0  1.008 u  0.100 nm  cos53.0

15.999 u  1.008 u  1.008 u
xCM  0.006 73 nm from the oxygen nucleus
x
A uniform piece of sheet steel is shaped as in Figure.
Compute the x and y coordinates of the center of
mass of the piece.
A uniform piece of sheet steel is shaped as in Figure.
Compute the x and y coordinates of the center of
mass of the piece.
A  A1  A 2  A 3
M  M 1M 2M
A 3  200 cm
3
A 2  100 cm 2
M1 M

A1
A
 A 1  300 cm
M1M   
 A  600 cm
A 1  300 cm 2
2
2
 A 2  100 cm
M2M 


 A  600 cm
2
 A 3  200 cm
M3M 

 A  600 cm
2
2
2
M
M 
2
M
M 
6
M
M 
3
2
A3
A  600 cm
A2
A1
2
A uniform piece of sheet steel is shaped as in Figure.
Compute the x and y coordinates of the center of
mass of the piece.
A  A1  A 2  A 3
M  M 1M 2M
A 3  200 cm
3
xCM
xCM
yCM 
1
2M
A3
A 2  100 cm 2
M1 M

A1
A
x1M 1  x2M 2  x3M

M
 11.7 cm
2
A2
A 1  300 cm 2
3

15.0 cm
A1
 12 M   5.00 cm  61 M   10.0 cm  13 M 
M
 5.00 cm   61 M  15.0 cm    13 M   25.0 cm 
yCM  13.3 cm
M
A  600 cm
 13.3 cm
2
IV. Systems with varying mass
Rocket Propulsion
The operation of a rocket depends upon the law of
conservation of linear momentum as applied to a system
of particles, where the system is the rocket plus its
ejected fuel.
The initial mass of the rocket
plus all its fuel is M + Δm at
time ti and velocity v
The initial momentum of the
system is pi = (M + Δm) v
Rocket Propulsion
At some time t + Δt, the rocket’s mass has been
reduced to M and an amount of fuel, Δm has been
ejected.
The rocket’s speed has increased by Δv
Rocket Propulsion
Because the gases are given some momentum
when they are ejected out of the engine, the
rocket receives a compensating momentum in
the opposite direction
Therefore, the rocket is accelerated as a result of
the “push” from the exhaust gases
In free space, the center of mass of the system
(rocket plus expelled gases) moves uniformly,
independent of the propulsion process
IV. Systems with varying mass
Example: most of the mass of a rocket on its launching is fuel
that gets burned during the travel.
System: rocket + exhaust products
Closed and isolated  mass of this
system does not change as the
rocket accelerates.
P=const  Pi=Pf
After dt
Mv   dM  u  ( M  dM )  (v  dv)
Linear momentum of
exhaust products
released during the
interval dt
Linear momentum
of rocket at the
end of dt
Velocity of rocket relative to frame = (velocity of rocket relative
to products)+ (velocity of products relative to frame)
(v  dv)  vrel  u  u  (v  dv)  vrel
Mv  dM  u  ( M  dM )  (v  dv)
Mv  dM[(v  dv)  vrel ]  ( M  dM )(v  dv)
Mv  vdM  dvdM  vrel dM  Mv  Mdv  vdM  dvdM  Mv  vrel dM  Mdv
dM
dv
Mdv  vrel dM  
vrel  M
dt
dt
R=Rate at which the rocket losses mass= -dM/dt = rate of fuel
consumption
dM
dv

vrel  M
dt
dt

R  vrel  Ma
First rocket
equation
dM
dv

vrel  M
dt
dt
vf

vi
Mf
dv  vrel

Mi
dM
dv  
vrel
M


dM
 vrel ln M f  ln M i
M
Mi
v f  vi  vrel ln
Mf

Second rocket equation

• The basic equation for rocket propulsion is
 Mi
v f  vi  ve ln 
Mf




• The increase in rocket speed is proportional to
the speed of the escape gases (ve)
– So, the exhaust speed should be very high
• The increase in rocket speed is also proportional
to the natural log of the ratio Mi / Mf
– So, the ratio should be as high as possible, meaning
the mass of the rocket should be as small as possible
and it should carry as much fuel as possible
Thrust
• The thrust on the rocket is the force exerted
on it by the ejected exhaust gases
dv
dM
 ve
Thrust = M
dt
dt
• The thrust increases as the exhaust speed
increases
• The thrust increases as the rate of change of
mass increases
– The rate of change of the mass is called the rate
of fuel consumption or burn rate
The first stage of a Saturn V space vehicle consumed fuel
and oxidizer at the rate of 1.50  104 kg/s, with an exhaust
speed of 2.60  103 m/s. (a) Calculate the thrust produced
by these engines. (b) Find the acceleration of the vehicle
just as it lifted off the launch pad on the Earth if the vehicle’s
initial mass was 3.00  106 kg. Note: You must include the
gravitational force to solve part (b).
The first stage of a Saturn V space vehicle consumed fuel
and oxidizer at the rate of 1.50  104 kg/s, with an exhaust
speed of 2.60  103 m/s. (a) Calculate the thrust produced
by these engines. (b) Find the acceleration of the vehicle
just as it lifted off the launch pad on the Earth if the vehicle’s
initial mass was 3.00  106 kg. Note: You must include the
gravitational force to solve part (b).
(a) Thrust  ve dM
dt
(b)


 Fy  Thrust M g  M a


 2.60  103 m s 1.50  104 kg s  3.90  107 N



3.90  107  3.00  106  9.80  3.00  106 a
a 3.20 m s2
Model rocket engines are sized by thrust, thrust duration, and
total impulse, among other characteristics. A size C5 model
rocket engine has an average thrust of 5.26 N, a fuel mass of
12.7 grams, and an initial mass of 25.5 grams. The duration
of its burn is 1.90 s. (a) What is the average exhaust speed of
the engine? (b) If this engine is placed in a rocket body of
mass 53.5 grams, what is the final velocity of the rocket if it is
fired in outer space? Assume the fuel burns at a constant rate.
Model rocket engines are sized by thrust, thrust duration, and
total impulse, among other characteristics. A size C5 model
rocket engine has an average thrust of 5.26 N, a fuel mass of
12.7 grams, and an initial mass of 25.5 grams. The duration
of its burn is 1.90 s. (a) What is the average exhaust speed of
the engine? (b) If this engine is placed in a rocket body of
mass 53.5 grams, what is the final velocity of the rocket if it is
fired in outer space? Assume the fuel burns at a constant rate.
(a)
The fuel burns at a rate:
dM
Trust  ve
dt
12.7 g
dM

 6.68  103 kg s
dt 1.90 s


5.26 N  ve 6.68  103 kg s
ve  787 m s
Model rocket engines are sized by thrust, thrust duration, and
total impulse, among other characteristics. A size C5 model
rocket engine has an average thrust of 5.26 N, a fuel mass of
12.7 grams, and an initial mass of 25.5 grams. The duration
of its burn is 1.90 s. (a) What is the average exhaust speed of
the engine? (b) If this engine is placed in a rocket body of
mass 53.5 grams, what is the final velocity of the rocket if it is
fired in outer space? Assume the fuel burns at a constant rate.
(b)
M 
vf  vi  ve ln  i 
 M f


53.5 g  25.5 g
vf  0   797 m s ln 

 53.5 g  25.5 g  12.7 g 
vf  138 m s
A rocket for use in deep space is to be capable of boosting a
total load (payload plus rocket frame and engine) of 3.00
metric tons to a speed of 10 000 m/s. (a) It has an engine
and fuel designed to produce an exhaust speed of 2 000 m/s.
How much fuel plus oxidizer is required? (b) If a different fuel
and engine design could give an exhaust speed of 5 000 m/s,
what amount of fuel and oxidizer would be required for the
same task?
A rocket for use in deep space is to be capable of boosting a
total load (payload plus rocket frame and engine) of 3.00
metric tons to a speed of 10 000 m/s. (a) It has an engine
and fuel designed to produce an exhaust speed of 2 000 m/s.
How much fuel plus oxidizer is required? (b) If a different fuel
and engine design could give an exhaust speed of 5 000 m/s,
what amount of fuel and oxidizer would be required for the
same task?
M
v  ve ln
M
(a)
i
f
M i  ev ve M
f
5

3

5
M i  e 3.00  10 kg  4.45  10 kg
The mass of fuel and oxidizer is:
M  M i  M
3

445

3.
00

10
kg  442 m etrictons

f 
A rocket for use in deep space is to be capable of boosting a total load (payload
plus rocket frame and engine) of 3.00 metric tons to a speed of 10 000 m/s. (a) It
has an engine and fuel designed to produce an exhaust speed of 2 000 m/s. How
much fuel plus oxidizer is required? (b) If a different fuel and engine design could
give an exhaust speed of 5 000 m/s, what amount of fuel and oxidizer would be
required for the same task?
M
v  ve ln
M
i
f
(b)
M  e2  3.00 m etrictons  3.00 m etrictons 19.2 m etrictons
Because of the exponential, a relatively small increase in fuel and/or
engine efficiency causes a large change in the amount of fuel and oxidizer
required.
A 60.0-kg person running at an initial speed of 4.00 m/s jumps onto a 120-kg cart
initially at rest. The person slides on the cart’s top surface and finally comes to rest
relative to the cart. The coefficient of kinetic friction between the person and the cart is
0.400. Friction between the cart and ground can be neglected. (a) Find the final
velocity of the person and cart relative to the ground. (b) Find the friction force acting
on the person while he is sliding across the top surface of the cart. (c) How long does
the friction force act on the person? (d) Find the change in momentum of the person
and the change in momentum of the cart. (e) Determine the displacement of the
person relative to the ground while he is sliding on the cart. (f) Determine the
displacement of the cart relative to the ground while the person is sliding. (g) Find the
change in kinetic energy of the person. (h) Find the change in kinetic energy of the
cart. (What kind of collision is this, and what accounts for the loss of mechanical
energy?)
(a)
60.0 kg 4.00 m
s 120  60.0 kgvf
ˆ
v f  1.33 m si
(b)
 Fy  0
n  60.0 kg 9.80 m s2  0 n  588N
fk  kn  0.400 588 N   235 N
fk  235 N ˆ
i
pi  I  pf
(c) For the person,
m vi  Ft m vf
 60.0 kg 4.00 m
s  235 N  t  60.0 kg 1.33 m s
t 0.680 s
(d) person:
cart:
(e)
(f)
ˆ
m v f  m vi  60.0 kg  1.33  4.00 m s  160 N  si
120 kg 1.33 m s  0  160 N  sˆ
i
xf  xi 


1
1
vi  vf t  4.00  1.33 m s 0.680 s  1.81 m
2
2


1
1
xf  xi  vi  vf t  0  1.33 m s 0.680 s  0.454 m
2
2
1 2 1 2 1
2 1
2
m vf  m vi  60.0 kg 1.33 m s  60.0 kg  4.00 m s  427 J
2
2
2
2
(g)
(h)
(i)
1 2 1 2 1
2
m vf  m vi  120.0 kg 1.33 m s  0  107 J
2
2
2
The force exerted by the person on the cart must equal in magnitude and
opposite in direction to the force exerted by the cart on the person. The
changes in momentum of the two objects must be equal in magnitude and
must add to zero. Their changes in kinetic energy are different in
magnitude and do not add to zero. The following represent two ways of
thinking about “way”. The distance the cart moves is different from the
distance moved by the point of application of the friction force to the cart.
The total change of mechanical energy for both objects together, -320J,
becomes +320J of additional internal energy in this perfectly inelastic
collision.
A 2.00-kg block situated on a frictionless incline is
connected to a spring of negligible mass having a spring
constant of 100 N/m. The pulley is frictionless. The block
is released from rest with the spring initially unstretched.
(a) How far does it move down the incline before coming
to rest? (b) What is its acceleration at its lowest point? Is
the acceleration constant?
A 2.00-kg block situated on a frictionless incline is
connected to a spring of negligible mass having a spring
constant of 100 N/m. The pulley is frictionless. The block
is released from rest with the spring initially unstretched.
(a) How far does it move down the incline before coming
to rest? (b) What is its acceleration at its lowest point? Is
the acceleration constant?
(a)
E  0 K  0
U i   m gsin  x
U i U
f
1 2
U f  kx
2
x
 2.00 9.80 sin 37.0  100
2
x  0.236 m
A 2.00-kg block situated on a frictionless incline is connected to a
spring of negligible mass having a spring constant of 100 N/m. The
pulley is frictionless. The block is released from rest with the spring
initially unstretched. (a) How far does it move down the incline
before coming to rest? (b) What is its acceleration at its lowest point?
Is the acceleration constant?
(b)
F  ma
 kx  m g sin   m a
x 0.236 m
.
(from (a))
a 5.90 m s2
The negative sign indicates a
is up the incline.
The acceleration dependson position
A 2.00-kg block situated on a rough incline is connected
to a spring of negligible mass having a spring constant of
100 N/m. The pulley is frictionless. The block is released
from rest when the spring is unstretched. The block
moves 20.0 cm down the incline before coming to rest.
Find the coefficient of kinetic friction between the block
and incline.
A 2.00-kg block situated on a rough incline is connected to a spring of
negligible mass having a spring constant of 100 N/m. The pulley is
frictionless. The block is released from rest when the spring is unstretched.
The block moves 20.0 cm down the incline before coming to rest. Find the
coefficient of kinetic friction between the block and incline.
n  m g cos
f k   k m g cos
1
  k m gxcos  K  kx2  m gxsin 
2
vi  v f  0
K  0
(100N / m)(0.2m) 2
 k (2kg )(9.8m / s )(cos37 )(0.2m) 
 (2kg )(9.8m / s 2 )(sin 370 )(0.2m)
2
 k  0.115
2
0
Example: Tennis Racket
A 50 g ball is struck by a racket. If the ball is
initially travelling at 5 m/s up and ends up
travelling at 5 m/s down after 0.1 s of contact what
is the (average) force exerted by the racket on the
ball? What is the impulse?
Example: Tennis Racket
A 50 g ball is struck by a racket. If the ball is initially
travelling at 5 m/s up and ends up travelling at 5 m/s
down after 0.1 s of contact what is the (average)
force exerted by the racket on the ball? What is the
impulse?
a = (vf - vi)/Δt = (5 - (-5))/0.1 = 100 m/s2 down
Fav = ma = 0.05 x 100 = 5 N down
Dp = FΔt = 5 x 0.1 = 0.5 Ns down
Example: Bowling balls
Two particles masses m and 3m are moving towards each
other at the same speed, but opposite velocity, and collide
elastically
vo
vo
m
3m
vfm
y
m
x
After collision, m moves off at right angles
What are vmf and v3mf? At what angle is 3m scattered?
By conservation of KE:
½mvim2 + ½3mvi3m2 = ½mvfm2 + ½3mvf3m2
By conservation of momentum:
In x direction:
mvim + 3mvi3m = mvfmcosθm + 3mvf3mcosθ3m
In y direction:
0 = mvfmsinθm + 3mvf3msinθ3m
so
½mvim2 + ½3mvi3m2 = ½mvfm2 + ½3mvf3m2
vim2=vi3m2=vo2
2mvo2 = ½m(vfm2 + 3vf3m2)
In x direction:
vi3m = -vim and m=90o
mv0  3mv0  mv fm 0  3mv f 3m cos3m
 2mv0  3mv f 3m cos3m
3
v0   v f 3m cos3m
2
In y direction:
m=90o
0  mv fm  3mv f 3m sin 3m
v f 3m  
v fm
3 sin 3m
3m  35 ,
0
v f 3m 
2
v0 ,
3
v fm  2v0
Sulfur dioxide (SO2) consist of two oxygen atoms (each
of mass 16u) and single sulfur atom (of mass 32u). The
center-to-center distance between the atoms is
0.143nm. Angle is given on the picture. Find the x- and
y-coordinate of center of the mass of this molecule.