Transcript Chapt 9 - AJRomanello

Chapter 9
Systems of Particles
Section 9.2:
Center of Mass in a Two Particle System

Center of Mass is the point at which all forces are
assumed to act. For a two body system, this force is
calculated using the equation:
xcm
1

(m1 x1  m2 x2 )
m1  m2
Equation 1
Where m represents the mass of each body and x
represents its position.
Velocity and Acceleration of
the Center of Mass

The Velocity and Acceleration of the center of
mass of a system is calculated using a similar
equation:
vc 
1
(m1v1  m2 v2 )
m1  m2
1
ac 
(m1a1  m2 a2 )
m1  m2
Equation 2
Equation 3
Section 9.2:
Many Particle Systems

Expanding equations
1
xcm 
(m1 x1  m2 x2  ...mn xn )
1,2,& 3 from a two
m1  m2  ..mn
particle system to a
many particle system
1
v

is simply a matter of c m  m  ..m (m1v1  m2 v2  ..mn vn )
1
2
n
expanding the terms
to as many particles
1
as the system holds, ac 
(m1a1  m2 a2  ..mn an )
m1  m2  ..mn
thus the general form
of these equations
are:
Analyzing Multi-particle Systems


Since we can calculate the center of mass of a
system and its velocity and acceleration, we can
then analyze the system using the following rule:
The overall translational motion of a system of
particles can be analyzed using Newton’s Laws as if
all the mass were concentrated at the center of
mass and the total external force were applied at
that point.
A corollary which follows is: If the net external force
on a system of particles is zero, then the center of
mass of the system move with a constant velocity.
The Center of Mass of Solid Objects


If an object has spherical symmetry – the
center of mass is at the geometric center
If an object is not symmetrical then you
calculate the center of mass along each axis
and this determines the center of mass of
that object for translational motion in that
direction.
Section 9.3:
Newton’s Second Law for a System of Particles

Fnet = Macm where:




Fnet – the sum of all external forces acting on the
system.
M – total mass of the system
acm – the acceleration of the center of mass of the
system.
This is equivalent to the three equations:

Fnet,x = Macm,x Fnet,y = Macm,y Fnet,z = Macm,z
Section 9.4
Linear Momentum of a Particle


Linear Momentum is the product of the mass
of an object and its linear velocity.
ρ = mv where ρ (roe) is the momentum, m is
the mass in kilograms, and v is velocity in
m/s.
F
Section 9.5:
Linear Momentum in System of Particles

In a system of particles:

Fnet  
So
dvcm
d
M
 Macm
dt
dt
d
dt
Section 9:6
Change in Momentum: Impulse

The rate of change of momentum of a body
(impulse) is equal to the resultant force acting
on the body and is the direction of that force.



Since ρ = mv and F = ma and a = Δv/t
Δρ =mΔv = FΔt
Hence change in momentum is a product of
the force acting on an object and the time
during which it acts.
Section 9.6
Conservation of Linear Momentum





Like energy – momentum in a system is also
conserved.
For example, prior to firing a shotgun, neither the
gun, nor the shell inside are moving, hence the
momentum of the system is zero.
mgvg + msvs = 0
After the gun is fired, the shell moves forward, and
by conservation of momentum, the gun moves
backward
mgvg = -(msvs)
Sample Problem


A 3 kg shotgun contains a .140 kg shell.
When the shotgun is fired, the shell leaves
the barrel with a velocity of 400 m/s. What is
the recoil velocity of the shotgun?
If the Shotgun impacts the hunter’s shoulder
for 0.03 seconds, what force does it exert on
her shoulder?
Solution
By Conservation of Momentum:
(mg  ms )V  mg vg  ms vs
Hence
(3 kg + 0.140 kg)(0 m/s) = (3 kg)(Vg) + (0.140 kg)(400 m/s)
-(3 kg)(Vg) = 56 kg•m/s
56kg  m / s
Vg 
 18.67m / s
3kg
Section 9.8:
Momentum and Kinetic Energy in Collisions
Two type of Collisions

Elastic



Two objects collide and there are two objects move away
In a perfectly elastic collision kinetic energy is conserved.
Inelastic collision



Two objects collide and stick together- hence they move off as
one object after the collision
The converse of an inelastic collision is when one object breaks
into two pieces and both move off after the breaking (often an
explosion)
While total energy is conserved in the collision, often it is
changed into heat energy or energy of deformation so kinetic
energy is not conserved.
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Section 9.9
Inelastic Collisions

Inelastic Collisions can be described
mathematically by the equation:
m1v1 + m2v2 = (m1 + m2 )v’
since kinetic energy is not conserved this is the only
equation for an inelastic collision
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Sample Problem
A car with a mass of 1000 kg moving at a velocity of
25 m/s, strikes a second car with a mass of 1500 kg
that is at rest. The bumpers lock and the cars travel
together after the collision. What is the velocity of the
two car system immediately after the collision?
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Sample Problem
Solution
m1v1 + m2v2 = (m1 + m2)v’
(1000 kg)(25 m/s) + (1500 kg)(0 m/s) = (2500
kg)(v’)
25000 kg m/s = (2500 kg)(v’)
V’ = 10 m/s
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Section 9.10
Elastic Collisions

Elastic Collisions can be described
mathematically by the following:


Conservation of Momentum
m 1v 1 + m 2v 2 = m 1v 1’ + m 2v 2’
Conservation of kinetic energy
v 1 - v 2 = v 2’ - v 1’
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Sample Problem
A ball with a mass of 0.5 kg rolling at a velocity of 10
m/s, strikes a second ball with a mass of 0.75 kg that
is at rest in an elastic collision. What is the velocity of
each ball after the collision?
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Sample Problem
Solution
m1v1 + m2v2 = m1v1’ + m2v2’
(0.5 kg)(10 m/s) + (0.75 kg)(0 m/s) = (0.5 kg)(v1’)
+ (0.75 kg)(v2’)
v1 - v2 = v2’ - v1’
10 m/s - 0 m/s = v2’ - v1’
V2’ = 10m/s + v1’
5 kg m/s = (0.5 kg)(v1’) + 7.5 kg m/s + 0.75v1’
-2.5 kg m/s = 1.25 kg(v1’)
V1’ = -2 m/s
V2’ = 10 m/s + -2 m/s = 8 m/s
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Section 9.11
Two Dimensional Collisions



When analyzing a collision in two
dimensions, the key is to remember that
momentum is conserved in each direction.
ρix = ρfx and ρiy = ρfy
Remembering that since ρ = mv and v is a
vector.
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Sample Problem

A gas molecule having a speed of 322 m/s
collides elastically with another molecule
having the same mass which is initially at
rest. After the collision, the first molecule
move at an angle of 30° with respect to its
initial direction. Find the velocity of each
molecule after the collision and the angle
made with the incident direction by the
recoiling target (second) molecule.
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Solution
m1v1x + m2v2x = m1v’1x + m2v’2x
and
θ
β
m1v1y + m2v2y = m1v’1y + m2v’2y
Since m1=m2 mass cancels so
you have
322 m/s + 0 m/s = v’1x + v’2x
0 m/s + 0 m/s = v’1y + v’2y
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Solution (Continued)
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From previous work with vectors we remember:
V’1x = V’1cos θ and V’1y = V’1sin θ
V’2x = V’2cos β and V’2y = V’2sin β
So the equations become:
V1x = V’1cos θ + V’2cos β
-V’1sin θ = V’2sin β
(negative sign denotes opposite direction)
Rearranging the first equation we get:
V1x - V’1cos θ = V’2cos β
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Solution (Continued)
Squaring both equations and adding them
together you get:
V1x2 -2V1xV’1cos θ + V’12 sin2θ + V’12 cos2 θ =
V’22 cos2β + V’22 sin2β
 Using a well know trig identity this becomes:
V1x2 -2V1xV’1cos θ + V’12 = V’22

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Solution (Continued)

Remember that this is an elastic collision where
energy is conserved so:
V12 = V’12 + V’22
Combined with
V1x2 -2V1xV’1cos θ + V’12 = V’22
And the fact that
V1x =v1
(since there was no velocity in the Y direction initially)
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Solution (Continued)
V’22 +V’12 -2V1V’1cos θ + V’12 = V’22
2V’12 = 2V1V’1cos θ
V’1 = V1cos θ = (322 m/s)(cos 30°)= 279 m/s
Remember: V12 = V’12 + V’22 so:
(322 m/s)2 = (279 m/s)2 + V’22
V’2 = 161 m/s
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Solution (continued)
Finally, since:
V’1sin θ = V’2sin β
Sin β = V’1sin θ/V’2
Sin β = 279 m/s sin 30°/161 m/s
Β = 60°
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Momentum and Energy Combined

Momentum is often combined in problems with
energy. One of the most common problems used to
illustrate this is the ballistic pendulum problem.
Ballistics is the study of projectiles and a ballistic
pendulum is often used to measure the speed of a
bullet. This is done by firing the bullet into a wooden
block suspended on a string (hence the pendulum)
and using the change in the pendulum’s height to
back calculate the bullet’s speed.
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Sample Problem
A bullet with a mass of 14 grams is fired
into a suspended wooden block with a
mass of 2 kg. If the block travels to a
maximum height of 0.75 meters above its
rest position, what was the velocity of the
bullet when it was fired?
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Sample Problem
Solution
In order to solve this problem you work backward. At the top of
the arc the pendulum-bullet combination has only potential
energy. At the bottom of the arc, immediately after the bullet is
embedded in the block the combination has only kinetic
energy. So:
½ (m1 + m2)v’2 = (m1 + m2)gh
v’ = SQRT (2gh) = SQRT(2 x 9.8 m/s2 x 0.75 m)
= 3.83 m/s
Since the bullet embedding itself into the block is an inelastic
collision the equation:
m1v1 + m2v2 = (m1 + m2)v’
applies, hence
(0.014 kg)(v1) + (2 kg)(0 m/s) = (2.014 kg)(3.83 m/s)
V1 = 551.5 m/s
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Problem Types

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Linear momentum of a single object
Impulse
Collisions

Elastic
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Conservation of Momentum and Kinetic Energy
Inelastic
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Conservation of Momentum but not Kinetic Energy


Locked Bumpers, explosions
Two dimensional Collisions

Energy Momentum Combinations
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