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Momentum and Collisions
Russ Ballard
Kentlake Science Department
April 22, 2003
Linear Momentum
• Objective – be able to
compute linear momentum
and components of
momentum.
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One particle systems
• Linear momentum is the product
•
•
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of the mass and the velocity.
 = mv
units = kg  m/s
it is a vector quantity in the
same direction as the velocity.
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More than one particle
• Total linear momentum is the
total momentum of the
system
total = 1 + 2 + 3 + … = ii
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Example 6.1
Momentum: Mass and Velocity
A 100 kg player runs straight down the
field with a velocity of 4.0 m/s. A 1.0
kg artillery shell leaves the barrel of
a gun with a muzzle velocity of 500
m/s. Which has the greater
momentum?
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Follow-up
• What would the speed of the
football payer need to have the
same momentum?
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Total momentum: a vector sum
• What is the total sum for
each of the systems of
particles (2systems).
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System One
 = 5.0 kgm/s
2 = 3.0 kgm/s
1 = 2.0 kgm/s
Total
Individual
momentum
momenta
of system
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System Two
53°
3 = 4.0 kgm/s
y = 4.0 kgm/s
2 = -8.0 
kgm/s
x = -3.0 kgm/s1 = 5.0 kgm/s
Total
Individual
momentum
momenta
of system
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The Conservation of Linear
Momentum
• Objective – to be able to
• Explain the conditions for the
conservation of linear
momentum
• Apply it to physical conditions
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Example 6.4 Before and After:
considerations of Momentum
• Two masses m1 = 1.0kg and m2 =
2.0kg, are held on either side of a
light compressed spring by a string
joining them. The string is burned
and the masses move apart on a
frictionless surface, with m1 having a
velocity of 1.8 m/s to the left. What
is the velocity of m2?
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Example 6.6 a glancing
collision: components of
momentum
• A moving shuffleboard puck has
a glancing collision with a
stationary one of the same
mass. Considering friction to be
negligible, what are the speeds
of the pucks after the collision?
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Path of Collision
Vi = .95m/s
50°
40°
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Impulse
• Objective to be able to relate
• Impulse and momentum
• Kinetic energy and momentum
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Impulse
• Objects in contact exert
force on each other.
• If you consider average
force, Newton’s Laws help.
Ft =  = f -I
Impulse = Ft
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Impulse
• If you increase the time you
decrease the impulse.
• Jump from height.
• Catch a ball.
• Air bag.
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Impulse
• Applied impulse.
• Following through
• Increase contact time
• Increases force applied
• Improve control
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Example 6.7 teeing off: the
impulse-momentum theorem
• A golfer drives a .10-kg ball from an
elevated tee, giving it an initial
horizontal speed of 40m/s (about
90mph). The club and the ball are
in contact for 1.0 ms (millisecond).
What is the average force exerted
by the club on the ball during this
time?
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Impulse
• A collision is an exchange of
momentum and energy.
• KE = ½ mv2 =
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(mv) 2
2m
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=
ρ2
2m
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Elastic and Inelastic
Collisions
• Objective to be able to
describe the conditions on
kinetic energy and
momentum in elastic and
inelastic collisions
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Elastic Collisions
• Kinetic energy is conserved.
• Some is transformed to
potential during the collision.
• Object is elastic it returns to
original shape.
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Inelastic Collisions
• Kinetic energy is not
conserved.
• Object do not return to
original shape.
• Isolated systems momentum
is always conserved.
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Center of Mass
• Objective to be able to
• Explain the concept of the
center of mass and compute
its location for simple systems
• Describe how the center of
mass and center of gravity are
related
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Example 6.12 Finding the
Center of Mass: a Summation
• Example 6.13 a Dumbbell:
Center of Mass Revisited
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Jet Propulsion and Rockets
• Objective to apply the
conservation of momentum
in the explanation of jet
propulsion and the operation
of rockets
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Important concepts
•  = mv
• total = 1 + 2 + 3 + … = ii
• F = /t
• Impulse = Ft =  = mv-mvo
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Conservation of Momentum
Perfect inelastic collision
m1v1i + m2v2i = (m1+ m2)vf
Elastic collision
m1v1i + m2v2i = m1v1f + m2v2f
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Elastic vs. Inelastic
• Elastic
• f = i
• Kf = Ki
• Inelastic
• f = i
• Kf < Ki
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The center of mass
• Is the point at which all of the
•
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mass of am object or system
may be considered to be
connected. (The center of
gravity is the point where all the
weight may be considered to be
concentrated.
XCM =
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