Transcript Chapter 6 - MrCrabtreesScience

Chapter 6
Momentum and Collisions
6-1 Momentum and Impulse
• Momentum(p) describes the tendency of an
object to continue moving (or not moving)
at a constant speed.
• The heavier or faster an object moves, the
greater its momentum.
• Momentum = Mass x Velocity
• p = mv
Example
• A 2250 kg truck has a velocity of 25m/s.
What is its momentum?
• p = mv = (2250kg)(25m/s)
• p = 56,000 kgm/s
• (Note the Units)
• On a separate sheet of paper, do problems 13 on page 209.
6-1 (cont)
• A change in momentum takes a force
applied over a time interval (Impulse)
• ∆p = F∆t
• A change in momentum can also be defined
as (pf - pi) or (mvf - mvi)
• ∆p = mvf - mvi
• F∆t = mvf - mvi
• F = (mvf - mvi)/∆t
Example
• A 1400kg car with a velocity of 15m/s
impacts a telephone pole stopping in 0.3s.
• How much force is exerted on the car
• (on Board)
• On a separate sheet of paper, do practice
problems 1-3 on page 211.
Stopping Distance
• Remember from Chapter 4 that…
– ∆x = 1/2 (vi +vf)∆t
• Sample Problem on Board
• On separate sheet of paper, do practice
problems 1- 3 on page 213.
Stopping Time and Force
• A change in motion over a longer time,
requires less force.
• For example a car dashboard stops a person
over a period of 0.026sec applies a
significantly greater force than an airbag
that stops a person over a period of .75
seconds.
• How much more force?
6-2 Conservation of Momentum
• Momentum is a conserved quantity
• Imagine a soccer ball traveling at some
velocity hits a stationary soccer ball.
• What would happen?
• It is likely that soccer ball one will slow
down and soccer ball two will accelerate.
Conservation of Momentum
• Ball 1
– Before Collision
• m = .47kg
• v = .84m/s
• p = .40kgm/s
– After Collision
• m = .47kg
• v = .04m/s
• p = .02kgm/s
• Ball 2
– Before Collision
• m = .47kg
• v = 0.0m/s
• p = 0.0kgm/s
– After Collision
• m = .47kg
• v = .80m/s
• p = .38kgm/s
Continued…
• Notice that before the collision
– p1 + p2 = (.40+0.0) = .40kgm/s
• After the collision
– p1 + p2 = (.02+.38) = .40kgm/s
• This illustrates the Law of Conservation of
Momentum.
• p1i + p2i = p1f + p2f
• m1v1i + m2v2i = m1v1f + m2v2f
Example Problem
• A 76kg boater is at rest in a 45kg canoe next
to a dock. The boater steps out of the canoe
at a velocity of 2.5m/s to the right. What
velocity will the boat move to the left at?
• Do Problems 1-3 on page 219 Separate
sheet of paper.
6-3 Elastic and Inelastic
Collisions
• There are two basic types of collisions.
• The difference is whether or not the two
objects separate (elastic) or stick together
(inelastic).
Inelastic Collisions
• Perfectly inelastic collisions occur when two
objects collide and stick together.
• After the collision we can treat the objects as
though they were one.
• Because vf is the same for both objects you can
factor it out
– m1v1i + m2v2i = m1v1f + m2v2f becomes
– m1v1i + m2v2i = (m1 + m2)vf
Inelastic Collisions (cont)
• During inelastic collisions, some of the
Kinetic energy gets converted to sound
energy, internal energy, as well as elastic
potential energy, as the objects deform from
the force of the collision.
Example
• 1850kg Car is rear-ended by a 975kg car,
the two cars entangle. If the 975kg car was
moving at 22 m/s what is the final velocity
of the entangled cars after the collisions?
• Practice problems 1-4 on own, problem 5
with class on board.
Example (cont)
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m1 = 1850
m2 = 975
v1i = 0 m/s
v2i = 22.0 m/s
vf = ??
(m1v1i + m2v2i) / (m1 + m2) = vf
(1850x0)+(975x22) / (1850+975) = vf
7.59m/s = vf
Elastic Collisions
• Elastic collisions result in the two objects
moving separately after a collision.
• With elastic collisions both momentum and
kinetic energy are conserved.
• To solve these problems you need to know
all but one of the variables.
Example
• Sample problem 6G on board.
• Practice problems 1a&b and 2-4a