Transcript Momentum & Collisions

Momentum & Collisions
Chapter 6
What is this chapter about?
• To this point, we have learned about the
motion of single objects.
• This chapter is about collisions, the
interaction of a moving object with other
objects.
Important Concepts
We will learn about…
• Momentum
– Its transfer
– Its conservation
• Impulse
– Relates force to changes in momentum
• Types of Collisions
– Elastic
– Inelastic
6.1 Momentum and Impulse
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Linear Momentum
• Momentum describes motion
• A vector quantity defined as the product of
an object’s mass and velocity
• p = mv
• Units: kg·m/s
• Momentum increases with increasing
mass and/or increasing velocity
Why is momentum important?
It allows us to discuss the interaction of two
objects…
Particularly collisions
collisions between moving objects
collision of a moving object with a
stationary object
An ostrich with mass of 145 kg
escapes its pen with a velocity
of +17m/s (~ 38 mi/h).
Find the momentum of the
bolting bird.
p = mv
= 145 kg x 17 m/s
= 2500 kg∙m/s
Impulse
• Changing momentum requires force and
time
• Impulse is the product of an average
external force and the time during which it
acts on an object
• Impulse causes a change in momentum
• Impulse·momentum theorem
F∆t = ∆p
F∆t = ∆p = mvf – mvi
Sometimes symbolized as J or I.
Impulse
• The quantity F∆t is the impulse of the force F
acting for the time interval ∆t
• Is related to Newton’s second law
• F = ma
• F = m(∆v/∆t) F∆t = m∆v
• F∆t = ∆p
• Shows that the longer a force is applied, or the
greater the force…
• the greater the change in momentum, e.g. follow
through in batting a baseball or in tennis.
Plane & Duck Collison
• Estimate the average impact
force between an airliner
traveling at 600 mi/hr (268 m/s)
and a 1 pound (0.454 kg) duck
whose collison time is 0.0011
sec.
Plane & Duck Collison
• Estimate the average impact
force between an airliner
traveling at 600 mi/hr (268 m/s)
and a 1 pound (0.454 kg) duck
whose collison time is 0.0011
sec.
Ft  p  m vf  m vi
mv f  vi 
0.454kg 268m / s  0
F

t
0.0011s
F  1.1105 N  12tons
Stopping Times and Distances
Impulse-momentum theorem can be used to
solve these problems
A force of 8410 N is applied to a car (2240 kg),
which slows uniformly from 20.0 m/s to 5.00
m/s.
How long does it take for the car to slow down?
FΔt = Δp
What distance is required to slow down?
vf2= vi2 + 2ax
Stopping Time and Distance
A force of 8410 N is applied to a car (2240 kg), which
slows uniformly from 20.0 m/s to 5.00 m/s. How long
does it take for the car to slow down?
Ft  p
p (m vf  m vi )
t 

 4.00s
F
F
How far did the car travel while slowing down?
vavg
x

t
 x  vavg t
What was the acceleration of the car?
v
a
t
6.2 Conservation of Momentum
Objectives
1. Describe the interaction between two
objects in terms of the change in
momentum of each object
2. Compare the total momentum of two
objects before and after they interact
3. State the law of conservation of
momentum
4. Predict the final velocities of objects after
collisions, given the initial velocities
Momentum is Conserved
•
•
•
•
Law of Conservation of Momentum
Total initial momentum = total final momentum
pi = pf
mvi = mvf
Total momentum of all objects interacting with
one another remains constant
• Momentum is conserved in collisions and when
objects push away from one another
m1v 1i  m2 v 2i  m1v 1f  m2 v 2f
Practice Problems 6D
An 85.0 kg fisherman jumps from a dock into a 135.0 kg
rowboat at rest. If the velocity of the fisherman is 4.30 m/s,
what is the final velocity of the fisherman and the boat?
What do you know?
m1 = 85.0 kg; v1 = 4.30 m/s; v2 = 0 m/s; m2 = 135 kg
What principle is involved in this problem?
Conservation of momentum; ∑pi = ∑pf
What is the unknown variable?
Velocity of the boat
How would you set up this problem?
p1i + p2i = p1f + p2f
m1v1i + m2v2i = m1v1f + m2v2f but…the two objects join, so
m1v1i + m2v2i = (m1+ m2)vf
What is your answer?
1.66 m/s
6.3 Elastic and Inelastic Collisions
Objectives
1. Identify different types of collisions
2. Determine the decrease in kinetic energy
during perfectly inelastic collisions
3. Compare conservation of momentum
and conservation of kinetic energy in
perfectly inelastic collisions
4. Find the final velocity of an object in
perfectly inelastic and elastic collisions
Types of Collisions
Collision
Momentum
Kinetic Energy
Perfectly
inelastic
Conserved
Not Conserved
Elastic
Conserved
Conserved
Perfectly Inelastic Collisions
• When two objects collide, they stick
together.
• If the objects stick together, then the
consolidated mass has a single velocity
• So, if m1v1i + m2v2i = m1v1f + m2v2f
• Then
m1v1i + m2v2i = (m1+ m2)vf
Practice 6E
A 2500 kg car traveling at 20.0 m/s runs into a 5000 kg truck which is
stopped at a traffic light. If the two vehicles stick together and move
together after a collision, what is the final velocity of the two-vehicle
mass?
m1v1i + m2v2i = (m1+ m2)vf
6.7 m/s
KE is not conserved during
inelastic collisions
6F #1
a.
A 0.25 kg arrow with velocity 12 m/s strikes and pierces a
6.8 kg target
What is the final velocity of the combined mass?
m1v1i + m2v2i = (m1+ m2)vf
(0.25kg)(12m/s) + (6.8kg)(0 m/s) = (0.25 + 6.8kg)(vf)
vf = 0.43 m/s
b.
What is the decrease in kinetic energy during the collision?
ΔKE = KEf-KEi
[KE = ½mv2]
ΔKE = KE combined mass – (KE arrow + KE target)
ΔKE = 17 J