Transcript + m 2 v 2

12/20 do now
• A ball with a momentum of +4.0 kg • m/s
hits a wall and bounces straight back
without losing any kinetic energy. What is
the change in the ball’s momentum?
Class work
• Page 213 practice 6C
1. 5.33 s; 53.3 m W
2. a. 14 m/s N;
b. 42 m N; c. 8.0 s
3. a. 1.22 x 104N E;
b. 53.3 m W
• Page 214 section review #1-5
2. a. 31 m/s ;
5. a. 2.5 kg∙m/s down field;
• Page 232 #16, 47
16. 0.010 s; 0.13 m
47. 400 N
b. 130 N downfield
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.
Homework – 6.2 essay – is due
Extra credit on chapter 6 test – do homework on time
6.1 essay rewrite – get average of the two essays – samples
are posted.
Homework assignments
• Castle learning assignments:
– Winter break reviews (5)
– 04 conservation of momentum 1
• Chapter 6 project
• Essay 6.3
The Law of Action-Reaction
demo
• In a collision between two objects, both objects experience
forces that are equal in magnitude and opposite in direction in
accord with Newton’s 3rd Law.
While the forces are equal in
magnitude and opposite in
direction, the accelerations of the
objects are not necessarily equal in
magnitude.
According to Newton's second law
of motion, the acceleration of an
object is dependent upon both
force and mass.
Bigger mass has smaller acceleration, smaller mass has
bigger acceleration
1.
2.
3.
4.
m1(v1’ – v1) = -m2(v2’– v2)
m1v1 + m2v2 = m1v1’ + m2v2’
p(before) = p(after)
Momentum Conservation Principle
• The law of momentum conservation can be stated as
follows:
– For a collision occurring between object 1 and
object 2 in an isolated system, the total
momentum of the two objects before the collision
is equal to the total momentum of the two objects
after the collision. That is, the momentum lost by
object 1 is equal to the momentum gained by
object 2.
Isolated Systems
• A system is a collection of two or
more objects.
• An isolated system is a system
which is free from the influence
of a net external force which
alters the momentum of the
system.
Collision
Description
1.
Two cars collide on a
gravel roadway on
which frictional forces
are large.
2.
Hans Full is doing the
annual vacuuming.
Hans is pushing the
Hoover vacuum
cleaner across the
living room carpet.
3.
Two air track gliders
collide on a frictionfree air track.
Isolated
System?
Yes or No
No
No
yes
If No, the
external force is
friction
The friction
between the carpet
and the floor and
the applied force
exerted by Hans
are both external
forces.
In reality
• Because of the inevitability of friction and air
resistance in any real collision, one might can
conclude that no system is ever perfectly isolated.
• In any real collision, resistance forces such as
friction and air resistance are inevitable. However,
compared to the impact force, they are very small,
and they can be ignored. Therefore, in most cases,
momentum can be approximate to be conserved for
every situations.
Consider a fullback plunges across the goal line and collides in
midair with the linebacker in a football game. The linebacker
and fullback hold each other and travel together after the
collision. Before the collision, the fullback possesses a
momentum of 100 kg∙m/s, East and the linebacker possesses
a momentum of 120 kg∙m/s, West. The total momentum of the
system before the collision is _____________________.
20 kg∙m/s, West
Therefore, the total momentum of the system after the collision
20 kg∙m/s, West.
must also be __________________
Vector diagram for the situation:
• consider a medicine ball is thrown to a clown who is at rest
upon the ice; the clown catches the medicine ball and
glides together with the ball across the ice.
• The momentum of the medicine ball is 80 kg∙m/s before
the collision. The momentum of the clown is 0 kg∙m/s
before the collision. The total momentum of the system
before the collision is ______________
80 kg∙m/s.
• Therefore, the total momentum of the system after the
80 kg∙m/s.
collision must also be ________________.
The clown and
the medicine ball move together as a single unit after the
collision with a combined momentum of 80 kg∙m/s.
Momentum is conserved in the collision.
Vector diagram
for the situation:
Using Equations for Algebraic
Problem-Solving
• Total system momentum is conserved for collisions
between objects in an isolated system.
pbefore = pafter
m1v1 + m2v2 = m1v1’ + m2v2’
Example 1
• A 2.0-kilogram ball traveling north at 5.0 meters
per second collides head-on with a 1.0 kilogram
ball traveling south at 8.0 meters per second.
What is the magnitude of the total momentum of
the two balls after collision?
m1v1 + m2v2 = m1v1’ + m2v2’
m1 = 2.0 kg; v1 = 5.0 m/s
pafter = ?
m2 = 1.0 kg; v2 = -8.0 m/s
m1v1 + m2v2 = pafter
(2.0 kg)(5.0 m/s) + (1.0 kg)(-8.0 m/s) = pafter
2 kg∙m/s, north = pafter
Example 2
• A 3000-kg truck moving with a velocity of 10 m/s hits a
1000-kg parked car. The impact causes the 1000-kg car to
be set in motion at 15 m/s. Assuming that momentum is
conserved during the collision, determine the velocity of the
truck immediately after the collision.
It is a collision problem, momentum is conserved.
m1v1 + m2v2 = m1v1’ + m2v2’
m1 = 3000 kg; v1 = 10 m/s
v1’ = ?
m2 = 1000 kg; v2 = 0
v2’ = 15 m/s
(3000 kg)∙(10 m/s) + 0 = (3000 kg)∙v + (1000 kg)∙(15 m/s)
30000 kg∙m/s = (3000 kg)∙v + 15000 kg∙m/s
v = 5 m/s
Example 3
• A 15-kg medicine ball is thrown at a velocity of 20 km/hr
to a 60-kg person who is at rest on ice. The person
catches the ball and subsequently slides with the ball
across the ice. Determine the velocity of the person and
the ball after the collision.
m1v1 + m2v2 = m1v1’ + m2v2’
m1 = 15 kg;
v1 = 20 km/hr
m2 = 60 kg;
v2 = 0
v1’ = v2’ = v
(15 kg)(20 km/hr) + (60 kg)(0)=(15 kg)·v + (60 kg)·v
300 kg∙km/hr = (75 kg)∙v
v = 4 km/hr
Example 4
• A 0.150-kg baseball moving at a speed of 45.0 m/s crosses
the plate and strikes the 0.250-kg catcher's mitt (originally at
rest). The catcher's mitt immediately recoils backwards (at
the same speed as the ball) before the catcher applies an
external force to stop its momentum. Determine the postcollision velocity of the mitt and ball.
m1v1 + m2v2 = m1v1’ + m2v2’
m1 = 0.150 kg; v1 = 45.0 m/s
m2 = 0.250 kg; v2 = 0
v1’ = v2’ = v
(0.15 kg)(45.0 m/s) + 0 =(0.15 kg)·v + (0.25 kg)·v
6.75 kg∙m/s = (0.40 kg)∙v
v = 16.9 m/s
Example 5 - Sample problem 6D
• A 76 kg boater, initially at rest in a
stationary 45 kg boat, steps out of the boat
and onto the dock. If the boater moves out
of the boat with a velocity of 2.5 m/s to the
right, what is the final velocity of the boat?
Class work
• Page 219 – practice 6D
1.
2.
3.
4.
1.90 m/s
1.66 m/s E
a. 12.0 m/s;
38 kg
b. 9.6 m/s
c. 11.7 m/s
• Page 232 #24-26, 46*, 52*, 53*, 57*, 59*
1.
2.
3.
4.
1.90 m/s
1.66 m/s E
a. 12.0 m/s;
38 kg
b. 9.6 m/s
c. 11.7 m/s
Jonathan
Lichtenfeld