Collisions
_________________ are a fact of life:
1. In _________: Hands, feet, heads, bats, rackets,
sports
clubs, collide with balls, nets, goals, posts, people,
Players
diving boards, etc. _________collide with each other.
2. __________ collide with other cars, buildings,
Cars
people, bicycles, etc.
3. __________ or parts of atoms collide with other
Atoms
atoms. matter
Light collides with ____________.
4. Planets, stars, and galaxies collide with
each other
__________________.
momentum (p = mv)
Physics uses ______________________ to study
collisions because it allows us to ignore the
forces
___________ between the objects, which can be
very __________________ during a collision.
complicated
A collection of objects that _____________ with
interact
system
each other is called a ____________ of objects:
SYSTEM
internal
_________
forces: 1
2
3
momentum
The total ______________ of a system of objects
impulse
will change if a net ____________ is applied to it:
impulse = change in momentum
J = DpTotal
Fnett = D (p1 + p2 + p3 + … )
NO net
But what if there is ________ impulse acting on the
system? This can only happen if the system is
isolated
________________, which means there is no net
force
____________ acting on it.
“external”
Any ___________ (outside)
no net
force exerts _________
SYSTEM force on the system
1
2
The “system” exerts no
3 friction
force, such as__________
on the “outside”
Dptotal
No Fnet no ____ no _____
J
remains constant
ptotal _____________________
The Law of Conservation of Linear Momentum:
total
The __________ momentum of an isolated
system of objects ____________________ . This
remains constant.
before
means that the total p _________ a collision (or an
after
explosion) equals the total p ________ the collision:
pbefore = pafter PhysRT
(In ___________)
If the “system” consists of 2 objects, this is written:
p1 + p2 = p1' + p2 '
m1v1 + m2v2 = m1v1' + m 2 v2 '
after
the prime symbol: ' represents “_______”
helmet construction:
Elastic
Ex 1: ___________ Collision. A 1.0-kg block and a 2.0-kg
block slide on a horizontal frictionless table as shown.
v1 = 6.0 m/s v2 = 1.5 m/s
1.0
2.0
kg kg
2
The "system" consists of _____ blocks.
friction
The system is "isolated," b/c there is no ___________.
interact
The two blocks collide and ____________ (exert
forces on each other). After the collision, they move
apart with the velocities shown below:
v1 = 4.0 m/s v2 = ?
1.0
2.0
kg kg
Conservation of momentum says:
pbefore = pafter
p1 + p2 = p1' + p 2'
m1v1 + m2v2 = m1v1' + m2v2'
(1)(6) + (2)(-1.5) = (1)(-4) + (2)v2'
negative
(Velocities have direction: left is_____________ )
6 + -3 = -4 + 2v2'
3 = -4 + 2v2'
7 = 2v2'
3.5 = v 2'
w/units: 3.5 m/s = v 2'
Inelastic
Ex 2: ______________ Collision. A 4.0-kg block and a 2.0-kg
block slide on a horizontal frictionless table as shown below.
v1 = 0.75 m/s v2 = 6.0 m/s
4.0 2.0
kg kg
stick together as one
After they collide, they ________________ and move ________.
What is the velocity of the "stuck together" mass?
v' = ?
4.0 2.0
kg kg
subscripts
Notice that the final v does not have _______________
have the same
1 or 2, because both masses ________________________
final velocity
____________________ .
car deer
collisions:
Conservation of momentum says:
pbefore = pafter
p1 + p2 = p1' + p2'
m1v1 + m2v2 = m1v1' + m2v2'
(4)(0.75) + (2)(-6.0) = (4)v1' + (2)v2'
3 + -12 = 4v' + 2v'
subscripts v'
(The ______________ on _____ are dropped.)
-9 = 6v'
combined
-1.5 = v' mass
w/units: -1.5 m/s = v'
left
The combined mass moves to the ________ .
Ex 3: ___________ /Spring Release. A 3.6-kg mass
Explosion
and a 1.2-kg mass are connected by a spring and
at rest
__________ on a horizontal frictionless table:
v=0
3.6 1.2
kg kg
When the spring _______________ , the 3.6-kg mass
is released
moves off to the left at the speed given. Determine
the speed of the 1.2-kg mass.
v1 = 0.70 m/s
v2 = ???
3.6 1.2
kg kg
Conservation of momentum says:
pbefore = pafter
p1 + p2 = p1' + p 2'
m1v1 + m2v2 = m1v1' + m2v2'
(3.6)(0) + (1.2)(0) =
(3.6)(-0.70) + (1.2)v2'
vi = 0
Both masses begin at rest ________ for both.
0 + 0 = -2.52 + 1.2v2'
0 = -2.52 + 1.2v2'
2.52 = 1.2v2'
2.1 = v 2'
w/units: 2.1 m/s = v 2'
Large Hadron Collider:
In this example:
before
1. The total momentum of the system _________
0
the spring is released equals ______ because both
at rest
masses begin _____________ .
after
2. The total momentum of the system ________
the spring is released equals ______ because of the
0
Conservation
Law of ___________________ of Linear Momentum.
Neither
3. __________ mass receives a greater force b/c of
Equal opposite
Newton's 3rd Law: ________ but ____________ forces.
4. The smaller mass moves ___________ because
faster
acceleration a = F/m is _____________ proportional
inversely
smaller
to mass, and its mass is _____________ .
3x
It has ____ less mass, so it gains ____ more speed
3x
pbefore = pafter
In sum: ________________ , is used in 3 cases:
Elastic
1. _____________ (bouncing):
Inelastic
2. _______________ (sticking): the same
v ____________
combined
_______________ mass
Explosion spring
3. ____________ /__________ release:
v=0
total p = 0
____ still 0
total p = ___________
In Textbook:
page 219: # 1, 2 and 3a
page 224: # 1 and 4