Your comments:
I
Could you please clarify when momentum is conserved and
when it is not? I think this pre-lecture was the first to
introduce a situation in which momentum is not conserved
and it confused me a little.
Remember Fnet, external = dP/dt.
Momentum will only be conserved it the net external force
on the system is zero.
forces within a system will conserve momentum
(eg collision forces – as long as both part of involved
in the collision are considered in the system.)
Mechanics Lecture 13, Slide 1
System 1: Momentum is not
conserved. There is an
outside from the light blue car.
System 2: Momentum is
conserved the force of
collision is within the system
and here are not external
forces.
Mechanics Lecture 13, Slide 2
Physics 211
Lecture 13
Today’s Concepts:
a) More on Elastic Collisions
b) Average Force during Collisions
Mechanics Lecture 13, Slide 3
I did not understand the concept at the end where one component of the kinetic
energy of a system depended on the reference frame of the observer and the
other did not. Isn't kinetic energy (or any energy for that matter) intrinsic?
Mechanics Lecture 13, Slide 4
More on Elastic Collisions
In CM frame, the speed of an object before an elastic collision is
the same as the speed of the object after.
v*1,i
v*1, f
v*2,i
v*2, f
|v*1,i - v*2,i| = |-v*1, f + v*2, f | = |-(v*1, f - v*2, f)| = |v*1, f - v*2, f |
So the magnitude of the difference of the two velocities is the
same before and after the collision.
But the difference of two vectors is the same in any reference
frame.
Just Remember This
Rate of approach before an elastic collision is the same as the rate
of separation afterward, in any reference frame!
Mechanics Lecture 13, Slide 5
Clicker Question
Consider the two elastic collisions shown below.
In 1, a golf ball moving with speed V hits a stationary
bowling ball head on. In 2, a bowling ball moving with
the same speed V hits a stationary golf ball.
In which case does the golf ball have the greater speed
after the collision?
A) 1
B) 2
V
1
C) same
V
2
Mechanics Lecture 13, Slide 6
Clicker Question
A small ball is placed above a much bigger ball,
and both are dropped together from a height H
above the floor. Assume all collisions are elastic.
What height do the balls bounce back to?
H
Before
A
After
B
C
Mechanics Lecture 13, Slide 7
Explanation
For an elastic collision, the rate of approach before
is the same as the rate of separation afterward:
m
v
v
video
M
3v
m
v
v
v
M
v
Rate of
approach
= 2V
Rate of
separation
= 2V
Mechanics Lecture 13, Slide 8
CheckPoint
A block slides to the right with speed V on a frictionless floor and
collides with a bigger block which is initially at rest. After the
collision the speed of both blocks is V/3 in opposite directions. Is
the collision elastic?
A) Yes
B) No
A) the velocities are equal so it is elastic
B) rate of approach does not equal rate of separation
V
Before Collision
V/3
V/3
After Collision
Mechanics Lecture 13, Slide 9
Forces during Collisions


dP
Ftot =
dt
Ftot = ma
t2



 Ftotdt = P(t2 ) - P(t1 )


Ftot dt = dP
F
t1

Favet

P
 
P = Favet
Fave
t2
ti
t
t
tf


 Ftotdt  Favet
t1
Impulse
Mechanics Lecture 13, Slide 10
Clicker Question
 
P = Favet
Two blocks, B having twice the mass of A, are initially at rest on
frictionless air tracks. You now apply the same constant force to both
blocks for exactly one second.
F
air track
A
F
air track
B
The change in momentum of block B is:
A) Twice the change in momentum of block A
B) The same as the change in momentum of block A
C) Half the change in momentum of block A
Mechanics Lecture 13, Slide 11
Clicker Question
 
P = Favet
Two boxes, one having twice the mass of the other, are initially at
rest on a horizontal frictionless surface. A force F acts on the
lighter box and a force 2F acts on the heavier box. Both forces act
for exactly one second.
Which box ends up with the biggest momentum?
A) Bigger box
F
B) Smaller box
M
C) same
2F
2M
Mechanics Lecture 13, Slide 12
Clicker Question
 
P = Favet
Two boxes, one having twice the mass of the other, are initially at
rest on a horizontal frictionless surface. A force F acts on the
lighter box and a force 2F acts on the heavier box. Both forces act
for exactly one second.
Which box ends up with the biggest speed?
A) Bigger box
F
B) Smaller box
M
C) same
2F
2M
Mechanics Lecture 13, Slide 13
Clicker Question
 
P = Favet
Two boxes, one having twice the mass of the other, are initially at
rest on a horizontal frictionless surface. A force F acts on both
boxes over exactly the same distance, d.
Which box ends up with the biggest kinetic energy?
A) Bigger box
F
B) Smaller box
M
C) same
F
2M
Mechanics Lecture 13, Slide 14
The experiment is repeated on a block with twice the mass
using a force that’s half as big. For how long would the force
have to act to result in the same final velocity?
F
A) Four times as long.
B) Twice as long.
C) The same length.
The momentum of the big block will be twice that of the smaller.
With the same force this would require twice as much time,
but with half the force you now need four times as much time to have the same final velocity.
Mechanics Lecture 13, Slide 15
CheckPoint
Identical balls are dropped from the same initial height
and bounce back to half the initial height. In Case 1 the ball bounces
off a cement floor and in Case 2 the ball bounces off a piece of
stretchy rubber. In which case is the average force acting on the ball
during the collision the biggest?
A) Case 1
B) Case 2
Egg Demo
Case 1
Case 2
Mechanics Lecture 13, Slide 16
CheckPoint Responses
In which case is the average force acting on the ball
during the collision the biggest?
A) Case 1
B) Case 2
Fave
Case 1
P
=
t
Case 2
Mechanics Lecture 13, Slide 17
Demo – similar concept
p = m ( v f - vi )
t = time between bounces
p
Fave =
t
H
h
vi
= reading on scale
vf
Scale
vi = - 2 gH
v f = 2 gh
Mechanics Lecture 13, Slide 18
HW Problem with demo
Mechanics Lecture 13, Slide 19
pi
q
q
pf
Another way to look at it:
P = pf -pi
-pi
pf
mv
|PX | = 2mv cos(q)cosq
|PY | = 0
Mechanics Lecture 13, Slide 20
|FAVE | = |P | /t = 2mv cos(q) /t
Mechanics Lecture 13, Slide 21
pi
pf
Another way to look at it:
P = pf -pi
pf
-pi
FAVE = P/t
|P| = |pf –pi| = m|vf –vi|
t = P/FAVE
K =
1 2 1 2
mv f - mvi
2
2 Mechanics Lecture 13, Slide 22
Mechanics Lecture 13, Slide 23