Your comments:
I really like how we have to explain our answers because on the checkpoint questions, I always
think I have the right answer, but when I explain how I got to that answer, I realize that my answer is
wrong.
Give me food, and I will live; give me water, and I will die. What am I?
I felt pretty confident with this material.
I'm panicking. I'm seriously panicking. What am I going to do.
Good pre-lecture, and I felt that it was easy to understand. Question: Why did you go into physics?
OH MY GOD! I understood this! It's a MIRACLE!
This was a very confusing prelecture.
This was a pretty easy topic, but I have one request. Please, never ever play those cartoons for us
in class ever again. Too painful to listen to.....
I'm a little lost on the reference frames. Could you explain the difference and maybe some
examples?
I think all of this makes sense. But I am somewhat worried about rotation...never understood that in
high school...
Could we possibly do a Harlem Shake: the physics 211 version?
I hope we're doing a demo of the two balls
On Friday, I'll have the unofficial impulse to elastically collide with the floor.
This is a haiku/ It is meaningless to you/ So do not read it.
I was more scared when he said that rotation is coming up next than with this material. I think I get
this stuff, but I know rotation kicked my fanny in high school
If "astro" means space and "naut" means none, then does "astronaut" mean no space?
Mechanics Lecture 13, Slide 1
Physics 211
Lecture 13
Today’s Concepts:
a) More on Elastic Collisions
b) Average Force during Collisions
Mechanics Lecture 13, Slide 2
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 3
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 4
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 5
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 6
Explanation
For an elastic collision, the rate of approach before
is the same as the rate of separation afterward:
v
v
video
3v
m
m
v
M
v
v
M
v
Rate of
approach
= 2V
Rate of
separation
= 2V
Mechanics Lecture 13, Slide 7
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 doesn not equal rate of seperation
V
Before Collision
V/3
V/3
After Collision
Mechanics Lecture 13, Slide 8
Forces during Collisions

Ftot =
Fto t = m a

dP
t2



 Ftot dt = P (t 2 ) - P (t1 )


Ftot dt = d P
dt
t1

F ave  t

P
F


 P = F ave  t
Fave
t2
t
t
ti
tf


 Ftot dt  Fave  t
t1
Impulse
Mechanics Lecture 13, Slide 9
Clicker Question


 P = F ave  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 10
Clicker Question


 P = F ave  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
C) same
2F
M
2M
Mechanics Lecture 13, Slide 11
Clicker Question


 P = F ave  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
C) same
2F
M
2M
Mechanics Lecture 13, Slide 12
Clicker Question


 P = F ave  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
C) same
F
M
2M
Mechanics Lecture 13, Slide 13
CheckPoint
A constant force acts for a time t on a block that is initially at rest
on a frictionless surface, resulting in a final velocity V. Suppose 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.
D) Half as long.
E) A quarter as long.
Lets try it again !
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.
A) 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.
B) Since we are using half the force, we need to apply it for twice the amount of time..
C) Ft=mv, the 2 and the one half cancel
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
A) It's like punching a brick wall and punching a pillow
B) The rubber contracts and expands exerting more force.
Mechanics Lecture 13, Slide 17
Prof. Selen, I have been trying to remember where I have
seen you before…
WhysGuy (2:30)
Mechanics Lecture 13, Slide 18
Demo – similar concept
 p = m ( v f - vi )
 t = time between bounces
H
Fave =
h
vi
p
t
= reading on scale
vf
Scale
vi = - 2 g H
vf =
2 gh
Mechanics Lecture 13, Slide 19
HW Problem with demo
Mechanics Lecture 13, Slide 20
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 21
|FAVE | = |P | /t = 2mv cos(q) /t
Mechanics Lecture 13, Slide 22
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 23