Transcript PPT

Physics 211
Lecture 13
Today’s Concepts:
a) More on Elastic Collisions
b) Average Force during Collisions
Mechanics Lecture 13, Slide 1
In the last slides of the prelecture, it talked about KE in terms of different reference
frames and in one of the reference frames the KE = 0 why?
Demo: throw
baton
Mechanics Lecture 13, Slide 2
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.
m1
v*1, f
m1
v*1,i
m2
v*2,i
m2
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 3
ACT
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
Demo: 283
2 ball pendula
B) 2
C) same
V
1
V
2
Mechanics Lecture 13, Slide 4
ACT
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 5
Explanation
For an elastic collision, the rate of approach before
is the same as the rate of separation afterward:
v
v
3v
m
m
v
M
Demo: 127 2 ball drop
v
v
M
v
Rate of
approach
= 2v
Rate of
separation
= 2v
Mechanics Lecture 13, Slide 6
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
m
V
M
Before Collision
V/3
V/3
m
M
After Collision
Mechanics Lecture 13, Slide 7
Comments
But they don’t have the same speed. Tell me more.
But that’s not nearly enough.
But that’s not enough.
No
But that’s not enough.
No, the speeds are not the same.
But that’s not enough.
!!!
Good
Mechanics Lecture 13, Slide 8
CheckPoint Response
Is the collision elastic?
A) Yes
B) No
m
V
M
Before Collision
V/3
V/3
m
M
After Collision
A) Yes, the blocks do not stick together.
B) No because the relative speed before the collision
is V and after it's 2V/3 and since those two do not
equal each other the collision is not elastic.
C) The masses of the blocks need to be known.
Mechanics Lecture 13, Slide 9
Forces during Collisions


dP
Ftot =
dt
Ftot = ma



 Ftotdt = P(t2 ) - P(t1 )
t2


Ftotdt = dP
F
t1

Fave t

P
 
P = Fave t
Fave


 Ftotdt  Favet
t2
t
t
ti
tf
t1
Impulse
Mechanics Lecture 13, Slide 10
ACT
 
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
ACT
 
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 bigger momentum?
A) Bigger box
F
B) Smaller box
C) same
2F
M
2M
Mechanics Lecture 13, Slide 12
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 13
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) Ft=mv. If the velocity is to remain constant between case 1 and case 2, then halving the
force and doubling the mass would require 4 times as much time to balance the equation.
B) the momentum is the same, and the force is halved, so it would take twice the given time to
get the same velocity
C) mv=Ft, so if you solve for t in each case using the given values then t will be the same.
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.
For the same final velocity we need to change the momentum by a factor of two since the
mass is twice as large, p = m v.
If the momentum change is twice as large, the Impulse must be twice as large.
With only half the first force, an Impulse that is twice as large requires a time that is four
times as long.
Impulse = (Force) (time ) = F t
A) Four times as long.
Mechanics Lecture 13, Slide 15
Comments
But you guessed wrong.
We do have to be concerned with mass since p = mv.
WHAT is squared?
But you guessed wrong also.
Is it any clearer now?
You began correctly but the force is only half as much and we need to account for that.
Good; what does that mean about the time?
Good.
How much more?
What about the time?
But you guessed wrong also.
Good.
Mechanics Lecture 13, Slide 16
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 bigger?
A) Case 1
B) Case 2
Let’s try it again !
Case 1
Case 2
Mechanics Lecture 13, Slide 17
CheckPoint Responses
In which case is the average force acting on the ball
during the collision bigger?
A) Case 1
B) Case 2
Fave
P
=
t
Case 2
Case 1
A) Because the same change in momentum
happens in a shorter time.
B) It’s in contact with the ball for longer
Demo: sheet + raw eggs
Mechanics Lecture 13, Slide 18
The impulse must be the same but what about the force; Impulse = force x time .
Tell me more.
???
And you even guessed correctly!
But your answer was 2) and that’s wrong.
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
“Inelastic” here means “not totally elastic”.
This is not the totally inelastic collision
where the ball sticks to the wall.
We tend to look at totally elastic collisions
or totally inelastic collisions. This one is
neither of our “usual” situations.
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 24
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 gH
v f = 2 gh
Demo
Mechanics Lecture 13, Slide 25
Parting Thoughts/Questions
That always happens for a totally elastic collision.
Has today’s lecture helped?
Make use of Office Hours! ! !
Mechanics Lecture 13, Slide 26