Transcript PPTX

Section Next Week Reminder
• You must complete at least 5 of the 6 labs to receive a passing grade
in this course.
•
If you missed a lab, you can make it up during one of two
Review/Lab make-up weeks:
– October 24-28 (labs 1-3)
– December 5-9 (labs 4-6)
• Even if you don’t need to make up a lab, you still must attend your
section those weeks for the Review Recitation.
• To make up a lab, contact your TA ahead of time.
You will need to arrange attending twice: (1) for lab make-up
and (2) for the review recitation. You can attend any other section
(in addition to your regular section), if you have that section’s TA
permission in advance.
Announcements
• CAPA Set #9 due Friday at 10 pm
• CAPA Set #10 now available, due next Friday
• Next week in Section: See next slide
• Reading – Finish Chapter 7 on Momentum
• Reading for next week – Chapter 8 on Rotation
• Next week Professor Uzdensky will give the
lectures on rotational motion.
• Nagle Monday office hours cancelled
• Wednesday Help Room 1:45 – 3:45 pm by Uzdensky
Clicker Question
Room Frequency BA
Consider a suspended blue sphere that collides
elastically with a suspended red sphere.
vA  vB
vA  0
m m
vB
'
m m
vB  0
'
After impact, is the situation at right possible: A) Yes B) No
Exactly as we derived for an elastic collision between
two equal mass objects.
Clicker Question
Room Frequency BA
Ballistic Pendulum
A bullet of mass m with unknown initial horizontal velocity
v0 is fired into a large suspended block of mass M.
Which of the following is true for the initial collision?
A) Only energy is conserved
B) Only momentum is conserved
C) Only kinetic Energy is conserved
D) Energy and momentum are conserved
E) Kinetic energy and momentum are conserved
Clicker Question
Room Frequency BA
h
For the initial collision where the bullet hits the
pendulum, assuming no external work, can there be zero
thermal energy generated?
A) Yes
B) No
Impossible
?
KE  PE  KE  PE
i
i
f
f
1 2
1
mv 0  0  ( M  m)v 2  0
2
2
mv 0  (M  m)v
m
v
v0
M m
m
v
v0
M m
Clicker Question
Room Frequency BA
h
Is the momentum of the system (M+m) conserved for the
second part?
A) Yes
B) No
No, because an external net force is acting (gravity)!
We say so often the momentum is conserved,
but it clearly cannot be if a
net external force is acting on the system.
h
However, gravity is a conservative force and so
mechanical energy is still conserved
(assuming no friction or air resistance).
KEi  PEi  KE f  PE f
1
( M  m)v 2  0  ( M  m) gh
2
v2
h
2g
Clicker Question
Room Frequency BA
A big ball of mass M = 10m and speed v strikes a small
ball of mass m at rest.
After the collision, could
the big ball come to a
complete stop and the
small ball take off with
speed 10 v?
A) Yes this can occur
B) No, because it violates
conservation of
momentum
C) No, because it violates
conservation of energy
pinitial  (10 m)v  10 mv
p final  m(10v)  10mv
1
KEinitial  (10 m)v 2  5mv 2
2
1
2
2
KE final  m(10 v)  50 mv
2
Clicker Question
Billiards
v
Room Frequency BA
450
450
What is true about the speed of the red and white ball after?
A) Equal Speeds, B) Red faster, C) White faster,
D) Cannot determine from the information given
p x ,initial  mv  px, final  mv1 cos(45)  mv2 cos(45)
p y,initial  0  p y, final  mv1 sin(45)  mv2 sin(45)
Impulse
Momentum is conserved (Δp = 0) if no net force operates
during the interaction of two objects.
But, what if we only consider a single object or
if Fnet is non-zero?


v

Fnet  ma  m
t



Fnet t  mv  p
Impulse:
 

I  Fnet t  p
= net force x time = change in momentum
Clicker Question
Room Frequency BA
I   p  F net t
A ball bounces off the floor as
shown. The direction of the impulse
of the ball,  p , is
v2
v1
A) Straight up,
θ θ
B) Straight down,
C) To the right,
D) To the left,
Note: the angle of incidence is
equal to the angle of reflection.
E) Indeterminate.
p1  ( px ,  py )
p1   p  p2
p 2  ( px , py )
p  p 2  p1  ( px , py )  ( px ,  py )
= (0, 2 py )
Clicker Question
I   p  F net t
Room Frequency BA
A fast-ball thrown at a batter has a momentum of
magnitude |pi| = (0.3kg)(40m/s) = 12 kg m/s.
The batter hits the ball back in a line drive with momentum
of magnitude |pf| = (0.3kg)(80m/s) = 24 kg m/s.
What is the magnitude of the impulse |Δp|?
A)
B)
C)
D)
E)
Pinitial = + 12 kg m/s
12 kg m/s
24 kg m/s
Pfinal = - 24 kg m/s
36 kg m/s
|Impulse|=|P| = |-24 –(+12)|=36 kg m/s
0 kg m/s
None of these.
Air Bags and Impulse
About 15 to 20 milliseconds after the collision occurs the crash
sensors decide whether or not the collision is serious enough to
inflate the airbag (usually 6 - 10 km/h).
It takes about 20 milliseconds to inflate the airbag.
Around 60 milliseconds the person has made contact with the
airbag and the airbag now starts to deflate.
The passenger continues to be acted on by the airbag as it is in
the deflation process which takes about 35 to 40 milliseconds.
A 75kg man is involved in a car accident.
He was traveling at 18 m/s (= 40 mph) when he hit a truck.
If he had no airbag in his car and he came to rest against the steering
wheel in 50 milliseconds (0.05 seconds) find the force on his body.
Impulse = F Δt = Δp = m Δv
F = (m Δv) / Δt
F = (75kg)(-18m/s) / (0.05s)
F = - 27,000 Newtons
If he had an airbag that inflated and deflated correctly, bringing him to
rest over a time of 0.78s, find the force on his body.
F = (m Δv) / Δt
F = (75kg)(-18m/s) / (0.78s)
F = - 1,700 Newtons
Which is only about 6% of the force felt without an airbag…
a definite improvement!
What is another benefit of the airbag in addition to
a reduced force?
Distributing the force over a larger area also reduced
injury (larger surface of contact with airbag compared to
just hitting the steering wheel).