Transcript Physics 207: Lecture 2 Notes

Lecture 13
Goals:
• Chapter 9
•
 Employ conservation of momentum in 1 D & 2D
 Examine forces over time (aka Impulse)
Chapter 10
 Understand the relationship between motion and energy
Assignments:
 HW5, due tomorrow
 For Wednesday, Read all of Chapter 10
Physics 207: Lecture 13, Pg 1
Momentum is conserved if no external force


pbefore  pafter
Physics 207: Lecture 13, Pg 2
Inelastic collision in 1-D: Example

A block of mass M is initially at rest on a frictionless horizontal
surface. A bullet of mass m is fired at the block with a muzzle
velocity (speed) v. The bullet lodges in the block, and the
block ends up with a final speed V.

In terms of m, M, and V :
What is the momentum of the bullet with speed v ?
x
v
V
before
after
Physics 207: Lecture 13, Pg 3
Inelastic collision in 1-D: Example

What is the momentum of the bullet with speed v ?
m
v
 Key question: Is there a net external x-dir force ?
 If not, then momentum in the x-direction is conserved!
P After
P Before
mv  M 0  (m  M )V
aaaa
v
V
after
x
before
v  (1  M / m)V
Physics 207: Lecture 13, Pg 4
Momentum is a vector


p  mv  p x iˆ  p y ˆj  p z kˆ
p x  mv x
p y  mv y
p z  mv z
Physics 207: Lecture 13, Pg 5
A perfectly inelastic collision in 2-D

Consider a collision in 2-D (cars crashing at a
slippery intersection...no friction).
V
v1
q
m1 + m2
m1
m2
v2
before
after
If no external force momentum is conserved.
 Momentum is a vector so px, py and pz

Physics 207: Lecture 13, Pg 6
A perfectly inelastic collision in 2-D
p
p1
V
v1
q
m1 + m2
m1
m2
v2
p
2
before
after
x-dir px : m1 v1 = (m1 + m2 ) V cos q
 y-dir py : m2 v2 = (m1 + m2 ) V sin q

Physics 207: Lecture 13, Pg 7
Exercise
Momentum is a Vector (!) quantity


A.
B.
C.
D.
A block slides down a frictionless ramp and then falls and
lands in a cart which then rolls horizontally without friction
In regards to the block landing in the cart is momentum
conserved?
Yes
No
Yes & No
Too little information given
Physics 207: Lecture 13, Pg 8
Elastic Collisions

Elastic means that the objects do not stick.

There are many more possible outcomes but, if no external
force, then momentum will always be conserved

Start with a 1-D problem.
Before
After
Physics 207: Lecture 13, Pg 12
Billiards

Consider the case where one ball is initially at rest.
after
before
pa q
pb
Pa f
F
The final direction of the red ball will
depend on where the balls hit.
Physics 207: Lecture 13, Pg 13
Billiards: Without external forces, conservation
of momentum (and energy Ch. 10 & 11)



Conservation of Momentum
x-dir Px : m vbefore = m vafter cos q + m Vafter cos f
y-dir Py :
0
= m vafter sin q + m Vafter sin f
after
before
pafter q
pb
F
Pafter f
Physics 207: Lecture 13, Pg 14
Explosions: A collision in reverse

A two piece assembly is hanging
vertically at rest at the end of a 2.0 m
long massless string. The mass of the
two pieces are 60 and 20 kg respectively.
Suddenly you observe that the 20 kg is
ejected horizontally at 30 m/s. The time
of the “explosion” is short compared to
the swing of the string.

Does the tension in the string increase or
decrease after the explosion?
Before
After
Physics 207: Lecture 13, Pg 17
Explosions: A collision in reverse
A two piece assembly is hanging
vertically at rest at the end of a 2.0 m
long massless string. The mass of the
two pieces are 60 and 20 kg respectively.
Suddenly you observe that the 10 kg is
ejected horizontally at 30 m/s.
 Decipher the physics:
1. The green ball recoils in the –x
direction (3rd Law) and, because there is
no net force in the x-direction the xmomentum is conserved.
2. The motion of the green ball is
Before
constrained to a circular path…there
must be centripetal (i.e., radial
acceleration)

After
Physics 207: Lecture 13, Pg 18
Explosions: A collision in reverse
A two piece assembly is hanging vertically at
rest at the end of a 20.0 m long massless string.
The mass of the two pieces are 60 & 20 kg
respectively. Suddenly you observe that the 20
kg mass is suddenly ejected horizontally at 30
m/s.
 Cons. of x-momentum
Px before= Px after = 0 = - M V + m v
V = m v / M = 20*30/ 60 = 10 m/s
Tbefore = Weight = (60+20) x 10 N = 800 N
Before

After
SFy = m ay (radial) = M V2/r = T – Mg
T = Mg + MV2 /r = 600 N + 60x(10)2/20 N = 900 N
Physics 207: Lecture 13, Pg 19
Impulse
(A variable force applied for a given time)

Gravity: At small displacements a “constant” force

Springs often provide a linear force (-k x) towards
its equilibrium position (Chapter 10)
Collisions often involve a varying force
F(t): 0  maximum  0
 We can plot force vs time for a typical collision. The
impulse, J, of the force is a vector defined as the
integral of the force during the time of the collision.

Physics 207: Lecture 13, Pg 20
Force and Impulse
(A variable force applied for a given time)

J a vector that reflects momentum transfer
 t

t
p 
J   F dt   (dp / dt )dt   dp
F
Impulse J = area under this curve !
(Transfer of momentum !)
t
t
Impulse has units of Newton-seconds
ti
tf
Physics 207: Lecture 13, Pg 21
Force and Impulse

Two different collisions can have the same impulse since
J depends only on the momentum transfer, NOT the
nature of the collision.
F
same area
F
t
t big, F small
t
t
t
t small, F big
Physics 207: Lecture 13, Pg 22
Average Force and Impulse
F
Fav
F
Fav
t
t big, Fav small
t
t
t
t small, Fav big
Physics 207: Lecture 13, Pg 23
Exercise
Force & Impulse

Two boxes, one heavier than the other, are initially at rest on
a horizontal frictionless surface. The same constant force F
acts on each one for exactly 1 second.
Which box has the most momentum after the force acts ?
F
A.
B.
C.
D.
light
F
heavy
heavier
lighter
same
can’t tell
Physics 207: Lecture 13, Pg 24
Boxing: Use Momentum and Impulse to estimate g “force”
Physics 207: Lecture 13, Pg 25
Back of the envelope calculation
 t

J   F dt  Favg t
(1) marm~ 7 kg
(2) varm~7 m/s (3) Impact time t ~ 0.01 s
Question: Are these reasonable?
 Impulse J = p ~ marm varm ~ 49 kg m/s
 Favg ~ J/t ~ 4900 N
(1) mhead ~ 7 kg
 ahead = F / mhead ~ 700 m/s2 ~ 70 g !


Enough to cause unconsciousness ~ 40% of a fatal blow
Only a rough estimate!
Physics 207: Lecture 13, Pg 26
Ch. 10 : Kinetic & Potential energies

Kinetic energy, K = ½ mv2, is defined to be the
large scale collective motion of one or a set of
masses

Potential energy, U, is defined to be the “hidden”
energy in an object which, in principle, can be
converted back to kinetic energy

Mechanical energy, EMech, is defined to be the sum
of U and K.
Physics 207: Lecture 13, Pg 28
Lecture 13
Assignment:
 HW6 up soon
 For Wednesday: Read all of chapter 10
Physics 207: Lecture 13, Pg 29