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Lecture 13
Goals:
• Chapter 9
•
 Employ conservation of momentum in 2D
 Examine forces over time (aka Impulse)
Chapter 10
 Understand the relationship between motion and energy
Assignments:
 HW5, due tomorrow, HW6 due Tuesday, Oct. 19th
 For Wednesday, Read all of Chapter 10
Physics 207: Lecture 13, Pg 1
Exercise: Momentum is a Vector (!) quantity
Let a 2 kg block start at rest on a 30° incline and slide vertically
a distance 5.0 m and fall a distance 7.5 m into the 10 kg cart
System….block and cart only (assume Earth has infinite mass)
Px is conserved once the block leaves the slide
Py is not conserved (if “system” is block and cart only)
2 kg
5.0 m
30°
10 kg
What is the final velocity of the cart? 7.5 m
Physics 207: Lecture 13, Pg 2
Exercise
Momentum is a Vector (!) quantity


1) ai = g sin 30°
= 5 m/s2
x-direction: No net force so Px is conserved
y-direction: vy of the cart + block will be zero
and we can ignore vy of the block when it
2) d = 5 m / sin 30°
lands in the cart.
j
= ½ ai Dt2
N
i 5.0 m
30° mg
30°
Initial
Final
Px: MVx + mvx = (M+m) V’x
M 0 + mvx = (M+m) V’x
V’x = m vx / (M + m)
= 2 (8.7)/ 12 m/s
V’x = 1.4 m/s
10 m = 2.5 m/s2 Dt2
2s = Dt
v = ai Dt = 10 m/s
vx= v cos 30°
= 8.7 m/s
7.5 m
y
x
Physics 207: Lecture 13, Pg 3
Home Exercise
Inelastic Collision in 1-D with numbers
Do not try this at home!
ice
(no friction)
Before: A 4000 kg bus, twice the mass of the car, moving
at 30 m/s impacts the car at rest.
What is the final speed after impact if they move together?
Physics 207: Lecture 13, Pg 4
Home exercise
Inelastic Collision in 1-D
M = 2m
m
initially
v=0
V0
ice
(no friction)
M
2m
MV0  (m  M )V or V 
V0 
V0
mM
2m  m
finally
2Vo/3 = 20 m/s
vf =?
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


If no external force momentum is conserved.
Momentum is a vector so px, py and pz are conseved
V
v1
q
m1 + m2
m1
m2
v2
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
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 8
Billiards

Consider the case where one ball is initially at rest.
after
before
pa q
pb
vcm
Pa f
F
The final direction of the red ball will
depend on where the balls hit.
Physics 207: Lecture 13, Pg 9
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 10
Explosions: A collision in reverse


A two piece assembly is hanging
vertically at rest at the end of a 3.0 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 11
Explosions: A collision in reverse
A two piece assembly is hanging
vertically at rest at the end of a 3.0 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.
 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 12
Explosions: A collision in reverse
A two piece assembly is hanging vertically at rest
at the end of a 3.0 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)2/3 N = 2900 N
Physics 207: Lecture 13, Pg 13
Impulse
(A variable force applied for a given time)

Gravity: At small displacements a “constant” force t

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 14
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
Dt
Impulse has units of Newton-seconds
ti
tf
Physics 207: Lecture 13, Pg 15
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
Dt
Dt big, F small
t
Dt
t
Dt small, F big
Physics 207: Lecture 13, Pg 16
Average Force and Impulse
F
Fav
F
Fav
Dt
Dt big, Fav small
t
Dt
t
Dt small, Fav big
Physics 207: Lecture 13, Pg 17
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 18
Boxing: Use Momentum and Impulse to estimate g “force”
Physics 207: Lecture 13, Pg 19
Back of the envelope calculation
 t

J   F dt  Favg Dt
(1) marm~ 7 kg
(2) varm~7 m/s (3) Impact time Dt ~ 0.01 s
Question: Are these reasonable?
 Impulse J = Dp ~ marm varm ~ 49 kg m/s
 Favg ~ J/Dt ~ 4900 N
(1) mhead ~ 7 kg
 ahead = F / mhead ~ 700 m/s2 ~ 70 g !


Enough to cause unconsciousness ~ 35% of a fatal blow
Only a rough estimate!
Physics 207: Lecture 13, Pg 20
Woodpeckers
During "collision" with a tree
abeak ~ 600 - 1500 g
How do they survive?
• Jaw muscles act as shock
absorbers
• Straight head trajectory
reduces damaging
rotations (rotational motion
is very problematic)
Physics 207: Lecture 13, Pg 21
Discussion Exercise

The only force acting on a 2.0 kg object moving along the xaxis. Notice that the plot is force vs time.
If the velocity vx is +2.0 m/s at 0 sec, what is vx at 4.0 s ?

Dp = m Dv = Impulse

m Dv = J0,1 + J1,2 + J2,4

m Dv = (-8)1 N s +
½ (-8)1 N s + ½ 16(2) N s

m Dv = 4 N s
Dv = 2 m/s
vx = 2 + 2 m/s = 4 m/s
Physics 207: Lecture 13, Pg 22
Newton’s Laws rearranged (Ch. 10)
From motion in the y-dir, Fy = m ay and let ay be constant


y(t) = y0 + vy0 Dt + ½ ay Dt2 
Dy = y(t)-y0= vy0 Dt + ½ ay Dt2
vy (t) = vy0 + ay Dt

Dt = (vy - vy0) / ay
and eliminating Dt yields

2 ay Dy= (vy2 - vy02 ) which is nothing new
or if ay= -g
-mg Dy= ½ m (vy2 - vy02 )
Physics 207: Lecture 13, Pg 23
“Energy” defined
-mg Dy= ½ m (vy2 - vy02 )
-mg (yf – yi) = ½ m ( vyf2 -vyi2 )
A relationship between
y- displacement and change in the y-speed squared
Rearranging to give initial on the left and final on the right
½ m vyi2 + mgyi = ½ m vyf2 + mgyf
We now define mgy as the “gravitational potential energy”
Physics 207: Lecture 13, Pg 24
Energy

Notice that if we only consider gravity as the external force then
the x and z velocities remain constant

To
½ m vyi2 + mgyi = ½ m vyf2 + mgyf

Add
½ m vxi2 + ½ m vzi2
and
½ m vxf2 + ½ m vzf2
½ m vi2 + mgyi = ½ m vf2 + mgyf

where
vi2 ≡ vxi2 +vyi2 + vzi2
½ m v2 terms are defined to be kinetic energies
(A scalar quantity of motion)
Physics 207: Lecture 13, Pg 25
Lecture 13
Assignment:
 HW6 up soon
 For Wednesday: Read all of chapter 10
Physics 207: Lecture 13, Pg 27