Transcript Physics 207: Lecture 2 Notes

Lecture 12
Goals:
•
•
Chapter 9: Momentum & Impulse
 Solve problems with 1D and 2D Collisions
 Solve problems having an impulse (Force vs. time)
Chapter 10
 Understand the relationship between motion and
energy
 Define Potential & Kinetic Energy
 Develop and exploit conservation of energy principle
Assignment:
 HW5 due Wednesday
 For Wednesday: Read all of chapter 10
Physics 207: Lecture 12, Pg 1
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 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 12, Pg 2
Inelastic collision in 1-D: Example
What is the momentum of the bullet with speed v ?

mv
 Key question: Is x-momentum conserved ?
Before
After
aaaa
mv  M 0  (m  M )V
v
V
after
x
before
Physics 207: Lecture 12, Pg 3
Exercise
Momentum Conservation

Two balls of equal mass are thrown horizontally with the
same initial velocity. They hit identical stationary boxes
resting on a frictionless horizontal surface.

The ball hitting box 1 bounces elastically back, while the ball
hitting box 2 sticks.
 Which box ends up moving fastest ?
A.
B.
C.
Box 1
Box 2
same
1
2
Physics 207: Lecture 12, Pg 4
Exercise Momentum Conservation

The ball hitting box 1 bounces elastically back, while the ball
hitting box 2 sticks.
 Which box ends up moving fastest ?
 Notice the implications from the graphical solution: Box 1’s
momentum must be bigger because the length of the
summed momentum must be the same.
 The longer the green vector the greater the speed
Before
After
Before
After
Ball 1
Ball 1
Ball 2
Ball 2
Box 1
Box 1
Box 2
Box 2
Box 2+Ball 2
Box 2+Ball 2
Box 1+Ball 1 Box 1+Ball 1
1
2
Physics 207: Lecture 12, 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 12, 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 12, 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 12, 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 12, Pg 9
Billiards: All that really matters is conservation
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 12, Pg 10
Force and 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 12, Pg 11
Force and Impulse
(A variable force applied for a given time)

J 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 12, Pg 12
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 12, Pg 13
Average Force and Impulse
F
Fav
F
Fav
t
t big, Fav small
t
t
t
t small, Fav big
Physics 207: Lecture 12, Pg 14
Exercise 2
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 12, Pg 15
Boxing: Use Momentum and Impulse to estimate g “force”
Physics 207: Lecture 12, Pg 16
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
 F ~ J/t ~ 4900 N
(1) mhead ~ 6 kg
 ahead = F / mhead ~ 800 m/s2 ~ 80 g !


Enough to cause unconsciousness ~ 40% of fatal blow
Only a rough estimate!
Physics 207: Lecture 12, Pg 17
Woodpeckers
During "collision" with a tree
ahead ~ 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 12, Pg 18
Home 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 ?

p = m v = Impulse

m v = J0,1 + J1,2 + J2,4

m v = (-8)1 N s +
½ (-8)1 N s + ½ 16(2) N s

m v = 4 N s
v = 2 m/s
vx = 2 + 2 m/s = 4 m/s
Physics 207: Lecture 12, Pg 19
Chapter 10: Energy

We need to define an “isolated system” ?

We need to define “conservative force” ?

Recall, chapter 9, force acting for a period of time
gives an
impulse or a change (transfer) of momentum

What if a force acting over a distance:
Can we identify another useful quantity?
Physics 207: Lecture 12, Pg 20
Energy
Fy = m ay and let the force be constant
 y(t) = y0 + vy0 t + ½ ay t2 
y = y(t)-y0= vy0 t + ½ ay t2
vy (t) = vy0 + ay t 
t = (vy - vy0) / ay
Eliminate t and regroup
 So y = vy0 (vy- vy0) / ay+ ½ ay (vy2 - 2vy vy0+vy02 ) / ay2

y = (vy0vy- vy02) / ay+ ½
(vy2 - 2vy vy0+vy02 ) / ay
y = (
(vy2 -
- vy02) / ay+ ½
+vy02 ) / ay
y = ½ (vy2 - vy02 ) / ay
Physics 207: Lecture 12, Pg 21
Energy
And now
y = ½ (vy2 - vy02 ) / ay
can be rewritten as:
may y = ½ m (vy2 - vy02 )
And if the object is falling under the influence of gravity
then
ay= -g
Physics 207: Lecture 12, Pg 22
Energy
-mg y= ½ m (vy2 - vy02 )
-mg (yf – yi) = ½ m ( vyf2 -vyi2 )
A relationship between y-displacement and change in the
y-speed
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 12, Pg 23
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 12, Pg 24
Energy

If only “conservative” forces are present, the total energy
(sum of potential, U, and kinetic energies, K) of a system is
conserved
Emech = K + U
Emech = K + U = constant

K and U may change, but Emech = K + U remains a fixed value.
Emech is called “mechanical energy”
Physics 207: Lecture 12, Pg 25
Example of a conservative system:
The simple pendulum.

Suppose we release a mass m from rest a distance h1 above
its lowest possible point.
 What is the maximum speed of the mass and where does
this happen ?
 To what height h2 does it rise on the other side ?
m
h1
h2
v
Physics 207: Lecture 12, Pg 26
Example: The simple pendulum.
 What is the maximum speed of the mass and
where does this happen ?
E = K + U = constant and so K is maximum when U
is a minimum.
y
y=h1
y=
0
Physics 207: Lecture 12, Pg 27
Example: The simple pendulum.
What is the maximum speed of the mass and
where does this happen ?
E = K + U = constant and so K is maximum when U
is a minimum
E = mgh1 at top
E = mgh1 = ½ mv2 at bottom of the swing
y
y=h1
y=0
h1
v
Physics 207: Lecture 12, Pg 28
Example: The simple pendulum.
To what height h2 does it rise on the other side?
E = K + U = constant and so when U is maximum
again (when K = 0) it will be at its highest point.
E = mgh1 = mgh2 or h1 = h2
y
y=h1=h2
y=0
Physics 207: Lecture 12, Pg 29
Cart Exercise Revisited: How does this help?
 1st
ai = g sin 30°
= 5 m/s2
Part: Find v at bottom of incline
j
N
i
Emech is conserved
Ki+ Ui = Kf
+ Uf
5.0 m
0 + mgyi = ½ mv2 + 0
(2gy)½ = v = (100) ½ m/s
vx= v cos 30°
= 8.7 m/s
One step, no FBG needed
mg
30°
d = 5 m / sin 30°
= ½ ai t2
10 m = 2.5 m/s2 t2
2s = t
v = ai t = 10 m/s
vx= v cos 30°
= 8.7 m/s
7.5 m
y
x
Physics 207: Lecture 12, Pg 30
Home exercise
A block is shot up a frictionless 40° slope with initial velocity
v. It reaches a height h before sliding back down. The
same block is shot with the same velocity up a frictionless
20° slope.
On this slope, the block reaches height
A.
B.
C.
D.
E.
2h
h
h/2
Greater than h, but we can’t predict an exact value.
Less than h, but we can’t predict an exact value.
Physics 207: Lecture 12, Pg 31
Exercise 3: U, K, E & Path
A ball of mass m, initially at rest, is released and follows three
difference paths. All surfaces are frictionless
1. Ball is dropped
2. Ball slides down a straight incline
3. Ball slides down a curved incline
After traveling a vertical distance h, how do the speeds compare?

1
2
3
h
(A) 1 > 2 > 3
(B) 3 > 2 > 1
(C) 3 = 2 = 1 (D) Can’t tell
Physics 207: Lecture 12, Pg 32