Transcript sy12_oct15_07

Physics 207, Lecture 12, Oct. 15
Agenda: Finish Chapter 9, start Chapter 10
• Chapter 9: Momentum &
 Collisions
 Momentum conservation in 2D
 Impulse
Assignment:
HW5 due Wednesday
 HW6 posted soon

Physics 207: Lecture 12, Pg 1
Impulse & Linear Momentum

Transition from forces to conservation laws
Newton’s Laws  Conservation Laws
Conservation Laws  Newton’s Laws
They are different faces of the same physics
phenomenon for special “cases”
Physics 207: Lecture 12, Pg 2
Lecture 12, Exercise 1
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.
 In which case does the box ends up moving fastest ?
 No external force then, vectorially, COM
A.
B.
C.
Box 1
Box 2
Same
1
2
Physics 207: Lecture 12, Pg 3
Lecture 12, Exercise 1
Momentum Conservation

Which box ends up moving fastest ?

Examine the change in the momentum of the ball.
In the case of box 1 the balls momentum changes sign
and so its net change is largest. Since momentum is
conserved the box must have the largest velocity to
compensate.
(A) Box 1
1
(B) Box 2
(C) same
2
Physics 207: Lecture 12, Pg 4
A perfectly inelastic collision in 2-D

Consider a collision in 2-D (cars crashing at a
slippery intersection...no friction).
V
v1
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 5
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 6
Elastic Collision in 1-D
m2
m1
before
v1b
v2b
x
m1
m2
after
v1a
v2a
Physics 207: Lecture 12, Pg 7
Force and Impulse
(A variable force applied for a given time)

Gravity: usually a constant force to an object

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 8
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 9
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 10
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 11
Example from last time
A 2 kg cart initially at rest on frictionless horizontal
surface is acted on by a 10 N horizontal force along
the positive x-axis for 2 seconds what is the final
velocity?



F is in the x-direction F = ma so a = F/m = 5 m/s2
v = v0 + a t = 0 m/s + 2 x 5 m/s = 10 m/s (+x-direction)
but mv = F t
[which is the area with respect to F(t) curve]
Physics 207: Lecture 12, Pg 12
Lecture 12, 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 13
Lecture 12, 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 ?
(A) heavier
F
(B)
light
lighter
F
(C) same
heavy
Physics 207: Lecture 12, Pg 14
Boxers:
Physics 207: Lecture 12, Pg 15
Back of the envelope calculation
 t

J   F dt  Favg t
(2) varm~7 m/s (3) Impact time t ~ 0.01 s
(1) marm~ 7 kg
 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
Physics 207: Lecture 12, Pg 16
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 17
Energy

What do we mean by an isolated system ?

What do we mean by a conservative force ?

If a force acting on an object act for a period of time then
we have an Impulse  change (transfer) of momentum

What if we consider this force acting over a distance:
Can we identify another useful quantity?
Physics 207: Lecture 12, Pg 18
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 = vy / ay
 So y = vy0 vy / ay
+ ½ ay (vy/ay)2
= (vy vy0 - vy02 ) / ay + ½ (vy2 - 2vy vy0+vy02 ) / ay
 2 ay y = (vy2 - vy02 )
 Finally:
If falling:
may y = ½ m (vy2 - vy02 )
-mg y = ½ m (vy2 - vy02 )
Physics 207: Lecture 12, Pg 19
Energy
-mg y= ½ m (vy2 - vy02 )
-mg (yf – yi) = ½ m ( vyf2 -vyi2 )
A relationship between y displacement and y speed
Rearranging
½ m vyi2 + mgyi = ½ m vyf2 + mgyf
We associate mgy with the “gravitational potential energy”
Physics 207: Lecture 12, Pg 20
Energy

Notice that if we only consider gravity as the external force
then
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 referred to as kinetic energy
Physics 207: Lecture 12, Pg 21
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 E = K + Umech remains a fixed
value.
Emech is called “mechanical energy”
Physics 207: Lecture 12, Pg 22
Another 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 23
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 24
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 25
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 26
Lecture 12, Exercise 3
Conservation of Mechanical Energy
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 27
Lecture 12, Example
The Loop-the-Loop … again


Ub=mgh

U=mg2R
To complete the loop the loop, how high do we
have to let the release the car?
Condition for completing the loop the loop:
Circular motion at the top of the loop (ac = v2 / R)
Use fact that E = U + K = constant !
Car has mass m
h?
R
(A) 2R
(B) 3R
(C) 5/2 R
Recall that “g” is the source of
the centripetal acceleration
and N just goes to zero is
the limiting case.
Also recall the minimum speed
at the top is
v  gR
(D) 23/2 R
Physics 207: Lecture 12, Pg 28
Lecture 12, Example
The Loop-the-Loop … again


Use E = K + U = constant
mgh + 0 = mg 2R + ½ mv2
mgh = mg 2R + ½ mgR = 5/2 mgR
h = 5/2 R (C)
v
gR
h?
R
(A) 2R
(B) 3R
(C) 5/2 R
(D) 23/2 R
Physics 207: Lecture 12, Pg 29
Inelastic collision in 1-D: Example 1

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.
 What is the initial energy of the system ?
 What is the final energy of the system ?
 Is energy conserved?
x
v
V
before
after
Physics 207: Lecture 12, Pg 30
Inelastic collision in 1-D: Example 1

mv
What is the momentum of the bullet with speed v ?
1   1 2
mv  v  mv
2
2
1
 What is the final energy of the system ?
(m  M )V 2
2
aaaa
 Is momentum conserved (yes)?
mv  M 0  (m  M )V
 What is the initial energy of the system ?
Examine Ebefore-Eafter
1
1
1
1
m
1
m
mv 2  [( m  M )V]V  mv 2  (mv)
v  mv 2 1 
2
2
2
2
mM
2
mM
(
v
No!
before
)
V
after
x
 Is energy conserved?
Physics 207: Lecture 12, Pg 31
Example – Fully Elastic Collision


Suppose I have 2 identical bumper cars.
One is motionless and the other is approaching it with
velocity v1. If they collide elastically, what is the final velocity
of each car ?
Identical means m1 = m2 = m
Initially vGreen = v1 and vRed = 0




COM  mv1 + 0 = mv1f + mv2f  v1 = v1f + v2f
COE  ½ mv12 = ½ mv1f2 + ½ mv2f2  v12 = v1f2 + v2f2
v12 = (v1f + v2f)2 = v1f2 +2v1fv2f + v2f2  2 v1f v2f = 0
Soln 1: v1f = 0 and v2f = v1 Soln 2: v2f = 0 and v1f = v1
Physics 207: Lecture 12, Pg 32
Lecture 12, Exercise for home
Elastic Collisions

I have a line of 3 bumper cars all touching. A fourth car
smashes into the others from behind. Is it possible to
satisfy both conservation of energy and momentum if
two cars are moving after the collision?
All masses are identical, elastic collision.
(A) Yes (B) No (C) Only in one special case
v
Before
v1
v2
After?
Physics 207: Lecture 12, Pg 33
Physics 207, Lecture 12, Oct. 15
Agenda: Finish Chapter 9, start Chapter 10
• Chapter 9: Momentum &
 Collisions
 Momentum conservation in 2D
 Impulse
Assignment:
HW5 due Wednesday
 HW6 posted soon
 Finish Chapter 10, Start 11 (Work)

Physics 207: Lecture 12, Pg 34
Lecture 12, Exercise for home
Elastic Collisions



COM  mv = mv1 + mv2 so v = v1 + v2
COE  ½ mv2 = ½ mv12 + ½ mv22
v2 = (v1 + v2)2 = v12 + v22  v1 v2 = 0
(A) Yes
(B) No (C) Only in one special case
Before
After?
Physics 207: Lecture 12, Pg 35
Example of the U - F relationship

For a Hooke’s Law spring,
 U(x) = (1/2)kx2

Notice that the derivative gives Hooke’s Law
 Fx = - d ( (1/2)kx2) / dx
 Fx = - 2 (1/2) kx
 Fx = -kx
Physics 207: Lecture 12, Pg 36
Equilibrium

Example
 Spring: Fx = 0 => dU / dx = 0 for x=0
The spring is in equilibrium position

In general: dU / dx = 0  for ANY function
establishes equilibrium
U
stable equilibrium
U
unstable equilibrium
Physics 207: Lecture 12, Pg 37
Hooke’s Law Springs:
Another kind of potential energy
Physics 207: Lecture 12, Pg 38
Elastic vs. Inelastic Collisions

A collision is said to be elastic when energy as well as
momentum is conserved before and after the collision.
Kbefore = Kafter
 Carts colliding with a perfect spring, billiard balls, etc.
vi

A collision is said to be inelastic when energy is not
conserved before and after the collision, but momentum is
conserved.
Kbefore  Kafter
 Car crashes, collisions where objects stick together, etc.
Physics 207: Lecture 12, Pg 39
Comment on Energy Conservation

We have seen that the total kinetic energy of a system
undergoing an inelastic collision is not conserved.
 Mechanical energy is lost:
 Heat (friction)
 Bending of metal and deformation

Kinetic energy is not conserved since negative work is by
a non-conservative force done during the collision !

Momentum along a certain direction is conserved when
there are no external forces acting in this direction.
 In general, easier to satisfy than energy conservation.
Physics 207: Lecture 12, Pg 40
Example of 2-D Elastic collisions:
Billiards

If all we are given is the initial velocity of the cue ball, we
don’t have enough information to solve for the exact paths
after the collision. But we can learn some useful things...
Physics 207: Lecture 12, Pg 41
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 42
Billiards: All that really matters is conservation
of energy and momentum



COE: ½ m vb2 = ½ m va2 + ½ m Va2
x-dir COM: m vb = m va cos q + m Vb cos f
y-dir COM:
0 = m va sin q + m Vb sin f
after
before
pa q
pb
vcm
F
Active Figure
 The
Pa f
final directions are separated by 90° : q – f = 90°
Physics 207: Lecture 12, Pg 43
Lecture 12 – Exercise 4
Pool Shark

Can I sink the red ball without scratching (sinking the
cue ball) ?
(Ignore spin and friction)
(A) Yes
(B) No
(C) More info needed
Physics 207: Lecture 12, Pg 44
Lecture 12 – Exercise 4
Pool Shark

Can I sink the red ball without scratching (sinking the
cue ball) ?
(Ignore spin and friction)
(A) Yes
(B) No
(C) More info needed
Physics 207: Lecture 12, Pg 45