Download Lecture 12 Inelastic collision in 1

Physics 207 – Lecture 12
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
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Physics 207 – Lecture 12
Inelastic collision in 1-D: Example
What is the momentum of the bullet with speed v ?
r
mv
Key question: Is x-momentum conserved ?
Before
After
aaaa
mv + M 0 = (m + M )V
v
V
x
before
after
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
2
1
Physics 207: Lecture 12, Pg 4
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Physics 207 – Lecture 12
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
2
1
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
θ
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
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Physics 207 – Lecture 12
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
θ
m1 + m2
m1
m2
v2
before
after
x-dir px : m1 v1 = (m1 + m2 ) V cos θ
y-dir py : m2 v2 = (m1 + m2 ) V sin θ
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
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Physics 207 – Lecture 12
Billiards
Consider the case where one ball is initially at rest.
after
before
pa θ
pb
vcm
Pa φ
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 θ + m Vafter cos φ
y-dir Py :
0
= m vafter sin θ + m Vafter sin φ
after
before
pafter θ
pb
F
Pafter φ
Physics 207: Lecture 12, Pg 10
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Physics 207 – Lecture 12
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
r tr
J = ∫ F dt = ∫t (dpr / dt )dt = ∫ p dpr
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
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Physics 207 – Lecture 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
∆t
∆t big, F small
t
∆t small, F big
Physics 207: Lecture 12, Pg 13
Average Force and Impulse
F
Fav
F
Fav
∆t
t
∆t
∆t big, Fav small
t
∆t small, Fav big
Physics 207: Lecture 12, Pg 14
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Physics 207 – 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 15
Boxing: Use Momentum and Impulse to estimate g “force”
Physics 207: Lecture 12, Pg 16
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Physics 207 – Lecture 12
Back of the envelope calculation
r tr
r
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
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Physics 207 – Lecture 12
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
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Physics 207 – Lecture 12
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
∆y = ½ (vy2 - vy02 ) / ay
+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
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Physics 207 – Lecture 12
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
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Physics 207 – Lecture 12
Energy
If only “conservative”
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
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Physics 207 – Lecture 12
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
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Physics 207 – Lecture 12
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?
ai = g sin 30°
= 5 m/s2
1st Part: Find v at bottom of incline
j
N
i
Emech is conserved
Ki+ Ui = Kf
+ Uf
5.0 m
2
0 + mgyi = ½ mv + 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 a i ∆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
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Physics 207: Lecture 12, Pg 30
Physics 207 – Lecture 12
Home exercise
A block is shot up a frictionless 40°slope with initial ve locity
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
3
2
h
(A) 1 > 2 > 3
(B) 3 > 2 > 1
(C) 3 = 2 = 1 (D) Can’t tell
Physics 207: Lecture 12, Pg 32
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