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Lecture 12
Goals:Goals:• Momentum & ImpulseMomentum & Impulse
Solve problems with 1D and 2D Collisions Solve problems having an impulse (Force vs. time)
• EnergyEnergy Understand the relationship between motion and energy Define Potential & Kinetic Energy Develop and exploit conservation of energy principle
Inelastic collision in 1-D: Example
Inelastic means that the objects stick.
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 ?
vV
before after
x
A perfectly inelastic collision in 1-D
Example
What is the momentum of the bullet with speed v ?
Key question: Is x-momentum conserved ?
vV
before after
x
aaaa
vm
V)( 0 M v Mmm BeforeBefore AfterAfter
Exercise Momentum Conservation
A. Box 1
B. Box 2
C. same
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 ?
1 2
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
1 2
Before After Before After
Ball 1 Ball 1 Ball 2 Ball 2
Box 1 Box 1 Box 2 Box 2
Box 1+Ball 1 Box 1+Ball 1 Box 2+Ball 2 Box 2+Ball 2
A perfectly inelastic collision in 2-D
Consider a collision in 2-D (cars crashing at a slippery intersection...no friction).
vv1
vv2
VV
before after
m1
m2
m1 + m2
If no external force momentum is conserved. Momentum is a vector so px, py and pz
A perfectly inelastic collision in 2-D
vv1
vv2
VV
before after
m1
m2
m1 + m2
x-dir px : m1 v1 = (m1 + m2 ) V cos y-dir py : m2 v2 = (m1 + m2 ) V sin
If no external force momentum is conserved. Momentum is a vector so px, py and pz are conseved
Elastic Collisions
Elastic means that the objects do not stick.
If no external force, then momentum will always be conserved
Start with a 1-D problem.
Before After
Billiards
Consider the case where one ball is initially at rest.
ppa
ppb
FF PPa
beforeafter
The final direction of the red ball will depend on where the balls hit.
vvcm
Billiards: All that really matters is conservation momentum (and energy Ch. 10 & 11)
Conservation of Momentum x-dir Px : m vbefore = m vafter cos + m Vafter cos y-dir Py : 0 = m vafter sin + m Vafter sin
ppafter
ppb
FF PPafter
before after
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 (-kx) towards its equilibrium position (Chapter 10)
Collisions often involve a varying force
F(t): 0 maximum 0We can plot force vs time for a typical collision.
The impulse, ΔΔρρ, of the force is a vector defined as the integral of the force during the time of the collision.
Force and Impulse (A variable force applied for a given time)
F
ΔΔρ ρ reflects momentum transfer
t
ti tf
t
Impulse ΔΔρρ = area under this curve !
(Transfer of momentum !)
Impulse has units of Newton-seconds
ΔΔρρ
Force and Impulse Two different collisions can have the same impulse since
ΔΔρρ depends only on the momentum transfer, NOT the nature of the collision.
t
F
t
F
tt
same area
t big, FF smallt small, FF big
Exercise Force & Impulse
A. heavier
B. lighter
C. same
D. can’t tell
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 F light heavy
Back of the envelope calculation
(1) marm~ 7 kg (2) varm~7 m/s (3) Impact time t ~ 0.01 s
Question: Are these reasonable? Impulse p ~ marm varm ~ 49 kg m/s F ~ Δρ/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!
الله سبحان
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)
Woodpeckers
Home Exercise The only force acting on a 2.0 kg object moving along the x-
axis. 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
mv =+ +
mv = (-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
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?
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 (vy
2 - 2vy vy0+vy0
2 ) / ay
2
y = (vy0vy- vy02) / ay+ ½ (vy
2 - 2vy vy0+vy0
2 ) / ay
y = ( - vy02) / ay+ ½ (vy
2 - +vy02
) / ay
y = ½ (vy2 - vy0
2 ) / ay
Energy
And now
y = ½ (vy2 - vy0
2 ) / ay
can be rewritten as:
may y = ½ m (vy2 - vy0
2 )
And if the object is falling under the influence of gravity
then
ay= -g
Energy
-mg y= ½ m (vy2 - vy0
2 )
-mg yf – yi) = ½ m ( vyf2 -vyi
2 )
Rearranging to give initial on the left and final on the right
½ m vyi2 + mgyi = ½ m vyf
2 + mgyf
We now define mgy as the “gravitational potential energy”
A relationship between y-displacement and change in the
y-speed
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 vyf
2 + mgyf
Add ½ m vxi2 + ½ m vzi
2 and ½ m vxf2 + ½ m vzf
2
½ m vi2 + mgyi = ½ m vf
2 + mgyf
where vi2 = vxi
2 +vyi2 + vzi
2
½ m v2 terms are defined to be kinetic energies
(A scalar quantity of motion)
Energy
If only “conservative” forces are present, the total energy If only “conservative” forces are present, the total energy
((sum of potential, U, and kinetic energies, K) of a system) of a system is is conservedconserved
Emech = K + U
K and U may change, but Emech = K + U remains a fixed value.
Emech = K + U = constant
Emech is called “mechanical energy”
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 ?
v
h1 h2
m
Example: The simple pendulum.
y
y=0
y=h1
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.
Example: The simple pendulum.
v
h1
y
y=h1
y=0
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
Example: The simple pendulum.
y
y=h1=h2
y=0
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
Cart Exercise Revisited: How does this help?
1st Part: Find v at bottom of incline
5.0 m
30°
7.5 m
N
mg
ai = g sin 30°
= 5 m/s2
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
i
j
x
y
Emech is conserved
Ki+ Ui = Kf + Uf
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
Exercise : 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?
(A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell
h
1 32