lecture 12 momentum, energy, and collisions. announcements exam: thursday, october 17 (chapter 6-9)...
TRANSCRIPT
Lecture 12Momentum, Energy, and
Collisions
Announcements
•EXAM: Thursday, October 17 (chapter 6-9)
•Equation sheet to be posted
•Practice exam to be posted
Linear Momentum
Impulse
With no net external force:
Center of MassThe center of mass of a system is the point where the
system can be balanced in a uniform gravitational field.
For two objects:
The center of mass is closer to the more massive object.
Think of it as the “average location of the mass”
Center of Mass
(1)
XCM
(2)
a) higher
b) lower
c) at the same place
d) there is no definable CM in this case
The disk shown below in (1) clearly has its center of mass at the center.
Suppose the disk is cut in half and the pieces arranged as shown in (2).Where is the center of mass of (2) as compared to (1) ?
(1)
XCM
(2)
CM
Center of Mass
The CM of each half is closer to the top of the semicircle than the bottom. The CM of the whole system is located at the midpoint of the two semicircle CMs, which is higher than the yellow line.
a) higher
b) lower
c) at the same place
d) there is no definable CM in this case
The disk shown below in (1) clearly has its center of mass at the center.
Suppose the disk is cut in half and the pieces arranged as shown in (2).
Where is the center of mass of (2) as compared to (1) ?
When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)?
a) it is much sharper than the gun
b) it is smaller and can penetrate your body
c) it has more kinetic energy than the gun
d) it goes a longer distance and gains speed
e) it has more momentum than the gun
Gun Control
When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)?
a) it is much sharper than the gun
b) it is smaller and can penetrate your body
c) it has more kinetic energy than the gun
d) it goes a longer distance and gains speed
e) it has more momentum than the gun
Even though it is true that the magnitudes of the momenta of the gun and the bullet are equal, the bullet is less massive and so it has a much higher velocity. Because KE is related to v2, the bullet has considerably more KE and therefore can do more damage on impact.
Gun Control
Rolling in the Rain
a) speeds up
b) maintains constant speed
c) slows down
d) stops immediately
An open cart rolls along a
frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)
Because the rain falls in vertically, it
adds no momentum to the box, thus
the box’s momentum is conserved.
However, because the mass of the box
slowly increases with the added rain,
its velocity has to decrease.
Rolling in the Rain
a) speeds up
b) maintains constant speed
c) slows down
d) stops immediately
An open cart rolls along a
frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)
Two objects collide... and stick
A completely inelastic collision: no “bounce back”
No external forces... so momentum of system is conservedinitial px = mv0
final px = (2m)vf
mv0 = (2m)vf
vf = v0 / 2
mass m mass m
Inelastic collision: What about energy?
mass m mass m vf = v0 / 2
Kinetic energy is lost! KEfinal = 1/2 KEinitial
initial
final
Collisions
This is an example of an “inelastic collision”
Collision: two objects striking one another
Elastic collision ⇔ “things bounce back”
⇔ energy is conserved
Inelastic collision: less than perfectly bouncy
⇔ Kinetic energy is lost
Time of collision is short enough that external forces may be ignored so momentum is conserved
Completely inelastic collision: objects stick together afterwards. Nothing “bounces back”. Maximal energy loss
Elastic vs. Inelastic
Inelastic collision: momentum is conserved but kinetic energy is not
Completely inelastic collision: colliding objects stick together, maximal loss of kinetic energy
Elastic collision: momentum and kinetic energy is conserved.
Completely Inelastic Collisions in One Dimension
Solving for the final momentum in terms of initial velocities and masses, for a 1-dimensional, completely inelastic collision between unequal masses:
Completely inelastic only(objects stick together, so have same final velocity)
KEfinal < KEinitial
Momentum Conservation:
Crash Cars I
a) I
b) II
c) I and II
d) II and III
e) all three
If all three collisions below
are totally inelastic, which
one(s) will bring the car on
the left to a complete halt?
Crash Cars I
In case I, the solid wall clearly stops the car.
In cases II and III, because
ptot = 0 before the collision,
then ptot must also be zero
after the collision, which means that the car comes to a halt in all three cases.
a) I
b) II
c) I and II
d) II and III
e) all three
If all three collisions below
are totally inelastic, which
one(s) will bring the car on
the left to a complete halt?
Crash Cars II
If all three collisions below are
totally inelastic, which one(s)
will cause the most damage (in
terms of lost energy)?
a) I
b) II
c) III
d) II and III
e) all three
Crash Cars II
a) I
b) II
c) III
d) II and III
e) all three
The car on the left loses the same KE in all three cases, but in case III, the car on the right loses the most KE because KE = mv2 and the car in case III has the largest velocity.
If all three collisions below are
totally inelastic, which one(s) will
cause the most damage (in
terms of lost energy)?
Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.
vf = m v0 / (m+M)
momentum conservation in inelastic collision
KE = 1/2 (mv0)2 / (m+M)
energy conservationafterwards
PE = (m+M) g h
hmax = (mv0)2 / [2 g (m+M)2]
Velocity of the ballistic pendulum
Pellet Mass (m): 2 gPendulum Mass (M): 3.81 kgWire length (L): 4.00 m
approximation
Inelastic Collisions in 2 Dimensions
Energy is not a vector equation: there is only 1 conservation of energy equation
Momentum is a vector equation: there is 1 conservation of momentum equation per dimension
For collisions in two dimensions, conservation of momentum is applied separately along each axis:
Elastic CollisionsIn elastic collisions, both kinetic
energy and momentum are conserved.
One-dimensional elastic collision:
Elastic Collisions in 1-dimension
solving for the final speeds:
Note: relative speed is conserved for head-on (1-D) elastic collision
We have two equations: conservation of momentum conservation of energy
and two unknowns (the final speeds).
For special case of v2i = 0
Limiting cases of elastic collisions
note: relative speed conserved
Limiting cases
note: relative speed conserved
Limiting cases
note: relative speed conserved
Toy Pendulum
Could two balls recoil and conserve both momentum and energy?
Incompatible!
Elastic Collisions I
v 2v
1at rest
at rest
a) situation 1
b) situation 2
c) both the same
Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on.
In which case does the golf ball have the greater speed after the collision?
Remember that the magnitude of the relative velocity has to be equal before and after the collision!
Elastic Collisions I
v1
In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v.
v 22v
In case 2 the bowling ball will keep going with speed close to v, hence the golf ball will rebound with speed close to 2v.
a) situation 1
b) situation 2
c) both the same
Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on.
In which case does the golf ball have the greater speed after the collision?
A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a 0.150 kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?
A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a 0.150 kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?
Our simplest formulas for speed after an elastic collision relied on one body being initially at rest.
So lets try a frame where one body (the elephant) is at rest!
In the frame, what is the final speed of the ball?
Back in the frame of the ground:
What is the speed of the ball relative to the elephant?
A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a 0.150 kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?
Our simplest formulas for speed after an elastic collision relied on one body being initially at rest.
So lets try a frame where one body (the elephant) is at rest!
In the frame, what is the final speed of the ball?
Back in the frame of the ground:
What is the speed of the ball relative to the elephant?
NOTE: Formulas for 1-D elastic scattering with non-zero initial velocities are given in end-of-chapter problem 88.
Elastic Collisions in 2-D
Two-dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object
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A proton collides elastically with another proton that is initially at rest.
The incoming proton has an initial speed of 3.5x105 m/s and makes
a glancing collision with the second proton. After the collision one
proton moves at an angle of 37o to the original direction of motion,
the other recoils at 53o to that same axis. Find the final speeds of the
two protons.
v0 = 3.5x105 m/s
initial
37o
53o
v2
v1
final
A proton collides elastically with another proton that is initially at rest.
The incoming proton has an initial speed of 3.5x105 m/s and makes
a glancing collision with the second proton. After the collision one
proton moves at an angle of 37o to the original direction of motion,
the other recoils at 53o to that same axis. Find the final speeds of the
two protons.
v0 = 3.5x105 m/s
initial
37o
53o
v2
v1
final
Momentum conservation:
if we’d been given only 1 angle, would have needed conservation of energy also!
Rotational Kinematics
Angular Position
Degrees and revolutions:
Angular Positionθ > 0
θ < 0
Arc Length
Arc length s, from angle measured in radians:
s = rθ
- Arc length for a full rotation (360o) of a radius=1m circle?
- What is the relationship between the circumference of a circle and its diameter?C / D = π
C = 2 π r
s = 2 π (1 m) = 2 π meters
1 complete revolution = 2 π radians
1 rad = 360o / (2π) = 57.3o
Angular Velocity
Instantaneous Angular Velocity
Period = How long it takes to go 1 full revolution
Period T:
SI unit: second, s
Linear and Angular Velocity
Greater translation for same rotation
Bonnie and Klyde II
ω
BonnieBonnieKlydeKlyde
a) Klyde
b) Bonnie
c) both the same
d) linear velocity is zero for both of them
Bonnie sits on the outer rim of amerry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one revolution every 2 seconds. Who has the larger linear (tangential) velocity?
a) Klyde
b) Bonnie
c) both the same
d) linear velocity is zero for both of them
Their linear speeds linear speeds vv will be
different because v = r v = r ωω
and Bonnie is located farther Bonnie is located farther
outout (larger radius r) than
Klyde.
ω
BonnieBonnie
KlydeKlyde
Bonnie and Klyde II
Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one revolution every 2 seconds. Who has the larger linear (tangential) velocity?
Angular Acceleration
Instantaneous Angular Acceleration
Rotational Kinematics, Constant Acceleration
If the acceleration is constant:
If the angular acceleration is constant:
v = v0 + at
Analogies between linear and rotational kinematics:
An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in the time t, through what angle did it rotate in the time t?
Angular Displacement IAngular Displacement I
½
a) θ
b) θ
c) θ
d) 2 θ
e) 4 θ
½
¼
¾
An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in the time t, through what angle did it rotate in the time t ?
a) θ
b) θ
c) θ
d) 2 θ
e) 4 θ
The angular displacement is θ = αt 2 (starting from rest), and
there is a quadratic dependence on time. Therefore, in half the half the
timetime, the object has rotated through one-quarter the angleone-quarter the angle.
Angular Displacement IAngular Displacement I½
¼
½
¾
Which child experiences a greater acceleration?(assume constant angular speed)
Larger r: - larger v for same ω - larger ac for same ω
ac is required for circular motion.
An object may have at as well, which implies angular acceleration
Angular acceleration and total linear acceleration