lecture 12 momentum, energy, and collisions. announcements assignment 6 due wednesday, oct 5...
TRANSCRIPT
Lecture 12
Momentum, Energy, and Collisions
Announcements
Assignment 6 due Wednesday, Oct 5 (11:59pm)
EXAM: October 13 (through Chapter 9)
Look for messages regarding special TA office hours
Reminder that formula sheet for Test 2 now on Collab under Midterms
Practice problems will be attached to posted slides (this afternoon, I promise)
Practice test from last year on Collab under Resources
Reading and Review
Linear Momentum
Impulse
With no net external force:
A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v along the floor, but at right angles to one another. Find the speed and direction of the third piece.
A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v along the floor, but at right angles to one another. Find the speed and direction of the third piece.
We know that px=0, py = 0 in initial stateand no external forces act in the horizontal
An 85-kg lumberjack stands at one end of a 380-kg floating log, as shown in the figure. Both the log and the lumberjack are at rest initially. (a) If the lumberjack now trots toward the other end of the log with a speed of 2.7 m/s relative to the log, what is the lumberjack’s speed relative to the shore? Ignore friction between the log and the water. (b) If the mass of the log had been greater, would the lumberjack’s speed relative to the shore be greater than, less than, or the same as in part (a)? Explain. (c) Check (b) by assuming log has mass of 450 kg.
Rolling in the RainRolling 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 RainRolling 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.)
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 ControlGun 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 ControlGun Control
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
What about energy?
mass m mass m vf = v0 / 2
Kinetic energy is lost! KEfinal = 1/2 KEinitial
initial
final
Inelastic Collisions
This is an example of an “inelastic collision”
Collision: two objects striking one another
“Elastic” collision <=> “things bounce back”
Completely inelastic collision: objects stick together afterwards... no thing “bounces back”
Time of collision is short enough that external forces may be ignored so momentum is conserved
Inelastic collision: momentum is conserved but kinetic energy is not
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:
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]
Binomial expansion
If <<1 then
(1+ )n ~1+n
Ex.
3
33
1.02 1.061208
1.02 1.00 0.02 1.00 3 0.02 1.06
correct with 0.1%
Velocity of the ballistic pendulum
Pellet Mass (m): 1.84x10-3 kgPendulum Mass (M): 3.81 kgWire length (L): 4.00 m
30 3.24 10 /v s x
Crash Cars ICrash 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 ICrash 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 IICrash 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 IICrash 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
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.
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
For given case of v2i = 0, 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).
Limiting cases
Limiting cases
Limiting cases
Toy Pendulum
Could two balls recoil and conserve both momentum and energy?
Incompatible!
Elastic Collisions IElastic 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 IElastic 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?
Elastic Collisions IIElastic Collisions II
v
v
m
M
Carefully place a small rubber ball (mass m) on
top of a much bigger basketball (mass M) and
drop these from the same height h so they
arrive at the ground with the speed v. What is
the velocity of the smaller ball after the
basketball hits the ground, reverses direction,
and then collides with the small rubber ball?
a) zero
b) v
c) 2v
d) 3v
e) 4v
• Remember that relative velocity has to be equal before and after collision! Before the collision, the basketball bounces up with v and the rubber ball is coming down with v, so their relative velocity is –2v. After the collision, it therefore has to be +2v!!
Elastic Collisions IIElastic Collisions II
v
v
v
v
3v
v
(a) (b) (c)
m
M
Carefully place a small rubber ball (mass m) on
top of a much bigger basketball (mass M) and
drop these from the same height h so they
arrive at the ground with the speed v. What is
the velocity of the smaller ball after the
basketball hits the ground, reverses direction,
and then collides with the small rubber ball?
a) zero
b) v
c) 2v
d) 3v
e) 4v
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!
Center of MassTreat extended mass as a bunch of small masses:
In a uniform gravitational field you can treat gravitational force as if it acts at the “Center of Mass”
g i iF m g m g M g
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.
Center of Mass
In general:
Symmetry often very useful in determining the
Center of Mass
1 1 1 1 1 1
1 1 1
i iCM
i
m r m rm r m r m rR
m m m m M
Center of Mass
The center of mass need not be within the object
Motion about the Center of MassThe center of mass of a complex or composite object follows a trajectory as if it were a single particle - with mass equal to the complex object, and experiencing a force equal to the sum of all external forces on that complex object
Motion of the center of mass
Action/Reaction pairs inside the system cancel out
The total mass multiplied by the acceleration of the center of mass is equal to the net external force
The center of mass accelerates just as though it were a point particle of mass M acted on by
Momentum of a composite object
Recoil SpeedRecoil Speed
a) 0 m/s
b) 0.5 m/s to the right
c) 1 m/s to the right
d) 20 m/s to the right
e) 50 m/s to the right
A cannon sits on a stationary
railroad flatcar with a total mass
of 1000 kg. When a 10-kg
cannonball is fired to the left at
a speed of 50 m/s, what is the
speed of the center of mass (of
the flatcar + cannonball)?
Recoil SpeedRecoil Speed
Because the initial momentum of the
system was zero, the final total
momentum must also be zero,
regardless of the release of internal
energy, internal forces, etc.
If no external forces act, the motion of
the center of mass does not change
a) 0 m/s
b) 0.5 m/s to the right
c) 1 m/s to the right
d) 20 m/s to the right
e) 50 m/s to the right
A cannon sits on a stationary
railroad flatcar with a total mass
of 1000 kg. When a 10-kg
cannonball is fired to the left at
a speed of 50 m/s, what is the
speed of the center of mass (of
the flatcar + cannonball)?
Recoil Speed IIRecoil Speed II
a) 0 m/s
b) 0.5 m/s to the right
c) 1 m/s to the right
d) 20 m/s to the right
e) 50 m/s to the right
A cannon sits on a stationary
railroad flatcar with a total
mass of 1000 kg. When a 10-
kg cannonball is fired to the
left at a speed of 50 m/s, what
is the recoil speed of the
flatcar?
Recoil Speed IIRecoil Speed II
Because the initial momentum of the
system was zero, the final total
momentum must also be zero. Thus, the
final momenta of the cannonball and the
flatcar must be equal and opposite.
pcannonball = (10 kg)(50 m/s) = 500 kg-
m/s
pflatcar = 500 kg-m/s = (1000 kg)(0.5
m/s)
a) 0 m/s
b) 0.5 m/s to the right
c) 1 m/s to the right
d) 20 m/s to the right
e) 50 m/s to the right
A cannon sits on a stationary
railroad flatcar with a total
mass of 1000 kg. When a 10-
kg cannonball is fired to the
left at a speed of 50 m/s, what
is the recoil speed of the
flatcar?
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) ?