Momentum

Exam II — A week today
➢ Chapters
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Application of Newton’s Laws
Work-energy theorem
Potential and Energy Conservation
Momentum; conservation and collisions
➢ How
A.
B.
C.
D.
E.
5, 6, 7 and 8
are you feeling for this one?
Should be a piece of cake
Expect I’ll do better than I did on the 1st midterm
I think I’ll do okay; about the same as before
I’ll pass, but not do as well as I wanted to
I’m f***ed!
➢ Exam
review this Sunday from 2-4 pm in this room
Kinetic energy vs. momentum
➢ Magnitude
➢ But
of KE and momentum related: 𝑝 = 2𝑚𝐾
one is a scalar, the other is a vector…
➢ Really
matters when we consider more than one
particle (or a system of particles)!!
A
+30𝑖 m/s
−30𝑖 m/s
1800 kg
1800 kg
A
𝐾=
1
𝑚𝑣 2
2
𝑝 = 𝑚𝑣
B
1
1800 30
2
2
B
= +810 kJ
1800 +30 = 54 × 103 𝑖 N⋅s
1
1800 −30
2
A+B
2
= +810 kJ
1800 −30 = −54 × 103 𝑖 N⋅s
1620 kJ
54 − 54 × 103 = 0
Internal versus external forces
Internal
External
Interactions within the system as you’ve
defined it
Interactions with objects not part of the
system as you’ve defined it
Example: A baseball has many
molecular forces holding the atoms of
the ball together; these forces are
internal to the “just-the-ball” system
Example: A bat hits a baseball; the
impulse it applies to the ball is external
from the “just-the-ball” system: the
momentum of the ball changes
Example: Rifle shoots a bullet, but if we
Example: External forces to the
define the system as “rifle+bullet”, then
“rifle+bullet” system include gravity and
the explosion of the bullet and the force it whatever is holding up the rifle
has with the rifle are all internal; the rifle
must recoil with a momentum equal to
the bullet’s momentum
Example: Bomb is initially at rest and
explodes into three pieces. The total
momentum of the bomb before is zero, so
the total momentum of the three pieces
after must also be—and therefore add up
to—zero
Example: If we consider just one piece of
the bomb, it feels a force from the
explosion (outside it’s system) and gains
momentum; momentum is not conserved
because there is an external force acting
on it
Conservation of momentum
➢ From
Newton’s 2nd Law:
➢ Here
“isolated” means
The total momentum of an
isolated system is constant
𝑑𝑝
𝑑𝑡
“no external forces” ⇒ ∑𝐹 = 0 =
⇒ 𝑝 doesn’t change with time; it
is constant or “conserved”
➢ Internal
forces can’t change the momentum of a
system of particles because of
Newton’s 3rd Law
𝑑𝑝𝐴
𝐹𝐵on𝐴 =
𝑑𝑡
and
𝑑𝑝𝐵
𝐹𝐴on𝐵 =
𝑑𝑡
Isolated systems
➢ King
Kong is standing on a sheet of ice and begins to
beat his chest. Does he go backwards?
A) Yes
B) No
➢ Duck
Dodgers and Marvin the Martian have a
showdown in outer space. They approach each other
with equal momentum and grab onto each other when
the collide:
What about kinetic energy?
More astronaut examples
➢ Now
Duck Dodgers and Marvin push off each other
More astronaut examples
➢ Now
Duck Dodgers and Marvin push off each other
Where did the KE come from?
➢ Dodgers
stationary, Marvin crashes into him:
Momentum conserved in 2 and 3D as well
➢ Momentum
is a vector quantity; conserved separately
in each direction
Understanding momentum conservation
➢ Consider
the collision of Example 8.6 on page 250:
Impules of 𝐴 is:
∆𝑝𝐴 = 𝑚𝐴 𝑣𝐴,𝑓 − 𝑣𝐴,𝑖
= 𝐹𝐵on𝐴 ∆𝑡
and note its acceleration is
not zero:
𝑣𝐴,𝑓 − 𝑣𝐴,𝑖
𝑎𝐴 =
≠0
∆𝑡
⇒ acceleration is caused
by force of 𝐵 on 𝐴
of 𝐴 alone is not conserved; 𝐵’s force is
external to 𝐴.
➢ Momentum
➢ But,
if we consider the combined system of 𝐴 and 𝐵…
Understanding momentum conservation
➢ Consider
the collision of Example 8.6 on page 250:
Impules of 𝐵 is:
∆𝑝𝐵 = 𝑚𝐵 𝑣𝐵,𝑓 − 𝑣𝐵,𝑖
= 𝐹𝐴on𝐵 ∆𝑡
and from Newton’s 3rd Law:
𝐹𝐴on𝐵 = −𝐹𝐵on𝐴
so 𝑚𝐴 𝑣𝐴,𝑓 − 𝑣𝐴,𝑖 =
−𝑚𝐵 𝑣𝐵,𝑓 − 𝑣𝐵,𝑖
⇒
𝑝tot,𝑖 = 𝑝tot,𝑓
or 𝑚𝐴 𝑣𝐴,𝑓 + 𝑚𝐵 𝑣𝐵,𝑓 =
𝑚𝐴 𝑣𝐴,𝑖 + 𝑚𝐵 𝑣𝐵,𝑖
and we again find that the
total momentum of an
isolated system is conserved
Concrete example of momentum conservation
➢A
bomb explodes into 3 pieces. One piece, with
𝑚 = 0.5 kg goes 40 m/s 45∘ above the 𝑥-axis and
another (𝑚 = 0.3 kg) goes at 60 m/s 10∘ above the
negative 𝑥-axis. If the shell of the bomb was 1.0 kg,
where and how fast did the 3rd piece go?
1.
Define your system, and draw a picture of before and
after
Concrete example of momentum conservation
2.
3.
Calculate the total
momentum in both cases:
Solve the 𝑖 and 𝑗 components using momentum
conservation
Types of collisions
1.
Elastic — if no permanent deformation occurs, the
objects don’t gain internal energy, and if no energy is
lost to friction
⇒ kinetic energy IS conserved as well as momentum
2.
Inelastic — not elastic: more realistic and more
common. Why do raquetball players whack the ball
against the wall before starting a game?
2. Completely inelastic — an extreme case of an
inelastic collision: the objects stick together (clearly
cannot be elastic in this case; this is the way to remember elastic versus
inelastic)
⇒ kinetic energy NOT conserved; but momentum still is
Collision PhET
➢ In
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“real world”, almost every collision is inelastic
car accident
slap shot
QB gets sacked
basketball dribbled
bug on a windshield
⋮
➢ Cases
where collisions are almost elastic?
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very hard plastic balls of billiards or bocce ball
steel balls of Newton’s cradle
spring-side of Pasco carts
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Coulomb repulsion of two ions
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Centre of mass
➢ Very
often, when considering systems of particles or
objects with finite size, the concept of the centre of
mass is extremely useful
➢ It
is simply the weighted average of a bunch of
positions (like your final grade is a weighted average of
exams, homework, etc.).
➢ Specifically,
for a collection of 𝑁 particles:
𝑥cm
𝑚1 𝑥1 + 𝑚2 𝑥2 + ⋯ + 𝑚𝑁 𝑥𝑁 ∑𝑁
𝑖=1 𝑚𝑖 𝑥𝑖
=
= 𝑁
∑𝑖=1 𝑚𝑖
𝑚1 + 𝑚2 + ⋯ + 𝑚𝑁
𝑦cm
𝑚1 𝑦1 + 𝑚2 𝑦2 + ⋯ + 𝑚𝑁 𝑦𝑁 ∑𝑁
𝑖=1 𝑚𝑖 𝑦𝑖
=
= 𝑁
∑𝑖=1 𝑚𝑖
𝑚1 + 𝑚2 + ⋯ + 𝑚𝑁
𝑧cm
𝑚1 𝑧1 + 𝑚2 𝑧2 + ⋯ + 𝑚𝑁 𝑧𝑁 ∑𝑁
𝑖=1 𝑚𝑖 𝑧𝑖
=
= 𝑁
∑𝑖=1 𝑚𝑖
𝑚1 + 𝑚2 + ⋯ + 𝑚𝑁
or, more succinctly, with
the total mass
𝑀 = 𝑚1 + 𝑚2 + ⋯ + 𝑚𝑁
= ∑𝑚𝑖 :
𝑟cm
∑𝑚𝑖 𝑟𝑖
=
𝑀
Solid bodies
➢ For
certain symmetrical shapes, the centre of mass is
in the obvious spot:
For arbitrary shapes, have
to find it either with
calculus or experimentally.
But point is that every
object has a centre of
mass. And the bulk motion
of that object can be
described as a point-like
object moving at its centre
of mass
Motion of the centre of mass
➢ As
defining the position centre-of-mass, we can define
the centre of mass velocity and acceleration of an
object. Not surprisingly, they are:
𝑣𝑐𝑚
𝑚1 𝑣1 + 𝑚2 𝑣2 + ⋯ + 𝑚𝑁 𝑣𝑁 ∑𝑚𝑖 𝑣𝑖
=
=
𝑚1 + 𝑚2 + ⋯ + 𝑚𝑁
𝑀
𝑎𝑐𝑚
𝑚1 𝑎1 + 𝑚2 𝑎2 + ⋯ + 𝑚𝑁 𝑎𝑁 ∑𝑚𝑖 𝑎𝑖
=
=
𝑚1 + 𝑚2 + ⋯ + 𝑚𝑁
𝑀
➢ Why
the heck are we bothering to define these?
Because look at the total momentum and sum of the
forces on the system of particles:
𝑀𝑣cm = 𝑚1 𝑣1 + 𝑚2 𝑣2 + ⋯ + 𝑚𝑁 𝑣𝑁
= 𝑝tot
and
𝑀𝑎cm = 𝑚1 𝑎1 + 𝑚2 𝑎2 + ⋯ + 𝑚𝑁 𝑎𝑁
= ∑𝐹ext
Newton’s 2nd and systems of particles
𝑑𝑝tot
whether for a finite-sized
∑𝐹ext =
object or system of particles
𝑑𝑡
If there are no external forces acting on a system, the
total momentum of that system is conserved
Q: When the bottom shell hits the ground,
does the CoM still follow the same parabolic
trajectory?
EOC 8.45 ( CP)
➢A
5.00-kg ornament is hanging by a 1.50-m wire when
it is suddenly hit by a 3.00-kg missile traveling
horizontally at 12.0 m/s. The missile embeds itself in
the ornament during the collision. What is the tension
in the wire immediately after the collision?
EOC 8.85 ()
➢ Movie
stuntman (80.0 kg) is 5 m above floor and
swings down using a rope to grab a 70.0 kg villain.
a)
With what speed do the two slide across the floor?
b)
If 𝜇𝑘 = 0.250, how far do they slide?
EOC 8.105 ()
➢ Bert
(30 kg) and Ernie (40 kg) are at opposite ends of
a 3.0-m long, 20.0-kg uniform log. Ernie walks to Bert;
how far has the log moved?