a number of further opportunities to solve equations of motion this
and next semester that are rst order, linear, inhomogeneous ordinary
dierential equations such as this one.
Given v(t) it isnt too d
if you like, to an actual chemical interaction between them) followed
by a hard core repulsion as the electron clouds are prevented from
interpenetrating by e.g. the Pauli exclusion principle. This se
c) Determine the speed v0 the ball must have at the bottom to arrive at
the top with this minimum speed. You may use either work or potential
energy for this part of the problem.
Week 3: Work and Ener
bushy tails or skin webs like those observed in the ying squirrel59
have a much lower terminal velocity than (say) humans and hence
have a much better chance of survival. One rather imagines that thi
v
m
Figure 40:
Here is a lovely problem so lovely that you will solve it ve or six
times, at least, in various forms throughout the semester, so be sure
that you get to where you understand it that re
m1
A block of mass m2 sits on a rough table. The coecients of friction
between the block and the table are s and k for static and kinetic
friction respectively. A much larger mass m1 (easily heavy eno
The energy in a given system is not, of course, usually constant in
time. Energy is added to a given mass, or taken away, at some rate. We
accelerate a car (adding to its mechanical energy). We brake
three dimensions. In two dimensional collisions we are a bit better o
we have three conservation equations (two momenta, one energy)
and four unknowns (four components of the nal velocity) and can
so
x
Hmax
Figure 41: A spring compressed an initial distance x res a mass m
across a smooth (k 0) oor to rise up a rough k 6= 0) incline.
How far up the incline does it travel before coming to rest?
In g
So any time a mass moves down a distance H, its change in potential
energy is mgH, and since total mechanical energy is conserved, its
change in kinetic energy is also mgH the other way. As one increa
out of an airplane, for example) might also be expected to behave
within reason like a single object independent of the motion of the
baseballs inside, or the motion of the atoms in the baseballs, or
m x
=
dm dx
= lim x0
m A
=
dm dA
= lim x0
m V
=
dm dV
In each of these expressions, m is the mass in a small chunk of the
material, one of length x, area A, or volume V . The mass
distribution of an
Example 3.5.1: Block Sliding Down a Rough Incline
Suppose a block of mass m slides down an incline of length L at an
incline with respect to the horizontal and with kinetic friction
(coecient k) actin
given and discussed in this textbook (or any other), but do it yourself
and without looking after studying.
Problem 3.
L
m
k
D?
at rest
A block of mass m slides down a smooth (frictionless) incline of
=
1 MtotX i
mi
d2~xi dt2
=
1 MtotX i
mi
d~vi dt
(348)
188 Week 4: Systems of Particles, Momentum and Collisions
and can integrate twice on both sides (as usual, but we only do the
integrals formally).
Eventually, the drag force will balance the gravitational force and the
object will no longer accelerate. It will fall instead at a constant speed.
This speed is called the terminal velocity.
It is ex
gasoline to drive the same distance at 100 kph ( 62 mph) than they
do at 50 kph, it isnt four times as much, or even twice as much. Why
not?
Things to think about include gears, engine eciency, fuel w
= g
bm
v2 = b mv2 mg b (188)much as before. Also as before, we divide
all of the stu with a v in it to the left, multiply the dt to the right,
integrate, solve for v(t), set the constant of integratio
evaluating the algebraic expressions for the x and y components of the
center of mass separately:
xcm =
1 MtotX i
mixi =
18
(20 + 21 + 32 + 12) = 1.25 m (355)
ycm =
1 MtotX i
miyi =
18
(20 + 30 + 21 +
K = Kf Ki =
P2 tot 2Mtot p2 1,i 2m1
+
p2 2,i 2m2!
In a partially inelastic collision, the particles collide but dont quite
stick together. One has three (momentum) conservation equations and
needs six
basically slowly unstable) they are classically bound for a while
but eventually escape to innity.
However, in nature pairs of atoms in the metastable conguration
have a chance of giving up some energ
To outline the solution, following a previous homework problem, we
write: F mg = ma (329)
or
a=
F mg m
(330)
We integrate twice to obtain (starting at y(0) = 0 and v(0) = 0):
v(t) = at =
F mg m
t (331
tanh1 (x/a) a (190) Now you know what those rarely used buttons
on your calculator are for. We substitute x > v, a mg/b, multiply
out thepmg/b and then take the hyperbolic tangent of both sides and
th
v0
L
m
L
H
Figure 39: Find the maximum angle through which the pendulum
swings from the initial conditions.
Heres the same problem, formulated a dierent way: A mass m is
hanging by a massless thread o
accelerations ~ai). From this, we construct a weighted average
acceleration of the system, in such a way that Newtons Second Law is
satised for the total mass.
internal equal and opposite force pairs
In the discussion below, we will concentrate on one-dimensional
potentials to avoid overstressing students calculus muscles while they
are still under development, but the ideas all generalize beautif
or
Fhaystack = ma + mg = m(a + g) = mg = mv2 t 2H
+ g (183) Lets suppose the haystack was H = 1.25 meter high and,
because you cleverly landed on it in a blu position to keep vt as
small as possible,
= |~ Favg|t and it usually acts along the line of the collision. Note
that this the impulse is directly related to the average force exerted by
a collision that lasts a very short time t:
~ Favg =
1 t
1 MX i
mi~xi (352) (with M =Pi mi). If we consider the location of the
system of particles to be the center of mass, then Newtons Second
Law will be satised for the system as if it were a particle, an
below, one that I commend to all students when confronted by
problems of this sort.
Example 2.2.3: Dropping the Ram
The UNC ram, a wooly beast of mass Mram is carried by some
naughty (but intellectual