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 dicult to integrate again and nd x(t), if we
care to, bu
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 second
potential energy well is often modelled by a Lenna
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 Energy 177
Problem 10.
H
m
vR
A block of mass M sits at the
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 this
provided a direct evolutionary path to actual ight fo
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 requires you to use three
dierent principles weve learned
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 enough to
overcome static friction) is suspended from a ma
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 a car
(turning its kinetic energy into heat). There are
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
solve the collision if we know one more number, say the a
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 gure 41 a mass m is released from rest from a position o
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 increases,
the other decreases, and vice versa!
This makes ki
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 the
motion of the electrons and quarks in the atoms.
We
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 object is in general a complicated
91Near the Earths s
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) acting against gravity. How fast is it going (released
from
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 length
L that makes an angle with the horizontal as sh
=
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). The rst integral is: d ~ X dt = ~ V = 1 MtotX i
mi~vi
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 extremely easy to compute the terminal velocity for a fal
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 wasted
idly, friction, streamlining (dropping to Fd = bv
= 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 integration, and answer the
questions.
Ill do the rst few steps i
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 + 12) = 0.5 m (356)
Hence the center of mass of this sys
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 nal velocities, so one in general must be given three
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 energy (by, for example, giving up a
photon or phonon, where
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)
y(t) =
12
at2 = 1 2
F mg m
t2 (332)
(333)
From this w
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
then multiply bypmg/b again to get the following result f
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 of length L and is given an initial speed
v0 to the righ
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 between particles that help
hold the system together (s
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 beautifully to
two or three (or in principle still more) dimen
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, you start at the top moving at only vt = 50 meters per
= |~ 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 tZ t 0
~ F(t) dt
An Elastic Collision is by denition a
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, and the
location in question will be exactly what we intu
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 intellectually curious) Duke students up in a helicopter
to a heigh