# Lecture 8 - Momentum

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```Lecture 8 - Momentum
Onwards ho! Let’s dive into David Morin’s book Introduction to Classical Mechanics with Problems and
Solutions.
A Puzzle...
An Experiment on Energy
The shortest configuration of string joining three given points is the one where all three angles at the point of
intersection equal 120°.
How could you prove this fact experimentally by cutting three holes in a table and making use of three equal
masses attached to the ends of strings (the other ends of which are connected), as shown below? (Assume that the
strings are long, but they don’t need any particular length and all three lengths can be different.)
Cut three holes in the table at the locations of the three given points. Drop the masses through the holes, and let the
system reach its equilibrium position. The equilibrium position is the one with the lowest potential energy of the
masses, that is, the one with the most string hanging below the table. In other words, it is the one with the least
string lying on the table. This is the desired minimum-length configuration.
What are the angles at the vertex of the string? The tensions in all three strings are equal to m g. The vertex of the
string is in equilibrium, so the net force on it must be zero. This implies that each string must bisect the angle
formed by the other two. Therefore, the angles between the strings must all be 120°. □
A Last Tidbit on Energy
An Important 1D Potential
Example (Spring)
In 1D, the potential V[x] can be calculated from a force F[x] using
x
V[x] = - ∫x0 F[x'] ⅆx'
Calculate the 1D potential of a spring aligned along the x-axis (F[x] = -k x) defining x = 0 as the relaxed spring
length (the reference point here is the x0 = 0).
Setting x0 = 0 and substituting into the above equation.
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Lecture 8 - 10-20-2016.nb
x
V[x] = - ∫0 (-k x') ⅆx'
=  12 k (x')2 x'=0
x'=x
=
1
2
k
(2)
x2
Example
A mass m lies on a frictionless floor attached to a spring with spring constant k. The system is initially in equilibrium, and then at t = 0 the mass is given a kick so that it obtains velocity v away from the spring.
1. How far will the spring extend?
2. After the spring reaches its maximum length, the mass will return back to its original location and then overshoot it, contracting the spring. How far will the spring contract?
Spring potential energy
Total energy
x
1. We define x = 0 as the location of the mass m at equilibrium. At time t = 0, the energy is completely kinetic
energy, E = 12 m v2 . When the spring reaches its maximum length, the mass has stopped moving and hence the
total energy equals the spring’s potential energy E =
1
2
k x2 . Thus we can solve to obtain xmax =
m
k
v.
2. When the mass returns to x = 0, it will once again have velocity v (by conservation of energy), but this time v
will be pointed in the -x direction. The spring will contract, and the next time that the mass is stationary we will
once again have the entire energy in the spring’s potential energy E =
1
2
k x2 , and therefore xmin = -
m
k
v where
this time we have chosen the negative root. □
An Important 3D Potential
Example (Gravity)
In 3D, the potential V[r] can be calculated from a force F using

r
V[r] = - ∫r0 F[r '] · ⅆr '
(3)
m 
Calculate the potential of gravity from a mass M centered at the origin (F[r] = - G M
r) using the reference point
r2
r = ∞.
0
First, note that the force is spherically symmetric, so that for any point r with magnitude r the potential must be
the same. Set r0 = ∞. Since gravity is a central force, we know that it permits a potential because this integral is
path-independent. Therefore, we can choose to integrate along the simplest path: namely, the straight line between
the point r and ∞. In spherical coordinates, it makes sense to use the radial line r r which would have the derivam
tive ⅆ r r . Therefore, the integrand becomes Fr '·ⅆ r ' = - G M
ⅆ r. We substitute this into the above integral and
r2

use the spherical symmetry to replace r with r to obtain
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Lecture 8 - 10-20-2016.nb
r
V[r] = ∫∞ G r'M2 m ⅆr'
r'=r
= - G Mr' m r'=∞
(4)
GM m
r
=-
Notice that at r = ∞, V[r] → 0 as it must. Furthermore, suppose a mass m is placed at rest infinitely far away from
a fixed mass M (and therefore its energy equals E = 0). The mass m would slowly approach M, picking up more
speed as it got closer. At a distance r from M, the mass m would have potential energy - G Mr m and since its total
energy equals 0 its kinetic energy must equal
1
2
or v =
2GM
r
m v2 =
GM m
r
(5)
.
Now consider if mass m started from rest a distance 2 R from the fixed mass M. What would its velocity be when
mass m reached a distance R from M? In that case its total energy (calculated at 2 R) equals E = - G2MRm so that at a
1
2
distance R away from M conservation of energy yields
v=
GM
R
m v2 -
GMm
R
= E = - G2MRm which we can use to solve for
which is independent of m (as it must be, because we know that any object - such as a feather or a
bowling ball - falls at the same rate). □
Momentum
Summary of the Book
For a single particle, we can rewrite F = m a as
F=
ⅆp
ⅆt
(16)
where
p = m v
(17)
is the momentum of the particle. Momentum of a system is conserved if there are no external forces acting upon it;
this makes momentum immensely useful when analyzing systems!
For a group of particles with masses m are positions r , the total force on all of the particles equals
j
j
F tot = M aavg
(18)
M = ∑j m j

r = ∑j mj rj
avg
M
v = ∑j mj vj
(19)
aavg =
(22)
where
avg
M
∑j mj aj
M
(20)
(21)
we call ravg the center of mass. Defining the total momentum
ptot = M vavg
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Lecture 8 - 10-20-2016.nb
we can rewrite the force equation as
F tot =
ⅆptot
ⅆt
(24)
In other words, we can treat all of the particles as one single particle of mass M and think about applying the force
to this single mass at the center of mass ravg . While this is true for any infinitesimal time step, for it to be true for a
length of time we cannot allow the particles to move relative to one another (in other words, the object cannot
deform). We call such objects that don’t deform rigid objects, and we will deal with them nearly exclusively
except in problems where we have inelastic collisions (coming shortly).
Conservation of Momentum
Example
A snowball is thrown against a wall. Where does its momentum go? Where does its energy go?
All of the snowball’s momentum goes into the earth, which then translates (and rotates) a tiny bit faster (or slower,
depending on which way the snowball was thrown).
Denote the masses of the Earth and the snowball as M and m, respectively. In Earth’s rest frame, suppose the
snowball was thrown with velocity v. Then conserving momentum the Earth’s final velocity would equal V ≈
mv
.
M
Hence, the Earth picks up kinetic energy
1
2
2
M  mMv  =
1
2
m
m v2  M
≪
1
2
m v2
(25)
There is also a rotational energy term of this same order of magnitude (that we will learn about next week). We
see that essentially none of the snowball’s energy goes into the Earth. It therefore must all go into the form of heat,
which melts some of the snow.
This is a general result for a small object hitting a large object: the large object picks up essentially all of the
momentum but essentially none of the energy. □
Elastic Collisions
Example
A ball of mass m and initial speed v0 bounces back and forth between a fixed wall and a block of mass M (with
M ≫ m). M is initially at rest. Assume that the ball bounces elastically and instantaneously. The coefficient of
kinetic friction between the block and the ground is μ. There is no friction between the ball and the ground and
also no friction between the ball and the mass M (this means that we can completely ignore the rotation of the ball
in our energy equation).
1. What is the speed of the ball after the nth bounce of the block?
2. How far does the block eventually move?
3. How much total time does the block actually spend in motion?
Work in the approximation where M ≫ m, and assume that μ is large enough so that the block comes to rest by the
time the next bounce occurs.
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Lecture 8 - 10-20-2016.nb
1. Let v j be the speed of the ball after the jth bounce, and let V j be the speed of the block right after the jth bounce.
Then conservation of momentum and energy give
1
2
(26)
(27)
m v j = M V j+1 - m v j+1
m v2j = 12 M V 2j+1 + 12 m v2j+1
Solving this system of equations yields (in terms of ϵ ≡
m
M
≪ 1)
2m
v ≈ 2 ϵ vj
M+m j
M-m
1-ϵ
v ≈ 1+ϵ v j ≈ (1 - 2 ϵ) v j
M+m j
(28)
V j+1 =
v j+1 =
Therefore, after the
nth
(29)
bounce,
vn = (1 - 2 ϵ)n v0
(30)
2. The total distance traveled by the block can be obtained by looking at the work done by the friction force F f .
Eventually, the ball has negligible energy, so all of its initial kinetic energy goes into heat from friction. Therefore,
1
m v20 = F f d = (μ M g) d, which gives
2
d=
m v20
2μM g
(31)
3. To find the total time, we can add up the times, tn , the block moves after each bounce. Since force times time
equals the change in momentum, we have F f tn = M Vn , and so (μ M g) tn = M(2 ϵ vn-1 ) = 2 M ϵ (1 - 2 ϵ)n-1 v0 .
Therefore,
t = ∑∞
n=1 tn
=
=
=
2 ϵ v0 ∞
(1 - 2 ϵ)n
μ g ∑n=0
2 ϵ v0
1
μ g 1-(1-2 ϵ)
v0
μg
(32)
This t is independent of the masses! Note that it is much larger than the result obtained in the case where the ball
sticks to the block on the first hit, in which case the answer is μmMv0g . The total time is proportional to the total
momentum that the block picks up, and the present answer is larger because the wall keeps transferring positive
momentum to the ball, which then transfers it to the block. □
Example
A tennis ball with a small mass m2 sits on top of a basketball with a large mass m1 . The bottom of the basketball is
a height h above the ground, and the bottom of the tennis ball is a height h + d above the ground. The balls are
dropped. To what height does the tennis ball bounce?
Note: Work in the approximation where m1 ≫ m2 , and assume that the balls bounce elastically. Also assume, for
the sake of having a nice clean problem, that the balls are initially separated by a small distance, and that the balls
bounce instantaneously.
Right before the basketball hits the ground, both balls move downward with speed (using
Printed by Wolfram Mathematica Student Edition
1
2
m v2 = m g h)
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Lecture 8 - 10-20-2016.nb
(33)
2gh
v=
Right after the basketball bounces off the ground, it moves upward with speed v, while the tennis ball still moves
downward with speed v. The relative speed is therefore 2 v. After the balls bounce off each other, the relative
speed is still 2 v (as can be seen by working in the heavy balls reference frame). Since the upward speed of the
basketball essentially stays equal to v, the upward speed of the tennis ball is 2 v + v = 3 v. By conservation of
energy, it will therefore rise to a height of H = d +
(3 v)2
.
2g
But v2 = 2 g h, so that
H = d+9h
(34)
That is a huge increase in height! We can extend this setup to more balls (such as n = 4 balls shown below where
m1 ≫ m2 ≫ m3 ≫ m4 ) and set the initial drop height to h = 1 meter.
A stack of n = 3 balls is shown in this YouTube video. With n = 5 balls, the smallest one will bounce more than 1
kilometer and with n = 12 balls the smallest ball will reach escape velocity for the Earth (although in that case the
elasticity assumption is absurd, not to mention that you could not find 12 objects that satisfy
m1 ≫ m2 ≫ · · · ≫ m12 ). □
Inelastic Collisions
Example
A mass M, initially moving at speed v, collides and sticks to a mass m, initially at rest. Assume M ≫ m. What are
the final energies of the two masses, and how much energy is lost to heat, in:
1. The lab frame
2. The frame in which M is initially at rest?
1. The initial energy equals Ei =
Mv
M+m
≈ 1 -
m
m 2
 v + O M
.
M
1
2
M v2 . By conservation of momentum, the final speed of the combined masses is
Therefore, the final energies will equal
1
2
Em =
EM =
1
2
m 2 2
 v ≈ 12 m v2
M
m 2 2
 v ≈ 12 M v2 - m v2
M
m 1 -
M 1 -
(35)
(36)
Therefore, the change in energy equals
Ei - (Em + EM ) =
1
2
m v2
(37)
which is the energy lost as heat.
2. In this frame, mass m has initial speed v, so its initial energy is Ei =
final speed of the combined masses is
mv
M+m
m
M
≈
Em =
1
2
EM =
1
2
m v2 . By conservation of momentum, the
2
m
v + O M
 and hence the final energies are
2
2
m
m
m M
 v2 ≈  M
 Ei ≈ 0
1
2
m 2 2
M M
 v
≈
m
M
Ei ≈ 0
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(39)
Lecture 8 - 10-20-2016.nb
7
This negligible final energy means that the change in energy equals
Ei - (Em + EM ) =
1
2
m v2
which is the energy lost as heat, in agreement with the result found in the lab frame. □
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