Path: Notes/Day 6: Momentum / Momentum.ipynb

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Kernel: VPython

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Momentum

PY345

Momentum

the total momentum $\vec{p}=m\vec{v}$ of an isolated system stays the same
a system is isolated if $\vec{F}_\text{net}=0$
Newton's Second Law is more accurately written as $\vec{F}_\text{net}=\frac{d\vec{p}}{dt}$

Practice

For each situation below, indicate for what choice of system momentum will be conserved (if any).

Two carts roll toward each other on a horizontal track and collide.
Alice and Bob are on rollerblades and initially at rest. Alice abruptly pushes Bob from behind.
Alice and Bob are on rollerblades and initially at rest. Alice and Bob both push away from one another with their hands.
A ball is thrown downward and hits the earth.
Two cars collide with each other.
A ball hits a wall.
A satellite slingshots around a comet.

Solution

By choosing the system to be both carts, the only external forces are the gravitational force and the normal force, which cancel out because there is no vertical acceleration. If a single cart is chosen as the system, then the system would not be isolated because the interaction between the two carts would be external.
Even though Alice and Bob both start from rest, the interaction diagram is the same as the colliding carts, so the reasoning is the same as before.
The interaction diagram is the same as before. It doesn't matter who is doing the pushing.
The only objects involved are the ball and the earth, and there are two types of interactions. The first interaction is through the gravitational field. The second interaction is the contact between the ball and the ground. The ball alone can't constitute an isolated system because both of those interactions with the earth would be considered external. The only choice of isolated system is ball + earth, in which case there are no external forces. This is why the force diagram has no force vectors on it. Note that we could have included interaction with air (drag) in this diagram (or any of these diagrams for that matter). That would make the system no longer isolated, but the effect of the drag is likely small enough to be negligible.
This is similar to the colliding carts, except this time there is likely more friction between the cars and the ground. Furthermore, the amount of friction on each car is likely to be different, so technically the system is not isolated because the net force acting on the system of two cars is not zero. However, if we restrict ourselves to a narrow window of time from just before the collision to just after, then we could say that the small net force will not have had enough time to appreciably change the momentum of the system. We say that the system is momentarily isolated. If we look at a much larger window of time then we can no longer say that the system is isolated.
In many collisions between two bodies, the two bodies together as a system is isolated. This is not one of those cases. The wall experience two (or arguably three) interactions with the earth. The first is gravitational, and the second is a contact force (both vertical and potentially horizontal) due to the fact that the wall is attached to the ground. Since the wall does not move when the ball hits it, that means that there must be a horizontal force on the wall from the ground. This is the force that makes ball + wall not isolated. The only choice of isolate system is ball + wall + earth.
There is only one interaction between the satellite and the comet, and that is through the gravitational field. There are no external forces on the system if the system is chosen to be the satellite and comet together.

Practice

Two cars collide at an intersection and stick together. Determine an expression for the final velocity of the two cars (just after they collide) in terms of their masses and initial velocities.

Solution

Assumptions

the masses of the cars are $m_1$ and $m_2$ , and their initial velocities are $\vec{v}_1$ and $\vec{v}_2$ , respectively
the cars are points masses
the final velocity of the two cars is $\vec{v}_f$
the two cars constitute a momentarily isolated system

Diagrams

Analysis

If we consider a very short window of time from just before the collision to just after the collision, then the system consisting of the two cars is momentarily isolated. That means that the total momentum of the cars before the collision is equal to the total momentum of the system after the collision.

$m_1\vec{v}_1+m_2\vec{v}_2=(m_1+m_2)\vec{v}_f$

Solve for $\vec{v}_f$ .

$\vec{v}_f=\frac{m_1\vec{v}_1+m_2\vec{v}_2}{m_1+m_2}$

Check

The units of our answer are $\frac{[kg][m/s]}{[kg]}=[m/s]$ , which are what we expect for a velocity.

Let's check limiting cases. If $m_1=0$ then we would expect the final velocity to be equal to $\vec{v}_2$ .

ParseError: KaTeX parse error: Expected group after '_' at position 5: \lim_̲\limits{m_1\to …

The same argument applies if $m_2\gg m_1$ .

Interpretation

Because the final velocity is a linear combination of the two initial velocities (with positive coefficients), that means that the final velocity will always be between the two initial velocities. For example, if one car travels north and the other east, then the final velocity will be somewhere between north and east.

Practice

A rocket shell is launched at an angle and travels along the usual parabolic path above flat ground. Right at the very peak of its path, it explodes into two pieces of equal mass. Just after the explosion, one of those pieces is observed to "stop dead" before falling straight to the ground. Determine the trajectory of the other piece.

Solution

Assumptions

the rocket shell is launched with speed $v_0$ at an angle of $\theta$ from the horizontal
the rocket shell has mass $m$ before the explosion, and each of the pieces has mass $\frac{m}{2}$
the only force acting on the shell(s) before and after the collision is gravity
the shell is a point mass
drag is negligible

Diagrams

Analysis

There are three phases to the motion: 1) the parabolic trajectory before the explosion, 2) the explosion itself, and 3) the trajectory after the explosion. Let's do one piece at a time.

According to the free body diagram before the explosion

$-F_{es}^g=m\ddot{y}$

$-\cancel{m}g=\cancel{m}\ddot{y}$

and

$0=m\ddot{x}$ .

By integrating these (which we have done previously), we arrive at the solutions

$x(t)=v_0\cos\theta t+x_0$

$v_x(t)=v_0\cos\theta$

$y(t)=-\frac{1}{2}gt^2+v_0\sin\theta t+y_0$

$v_y(t)=-gt+v_0\sin\theta$ .

At the top of the trajectory $v_y(t_\text{top})=0$ .

$v_y(t_\text{top})=-gt+v_0\sin\theta=0\Rightarrow t_\text{top}=\frac{v_0\sin\theta}{g}$

The x- and y-coordinates of the shell at this time are

$x_\text{top}=v_0\cos\theta \left (\frac{v_0\sin\theta}{g}\right )+x_0=\frac{v_0^2\sin\theta\cos\theta}{g} +x_0$

and

$y_\text{top}=-\frac{1}{2}g\left (\frac{v_0\sin\theta}{g}\right )^2+v_0\sin\theta\left (\frac{v_0\sin\theta}{g}\right )+y_0=\frac{v_0^2\sin^2\theta}{2g}+y_0$

The two pieces of the shell are not truly isolated during the explosion, but it is reasonable to approximate that they are momentarily isolated. Therefore momentum is conserved. Since the explosion occurs at the top of the trajectory, we know that there is no y-component to the momentum of the shell before the explosion. This means that there is also no y-component to the momentum after the collision.

$mv_0\cos\theta\hat{x}+0\hat{y}=\frac{m}{2}(0)\hat{x}+\frac{m}{2}(0)\hat{y}+\frac{m}{2}v_{fx}\hat{x}+\frac{m}{2}v_{fy}\hat{y}$

Separating this into two equation, we get

$mv_0\cos\theta=\frac{m}{2}v_{fx}\Rightarrow v_{fx}=2v_0\cos\theta$

and

$0=\frac{m}{2}v_{fy}\Rightarrow v_{fy}=0$ .

The third phase of the motion is back to parabolic projectile motion. I am going to reset the clock to $t=0$ and set the initial coordinates of the motion to be equal to the final coordinates of the motion from the first phase of motion. The initial velocities are taken from the final velocities of phase 2.

$x(t)=(2v_0\cos\theta)t+\frac{v_0^2\sin\theta\cos\theta}{g}+x_0$

$y(t)=-\frac{1}{2}gt^2+\frac{v_0\sin^2\theta}{2g}+y_0$

When the shell lands, $y(t)=y_0$ .

$y(t)=-\frac{1}{2}gt^2+\frac{v_0^2\sin^2\theta}{2g}+y_0=y_0$

$\frac{1}{2}gt^2=\frac{v_0^2\sin^2\theta}{2g}\Rightarrow t=\frac{v_0\sin\theta}{g}$

We can use this time to figure out the final x-coordinate.

$x(t_\text{land})=2v_0\cos\theta\left (\frac{v_0\sin\theta}{g}\right )+\frac{v_0^2\sin\theta\cos\theta}{g}+x_0$

$x(t_\text{land})=\frac{3v_0^2\sin\theta\cos\theta}{g}+x_0$

Check

$\frac{v_0^2}{g}$ has units of $\frac{[m/s]^2}{[m/s^2]}=[m]$ , which is what we expect for a distance.

Let's explore the limits of $\theta=0$ and $\theta=90^\circ$ . In the former case, the shell should immediately hit the ground. In the later case, it should land back where it started.

ParseError: KaTeX parse error: Expected group after '_' at position 5: \lim_̲\limits{\theta\…

Interpretation

Had the shell not exploded, the projectile would have landed at $\frac{2v_0^2\sin\theta\cos\theta}{g}+x_0$ . This projectile makes it farther than that because the explosion doubles the x-component of the velocity.

In [1]:

from __future__ import division, print_function
from vpython import *

#initial conditions
x0=0 #x-coordinate
y0=0 #y-coordinate
v0=20 #speed
theta=45*pi/180 #angle converted to radians
m=1 #mass of shell
g=9.8 #gravitional field strength
Fg=vec(0,-m*g,0) #gravitational force
a=Fg/m #acceleration
t=0
dt=0.01 #time step

#draw objects
scene=canvas(title="exploding shell",center=vec(0.5*(3*v0**2*sin(theta)*cos(theta)/g+x0),5,0))
shell=sphere(pos=vec(x0,y0,0),v=vec(v0*cos(theta),v0*sin(theta),0), radius=1,color=color.red, make_trail=True)
ground=box(pos=vec(0.5*(3*v0**2*sin(theta)*cos(theta)/g+x0),0,-0.1),length=(3*v0**2*sin(theta)*cos(theta)/g+x0), width=5, height=0.1, color=color.green)

#continue until shell reaches top of trajectory
while shell.v.y>=0:
    rate(1/dt) #ensure that looping rate does not exceed real time
    shell.v=shell.v+a*dt #update velocity
    shell.pos=shell.pos+shell.v*dt #update position
    t=t+dt #increment time

    
shell.visible=False #make the initial shell disappear
m1=m/2 #mass of piece 1
m2=m/2 #mass of piece 2
F1g=vec(0,-m1*g,0) #gravitational force on piece 1
F2g=vec(0,-m2*g,0) #gravitational force on piece 2
a1=F1g/m1 #acceleration of piece 1
a2=F2g/m2 #acceleration of piece 2

#draw the two shell pieces
#make velocity of piece 1 zero while doubling velocity of piece 2
piece1=sphere(pos=shell.pos, v=vec(2*shell.v.x,0,0),radius=0.5,color=color.blue, make_trail=True)
piece2=sphere(pos=shell.pos, v=vec(0,0,0),radius=0.5,color=color.yellow, make_trail=True)

#continue until pieces hit the ground
while piece1.pos.y>=0:
    rate(1/dt) 
    piece1.v=piece1.v+a1*dt
    piece2.v=piece2.v+a2*dt
    piece1.pos=piece1.pos+piece1.v*dt
    piece2.pos=piece2.pos+piece2.v*dt
    t=t+dt

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Practice

Rockets accelerate basically by expelling bits of mass out the back. Determine an expression for the velocity of a rocket (that starts from rest) as a function of time.

Solution

Assumptions

the rocket starts with mass $m_0$
the motion is one-dimensional
the initial velocity of the rocket relative to the observer is $v_0=0$
the speed of the exhaust relative to the rocket is $v_{ex}$ , which means that the velocity of the exhaust relative to the observer is $v-v_{ex}$ (since the rocket is moving forward and the exhaust is moving backward relative to it)
in a time $dt$ , the mass of the rocket changes by $dm$ and the speed of the rocket changes by $dv$
$dt$ , $dm$ , $dv$ , $v$ , and $v_{ex}$ are all assumed to be positive quantities (they aren't all positive, but the math will take care of that)

Diagrams

To be clear, I am assuming that $dm$ , $dv$ , and $v_{ex}$ are all positive quantities.

Analysis

At a given initial moment before a bit is expelled, the total momentum of the system is

$p(t)=mv$ .

(Note that because the motion is one-dimesional, I am dropping vector signs.) After a time $dt$ has elapsed and a bit has been expelled, the total momentum of the system is

$p(t+dt)=(m+dm)(v+dv)+(-dm)(v-v_{ex})$ .

The change in momentum of the system is

$dp=p(t+dt)-p(t)=(m+dm)(v+dv)+(-dm)(v-v_{ex})-mv$

$dp=\cancel{mv}+mdv+\cancel{vdm}+(dm)(dv)-\cancel{vdm}+v_{ex}dm-\cancel{mv}$

If $dm$ and $dv$ are each infinitesimally small quantities, then the product $(dm)(dv)$ is unimaginably small, so we will ignore it.

ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 8: dp=mdv+\̲c̲a̲n̲c̲e̲l̲t̲o̲{0}{(dm)(dv)}+v…

Since the system of rocket + bit is isolated, the total momentum of the system stays the same. That means that

$dp=0=mdv+v_{ex}dm$

This is a separable differential equation. Move v's to one side and m's to the other ( $v_{ex}$ is a constant so it doesn't matter what side it is on).

$\frac{v_{ex}dm}{m}=dv$

Integrate both sides.

$-v_{ex}\int\limits_{m_0}^{m(t)}dm=\int\limits_{v_0}^{v(t)}dv$

$-v_{ex}\ln\left (\frac{m(t)}{m_0}\right )=v(t)-v_0$

$v(t)=v_0-v_{ex}ln\left (\frac{m(t)}{m_0}\right )=v_0+v_{ex}\ln \left (\frac{m_0}{m(t)}\right )$

Check

$\frac{m_0}{m(t)}$ has no units, which is good because the argument of a natural log function can't have units.

If $m(t)=m_0$ (no mass loss, in other words), then ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 16: v(t)=v_0+v_{ex}\̲c̲a̲n̲c̲e̲l̲t̲o̲{0}\ln(0)=v_0. This makes sense, since no mass loss means that the rocket isn't actually expelling any fuel.

Interpretation

The relative magnitude of the exhaust speed relative to the speed of the rocket doesn't change whether the rocket speeds up or slows down. As long as the exhaust is backwards relative to the rocket, the rocket will speed up. We can also see from the final equation that if the exhaust is propelled forward (which means that we switch its sign), then the rocket will slow down with time.

In [2]:

from __future__ import division, print_function
from vpython import *
import matplotlib.pyplot as plt
%matplotlib inline

scene=canvas(range=50) #sets up the drawing area
#draw rocket
body=cylinder(pos=vec(0,0,0),axis=vec(3,0,0)) #body of rocket
head=cone(pos=vec(3,0,0),axis=vec(1,0,0)) #head of rocket
fin1=triangle(v0=vertex(pos=vec(0,-1,0)),v1=vertex(pos=vec(0,-2,0)),v2=vertex(pos=vec(1,-1,0))) #rocket fin
fin2=triangle(v0=vertex(pos=vec(0,1,0)),v1=vertex(pos=vec(0,2,0)),v2=vertex(pos=vec(1,1,0))) #rocket fin
rocket=compound([body,head,fin1,fin2]) #treat all of the rocket pieces as one object

#initial conditions
v0=vec(0,0,0)
m0=1
rocket.m=m0 #mass of rocket
rocket.v=v0 #initial velocity of rocket
vex=vec(50,0,0)
t=0 #initialize time
dt=0.01 #time step
dm=-0.002

#create graph
tdata=[]
numericaldata=[] 
exactdata=[]
 
while rocket.m>=m0/2: #assume that half of the rocket's initial mass is fuel
    rate(1/dt) #100 calculations per second so we can see it
    rocket.v=rocket.v-vex*dm/rocket.m
    rocket.pos=rocket.pos+rocket.v*dt #calculate new position of rocket
    tdata.append(t)
    numericaldata.append(rocket.v.x) #plot the velocity and time
    exactdata.append(v0.x+vex.x*log(m0/rocket.m)) #plot exact result
    t=t+dt #increment the time
    rocket.m=rocket.m+dm #calculate the new mass of the rocket
    
plt.title("velocity vs. time")
plt.xlabel("t (s)")
plt.ylabel("v (m/s)")
plt.plot(tdata,numericaldata, "ro", tdata, exactdata, "b.")
plt.legend(["numerical","exact"])

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Practice

A cart full of water is propelled by a powerful fan. The cart is leaking at a constant rate. Determine the velocity of the cart as a function of time.

Solution

Assumptions

the fan exerts a constant force $F_{ac}^c$
the mass of the cart system changes by $dm$ whenever a water droplet drops, and the velocity of the drops relative to the cart is 0
the mass of the cart is $m(t)=m_0-bt$ , where $b$ is a constant
the motion is one dimensional

Diagrams

Analysis

At any given time, the momentum of the cart is

$p(t)=mv$ .

At a time $dt$ later, the momentum of the cart + drop is

ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 45: …v+\cancel{vdm}+\̲c̲a̲n̲c̲e̲l̲t̲o̲{0}{(dm)(dv)}-\….

The change in momentum during that time is

$dp=p(t+dt)-p(t)=mv+mdv-mv=mdv$ .

According to Newton's Second Law and the free body diagram

$F_{a(c+d)}^c=\frac{dp}{dt}=\frac{mdv}{dt}=(m_0-bt)\frac{dv}{dt}$ .

As before, we put terms with $t$ on one side and terms with $v$ on the other side. $F_{a(c+d)}^c$ is a constant, so it doesn't matter what side we put it on.

$F_{a(c+d)}^c\frac{dt}{m_0-bt}=dv$

Integrate both sides.

$F_{a(c+d)}^c\int\limits_{0}^t\frac{dt}{m_0-bt}=\int\limits_{v_0}^{v(t)}$

If the $m_0$ were not there, we would easily recognize the integral on the left hand side as a natural log. We can make it look like a natural log integral by making a change of variables.

Let $u=m_0-bt$ , then $du=-bdt\Rightarrow dt=\frac{-1}{b}du$ .

$\frac{-F_{a(c+d)}^c}{b}\int\limits_{u_0}^{u(t)}\frac{du}{u}=v(t)-v_0$

$-\frac{F_{a(c+d)}^c}{b}\ln\left (\frac{u(t)}{u_0}\right )=v(t)-v_0$

$-\frac{F_{a(c+d)}^c}{b}\ln\left (\frac{m_0-bt}{m_0}\right )=v(t)-v_0$

$v(t)=v_0-\frac{F_{a(c+d)}^c}{b}\ln\left (\frac{m_0-bt}{m_0}\right )=v_0-\frac{F_{a(c+d)}^c}{b}\ln\left (1-\frac{bt}{m_0}\right )$

Check

In order for $m(t)=m_0-bt$ to have units of $[kg]$ , $bt$ must have units of $[kg]$ as well. This means that $b$ has units of $\frac{[kg]}{[s]}$ . $\frac{F_{a(c+d)}^c}{b}$ then has units of $\frac{[N]}{[kg]/[s]}=\frac{[kg][m]/[s]^2}{[kg]/[s]}=\frac{[m]}{[s]}$ , which is what we expect since the natural log function has no units.

Let's check limits, starting with $t=0$ .

ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 32: …_{a(c+d)^c}}{b}\̲c̲a̲n̲c̲e̲l̲t̲o̲{0}{\ln (1)}=v_…

As time goes on, the argument of the natural log function becomes closer and closer to zero, which means that the natural log function itself evaluates to larger and larger negative numbers. Combined with the negative sign in front of the coefficient, this makes the velocity increase.

We could also check what happens when $b=0$ . In this case we would expect that the velocity would increase linearly with time, since a constant force is applied to a constant mass. The problem is that there is a $b$ in the denominator. One way around this is to take the limit as $b$ approaches zero.

ParseError: KaTeX parse error: Expected group after '_' at position 5: \lim_̲\limits{b\to 0}…

How do we evaluate this limit? My preferred method to to perform a Taylor Expansion for the natural log function. In a Taylor Series table, I find that

$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$ .

Therefore

ParseError: KaTeX parse error: Expected group after '_' at position 5: \lim_̲\limits{b\to 0}…

ParseError: KaTeX parse error: Expected group after '_' at position 5: \lim_̲\limits{b\to 0}…,

which is exactly what we expected.

Interpretation

This shows a non-linearly increase velocity. In reality, the mass would not keep decreasing forever. Once the water runs out, the mass would stop decreasing. To make this more realistic, we would have to turn this into a piecewise problem where the mass becomes constant again after a while.

In [3]:

from __future__ import division, print_function
from vpython import *
import matplotlib.pyplot as plt
%matplotlib inline

#initial conditions
x0=0 #x-coordinate
v0=0 #initial speed
F=vec(1,0,0) #force from fan
m0=1 #starting mass (used for exact solution)
m=m0 #starting mass (used for numerical solution)
v=vec(v0,0,0) #initial vector velocity
mcart=0.5 #mass of cart without water
b=0.05 #how fast is it leaking?
t=0
dt=0.01 #time step

#draw cart
scene=canvas(title="leaking cart",range=20)
cart=box(pos=vec(x0,0,0),length=1,width=0.5, height=0.5)

#set up arrays for plotting
tdata=[]
numericaldata=[]
exactdata=[]

while m>=mcart: #continue looping until water runs out
    rate(1/dt) #makes sure that looping rate does not exceed real time
    a=F/m #calculate acceleration
    v=v+a*dt #update velocity
    cart.pos=cart.pos+v*dt #update position
    #append data for plotting
    tdata.append(t) 
    numericaldata.append(v.x)
    exactdata.append(v0-F.x/b*log(1-b*t/m0))
    t=t+dt #increment time
    m=m0-b*t #decrement mass

#plot all the things
plt.title("velocity vs. time")
plt.xlabel("t (s)")
plt.ylabel("v (m/s)")
plt.plot(tdata,numericaldata,"ro",tdata,exactdata,"b.")
plt.legend(["numerical","exact"])

<IPython.core.display.Javascript object>

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In [ ]:

Momentum

PY345

Momentum

Practice

Solution

Practice

Solution

Assumptions

Diagrams

Analysis

Check

Interpretation

Practice

Solution

Assumptions

Diagrams

Analysis

Check

Interpretation

Practice

Solution

Assumptions

Diagrams

Analysis

Check

Interpretation

Practice

Solution

Assumptions

Diagrams

Analysis

Check

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