Physics 325 – Homework #2
due in 325 homework box1 by Friday, 1 pm
Homework rules are at the top of Homework #1, e.g. NO WORK = NO POINTS : answers or solution steps
given without explanation get zero points. However you may use without proof any relation from the 3D- and
1D-math formula sheets on our website. Write your NAME and DISCUSSION SECTION on your solution.
Problem 1 : A Force with Mixed Dependences 2
!
A particle of mass m moves under the influence of the following force: F(t, v) = −k t v v̂ , where k is a positive
constant. At time t = 0 the particle passes through the origin with velocity v0 x̂ , where v0 is a positive constant.
!
(a) Calculate the particle’s velocity v(t) for times t > 0.
(b) How much time does it take for the particle to come to a stop?
(c) What is the maximum x-position reached by the particle? Hint: you will need to look up
∫
∞
0
(something)dt .
I urge you to use http://wolframalpha.com , which is an extremely valuable tool that you should learn to use!
You may use WolframAlpha.com to perform any integrals on this homework. ☺︎
Please read the brief Appendix 2 for some crucial tips about using this tool.
Problem 2 : A Position-dependent Force
A particle
! of mass m is constrained to move along the positive x-axis and is under the influence of the following
force: F(x) = − ( K / x 2 ) x̂ , where K is a positive constant. At time t = 0 the particle is released from rest at
position x = x0, where x0 is positive. Calculate the time at which the particle reaches the origin.
CAUTION: When you calculate velocity, you will have to take a square root … and every time you do that,
you must think carefully — physically! — about which sign to choose for the result.
INTEGRATION: The final integral you must perform is a tough one. If you would like to integrate it by hand,
a clever substitution is provided in the footnote3. If you use WolframAlpha – and I suggest you do – there is a
good chance you will type something like this into the program:
integrate 1/sqrt(1/x-1/a) dx
This produces a terrifying expression involving imaginary numbers (!!!) This is actually a feature not a bug,
and it is explained in Appendix 3: Mathematica Oddities. ☞ Whichever method you use, please
read Appendix 3 and, for your own education, figure out how to fix the output of the above expression
using the algebraic rearrangement trick described as “Tactic #1” on the last page.
Problem 3 : Drag to Any Power
A block with mass m slides on a horizontal surface. At time t = 0, the block is at position x = 0 and has speed v0
in the +x direction. (Note: “speed” means “magnitude of velocity”, so v0 is! implicitly a positive number.)
Let’s explore a generic form for the drag force that the block experiences: F(v) = −k v β v̂ , where k and β are
positive constants. The air-resistance cases β = 1 and β = 2 are derived in Taylor; let’s try other options.
1 All
physics homework boxes are in the 2nd floor overpass between Loomis and MRL (to the north of Loomis).
2 Thank you to Richard Weaver, from whom problems #1–3 are adapted.
3
Integration assistance for Problem 2: The trig substitution x = x0 sin2 α will help, as will the identity sin θ = (1 − cos 2θ ) / 2 .
That last identity is very useful, by the way; I strongly recommend you derive it, just for your own knowledge! There’s no need to
memorize it → just know that switching from trig(θ) to trig(2θ) is a helpful trick, then derive the switch when you need it.
2
(a) At what position does the block stop and how long does it take to get there in the case β = 3 / 2?
(b) Consider the case β = 1 / 2, and change the experiment so that the block is now dropped vertically. It will
now experience a downward gravitational force as well as the drag force. Experimenters drop the block from
rest at t = 0 and determine that the block's terminal velocity is vTER = 50 m/s. Using g = 10 m/s2, calculate the
times (in seconds, to two significant digits only) at which the block reached these speeds: 50%, 70%, 90%,
99%, and 99.98% of vTER. Recall: Appendix 2 shows how to use WolframAlpha as a batch-processing
calculator that can perform five calculations in one line. ☺︎
Problem 4 : Parking Orbit with Tractor Beam Assistance
Space pilot Mako is preparing to put her ship into a parking orbit around space station DS9. Mako uses a
coordinate system with DS9 at the origin and her ship in the xy-plane. Mako must calculate a trajectory
{ s(t), φ(t) } that follows the station’s strict rules once her ship passes into station-controlled space at s ≤ R0 :
● Her ship must maintain a constant angular speed ω around the station → she must maintain φ(t) = ωt
● Her ship’s final parking orbit must be a circle of radius R (< R0) → she must ensure that s(t) → R as t → ∞
● Her ship’s final orbit must maintain a constant radius R → she must ensure that ds / dt = 0 when s = R
DS9’s artificial-gravity generator provides a tractor beam that acts on all ships at s ≤ R0 to help them park :
!
● DS9’s tractor beam exerts a constant attractive force of F T = − ŝ mRω 2 on any ship of mass m
Mako completes her calculations; at time t = 0 she starts her parking run.
● Mako starts her ship at initial position s t=0 = R0 and φ t=0 = 0
● Mako gives her ship an initial angular speed φ! t=0 = ω (duh! she must maintain φ = ωt at all times)
● Mako gives her ship an initial radial speed s! t=0 = A
!
● Mako programs her engines to supply a position-dependent azimuthal force F eng (s) = φ̂ m f eng (s)
Calculate the following quantities, in no particular order, in terms of the given constants R, R0, ω, and/or m:
➤ the initial radial speed A,
➤ the azimuthal force-per-unit-mass f eng (s) provided by the engines, and
➤ the radial component s(t) of the ship’s trajectory.
➤ Naturally, you will be unable to leave this problem without making a little sketch of Mako’s trajectory. ☺︎
IMPORTANT: Please use the separate & integrate method to solve the differential equations you encounter in
this problem. Normally you are free to select any method you want, but this problem is designed to help you
practice one of our new separate & integrate techniques.
Now read the small Appendix 1 : Trig Triangles → it may save your life in the next problem ;-)
Problem 5 : What Goes Up Must Come Down
A! baseball of mass m travels through the air under the influence of uniform gravity g and a quadratic drag force
F(v) = −c v 2 v̂ due to air resistance. The constant c is positive. (We will take c as a known quantity; as we
learned in Discussion 2, it depends on the properties of the ball and the air.)
(a) What is the terminal velocity vT of the baseball when it is dropped vertically through the air? The answer
is in your lecture notes and textbook, but, as always, you must explain where it comes from, i.e. you must derive
the expression. (“No Work, No Points!”).
We now consider a specific experiment. The baseball is thrown vertically upwards from the ground (y = 0) with
initial speed v0. (Remember, the word speed means magnitude of velocity, so speed is always a positive
number.) The ball will travel upwards until it reaches some maximum height, above the ground, then it will
turn around and fall back down to the ground. The ground is at y = 0, but I leave it to you to decide whether
you want the +y direction to point up or down … just be sure to clearly indicate the choice you make.
Hint : As always, your answers must be expressed in terms of the given quantities, which are m, g, v0, c ... and
now the terminal velocity vT from (a). Your algebra will simplify if you use the symbol vT as much as possible.
(b) Calculate these two quantities about the ball’s upward trip:
● the maximum height, H, that the ball reaches above the ground
● the time Tup that the ball takes to get from the ground to the maximum height H
(c) Perform a limiting-case check on the maximum height H you just calculated: take the limit of zero drag
and explain clearly why the result you get is exactly what you must get in the absence of air resistance.
(d) Now for the return trip back to the ground! Calculate v(h) = the speed v of the ball as a function of the
ball’s height h above the ground. (I’m asking for v(h) instead of v(y) as you may have chosen +y to be
downwards; use whatever y-convention you chose, then translate your result into the form v(h) at the end.)
Hint : To find v as a function of position rather than time, recall the “Playing with Differentials” technique we
learned for dealing with position-dependent forces, as in Problem 2.
(e) Calculate these two quantities about the ball’s return trip to the ground:
● the speed vf of the ball when it hits the ground (“f” stands for “final”)
● the time Tdown that the ball takes to get from the maximum height H back to the ground;
to get Tdown you may use without proof the v(t) formula derived in Taylor for this situation
(or derive it yourself! great practice ☺︎)
(f) Perform a limiting-case check on the final speed vf you just calculated: Take the limit of zero drag and
make sure you get the answer you know you must get in the absence of air resistance.
____________________________________________________________________________________
APPENDIX 1 : TRIG TRIANGLES
Suppose you perform an integral and the result is tan(x/a) … but you want a different trig function, e.g. cos(x/a).
Or suppose you encounter this construction: sin [ tan–1 (x/a) ] … how to make progress? Must you resort to a
mess of obscure trig relations? Nope! You can always translate any trig or arc-trig function into any other with
ease using this secret weapon : The most powerful way to represent any trig result is with a triangle.
Suppose you know that sin(α) = x / b, but you need tan(α). Well sine is opposite-over-hypotenuse, so writing
sin(α) = x / b is completely! equivalent! to drawing the left-hand triangle below :
b
x
α
You can now invoke
Pythagoras to fill in
the length of the
bottom side. →
b
x
α
b !x
2
2
This triangle allows you to read off all other trig functions: cos(α ) = b 2 − x 2 / b , tan(α ) = x / b 2 − x 2 .
ky
1+ k 2 y2
α
1
You can do the exact same thing with arc-trig functions. Suppose you must
evaluate sin [ tan–1 (ky) ]. Define α ≡ tan–1 (ky). That statement is completely
summarized by the triangle at left … which again allows you to read off
whatever you want:
(
)
(
α = tan −1 (ky) = sin −1 ky / 1+ k 2 y 2 = cos −1 1 / 1+ k 2 y 2
… and sin ⎡⎣ tan −1 ( ky ) ⎤⎦ = sin [α ] = ky / 1+ k 2 y 2
)
APPENDIX 2 : WolframAlpha BY EXAMPLES
WolframAlpha.com has a compact set of examples that will get you started immediately, but there are some
really useful features that the examples don’t cover, and some idiosyncrasies that aren’t explained. Below are
further examples, in monospaced font … please try them out to learn what they do!
● Basic Syntax
1+a^2
d/da a^y
1+e^pi
✓ a is a single letter with no value so it’s taken as a variable → this expr. is plotted vs a
✓ here a and y are variables = single letters without values → produces ∂(a y ) / ∂a = y a y−1
✓ e, i, & pi are reserved words with values → this expression is evaluated just as a
calculator would, producing 1+eπ = 24.14069
1+3^n for n=1,2,3,4
✓ comma-separated lists can be used in many contexts
→ produces {4, 10, 28, 82} : a batch-processing calculator! ☺︎
d/dx x^n where n=1,2,3
✓ “for”, “with”, “where” are synonymous → produces { 1, 2x, 3x2 }
● produces two values, not four → { 8, 64 }
(x+y)^n for n=3 for x=1,2 for y=1,2
log(e), ln(e), log10(e) ✓ “log” means “ln” !!! → this produces {1, 1, 0.434}
3*a^2b sqrt(c)
✓ flexible syntax: three ways of multiplying → produces 3 a2 b √c
e^x/(e**-x + 2exp(x))
✓ flexible syntax: three ways of doing ex → produces ex / ( e–x + 2 ex )
✓ 2nd derivatives: two syntaxes →
d^2/dy^2 ln(y), d2/dy2 log(y)
{−1 / y , − 1 / y }
2
2
● Variable names must be one letter long and are case sensitive
✓ what you expect → produces (a 2 + x 2 )
a^2+x^2
d/dx ax+Ax
ab^2+x^2
a5^2+x^2
● “a” and “A” are not the same variable → produces (a + A) , not 2a or 2A
✘ “ab” is interpreted as the product of 2 variables → produces (a b 2 + x 2 )
✘ “a5” is interpreted as a5 (!?!) → produces ( a10 + x 2 )
● Variable names with a numerical _subscript seem to be possible … but beware
a0 x
a
int x/a_0 from x = 0 to a_0
✓ produces ∫
dx = 0 as expected
0 a
2
0
x0 x
int x/x_0 from x = 0 to x_0
✘ writes ∫
dx and STOPS! x vs x0 confuses it … !
0 x
0
int x/a_b from x = 0 to a_b
✘ FAILS, numerical subscripts only
● When definite integrals fail, try the indefinite version
integrate arctanh(1-sqrt(x)+x) dx
✓ standard integral syntax
int atanh(1-sqrt(x)+x) dx
✓ “int”, “atanh”, and many other abbreviations work fine
int atanh(1-sqrt(v)+v)
✓ there’s only 1 variable ∵ no need to specify “dv”
int atanh(1-sqrt(v)+v) from v=e to pi
✘ FAILS: the definite integral times out
☞ When a definite integral fails, evaluate the indefinite integral and put in your bounds by hand.
● 1D Plotting : superimpose several curves using lists of variable values, and specify plotting ranges
plot y^n for n=2,3,2*2
✓ superimpose several curves on one plot via a list of variable values
✓ same plot produced via an list of expressions
plot y^2 from y=-1 to 3 ✓ specify a range for the horizontal axis
plot y^2,y^3,y^4
plot y^n for n=2,3,4 from y=0 to 3 ✓ supply both a variable-value list and a plotting range
plot y^n with n=2,3,4 for y=0 to 3 ✓ for, with, where … WolframAlpha doesn’t care ☺︎
plot y^n for y=0 to 3 with n=2,3,4 ✘ FAILS! variable values must precede plotting ranges!!
APPENDIX 3 : MATHEMATICA ODDITIES
As you probably know, wolframalpha.com is a user-friendly front-end to Mathematica, the industry-standard
software package for analytical math. The best way to learn how to use WolframAlpha is the Examples pages:
unlike Mathematica, the WolframAlpha front-end is incredibly flexible in the syntax it accepts and almost
anything you type (or mistype!) will be correctly interpreted. Now about the back-end. The Mathematica
software behind WolframAlpha is extremely powerful, but it has some idiosyncrasies that can produce some
surprising / confusing output. Here are two oddities that you should be aware of as they appear quite often.
(1) Mathematica doesn’t simplify expressions as much as you would.
Here’s an example of WolframAlpha / Mathematica behaving strangely:
!
Input: !
integrate sqrt(a/(a-x^2)) dx
Output:
∫
a
dx =
a − x2
⎛
a
x ⎞
a − x 2 tan −1 ⎜
+ constant
2
a− x
⎝ a − x 2 ⎟⎠
!?! The program refuses to cancel those a − x 2 terms in the numerator and denominator! This oddity arises
because Mathematica keeps very careful track of all possible values that the arguments of special functions can
have. In the above case, those square roots might have negative arguments, which would make the results
In[23]:= Integrate@1 ê Sqrt@1 - x ^ 2D, xD
imaginary. Mathematica is also considering the possibility of complex values for all variables. To illustrate,
Out[23]= ArcSin@xD
ê Sqrt@1
- x ^ 2D,
xD
In[23]:= Integrate@1
here is full-blown
Mathematica
refusing
to perform
an “obvious” cancellation:
Out[23]=
In[24]:=
ArcSin@xD
Integrate@Sqrt@1
ê H1 - x ^ 2LD, xD
Integrate@Sqrt@1
ê H1 - x ^ 2LD, xD
1
1 - x2 ArcSin@xD
1 -1x2
1 - x2 ArcSin@xD
Out[24]=
2
1you
- x have to do to
Integrate@Sqrt@1
ê H1
x, Assumptions
Ø 8x œ Reals , Abs@xD < 1<D
here In[21]:=
is what
get- itx ^to2LD,
perform
said cancellation:
In[24]:=
Out[24]=
and
Out[21]=
In[21]:=
ArcSin@xD
Integrate@Sqrt@1 ê H1 - x ^ 2LD, x, Assumptions Ø 8x œ Reals , Abs@xD < 1<D
Out[21]=
ArcSin@xD
You must add both of those “Assumption” clauses to get the cancellation. Assumption clauses aren’t available
in WolframAlpha, and they are sufficiently awkward to apply in Mathematica that one usually doesn’t bother ...
so the message is:
☞
Keep your eyes open for TERMS YOU CAN CANCEL.
(2) Mathematica’s output can change drastically with slight input changes … and watch out for i’s !
Here is what WolframAlpha does with an integral handed to it in two totally equivalent forms:
!
Form 1: !
1
a ex
e− x/2
a ex
dx
=
1− a e x sin −1
x
x
∫
2 1− a e
a 1− a e
Form 2:
1
a
e− x/2
dx
=
2 ∫ e− x − a
a
(
a e x/2
)
+ constant
(
)
a
a e x − 1 ln a e x/2 + a a e x − 1 + constant
e −a
−x
This happens a lot: a seemingly insignificant change in the input can produce substantial changes in the output.
(For an example related to point 1, try integrate 1/sqrt(1-x^2) and integrate sqrt(1/(1-x^2)) … !?!)
The reason is the same as before: Mathematica is accounting for values of the variables that we don’t care about
(e.g. complex values), so slight changes in input form can remove or add possibilities that we don’t need to
consider. In this case, however, we really seem to have a problem: does this integral give an arcsin or a log??
To figure it out, you must first address Issue #1 → there are many “obvious” cancellations you can perform:
Reduced Form 1: sin −1
(
)
a e x/2 + const
Now look at that Form 2: the first factor is
(
)
a − e− x
ln a e x/2 + a a e x − 1 + const
e− x − a
Reduced Form 2:
−1 = i ! To be exact, it’s ±i, as every square root has an implicit ±
accompanying it. Are these expressions really equivalent?? Let’s investigate further by defining u ≡ a e x/2 :
(
Form 1: sin −1 ( u ) + const
)
Form 2: ± i ln u + u 2 − 1 ± i ln( a ) + const
The second term in Form 2, ± i ln( a ) , can be absorbed into the constant of integration … and if you are very
experienced with complex numbers, you can now see that these forms actually are the same. Here are some
relations that are often useful for massaging Mathematica’s unpredictable output into a more helpful form:
(
)
☞ arcsin(u) = −i ln u + u 2 − 1 +
π
2
(
)
arccos(u) = i ln u + u 2 − 1 arctan(u) =
i ⎛ 1− i u ⎞
ln
2 ⎜⎝ 1+ i u ⎟⎠
They are all easy to prove with Euler’s theorem (see Unit 13 from PHYS 225 for tons more about that), and you
can find them all in Wikipedia (just search for arcsin and you’ll get the whole article), but they are not very
pleasant to work with. So what’s the final message? Something like this:
☞
If Mathematica gives you an expression that has an imaginary i in it and your problem doesn’t
involve complex numbers at all, there is undoubtedly an equivalent real-valued expression.
To find it:
TACTIC #1 : Try REARRANGING YOUR INPUT : move terms from numerators to
denominators, break up or combine square roots, etc. Amazingly, this almost always works!
There is almost always some simple algebraic rearrangement you can apply to e.g. an integrand
that produces a real-valued result instead of a complex-valued one.
TACTIC #2 : If rearrangement fails, try the three relations above to turn Mathematica’s complex
output into its real-valued equivalent. (In my experience it’s usually one of those that you need.)