Created by Jerald Izatt, U. of Alabama

A. Free-fall exercise.
If the vertical direction is given by y, the instantaneous position of a body that starts from height yo with initial velocity vo is given by
(1)    y(t) = yo + vo t -1/2 g t^2

Newton's 2nd law can be written for this case as follows
(2)     F = ma = m dv/dt = -mg .

Task 1 - Demonstrate that the function given by equation (1) is a solution to equation (2).
Hint: Differentiate (1) to obtain v(t). Then find dv/dt and see if it satisfies equation (2).

Task 2. - Use Maple to plot v(t) over the range t=0 to t=10 sec for the special case vo=0.

B. Falling body in the presence of a resistive force proportional to v(t).
In this case Newton's law becomes
(3)    F = ma = m dv/dt = -mg -bv ,          where b is a constant.

Task 3 - Determine the dimensions of b and of the quantity mg/b .

Task 4 - By reference to equation (3), describe what you think will happen to the body accelerating downward as v approaches mg/b. Consider the function (4) v(t) = mg/b (exp(-bt/m)-1).

Task 5 - See if (4) is a solution to equation (3), doing something similar to Task 1.

Task 6 - Use Maple to plot eq (4) from t=0 to t=10 sec for the special case b/m = 1.
Repeat this for b/m = 2, and for b/m = 3.
Display these curves on a single graph.
Compare these graphs to the one you made in Task 2.

Task 7 - Is the conclusion you drew in Task 4 borne out by these graphs?
Would 'terminal velocity' be an apt name for the quantity mg/b? Why or why not?

airdrag.mws     

restart;with(plots):
y:=yo+vo*t-1/2*g*t^2;
vy:=diff(y,t);
plot(subs(vo=0,g=98/10,vy),t=0..10);
a:=diff(vy,t); > -m*g =m*diff(y,t,t);
restart;
v:=m*g/b*(exp(-b*t/m)-1);
eq1:=-m*g-b*v = m*diff(v,t);
simplify(eq1);
plot(subs(b=1,m=1,g=98/10,v),t=0..10);
plot({subs(b=1,m=1,g=98/10,v),subs(b=2,m=1,g=98/10,v),subs(b=3,m=1,g=98/10,v)},t=0..10);

Download airdrag.mws      (Use 'Save File'')

monkey.mws     monkey and hunter simulation. MJM 12 June 1998

Students are given a simple plot of two circles, representing the monkey and the projectile. They are also given an animation of the projectile moving with initial velocity, but no gravity, such that the projectile misses the monkey.

The plan is to have students run the animation and see the projectile miss the monkey. Then they must

1) revise the worksheet to hit the monkey
2) predict what will happen when both monkey and projectile are subject to gravity (when gravity is 'turned on')
3) turn gravity on and see what happens in the animation
4) work out (without running any more simulations) what initial velocity is needed so that the projectile hits the monkey at ground level.
5) revise the simulation and run it to see if the falling monkey is struck by the projectile at ground level.

restart; with(plots):
Function (parametric plot) for a circle of radius 0.2, centered at x=a, y=b).
circ:=[0.2*cos(theta)+a,0.2*sin(theta)+b,theta=0..2*Pi];
plot(subs(a=5,b=3,circ)); # Parametric plot of a circle
Now plot the projectile at the origin and target on the same plot, to 1:1 scale plot({subs(a=0,b=0,circ),subs(a=5,b=3,circ)},color=black,scaling=constrained);
Now for the animation command, to illustrate time duration and number of frames animate({subs(a=10*t,b=8*t,circ),subs(a=5,b=3,circ)},t=0..1/2,frames=10,color=black,scaling=constrained);
Students:
1) modify the animation to make the projectile 'hit' the target
2) modify the animation for an initial velocity and angle for the projectile
3) change the projectile's angle so it 'hits' the target
4) with gravity present for both projectile and target, predict if the projectile will pass under, over, or through the target
5) put gravity in the simulation and see what happens (test your hypothesis)
6) work out what initial velocity will be needed for the projectile to hit the target at y=0
7) adjust the initial velocity and see if you got the prediction correct.

Download monkey.mws (Use'Save File'')

Policeman-and-speeder activity for students

A speeder passes a stationary policeman traveling 24 m/s.
At the instant the speeder goes past, the cop accelerates at a constant 3.0 m/s^2.

Students:
1) write an expression 'sp' for the speeder's position as a function of time
2) write an expression 'cop' for the cop's position as a function of time
3) before plotting, figure out what time the cop reaches a speed of 24 m/s.
4) plot both expressions on the same graph {sp,cop} for t=0..2 sec
5) change the time interval to show when the cop reaches the speeder's speed
6) Sketch the graph. Show how you 'eyeballed' that the cop and speeder have the same speed
7) Change the time interval to show when the cop overtakes the speeder. Record this time

speeder.mws               

restart; with(plots):
sp:= 24*t; cop:=3/2*t^2;
plot({sp,cop},t=0..12,color=black);

Download speeder.mws (Use 'Save File'')

Energy diagrams, forces, and equilibrium (J. W. Harrell, U of Alabama)

A particle is subject to a conservative force for which the potential energy function U(x) is given by

1. Use Maple to plot U(x) from x= -1 to x= +5 .

2. Examine the plot and estimate the following

3. a) At what points would the particle be in stable equilibrium if placed there at rest?

   b) At what points would the particle be in unstable equilibrium if placed there at rest?

4. Now use Maple to calculate the force function F(x) = -dU(x)/dx.

5. On a single graph, plot both U(x) and F(x). Is your graph consistent with your answers to questions 2 and 3? If not, then reconcile your answers.

6. Now suppose the particle has an energy of 10 units.

On a single graph, plot both U(x) and total energy. (Don't use E for energy; Maple has claimed E for something else.)

7. If no other forces act on the particle, determine from your graph

Notes:

equil.mws

restart; with(plots):
U := 11*x^2-7*x^3+x^4;
plot(U,x=-1..5);
F:=diff(U,x);
plot({U,F},x=-1..5,color=black);

Download equil.mws (Use 'Save File'')

collide.mws           Created by Mike Moloney June 10, 1998

A 2-kg mass m1 moves at 5 m/s in one dimension. A massless spring of unstretched length 0.5 m and spring constant 10 N/m travels with the moving mass. The moving mass and spring approach a stationary 2-kg mass m2.

The questions below involve linear momentum, kinetic and potential energy, and ability to interpret a graph.

For the 'collision' which follows make your best estimate of
   1) the 'final' speeds of m1 and m2, after the collision is over.
   2) the speed of m1 and m2 at that instant when they have the same speed
   3) the distance between m1 and m2 when they have the same speed

Run through the worksheet below and obtain the graph of position vs. time for each mass during the collision. (Work from the screen, or print a copy of this graph.). Then
   1) verify the initial speed of m1 to be 5 m/s
   2) verify the final speed of m2
   3) find the point where m1 & m2 have the same speed
   4) Find the distance between the masses when they have the same speed
   5) Calculate the total KE and PE when the masses have the same speed.
       Compare this to the initial amount of energy before the collision.

Change the mass of m2 to 3 kg and repeat all of the work above.

restart;with(plots):
d:=-x2(t)+x1(t)+x0; # spring compression
eq1:=-k*d*Heaviside(d) = m1*diff(x1(t),t,t);
If d<0 then the spring force will be zero (x2-x1>x0) because spring not yet compressed
eq2:=+k*(d)*Heaviside(d) = m2*diff(x2(t),t,t);
values:={m1=1,m2=1,k=10,x0=1/2};
eqs:=subs(values,eq1),subs(values,eq2);
s:=dsolve({eqs,x1(0)=0,D(x1)(0)=5,x2(0)=1,D(x2)(0)=0},{x1(t),x2(t)},numeric);
h:=odeplot(s,[t,x1(t)],0..1,color=red):
j:=odeplot(s,[t,x2(t)],0..1,color=blue):
display({h,j},title=`x1(t)[red], x2(t)[blue] vs t`);

Download collide.mws (Use 'Save File')