Worksheet #7 – PHY102 (Spring 2011)CollisionsIn this worksheet, we will return to solving equations and solving differential equations.Often there are multiple ways to accomplish something in Mathematica. Usually one way is easierthan another but less elegant. Why might you want to use the elegant method rather than the“easy” one? Because it can often save trouble later on in your Mathematica session. Here isan example. Let’s say you want to know the distance a mass of 50 kg falls in 30 s, after fallingout of an airplane. Obviously you want t o use y = v0t +12at2, where v0= 0, t = 30 s anda = −g = −9.81 m/s2. The simplest way is to type directly into Mathematica:y = -9.81*30ˆ2/2A more elegant route is to type:v0 = 0;a = -9.81;t = 30;y = v0*t + a*tˆ2/2Or a more space-saving way would be:{v0,a,t}={0,-9.81,30};y=v0*t + a*tˆ2/2A problem with these approaches may arise later, because you have permanently defined the vari-ables v0, a and t to these values, so wherever they appear later in your Mathematica notebook,those numerical values will be substituted—possibly leading to undesired results. It is to clean upmesses like this that we often useRemove["Global`*"]Furthermore, whenever you are exploring a physics problem, you will often find yourself wantingto see what happens if you change the initial assumptions, such as the values of v0, a, and t inthis example.A more elegant solution to this problem is to define the equation in algebraic form, and then obtainthe particular solution using substitutions. O ne way to do this is to writey = v0*t + a*tˆ2/2sol1 = y /. {v0 → 0, a → -9.81, t → 30}(To make the → symbol, just type “-” followed by “>” and Mathematica will automatically producethe desired arrow.)1Or similarly, you can writey[t] := v0*t + a*tˆ2/2sol1 = y[t] /. {v0 → 0, a → -9.81, t → 30}In this method, you could also replace the second line bysol1 = y[30] /. {v0 → 0, a → -9.81}These solutions are elegant because the quantity of interest is defined as a function, so we canoperate on it (for example, find its derivative etc.). We found the particular solution we werelooking for (i.e., got the same result as we did when we used the “easy” methods shown above)but didn’t permanently reset the values of any internal variables in Mathematica. Try it. Enterthe different commands above. Then check what Mathematica thinks the variables (e.g., a andv0) are after each case. Use a Remove["Global`*"] in between each test. You may have alreadybeen using this “substitution” technique, if you have been using the DSolve example given outwith worksheet 4. When you use DSolve or Solve, for example, the solutions t o the equation arereturned as a list of su bstitut ions. You can see this by entering the following code:(* This solves two simultaneous linear equations *)f1[x,y ]:= a*x + b*y + cf2[x,y ]:= d*x + e*y + fsol = Solve[{f1[x,y]==0, f2[x,y]==0}, {x,y}]{x,y}={x,y}/. sol[[1]](* This checks to see that the solutions are correct *)Simplify[f1[x,y]]Simplify[f2[x,y]]The definition f1[x_,y_]:=. . . defines a function whose two arguments can later be called anythingyou like. It is often simpler to specify a function in the following way, which is equivalent to theabove:(* Alternative solution for the two simultaneous equations *)Remove[”Global‘*”]f1 = a*x + b*y - c ;f2 = c*x + d*y - e ;sol = Solve[f1 == 0, f2 == 0, x, y]x, y = x, y /. sol[[1]](*This checks to see that the solutions are correct*)Simplify[f1]Simplify[f2]2Problem 1Use Mathematica to solve the following problem (using the examples above to see how to solve theequations). A ball of mass m moving horizontally with a velocity ui undergoes a h ead-on elasticcollision with another ball of mass M traveling at velocity Ui. Apply conservation of momentumand energy to find expressions for the final velocities uf and Uf of these two particles as a functionof m, M, ui and Ui. Verify your solutions by confirming that they preserve energy and momentumconservation.Now find the final velocities for each of the following special cases:(i) m = M(ii) m = 2M(iii) m = 0Problem 2A particle of mass m traveling with speed v in the horizontal direction strikes a pendulum, whichconsists of a thin uniform rod of length A and mass M which is initially hanging vertically at rest.The particle hits the very bottom of the pendulum and sticks to it there. As a result, the center ofmass of the pendulum+particle system rises to a maximum vertical distance H above its originalvalue. After that, if falls back and continues to oscillate forever since we ignore friction.Note that Energy is not conserved during the collision: the collision is inelastic or the projectilewouldn’t stick. Also Momentum is not conserved, because the pivot point of the pendulum suppliesa force.The quantity that is conserved during the collision is the angular momentum, so that is the con-servation law you need to find the initial conditions at t = 0, where the angle of the pendulumis 0. After you use angular momentum conservation to find the initial angular velocity of therod+projectile system, you can find the kinetic energy of that system. Then use energy conser-vation to compute the subsequent motion. Use the angle of the pendulum with respect to thedownward direction as the coordinate.For anyone rusty on their mechanics knowledge—or who has not yet encountered this material inother physics classes, the moment of inertia of the uniform rod pivoted about its end is M A2/3.(You might enjoy using Mathematica to derive that result.) The moment of inertia contributed bythe original projectile is m A2since all of that mass is located at radius A. The angular momentumof the mass+rod system is given by L = I ω. The kinetic energy of the m ass+rod system is givenby KE = (1/2) I ω2. Here, ω = dθ/dt and I is the total moment of inertia. Note, you mustchoose some other name for the moment of inertia in your Mathematic code, becauseMathematica insists on using I to denote the imaginary number√−1.3(i) Find the maximum angle by which the pendulum swings (as a function of H).(ii) Find the initial speed of the p article (as a function of H).(iii) Now suppose m = 1 kg, M = 9 kg, g = 10.0m/s2, H = 4 m, A = 10 m. Find and plot thependulum angle as a function of time.There are two ways to find equations of motion for this problem: you can write a differentialequation for the motion using the torque due to gravity on the system; or you can use