Worksheet #7 – PHY102 (Spring 2010)CollisionsIn this worksheet, we will return to solving equations and solving differentialequations.Often there are multiple ways of accomplishing something in Mathemat-ica. Usually one way is easier than another but less elegant. Why might youwant to use the elegant method rather than the “easy” one? Because it canoften save trouble later on in your Mathematica session. Here is an example.Let’s say you want to know the distance a mass of 50 kg falls in 30 s, afterfalling out of an airplane. Obviously you want to use y = v0t +12at2, wherev0= 0, t = 30 s and a = −g = −9.81 m/s2. The simplest way is to typedirectly 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 perma-nently defined the variables v0, a and t to these values, so wherever theyappear later in your Mathematica notebook, those numerical values will besubstituted—possibly leading to undesired results. It is to clean up messeslike this that we often use Remove["Global`*"]. Furthermore, whenever youare exploring a physics problem, you will often find yourself wanting to seewhat happens if you change the initial assumptions, such as the values of v0,a, and t in this example.1The most elegant solution to this problem is to define the equation in alge-braic form, and then obtain the particular solution using “substitutions”:y[t ] := v0*t + a*tˆ2/2sol1 = y[t] /. {v0 → 0, a → -9.81, t → 30}This solution is elegant because the function of interest is defined as a func-tion so we can operate on it (for example, find its derivative etc.). We foundthe particular solution we were looking for (i.e., got the same result as wedid when we used the “easy” methods shown above) but didn’t permanentlyreset the values of any internal variables in Mathematica. Try it. Enterthe different commands above. Then check what Mathematica thinks thevariables (e.g., a and v0) are after each case. Use a Remove["Global`*"]in between each test. You may have already been using this “substitution”technique, if you have been using the DSolve example given out with work-sheet 4. When you use DSolve or Solve, for example, the solutions to theequation are returned as a list of substitutions. You can see this by enteringthe following code:(* You can structure your equation solvers like this *)(* This solves two simultaneous linear equations *)f1[x ,y ]:= a*x + b*y - cf2[x ,y ]:= c*x + d*y - esol = 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]]2The definition f1[x_,y_]:=. . . defines a function whose two argumentscan later be called anything you like. It is often simpler to specify a functionin the following way, which is equivalent to the above:(* 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]Problem 1Use Mathematica to solve the following problem (see the example above).A ball of mass m moving horizontally with a velocity ui undergoes a head-on elastic collision with another ball of mass M travelling at velocity Ui.Apply conservation of momentum and energy to find expressions for thefinal velocities uf and Uf of these two particles as a function of m, M, uiand Ui. Verify your solutions by confirming that they preserve energy andmomentum conservation.Now find the final velocities for the following special cases:(i) m = M(ii) m = 2M(iii) m = 0Problem 2A particle of mass m travelling with speed v in the horizontal directionstrikes a pendulum, which consists of a thin uniform rod of length A andmass M which is initially hanging vertically at rest. The particle hits thevery 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 distanceH above its original value. After that, if falls back and continues to oscillate3forever since we ignore friction.Note that Energy is not conserved during the collision: the collision is inelas-tic or the projectile wouldn’t stick. Also the Momentum is not conserved,because the pivot point of the pendulum supplies a force.The quantity that is conserved during the collision is the angular momentum,so that is the conservation law you need to find the initial conditions at t =0, where the angle of the pendulum is 0. After you use angular momentumconservation to find the initial angular velocity of the rod+projectile system,you can find the kinetic energy of that system. Then use energy conserva-tion to compute the subsequent motion. Use the angle of the pendulum withrespect to the downward direction as the coordinate.(For anyone rusty on their mechanics knowledge, the moment of inertia of therod pivoted about its end is M A2/3. You might enjoy using Mathematicato derive that result.)(i) Find the initial speed of the particle.(ii) Find the maximum angle by which the pendulum swings.(iii) Now suppose m = 1 kg, M = 9 kg, g = 10.0m/s2, H = 4 m,A = 10 m. Find and plot the pendulum angle as a function of time.There are two ways to find equations of motion for this problem: you canwrite a differential equation for the motion using the torque due to gravityon the system; or you can use energy conservation. The energy conservationmethod is probably better, because it leads to a first-order differential equa-tion instead of second order; and it lets you build in the boundary conditioncorresponding to the initial condition of known total energy directly.Note: when your instructor solved this problem, Mathematica was notable to solve the equation when the initial condition theta[0]==0 was in-cluded in DSolve. But without that condition, it did give a solution, whichcontained an unknown integration constant. You can set that constant to zeroby /. C[1] -> 0.4(iv) Now solve the linear pendulum problem using the same parameters.(The “linear” approximation for the pendulum corresponds to approximat-ing sin θ by θ in the torque equation; or equivalently approximating cos θ by1 − θ2/2 in the potential energy.)Plot the time dependent oscillations of the full solution and the linear