Interactive work with Computer Algebra Systems when solving problems on physics

Transcription

1 1 Interactive work with Computer Algebra Systems when solving problems on physics Leon Magiera, Grażyna Mulak Institute of Physics Wrocław University of Technology, POLAND Abstract It is an obvious fact that using Computer Algebra System (CAS) for solving physics problem is not always straightforward. Very often an interactive human contribution is necessary, not only on conceptual level but also on the level of technical side of the action that leads to the solution. One should note however, that even if the human interference is sometimes necessary one can not overestimate the help provided by CAS. As the first example it was shown how to obtain the general expressions for the radial and transversal components of velocity and acceleration. In the second example the general analysis of two bodies collision in 1D has been done in which the completely elastic and completely inelastic collisions appear as two extreme cases. The analysis has been generalized on D case of the Compton effect. For the analysis the Derive 6.1 and the Mathematica 5. were used. Problem I Velocity and acceleration in the plane polar coordinates and their radial and transversal components It is often convenient to employ polar coordinates r, φ to express the position of a particle moving in the plane. The position vector of the particle can be written as r ( x, y) with x r cos( φ), y r sin( φ) The velocity vector v v is the time derivative of the radial vector r v v v dr dt The acceleration vector a v is the time derivative of the velocity vector v v dv dt Velocity and acceleration vectors can be decomposed into components. If we apply the local coordinate system of mutually perpendicular axes, one parallel to the radial vector r v and the second one perpendicular to it, we get the radial and transversal components, respectively. In physics books we can find the following results for these components: and v r r ar r φ, for the radial components

2 vφ r r φ, ar r φ+ r φ for the transversal components. where the point over any quantity denotes the derivative with respect to time. We will try to obtain these expressions by applying Derive and Mathematica. As the first step we enter the definition of a position vector: (remark: vector quantities are denoted in Derive by characters together with the underline symbol _ ) and definitions of velocity and acceleration. To calculate the components we need also to enter unit vectors: radial unit vector er_ and transversal unit vector eφ_: The above unit vectors are mutually perpendicular, so their the dot product is zero: The radial component of velocity is the dot product of the velocity vector v_ and the unit vector er_ (projection on the direction defined by er_): We see that Derive returns (#8) which is the expected result. Analogously, transversal component of velocity can be computed as the dot product of the velocity vector v_ and the transversal unit vector eφ_

3 3 We have got rather complicated result, which can be simplified via substituting 1 for each sign function of the squared trigonometrical functions: Unfortunately Derive was not able to make the final simplification step, i.e. substitution of SIGN function of the product of squared trigonometrical functions. However, it is easy to see the final result now: In a similar way we calculate the components of acceleration: the radial component, transversal component,

4 4 It can be easily seen that in #15 the denominator is equal to 1, therefore we can rewrite it in the form Having obtained one type of exponents lets say radial, the remaining transversal ones can be calculated via Pythagorean theorem: Unfortunatelly, the declaration of domain for functions is not possible in Derive. The results #17, #18 should be understood as follows: v φ and a φ r φ r φ φ r+ r φ The solution of the above problem with Mathematica is rather straightforward. In the first step we enter the needed definitions: (remark: here, the position vector is denoted as r and its radial component as ρ[t]) The defined unit vectors are mutually perpendicular:

5 5 Then we calculate the needed components via Simplify command: In the above example we see the profitable feature of Mathematica: the possibility to declare a function as nonnegative. It sufficed to assume the functions ρ(t) and ϕ (t) as nonnegative. Without these assignments the calculation results would be messy. Problem II Collisions 1. One dimensional collision Consider a simple head-on collision of two bodies of masses m, m 1 and initial velocities v, v 1, respectively. We assume that this two-bodies system is closed (no mass enters or leaves it) and isolated (no net external force acts on it). The linear momentum of a closed, isolated system is always conserved. For a completely inelastic collision we can write the following linear equation: m1 v1 + mv ( m1 + m )u where u denotes the velocity after the collision. In completely elastic collision the conservation of linear momentum states: m + 1v1 + mv m1u1 mu where u, u denote velocities of masses m, m 1 1 after the collision, respectively. In elastic collision the kinetic energy is also conserved: 1 1 mv m1u1 u m v + m + By doing simple algebra we can solve the above equations and calculate the velocities after the collision.

6 6 If the collision is neither completely elastic nor completely inelastic a part of kinetic energy is transferred to some other form of energy. Then the conservation laws for the linear momentum and energy take the forms: and m + 1v1 + mv m1u1 mu u m v mv + m1u m + + ε, where ε is the kinetic energy loss during the collision. Let us try to solve this equation system with Derive. We enter the above equations: where a. Completely elastic collision. In the case of the completely elastic collision we set ε and get: One of the obtained solutions, u1v1 uv, denotes no collision and should be rejected. b. Completely elastics collision of identical masses. For elastic collision of identical masses ( m 1 m m ) Derives returns:

7 7 which means the bodies exchange their velocities. c. Completely elastics collision of identical masses and static target In this case we set ε, m 1 m m and v: We see that after the collision the body 1 stops (u1) and the body takes off with the initial speed of body 1 (uv1). d. Completely elastics collision (ε) and static target (v) We see that the body 1 rebounds if m1<m. e. Massive target m-> In this case the solution reduces to: The velocity of body is unchanged. The body 1 bounds back if v<v1.

8 8 If we try to calculate final velocities in the one dimensional collision but for general case Derive returns large formula. We display only a part of the whole expression to see that the terms contain SQRT functions: It is obvious that an argument of SQRT function should be nonnegative. The solution of this inequality provides the condition for the kinetic energy loss: We calculate the velocities for the case of the maximum kinetic energy loss: Then we get: From #19 we see that in this case velocities after the collision are identical. It means that the maximum kinetic energy loss exists in completely inelastic collision (try to explain this fact by discussing the collision in different reference frames).

9 9. Two dimensional collision As an example of two dimensional collision we consider the Compton effect. We consider a collision between a photon and a free electron in the target. The principle of energy conservation takes the form: h + ν + mc hν mc (1) where: h - Planck constant ν -photon frequency before the collision m - rest mass of electron c -light speed ν -photon frequency after the collision, m - mass of the recoiling electron; m m, V- speed of the electron V 1 c Graphical representation of the principle of linear momentum conservation is shown in the following picture where p photon momentum before collision p - momentum of scattered photon p e - momentum of scattered electron The principle of linear momentum conservation takes the form: i.e. p and p cos p cos ( φ) + p cos( θ ) (x component) (a) ( φ) p cos( θ ) (y component) (b) e e

10 1 hν hν cos c c and ( φ) + mv cos( θ ) (x component) hν cos( φ) mv cos( θ ) (y component) (3b) c In the above equations we can eliminate ν, ν frequencies by using the general relation between frequency and wavelength cλν Our aim is to find the wavelength shift ( λ λ λ ) of the scattered photon. In physics books we can find the following result h λ ( 1 cosφ) mc Let us try to obtain this result with Derive. We have three equations (1,a, and b) but for unknowns (λ, φ, V, and θ). By using cosine theorem we can eliminate the θ unknown: (3a) p e p + p pp cos( φ) We enter all the above formulas: (remark: Instead of V we entered parameter β) Then we try to solve the entered equations The simplification result (not displayed here) contains in many places the expression 1 β :

11 11 To get rid of this we enter the new parameter A: and try to solve our equations. Simplification of #1 returns no solution. We try to inspect the problem in details. In the first step we simplify only equations in #1: and in the second step we simplify SOLVE function: Surprisingly we have got the solution:

12 1 The solution way with Mathematica is not so troublesome. We enter the needed formulas: If we try to solve the equation we get no result: However, if we repeat the solution procedure with additional information (entered ranges for parameters) we get the solution: Therefore for λ we get: The above examples show that using the CAS is not always straightforward, very often the human interference can not be avoided.