TI1-1.2_Xocs_elastics.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Xocs elàstics"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "3. En un joc de hoquei taula un jugador deixa anar la seva maça de 150 g que dona un cop a 0,6 m/s contra el disc de 39 g que anava a 1,6  m/s en sentit contrari. Calcula las velocitats finals de la maça i del disc."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Solució\n",
    "\n",
    "Es tracta d'un xoc elàstic, on és compleix la conservació de l'energia cinètica, a més a més de la conservació del moment. Encara que, en general, aquest tipus de problema l'hauríem de plantejar vectorialment, com les velocitats són oposades el podem escriure de forma escalar:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$$\n",
    "\\frac{1}{2}\\cdot m_{1} \\cdot v_{1i}^{2} + \\frac{1}{2}\\cdot m_{2} \\cdot v_{2i}^{2} = \\frac{1}{2}\\cdot m_{1} \\cdot v_{1f}^{2} + \\frac{1}{2}\\cdot m_{2} \\cdot v_{1f}^{2} \\\\\n",
    "m_{1} \\cdot v_{1i} + m_{2} \\cdot v_{2i} = m_{1} \\cdot v_{1f} + m_{2} \\cdot v_{2f}\n",
    "$$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "posem les dades en unitats S.I., encara que ometrem les unitats per simplificar la feina:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$$\n",
    "\\frac{1}{2}\\cdot 0,15 \\cdot 0,6^{2} + \\frac{1}{2}\\cdot 0,039 \\cdot 1,6^{2} = \\frac{1}{2}\\cdot 0,15 \\cdot v_{1f}^{2} + \\frac{1}{2}\\cdot 0,039 \\cdot v_{1f}^{2} \\\\\n",
    "0,15 \\cdot 0,6 + 0,039 \\cdot 1,6 = 0,15 \\cdot v_{1f} + 0,039 \\cdot v_{2f}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Es tracta d'un sistema **no linial** de 2 equacions amb 2 incògnites. \n",
    "\n",
    "Podem calcular el primer membre de cada equació, ja que coneixem les condicions inicials:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.07692000000000002"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "1/2*0.15*0.6**2+1/2*0.039*1.6**2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.1524"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "0.15*0.6+0.039*1.6"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Per resoldre aquest sistema podem utilitzar el paquet de calcul simbòlic *sympy*"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[(0.600000000000000, 1.60000000000000), (1.01269841269841, 0.0126984126984127)]"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from sympy import *\n",
    "x, y = symbols(\"x y\")\n",
    "\n",
    "solve([Eq(0.15/2*x**2+0.039/2*y**2, 0.07692), Eq(0.15*x+0.039*y, 0.1524)],[x,y])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Conclusions\n",
    "\n",
    "Fixeu-vos que el sistema admet dues solucions:\n",
    "\n",
    "Aquesta solució correspon a abans del xoc:\n",
    "$\n",
    " \\left\\{ \\begin{array}{cl}\n",
    "v_{1}=0,6  \\\\\n",
    " v_{2}=1,6\n",
    "\\end{array} \\right.\n",
    "$\n",
    "\n",
    "Aquesta solució correspon a després del xoc:\n",
    "$\n",
    " \\left\\{ \\begin{array}{cl}\n",
    "v_{1}=1,0127  \\\\\n",
    " v_{2}=0,0127\n",
    "\\end{array} \\right.\n",
    "$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Potència de *sympy*\n",
    "\n",
    "L'ajut de la llibreria *sympy* és en aquest cas espectacular. Sense ella hauríem de resoldre a ma el sistema, operació extremadament pesada ja que es tracta d'un sistema no linial. La millor forma de resoldre aquests sistemes és el mètode de substitució:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Aïllem $v_{2f}$ a l'equació de conservació del moment:\n",
    "\n",
    "$$\n",
    "0,1524 = 0,15 \\cdot v_{1f} + 0,039 \\cdot v_{2f} \\Rightarrow v_{2f} = \\frac{0,1524 - 0,15 \\cdot v_{1f}}{0,039} \\\\\n",
    "0,07692 = 0,075 \\cdot v_{1f}^{2} + 0,0195 \\cdot v_{1f}^{2}\n",
    "$$\n",
    "\n",
    "i substituim la seva expressió a l'equació de conservació de l'energia:\n",
    "\n",
    "$$\n",
    "0,07692 = \\frac{1}{2}\\cdot 0,15 \\cdot v_{1f}^{2} + \\frac{1}{2}\\cdot 0,039 \\cdot v_{1f}^{2} \\Rightarrow 0,07692 = 0,075 \\cdot v_{1f}^{2} + 0,0195 \\cdot  \\left(\\frac{0,1524 - 0,15 \\cdot v_{1f}}{0,039}\\right)^2\n",
    "$$\n",
    "\n",
    "Desenvolupem el quadrat:\n",
    "\n",
    "$$\n",
    "0,07692 = 0,075 \\cdot v_{1f}^{2} + 0,0195 \\cdot  \\left( \\frac{0,1524^2 - 2 \\cdot 0,1524 \\cdot 0,15 \\cdot v_{1f} + 0,15^2 \\cdot v_{1f}^2}{0,039^2} \\right)\n",
    "$$\n",
    "\n",
    "Simplifiquem l'expressió:\n",
    "\n",
    "$$\n",
    "0,07692 = 0,075 \\cdot v_{1f}^{2} + 0,2978 - 0,5862 \\cdot v_{1f} + 0,2885 \\cdot v_{1f}^2\n",
    "$$\n",
    "\n",
    "$$\n",
    "(0,075 +0,2885) \\cdot v_{1f}^2 - 0,5862 \\cdot v_{1f}  + 0,2978 - 0,07692 =  0 \n",
    "$$\n",
    "\n",
    "Arribem a una eqüació de 2n grau:\n",
    "\n",
    "$$\n",
    "0,3635 \\cdot v_{1f}^2 - 0,5862 \\cdot v_{1f}  + 0,2209 =  0 \n",
    "$$\n",
    "\n",
    "que té com a solucions:\n",
    "\n",
    "$\n",
    "v_{1f} = \\left\\{ \\begin{array}{cl}\n",
    "1,0127  \\\\\n",
    "0,6\n",
    "\\end{array} \\right.\n",
    "$\n",
    "\n",
    "Calculem $v_{2f}$ a partir de $v_{1f}$. Ems dona, respectivament,\n",
    "\n",
    "$\n",
    "v_{2f} = \\left\\{ \\begin{array}{cl}\n",
    "0,0127  \\\\\n",
    "1,6\n",
    "\\end{array} \\right.\n",
    "$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.7.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 4
}