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+ "metadata": {
+ "name": "chapter18.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 18: Impact:Collision Of Elastic Bodies"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-1,Page No:474"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "# Initilization of variables\n",
+ "m_a=1 # kg # mass of the ball A\n",
+ "v_a=2 # m/s # velocity of ball A\n",
+ "m_b=2 # kg # mass of ball B\n",
+ "v_b=0 # m/s # ball B at rest\n",
+ "e=1/2 # coefficient of restitution\n",
+ "\n",
+ "# Calculations\n",
+ "# Solving eqn's 1 & 2 using matrix for v'_a & v'_b,\n",
+ "A=np.array([[1 ,2],[-1 ,1]])\n",
+ "B=np.array([2,1])\n",
+ "C=np.linalg.solve(A,B)\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The velocity of ball A after impact is \",round(C[0]),\"m/s\"\n",
+ "print\"The velocity of ball B after impact is \",round(C[1]),\"m/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of ball A after impact is 0.0 m/s\n",
+ "The velocity of ball B after impact is 1.0 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-2,Page No:480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "# Initilization of variables\n",
+ "m_a=2 # kg # mass of ball A\n",
+ "m_b=6 # kg # mass of ball B\n",
+ "m_c=12 # kg # mass of ball C\n",
+ "v_a=12 # m/s # velocity of ball A\n",
+ "v_b=4 # m/s # velocity of ball B\n",
+ "v_c=2 # m/s # velocity of ball C\n",
+ "e=1 # coefficient of restitution for perfectly elastic body\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# (A)\n",
+ "# Solving eq'n 1 & 2 using matrix for v'_a & v'_b,\n",
+ "A=np.array([[2 ,6],[-1, 1]])\n",
+ "B=np.array([48,8])\n",
+ "C=np.linalg.solve(A,B)\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# (B)\n",
+ "# Solving eq'ns 3 & 4 simultaneously using matrix for v'_b & v'_c\n",
+ "P=np.array([[1 ,2],[-1, 1]])\n",
+ "Q=np.array([12,6])\n",
+ "R=np.linalg.solve(P,Q)\n",
+ "\n",
+ "# Results (A&B)\n",
+ "\n",
+ "print\"The velocity of ball A after impact on ball B is \",round(C[0]),\"m/s\" # here the ball of mass 2 kg is bought to rest\n",
+ "print\"The velocity of ball B after getting impacted by ball A is \",round(C[1]),\"m/s\"\n",
+ "print\"The final velocity of ball B is \",round(R[0]),\"m/s\" # here the ball of mass 6 kg is bought to rest\n",
+ "print\"The velocity of ball C after getting impacted by ball B is \",round(R[1]),\"m/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of ball A after impact on ball B is 0.0 m/s\n",
+ "The velocity of ball B after getting impacted by ball A is 8.0 m/s\n",
+ "The final velocity of ball B is 0.0 m/s\n",
+ "The velocity of ball C after getting impacted by ball B is 6.0 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-3,Page No:481"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "h_1=9 # m # height of first bounce\n",
+ "h_2=6 # m # height of second bounce\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# From eq'n (5) we have, Coefficient of restitution between the glass and the floor is,\n",
+ "e=(h_2*h_1**-1)**0.5\n",
+ "# From eq'n 3 we get height of drop as,\n",
+ "h=h_1/e**2 # m\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The ball was dropped from a height of \",round(h,1),\"m\"\n",
+ "print\"The coefficient of restitution between the glass and the floor is \",round(e,1)\n",
+ "# Here we use h`=h_1 & h``=h_2 because h` & h`` could not be defined in Scilab.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ball was dropped from a height of 13.5 m\n",
+ "The coefficient of restitution between the glass and the floor is 0.8\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-4,Page No:484"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "e=0.90 # coefficient o restitution\n",
+ "v_a=10 # m/s # velocity of ball A\n",
+ "v_b=15 # m/s # velocity of ball B\n",
+ "alpha_1=30 # degree # angle made by v_a with horizontal\n",
+ "alpha_2=60 # degree # angle made by v_b with horizontal\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# The components of initial velocity of ball A:\n",
+ "v_a_x=v_a*cos(alpha_1*(pi/180)) # m/s\n",
+ "v_a_y=v_a*sin(alpha_1*(pi/180)) # m/s\n",
+ "\n",
+ "# The components of initial velocity of ball B:\n",
+ "v_b_x=-v_b*cos(alpha_2*(pi/180)) # m/s\n",
+ "v_b_y=v_b*sin(alpha_2*(pi/180)) # m/s\n",
+ "\n",
+ "# From eq'n 1 & 2 we get,\n",
+ "v_ay=v_a_y # m/s # Here, v_ay=(v'_a)_y\n",
+ "v_by=v_b_y # m/s # Here, v_by=(v'_b)_y\n",
+ "\n",
+ "# On adding eq'n 3 & 4 we get,\n",
+ "v_bx=((v_a_x+v_b_x)+(-e*(v_b_x-v_a_x)))*0.5 # m/s # Here. v_bx=(v'_b)_x\n",
+ "\n",
+ "# On substuting the value of v'_b_x in eq'n 3 we get,\n",
+ "v_ax=(v_a_x+v_b_x)-(v_bx) # m/s # here, v_ax=(v'_a)_x\n",
+ "\n",
+ "# Now the eq'n for resultant velocities of balls A & B after impact are,\n",
+ "v_A=(v_ax**2+v_ay**2)**0.5 # m/s\n",
+ "v_B=(v_bx**2+v_by**2)**0.5 # m/s\n",
+ "\n",
+ "# The direction of the ball after Impact is,\n",
+ "theta_1=arctan(-(v_ay/v_ax))*(180/pi) # degree\n",
+ "theta_2=arctan(v_by/v_bx)*(180/pi) # degree\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The velocity of ball A after impact is \",round(v_A,2),\"m/s\"\n",
+ "print\"The velocity of ball B after impact is \",round(v_B,2),\"m/s\"\n",
+ "print\"The direction of ball A after impact is \",round(theta_1,2),\"degree\"\n",
+ "print\"The direction of ball B after impact is \",round(theta_2,2),\"degree\"\n",
+ "# Her we use, (1) v'_a & v'_b as v_A & v_B.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of ball A after impact is 8.35 m/s\n",
+ "The velocity of ball B after impact is 15.18 m/s\n",
+ "The direction of ball A after impact is 36.77 degree\n",
+ "The direction of ball B after impact is 58.85 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-5,Page No:485"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initiization of variables\n",
+ "theta=30 # degrees # ange made by the ball against the wall\n",
+ "e=0.50\n",
+ "# Calculations\n",
+ "# The notations have been changed\n",
+ "# Resolving the velocity v as,\n",
+ "v_x=cos(theta*(pi/180))\n",
+ "v_y=sin(theta*(pi/180))\n",
+ "V_y=v_y\n",
+ "# from coefficient of restitution reation\n",
+ "V_x=-e*v_x\n",
+ "# Resultant velocity\n",
+ "V=sqrt(V_x**2+V_y**2)\n",
+ "theta=arctan(V_y*(-V_x)**-1)*(180/pi) # taking +ve value for V_x\n",
+ "# NOTE: Here all the terms are multiplied with velocity i.e (v).\n",
+ "# Results\n",
+ "\n",
+ "print\"The velocity of the ball is \",round(V,3),\"v\"\n",
+ "print\"The direction of the ball is \",round(theta,1),\"degree\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of the ball is 0.661 v\n",
+ "The direction of the ball is 49.1 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-6,Page No:488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "# Initilization of variables\n",
+ "e=0.8 # coefficient of restitution\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calcuations\n",
+ "\n",
+ "# Squaring eqn's 1 &2 and Solving eqn's 1 & 2 using matrix for the value of h\n",
+ "A=np.array([[-1 ,(2*g)],[-1 ,-(1.28*g)]])\n",
+ "B=np.array([0.945**2,(-0.4*9.81)])\n",
+ "C=np.linalg.solve(A,B) # m\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The height from which the ball A should be released is \",round(C[1],3),\"m\"\n",
+ "# The answer given in the book i.e 0.104 is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The height from which the ball A should be released is 0.15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-7,Page No:490"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "theta_a=60 # degree # angle made by sphere A with the verticle\n",
+ "e=1 # coefficient of restitution for elastic impact\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# theta_b is given by the eq'n cosd*theta_b=0.875, hence theta_b is,\n",
+ "theta_b=arccos(0.875)*(180/pi) # degree\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The angle through which the sphere B will swing after the impact is \",round(theta_b,2),\"degree\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle through which the sphere B will swing after the impact is 28.96 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-8,Page No:491"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "m_a=0.01 # kg # mass of bullet A\n",
+ "v_a=100 # m/s # velocity of bullet A\n",
+ "m_b=1 # kg # mass of the bob\n",
+ "v_b=0 # m/s # velocity of the bob\n",
+ "l=1 # m # length of the pendulum\n",
+ "v_r=-20 # m/s # velocity at which the bullet rebounds the surface of the bob # here the notation for v'_a is shown by v_r\n",
+ "v_e=20 # m/s # velocity at which the bullet escapes through the surface of the bob # here the notation for v_a is shown by v_e\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Momentum of the bullet & the bob before impact is,\n",
+ "M=(m_a*v_a)+(m_b*v_b) # kg.m/s......(eq'n 1)\n",
+ "\n",
+ "# The common velocity v_c ( we use v_c insted of v' for notation of common velocity) is given by equating eq'n 1 & eq'n 2 as,\n",
+ "\n",
+ "# (a) When the bullet gets embedded into the bob\n",
+ "v_c=M/(m_a+m_b) # m/s\n",
+ "# The height h to which the bob rises is given by eq'n 3 as,\n",
+ "h_1=(0.5)*(v_c**2/g) # m\n",
+ "# The angle (theta_1) by which the bob swings corresponding to the value of height h_1 is,\n",
+ "theta_1=arccos((l-h_1)/l)*(180/pi) # degree\n",
+ "\n",
+ "# (b) When the bullet rebounds from the surface of the bob\n",
+ "# The velocity of the bob after the rebound of the bullet from its surface is given by equating eq'n 1 & eq'n 4 as,\n",
+ "v_bob_rebound=M-(m_a*v_r) # m/s # here v_bob_rebound=v'_b\n",
+ "# The equation for the height which the bob attains after impact is,\n",
+ "h_2=(v_bob_rebound**2)/(2*g) # m\n",
+ "# The corresponding angle of swing \n",
+ "theta_2=arccos((l-h_2)/l)*(180/pi) # degree\n",
+ "\n",
+ "# (c) When the bullet pierces and escapes through the bob\n",
+ "# From eq'n 1 & 5 the velocity attained by the bob after impact is given as,\n",
+ "v_b_escape=M-(m_a*v_e) # m/s # here we use, v_b_escape insted of v'_b\n",
+ "# The equation for the height which the bob attains after impact is,\n",
+ "h_3=(v_b_escape**2)/(2*g) # m\n",
+ "# The corresponding angle of swing \n",
+ "theta_3=arccos((l-h_3)/(l))*(180/pi) # degree\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"(a) The maximum angle through which the pendulum swings when the bullet gets embeded into the bob is \",round(theta_1,1),\"degree\"\n",
+ "print\"(b) The maximum angle through which the pendulum swings when the bullet rebounds from the surface of the bob is \",round(theta_2,2),\"degree\"\n",
+ "print\"(c) The maximum angle through which the pendulum swings when the bullet escapes from other end of the bob the bob is \",round(theta_3,1),\"degree\"\n",
+ "# IN THIS SUM WE HAVE USED DIFFERENT NOTATIONS CONSIDERING DIFFERENT CASES BECAUSE IN THE TEXT BOOK WE HAD 3 VARIABLES WITH SAME NOTATION BUT WITH A DIFFERENT VALUE WHICH COULD NOT BE EXECUTED INTO SCILAB.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The maximum angle through which the pendulum swings when the bullet gets embeded into the bob is 18.2 degree\n",
+ "(b) The maximum angle through which the pendulum swings when the bullet rebounds from the surface of the bob is 22.09 degree\n",
+ "(c) The maximum angle through which the pendulum swings when the bullet escapes from other end of the bob the bob is 14.7 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-9,Page No:493"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "W_a=50 # N # falling weight\n",
+ "W_b=50 # N # weight on which W_a falls\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "m_a=W_a/g # kg # mass of W_a\n",
+ "m_b=W_b/g # kg # mass of W_b\n",
+ "k=2*10**3 # N/m # stiffness of spring\n",
+ "h=0.075 # m # height through which W_a falls\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "# The velocity of weight W_a just before the impact and after falling from a height of h is given from the eq'n, ( Principle of conservation of energy)\n",
+ "v_a=(2*g*h)**0.5 # m/s\n",
+ "\n",
+ "# Let the mutual velocity after the impact be v_m (i.e v_m=v'), (by principle of conservation of momentum)\n",
+ "v_m=(m_a*v_a)/(m_a+m_b) # m/s\n",
+ "\n",
+ "# Initial compression of the spring due to weight W_b is given by,\n",
+ "delta_st=(W_b/k)*(10**2) # cm\n",
+ "\n",
+ "# Let the total compression of the spring be delta_t, Then delta_t is found by finding the roots from the eq'n........ delta_t^2-0.1*delta_t-0.000003=0. In this eq'n let,\n",
+ "a=1\n",
+ "b=-0.1\n",
+ "c=-0.000003\n",
+ "delta_t=((-b+((b**2-(4*a*c))**0.5))/2*a)*(10**2) # cm # we consider the -ve value\n",
+ "delta=delta_t-delta_st # cm\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The compression of the spring over and above caused by the static action of weight W_a is \",round(delta),\"cm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The compression of the spring over and above caused by the static action of weight W_a is 10.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-10,Page No:494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "v_a=600 # m/s # velocity of the bullet before impact\n",
+ "v_b=0 # m/s # velocity of the block before impact\n",
+ "w_b=0.25 # N # weight of the bullet\n",
+ "w_wb=50 # N # weight of wodden block\n",
+ "mu=0.5 # coefficient of friction between the floor and the block\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "m_a=w_b/g # kg # mass of the bullet\n",
+ "m_b=w_wb/g # kg # mass of the block\n",
+ "\n",
+ "# Let the common velocity be v_c which is given by eq'n (Principle of conservation of momentum)\n",
+ "v_c=(w_b*v_a)/(w_wb+w_b) # m/s\n",
+ "\n",
+ "# Let the distance through which the block is displaced be s, Then s is given by eq'n\n",
+ "s=v_c**2/(2*g*mu) # m\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The distance through which the block is displaced from its initial position is \",round(s,2),\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance through which the block is displaced from its initial position is 0.91 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18-11,Page No:495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "M=750 # kg # mass of hammer\n",
+ "m=200 # kg # mass of the pile\n",
+ "h=1.2 # m # height of fall of the hammer\n",
+ "delta=0.1 # m # distance upto which the pile is driven into the ground\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Caculations\n",
+ "\n",
+ "# The resistance to penetration to the pile is given by eq'n,\n",
+ "R=(((M+m)*g)+((M**2*g*h)/((M+m)*delta)))*(10**-3) # kN \n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The resistance to penetration to the pile is \",round(R),\"KN\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance to penetration to the pile is 79.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |