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   "source": [
    "# Chapter 17 Kinetics of a Particle Impulse and Momentum"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Example 17.1 Principle of impulse and momentum"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
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   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The average impules force exerted by the bat on the ball is 408.633978 N\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "#Initilization of variables\n",
    "m=0.1 # kg # mass of ball\n",
    "# Calculations\n",
    "# Consider the respective F.B.D.\n",
    "# For component eq'n in x-direction\n",
    "delta_t=0.015 # seconds # time for which the ball &the bat are in contact\n",
    "v_x_1=-25 # m/s \n",
    "v_x_2=40*math.cos(40*math.pi/180) # m/s\n",
    "F_x_average=((m*(v_x_2))-(m*(v_x_1)))/(delta_t) # N\n",
    "# For component eq'n in y-direction\n",
    "delta_t=0.015 # sceonds\n",
    "v_y_1=0 # m/s\n",
    "v_y_2=40*math.sin(40*math.pi/180) # m/s\n",
    "F_y_average=((m*v_y_2)-(m*(v_y_1)))/(delta_t) # N\n",
    "F_average=math.sqrt(F_x_average**2+F_y_average**2) # N\n",
    "# Results\n",
    "print('The average impules force exerted by the bat on the ball is %f N'%F_average)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Example 17.2 Principle of impulse and momentum"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The recoil velocity of gun is -5 m/s\n",
      "The Force required to stop the gun is 62500.000000 N\n",
      "The time required to stop the gun is 0.240000 seconds\n"
     ]
    }
   ],
   "source": [
    "# Initiliation of variables\n",
    "m_g=3000 # kg # mass of the gun\n",
    "m_s=50 # kg # mass of the shell\n",
    "v_s=300 # m/s # initial velocity of shell\n",
    "s=0.6 # m # distance at which the gun is brought to rest\n",
    "v=0 # m/s # initial velocity of gun\n",
    "# Calculations\n",
    "# On equating eq'n 1 & eq'n 2 we get v_g as,\n",
    "v_g=(m_s*v_s)/(-m_g) # m/s\n",
    "# Using v^2-u^2=2*a*s to find acceleration,\n",
    "a=(v**2-v_g**2)/(2*s) # m/s**2\n",
    "# Force required to stop the gun,\n",
    "F=m_g*(-a) # N # here we make a +ve to find the Force\n",
    "# Time required to stop the gun, using v=u+a*t:\n",
    "t=(-v_g)/(-a)  # seconds # we take -a to consider +ve value of acceleration\n",
    "# Results\n",
    "print('The recoil velocity of gun is %d m/s'%v_g)\n",
    "print('The Force required to stop the gun is %f N'%F)\n",
    "print('The time required to stop the gun is %f seconds'%t)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Example 17.3 Principle of impulse and momentum"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(a) The velocity of boat as observed from the ground is 0.166667 m/s\n",
      "(b) The distance by which the boat gets shifted is 0.833333 m\n"
     ]
    }
   ],
   "source": [
    "# Initilization of variables\n",
    "m_m=50 # kg # mass of man\n",
    "m_b=250 # kg # mass of boat\n",
    "s=5 # m # length of the boat\n",
    "v_r=1 # m/s # here v_r=v_(m/b)= relative velocity of man with respect to boat\n",
    "# Calculations\n",
    "# Velocity of man is given by, v_m=(-v_r)+v_b\n",
    "# Final momentum of the man and the boat=m_m*v_m+m_b*v_b. From this eq'n v_b is given as\n",
    "v_b=(m_m*v_r)/(m_m+m_b) # m/s # this is the absolute velocity of the boat\n",
    "# Time taken by man to move to the other end of the boat is,\n",
    "t=s/v_r # seconds\n",
    "# The distance travelled by the boat in the same time is,\n",
    "s_b=v_b*t # m to right from O\n",
    "# Results\n",
    "print('(a) The velocity of boat as observed from the ground is %f m/s'%v_b)\n",
    "print('(b) The distance by which the boat gets shifted is %f m'%s_b)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Example 17.5 Principle of impulse and momentum"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(a) The Final velocity of boat when two men dive simultaneously is 1.333333 m/s\n",
      "(b) The Final velocity of boat when the man of 75 kg dives first and 50 kg dives second is 1.466667 m/s\n",
      "(c) The Final velocity of boat when the man of 50kg dives first followed by the man of 75 kg is 1.456410 m/s\n"
     ]
    }
   ],
   "source": [
    "# Initilization of variables\n",
    "M=250 # kg # mass of the boat\n",
    "M_1=50 # kg # mass of the man\n",
    "M_2=75 # kg # mass of the man\n",
    "v=4 # m/s # relative velocity of man w.r.t boat\n",
    "# Calculations \n",
    "# (a)\n",
    "# Let the increase in the velocity or the final velocity of the boat when TWO MEN DIVE SIMULTANEOUSLY is given by eq'n,\n",
    "deltaV_1=((M_1+M_2)*v)/(M+(M_1+M_2)) # m/s\n",
    "# (b) # The increase in the velocity or the final velocity of the boat when man of 75 kg dives 1st followed by man of 50 kg\n",
    "# Man of 75 kg dives first, So let the final velocity is given as\n",
    "deltaV_75=(M_2*v)/((M+M_1)+M_2) # m/s\n",
    "# Now let the man of 50 kg jumps  next, Here\n",
    "deltaV_50=(M_1*v)/(M+M_1) # m/s\n",
    "# Let final velocity of boat is,\n",
    "deltaV_2=0+deltaV_75+deltaV_50 # m/s\n",
    "# (c) \n",
    "# The man of 50 kg jumps first,\n",
    "delV_50=(M_1*v)/((M+M_2)+(M_1)) # m/s\n",
    "# the man of 75 kg jumps next,\n",
    "delV_75=(M_2*v)/(M+M_2) # m/s\n",
    "# Final velocity of boat is,\n",
    "deltaV_3=0+delV_50+delV_75 # m/s\n",
    "# Results\n",
    "print('(a) The Final velocity of boat when two men dive simultaneously is %f m/s'%deltaV_1)\n",
    "print('(b) The Final velocity of boat when the man of 75 kg dives first and 50 kg dives second is %f m/s'%deltaV_2)\n",
    "print('(c) The Final velocity of boat when the man of 50kg dives first followed by the man of 75 kg is %f m/s'%deltaV_3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Example 17.6 Principle of impulse and momentum"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The velocity of the canoe is 0.108333 m/s\n"
     ]
    }
   ],
   "source": [
    "# Initilization of variables\n",
    "m_m=70 # kg # mass of man\n",
    "m_c=35 # kg # mass of canoe\n",
    "m=25/1000 # kg # mass of bullet\n",
    "m_wb=2.25 # kg # mass of wodden block\n",
    "V_b=5 # m/s # velocity of block\n",
    "# Calculations\n",
    "# Considering Initial Momentum of bullet=Final momentum of bullet & the block we have,Velocity of  bullet (v) is given by eq'n,\n",
    "v=(V_b*(m_wb+m))/(m) # m/s \n",
    "# Considering, Momentum of the bullet=Momentum of the canoe & the man,the velocity on canoe is given by eq'n\n",
    "V=(m*v)/(m_m+m_c) # m/s\n",
    "# Results\n",
    "print('The velocity of the canoe is %f m/s'%V)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Example 17.8 Principle of conservation of angular momentum"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The new speed of the particle is 40 m/s\n",
      "The tension in the string is 6400 N\n"
     ]
    }
   ],
   "source": [
    "# Initilization of variables\n",
    "m=2 # kg # mass of the particle\n",
    "v_0=20 # m/s # speed of rotation of the mass attached to the string\n",
    "r_0=1 # m # radius of the circle along which the particle is rotated\n",
    "r_1=r_0/2 # m\n",
    "# Calculations\n",
    "# here, equating (H_0)_1=(H_0)_2 i.e  (m*v_0)*r_0=(m*v_1)*r_1 (here, r_1=r_0/2). On solving we get v_1 as,\n",
    "v_1=2*v_0 # m/s\n",
    "# Tension is given by eq'n,\n",
    "T=(m*v_1**2)/r_1 # N\n",
    "# Results\n",
    "print('The new speed of the particle is %d m/s'%v_1)\n",
    "print('The tension in the string is %d N'%T)"
   ]
  }
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