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"source": [
"# Chapter 17 Kinetics of a Particle Impulse and Momentum"
]
},
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"cell_type": "markdown",
"metadata": {},
"source": [
"# Example 17.1 Principle of impulse and momentum"
]
},
{
"cell_type": "code",
"execution_count": 3,
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"name": "stdout",
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"text": [
"The average impules force exerted by the bat on the ball is 408.633978 N\n"
]
}
],
"source": [
"from __future__ import division\n",
"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,
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"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": [
"from __future__ import division\n",
"# 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": {
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"outputs": [
{
"name": "stdout",
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"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": [
"from __future__ import division\n",
"# 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": {
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"outputs": [
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"name": "stdout",
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"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": [
"from __future__ import division\n",
"# 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
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"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The velocity of the canoe is 0.108333 m/s\n"
]
}
],
"source": [
"from __future__ import division\n",
"# 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": [
"from __future__ import division\n",
"# 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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