PKCIDMM>Physics Textbook Part-I for class XI by NCERT/Chapter2_3.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 2 : Units and Measurement"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.1 , page : 19"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) One degree = 0.017453292519943295 rad\n",
"(b) One minute = 0.0002908882086657216 rad\n",
"(c) One second = 4.84813681109536e-06 rad\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
"\n",
"three_sixty_degree=2*math.pi\n",
"\n",
"# Calculation\n",
"\n",
"# Since 360° = 2π rad \n",
"one_degree=three_sixty_degree/360\n",
"# Since 1° = 60′ \n",
"one_minute=one_degree/60\n",
"# Since 1′ = 60″\n",
"one_second=one_minute/60\n",
"\n",
"# Result\n",
"\n",
"print(\"(a) One degree =\",one_degree,\"rad\")\n",
"print(\"(b) One minute =\",one_minute,\"rad\")\n",
"print(\"(c) One second =\",one_second,\"rad\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.2 , page : 19"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The distance of the tower C from his original position A = 119.0 m\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
" \n",
"θ=40 # Parallax angle in degree\n",
"AB=100 # Distance between A and Bin m\n",
"\n",
"# Calculation\n",
"\n",
"# AB = AC tan θ \n",
"AC=AB/math.tan(math.radians(θ))\n",
"\n",
"# Result\n",
"\n",
"print(\"The distance of the tower C from his original position A =\",round(AC,0),\"m\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.3 , page : 19"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The distance of the moon from the Earth = 0.0003846385723759571 m\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
"\n",
"# Since, 1°54′ = 114′ \n",
"θ=114 # The angle θ subtended at the moon by the two directions of observation \n",
"b=1.276*10**7 # Diameter of the Earth\n",
"\n",
"# Calculation\n",
"\n",
"θ=114*60*4.85*10**6\n",
"D=b/θ # The earth-moon distance\n",
"\n",
"# Result\n",
"\n",
"print(\"The distance of the moon from the Earth =\",D,\"m\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.4 , page : 19"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Sun's diameter = 1393075199.9999998 m\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"α=1920 # Sun's angular diameter in minutes\n",
"D=1.496*10**11 # The distance of the Sun from the Earth \n",
"\n",
"# Calculation\n",
"\n",
"α=1920*4.85*10**-6 # Sun's angular diameter in radians\n",
"d=α*D # Sun's diameter\n",
"\n",
"#Result\n",
"\n",
"print(\"Sun's diameter =\",d,\"m\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.5 , page : 20"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long\n"
]
}
],
"source": [
"# Calculation\n",
"\n",
"# The size of a nucleus is in the range of 10**-15 m and 10**-14 m.\n",
"# The tip of a sharp pin is taken to be in the range of 10**5 m and 10**4 m.\n",
"# Thus we are scaling up by a factor of 1010. An atom roughly of size 1010 m will be scaled up to a size of 1 m.\n",
"\n",
"# Result\n",
"\n",
"print(\"Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.6 , page : 25"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Clock 2 is to be preferred to clock 1\n"
]
}
],
"source": [
"# Calculations\n",
"\n",
"# The range of variation over the seven days of observations is 162 s for clock 1, and 31 s for clock 2.\n",
"# The average reading of clock 1 is much closer to the standard time than the average reading of clock 2.\n",
"# The important point is that a clocks zero error is not as significant for precision work as its variation, because a zero-error can always be easily corrected.\n",
" \n",
"\n",
"# Result\n",
"\n",
"print(\"Clock 2 is to be preferred to clock 1\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.7 , page : 25"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The absolute errors = [0.005999999999999783, 0.06400000000000006, 0.20400000000000018, 0.08599999999999985, 0.1759999999999997]\n",
"A more correct way will be to write, T = 2.6 ± 0.1 s \n",
"The relative error or the percentage error = 3.8461538461538463 ≈ 4 %\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
"\n",
"p=[2.63,2.56,2.42,2.71,2.80] # The readings of period of oscillation of a simple pendulum\n",
"\n",
"# Calculation\n",
"\n",
"T=sum(p)/len(p)\n",
"p[:] = [x - T for x in p]\n",
"q=[abs(x) for x in p]\n",
"DT=sum(q)/len(p)\n",
"δa=(round(DT,1)/round(T,1))*100\n",
"\n",
"# Result\n",
"\n",
"print(\"The absolute errors =\",q)\n",
"print(\"A more correct way will be to write, T =\",round(T,1),\"±\", round(DT,1),\" s \")\n",
"\n",
"print(\"The relative error or the percentage error =\",δa,\"≈ 4 %\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.8 , page : 26"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The temperature difference wth the error = 30 ° C ± 1.0 ° C\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
"\n",
"t1=20 # Temperature of first body in degree Celsius\n",
"Δt1=.5 # Error in temperature of first body degree Celsius\n",
"t2=50 # Temperature of second body degree Celsius\n",
"Δt2=.5 # Error in temperature of first body degree Celsius\n",
"\n",
"# Calculation\n",
"\n",
"t=t2 - t1\n",
"Δt=max((Δt1 + Δt2),(Δt1 - Δt2))\n",
"\n",
"\n",
"# Result\n",
"\n",
"print(\"The temperature difference wth the error =\",t,\"\\u00b0 C ±\",Δt,\"\\u00b0 C\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.9 , page : 27"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The total error in Resistance = 7 %\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
" \n",
"V=5 # The percentage error in voltage \n",
"I=2 # The percentage error in current\n",
"\n",
"# Calculation\n",
"\n",
"R=V+I\n",
"\n",
"# Result\n",
"\n",
"print(\"The total error in Resistance =\",R,\"%\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.10 , page : 27"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The equivalent resistance of the series combination = 300 ± 7 ohm\n",
"The equivalent resistance of the parallel combination = 66.7 ± 1.8 ohm\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
"\n",
"R1=100 # Resistance of first resistor in ohm\n",
"ΔR1=3 # Error in Resistance of first resistor in ohm\n",
"R2=200 # Resistance of second resistor wth error term in ohm\n",
"ΔR2=4 # Error in Resistance of second resistor in ohm\n",
"\n",
"# Calculation\n",
"\n",
"R=R1+R2\n",
"ΔR=ΔR1+ΔR2\n",
"R_prim=R1*R2/(R1+R2)\n",
"ΔR_prim=(R_prim/R1)**2*ΔR1 + (R_prim/R2)**2*ΔR2 \n",
"\n",
"# Result\n",
"\n",
"print(\"The equivalent resistance of the series combination =\",R,\"±\",ΔR,\"ohm\")\n",
"print(\"The equivalent resistance of the parallel combination =\",round(R_prim,1),\"±\",round(ΔR_prim,1),\"ohm\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.11 , page : 27"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The relative error in Z is ΔZ/Z = 4(ΔA/A)+(1/3)(ΔB/B)+(ΔC/C)+(3/2)(ΔD/D)\n"
]
}
],
"source": [
"# Since the relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity.\n",
"\n",
"# Result\n",
"\n",
"print(\"The relative error in Z is ΔZ/Z = 4(ΔA/A)+(1/3)(ΔB/B)+(ΔC/C)+(3/2)(ΔD/D)\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.12 , page : 27"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The percentage error in g = 2.7222222222222223 ≈ 3\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
"\n",
"L=20 # Length in cm\n",
"ΔL=1 # Eror in lengthin mm\n",
"t=90 # Total time in s\n",
"Δt=1 # Error in time in s\n",
"n=100 # Number of oscillations\n",
"\n",
"# Calculation\n",
"\n",
"# The period of oscillation of a simple pendulum is, T = 2Π √(L/g)\n",
"# Hence, g = 4π²L/T²\n",
"\n",
"ΔL=ΔL*10**-1\n",
"T=n/t\n",
"ΔT_div_T=Δt/t # Error in T\n",
"Δg_div_g= (ΔL/L) + 2*(ΔT_div_T) # Error in g\n",
"per_g= 100*(ΔL/L) + 2*100*(ΔT_div_T ) \n",
"\n",
"# Result\n",
"\n",
"print(\"The percentage error in g =\",per_g,\"≈ 3\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.13 , page : 30"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Surface area of the cube = 311.3 m²\n",
"Volume of the cube = 373.7 m^3\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
"\n",
"a=7.203 # Side of the cube in m\n",
"\n",
"# Calculation\n",
"# The number of significant figures in the measured length is 4.\n",
"# Hence the calculated area and the volume should therefore be rounded off to 4 significant figures. \n",
"\n",
"SA=6*a**2 # Surface area of the cube\n",
"V=a**3 # Volume of the cube \n",
"\n",
"# Result\n",
"\n",
"print(\"Surface area of the cube =\",round(SA,1),\"m²\")\n",
"print(\"Volume of the cube =\",round(V,1),\"m^3\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.14 , page : 30"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Density = 4.8 g/cm^3\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delaration\n",
"\n",
"m=5.74 # Mass of the substancein g\n",
"v=1.2 # Volume of the substance in cm^3\n",
"\n",
"# Calculation\n",
"\n",
"# There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume.\n",
"# Hence the density should be expressed to only 2 significant figures.\n",
"D=m/v\n",
"\n",
"# Result\n",
"\n",
"print(\"Density =\",round(D,1),\"g/cm^3\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.15 , page : 33"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The dimensions of LHS are :[M][L T-1 ]² = [M][L² T-²] = [M L² T-²]\n",
"The dimensions of RHS are :[M][L T-²][L] = [M][L² T-²] = [M L² T-²]\n",
"The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct\n"
]
}
],
"source": [
"# Consider the equation (1/2)mv²=mgh\n",
"# where\n",
"# m : Mass of the body\n",
"# v : Velocity of the body\n",
"# g : Acceleration due to gravity \n",
"# h : Height\n",
"\n",
"#Calculation\n",
"\n",
"print(\"The dimensions of LHS are :[M][L T-1 ]² = [M][L² T-²] = [M L² T-²]\")\n",
"print(\"The dimensions of RHS are :[M][L T-²][L] = [M][L² T-²] = [M L² T-²]\")\n",
"\n",
"# Result\n",
"\n",
"print(\"The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.16 , page : 33"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The correct formula for kinetic energy is, K = (1/2)mv²\n"
]
}
],
"source": [
"# Calculation\n",
"\n",
"# Every correct formula or equation must have the same dimensions on both sides of the equation.\n",
"# Also, only quantities with the same physical dimensions can be added or subtracted.\n",
"# The dimensions of the quantity on the right side are \n",
"# [M² L^3 T^-3] for (a) \n",
"# [M L² T-²] for (b)and (d)\n",
"# [M L T-²] for (c)\n",
"# The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added.\n",
"# Since the kinetic energy K has the dimensions of [M L2 T2], formulas (a), (c) and (e) are ruled out.\n",
"# Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula. \n",
"# For this, one must turn to the actual definition of kinetic energy.\n",
"# The correct formula for kinetic energy is given by (b). \n",
"\n",
"# Result\n",
"\n",
"print(\"The correct formula for kinetic energy is, K = (1/2)mv²\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2.17 , page : 33"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The expression for the time period of a simple pendulum is, T = 2π √(l/g)\n"
]
}
],
"source": [
"# Calculation\n",
"\n",
"# The dependence of time period T on the quantities l, g and m as a product may be written as :\n",
"# T = k l^x g^y m^z \n",
"# where k is dimensionless constant and x, y and z are the exponents. \n",
"# By considering dimensions on both sides, we have [L^0 M^0 T^1] = [L^1]^x [L^1 T^-2]^y [M^1]^z = L^(x+y) T^(-2y) M^z\n",
"# On equating the dimensions on both sides, we have: x + y = 0; 2y = 1; and z = 0\n",
"# So that x = (1/2); y = -(1/2) and z = 0\n",
"# T = k * l^(1/2) * g^-(1/2)\n",
"# In other word, T = k √(l/g)\n",
"# The value of constant k can not be obtained by the method of dimensions.\n",
"# Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.\n",
"# Actually, k = 2π so that T = 2π √(l/g)\n",
"\n",
"# Result\n",
"\n",
"print(\"The expression for the time period of a simple pendulum is, T = 2π √(l/g)\")\n"
]
}
],
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PKCI˞_D9D9>Physics Textbook Part-I for class XI by NCERT/Chapter3_3.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 3 : Motion in a Straight Line"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.1 , page : 43"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a)\n",
"The average velocity of the car in going from O to P = 20.0 m/s\n",
"The average speed of the car in going from O to P = 20.0 m/s\n",
"(b)\n",
"The average velocity of the car in going from O to P and back to Q = 10.0 m/s\n",
"The average speed of the car in going from O to P and back to Q = 20.0 m/s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"l1=360 # The distance from O to P in m\n",
"l2=120 # The distance from P to Q in m\n",
"t1=18 # The time taken to travel OP in s\n",
"t2=6 # The time taken to travel PQ in s\n",
"d1=360 # The displacement from O to P in m\n",
"d2=(l1-l2) # The displacement from P to Q in m\n",
"p1=360 # The total pathlength from O to P in m\n",
"p2=(l1+l2) # The total pathlength from O to P and p to Q in m\n",
"\n",
"# Calculation\n",
"\n",
"#(a)\n",
"a_v1=d1/t1\n",
"a_s1=p1/t1\n",
"\n",
"#(b)\n",
"a_v2=d2/(t1+t2)\n",
"a_s2=p2/(t1+t2)\n",
"\n",
"# Result\n",
"\n",
"print(\"(a)\")\n",
"print(\"The average velocity of the car in going from O to P =\",a_v1,\"m/s\")\n",
"print(\"The average speed of the car in going from O to P =\",a_s1,\"m/s\")\n",
"print(\"(b)\")\n",
"print(\"The average velocity of the car in going from O to P and back to Q =\",a_v2,\"m/s\")\n",
"print(\"The average speed of the car in going from O to P and back to Q =\",a_s2,\"m/s\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.2 , page : 45"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The velocity at t = 0.0 s = 0 m/s\n",
"The velocity at t = 2.0 s = 10 m/s\n",
"The average velocity between t = 2.0 s and t = 4.0 s = 15.0 m/s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"import numpy as np\n",
"\n",
"# Variable declaration\n",
" \n",
"a=8.5 # Distance in m\n",
"b=2.5 # Acceleration in m/s²\n",
"\n",
"# Calculation\n",
"\n",
"# In notation of differential calculus, the velocity is v = dx/dt = d (a+bt²)/dt = 2bt = 5.0t\n",
"\n",
"p0=np.polyval([0,5,0],0) # Velocity at t= 1.0 s\n",
"p2=np.polyval([0,5,0],2) # Velocity at t= 2.0 s\n",
"p4=np.polyval([0,5,0],4) # Velocity at t= 4.0 s\n",
"avg_v=(p2+p4)/2\n",
"\n",
"# Result\n",
"\n",
"print(\"The velocity at t = 0.0 s =\",p0,\"m/s\")\n",
"print(\"The velocity at t = 2.0 s =\",p2,\"m/s\")\n",
"print(\"The average velocity between t = 2.0 s and t = 4.0 s =\",avg_v,\"m/s\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.3 , page : 48 "
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The equations of motion for constant acceleration using method of calculus are as follows.\n",
"v = v_o + at\n",
"x = x_o + v_ot + (1/2)at²\n",
"v²= v²_o + 2a(x - x_o)\n"
]
}
],
"source": [
"# Calculation\n",
"\n",
"# By definition, a = dv/dt\n",
"# .i.e dv = adt\n",
"# Integrating on both sides we get, v - v_o = at or v = v_o + at\n",
"# Further we know that, v = dx/dt\n",
"# .i.e dx = vdt\n",
"# Integrating on both sides we get, x - x_o = v_ot + (1/2)at² or x = x_o + v_ot + (1/2)at² \n",
"# Now we can write, a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx) \n",
"# .i.e. v dv = a dx\n",
"# Integrating on both sides we get, (v² - v²_o) = a(x - x_o) or v² = v²_o + 2a(x - x_o)\n",
"\n",
"# Result\n",
"\n",
"print(\"The equations of motion for constant acceleration using method of calculus are as follows.\")\n",
"print(\"v = v_o + at\")\n",
"print(\"x = x_o + v_ot + (1/2)at²\")\n",
"print(\"v²= v²_o + 2a(x - x_o)\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.4 , page : 48 "
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) The height at which the ball has risen = 20.0 m\n",
"(b) The time taken before the ball to hit the ground = 5.0 s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import sympy\n",
"\n",
"# Variable declaration\n",
"\n",
"t = sympy.symbols('t')\n",
"v_o=20 # Initial velocity in m/s\n",
"y_o=25 # Height of the initial point from ground in m\n",
"a=-10 # Acceleration due to gravity\n",
"v=0\n",
"y=0\n",
"\n",
"# Calculation\n",
"\n",
"#(a)\n",
"# Let us take the y-axis in the vertically upward direction with zero at the ground\n",
"# Since v=(v_o)²+2ah\n",
"h=(-(v_o)**2)/(2*a)\n",
"\n",
"#(b)\n",
"# The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen\n",
"# and using equation (y - y_0) = v_ot + (1/2)at² \n",
"# Substituting the values in the above equation we get the quadratic equation for t as , 5t² - 20t - 25 = 0 \n",
"t = round(max(sympy.solve(5*t**2 - 20*t -y_o,t)),0)\n",
"\n",
"# Result\n",
"\n",
"print(\"(a) The height at which the ball has risen =\",h,\"m\")\n",
"print(\"(b) The time taken before the ball to hit the ground =\",t,\"s\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.5 , page : 49 "
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The motion of an object under free fall can be explained by the following equations\n",
"v = 0 - g t = 9.8t m/s\n",
"y = 0 - (1/2)gt² = 4.9t² m\n",
"v² = 0 - 2gy = -19.6 y m²/s²\n"
]
}
],
"source": [
"# Calculation\n",
"\n",
"# If air resistance is neglected, the object is said to be in free fall.\n",
"# If the height through which the object falls is small compared to the earths radius, g can be taken to be constant, equal to 9.8 ms².\n",
"# Free fall is thus a case of motion with uniform acceleration. \n",
"# We assume that the motion is in y-direction, more correctly in y-direction because we choose upward direction as positive.\n",
"# Since the acceleration due to gravity is always downward, it is in the negative direction.\n",
"# Then we have, a = g = 9.8 ms²\n",
"# The object is released from rest at y = 0. Therefore, v_0 = 0 and the equations of motion become as follows\n",
"# v = 0 - g t = 9.8t m/s \n",
"# y = 0 - (1/2)gt² = 4.9t² m\n",
"# v² = 0 - 2gy = -19.6 y m²/s²\n",
"\n",
"# Result\n",
"\n",
"print(\"The motion of an object under free fall can be explained by the following equations\")\n",
"print(\"v = 0 - g t = 9.8t m/s\")\n",
"print(\"y = 0 - (1/2)gt² = 4.9t² m\")\n",
"print(\"v² = 0 - 2gy = -19.6 y m²/s²\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.6 , page : 50 "
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distancestraversed during successive intervals of time.\n",
"Since initial velocity is zero, we have\n",
"Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ which are given in second column of Table 3.2. If we take (1/ 2) gτ2 as y0 the position coordinate after first time interval τ, then third column gives the positions in the unit of yo. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11 as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.\n",
"Hence the proof.\n"
]
}
],
"source": [
"# Result\n",
"\n",
"print(\"Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distancestraversed during successive intervals of time.\")\n",
"print(\"Since initial velocity is zero, we have\")\n",
"print(\"Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ which are given in second column of Table 3.2. If we take (1/ 2) gτ2 as y0 the position coordinate after first time interval τ, then third column gives the positions in the unit of yo. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11 as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.\")\n",
"print(\"Hence the proof.\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.7 , page : 50 "
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The stopping distance, d_s = -v²_o / 2a\n"
]
}
],
"source": [
"# Calculation\n",
"\n",
"# Let the distance travelled by the vehicle before it stops be d_s.\n",
"# Then, using equation of motion v² = v²_o + 2ax, and noting that v = 0, we have the stopping distance as given below,\n",
"# The stopping distance, d_s = -v²_o / 2a\n",
"\n",
"# Result\n",
"\n",
"print(\"The stopping distance, d_s = -v²_o / 2a\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.8 , page : 51 "
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Reaction Time = 0.2 s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
" \n",
"v0=0 # Initial velocity in m\n",
"g=9.8 # Acceleration due to gravity\n",
"d=21 # Distance travelled in cm \n",
"\n",
"# Calculation\n",
"\n",
"d=21*10**-2\n",
"t=math.sqrt((2*d)/g) # Reaction time = √(2d/g)\n",
"\n",
"# Result\n",
"\n",
"print(\"Reaction Time =\",round(t,1),\"s\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.9 , page : 52 "
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) Relative velocity of train B with respect to tain A = -40.0 m/s\n",
"(b) Relative velocity of ground with respect to train B = 25.0 m/s\n",
"(c) Speed of the monkey = 10.0 m/s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"# Choose the positive direction of X-axis to be from South to North\n",
"V_A=54 # The speed of train A in km/h\n",
"V_B=-90 # The speed of train B in km/h\n",
"V_MA=-18 # The relative speed of monkey km/h\n",
"\n",
"# Calculation\n",
"\n",
"V_A=54*(5/18) # The speed of train A in m/s\n",
"V_B=-90*(5/18) # The speed of train B in m/s\n",
"V_MA=-18*(5/18) # The relative speed of monkey m/s\n",
"\n",
"#(a)\n",
"V_BA=V_B-V_A # Relative velocity of train B with respect to A\n",
"\n",
"#(b)\n",
"V_GB=0-V_B # Relative velocity of ground with respect to train B\n",
"\n",
"#(c)\n",
"# Since V_MA = V_M - V_A\n",
"V_M=V_MA+V_A\n",
"\n",
"# Result\n",
"\n",
"print(\"(a) Relative velocity of train B with respect to tain A =\",V_BA,\"m/s\")\n",
"print(\"(b) Relative velocity of ground with respect to train B =\",V_GB,\"m/s\")\n",
"print(\"(c) Speed of the monkey =\",V_M,\"m/s\")\n"
]
}
],
"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": "3.4.3"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
PKCIh>>>Physics Textbook Part-I for class XI by NCERT/Chapter4_3.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 4 : Motion in a Plane"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.1 , page : 69"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Therefore,the boy should hold his umbrella in the vertical plane at an angle of about 19.0 ° with the vertical towards the east. \n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable delcaration\n",
"\n",
"v_r=35 # Velocity vector of the rain which falls vertically in m/s \n",
"v_w=12 # Velocity vector of the wind blowing in east to west direction in m/s\n",
"\n",
"# Calculation\n",
"\n",
"R=math.sqrt(v_r**2+v_w**2) # The magnitude of the resultant vector \n",
"tanθ=v_w/v_r\n",
"θ=math.degrees(math.atan(tanθ)) # The direction that R makes with the vertical \n",
"\n",
"# Result\n",
"\n",
"print(\"Therefore,the boy should hold his umbrella in the vertical plane at an angle of about\",round(θ,0),\"° with the vertical towards the east. \")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.2 , page : 71 "
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The magnitude of thr resultant vector is given by the equation, R = square root(A²+ B²+ 2ABcosθ)\n",
"The direction of the resultant vector is given by the equation, tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)\n"
]
}
],
"source": [
"# Calculation\n",
"\n",
"# Let OP and OQ represent the two vectors A and B making an angle θ.\n",
"# Then, using the parallelogram method of vector addition, OS represents the resultant vector R such that R = A + B \n",
"# SN is normal to OP and PM is normal to OS.\n",
"# From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + Bcosθ\n",
"# SN = Bsin θ and OS² = (A + Bcosθ )² + (Bsinθ )² or, R² = A² + B² + 2ABcosθ\n",
"# .i.e R = square root(A²+ B²+ 2ABcosθ) \n",
"# In ΔOSN, SN = OSsin α = Rsin α , and in ΔPSN, SN = PSsinθ = Bsinθ\n",
"# Therefore, Rsinα = Bsinθ ..............eqn 1\n",
"# Similarly, PM = Asinα = Bsinβ .........eqn 2\n",
"# From eqns 1 and 2 we get,\n",
"# sin α = (B/R) sin θ or tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)\n",
"\n",
"# Result\n",
"\n",
"print(\"The magnitude of thr resultant vector is given by the equation, R = square root(A²+ B²+ 2ABcosθ)\")\n",
"print(\"The direction of the resultant vector is given by the equation, tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.3 , page : 72 "
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The magnitude of the resultant vector = 22.0 km/h\n",
"The direction of the resultant vector = 23.4 °\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"v_b=25 # Velocity of the motor boat racing towards north in km/h\n",
"v_c=10 # Velocity of the water current in the direction of 60° east of south in km/h\n",
"θ=60 # The angle in degree\n",
"\n",
"# Calculation\n",
"\n",
"# Using the parallelogram method of vector addition we can obtain the resultant vector\n",
"# We can obtain the magnitude of resultant vector using the Law of cosine \n",
"R=math.sqrt(v_b**2+v_c**2+(2*v_b*v_c*(math.cos(math.radians(2*θ)))))\n",
"# We can obtain the direction of resultant vector using the Law of sines \n",
"sinφ=(v_c*math.sin(math.radians(θ)))/R\n",
"φ=math.degrees(math.asin(sinφ))\n",
"\n",
"# Result\n",
"\n",
"print(\"The magnitude of the resultant vector =\",round(R,0),\"km/h\")\n",
"print(\"The direction of the resultant vector =\",round(φ,1),\"°\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.4 , page : 75 "
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) The velocity vector, v(t) = 3.0i+4.0tj m/s\n",
" The acceleration vector, a(t) = 4.0j m/s²\n",
"(b) Magnitude of v(t) at t=1 s = 5.0 m/s\n",
" Direction of v(t) at t=1 s = 53.0 ° with x-axis\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"t=1 # Time in s\n",
"\n",
"# Calculation\n",
"\n",
"# The position vector of the particle is r(t)=3.0ti+2.0t²j+5.0k\n",
"# On differentiating r(t) with respect to t we get the velocity vector, v(t)=3.0i+4.0tj\n",
"# On differentiating v(t) with respect to t we get the acceleration vector, a(t)=4.0j\n",
"V_x=3 # X component of v(t)\n",
"V_y=4 # Y component of v(t)\n",
"V=math.sqrt(V_x**2 + V_y**2)\n",
"tanθ=V_y/V_x\n",
"θ=math.degrees(math.atan(tanθ)) \n",
"\n",
"# Result\n",
"\n",
"print(\"(a) The velocity vector, v(t) = 3.0i+4.0tj m/s\")\n",
"print(\" The acceleration vector, a(t) = 4.0j m/s²\")\n",
"print(\"(b) Magnitude of v(t) at t=1 s =\",V,\"m/s\")\n",
"print(\" Direction of v(t) at t=1 s =\",round(θ,0),\"° with x-axis\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.5 , page : 76 "
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) The y-coordinate of the particle at the instant its x-coordinate is 84 m = 36.0 m\n",
"(b) Speed of the particle at the instant its x-coordinate is 84 m = 26.0 m/s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import sympy\n",
"\n",
"# Variable declaration\n",
"\n",
"x=84 # X component of the position vector in m\n",
"t= sympy.symbols('t')\n",
"\n",
"# Calculation\n",
"\n",
"# Velocity vector is given as, v(t)= 5.0i m/s\n",
"# Acceleration vector is given as, a(t)= 3.0i+2.0j m/s²\n",
"# By the equation r(t)=v(t)+(a(t)t²)/2 we get the positon vector of the particle as follows\n",
"# r(t) = 5.0ti + 1.5t²i + 1.0t²j\n",
"t=round((max(sympy.solve(1.5*t**2 + 5*t -x,t))),0)\n",
"y=1.0*t**2\n",
"# Now the velocity vector can be obtained by differentiating r(t) with respect to t \n",
"# Then we get, v(t) = 5.0i +3ti+2tj m/s\n",
"v_x=5.0+3*t # X component of v(t) at time = t\n",
"v_y=2*t # Y component of v(t) at time = t\n",
"V=math.sqrt(v_x**2 + v_y**2)\n",
"\n",
"# Result\n",
"\n",
"print(\"(a) The y-coordinate of the particle at the instant its x-coordinate is 84 m =\",y,\"m\")\n",
"print(\"(b) Speed of the particle at the instant its x-coordinate is 84 m =\",round(V,0),\"m/s\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.6 , page : 76 "
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Therefore, the woman should hold her umbrella at an angle of about 19.0 ° with the vertical towards the west.\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"v_r=35 # Velocity vector of the rain which falls vertically in m/s\n",
"v_b=12 # Velocity vector of the bicycle riding in east to west direction in m/s\n",
"\n",
"# Calculation\n",
"\n",
"v_rb=v_r - v_b\n",
"tanθ=v_b/v_r\n",
"θ=math.degrees(math.atan(tanθ)) \n",
"\n",
"# Result\n",
"\n",
"print(\"Therefore, the woman should hold her umbrella at an angle of about\",round(θ,0),\"° with the vertical towards the west.\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.7 , page : 78 "
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"For a projectile launched with velocity v_0 at an angle θ_0 , the range is given by, R = (v_0²sin2θ_0)/g\n",
"Now,for angles,(45°+α ) and (45°- α ), 2θ_0 is (90° + 2α) and (90° - 2α), respectively.\n",
"The values of sin (90° + 2 α ) and sin (90° - 2α) are the same, equal to that of cos 2α.\n",
"Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.\n"
]
}
],
"source": [
"# Variable declaration\n",
"\n",
"v_0=1 # For convenience, velocity at whch the projectile launched is assumed to be unity \n",
"θ_0=1 # For convenience, angle at whch the projectile launched in degree is assumed to be unity\n",
"\n",
"# Result\n",
"\n",
"print(\"For a projectile launched with velocity v_0 at an angle θ_0 , the range is given by, R = (v_0²sin2θ_0)/g\")\n",
"print(\"Now,for angles,(45°+α ) and (45°- α ), 2θ_0 is (90° + 2α) and (90° - 2α), respectively.\")\n",
"print(\"The values of sin (90° + 2 α ) and sin (90° - 2α) are the same, equal to that of cos 2α.\")\n",
"print(\"Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.\") \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.8 , page : 78 "
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The time taken by the stone to reach the ground = 10.0 s\n",
"The speed with which the stone hits the ground = 99.0 m/s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"# We choose the origin of the x-,and y- axis at the edge of the cliff and t = 0 s at the instant the stone is thrown\n",
"# Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction\n",
"# The equations of motion are : x(t)=x_0 = v_0xt and y(t) = y_0+v_0yt+(1/2)a_y(t^2)\n",
"\n",
"g=9.8 # Acceleration due to gravity\n",
"x_0=0\n",
"y_0=0\n",
"v_oy=0\n",
"a_y=g\n",
"v_ox=15\n",
"y_t=490\n",
"\n",
"# Calculation\n",
"\n",
"# The stone hits the ground when y(t) = 490 m ,i.e. 490 = (1/2)(9.8)t\n",
"t=math.sqrt((-y_t*-2)/a_y)\n",
"# The velocity components are v_x = v_ox and v_y = v_oy - g t\n",
"v_x=v_ox\n",
"v_y=v_oy-(g*t)\n",
"V=math.sqrt(v_x**2+v_y**2)\n",
"\n",
"# Result\n",
"\n",
"print(\"The time taken by the stone to reach the ground =\",t,\"s\")\n",
"print(\"The speed with which the stone hits the ground =\",round(V,0),\"m/s\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.9 , page : 79 "
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The maximum height = 10.0 m\n",
"The time taken to return to the same level = 2.9 s\n",
"The distance from the thrower to the point where the ball returns to the same level = 69.0 m\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"v_0=28 # The initial velocity of the ball in m/s\n",
"θ=30 # The angle of inclination of the ball above the horizontal in degree\n",
"g=9.8 # Acceleration due to gravity in m/s²\n",
"\n",
"# Calculation\n",
"\n",
"h_m=(v_0*(math.sin(math.radians(θ))))**2/(2*g)\n",
"T_f=(2*v_0*math.sin(math.radians(θ)))/g\n",
"R=((v_0**2)*(math.sin(2*math.radians(θ))))/g \n",
" \n",
"# Result\n",
"\n",
"print(\"The maximum height =\",round(h_m,0),\"m\")\n",
"print(\"The time taken to return to the same level =\",round(T_f,1),\"s\")\n",
"print(\"The distance from the thrower to the point where the ball returns to the same level =\",round(R,0),\"m\") \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4.10 , page : 81 "
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) Angular speed = 0.44 rad/s\n",
" Linear speed = 5.3 cm/s\n",
"(b) Since the direction changes continuously, acceleration here is not a constant vector\n",
" Magnitude of acceleration = 2.3 cm/s²\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
" \n",
"R=12 # Radius of the circular groove in cm\n",
"n=7 # Total number of revolutions\n",
"t=100 # Time taken for all 7 revolutions\n",
"\n",
"# Calculation\n",
"\n",
"T=t/n # Time taken for one revolution\n",
"#(a)\n",
"w=2*math.pi/T # Angular speed in rad/s\n",
"v=w*R # Linear speed in cm/s\n",
"a=pow(w,2)*R # Magnitude of acceleration in cm/s²\n",
"\n",
"# Result\n",
"\n",
"print(\"(a) Angular speed =\",round(w,2),\"rad/s\")\n",
"print(\" Linear speed =\",round(v,1),\"cm/s\")\n",
"print(\"(b) Since the direction changes continuously, acceleration here is not a constant vector\")\n",
"print(\" Magnitude of acceleration =\",round(a,1),\"cm/s²\")\n"
]
}
],
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"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",
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}
PKCIod`}K}K>Physics Textbook Part-I for class XI by NCERT/Chapter5_3.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 5 : Laws of Motion"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.1 , page : 93"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The acceleration of the astranaunt = 0 m/s²\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"a=100 # Constant acceleration of the inter stellar space in m/s²\n",
"\n",
"# Calculation\n",
"\n",
"# Since there are no nearby stars to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero.\n",
"# By the first law of motion the acceleration of the astronaut is zero. \n",
"\n",
"# Result\n",
"\n",
"print(\"The acceleration of the astranaunt =\",0,\"m/s²\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.2 , page : 95 "
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The average resistive force exerted by the block on the bullet = 270.0 N\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
" \n",
"m=0.04 # Mass of the bullet in kg\n",
"u=90 # Speedof the bullet in m/s\n",
"s=60 # Thickness of the wooden block at which the bullet stops in cms\n",
"\n",
"# Calculation\n",
"\n",
"s=s*10**-2\n",
"a=u**2/(2*s) # Retardation of the bullet in m/s²\n",
"F=m*a # Retarding force in N\n",
"\n",
"# Result\n",
"\n",
"print(\"The average resistive force exerted by the block on the bullet =\",F,\"N\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.3 , page : 96 "
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Force acting on the particle, F = g*m\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import sympy\n",
"\n",
"# Variable declaration\n",
"\n",
"g = sympy.symbols('g')\n",
"m = sympy.symbols('m')\n",
"u = sympy.symbols('u')\n",
"ut = sympy.symbols('ut')\n",
"t = sympy.symbols('t')\n",
"gt = sympy.symbols('gt')\n",
"\n",
"# Calculation\n",
"\n",
"# The motion of a particle of mass m is given as follows\n",
"y = ut + (gt**2)/2\n",
"# Differentiating the equation of motion with respect to time t we get,\n",
"v = u + gt\n",
"# Again differentiating the above equation with respect to time t we get, \n",
"a = g\n",
"F = m*a # Force aciting on the particle\n",
"# Thus the given equation describes the motion of a particle under acceleration due to gravity and y is the position coordinate in the direction of g.\n",
"\n",
"# Result\n",
"\n",
"\n",
"print(\"Force acting on the particle, F = \",F)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.4 , page : 96"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The impulse in the direction from the batsman to the bowler = 3.6 Ns\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=0.15 # Mass of the ball in kg\n",
"u=12 # Initial speed of the ball in m/s\n",
"\n",
"# Calculation\n",
"\n",
"Delta_v=u-(-u) # Change in velocity= Final velocity-Initial velocity\n",
"I=m*Delta_v\n",
"\n",
"# Result\n",
"\n",
"print(\"The impulse in the direction from the batsman to the bowler =\",round(I,1),\"Ns\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.5 , page : 98 "
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(i) The direction of impulse (and force) is the same in both case and is normal to the wall along the negative x direction\n",
"(ii)The ratio of the magnitudes of the impulses imparted to the ball in two cases = 1.2\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"θ=30 \n",
"\n",
"# Calculation\n",
" \n",
"# Consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer \n",
" \n",
"# Case(a) \n",
"# Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball.\n",
"# Let us consider, initial momentum in x direction, P_x_in = mu and final momentum in x direction,P_x_fi = - mu\n",
"# Let us consider, initial momentum in y direction, P_y_in = 0 and final momentum in y direction,P_y_fi = 0\n",
"# Impulse is the change in momentum vector. Therefore, \n",
"# x-component of impulse = 2mu and y-component of impulse = 0\n",
"# Impulse and force are in the same direction. Clearly, the force on the ball due to the wall is normal to the wall, along the negative x-direction.\n",
"# Using Newtons third law of motion, the force on the wall due to the ball is normal to the wall along the positive x-direction. \n",
" \n",
"# Case(b)\n",
"# Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball.\n",
"# Let us consider, initial momentum in x direction, P_x_in = mu cos 30° and final momentum in x direction,P_x_fi = -mu cos 30°\n",
"# Let us consider, initial momentum in y direction, P_y_in = mu sin 30° and final momentum in y direction,P_y_fi = mu sin 30°\n",
"# x-component of impulse = 2mucos30 and y-component of impulse = 0\n",
"# Using Newtons third law, the force on the wall due to the ball is normal to the wall along the positive x direction.\n",
" \n",
"# The ratio of the magnitudes of the impulses imparted to the balls in (a) and (b) is 2mu/2mu cos 30°= 1/cos 30°\n",
"r=1/math.cos(math.radians(θ)) \n",
"\n",
"# Result\n",
" \n",
"print(\"(i) The direction of impulse (and force) is the same in both case and is normal to the wall along the negative x direction\")\n",
"print(\"(ii)The ratio of the magnitudes of the impulses imparted to the ball in two cases =\",round(r,1)) \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.6 , page : 99 "
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The angle that the rope makes with the vertical in equilibrium = 40.0 °\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=6 # The amount of mass suspended in kg\n",
"l=2 # Length of the rope in m\n",
"F=50 # The Force applied at the mid-point of the rope in the horizontal direction in N\n",
"g=10 # Acceleration due to gravity\n",
"\n",
"# Calculation\n",
"\n",
"T2=m*g\n",
"T1cosθ=T2\n",
"T2sinθ=F\n",
"tanθ=T2sinθ/T1cosθ\n",
"θ=math.degrees(math.atan(tanθ)) \n",
"\n",
"# Result\n",
"\n",
"print(\"The angle that the rope makes with the vertical in equilibrium =\",round(θ,0),\"°\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.7 , page : 102"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The maximum acceleration of the train = 1.5 m/s²\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"u=.15 # The co-efficient of static friction between the box and the trains floor\n",
"g=10 # Acceleration due to gravity\n",
"\n",
"# Calculation\n",
"\n",
"# Since the acceleration of the box is due to the static friction, ma = f ≤ μN = μmg, i.e. a ≤ μg \n",
"a_max=u*g\n",
"\n",
"# Result\n",
"\n",
"print(\"The maximum acceleration of the train =\",a_max,\"m/s²\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.8 , page : 102 "
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The coefficient of static friction between the block and the surface = 0.27\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=4 # Mass in kg\n",
"θ=15 # Angle of inclination of the plane with the horizontal\n",
"\n",
"# Calculation\n",
"\n",
"# Resolving the weight mg along the two directions shown, we have mgsin θ = fs, mgcosθ=N\n",
"# As θ increases, the self-adjusting frictional force fs increases until at θ = θ max, \n",
"# fs achieves its maximum value, max_f = μ s N\n",
"# Therefore, tan θ max = μ or θ_max = (tan)^-1 μ \n",
"tanθ=math.tan(math.radians(θ))\n",
"\n",
"# Result\n",
"\n",
"print(\"The coefficient of static friction between the block and the surface =\",round(tanθ,2))\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.9 , page : 102 "
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The acceleration of the block and trolley system = 0.96 m/s²\n",
"The tension on the string = 27.1 N\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m1=20 # Mass of the block in kg\n",
"m2=3 # Mass of the block in kg\n",
"g=10 # Acceleration due to gravity\n",
"u_k=0.04 # The coefficient of kinetic friction between the trolley and the surface\n",
"\n",
"# Calculation\n",
"\n",
"N=m1*g\n",
"f_k=u_k*N\n",
"# Applying second law to motion of the block , 30-T = 3a\n",
"# Applying the second law to motion of the trolley, T-fk = 20a.\n",
"a=22/23\n",
"T=(20*a)+f_k\n",
"\n",
"# Result\n",
"\n",
"print(\"The acceleration of the block and trolley system =\",round(a,2),\"m/s²\")\n",
"print(\"The tension on the string =\",round(T,1),\"N\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.10 , page : 105 "
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The cyclist will slip while taking the circular turn.\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"v=18 # Speed of the cycle in km/h\n",
"R=3 # Radius of the circular turn in m\n",
"µ_s=0.1 # Coefficient of static friction between the tyre and the road \n",
"g=9.8 # Acceleration due to gravity in m/s²\n",
"\n",
"# Calculation\n",
"\n",
"v=v*(5/18) # Speed of the cycle in m/s\n",
"v_sq=pow(v,2)\n",
"#The condition for the cyclist not to slip is given b, v2 ≤ μ_sRg\n",
"μ_sRg=μ_s*R*g\n",
"\n",
"# Result\n",
"\n",
"if v_sq < μ_sRg:\n",
" print(\"The cyclist will not slip while taking the circular turn.\")\n",
"else:\n",
" print(\"The cyclist will slip while taking the circular turn.\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.11 , page : 105 "
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The optimum speed of the race-car to avoid wear and tear on its tyres = 28.1 m/s\n",
"The maximum permissible speed to avoid slipping = 38.1 m/s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"R=300 # Radius of the circulas race track in m\n",
"θ=15 # Angle at which the road banked in degree\n",
"u=0.2 # The coefficient of friction between the wheels of a race-car and the road\n",
"g=9.8 # Acceleration due to gravity\n",
"\n",
"# Calculation\n",
"\n",
"#(a)\n",
"v_o=math.sqrt(R*g*(math.tan(math.radians(θ))))\n",
"#(b)\n",
"v_max=math.sqrt(R*g*(u+(math.tan(math.radians(θ))))/(1-u*math.tan(math.radians(θ))))\n",
"\n",
"\n",
"# Result\n",
"\n",
"print(\"The optimum speed of the race-car to avoid wear and tear on its tyres =\",round(v_o,1),\"m/s\")\n",
"print(\"The maximum permissible speed to avoid slipping =\",round(v_max,1),\"m/s\")\n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5.12 , page : 106 "
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) The action of the block is equal to 20 N and directed vertically downwards.\n",
"(b) The action of the system on the floor is equal to 267.3 N vertically downward.\n",
"For(a):\n",
"(i) The force of gravity 20 N on the block by the earth (action);the force of gravity on the earth by the block (reaction) 20 N directed upwards.\n",
"(ii) The force on the floor by the block (action); the force on the block by the floor (reaction).\n",
"For(b):\n",
"(i) The force of gravity 270 N on the system by the earth (action);the force of gravity on the earth by the system (reaction), equal to 270 N, directed upwards.\n",
"(ii) the force on the floor by the system (action); the force on the system by the floor (reaction).\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
" \n",
"m1=2 # Mass of the wooden block in kg\n",
"m2=25 # Mass of the iron cylinder in kg\n",
"a=0.1\n",
"g=10 # Acceleration due to gravity\n",
"\n",
"# Calculation\n",
"\n",
"#(a)\n",
"# The block is at rest on the floor. Its free-body has two forces on the block, the force of gravitational attraction by the earth and the normal force R of the floor on the block.\n",
"# By the First Law, the net force on the block must be zero, hence\n",
"e1=m1*g # The force of gravitational attraction by the earth\n",
"R=e1 # The normal force of the floor on the block\n",
"\n",
"#(b)\n",
"# The system (block + cylinder) accelerates downwards with 0.1 m/s².The free-body diagram of the system shows two forces on the system : the force of gravity due to the earth; and the normal force R′ by the floor. \n",
"e2=(m1+m2)*g # The force of gravitational attraction by the earth\n",
"R_prim=e2-(m1+m2)*a # The normal force of the floor on the block\n",
"\n",
"# Result\n",
"\n",
"print(\"(a) The action of the block is equal to\",R,\"N and directed vertically downwards.\")\n",
"print(\"(b) The action of the system on the floor is equal to\",R_prim,\"N vertically downward.\")\n",
"\n",
"print(\"For(a):\")\n",
"print(\"(i) The force of gravity\",e1,\"N on the block by the earth (action);the force of gravity on the earth by the block (reaction)\",R,\"N directed upwards.\")\n",
"print(\"(ii) The force on the floor by the block (action); the force on the block by the floor (reaction).\")\n",
"print(\"For(b):\")\n",
"print(\"(i) The force of gravity\",e2,\"N on the system by the earth (action);the force of gravity on the earth by the system (reaction), equal to 270 N, directed upwards.\")\n",
"print(\"(ii) the force on the floor by the system (action); the force on the system by the floor (reaction).\")\n"
]
}
],
"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",
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PKCI*O*O>Physics Textbook Part-I for class XI by NCERT/Chapter6_3.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 6 : Work, Energy and Power"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.1 , page : 115"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The angle between the force F and the displacement d = 0.32 °\n",
"The projection of F on d = 71.34\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"F = (3,4,-5) # Force vector \n",
"d = (5,4,3) # Displacement vector\n",
"\n",
"# Calculation\n",
"\n",
"Fd=sum(p*q for p,q in zip(F,d))\n",
"FF=sum(p*q for p,q in zip(F,F))\n",
"dd=sum(p*q for p,q in zip(d,d))\n",
"cosθ=Fd/math.sqrt(FF*dd)\n",
"θ=math.degrees(math.acos(cosθ))\n",
"\n",
"# Result\n",
"\n",
"print(\"The angle between the force F and the displacement d =\",cosθ,\"°\")\n",
"print(\"The projection of F on d =\",round(θ,2))\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.2 , page : 116"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) The work done by the gravitational force = 10.0 J\n",
"(b) The work done by the unknown resistive force -8.75 J\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=1 # Mass of the drop in g\n",
"h=1 # The height at which the drop is falling in km\n",
"v=50 # Speed at which the drop hits the ground in m/s\n",
"g=10 # Accelaration due to gravity\n",
"\n",
"# Calculation\n",
"\n",
"m=m*10**-3 # Mass of the drop in kg\n",
"h=h*10**3 # The height at which the drop is falling in m\n",
"\n",
"#(a)\n",
"# We have assumed that the drop is initially at rest\n",
"\n",
"K=(m*pow(v,2))/2 # Change in kinetic energy of the drop\n",
"W_g= m*g*h # The work done by the gravitational force on the drop in J\n",
"\n",
"#(b)\n",
"# From the work-energy theorm, K=W_g+W_r where Wr is the work done by the resistive force on the raindrop\n",
"W_r=K-W_g\n",
"\n",
"# Result\n",
"\n",
"print(\"(a) The work done by the gravitational force =\",W_g,\"J\")\n",
"print(\"(b) The work done by the unknown resistive force\",W_r,\"J\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.3 , page : 117 "
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) Work done by the road on the cycle = -2000.0 J\n",
"(b) Work done by cycle on the road = 0 J\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"d=10 # The distance in m\n",
"F=200 # The force on cycle due to the road in N\n",
"θ=180 # The angle the stopping force and the displacement make with each other in degrees\n",
"\n",
"# Calculation\n",
"\n",
"W_r=F*d*math.cos(math.radians(θ))\n",
"# From Newton’s Third Law an equal and opposite force acts on the road due to the cycle.\n",
"# Its magnitude is 200 N. However, the road undergoes no displacement.\n",
"# Thus, work done by cycle on the road is zero. \n",
"\n",
"# Result\n",
"\n",
"print(\"(a) Work done by the road on the cycle =\",W_r,\"J\")\n",
"print(\"(b) Work done by cycle on the road = 0 J\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.4 , page : 118 "
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The emergent speed of the bullet = 63.2 m/s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=50 # The mass of the bullet in g\n",
"in_v=200 # The initial velocity of the bullet in m/s\n",
"d=2 # The thickness of the plywood in cm\n",
"mp=1 # For convenience,mass is assumed to be unity \n",
"\n",
"# Calculation\n",
"\n",
"m=m*10**-3 # The mass of the bullet in kg\n",
"in_ke=(m*in_v**2)/2\n",
"# Since the bullet emerges with only 10% of its initial kinetic energy\n",
"fin_ke=0.1*in_ke\n",
"#If v_f is the emergent speed of the bullet,then mv_f²/2 = final kinetic energy\n",
"v_f=math.sqrt((2*fin_ke)/m) \n",
"\n",
"# Result\n",
"\n",
"print(\"The emergent speed of the bullet =\",round(v_f,1),\"m/s\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.5 , page : 119 "
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Work done by the women = 1750.0 J\n",
"Work done by the frictional force = -1000 J\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"F1=100 # Initial force in J\n",
"F2=50 # Final force in J\n",
"d1=10 # Initial distance covered in m\n",
"d2=10 # Final distance covered in m\n",
"\n",
"# Calculation\n",
"\n",
"d=d1+d2 # Total distance covered\n",
"A_rec1=F1*d1 # Area of the rectangle ABCD\n",
"A_tra=((F1+F2)/2)*d2\n",
"W_f=A_rec1+A_tra # Work done by the women\n",
"A_rec2=-F2*d # Area of the rectangle AGHI\n",
"\n",
"# Result\n",
"\n",
"print(\"Work done by the women =\",W_f,\"J\")\n",
"print(\"Work done by the frictional force =\",A_rec2,\"J\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.6 , page : 120 "
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The final kinetic energy of the block = 0.5 J\n",
"The final speed of the block = 1.0 m/s\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=1 # Mass of the block in kg\n",
"v_t=2 # Speed of the block in m/s\n",
"x1=0.10 # Initial point\n",
"x2=2.01 # Final point\n",
"k=0.5 # Proportionality ratio in J\n",
"\n",
"# Calculation\n",
"\n",
"k_t=(m*pow(v_t,2))/2\n",
"k_f=k_t-(k*math.log(x2/x1))\n",
"#Since Kinetic energy =mv²/2\n",
"v_f=math.sqrt(2*k_f/m)\n",
"\n",
"# Result\n",
"\n",
"print(\"The final kinetic energy of the block =\",round(k_f,2),\"J\")\n",
"print(\"The final speed of the block =\",round(v_f,2),\"m/s\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.7 , page : 122 "
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(i) v_o = √(5gL)\n",
"(ii) The speed at point B, v_b = √3gL\n",
" The speed at point C, v_c = √gL\n",
"(iii) The ratio of the kinetic energies (KB/KC) at B and C= 3.0\n",
"At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution.\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Calculation\n",
"\n",
"#(i)\n",
"# The total mechanical energy E of the system is conserved.We take the potential energy of the system to be zero at the lowest point A.\n",
"# Thus, at A : E = (1/2)mv²_o\n",
"# By Newton’s Second Law Ta-m = mv²_o/L where TA is the tension in the string at A. \n",
"# At the highest point C, the string slackens, as the tension in the string (TC) becomes zero. \n",
"# Thus, at C, E = mgl(5/2)\n",
"# Equating this to the energy at A we get, v_o = √(5gL)\n",
"\n",
"#(ii)\n",
"# We know that v_c= √(gL)\n",
"# At B, the energy is E =(1/2)mv²_b + mgL\n",
"# Equating this to the energy at A and employing the result namely v²_o = (5gL), we get v_b = √3gL\n",
"\n",
"#(iii)\n",
"# The ratio of the kinetic energies at B and C is; K_B/K_C = (1/2)mv²_b/(1/2)mv²_c = 3/1\n",
"\n",
"# Result\n",
"\n",
"print(\"(i) v_o = √(5gL)\")\n",
"print(\"(ii) The speed at point B, v_b = √3gL\")\n",
"print(\" The speed at point C, v_c = √gL\")\n",
"print(\"(iii) The ratio of the kinetic energies (KB/KC) at B and C=\",3/1)\n",
"print(\"At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution.\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.8 , page : 124 "
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The maximum compression of spring = 2.0 m\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=1000 # Mass of the car in kg\n",
"v=18 # Speed of the car in km/h\n",
"k=6.25*10**3 # Spring constant in N/m\n",
"\n",
"# Calculation\n",
"\n",
"v=v*(5/18) # Speed of the car in m/s\n",
"KE=m*v**2/2 # Kinetic energy of the car\n",
"# At maximum compression Xm, the potential energy V of the spring is equal to the kinetic energy KE of the moving car from the principle of conservation of mechanical energy.\n",
"Xm=math.sqrt((2*KE)/k)\n",
"\n",
"# Result\n",
"\n",
"print(\"The maximum compression of spring =\",Xm,\"m\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.9 , page : 125 "
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The maximum compression of the spring = 1.03 m\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=1000 # Mass of the car in kg\n",
"µ=0.5 # The coefficient of friction \n",
"g=10 # Acceleration due to gravity\n",
"k=6.25*10**3 # Spring constant in N/m\n",
"v=18 # Speed of the car in km/h\n",
"\n",
"# Calculation\n",
"\n",
"v=v*(5/18) # Speed of the car in m/s\n",
"# In presence of friction, both the spring force and the frictional force act so as to oppose the compression of the spring \n",
"# The change in kinetic energy is ∆K = K_f - K_i = 0 -(1/2)mv² ..........eqn 1\n",
"# The work done by the net force is W = (1/2)mv² + µmgX_m .............eqn 2\n",
"# by equating the above two equations and rearranging we obtain the following quadratic equation in the unknown X_m as \n",
"# kX²_m + 2µmgX_m + mv² = 0\n",
"t1=((-2*µ*m*g)+math.sqrt(abs(((2*µ*m*g)**2)-(4*k*m*v**2))))/(2*k)\n",
"t2=((-2*µ*m*g)-math.sqrt(abs(((2*µ*m*g)**2)-(4*k*m*v**2))))/(2*k)\n",
"X_m=max(t1,t2)\n",
"\n",
"# Result\n",
"\n",
"print(\"The maximum compression of the spring =\",round(X_m,2),\"m\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.10 , page : 127 "
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) Energy required to break one bond of DNA in eV = 6.25 ≈ 0.06 eV\n",
"(b) The kinetic energy of an air molecule in eV = 62.5 ≈ 0.0062 eV\n",
"(c) The average human consumption in a day in kcal = 2380.9523809523807 ≈ 2400 kcal\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"E=pow(10,20) # Energy required to break one bond of DNA in J\n",
"KE=pow(10,21) # The kinetic energy of an air molecule in J\n",
"Con=pow(10,7) # The average human consumption in a day in J\n",
" \n",
"# Calculation\n",
"#(a)\n",
"E=(10**20)/(1.6*10**19 ) # Energy required to break one bond of DNA in eV\n",
"#(b) \n",
"KE=(10**21)/(1.6*10**19) # The kinetic energy of an air molecule in eV\n",
"#(c) \n",
"consum=(10**7)/(4.2*10**3) # The average human consumption in a day in kcal\n",
"\n",
"# Result\n",
"\n",
"print(\"(a) Energy required to break one bond of DNA in eV =\",E,\"≈ 0.06 eV\")\n",
"print(\"(b) The kinetic energy of an air molecule in eV =\",KE,\"≈ 0.0062 eV\")\n",
"print(\"(c) The average human consumption in a day in kcal =\",consum,\"≈ 2400 kcal\")\n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.11 , page : 128 "
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The minimum power delivered by the motor to the elevator in watts = 44000 watts\n",
"The minimum power delivered by the motor to the elevator in hp = 59.0 hp\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=1800 # Maximum load the elevator can carry in kg\n",
"F_f=4000 # Frictional force appearing in N\n",
"v=2 # Speed of the elevator in m/s\n",
"g=10 # Acceleration due to gravity\n",
"\n",
"# Calculation\n",
"\n",
"F=m*g+F_f # The downward force on the elevator in N\n",
"P=F*v\n",
"# Since, 1 hp = 7.457*10**2 W\n",
"P_hp=P/(7.457*10**2)\n",
"\n",
"# Result\n",
"\n",
"print(\"The minimum power delivered by the motor to the elevator in watts =\",P,\"watts\")\n",
"print(\"The minimum power delivered by the motor to the elevator in hp =\",round(P_hp,2),\"hp\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.12 , page : 130 "
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"For deuterium the fractional kinetic energy lost is = 0.1111111111111111\n",
"For deuterium the fractional kinetic energy gained by the moderating nuclei = 0.8888888888888888\n",
"Hence we conclude that almost 90% of the neutron’s energy is transferred to deuterium.\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Calculation\n",
"\n",
"# The initial kinetic energy of the neutron is K_1i = (1/2)*m1V²_1i\n",
"# Its final kinetic energy is given by, K_1f = (1/2)*m1V²_1f =(1/2)*m1*(m1-m2/m1+m2)²*V²_1f\n",
"# The fractional kinetic energy lost is f1 = K_1f/K_1i = (m1-m2/m1+m2)²\n",
"# The fractional kinetic energy gained by the moderating nuclei is f2 = 1-f1 = (4*m1*m2)/(m1+m2)²\n",
"# For deuterium m2 = 2m1 \n",
"m1=1\n",
"m2=2*m1\n",
"f1=((m1-m2)/(m1+m2))**2\n",
"f2 = 1-f1\n",
"\n",
"# Result\n",
"\n",
"print(\"For deuterium the fractional kinetic energy lost is =\",f1)\n",
"print(\"For deuterium the fractional kinetic energy gained by the moderating nuclei =\",f2)\n",
"print(\"Hence we conclude that almost 90% of the neutron’s energy is transferred to deuterium.\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.13 , page : 131 "
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The angle at which the player has to strike the cue = 53.0 °\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"θ=37 # The angle of the corner pocket in degreeac\n",
"\n",
"# Calculation\n",
"\n",
"# Here m1 = m2, where m1 is the mass of the cue and m2 is the mass of the target\n",
"# From momentum conservation, since the masses are equal ; V_1i = V_1f + V_2f\n",
"# Or V_1f² = V_2f² + 2*V_1f*V_2f\n",
"# Since the collision is elastic and m1 = m2 it follows from conservation of kinetic energy that V_1i² = V_1f² + V_2f²\n",
"# Hence we get, cos(θ1+θ) = 0\n",
"θ1=math.degrees(math.acos(0))- θ\n",
"\n",
"# Result\n",
"\n",
"print(\"The angle at which the player has to strike the cue =\",θ1,\"°\")\n"
]
}
],
"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": "3.4.3"
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}
PKCI?$$>Physics Textbook Part-I for class XI by NCERT/Chapter7_3.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 7 : Systems of Particles and Rotational Motion"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7.1 , page : 146"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The centre of mass = ( 0.28 m, 0.11 m )\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"# With the x–and y–axes chosen as the coordinates of points O, A and B forming the equilateral triangle are respectively (0,0),(0.5,0),(0.25,0.25√3 ).\n",
"# Let the masses 100 g, 150g and 200g be located at O, A and B be respectively.\n",
"\n",
"m1=100 # Mass of the first particle in g\n",
"m2=150 # Mass of the second particle in g\n",
"m3=200 # Mass of the third particle in g\n",
"x1=0 # x-coordinate of the first particle \n",
"x2=0.5 # x-coordinate of the second particle \n",
"x3=0.25 # x-coordinate of the third particle \n",
"y1=0 # y-coordinate of the first particle \n",
"y2=0 # y-coordinate of the second particle \n",
"y3=0.25 # y-coordinate of the third particle \n",
"\n",
"# Calculation\n",
"\n",
"X=(m1*x1+m2*x2+m3*x3)/(m1+m2+m3)\n",
"Y=(m1*y1+m2*y2+m3*y3)/(m1+m2+m3)\n",
"\n",
"# Result\n",
"\n",
"print(\"The centre of mass = (\",round(X,2),\"m,\",round(Y,2),\"m\",\")\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7.2 , page : 147 "
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The centre of mass of a triangular lamina lies on the centroid of the triangle\n"
]
}
],
"source": [
"# The lamina (∆LMN) may be subdivided into narrow strips each parallel to the base (MN). \n",
"# By symmetry each strip has its centre of mass at its midpoint.\n",
"# If we join the midpoint of all the strips we get the median LP.\n",
"# The centre of mass of the triangle as a whole therefore, has to lie on the median LP.\n",
"# Similarly, we can argue that it lies on the median MQ and NR. \n",
"# This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle.\n",
"\n",
"# Result\n",
"\n",
"print(\"The centre of mass of a triangular lamina lies on the centroid of the triangle\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7.3 , page : 147 "
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The centre of mass = ( 0.83 m, 0.83 m )\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"# Letus choose the X and Y axes as the coordinates of the vertices of the L-shaped lamina.\n",
"# We can think of the L-shape to consist of 3 squares each of length 1m.\n",
"# The mass of each square is 1kg, since the lamina is uniform.\n",
"# The centres of mass C1, C2 and C3 of the squares are, by symmetry, their geometric centres and have coordinates (1/2,1/2), (3/2,1/2), (1/2,3/2) respectively.\n",
"# We take the masses of the squares to be concentrated at these points.\n",
"# hence the centre of mass of the whole L shape (X, Y) is the centre of mass of these mass points.\n",
"\n",
"\n",
"m1=1\n",
"m2=1\n",
"m3=1\n",
"x1=1/2\n",
"x2=3/2\n",
"x3=1/2\n",
"y1=1/2\n",
"y2=1/2\n",
"y3=3/2\n",
"\n",
"# Calculation\n",
"\n",
"X=(m1*x1+m2*x2+m3*x3)/(m1+m2+m3)\n",
"Y=(m1*y1+m2*y2+m3*y3)/(m1+m2+m3)\n",
"\n",
"# Result\n",
"\n",
"print(\"The centre of mass = (\",round(X,2),\"m,\",round(Y,2),\"m\",\")\")\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7.4 , page : 152 "
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Scalar product = -25\n",
"Vector product a × b = [ 7 -1 -5]\n",
"Vector product b × a = [-7 1 5]\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"import numpy as np\n",
"\n",
"# Variable declaration\n",
"\n",
"a = (3,-4,5) # Vector a\n",
"b = (-2,1,-3) # Vector b\n",
"\n",
"# Calculation\n",
"\n",
"s=sum(p*q for p,q in zip(a,b))\n",
"a1 = np.array([3,-4,5]) \n",
"b1 = np.array([-2,1,-3]) \n",
"v1=np.cross(a1,b1)\n",
"v2=np.cross(b1,a1)\n",
"\n",
"# Result\n",
"\n",
"print(\"Scalar product =\",s)\n",
"print(\"Vector product a × b =\",v1)\n",
"print(\"Vector product b × a =\",v2)"
]
}
],
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"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
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"codemirror_mode": {
"name": "ipython",
"version": 3
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PKCI/a^811>Physics Textbook Part-I for class XI by NCERT/Chapter8_3.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"# Chapter 8 : Gravitation"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## Example 8.1 , page : 185"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" The planet will take a longer time to traverse BAC than CPB\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"mp=1 # For convenience,mass is assumed to be unity \n",
"rp=1 # For convenience,sun-planet distance at perihelton is assumed to be unity \n",
"vp=1 # For convenience,speed of the planet at perihelton is assumed to be unity \n",
"ra=1 # For convenience,sun-planet distance at aphelton is assumed to be unity \n",
"va=1 # For convenience,speed of the planet at aphelton is assumed to be unity \n",
"Lp=mp*rp*vp # Angular momentum at perihelton\n",
"La=mp*ra*va # Angular momentum at ahelton\n",
"\n",
"# Result\n",
"\n",
"# From angular momentum conservation, mp*rp*vp = mp*ra*va or vp/va = rp/ra\n",
"# From Kepler’s second law, equal areas are swept in equal times\n",
"print(\" The planet will take a longer time to traverse BAC than CPB\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.2 , page : 187 "
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(a) The force acting = [0.0, 2.5849394142282115e-26, 0.0] ≈ 0\n",
"(b) The force acting = 2 Gm²\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"G=6.67*pow(10,-11) # Gravitational constant\n",
"m=1 # For convenience,mass is assumed to be unity \n",
"x=30 # The angle between GC and the positive x-axis is 30° and so is the angle between GB and the negative x-axis\n",
"y=math.radians(x) # The angle in radians\n",
"a=math.cos(y)\n",
"b=math.sin(y)\n",
"v1=(0,1,0)\n",
"v2=(-a,-b,0)\n",
"v3=(a,-b,0)\n",
"c=(2*G*pow(m,2))/1 # 2Gm²/1\n",
"\n",
"# Calculation\n",
"\n",
"#(a)\n",
"F1=[y * c for y in v1] # F(GA)\n",
"F2=[y * c for y in v2] # F(GB)\n",
"F3=[y * c for y in v3] # F(GC)\n",
"# From the principle of superposition and the law of vector addition, the resultant gravitational force FR on (2m) is given by\n",
"Fa=[sum(x) for x in zip(F1,F2,F3)]\n",
"\n",
"#(b)\n",
"# By symmetry the x-component of the force cancels out and the y-component survives\n",
"Fb=4-2 # 4Gm² j - 2Gm² j\n",
"\n",
"# Result\n",
"\n",
"print(\"(a) The force acting =\",Fa,\"≈ 0\")\n",
"print(\"(b) The force acting =\",Fb,\"Gm²\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.3 , page : 192 "
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Potential energy of a system of four particles = -5.414213562373095 Gm²/l\n",
"The gravitational potential at the centre of the square = -5.65685424949238 Gm²/l\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"G=6.67*pow(10,-11) # Gravitational constant\n",
"m=1 # For convenience,mass is assumed to be unity \n",
"l=1 # For convenience,side of the square is assumed to be unity \n",
"c=(G*pow(m,2))/l\n",
"n=4 # Number of particles\n",
"\n",
"# Calculation\n",
"\n",
"d=math.sqrt(2)\n",
"# If the side of a square is l then the diagonal distance is √2l\n",
"# We have four mass pairs at distance l and two diagonal pairs at distance √2l \n",
"# Since the Potential Energy of a system of four particles is -4Gm²/l) - 2Gm²/dl\n",
"w=(-n-(2/d)) \n",
"# If the side of a square is l then the diagonal distance from the centre to corner is \n",
"# Since the Gravitational Potential at the centre of the square\n",
"u=-n*(2/d)\n",
"\n",
"# Result\n",
"\n",
"print (\"Potential energy of a system of four particles =\",w,\"Gm²/l\")\n",
"print(\"The gravitational potential at the centre of the square =\",u,\"Gm²/l\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.4 , page : 193 "
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Minimum speed of the projectile to reach the surface of the second sphere = ( 0.6 GM/R ) ^(1/2)\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"R=1 # For convenience,radii of both the spheres is assumed to be unity \n",
"M=1 # For convenience,mass is assumed to be unity \n",
"m1=M # Mass of the first sphere\n",
"m2=6*M # Mass of the second sphere\n",
"m=1 # Since the mass of the projectile is unknown,take it as unity\n",
"d=6*R # Distance between the centres of both the spheres\n",
"r=1 # The distance from the centre of first sphere to the neutral point N\n",
"\n",
"G=6.67*pow(10,-11) # Gravitational constant\n",
"\n",
"# Calculation\n",
"\n",
"# Since N is the neutral point; GMm/r² = 4GMm/(6R-r)² and we get\n",
"r=2*R\n",
"# The mechanical energy at the surface of M is; Et = m(v^2)/2 - GMm/R - 4GMm/5R\n",
"# The mechanical energy at N is; En = -GMm/2R - 4GMm/4R\n",
"# From the principle of conservation of mechanical energy; Et = En and we get\n",
"v_sqr=2*((4/5)-(1/2))\n",
"\n",
"# Result\n",
"\n",
"print(\"Minimum speed of the projectile to reach the surface of the second sphere =\",\"(\",round(v_sqr,5),\"GM/R\",\")\",\"^(1/2)\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.5 , page : 195 "
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(i) Mass of Mars = 6.475139697520706e+23 kg\n",
"(ii) Period of revolution of Mars = 684.0033777694376 days\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"π=3.14 # Constant pi\n",
"G=6.67*pow(10,-11) # Gravitational constant\n",
"R=9.4*pow(10,3) # Orbital radius of Mars in km\n",
"T=459*60\n",
"Te=365 # Period of revolution of Earth\n",
"r=1.52 # Ratio of Rms/Res, where Rms is the mars-sun distance and Res is the earth-sun distance. \n",
"\n",
"# Calculation\n",
"\n",
"# (i) \n",
"R=R*pow(10,3)\n",
"# Using Kepler's 3rd law:T²=4π²(R^3)/GMm\n",
"Mm=(4*pow(π,2)*pow(R,3))/(G*pow(T,2))\n",
"\n",
"# (ii)\n",
"# Using Kepler's 3rd law: Tm²/Te² = (Rms^3/Res^3)\n",
"Tm=pow(r,(3/2))*365\n",
"\n",
"\n",
"# Result\n",
"\n",
"print(\"(i) Mass of Mars =\",Mm,\"kg\")\n",
"print(\"(ii) Period of revolution of Mars =\",Tm,\"days\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.6 , page : 195 "
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Mass of the Earth = 5.967906881559221e+24 kg\n",
"Mass of the Earth = 6.017752855396305e+24 kg\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"g=9.81 # Acceleration due to gravity\n",
"G=6.67*pow(10,-11) # Gravitational constant\n",
"Re=6.37*pow(10,6) # Radius of Earth in m\n",
"R=3.84*pow(10,8) # Distance of Moon from Earth in m\n",
"T=27.3 # Period of revolution of Moon in days\n",
"π=3.14 # Constant pi\n",
"\n",
"# Calculation\n",
"\n",
"# I Method\n",
"# Using Newton's 2nd law of motion:g = F/m = GMe/Re²\n",
"Me1=(g*pow(Re,2))/G\n",
"\n",
"# II Method\n",
"# Using Kepler's 3rd law: T²= 4π²(R^3)/GMe\n",
"T1=T*24*60*60\n",
"Me2=(4*pow(π,2)*pow(R,3))/(G*pow(T1,2))\n",
"\n",
"#Result\n",
"\n",
"print(\"Mass of the Earth =\",Me1,\"kg\")\n",
"print(\"Mass of the Earth =\",Me2,\"kg\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.7 , page : 195 "
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Period of revolution of Moon = 27.5 days\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"k=pow(10,-13) # A constant = 4π² / GME\n",
"Re=3.84*pow(10,5) # Distance of the Moon from the Earth in m\n",
"\n",
"# Calculation\n",
"\n",
"k=pow(10,-13)*(pow(1/(24*60*60),2))*(1/pow((1/1000),3))\n",
"T2=k*pow(Re,3)\n",
"T=math.sqrt(T2) # Period of revolution of Moon in days\n",
"\n",
"# Result\n",
"\n",
"print(\"Period of revolution of Moon =\",round(T,1),\"days\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8.8 , page : 196 "
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Change in Kinetic Energy = 3124485000.0 J\n",
"Change in Potential Energy = 6248970000.0 J\n"
]
}
],
"source": [
"# Importing module\n",
"\n",
"import math\n",
"\n",
"# Variable declaration\n",
"\n",
"m=400 # Mass of satellite in kg\n",
"Re=6.37*pow(10,6) # Radius of Earth in m\n",
"g=9.81 # Acceleration due to gravity\n",
"\n",
"# Calculation\n",
"\n",
"# Change in energy is E=Ef-Ei\n",
"ΔE=(g*m*Re)/8 # Change in Total energy\n",
"# Since Potential Energy is twice as the change in Total Energy (V = Vf - Vi)\n",
"ΔV=2*ΔE # Change in Potential Energy in J\n",
"\n",
"# Result\n",
"\n",
"print(\"Change in Kinetic Energy =\",round(ΔE,4),\"J\")\n",
"print(\"Change in Potential Energy =\",round(ΔV,4),\"J\")"
]
}
],
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"display_name": "Python 3",
"language": "python",
"name": "python3"
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