'''
Created on 01-Apr-2021
@author: rishi_priyansh
'''
import random
import inspect
import traceback
from sympy import (symbols, Symbol)

from xv.km.util import latex_fraction, latex_atom
from .....helper import BaseHelper

from xv.km.managers import BaseManager
from xv.km.kelements import KProblemTemplate
        
import os

# _dynamics_manager
# DynamicsManager

class DynamicsManager(BaseManager):
    '''
    DynamicsManager
    Example:
        ke = DynamicsManager()
        ke.getRandomProblem()
        ke.getRandomProblem(problem_type = 0)
        ...
        
        ke.printProblem()
        ke.printAnswer()
        ke.printSolution()
        
        ke.printProblemTypes()
        
    '''    
    _OPR_TYPE_COUNT = 10
    
    def __new__(cls, *args, **kwargs):
        instance = super(DynamicsManager, cls).__new__(cls, *args, **kwargs)
        for base_class in DynamicsManager.__bases__:
            base_class.__init__(instance, *args, **kwargs)
        return instance
    
    def __init__(self, *args, **kwargs):
        pass

    def _init_problem_types(self, *args, **kwargs):
        self._append_problem_types([
            '_problem_1_59',
            '_problem_1_60',
            '_problem_1_61',
            '_problem_1_62',
            '_problem_1_63',
            '_problem_1_64',
            '_problem_1_65',
            '_problem_1_66',
            '_problem_1_67',
            '_problem_1_68',
            '_problem_1_69',
            '_problem_1_70',
            '_problem_1_71',
            '_problem_1_72',
            '_problem_1_73',
            '_problem_1_74',
            '_problem_1_75',
            '_problem_1_76',
            '_problem_1_77',
            '_problem_1_78',
            '_problem_1_79',
            '_problem_1_80',
            '_problem_1_81',
            '_problem_1_82',
            '_problem_1_83',
            '_problem_1_84',
            '_problem_1_85',
            '_problem_1_86',
            '_problem_1_87',
            '_problem_1_88',
            '_problem_1_89',
            '_problem_1_90',
            '_problem_1_91',
            '_problem_1_92',
            '_problem_1_93',
            '_problem_1_94',
            '_problem_1_95',
            '_problem_1_96',
            '_problem_1_97',
            '_problem_1_98',
            '_problem_1_99',
            '_problem_1_100',
            '_problem_1_101',
            '_problem_1_102',
            '_problem_1_103',
            '_problem_1_104',
            '_problem_1_105',
            '_problem_1_106',
            '_problem_1_107',
            '_problem_1_108',
            '_problem_1_109',
            '_problem_1_110',
            '_problem_1_111',
            '_problem_1_112',
            '_problem_1_113',
            '_problem_1_114',
            '_problem_1_115',
            '_problem_1_116',
            '_problem_1_117',
            ])
        pass
            
    def _problem_1_59(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An aerostat of mass m starts coming down with a constant
        acceleration w. Determine the ballast mass to be dumped for the
        aerostat to reach the upward acceleration of the same magnitude.
        The air drag is to be neglected.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Let $R$ be the constant upward thurst on the aerostat of mass $m$, coming down with a constant acceleration w. Applying Newton's second law of motion for the aerostat in projection form
        $\displaystyle \\begin{{aligned}} F_{{y}} &=m w_{{y}} \\\\ m g-R &=m w \\quad (1) \\end{{aligned}}$
        Now, if $\\Delta m$ be the mass, to be dumped, then using the Eq. $F_{{y}}=m w_{{y}}$
        $R-(m-\\Delta m) g=(m-\\Delta m) w \\quad (2)$
        From Eqs. $(1)$ and $(2)$, we get, $\\quad \\Delta m=\\frac{{2 m w}}{{g+w}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_60(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the arrangement of Fig. 1.9 the masses $m_o$, $m_1$, and $m_2$
        of bodies are equal, the masses of the pulley and the threads are
        negligible, and there is no friction in the pulley. Find the
        acceleration w with which the body $m_o$ comes down, and the tension of
        the thread binding together the bodies $m_1$ and $m_2$, if the coefficient
        of friction between these bodies and the horizontal surface is equal
        to k. Consider possible cases.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-9.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us write the fundamental equation of dynamics for all the three blocks in terms of projections, having taken the positive direction of $x$ and $y$ axes as shown in Fig; and using the fact that kinematical relation between the accelerations is such that the blocks move with same value of acceleration (say $w$ )
        $\\begin{{aligned}} m_{{0}} g-T_{{1}} &=m_{{0}} w \\qquad (1) \\\\ T_{{1}}-T_{{2}}-k m_{{1}} g &=m_{{1}} w \\qquad (2) \\\\ \\text {{ and }} \\quad T_{{2}}-k m_{{2}} g &=m_{{2}} w \\qquad (3) \\end{{aligned}}$
        The simultaneous solution of Eqs. (1), (2) and (3) yields,
        $\\begin{{aligned}} &w=g \\frac{{\\left[m_{{0}}-k\\left(m_{{1}}+m_{{2}}\\right)\\right]}}{{m_{{0}}+m_{{1}}+m_{{2}}}} \\\\ &T_{{2}}=\\frac{{(1+k) m_{{0}}}}{{m_{{0}}+m_{{1}}+m_{{2}}}} m_{{2}} g \\end{{aligned}}$
        As the block $m_{{0}}$ moves down with acceleration $w$, so in vector form
        $\\vec{{w}}=\\frac{{\\left[m_{{0}}-k\\left(m_{{1}}+m_{{2}}\\right)\\right] \\vec{{g}}}}{{m_{{0}}+m_{{1}}+m_{{2}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_61(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        1.61. Two touching bars 1 and 2 are placed on an inclined plane
        forming an angle a with the horizontal (Fig. 1.10). The masses of
        the bars are equal to m1 and m2, and the coefficients of friction 
        between the inclined plane and these bars are equal to k1 and k2 respectively, with k1 > k2. Find:
        (a) the force of interaction of the bars in the process of motion;
        (b) the minimum value of the angle a at which the bars start sliding down.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-10.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us indicate the positive direction of $x$-axis along the incline (Fig.). Figures show the force diagram for the blocks.
        Let, $R$ be the force of interaction between the bars and they are obviously sliding down with the same constant acceleration $w$.
        Newton's second law of motion in projection form along $x$-axis for the blocks gives:
        $\\begin{{aligned}} &m_{{1}} g \\sin \\alpha-k_{{1}} m_{{1}} g \\cos \\alpha+R=m_{{1}} w \\quad (1)\\\\ &m_{{2}} g \\sin \\alpha-R-k_{{2}} m_{{2}} g \\cos \\alpha=m_{{2}} w \\quad (2) \\end{{aligned}}$
        Solving Eqs. (1) and (2) simultaneously, we get
        $\\begin{{gathered}} w=g \\sin \\alpha-g \\cos \\alpha \\frac{{k_{{1}} m_{{1}}+k_{{2}} m_{{2}}}}{{m_{{1}}+m_{{2}}}} \\text {{ and }} \\\\ R=\\frac{{m_{{1}} m_{{2}}\\left(k_{{1}}-k_{{2}}\\right) g \\cos \\alpha}}{{m_{{1}}+m_{{2}}}} \\quad (3) \\end{{gathered}}$
        (b) when the blocks just slide down the plane, $w=0$, so from Eqn. (3)
        $\\begin{{gathered}} g \\sin \\alpha-g \\cos \\alpha \\frac{{k_{{1}} m_{{1}}+k_{{2}} m_{{2}}}}{{m_{{1}}+m_{{2}}}}=0 \\\\ \\text{{or, }} \\quad \\left(m_{{1}}+m_{{2}}\\right) \\sin \\alpha=\\left(k_{{1}} m_{{1}}+k_{{2}} m_{{2}}\\right) \\cos \\alpha \\end{{gathered}}$
        Hence
        $\\tan \\alpha=\\frac{{\\left(k_{{1}} m_{{1}}+k_{{2}} m_{{2}}\\right)}}{{m_{{1}}+m_{{2}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_62(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A small body was launched up an inclined plane set at an
        angle a = $15^{{\\circ}}$ against the horizontal. Find the coefficient of friction,
        if the time of the ascent of the body is $\\eta$ = 2.0 times less than the
        time of its descent
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Case 1 . When the body is launched up :
        Let $k$ be the coefficeint of friction, $u$ the velocity of projection and $l$ the distance traversed along the incline. Retarding force on the block $=m g \\sin \\alpha+k m g \\cos \\alpha$ and hence the retardation $=g \\sin \\alpha+k g \\cos \\alpha$
        Using the equation of particle kinematics along the incline,
        $0=u^{{2}}-2(g \\sin \\alpha+k g \\cos \\alpha) l$
        or,
        $l=\\frac{{u^{{2}}}}{{2(g \\sin \\alpha+k g \\cos \\alpha)}} \\quad (1)$
        and
        $0=u-(g \\sin \\alpha+k g \\cos \\alpha) t u=(g \\sin \\alpha+k g \\cos \\alpha) t \\quad (2)$
        or,
        $u=(g \\sin \\alpha+k g \\cos \\alpha) t \\quad (3)$

        Using (2) in (1) $l=\\frac{{1}}{{2}}(g \\sin \\alpha+k g \\cos \\alpha) t^{{2}}$
        Case (2). When the block comes downward, the net force on the body $=m g \\sin \\alpha-k m g \\cos \\alpha$ and hence its acceleration $=g \\sin \\alpha-k g \\cos \\alpha$ Let, $t$ be the time required then,
        $l=\\frac{{1}}{{2}}(g \\sin \\alpha-k g \\cos \\alpha) t^{{\\prime 2}} \\quad (4)$
        From Eqs. (3) and (4)
        $\\begin{{aligned}} &\\frac{{t^{{2}}}}{{t^{{\\prime 2}}}}=\\frac{{\\sin \\alpha-k \\cos \\alpha}}{{\\sin \\alpha+k \\cos \\alpha}} \\\\ &\\text {{ But }} \\frac{{t}}{{t^{{\\prime}}}}=\\frac{{1}}{{\\eta}} \\quad \\text{{(according to the question),}} \\end{{aligned}}$

        Hence on solving we get
        $k=\\frac{{\\left(\\eta^{{2}}-1\\right)}}{{\\left(\\eta^{{2}}+1\\right)}} \\tan \\alpha=0.16$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_63(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The following parameters of the arrangement of Fig. 1.11
        are available: the angle $\\alpha$ which the inclined plane forms with the
        horizontal, and the coefficient of friction k between the body $m_1$
        and the inclined plane. The masses of the pulley and the threads,
        as well as the friction in the pulley, are negligible. Assuming both
        bodies to be motionless at the initial moment, find the mass ratio
        $m_2/m_1$ at which the body $m_2$
        (a) starts coming down;
        (b) starts going up;
        (c) is at rest. 
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-11.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        At the initial moment, obviously the tension in the thread connecting $m_{{1}}$ and $m_{{2}}$ equals the weight of $m_{{2}}$
        (a) For the block $m_{{2}}$ to come down or the block $m_{{1}}$ to go up, the conditions is
        $m_{{2}} g-T \\geq 0 \\text {{ and }} T-m_{{1}} g \\sin \\alpha-f r \\geq 0$
        where $T$ is tension and $f_{{r}}$ is friction which in the limiting case equals $k m_{{1}} g \\cos \\alpha .$ Then
        or $m_{{2}} g-m_{{1}} \\sin \\alpha>k m_{{1}} g \\cos \\alpha$
        or $\\frac{{m_{{2}}}}{{m_{{1}}}}>(k \\cos \\alpha+\\sin \\alpha)$
        (b) Similarly in the case
        $m_{{1}} g \\sin \\alpha-m_{{2}} g>f r_{{1 \\mathrm{{~m}}}}$
        or, $m_{{1}} g \\sin \\alpha-m_{{2}} g>k m_{{1}} g \\cos \\alpha$
        or, $\\frac{{m_{{2}}}}{{m_{{1}}}}<(\\sin \\alpha-k \\cos \\alpha)$
        (c) For this case, neither kind of motion is possible, and $f r$ need not be limiting. Hence,
        $(k \\cos \\alpha+\\sin \\alpha)>\\frac{{m_{{2}}}}{{m_{{1}}}}>(\\sin \\alpha-k \\cos \\alpha)$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_64(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The inclined plane of Fig. 1.11 forms an angle a = $30^{{\\circ}}$ with
        the horizontal. The mass ratio $m_2/m_1$ = $\\eta$ = 2/3. The coefficient of
        friction between the body m1 and the inclined plane is equal to 
        k = 0.10. The masses of the pulley and the threads are negligible.
        Find the magnitude and the direction of acceleration of the body $m_2$
        when the formerly stationary system of masses starts moving. 
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-11.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        From the conditions, obtained in the previous problem, first we will check whether the mass $m_{{2}}$ goes up or down.
        Here, $m_{{2}} / m_{{1}}=\\eta>\\sin \\alpha+k \\cos \\alpha$, (substituting the values). Hence the mass $m_{{2}}$ will come down with an acceleration (say $w$ ). From the free body diagram of previous problem,
        $m_{{2}}-g-T=m_{{2}} w \\quad (1)$
        and
        $T-m_{{1}} g \\sin \\alpha-k m_{{1}} g \\cos \\alpha=m_{{1}} w \\quad (2)$
        Adding (1) and (2), we get,
        $\\begin{{gathered}} m_{{2}} g-m_{{1}} g \\sin \\alpha-k m_{{1}} g \\cos \\alpha=\\left(m_{{1}}+m_{{2}}\\right) w \\\\ w=\\frac{{\\left(m_{{2}} / m_{{1}}-\\sin \\alpha-k \\cos \\alpha\\right) g}}{{\\left(1+m_{{2}} / m_{{1}}\\right)}}=\\frac{{(\\eta-\\sin \\alpha-k \\cos \\alpha) g}}{{1+\\eta}} \\end{{gathered}}$
        Substituting all the values, $w=0.048 \\mathrm{{~g}} \\approx 0.05 \\mathrm{{~g}}$
        As $m_{{2}}$ moves down with acceleration of magnitude $w=0.05 g>0$, thus in vector form acceleration of $m_{{2}}$ :
        $\\vec{{w}}_{{2}}=\\frac{{(\\eta-\\sin \\alpha-k \\cos \\alpha) \\vec{{g}}}}{{1+\\eta}}=0.05 \\vec{{g}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_65(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A plank of mass $m_1$ with a bar of mass $m_2$ placed on it lies on
        a smooth horizontal plane. A horizontal force growing with time t
        as F = at (a is constant) is applied to the bar. Find how the accelerations
        of the plank $w_1$ and of the bar $w_2$ depend on t, if the coefficient
        of friction between the plank and the bar is equal to k. Draw the approximate plots of these dependences. 

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Let us write the Newton's second law in projection form along positive $x$-axis for the plank and the bar
        $f r=m_{{1}} w_{{1}}, f r=m_{{2}} w_{{2}} \\quad (1)$
        At the initial moment, $f r$ represents the static friction, and as the force $F$ grows so does the friction force $f r$, but up to it's limiting value i.e. $f r=f r_{{s(\\max )}}=k N=k m_{{2}} g$.
        Unless this value is reached, both bodies moves as a single body with equal acceleration. But as soon as the force $f$ reaches the limit, the bar starts sliding over the plank i.e. $w_{{2}} \\geq w_{{1}}$
        Substituting here the values of $w_{{1}}$ and $w_{{2}}$ taken from Eq. (1) and taking into account that $f_{{r}}=k m_{{2}} g$, we obtain, $\\left(a t-k m_{{2}} g\\right) / m_{{2}} \\geq \\frac{{k m_{{2}}}}{{m_{{1}}}} g$, were the sign $"="$ corresponds to the moment $t=t_{{0}}$ (say)
        Hence,
        $t_{{0}}=\\frac{{k g m_{{2}}\\left(m_{{1}}+m_{{2}}\\right)}}{{a m_{{1}}}}$
        If
        $\\begin{{gathered}} t \\leq t_{{0}} \\text {{, then }} w_{{1}}=\\frac{{k m_{{2}} g}}{{m_{{1}}}} \\text {{ (constant). and }} \\\\ \\qquad w_{{2}}=\\left(a t-k m_{{2}} g\\right) / m_{{2}} \\end{{gathered}}$
        On this basis $w_{{1}}(t)$ and $w_{{2}}(t)$, plots are as shown in the figure of answersheet.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_66(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A small body A starts sliding down from the top of a wedge
        (Fig. 1.12) whose base is equal to 1 = 2.10 m. The coefficient of
        friction between the body and the wedge surface is k = 0.140. At
        what value of the angle a will the time of sliding be the least? What
        will it be equal to?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-12.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us designate the $x$-axis (Fig.) and apply $F_{{x}}=m w_{{x}}$ for body $A:$
        $m g \\sin \\alpha-k m g \\cos \\alpha=m w$
        or,
        $w=g \\sin \\alpha-k g \\cos \\alpha$
        Now, from kinematical equation :
        $\\begin{{aligned}} & l \\sec \\alpha=0+(1 / 2) w t^{{2}} \\\\ \\text{{or, }}\\quad t=& \\sqrt{{2 l \\sec \\alpha /(\\sin \\alpha-k \\cos \\alpha) g}} \\\\ =& \\sqrt{{2 l /\\left(\\sin 2 \\alpha / 2-k \\cos ^{{2}} \\alpha\\right) g}} \\end{{aligned}}$
        (using Eq. (1)).
        for $t_{{\\min }}$
        $\\frac{{d\\left(\\frac{{\\sin 2 \\alpha}}{{2}}-k \\cos ^{{2}} \\alpha\\right)}}{{d \\alpha}}=0$
        i.e. $\\frac{{2 \\cos 2 \\alpha}}{{2}}+2 k \\cos \\alpha \\sin \\alpha=0$
        or,
        $\\tan 2 \\alpha=-\\frac{{1}}{{k}} \\implies \\alpha=49^{{\\circ}}$
        and putting the values of $\\alpha, k$ and $l$ in Eq. (2) we get $t_{{\\min }}=1 \\mathrm{{~s}}$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_67(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A bar of mass m is pulled by means of a thread up'an inclined
        plane forming an angle $\\alpha$ with the horizontal (Fig. 1.13). The 
        coefficient of friction is equal to k. Find the angle $\\beta$ which the thread
        must form with the inclined plane for the tension of the thread to be
        minimum. What is it equal to? 
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-13.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us fix the $x-y$ co-ordinate system to the wedge, taking the $x$-axis up, along the incline and the $y$-axis perpendicular to it (Fig.).
        Now, we draw the free body diagram for the bar.

        Let us apply Newton's second law in projection form along $x$ and $y$ axis for the bar :
        $T \\cos \\beta-m g \\sin \\alpha-f r=0$
        $T \\sin \\beta+N-m g \\cos \\alpha=0$
        or, $N=m g \\cos \\alpha-T \\sin \\beta$ But $f_{{r}}=k N$ and using $(2)$ in (1), we get
        $T=m g \\sin \\alpha+k m g \\cos \\alpha /(\\cos \\beta+k \\sin \\beta)$
        For $T_{{\\operatorname{{ma}}}}$ the value of $(\\cos \\beta+k \\sin \\beta)$ should be maximum
        So
        $\\displaystyle \\frac{{d(\\cos \\beta+k \\sin \\beta)}}{{d \\beta}}=0 \\text {{ or }} \\tan \\beta=k$
        Putting this value of $\\beta$ in Eq. (3) we get,
        $\\displaystyle T_{{\\min }}=\\frac{{m g(\\sin \\alpha+k \\cos \\alpha)}}{{1 / \\sqrt{{1+k^{{2}}}}+k^{{2}} / \\sqrt{{1+k^{{2}}}}}}=\\frac{{m g(\\sin \\alpha+k \\cos \\alpha)}}{{\\sqrt{{1+k^{{2}}}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_68(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        At the moment t = 0 the force F = at is applied to a small
        body of mass m resting on a smooth horizontal plane (a is a constant).
        The permanent direction of this force forms an angle $\\alpha$ with the horizontal (Fig. 1.14). Find:
        (a) the velocity of the body at the moment of its breaking off the
        plane;
        (b) the distance traversed by the body up to this moment.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-14.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        First of all let us draw the free body diagram for the small body of mass $m$ and indicate $x$ - axis along the horizontal plane and $y$-axis, perpendicular to it, as shown in the figure. Let the block breaks off the plane at $t=t_{{0}}$ i.e. $N=0$
        So, $N=m g-a t_{{0}} \\sin \\alpha=0$ or,
        $t_{{0}}=\\frac{{m g}}{{a \\sin \\alpha}} \\quad (1)$
        From $F_{{x}}=m w_{{x}}$, for the body under investigation :
        $m d \\psi / d t=a t \\cos \\alpha ;$ Integrating within the limits for $v(t)$
        $m \\int_{{0}}^{{v}} d v_{{x}}=a \\cos \\alpha \\int_{{0}} t d t \\quad \\text {{ (using Eq. 1) }}$
        So, $v=\\frac{{d s}}{{d t}}=\\frac{{a \\cos \\alpha}}{{2 m}} t^{{2}} \\quad (2)$
        Integrating, Eqn. (2) for $s(t)$
        $s=\\frac{{a \\cos \\alpha}}{{2 m}} \\frac{{t^{{3}}}}{{3}} \\quad (3) $
        Using the value of $t=t_{{0}}$ from Eq. (1), into Eqs. (2) and (3)
        $v=\\frac{{m g^{{2}} \\cos \\alpha}}{{2 a \\sin ^{{2}} \\alpha}} \\text {{ and }} s=\\frac{{m^{{2}} g^{{3}} \\cos \\alpha}}{{6 a^{{2}} \\sin ^{{3}} \\alpha}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
        
    def _problem_1_69(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A bar of mass $m$ resting on a smooth horizontal plane starts moving due to the force $F=m g / 3$ of constant magnitude. In the process of its rectilinear motion the angle $\\alpha$ between the direction of this force and the horizontal varies as $\\alpha=a s$, where $a$ is a constant, and $s$ is the distance traversed by the bar from its initial position. Find the velocity of the bar as a function of the angle $\\alpha$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Newton's second law of motion in projection form, along horizontal or $x$-axis i.e. $F_x=m w_x$ gives.
        $\\displaystyle F \\cos (a s)=m v \\frac{{d v}}{{d s}} \\quad(\\text{{as }}\\quad \\alpha=a s)$
        or, $F \\cos (a s) d s=m v dv$ Integrating, over the limits for $v(s)$
        $\\displaystyle \\begin{{gathered}} \\frac{{F}}{{m}} \\int_0^{{\\infty}} \\cos (a s) d s=\\frac{{v^2}}{{2}} \\\\ \\text {{ or }} \\quad v=\\sqrt{{\\frac{{2 F \\sin \\alpha}}{{m a}}}} \\\\ =\\sqrt{{2 g \\sin \\alpha / 3 a}} \\text {{ (using }} F=\\frac{{m g}}{{3}} \\text {{ ) }} \\end{{gathered}}$
        $\\displaystyle =\\sqrt{{2 g \\sin \\alpha / 3 a}} \\text {{ (using }} F=\\frac{{m g}}{{3}} \\text {{ ) }}$
        which is the sought relationship.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
        
    def _problem_1_70(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A horizontal plane with the coefficient of friction $k$ supports two bodies: a bar and an electric motor with a battery on a block. A thread attached to the bar is wound on the shaft of the electric motor. The distance between the bar and the electric motor is equal to $l$. When the motor is switched on, the bar, whose mass is twice as great as that of the other body, starts moving with a constant acceleration $w$. How soon will the bodies collide?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        From the Newton's second law in projection from :
        For the bar,
        $T-2 k m g=(2 m) w \\qquad (1)$
        For the motor,
        $T-k m g=m w^{{\\prime}} \\qquad (2)$
        Now, from the equation of kinematics in the frame of bar or motor :
        $\\displaystyle l=\\frac{{1}}{{2}}\\left(w+w^{{\\prime}}\\right) t^2 \\qquad (3)$
        From (1), (2) and (3) we get on eliminating $T$ and $w^{{\\prime}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_71(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A pulley fixed to the ceiling of an elevator car carries a thread whose ends are attached to the loads of masses $m_1$ and $m_2$. The car starts going up with an acceleration $\\mathbf{{w}}_0$. Assuming the masses of the pulley and the thread, as well as the friction, to be negligible find:
        (a) the acceleration of the load $m_1$ relative to the elevator shaft and relative to the car;
        (b) the force exerted by the pulley on the ceiling of the car.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Let us write Newton's second law in vector from $\\vec{{F}}=m \\vec{{w}}$, for both the blocks (in the frame of ground).
        $\\begin{{aligned}} & \\vec{{T}}+m_1 \\vec{{g}}=m_1 \\overrightarrow{{w_1}} \\qquad (1) \\\\ & \\vec{{T}}+m_2 \\vec{{g}}=m_2 \\overrightarrow{{w_2}} \\qquad (2) \\end{{aligned}}$
        These two equations contain three unknown quantities $\\vec{{w}}_1, \\overrightarrow{{w_2}}$ and $\\vec{{T}}$. The third equation is provided by the kinematic relationship between the accelerations :
        $\\overrightarrow{{w_1}}=\\overrightarrow{{w_0}}+\\vec{{w}}, \\overrightarrow{{w_2}}=\\overrightarrow{{w_0}}-\\vec{{w}}^{{\\prime}} \\qquad (3)$
        where $\\vec{{w}}^{{\\prime}}$ is th acceleration of the mass $m_1$ with respect to the pulley or elevator car. Summing up termwise the left hand and the right-hand sides of these kinematical equations, we get
        $\\overrightarrow{{w_1}}+\\overrightarrow{{w_2}}=2 \\overrightarrow{{w_0}} \\qquad (4)$
        The simultaneous solution of Eqs•(1), (2) and (4) yields
        $\\displaystyle \\vec{{w}}_1=\\frac{{\\left(m_1-m_2\\right) \\vec{{g}}+2 m_2 \\vec{{w}}_0}}{{m_{{1+}} m_2}}$
        Using this result in Eq. (3), we get,
        $\\displaystyle \\vec{{w}}=\\frac{{m_1-m_2}}{{m_1+m_2}}\\left(\\vec{{g}}-\\vec{{w}}_0\\right) \\quad \\text {{ and }} \\quad \\vec{{T}}=\\frac{{2 m_1 m_2}}{{m_1+m_2}}\\left(\\overrightarrow{{w_0}}-\\vec{{g}}\\right)$
        Using the results in Eq. (3) we get $\\displaystyle \\vec{{w}}^{{\\prime}}=\\frac{{m_1-m_2}}{{m_1+m_2}}\\left(\\vec{{g}}-\\vec{{w}}_0\\right)$
        (b) obviously the force exerted by the pulley on the celing of the car
        $\\displaystyle \\vec{{F}}=-2 \\vec{{T}}=\\frac{{4 m_1 m_2}}{{m_1+m_2}}\\left(\\vec{{g}}-\\overrightarrow{{w_0}}\\right)$
        Note : one could also solve this problem in the frame of elevator car.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_72(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the acceleration $w$ of body 2 in the arrangement shown in Fig. 1.15, if its mass is $\\eta$ times as great as the mass of bar 1 and the angle that the inclined plane forms with the horizontal is equal to $\\alpha$. The masses of the pulleys and the threads, as well as the friction, are assumed to be negligible. Look into possible cases.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-15.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us write Newton's second law for both, bar 1 and body 2 in terms of projection having taken the positive direction of $x_1$ and $x_2$ as shown in the figure and assuming that body 2 starts sliding, say, upward along the incline
        $\\begin{{aligned}} T_1-m_1 g \\sin \\alpha & =m_1 w_1 \\\\ m_2 g-T_2 & =m_2 w \\end{{aligned}}$
        For the pulley, moving in vertical direction from the equation $F_x=m w_x$
        $2 T_2-T_1=\\left(m_p\\right) w_1=0$
        (as mass of the pulley $m_p=0$ )
        or $T_1=2 T_2$
        As the length of the threads are constant, the kinematical relationship of accelerations becomes
        $w=2 w_1$
        Simultaneous solutions of all these equations yields :
        $\\displaystyle w=\\frac{{2 g\\left(2 \\frac{{m_2}}{{m_1}}-\\sin \\alpha\\right)}}{{\\left(4 \\frac{{m_2}}{{m_1}}+1\\right)}}=\\frac{{2 g(2 \\eta-\\sin \\alpha)}}{{(4 \\eta+1)}}$
        As $\\eta>1, w$ is directed vertically downward, and hence in vector form
        $\\displaystyle \\vec{{w}}=\\frac{{2 \\vec{{g}}(2 \\eta-\\sin \\alpha)}}{{4 \\eta+1}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_73(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the arrangement shown in Fig. 1.16 the bodies have masses $m_0, m_1, m_2$, the friction is absent, the masses of the pulleys and the threads are negligible. Find the acceleration of the body $m_1$. Look into possible cases.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-16.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us write Newton's second law for masses $m_1$ and $m_2$ and moving pully in vertical direction along positive $x$-axis (Fig.) :
        $\\begin{{gathered}} m_1 g-T=m_1 w_{{1 x}} \\\\ m_2 g-T=m_2 w_{{2 x}} \\\\ T_1-2 T=0(\\text {{ as }} m=0) \\end{{gathered}}$
        or $T_1=2 T$
        Again using Newton's second law in projection form for mass $m_0$ along positive $x_1$ direction (Fig.), we get
        $T_1=m_0 w_0$
        The kinematical relationship between the accelerations of masses gives in terms of projection on the $x$-axis
        $w_{{1 x}}+w_{{2 x}}=2 w_0$
        Simultaneous solution of the obtained five equations yields :
        $\\displaystyle w_1=\\frac{{\\left[4 m_1 m_2+m_0\\left(m_1-m_2\\right)\\right] g}}{{4 m_1 m_2+m_0\\left(m_1+m_2\\right)}}$
        In vector form
        $\\displaystyle \\overrightarrow{{w_1}}=\\frac{{\\left[4 m_1 m_2+m_0\\left(m_1-m_2\\right)\\right] \\vec{{g}}}}{{4 m_1 m_2+m_0\\left(m_1+m_2\\right)}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_74(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the arrangement shown in Fig. 1.17 the mass of the rod $M$ exceeds the mass $m$ of the ball. The ball has an opening permitting it to slide along the thread with some friction. The mass of the pulley and the friction in its axle are negligible. At the initial moment the ball was located opposite the lower end of the rod. When set free, both bodies began moving with constant accelerations. Find the friction force between the ball and the thread if $t$ seconds after the beginning of motion the ball got opposite the upper end of the rod. The rod length equals $l$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-17.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        As the thread is not tied with $m$, so if there were no friction between the thread and the ball $m$, the tension in the thread would be zero and as a result both bodies will have free fall motion. Obviously in the given problem it is the friction force exerted by the ball on the thread, which becomes the tension in the thread. From the condition or language of the problem $w_M>w_m$ and as both are directed downward so, relative acceleration of $M=w_M-w_m$ and is directed downward. Kinematical equation for the ball in the frame of rod in projection form along upward direction gives :
        $\\displaystyle l=\\frac{{1}}{{2}}\\left(w_M-w_m\\right) t^2 \\qquad (1)$
        Newton's second law in projection form along vertically down direction for both, rod and ball gives,
        $\\begin{{aligned}} & M g-f r=M w_M \\qquad (2) \\\\ & m g-f r=m w_m  \\qquad (3) \\end{{aligned}}$
        Multiplying Eq. (2) by $m$ and Eq. (3) by $M$ and then subtracting Eq. (3) from (2) and after using Eq. (1) we get
        $\\displaystyle f r=\\frac{{2 l M m}}{{(M-m) t^2}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_75(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the arrangement shown in Fig. 1.18 the mass of ball 1 is $\\eta=1.8$ times as great as that of rod 2 . The length of the latter is $l=100 \\mathrm{{~cm}}$. The masses of the pulleys and the threads, as well as the friction, are negligible. The ball is set on the same level as the lower end of the rod and then released. How soon will the ball be opposite the upper end of the rod?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-18.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Suppose, the ball goes up with accleration $w_1$ and the rod comes down with the acceleration $w_2$.
        As the length of the thread is constant, $2 w_1=w_2 \\qquad (1)$
        From Newton's second law in projection form along vertically upward for the ball and vertically downward for the rod respectively gives,
        $T-m g=m w_1 \\qquad (2)$
        and $M g-T^{{\\prime}}=M w_2 \\qquad (3)$
        but $T=2 T$ (because pulley is massless) $\\qquad (4)$
        From Eqs. (1), (2), (3) and (4)
        $\\displaystyle w_1=\\frac{{(2 M-m) g}}{{m+4 M}}=\\frac{{(2-\\eta) g}}{{\\eta+4}} \\text {{ (in upward direction) }}$
        and $w_2=\\frac{{2(2-\\eta) g}}{{(\\eta+4)}}$ (downwards)
        From kinematical equation in projection form, we get
        $\\displaystyle l=\\frac{{1}}{{2}}\\left(w_1+w_2\\right) t^2$
        as, $w_1$ and $w_2$ are in the opposite direction.
        Putting the values of $w_1$ and $w_2$, the sought time becomes
        $t=\\sqrt{{2 l(\\eta+4) / 3(2-\\eta) g}}=1.4 \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_76(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the arrangement shown in Fig. 1.19 the mass of body $1$ is $\\eta=4.0$ times as great as that of body 2 . The height $h=20 \\mathrm{{~cm}}$. The masses of the pulleys and the threads, as well as the friction, are negligible. At a certain moment body 2 is released and the arrangement set in motion. What is the maximum height that body 2 will go up to?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-19.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Using Newton's second law in projection form along $x$-axis for the body 1 and along negative $x$ - axis for the body 2 respectively, we get
        $\\begin{{aligned}} & m_1 g-T_1=m_1 w_1 \\quad (1) \\\\ & T_2-m_2 g=m_2 w_2  \\quad (2) \\end{{aligned}}$

        For the pulley lowering in downward direction from Newton's law along $x$ axis,
        $T_1-2 T_2=0 \\text {{ (as pulley is mass less) }}$
        or, $T_1=2 T_2 \\quad (3)$

        As the length of the thread is constant so,
        $w_2=2 w_1 \\quad (4)$
        The simultaneous solution of above equations yields,
        $\\displaystyle w_2=\\frac{{2\\left(m_1-2 m_2\\right) g}}{{4 m_2+m_1}}=\\frac{{2(\\eta-2)}}{{\\eta+4}} \\left(\\text {{ as }} \\frac{{m_1}}{{m_2}}=\\eta\\right) \\quad (4)$

        Obviously during the time interval in which the body 1 comes to the borizontal floor covering the distance $h$, the body 2 moves upward the distance $2 h$. At the moment when the body 2 is at the beight $2 h$ from the floor its velocity is given by the expression :
        $\\displaystyle v_2^2=2 w_2(2 h)=2\\left[\\frac{{2(\\eta-2) g}}{{\\eta+4}}\\right] 2 h=\\frac{{8 h(\\eta-2) g}}{{\\eta+4}}$

        After the body $m_1$ touches the floor the thread becomes slack or the tension in the thread zero, thus as a result body 2 is only under gravity for it's subsequent motion.
        Owing to the velocity $v_2$ at that moment or at the height $2 h$ from the floor, the body 2 further goes up under gravity by the distance,
        $\\displaystyle h^{{\\prime}}=\\frac{{v_2^2}}{{2 g}}=\\frac{{4 h(\\eta-2)}}{{\\eta+4}}$
        Thus the sought maximum height attained by the body 2 :
        $\\displaystyle H=2 h+h^{{\\prime}}=2 h+\\frac{{4 h(\\eta-2)}}{{(\\eta+4)}}=\\frac{{6 \\eta h}}{{\\eta+4}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_77(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the accelerations of rod $A$ and wedge $B$ in the arrangement shown in Fig. 1.20 if the ratio of the mass of the wedge to that of the rod equals $\eta$, and the friction between all contact surfaces is negligible.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-20.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us draw free body diagram of each body, i.e. of $\\operatorname{{rod}} A$ and of wedge $B$ and also draw the kinemetical diagram for accelerations, after analysing the directions of motion of $A$ and $B$. Kinematical relationship of accelarations is :
        $\\displaystyle \\tan \\alpha=\\frac{{w_A}}{{w_B}}$
        Let us write Newton's second law for both bodies in terms of projections having taken positive directions of $y$ and $x$ axes as shown in the figure.
        $m_A g-N \\cos \\alpha=m_A w_A$
        and $N \\sin \\alpha=m_B w_B$
        Simultaneous solution of (1), (2) and (3) yields :
        $\\displaystyle w_A=\\frac{{m_A g \\sin \\alpha}}{{m_A \\sin \\alpha+m_B \\cot \\alpha \\cos \\alpha}}=\\frac{{g}}{{\\left(1+\\eta \\cot ^2 \\alpha\\right)}} \\text {{ and }}$
        $\\displaystyle w_B=\\frac{{w_A}}{{\\tan \\alpha}}=\\frac{{g}}{{(\\tan \\alpha+\\eta \\cot \\alpha)}}$
        Note : We may also solve this problem using conservation of mechanical energy instead of Newton's second law.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_78(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the arrangement shown in Fig. 1.21 the masses of the wedge $M$ and the body $m$ are known. The appreciable friction exists only between the wedge and the body $m$, the friction coefficient being equal to $k$. The masses of the pulley and the thread are negligible. Find the acceleration of the body $m$ relative to the horizontal surface on which the wedge slides.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-21.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us draw free body diagram of each body and fix the coordinate system, as shown in the figure. After analysing the motion of $M$ and $m$ on the basis of force diagrams, let us draw the kinematical diagram for accelerations (Fig.).
        As the length of threads are constant so,
        $d s_{{m M}}=d s_M$ and as $\\overrightarrow{{v_{{m M}}}}$ and $\\overrightarrow{{v_M}}$ do not change their directions that why
        $\\begin{{aligned}} & \\left|\\vec{{w}}_{{m M}}\\right|=\\left|\\vec{{w}}_M\\right|=w \\text {{ (say) and }} \\\\ & \\vec{{w}}_{{m M}} \\uparrow \\uparrow \\overrightarrow{{v_M}} \\text {{ and }} \\vec{{w}}_M \\uparrow \\uparrow \\overrightarrow{{v_M}} \\end{{aligned}}$
        As $\\vec{{w}}_m=\\vec{{w}}_{{m M}}+\\vec{{w}}_M$
        so, from the triangle law of vector addition
        $w_m=\\sqrt{{2}} w \\qquad (1)$
        From the Eq. $F_x=m w_x$, for the wedge and block :
        $\\begin{{aligned}} T-N & =M w, \\qquad (2)\\\\ \\text{{and }} \\quad N & =m w \\qquad (3)\\end{{aligned}}$
        Now, from the Eq. $F_y=m w_y$, for the block
        $m g-T-k N=m w \\qquad (4)$
        Simultaneous solution of Eqs. (2), (3) and (4) yields :
        $\\displaystyle w=\\frac{{m g}}{{(k n+2 m+M)}}=\\frac{{g}}{{(k+2+M / m)}}$
        Hence using Eq. (1)
        $\\displaystyle w_m=\\frac{{g \\sqrt{{2}}}}{{(2+k+M / m)}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_79(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the minimum acceleration with which bar $A$ (Fig. 1.22) should be shifted horizontally to keep bodies 1 and 2 stationary relative to the bar? The masses of the bodies are equal, and the coefficient of friction between the bar and the bodies is equal to $k$. The masses of the pulley and the threads are negligible, the friction in the pulley is absent.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-22.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Using Newton's second law in projection form along $x$-axis for the body 1 and along negative $x$ - axis for the body 2 respectively, we get
        $\\begin{{aligned}} & m_1 g-T_1=m_1 w_1 \\quad (1) \\\\ & T_2-m_2 g=m_2 w_2  \\quad (2) \\end{{aligned}}$

        For the pulley lowering in downward direction from Newton's law along $x$ axis,
        $T_1-2 T_2=0 \\text {{ (as pulley is mass less) }}$
        or, $T_1=2 T_2 \\quad (3)$

        As the length of the thread is constant so,
        $w_2=2 w_1 \\quad (4)$
        The simultaneous solution of above equations yields,
        $\\displaystyle w_2=\\frac{{2\\left(m_1-2 m_2\\right) g}}{{4 m_2+m_1}}=\\frac{{2(\\eta-2)}}{{\\eta+4}} \\left(\\text {{ as }} \\frac{{m_1}}{{m_2}}=\\eta\\right) \\quad (4)$

        Obviously during the time interval in which the body 1 comes to the borizontal floor covering the distance $h$, the body 2 moves upward the distance $2 h$. At the moment when the body 2 is at the beight $2 h$ from the floor its velocity is given by the expression :
        $\\displaystyle v_2^2=2 w_2(2 h)=2\\left[\\frac{{2(\\eta-2) g}}{{\\eta+4}}\\right] 2 h=\\frac{{8 h(\\eta-2) g}}{{\\eta+4}}$

        After the body $m_1$ touches the floor the thread becomes slack or the tension in the thread zero, thus as a result body 2 is only under gravity for it's subsequent motion.
        Owing to the velocity $v_2$ at that moment or at the height $2 h$ from the floor, the body 2 further goes up under gravity by the distance,
        $\\displaystyle h^{{\\prime}}=\\frac{{v_2^2}}{{2 g}}=\\frac{{4 h(\\eta-2)}}{{\\eta+4}}$
        Thus the sought maximum height attained by the body 2 :
        $\\displaystyle H=2 h+h^{{\\prime}}=2 h+\\frac{{4 h(\\eta-2)}}{{(\\eta+4)}}=\\frac{{6 \\eta h}}{{\\eta+4}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_80(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Prism 1 with bar 2 of mass $m$ placed on it gets a horizontal acceleration $w$ directed to the left (Fig. 1.23). At what maximum value of this acceleration will the bar be still stationary relative to the prism, if the coefficient of friction between them $k<\\cot \\alpha$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-23.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        On the basis of the initial argument of the solution of 1.79 , the tendency of bar 2 with respect to 1 will be to move up along the plane.
        Let us fix $(x-y)$ coordinate system in the frame of ground as shown in the figure.
        From second law of motion in projection form along $y$ and $x$ axes :
        $m g \\cos \\alpha-N=m w \\sin \\alpha$
        or, $\\quad N=m(g \\cos \\alpha-w \\sin \\alpha)$
        $m g \\sin \\alpha+f r=m w \\cos \\alpha$
        or, $f r=m(w \\cos \\alpha-g \\sin \\alpha)$
        but $f r \\leq k N$, so from (1) and (2)
        $(w \\cos \\alpha-g \\sin \\alpha) \\leq k(g \\cos \\alpha+w \\sin \\alpha)$
        or, $w(\\cos \\alpha-k \\sin \\alpha) \\leq g(k \\cos \\alpha+\\sin \\alpha)$
        or, $\\displaystyle \\quad w \\leq g \\frac{{(\\cos \\alpha+\\sin \\alpha)}}{{\\cos \\alpha-k \\sin \\alpha}}$,
        So, the sought maximum acceleration of the wedge :
        $\\displaystyle w_{{\\max }}=\\frac{{(k \\cos \\alpha+\\sin \\alpha) g}}{{\\cos \\alpha-k \\sin \\alpha}}=\\frac{{(k \\cot \\alpha+1) g}}{{\\cot \\alpha-k}} \\text {{ where }} \\cot \\alpha>k$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_81(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Prism $1$ of mass $m_1$ and with angle $\\alpha$ (see Fig. 1.23) rests on a horizontal surface. Bar 2 of mass $m_2$ is placed on the prism. Assuming the friction to be negligible, find the acceleration of the prism.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-23.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us draw the force diagram of each body, and on this basis we observe that the prism moves towards right say with an acceleration $\\overrightarrow{{w_1}}$ and the bar 2 of mass $m_2$ moves down the plane with respect to 1 , say with acceleration $\\vec{{w}}_{{21}}$, then, $\\overrightarrow{{w_2}}=\\vec{{w}}_{{21}}+\\vec{{w}}_1$ (Fig.)
        Let us write Newton's second law for both bodies in projection form along positive $y_2$ and $x_1$ axes as shown in the Fig.
        $m_2 g \\cos \\alpha-N=m_2 w_{{2\\left(y_2\\right)}}=m_2\\left[w_{{21\\left(y_2\\right)}}+w_{{1\\left(y_2\\right)}}\\right]=m_2\\left[0+w_1 \\sin \\alpha\\right]$
        or, $m_2 g \\cos \\alpha-N=m_2 w_1 \\sin \\alpha$
        and $N \\sin \\alpha=m_1 w_1$
        Solving (1) and (2), we get
        $\\displaystyle w_1=\\frac{{m_2 g \\sin \\alpha \\cos \\alpha}}{{m_1+m_2 \\sin ^2 \\alpha}}=\\frac{{g \\sin \\alpha \\cos \\alpha}}{{\\left(m_1 / m_2\\right)+\\sin ^2 \\alpha}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_82(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the arrangement shown in Fig. 1.24 the masses $m$ of the bar and $M$ of the wedge, as well as the wedge angle $\\alpha$, are known.The masses of the pulley and the thread are negligible. The friction is absent. Find the acceleration of the wedge $M$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-24.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        To analyse the kinematic relations between the bodies, sketch the force diagram of each body as shown in the figure.
        On the basis of force diagram, it is obvious that the wedge $M$ will move towards right and the block will move down along the wedge. As the length of the thread is constant, the distance travelled by the block on the wedge must be equal to the distance travelled by the wedge on the floor. Hence $d s_{{m M}}=d s_M$ As $\\vec{{v}}_{{m M}}$ and $\\vec{{v}}_M$ do not change their directions and acceleration that's why $\\vec{{w}}_{{m M}} \\uparrow \\uparrow \\vec{{v}}_{{m M}}$ and $\\vec{{w}}_M \\uparrow \\uparrow \\vec{{v}}_M$ and $w_{{m M}}=w_M=w$ (say) and accordingly the diagram of kinematical dependence is shown in figure.

        As $\\vec{{w}}_m=\\vec{{w}}_{{m M}}+\\vec{{w}}_M$, so from triangle law of vector addition.
        $w_m=\\sqrt{{w_M^2+w_{{m M}}^2-2 w_{{m M}} w_M \\cos \\alpha}}=w \\sqrt{{2(1-\\cos \\alpha)}}$
        From $F_x=m w_x$, (for the wedge),
        $T=T \\cos \\alpha+N \\sin \\alpha=M w$
        For the bar $m$ let us fix $(x-y)$ coordinate system in the frame of ground Newton's law in projection form along $x$ and $y$ axes (Fig.) gives

        $\\begin{{aligned}} & m g \\sin \\alpha-T=m w_{{m(x)}}=m\\left[w_{{m M(x)}}+w_{{M(x)}}\\right] \\\\ & =m\\left[w_{{m M}}+w_M \\cos (\\pi-\\alpha)\\right]=m w(1-\\cos \\alpha) \\\\ & m g \\cos \\alpha-N=m w_{{m(y)}}=m\\left[w_{{m M(y)}}+w_{{M(y)}}\\right]=m[0+w \\sin \\alpha] \\\\ &\\end{{aligned}}$
        Solving the above Eqs. simultaneously, we get
        $\\displaystyle w=\\frac{{m g \\sin \\alpha}}{{M+2 m(1-\\cos \\alpha)}}$
        Note : We can study the motion of the block $m$ in the frame of wedge also, alternately we may solve this problem using conservation of mechanical energy.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_83(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A particle of mass $m$ moves along a circle of radius $R$. Find the modulus of the average vector of the force acting on the particle over the distance equal to a quarter of the circle, if the particle moves
        (a) uniformly with velocity $v$;
        (b) with constant tangential acceleration
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Let us sketch the diagram for the motion of the particle of mass $m$ along the circle of radius $R$ and indicate $x$ and $y$ axis, as shown in the figure.
        (a) For the particle, change in momentum $\\Delta \\vec{{p}}=m v(-\\vec{{i}})-m v(\\vec{{j}})$
        so, $\\quad|\\Delta \\vec{{p}}|=\\sqrt{{2}} m v$
        and time taken in describing quarter of the circle,
        $\\displaystyle \\Delta t  =\\frac{{\\pi R}}{{2 v}}$
        Hence, $\\displaystyle \\langle\\vec{{F}}\\rangle =\\frac{{|\\Delta \\vec{{p}}|}}{{\\Delta t}}=\\frac{{\\sqrt{{2}} m v}}{{\\pi R / 2 v}}=\\frac{{2 \\sqrt{{2}} m v^2}}{{\\pi R}}$
        (b) In this case
        $\\displaystyle\\overrightarrow{{p_i}}=0 \\text {{ and }} \\overrightarrow{{p_f}}=m w_t t(-\\vec{{i}}),$
        so $|\\Delta \\vec{{p}}|=m w_t t$
        Hence, $\\displaystyle|<\\vec{{F}}>|=\\frac{{\\mid \\Delta \\overrightarrow{{p \\mid}}}}{{t}}=m w_t$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_84(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An aircraft loops the loop of radius $R=500 \\mathrm{{~m}}$ with a constant velocity $v=$ $=360 \\mathrm{{~km}}$ per hour. Find the weight of the flyer of mass $m=70 \\mathrm{{~kg}}$ in the lower, upper, and middle points of the loop.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        While moving in a loop, normal reaction exerted by the flyer on the loop at different points and uncompensated weight if any contribute to the weight of flyer at those points.
        (a) When the aircraft is at the lowermost point, Newton's second law of motion in projection form $F_n=m w_n$ gives
        $\\displaystyle N-m g=\\frac{{m v^2}}{{R}}$
        or, $\\displaystyle N=m g+\\frac{{m v^2}}{{R}}=2.09 \\mathrm{{kN}}$
        (b) When it is at the upper most point, again from $F_n=m w_n$ we get
        $\\displaystyle N^{{\\prime \\prime}}+m g=\\frac{{m v^2}}{{R}}$
        $\\displaystyle N^{{\\prime \\prime}}=\\frac{{m v^2}}{{R}}-m g=0.7 \\mathrm{{kN}}$
        (c) When the aircraft is at the middle point of the loop, again from $F_n=m w_n$
        $\\displaystyle N^{{\\prime}}=\\frac{{m v^2}}{{R}}=1.4 \\mathrm{{kN}}$
        The uncompensated weight is $m g$. Thus effective weight $=\\sqrt{{{{N^{{\\prime}}}}^2+m^2 g^2}}=1.56 \\mathrm{{kN}}$ acts obliquely.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_85(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A small sphere of mass $m$ suspended by a thread is first taken aside so that the thread forms the right angle with the vertical and then released. Find:
        (a) the total acceleration of the sphere and the thread tension as a function of $\\theta$, the angle of deflection of the thread from the vertical;
        (b) the thread tension at the moment when the vertical component of the sphere's velocity is maximum;
        (c) the angle $\\theta$ between the thread and the vertical at the moment when the total acceleration vector of the sphere is directed horizontally.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Let us depict the forces acting on the small sphere $m$, (at an arbitrary position when the thread makes an angle $\\theta$ from the vertical) and write equation $\\vec{{F}}=m \\vec{{w}}$ via projection on the unit vectors $\\hat{{u}}_t$ and $\\hat{{u}}_n$. From $F_t=m w_t$, we have
        $\\displaystyle \\begin{{gathered}} m g \\sin \\theta=m \\frac{{d v}}{{d t}} \\\\ =m \\frac{{v d v}}{{d s}}=m \\frac{{v d v}}{{l(-d \\theta)}} \\end{{gathered}}$
        (as vertical is refrence line of angular position)
        or $v d v=-g l \\sin \\theta d \\theta$
        Integrating both the sides :
        $\\displaystyle \\int_0^v v d v=-g l \\int_{{\\pi / 2}}^\\theta \\sin \\theta d \\theta$
        or, $\\displaystyle \\frac{{v^2}}{{2}}=g l \\cos \\theta$
        Hence $\\displaystyle \\frac{{v^2}}{{l}}=2 g \\cos \\theta=w_n \\qquad (1)$
        (Eq. (1) can be easily obtained by the conservation of mechanical energy).
        From $\\quad F_n=m w_n$
        $\\displaystyle T-m g \\cos \\theta=\\frac{{m v^2}}{{l}}$
        Using (1) we have
        $T=3 m g \\cos \\theta \\qquad (2)$
        Again from the Eq. $F_t=m w_t$ :
        $m g \\sin \\theta=m w_t$ or $w_t=g \\sin \\theta \\qquad (3)$
        Hence $w=\\sqrt{{w_t^2+w_n^2}}=\\sqrt{{(g \\sin \\theta)^2+(2 g \\cos \\theta)^2}}$ ( using 1 and 3)
        $=g \\sqrt{{1+3 \\cos ^2 \\theta}}$
        (b) Vertical component of velocity, $v_y=v \\sin \\theta$
        So, $v_y^2=v^2 \\sin ^2 \\theta=2 g l \\cos \\theta \\sin ^2 \\theta \\text {{ (using 1) }}$
        For maximum
        $\\displaystule \\begin{{gathered}} v_y \\text {{ or }} v_y^2, \\frac{{d\\left(\\cos \\theta \\sin ^2 \\theta\\right)}}{{d \\theta}}=0 \\\\ \\cos \\theta=\\frac{{1}}{{\\sqrt{{3}}}} \\end{{gathered}}$
        which yields
        Therefore from (2) $T=3 m g \\frac{{1}}{{\\sqrt{{3}}}}=\\sqrt{{3}} \\mathrm{{mg}}$
        (c) We have $\\vec{{w}}=w_t \\hat{{u}}_t+w_n \\hat{{u}}_n$ thus $w_y=w_{{n(y)}}+w_{{n(y)}}$
        But in accordance with the problem $w_y=0$
        So, $w_{{r(y)}}+w_{{n(y)}}=0$
        or, $g \\sin \\theta \\sin \\theta+2 g \\cos ^2 \\theta(-\\cos \\theta)=0$
        or, $\\cos \\theta=\\frac{{1}}{{\\sqrt{{3}}}} \\text {{ or, }} \\theta=54.7^{{\\circ}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_86(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A ball suspended by a thread swings in a vertical plane so that its acceleration values in the extreme and the lowest position are equal. Find the thread deflection angle in the extreme position.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The ball has only normal acceleration at the lowest position and only tangential acceleration at any of the extreme position. Let $v$ be the speed of the ball at its lowest position and $l$ be the length of the thread, then according to the problem
        $\\displaystyle \\frac{{v^2}}{{l}}=g \\sin \\alpha \\qquad (1)$
        where $\\alpha$ is the maximum deflection angle
        From Newton's law in projection form : $F_t=m w_t$
        $\\displaystyle -m g \\sin \\theta=m v \\frac{{d v}}{{l d \\theta}}$
        or, $-g l \\sin \\theta d \\theta=v d v$
        On integrating both the sides within their limits.
        $\\displaystyle -g l \\int_0^\\alpha \\sin \\theta d \\theta=\\int_v^0 v d v$
        or, $v^2=2 g l(1-\\cos \\alpha) \\qquad (2)$
        Note : Eq. (2) can easily be obtained by the conservation of mechanical energy of the ball in the uniform field of gravity.
        From Eqs. (1) and (2) with $\\theta=\\alpha$
        $\\displaystyle 2 g l(1-\\cos \\alpha)=l g \\cos \\alpha$
        or, $\\displaystyle \\cos \\alpha=\\frac{{2}}{{3}} \\text {{ so, }} \\alpha=53^{{\\circ}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_87(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A small body $A$ starts sliding off the top of a smooth sphere of radius $R$. Find the angle $\theta$ (Fig. 1.25) corresponding to the point at which the body breaks off the sphere, as well as the break-off velocity of the body.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-25.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Let us depict the forces acting on the body $\\vec{{A}}$ (which are the force of gravity $m \\vec{{g}}$ and the normal reaction $\\vec{{N}}$ ) and write equation $\\vec{{F}}=m \\vec{{w}}$ via projection on the unit vectors $\\hat{{u}}_t$ and $\\hat{{u}}_n$ (Fig.)
        From $F_t=m w_t$
        $\\displaystyle \\begin{{aligned}} m g \\sin \\theta & =m \\frac{{d v}}{{d t}} \\\\ & =m \\frac{{v d v}}{{d s}}=m \\frac{{v d v}}{{R d \\theta}} \\end{{aligned}}$
        or, $\\quad g R \\sin \\theta d \\theta=v d v$
        Integrating both side for obtaining $v(\\theta)$
        $\\displaystyle \\int_0^\\theta g R \\sin \\theta d \\theta=\\int_0^v v d v$
        or, $v^2=2 g R(1-\\cos \\theta) \\qquad (1)$
        From $F_n=m w_n$
        $\\displaystyle m g \\cos \\theta-N=m \\frac{{v^2}}{{R}} \\qquad (2)$
        At the moment the body loses contact with the surface, $N=0$ and therefore the Eq. (2) becomes
        $v^2=g R \\cos \\theta \\qquad (3)$
        where $v$ and $\\theta$ correspond to the moment when the body loses contact with the surface.
        Solving Eqs. (1) and (3) we obtain $\\displaystyle \\cos \\theta=\\frac{{2}}{{3}}$ or, $\\displaystyle \\theta=\\cos ^{{-1}}(2 / 3)$ and $v=\\sqrt{{2 g R / 3}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_88(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A device (Fig. 1.26) consists of a smooth L-shaped rod located in a horizontal plane and a sleeve $A$ of mass $m$ attached by a weight-less spring to a point $B$. The spring stiffness is equal to $x$. The whole system rotates with a constant angular velocity $\\omega$ about a vertical axis passing through the point $O$. Find the elongation of the spring. How is the result affected by the rotation direction?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-26.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        At first draw the free body diagram of the device as, shown. The forces, acting on the sleeve are it's weight, acting vertically downward, spring force, along the length of the spring and normal reaction by the rod, perpendicular to its length.
        Let $F$ be the spring force, and $\\Delta l$ be the elongation.
        From, $F_n=m w_n:$
        $N \\sin \\theta+F \\cos \\theta=m \\omega^2 r \\qquad (1)$
        where $r \\cos \\theta=\\left(l_0+\\Delta l\\right)$.
        Similarly from $F_t=m w_t$
        $N \\cos \\theta-F \\sin \\theta=0 \\text {{ or, }} N=F \\sin \\theta / \\cos \\theta \\qquad (2)$
        From (1) and (2)
        $\\begin{{gathered}} F(\\sin \\theta / \\cos \\theta) \\cdot \\sin \\theta+F \\cos \\theta=m \\omega^2 r \\\\ =m \\omega^2\\left(l_0+\\Delta l\\right) / \\cos \\theta \\end{{gathered}}$
        On putting $F=\\kappa \\Delta l$
        $\\kappa \\Delta l \\sin ^2 \\theta+\\kappa \\Delta l \\cos ^2 \\theta=m \\omega^2\\left(l_0+\\Delta l\\right)$
        on solving, we get,
        $\\displaystyle \\Delta l=m \\omega^2 \\frac{{l_0}}{{\\kappa-m \\omega^2}}=\\frac{{l_0}}{{\\left(\\kappa / m \\omega^2-1\\right)}}$
        and it is independent of the direction of rotation.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_89(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A cyclist rides along the circumference of a circular horizontal plane of radius $R$, the friction coefficient being dependent only on distance $r$ from the centre $O$ of the plane as $k=k_0(1-r / R)$, where $k_0$ is a constant. Find the radius of the circle with the centre at the point along which the cyclist can ride with the maximum velocity. What is this velocity?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        According to the question, the cyclist moves along the circular path and the centripetal force is provided by the frictional force. Thus from the equation $F_n=m w_n$
        $\\displaystyle f r=\\frac{{m v^2}}{{r}} \\text {{ or }} k m g=\\frac{{m v^2}}{{r}}$
        or $\\displaystyle k_0\\left(1-\\frac{{r}}{{R}}\\right) g=\\frac{{v^2}}{{r}} \\text {{ or }} v^2=k_0\\left(r-r^2 / R\\right) g$
        For $v_{{\\max }}$, we should have $\\displaystyle \\frac{{d\\left(r-\\frac{{r^2}}{{R}}\\right)}}{{d r}}=0$ or,
        $\\displaystyle 1-\\frac{{2 r}}{{R}}=0, \\text {{ so }} r=R / 2$
        Hence $\\displaystyle v_{{\\max }}=\\frac{{1}}{{2}} \\sqrt{{k_0 g R}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_90(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A car moves with a constant tangential acceleration $w_\\tau=$ $=0.62 \\mathrm{{~m}} / \\mathrm{{s}}^2$ along a horizontal surface circumscribing a circle of radius $R=40 \\mathrm{{~m}}$. The coefficient of sliding friction between the wheels of the car and the surface is $k=0.20$. What distance will the car ride without sliding if at the initial moment of time its velocity is equal to zero?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        As initial velocity is zero thus
        $v^2=2 w_t s$
        As $w_t>0$ the speed of the car increases with time or distance. Till the moment, sliding starts, the static friction provides the required centripetal acceleration to the car.
        Thus $f r=m w, \\text {{ but }} f r \\leq k m g$
        $\\begin{{aligned}} &\\text{{So, }} \\qquad &w^2 \\leq k^2 g^2 \\text {{ or, }} w_t^2+\\frac{{v^2}}{{R}} \\leq k^2 g^2 \\\\ &\\text{{or, }} \\qquad &v^2 \\leq\\left(k^2 g^2-w_t^2\\right) R \\\\ &\\text{{Hence, }} \\qquad &v_{{\\max }}=\\sqrt{{\\left(k^2 g^2-w_t^2\\right) R}} \\end{{aligned}}$
        so, from Eqn. (1), the sought distance $s=\\frac{{v_{{\\max }}^2}}{{2 w_t}}=\\frac{{1}}{{2}} \\sqrt{{\\left(\\frac{{k g}}{{w_t}}\\right)^2-1}}=60 \\mathrm{{~m}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_91(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A car moves uniformly along a horizontal sine curve $y=$ $=a \\sin (x / \\alpha)$, where $a$ and $\\alpha$ are certain constants. The coefficient of friction between the wheels and the road is equal to $k$. At what velocity will the car ride without sliding?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Since the car follows a curve, so the maximum velocity at which it can ride without sliding at the point of minimum radius of curvature is the sought velocity and obviously in this case the static friction between the car and the road is limiting.
        Hence from the equation $F_n=m w$
        $\\displaystyle \\begin{{gathered}} k m g \\geq \\frac{{m v^2}}{{R}} \\text {{ or }} v \\leq \\sqrt{{k R g}} \\\\ v_{{\\max }}=\\sqrt{{k R_{{\\min }} g}}  \\qquad (1) \\end{{gathered}}$
        We know that, radius of curvature for a curve at any point $(x, y)$ is given as,
        $\\displaystyle R=\\left|\\frac{{\\left[1+(d y / d x)^2\\right]^{{3 / 2}}}}{{\\left(d^2 y\\right) / d x^2}}\\right| \\qquad (2)$
        For the given curve,
        $\\displaystyle \\frac{{d y}}{{d x}}=\\frac{{a}}{{\\alpha}} \\cos \\left(\\frac{{x}}{{\\alpha}}\\right) \\text {{ and }} \\frac{{d^2 y}}{{d x^2}}=\\frac{{-a}}{{\\alpha^2}} \\sin \\frac{{x}}{{\\alpha}}$
        Substituting this value in (2) we get,
        $\\displaystyle R=\\frac{{\\left[1+\\left(a^2 / \\alpha^2\\right) \\cos ^2(x / \\alpha)\\right]^{{3 / 2}}}}{{\\left(a / \\alpha^2\\right) \\sin (x / \\alpha)}}$
        For the minimum $\\displaystyle R, \\frac{{x}}{{\\alpha}}=\\frac{{\\pi}}{{2}}$ and therefore, corresponding radius of curvature
        $\\displaystyle R_{{\\min }}=\\frac{{\\alpha^2}}{{a}} \\qquad (3)$
        Hence from (1) and (2)
        $v_{{\\max }}=\\alpha \\sqrt{{\\mathrm{{kg}} / \\mathrm{{a}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_92(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A chain of mass $m$ forming a circle of radius $R$ is slipped on a smooth round cone with half-angle $\\theta$. Find the tension of the chain if it rotates with a constant angular velocity $\\omega$ about a vertical axis coinciding with the symmetry axis of the cone.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The sought tensile stress acts on each element of the chain. Hence divide the chain into small, similar elements so that each element may be assumed as a particle. We consider one such element of mass $d m$, which subtends angle $d \\alpha$ at the centre. The chain moves along a circle of known radius $R$ with a known angular speed $\\omega$ and certain forces act on it. We have to find one of these forces.
        From Newton's second law in projection form, $F_x=m w_x$ we get
        $2 T \\sin (d \\alpha / 2)-d N \\cos \\theta=d m \\omega^2 R$
        and from $F_z=m w_z$ we get
        $d N \\sin \\theta=g d m$
        Then putting $d m=m d \\alpha / 2 \\pi$ and $\\sin (d \\alpha / 2)=d \\alpha / 2$ and solving, we get,
        $\\displaystyle T=\\frac{{m\\left(\\omega^2 R+g \\cot \\theta\\right)}}{{2 \\pi}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_93(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A fixed pulley carries a weightless thread with masses $m_1$ and $m_2$ at its ends. There is friction between the thread and the pulley. It is such that the thread starts slipping when the ratio $m_2 / m_1=$ $=\\eta_0$. Find:
        (a) the friction coefficient;
        (b) the acceleration of the masses when $m_2 / m_1=\\eta>\\eta_0$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Let, us consider a small element of the thread and draw free body diagram for this element. (a) Applying Newton's second law of motion in projection form, $F_n=m w_n$ for this element, $(T+d T) \\sin (d \\theta / 2)+T \\sin (d \\theta / 2)-d N=d m \\omega^2 R=0$
        or, $\\displaystyle 2 T \\sin (d \\theta / 2)=d N$, [negelecting the $\\operatorname{{term}}(d T \\sin d \\theta / 2)$ ] $\\qquad (1)$
        or, $\\displaystyle T d \\theta=d N, \\text {{ as }} \\sin \\frac{{d \\theta}}{{2}}=\\frac{{d \\theta}}{{2}} \\qquad (2)$
        Also, $d f r=k d N=(T+d T)-T=d T$
        From Eqs. (1) and (2),
        $\\displaystyle k T d \\theta=d T \\text {{ or }} \\frac{{d T}}{{T}}=k d \\theta$
        In this case $Q=\\pi$ so, 
        or, or, $\\displaystyle \\ln \\frac{{T_2}}{{T_1}}=k \\pi \\qquad (3)$
        So, $\\displaystyle \\quad k=\\frac{{1}}{{\\pi}} \\ln \\frac{{T_2}}{{T_1}}=\\frac{{1}}{{\\pi}} \\ln \\eta_0$
        as $\\displaystyle \\quad \\frac{{T_2}}{{T_1}}=\\frac{{m_2 g}}{{m_1 g}}=\\frac{{m_2}}{{m_1}}=\\eta_0$
        (b) When $\\displaystyle \\frac{{m_2}}{{m_1}}=\\eta$, which is greater than $\\eta_0$, the blocks will move with same value of acceleration. (say $w$ ) and clearly $m_2$ moves downward. From Newton's second law in projection form (downward for $m_2$ and upward for $m_1$ ) we get :
        $\\displaystyle \\begin{{aligned}} & m_2 g-T_2=m_2 w \\qquad (4)\\\\ \\text{{and }} \\quad & T_1-m_1 g=m_1 w \\qquad (5)\\end{{aligned}}$
        Also $\\displaystyle \\frac{{T_2}}{{T_1}}=\\eta_0$
        Simultaneous solution of Eqs. (4), (5) and (6) yields :
        $\\displaystyle w=\\frac{{\\left(m_2-\\eta_0 m_1\\right) g}}{{\\left(m_2+\\eta_0 m_1\\right)}}=\\frac{{\\left(\\eta-\\eta_0\\right)}}{{\\left(\\eta+\\eta_0\\right)}} g\\left(\\text {{ as }} \\frac{{m_2}}{{m_1}}=\\eta\\right)$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_94(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A particle of mass $m$ moves along the internal smooth surface of a vertical cylinder of radius $R$. Find the force with which the particle acts on the cylinder wall if at the initial moment of time its velocity equals $v_0$ and forms an angle $\\alpha$ with the horizontal.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The force with which the cylinder wall acts on the particle will provide centripetal force necessary for the motion of the particle, and since there is no acceleration acting in the horizontal direction, horizontal component of the velocity will remain constant througout the motion.
        So $\\displaystyle v_x=v_0 \\cos \\alpha$
        Using, $F_n=m w_n$, for the particle of mass $m$,
        $\\displaystyle N=\\frac{{m v_x^2}}{{R}}=\\frac{{m v_0^2 \\cos ^2 \\alpha}}{{R}}$
        which is the required normal force.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_95(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the magnitude and direction of the force acting on the particle of mass $m$ during its motion in the plane $x y$ according to the law $x=a \\sin \\omega t, y=b \\cos \\omega t$, where $a, b$, and $\\omega$ are constants.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Obviously the radius vector describing the position of the particle relative to the origin of coordinate is
        $\\vec{{r}}=x \\vec{{i}}+y \\vec{{j}}=a \\sin \\omega t \\vec{{i}}+b \\cos \\omega t \\vec{{j}}$
        Differentiating twice with respect the time :
        $\\displaystyle \\begin{{aligned}} & \\vec{{w}}=\\frac{{d^2 \\vec{{r}}}}{{d t^2}}=-\\omega^2(a \\sin \\omega t \\vec{{i}}+b \\cos \\omega t \\vec{{j}})=-\\omega^2 \\vec{{r}} \\\\ \\text{{Thus }} \\qquad & \\vec{{F}}=m \\vec{{w}}=-m \\omega^2 \\vec{{r}} \\end{{aligned}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_96(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A body of mass $m$ is thrown at an angle to the horizontal with the initial velocity $\\mathbf{{v}}_0$. Assuming the air drag to be negligible, find:
        (a) the momentum increment $\\Delta \\mathbf{{p}}$ that the body acquires over the first $t$ seconds of motion;
        (b) the modulus of the momentum increment $\\Delta \\mathbf{{p}}$ during the total time of motion.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\begin{{aligned}}\\text{{(a) We have }}\\Delta \\vec{{p}}&=\\int \\vec{{F}} d t \\\\ &=\\int_0^t m \\vec{{g}} d t=m \\vec{{g}} t \\end{{aligned}}$
        (b) Using the solution of problem $1.28(\\mathrm{{~b}})$, the total time of motion, $\\tau=-\\frac{{2\\left(\\overrightarrow{{v_0 \\cdot g}}\\right)}}{{g^2}}$
        $\\begin{{gathered}} \\text{{Hence using }} \\qquad t=\\tau \\text {{ in (1) }} \\\\ |\\Delta \\vec{{p}}|=m g \\tau \\\\ =-2 m\\left(\\overrightarrow{{v_0}} \\cdot \\vec{{g}}\\right) / g\\left(\\overrightarrow{{v_0}} \\cdot \\vec{{g}} \\text {{ is }}-v e\\right) \\end{{gathered}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_97(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        At the moment $t=0$ a stationary particle of mass $m$ experiences a time-dependent force $\\mathbf{{F}}=\\mathbf{{a}}t(\\tau-t)$, where $\\mathbf{{a}}$ is a constant vector, $\\tau$ is the time during which the given force acts. Find:
        (a) the momentum of the particle when the action of the force discontinued;
        (b) the distance covered by the particle while the force acted.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        From the equation of the given time dependence force $\\vec{{F}}=\\vec{{a}} t(\\tau-t)$ at $t=\\tau$, the force vanishes,
        (a) Thus $\\displaystyle \\Delta \\vec{{p}}=\\vec{{p}}=\\int_0^\\tau \\vec{{F}} d t$
        $\\displaystyle \\begin{{aligned}} &\\text{{or, }} \\qquad \\vec{{p}}=\\int_0^\\tau \\overrightarrow{{a}} t(\\tau-t) d t \\frac{{\\overrightarrow{{a}}\\tau^3}}{{6}} \\\\ &\\text{{but }} \\qquad \\vec{{p}}=m \\overrightarrow{{v}} \\text {{ so }} \\vec{{v}}=\\frac{{\\overrightarrow{{a}}\\tau^3}}{{6 m}} \\end{{aligned}}$
        (b) Again from the equation $\\vec{{F}}=m \\vec{{w}}$
        $\\displaystyle \\qquad \\overrightarrow{{a t}}(\\tau-t)=m \\frac{{d \\vec{{v}}}}{{d t}}$
        or, $\\displaystyle \\qquad \\vec{{a}}\\left(t \\tau-t^2\\right) d t=m d \\vec{{v}}$
        Integrating within the limits for $\\vec{{v}}(t)$,
        $\\displaystyle \\begin{{gathered}} \\int_0^t \\vec{{a}}\\left(t \\tau-t^2\\right) d t=m \\int_0^{{\\vec{{v}}}} d \\vec{{v}} \\\\ \\vec{{v}}=\\frac{{\\vec{{a}}}}{{m}}\\left(\\frac{{\\tau t^2}}{{2}}-\\frac{{t^3}}{{3}}\\right)=\\frac{{\\vec{{a}} t^2}}{{m}}\\left(\\frac{{\\tau}}{{2}}-\\frac{{t}}{{3}}\\right) \\\\ v=\\frac{{a t^2}}{{m}}\\left(\\frac{{\\tau}}{{2}}-\\frac{{t}}{{3}}\\right) \\text {{ for }} t \\leq \\tau \\end{{gathered}}$
        Thus $\\displaystyle v=\\frac{{a t^2}}{{m}}\\left(\\frac{{\\tau}}{{2}}-\\frac{{t}}{{3}}\\right) \\text {{ for }} t \\leq \\tau$
        Hence distance covered during the time interval $t=\\tau$,
        $\\displaystyle \\begin{{aligned}} s & =\\int_0^{{\\tau}} v d t \\\\ & =\\int_0^\\tau \\frac{{a t^2}}{{m}}\\left(\\frac{{\\tau}}{{2}}-\\frac{{t}}{{3}}\\right) d t=\\frac{{a}}{{m}} \\frac{{\\tau^4}}{{12}}\\end{{aligned}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_98(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        At the moment $t=0$ a particle of mass $m$ starts moving due to a force $F=F_0 \\sin \\omega t$, where $F_0$ and $\\omega$ are constants. Find the distance covered by the particle as a function of $t$. Draw the approximate plot of this function.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        We have $F=F_0 \\sin \\omega t$
        or $\\displaystyle m \\frac{{d \\vec{{v}}}}{{d t}}=\\vec{{F}}_0 \\sin \\omega t \\text {{ or }} m d \\vec{{v}}=\\vec{{F}}_0 \\sin \\omega t d t$
        On integrating,
        $\\displaystyle \\overrightarrow{{m v}}=\\frac{{-\\overrightarrow{{F_0}}}}{{\\omega}} \\cos \\omega t+C,(\\text {{ where }} C \\text {{ is integration constant) }}$
        When $\\displaystyle t=0, v=0, \\text {{ so }} C=\\frac{{\\overrightarrow{{F_0}}}}{{m \\omega}}$
        Hence, $\\displaystyle \\vec{{v}}=\\frac{{-\\overrightarrow{{F_0}}}}{{m \\omega}} \\cos \\omega t+\\frac{{\\overrightarrow{{F_0}}}}{{m \\omega}}$
        As $\\mid \\cos \\omega t \\leq 1$ so, $v=\\frac{{F_0}}{{m \\omega}}(1-\\cos \\omega t)$
        $\\begin{{aligned}} \\text{{Thus }} \\qquad s&=\\int_0^t v d t \\\\ &=\\frac{{F_0 t}}{{m \\omega}}-\\frac{{F_0 \\sin \\omega t}}{{m \\omega^2}}=\\frac{{F_0}}{{m \\omega^2}}(\\omega t-\\sin \\omega t) \\end{{aligned}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_99(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        At the moment $t=0$ a particle of mass $m$ starts moving due to a force $\\mathbf{{F}}=\\mathbf{{F}}_0 \\cos \\omega t$, where $\\mathbf{{F}}_0$ and $\\omega$ are constants. How long will it be moving until it stops for the first time? What distance will it traverse during that time? What is the maximum velocity of the particle over this distance?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        According to the problem, the force acting on the particle of mass $m$ is, $\\vec{{F}}=\\overrightarrow{{F_0}} \\cos \\omega t$
        So, $\\displaystyle m \\frac{{d \\vec{{v}}}}{{d t}}=\\overrightarrow{{F_0}} \\cos \\omega t \\quad \\text {{ or }} d \\vec{{v}}=\\frac{{\\overrightarrow{{F_0}}}}{{m}} \\cos \\omega t d t$
        Integrating, within the limits.
        $\\displaystyle \\int_0^{{\\vec{{v}}}} d \\vec{{v}}=\\frac{{F_0}}{{m}} \\int_0^t \\cos \\omega t d t \\text {{ or }} \\vec{{v}}=\\frac{{\\overrightarrow{{F_0}}}}{{m \\omega}} \\sin \\omega t$
        It is clear from equation (1), that after starting at $t=0$, the particle comes to rest fro the first time at $t=\\frac{{\\pi}}{{\\omega}}$.
        From Eq. (1), $\\displaystyle v=|\\overrightarrow{{v}}|=\\frac{{F_0}}{{m \\omega}} \\sin \\omega t$ for $t \\leq \\frac{{\\pi}}{{\\omega}}$
        Thus during the time interval $t=\\pi / \\omega$, the sought distance
        $\\displaystyle s=\\frac{{F_0}}{{m w}} \\int_0^{{\\pi / \\omega}} \\sin \\omega t d t=\\frac{{2 F}}{{m \\omega^2}}$
        From Eq. (1)
        $\\displaystyle v_{{\\max }}=\\frac{{F_0}}{{m \\omega}} \\text {{ as }}|\\sin \\omega t| \\leq 1$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_100(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A motorboat of mass $m$ moves along a lake with velocity $v_0$. At the moment $t=0$ the engine of the boat is shut down. Assuming the resistance of water to be proportional to the velocity of the boat $\\mathbf{{F}}=-r \\mathbf{{v}}$, find
        (a) how long the motorboat moved with the shutdown engine;
        (b) the velocity of the motorboat as a function of the distance covered with the shutdown engine, as well as the total distance covered till the complete stop;
        (c) the mean velocity of the motorboat over the time interval (beginning with the moment $t=0$ ), during which its velocity decreases $\\eta$ times.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        (a) From the problem $\\displaystyle \\vec{{F}}=-\\overrightarrow{{r v}} \\text {{ so }} m \\frac{{d \\vec{{v}}}}{{d t}}=-\\overrightarrow{{r v}}$
        Thus $\\displaystyle m \\frac{{d v}}{{d t}}=-r \\nu[\\text {{ as }} d \\overrightarrow{{v \\uparrow \\downarrow}} \\overrightarrow{{v]}}$
        or, $\\displaystyle \\frac{{d v}}{{v}}=-\\frac{{r}}{{m}} d t$
        On integrating $\\displaystyle \\ln v=-\\frac{{r}}{{m}} t+C$
        But at $t=0, v=v_0, \\text {{ so, }} C=\\ln v_0$
        or, $\\displaystyle \\ln \\frac{{v}}{{v_0}}=-\\frac{{r}}{{m}} t \\text {{ or, }} v=v_0 e^{{-\\frac{{r}}{{m}} t}}$
        Thus for $t \\rightarrow \\infty, v=0$
        (b) We have $\\displaystyle m \\frac{{d v}}{{d t}}=-r v$ so $d v=\\frac{{-r}}{{m}} d s$
        Integrating within the given limits to obtain $v(s)$ :
        or, $\\displaystyle \\int_{{v_0}}^v d v=-\\frac{{r}}{{m}} \\int_0^s d s \\text {{ or }} v=v_0-\\frac{{r s}}{{m}}$
        Thus for $\\displaystyle v=0, s=s_{{\\text {{total }}}}=\\frac{{m v_0}}{{r}}$
        (c) Let we have $\\displaystyle \\frac{{m d v}}{{v}}=-r v$ or $\\displaystyle \\frac{{d v}}{{v}}=\\frac{{-r}}{{m}} d t$
        or, $\\displaystyle \\int_0^{{v o / \\eta}} \\frac{{d v}}{{v}}=\\frac{{-r}}{{m}} \\int_0^t d t, \\text {{ or, }} \\ln \\frac{{v_0}}{{\\eta v_0}}=-\\frac{{r}}{{m}} t$
        So $\\displaystyle t=\\frac{{-m \\ln (1 / \\eta)}}{{r}}=\\frac{{m \\ln \\eta}}{{r}}$
        Now, average velocity over this time interval,
        $\\displaystyle \\langle v\\rangle=\\frac{{\\int v d t}}{{\\int d t}}=\\frac{{\\int_0^{{\\frac{{m}}{{r}} \\ln \\eta}} v_0 e^{{-\\frac{{n}}{{m}}}} d t}}{{\\frac{{m}}{{r}} \\ln \\eta}}=\\frac{{v_0(\\eta-1)}}{{\\eta \\ln \\eta}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_101(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Having gone through a plank of thickness $h$, a bullet changed its velocity from $v_0$ to $v$. Find the time of motion of the bullet in the plank, assuming the resistance force to be proportional to the square of the velocity.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        According to the problem
        $\\displaystyle m \\frac{{d v}}{{d t}}=-k v^2 \\text {{ or, }} m \\frac{{d v}}{{v^2}}=-k d t$
        Integrating, withing the limits,
        $\\displaystyle \\int_{{v_0}}^v \\frac{{d v}}{{v^2}}=-\\frac{{k}}{{m}} \\int_0^t d t \\text {{ or, }} t=\\frac{{m}}{{k}} \\frac{{\\left(v_0-v\\right)}}{{v_0 v}} \\qquad (1)$
        To find the value of $k$, rewrite
        $\\displaystyle m v \\frac{{d v}}{{d s}}=-k v^2 \\text {{ or, }} \\frac{{d v}}{{v}}=-\\frac{{k}}{{m}} d s$
        On integrating
        $\\displaystyle \\int_{{v_0}}^v \\frac{{d v}}{{v}}=-\\frac{{k}}{{m}} \\int_0^h d s$
        So, $\\displaystyle k=\\frac{{m}}{{h}} \\ln \\frac{{v_0}}{{v}} \\qquad (2)$
        Putting the value of $k$ from (2) in (1), we get
        $\\displaystyle t=\\frac{{h\\left(v_0-v\\right)}}{{v_0 v \\ln \\frac{{v_0}}{{v}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_102(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A small bar starts sliding down an inclined plane forming an angle $\\alpha$ with the horizontal. The friction coefficient depends on the distance $x$ covered as $k=a x$, where $a$ is a constant. Find the distance covered by the bar till it stops, and its maximum velocity over this distance.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        From Newton's second law for the bar in projection from, $F_x=m w_x$ along $x$ direction we get
        $m g \\sin \\alpha-k m g \\cos \\alpha=m w$
        or, $\\displaystyle v \\frac{{d v}}{{d x}}=g \\sin \\alpha-a x g \\cos \\alpha \\text {{, (as }} k=a x \\text {{ ), }}$
        or, $\\displaystyle v d v=(g \\sin \\alpha-a x g \\cos \\alpha) d x$
        or, $\\displaystyle \\int_0^v v d v=g \\int_0^x(\\sin \\alpha-x \\cos \\alpha) d x$
        So, $\\displaystyle \\quad \\frac{{v^2}}{{2}}=g\\left(\\sin \\alpha x-\\frac{{x^2}}{{2}} a \\cos \\alpha\\right) \\qquad (1)$
        From (1) $v=0$ at either
        $x=0, \\text {{ or }} x=\\frac{{2}}{{a}} \\tan \\alpha$
        As the motion of the bar is unidirectional it stops after going through a distance of $\\frac{{2}}{{a}} \\tan \\alpha$.
        From (1), for $v_{{\\max }}$
        $\\displaystyle \\frac{{d}}{{d x}}\\left(\\sin \\alpha x-\\frac{{x^2}}{{2}} a \\cos \\alpha\\right)=0 \\text {{, which yields }} x=\\frac{{1}}{{a}} \\tan \\alpha$
        Hence, the maximum velocity will be at the distance, $x=\\tan \\alpha / a$ Putting this value of $x$ in (1) the maximum velocity,
        $\\displaystyle v_{{\\max }}=\\sqrt{{\\frac{{g \\sin \\alpha \\tan \\alpha}}{{a}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_103(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A body of mass $m$ rests on a horizontal plane with the friction coefficient $k$. At the moment $t=0$ a horizontal force is applied to it, which varies with time as $\\mathbf{{F}}=\\mathbf{{a}} t$, where $\\mathbf{{a}}$ is a constant vector. Find the distance traversed by the body during the first $t$ seconds after the force action began.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Since, the applied force is proportional to the time and the frictional force also exists, the motion does not start just after applying the force. The body starts its motion when $F$ equals the limiting friction.
        Let the motion start after time $t_0$, then
        $\\displaystyle F=a t_0=k m g \\text {{ or, }} t_0=\\frac{{k m g}}{{a}}$
        So, for $t=s t_0$, the body remains at rest and for $t>t_0$ obviously
        $\\displaystyle \\frac{{m d v}}{{d t}}=a\\left(t-t_0\\right) \\quad \\text {{ or, }} \\quad m d v=a\\left(t-t_0\\right) d t$
        Integrating, and noting $v=0$ at $t=t_0$, we have for $t>t_0$
        $\\displaystyle \\int_0^v m d v=a \\int_{{t_0}}^t\\left(t-t_0\\right) d t \\text {{ or }} v=\\frac{{a}}{{2 m}}\\left(t-t_0\\right)^2$
        Thus $\\displaystyle s=\\int v d t=\\frac{{a}}{{2 m}} \\int_{{t_0}}^t\\left(t-t_0\\right)^2 d t=\\frac{{a}}{{6 m}}\\left(t-t_0\\right)^3$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_104(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A body of mass $m$ is thrown straight up with velocity $v_0$. Find the velocity $v^{{\\prime}}$ with which the body comes down if the air drag equals $k v^2$, where $k$ is a constant and $v$ is the velocity of the body.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        While going upward; from Newton's second law in vertical direction :
        $\\displaystyle m \\frac{{v d v}}{{d s}}=-\\left(m g+k v^2\\right) \\text {{ or }} \\frac{{v d v}}{{\\left(g+\\frac{{k v^2}}{{m}}\\right)}}=-d s$
        At the maximum height $h$, the speed $v=0$, so
        $\\displaystyle \\int_{{v_0}}^0 \\frac{{v d v}}{{g+\\left(k v^2 / m\\right)}}=-\\int_0^h d s$
        Integrating and solving, we get,
        $\\displaystyle h=\\frac{{m}}{{2 k}} \\ln \\left(1+\\frac{{k v_o^2}}{{m g}}\\right)$
        When the body falls downward, the net force acting on the body in downward direction equals $\\left(m g-k v^2\\right)$
        Hence net acceleration, in downward direction, according to second law of motion
        $\\displaystyle\\frac{{v d v}}{{d s}}=g-\\frac{{k v^2}}{{m}} \\text {{ or, }} \\frac{{v d v}}{{g-\\frac{{k v^2}}{{m}}}}=d s$
        Thus $\\displaystyle \\int_0^{{v^{{\\prime}}}} \\frac{{v d v}}{{g-k v^2 / m}}=\\int_0^h d s$
        Integrating and putting the value of $h$ from (1), we get,
        $v^{{\\prime}}=v_0 / \\sqrt{{1+k v_0^2 / m g}} .$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_105(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A particle of mass $m$ moves in a certain plane $P$ due to a force $F$ whose magnitude is constant and whose vector rotates in that plane with a constant angular velocity $\omega$. Assuming the particle to be stationary at the moment $t=0$, find:
        (a) its velocity as a function of time;
        (b) the distance covered by the particle between two successive stops, and the mean velocity over this time.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Let us fix $x-y$ co-ordinate system to the given plane, taking $x$-axis in the direction along which the force vector was oriented at the moment $t=0$, then the fundamental equation of dynamics expressed via the projection on $x$ and $y$-axes gives,
        $\\displaystyle \\begin{{aligned}} & F \\cos \\omega t=m \\frac{{d v_x}}{{d t}} \\\\ & F \\sin \\omega t=m \\frac{{d v_y}}{{d t}} \\end{{aligned}}$
        (a) Using the condition $v(0)=0$, we obtain 
        $\\displaystyle \\begin{{aligned}} & v_x=\\frac{{F}}{{m \\omega}} \\sin \\omega t \\\\ & v_y=\\frac{{F}}{{m \\omega}}(1-\\cos \\omega t) \\end{{aligned}}$
        Hence, $\\displaystyle v=\\sqrt{{v_x^2+v_y^2}}=\\left(\\frac{{2 F}}{{m \\omega}}\\right)\\left|\\sin \\left(\\frac{{\\omega t}}{{2}}\\right)\\right|$
        (b) It is seen from this that the velocity $v$ turns into zero after the time interval $\\Delta t$, which can be found from the relation, $\\displaystyle\\omega \\frac{{\\Delta t}}{{2}}=\\pi$. Consequentely, the sought distance, is
        $\\displaystyle s=\\int_0^{{\\Delta t}} v d t=\\frac{{8 F}}{{m \\omega^2}}$
        Average velocity, $\\displaystyle \\langle v\\rangle=\\frac{{\\int v d t}}{{\\int d t}}$
        So, $\\displaystyle \\langle v\\rangle=\\int_0^{{2 \\pi / \\omega}} \\frac{{2 F}}{{m \\omega}} \\sin \\left(\\frac{{\\omega t}}{{2}}\\right) d t /(2 \\pi \\omega)=\\frac{{4 F}}{{\\pi m \\omega}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_106(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A small disc $A$ is placed on an inclined plane forming an angle $\\alpha$ with the horizontal (Fig. 1.27) and is imparted an initial velocity $v_0$. Find how the velocity of the disc depends on the angle $\\varphi$ if the friction coefficient $k=\\tan \\alpha$ and at the initial moment $\\varphi_0=$ $=\\pi / 2$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-27.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The acceleration of the disc along the plane is determined by the projection of the force of gravity on this plane $F_x=m g \\sin \\alpha$ and the friction force $f r=k m g \\cos \\alpha$. In our case $k=\\tan \\alpha$ and therefore
        $f r=F_x=m g \\sin \\alpha$
        Let us find the projection of the acceleration on the derection of the tangent to the trajectory and on the $x$-axis :
        $\\begin{{aligned}} & m w_t=F_x \\cos \\varphi-f r=m g \\sin \\alpha(\\cos \\varphi-1) \\\\ & m w_x=F_x-f r \\cos \\varphi=m g \\sin \\alpha(1-\\cos \\varphi) \\end{{aligned}}$
        It is seen fromthis that $w_t=-w_x$, which means that the velocity $v$ and its projection $v_x$ differ only by a constant value $C$ which does not change with time, i.e.
        $v=v_x+C$
        where $v_x=v \\cos \\varphi$. The constant $C$ is found from the initial condition $v=v_0$, whence $C=v_0$ since $\\displaystyle \\varphi=\\frac{{\\pi}}{{2}}$ initially. Finally we obtain
        $v=v_0 /(1+\\cos \\varphi)$
        In the cource of time $\\varphi \\rightarrow 0$ and $v \\rightarrow v_0 / 2$. (Motion then is unaccelerated.)
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_107(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A chain of length $l$ is placed on a smooth spherical surface of radius $R$ with one of its ends fixed at the top of the sphere. What will be the acceleration $w$ of each element of the chain when its upper end is released? It is assumed that the length of the chain $l<\\frac{{1}}{{2}} \\pi R$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Let us consider an element of length $d s$ at an angle $\\varphi$ from the vertical diameter. As the speed of this element is zero at initial instant of time, it's centripetal acceleration is zero, and hence, $d N-\\lambda d s \\cos \\varphi=0$, where $\\lambda$ is the linear mass density of the chain Let $T$ and $T+d T$ be the tension at the upper and the lower ends of $d s$. we have from, $F_t=m w_t$
        $\\qquad (T+d T)+\\lambda d s g \\sin \\varphi-T=\\lambda d s w_t$
        or, $\\quad d T+\\lambda R d \\varphi g \\sin \\varphi=\\lambda d s w_t$
        If we sum the above equation for all elements, the term $\\displaystyle \\int d T=0$ because there is no tension at the free ends, so
        $\\displaystyle \\begin{{aligned}} & \\lambda g R \\int_0^{{l / R}} \\sin \\varphi d \\varphi=\\lambda w_t \\int d s=\\lambda l w_t \\\\ & \\text {{Hence }} \\quad w_t=\\frac{{g R}}{{l}}\\left(1-\\cos \\frac{{l}}{{R}}\\right) \\end{{aligned}}$
        As $w_n=a$ at initial moment
        So, $\\displaystyle w=\\left|w_t\\right|=\\frac{{g R}}{{l}}\\left(1-\\cos \\frac{{l}}{{R}}\\right)$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_108(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A small body is placed on the top of a smooth sphere of radius $R$. Then the sphere is imparted a constant acceleration $w_0$ in the horizontal direction and the body begins sliding down. Find:
        (a) the velocity of the body relative to the sphere at the moment of break-off;
        (b) the angle $\\theta_0$ between the vertical and the radius vector drawn from the centre of the sphere to the break-off point; calculate $\\theta_0$ for $w_0=g$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        In the problem, we require the velocity of the body, realtive to the sphere, which itself moves with an acceleration $w_0$ in horizontal direction (say towards left). Hence it is advisible to solve the problem in the frame of sphere (non-inertial frame).
        At an arbitary moment, when the body is at an angle $\\theta$ with the vertical, we sketch the force diagram for the body and write the second law of motion in projection form $F_n=m w_n$
        or, $\\displaystyle m g \\cos \\theta-N-m w_0 \\sin \\theta=\\frac{{m v^2}}{{R}} \\qquad (1)$
        At the break off point, $N=0, \\theta=\\theta_0$ and let $v=v_0$ so the Eq. (1) becomes,
        $\\displaystyle \\frac{{v_0^2}}{{R}}=g \\cos \\theta_0-w_0 \\sin \\theta_0 \\qquad (2)$
        From, $F_t=m w_t$
        $\\displaystyle m g \\sin \\theta-m w_0 \\cos \\theta=m \\frac{{v d v}}{{d s}}=m \\frac{{v d v}}{{R d \\theta}}$
        or, $v d v=R\\left(g \\sin \\theta+w_0 \\cos \\theta\\right) d \\theta$
        $\\begin{{aligned}} \\text {{ Integrating, }} \\int_0^{{v_0}} v d v&=\\int_0^{{\\theta_0}} R\\left(g \\sin \\theta+w_0 \\cos \\theta\\right) d \\theta \\\\ \\frac{{v_0^2}}{{2 R}}&=g\\left(1-\\cos \\theta_0\\right)+w_0 \\sin \\theta_0  \\qquad (3) \\end{{aligned}}$
        Note that the Eq. (3) can also be obtained by the work-energy theorem $A=\\Delta T$ (in the frame of sphere)
        therefore, $\\displaystyle \\quad m g R\\left(1-\\cos \\theta_0\\right)+m w_0 R \\sin \\theta_0=\\frac{{1}}{{2}} m v_0^2$
        [here $m w_0 R \\sin \\theta_0$ is the work done by the pseudoforce $\\left(-m \\vec{{w}}_o\\right)$ ]
        or, $\\displaystyle \\frac{{v_0^2}}{{2 R}}=g\\left(1-\\cos \\theta_0\\right)+w_0 \\sin \\theta_0$
        Solving Eqs. (2) and (3) we get,
        $\\displaystyle v_0=\\sqrt{{2 g R / 3}}$ and $\\theta_0=\\cos ^{{-1}}\\left[\\frac{{2+k \\sqrt{{5+9 k^2}}}}{{3\\left(1+k^2\\right)}}\\right]$, where $k=\\frac{{w_0}}{{g}}$
        Hence $\\left.\\theta_0\\right|_{{w_0-z}}=17^{{\\circ}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_109(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A particle moves in a plane under the action of a force which is always perpendicular to the particle's velocity and depends on a distance to a certain point on the plane as $1 / r^n$, where $n$ is a constant. At what value of $n$ will the motion of the particle along the circle be steady?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This is not central force problem unless the path is a circle about the said point. Rather here $F_t$ (tangential force) vanishes. Thus equation of motion becomes,
        $\\begin{{gathered}} v_t=v_0=\\text {{ constant }} \\\\ \\text{{and, }} \\qquad \\frac{{m v_0^2}}{{r}}=\\frac{{A}}{{r^2}} \\text {{ for }} r=r_0 \\end{{gathered}}$
        We can consider the latter equation as the equilibrium under two forces. When the motion is perturbed, we write $r=r_0+x$ and the net force acting on the particle is,
        $\\displaystyle -\\frac{{A}}{{\\left(r_0+x\\right)^n}}+\\frac{{m v_0^2}}{{r_0+x}}=\\frac{{-A}}{{r_0^n}}\\left(1-\\frac{{n x}}{{r_0}}\\right)+\\frac{{m v_0^2}}{{r_0}}\\left(1-\\frac{{x}}{{r_0}}\\right)=-\\frac{{m v_0^2}}{{r_0^2}}(1-n) x$
        This is opposite to the displacement $x$, if $n<1 \\cdot\\left(\\frac{{m v_0^2}}{{r}}\\right.$ is an outward directed centrifugul force while $\\frac{{-A}}{{r^n}}$ is thé inward directed external force).
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_110(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A sleeve $A$ can slide freely along a smooth rod bent in the shape of a half-circle of radius $R$ (Fig. 1.28). The system is set in rotation with a constant angular velocity $\\omega$ about a vertical axis $O O^{{\\prime}}$. Find the angle $\\theta$ corresponding to the steady position of the sleeve.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'problems-in-general-physics-fig-1-28.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        There are two forces on the sleeve, the weight $F_1$ and the centrifugal force $F_2$. We resolve both forces into tangential and normal component then the net downward tangential force on the sleeve is,
        $\\displaystyle m g \\sin \\theta\\left(1-\\frac{{\\omega^2 R}}{{g}} \\cos \\theta\\right)$
        This vanishes for $\\theta=0$ and for $\\displaystyle \\theta=\\theta_0=\\cos ^{{-1}}\\left(\\frac{{g}}{{\\omega^2 R}}\\right)$, which is real if $\\omega^2 R>g$. If $\\omega^2 R<g$, then $\\displaystyle 1-\\frac{{\\omega^2 R}}{{g}} \\cos \\theta$ is always positive for small values of $\\theta$ and hence the net tangential force near $\\theta=0$ opposes any displacement away from it. $\\theta=0$ is then stable.
        If $\\displaystyle \\omega^2 R>g, 1-\\frac{{\\omega^2 R}}{{g}} \\cos \\theta$ is negative for small $\\theta$ near $\\theta=0$ and $\\theta=0$ is then unstable.
        However $\\theta=\\theta_0$ is stable because the force tends to bring the sleeve near the equilibrium position $\\theta=\\theta_0$
        If $\\displaystyle \\omega^2 R=g$, the two positions coincide and becomes a stable equilibrium point.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_111(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A rifle was aimed at the vertical line on the target located precisely in the northern direction, and then fired. Assuming the air drag to be negligible, find how much off the line, and in what direction, will the bullet hit the target. The shot was fired in the horizontal direction at the latitude $\\varphi=60^{{\\circ}}$, the bullet velocity $v=900 \\mathrm{{~m}} / \\mathrm{{s}}$, and the distance from the target equals $s=1.0 \\mathrm{{~km}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Define the axes as shown with $z$ along the local vertical, $x$ due east and $y$ due north. (We assume we are in the northern hemisphere). Then the Coriolis force has the components.
        $\\begin{{gathered}} \\vec{{F}}_{{c o r}}=-2 m(\\vec{{\\omega}} \\times \\vec{{v}}) \\\\ \\left.=2 m \\omega\\left[v_y \\cos \\theta-v_z \\sin \\theta\\right) \\vec{{i}}-v_x \\cos \\theta \\vec{{j}}+v_x \\cos \\theta \\vec{{k}}\\right]=2 m \\omega\\left(v_y \\cos \\theta-v_z \\sin \\theta\\right) \\vec{{i}} \\\\ \\text {{ since }} v_x \\text {{ is small when the direction in which the gun is fired is due north }}\\end{{gathered}}$
        since $v_x$ is small when the direction in which the gun is fired is due north. Thus the equation of motion (neglecting centrifugal forces) are
        $\\ddot{{x}}=2 m \\omega\\left(v_y \\sin \\varphi-v_z \\cos \\varphi\\right), \\ddot{{y}}=0 \\text {{ and }} \\ddot{{z}}=-g$
        Integrating we get $\\dot{{y}}=v$ (constant), $\\dot{{z}}=-g t$ and $\\dot{{x}}=2 \\omega v \\sin \\varphi t+\\operatorname{{\\omega og}} t^2 \\cos \\varphi$
        Finally,
        $\\displaystyle x=\\omega v t^2 \\sin \\varphi+\\frac{{1}}{{3}} \\omega g t^3 \\cos \\varphi$
        Now $v>>t$ in the present case. so,
        $\\begin{{aligned}} x &\\approx \\omega v \\sin \\varphi\\left(\\frac{{s}}{{v}}\\right)^2=\\omega \\sin \\varphi \\frac{{s^2}}{{v}} \\\\ &\\approx 7 \\mathrm{{~cm}} \\text{{(to the east).}} \\end{{aligned}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_112(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A horizontal disc rotates with a constant angular velocity $\\omega=6.0 \\mathrm{{rad}} / \\mathrm{{s}}$ about a vertical axis passing through its centre. A small body of mass $m=0.50 \\mathrm{{~kg}}$ moves along a diameter of the disc with a velocity $v^{{\\prime}}=50 \\mathrm{{~cm}} / \\mathrm{{s}}$ which is constant relative to the disc. Find the force that the disc exerts on the body at the moment when it is located at the distance $r=30 \\mathrm{{~cm}}$ from the rotation axis.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The disc exerts three forces which are mutually perpendicular. They are the reaction of the weight, $m g$, vertically upward, the Coriolis force $2 m v^{{\\prime}} \\omega$ perpendicular to the plane of the vertical and along the diameter, and $m \\omega^2 r$ outward along the diameter. The resultant force is,
        $F=m \\sqrt{{g^2+\\omega^4 r^2+\\left(2 v^{{\\prime}} \\omega\\right)^2}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_113(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A horizontal smooth rod $A B$ rotates with a constant angular velocity $\\omega=2.00 \\mathrm{{rad}} / \\mathrm{{s}}$ about a vertical axis passing through its end $A$. A freely sliding sleeve of mass $m=0.50 \\mathrm{{~kg}}$ moves along the rod from the point $A$ with the initial velocity $v_0=1.00 \\mathrm{{~m}} / \\mathrm{{s}}$. Find the Coriolis force acting on the sleeve (in the reference frame fixed to the rotating rod) at the moment when the sleeve is located at the distance $r=50 \\mathrm{{~cm}}$ from the rotation axis.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The sleeve is free to slide along the rod $A B$. Thus only the centrifugal force acts on it. The equation is,
        $\\displaystyle m \\dot{{v}}=m \\omega^2 r \\text {{ where }} v=\\frac{{d r}}{{d t}} .$
        $\\displaystyle \\text {{ But }} \\dot{{v}}=v \\frac{{d v}}{{d r}}=\\frac{{d}}{{d r}}\\left(\\frac{{1}}{{2}} v^2\\right)$
        $\\displaystyle \\begin{{aligned}} &\\text{{So, }} \\qquad \\frac{{1}}{{2}} v^2 =\\frac{{1}}{{2}} \\omega^2 r^2+\\text {{ constant }} \\\\ &\\text{{or, }} \\qquad v^2 =v_0^2+\\omega^2 r^2 \\end{{aligned}}$
        $v_0$ being the initial velocity when $r=0$. The Coriolis force is then,
        $\\begin{{aligned}} 2 m \\omega & \\sqrt{{v_0^2+\\omega^2 r^2}}=2 m \\omega^2 r \\sqrt{{1+v_0^2 / \\omega^2 r^2}} \\\\ & =2.83 \\mathrm{{~N}} \\text {{ on putting the values. }} \\end{{aligned}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_114(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A horizontal disc of radius $R$ rotates with a constant angular velocity $\\omega$ about a stationary vertical axis passing through its edge. Along the circumference of the disc a particle of mass $m$ moves with a velocity that is constant relative to the disc. At the moment when the particle is at the maximum distance from the rotation axis, the resultant of the inertial forces $F_{{i n}}$ acting on the particle in the reference frame fixed to the disc turns into zero. Find:
        (a) the acceleration $w^{{\\prime}}$ of the particle relative to the disc;
        (b) the dependence of $F_{{\\text {{in }}}}$ on the distance from the rotation axis.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The disc $O B A C$ is rotating with angular velocity $\\omega$ about the axis $O O^{{\\prime}}$ passing through the edge point $O$. The equation of motion in rotating frame is,
        $m \\vec{{w}}=\\vec{{F}}+m \\omega^2 \\vec{{R}}+2 m \\vec{{v}} \\times \\vec{{\\omega}}=\\vec{{F}}+\\vec{{F}}_{{i n}}$
        where $\\vec{{F}}_{{\\text {{in }}}}$ is the resultant inertial forcu (pseudo force) which is the vector sum of centrifugal and Coriolis forces.
        (a) At $A, F_{{i n}}$ vanishes. Thus $0=-2 m \\omega^2 R \\hat{{n}}+2 m v^{{\\prime}} \\omega \\hat{{n}}$
        where $\\hat{{n}}$ is the inward drawn unit vector to the centre from the point in question, here $A$. 
        Thus, $v^{{\\prime}}=\\omega R$
        so, $\\displaystyle w=\\frac{{v^{{\\prime 2}}}}{{\\rho}}=\\frac{{v^{{\\prime 2}}}}{{R}}=\\omega^2 R .$
        (b) At $B$
        $\\vec{{F}}_{{i n}}=m \\omega^2 \\overrightarrow{{O C}}+m \\omega^2 \\overrightarrow{{B C}}$
        its magnitude is $m \\omega^2 \\sqrt{{4 R^2-r^2}}$, where $r=O B$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_115(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A small body of mass $m=0.30 \\mathrm{{~kg}}$ starts sliding down from the top of a smooth sphere of radius $R=1.00 \\mathrm{{~m}}$. The sphere rotates with a constant angular velocity $\\omega=6.0 \\mathrm{{rad}} / \\mathrm{{s}}$ about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The equation of motion in the rotating coordinate system is,
        $m \\vec{{w}}=\\vec{{F}}+m \\omega^2 \\vec{{R}}+2 m(\\vec{{v}} \\times \\vec{{\\omega}})$
        Now, $\\cdot \\vec{{v}}=R \\theta \\overrightarrow{{e_\\theta}}+R \\sin \\theta \\dot{{\\varphi}} \\overrightarrow{{e_{{\\varphi}}}}$
        and $\\quad \\vec{{w}}=w^{{\\prime}} \\cos \\theta \\overrightarrow{{e_r}}-w^{{\\prime}} \\sin \\theta \\overrightarrow{{e_\\theta}}$
        $\\frac{{1}}{{2 m}} \\vec{{F}}_{{c o r}}=\\left|\\begin{{array}}{{ccc}}
        \\overrightarrow{{e_r}} & \\overrightarrow{{e_\\theta}} & \\overrightarrow{{e_{{\\varphi}}}} \\\\
        0 & R \\dot{{\\theta}} & R \\sin \\theta \\dot{{\\varphi}} \\\\
        \\omega \\cos \\theta & -\\omega \\sin \\theta & 0
        \\end{{array}}\\right|$
        $=\\overrightarrow{{e_r}}\\left(\\omega R \\sin ^2 \\theta \\dot{{\\varphi}}\\right)+\\omega R \\sin \\theta \\cos \\theta \\dot{{\\varphi}} \\overrightarrow{{e_\\theta}}-\\omega R \\theta \\cos \\theta \\overrightarrow{{e_\\theta}}$
        Now on the sphere,
        $\\begin{{aligned}}
        & \\vec{{v}}=\\left(-R \\dot{{\\theta}}^2-R \\sin ^2 \\theta \\varphi^2\\right) \\overrightarrow{{e_r}} \\\\
        & +\\left(R \\dot{{\\theta}}^*-R \\sin \\theta \\cos \\theta \\dot{{\\varphi}}^2\\right) \\overrightarrow{{e_\\theta}} \\\\
        & +(R \\sin \\theta \\ddot{{\\varphi}}+2 R \\cos \\theta \\dot{{\\theta}} \\dot{{\\varphi}}) \\overrightarrow{{e_{{\\varphi}}}}
        \\end{{aligned}}$
        Thus the equation of motion are, $m\\left(-R \\dot{{\\theta}}^2-R \\sin ^2 \\theta \\dot{{\\varphi}}^2\\right)=N-m g \\cos \\theta+m \\omega^2 R \\sin ^2 \\theta+2 m \\omega R \\sin ^2 \\theta \\dot{{\\varphi}}$
        $m\\left(R \\dot{{\\theta}}^{{\\circ}}-R \\sin \\theta \\cos \\theta \\dot{{\\varphi}}^2\\right)=m g \\sin \\theta+m \\omega^2 R \\sin \\theta \\cos \\theta+2 m \\omega R \\sin \\theta \\cos \\theta \\dot{{\\varphi}}$
        $m(R \\sin \\theta \\ddot{{\\varphi}}+2 R \\cos \\theta \\dot{{\\theta}} \\dot{{\\varphi}})=-2 m \\omega R \\dot{{\\theta}} \\cos \\theta$
        From the third equation, we get, $\\dot{{\\varphi}}=-\\omega$
        A result that is easy to understant by considering the motion in non-rotating frame. The eliminating $\\dot{{\\varphi}}$ we get,
        $\\begin{{aligned}}
        & m R \\dot{{\\theta}}^2=m g \\cos \\theta-N \\\\
        & m R \\dot{{\\theta}}^{{\\circ}}=m g \\sin \\theta
        \\end{{aligned}}$
        Integrating the last equation,
        $\\frac{{1}}{{2}} m R \\dot{{\\theta}}^2=m g(1-\\cos \\theta)$
        Hence $N=(3-2 \\cos \\theta) m g$
        So the body must fly off for $\\theta=\\theta_0=\\cos ^{{-1}} \\frac{{2}}{{3}}$, exactly as if the sphere were nonrotating.
        Now, at this point $F_{{c f}}=$ centrifugal force $=m \\omega^2 R \\sin \\theta_0=\\sqrt{{\\frac{{5}}{{9}}}} m \\omega^2 R$
        $\\begin{{aligned}}
        F_{{c o r}} & =\\sqrt{{\\omega^2 R^2 \\theta^2 \\cos ^2 \\theta+\\left(\\omega^2 R^2\\right)^2 \\sin ^2 \\theta}} \\times 2 m \\\\
        & =\\sqrt{{\\frac{{5}}{{9}}\\left(\\omega^2 R\\right)^2+\\omega^2 R^2 \\times \\frac{{4}}{{9}} \\times \\frac{{2 g}}{{3 R}}}} \\times 2 m=\\frac{{2}}{{3}} m \\omega^2 R \\sqrt{{5+\\frac{{8 g}}{{3 \\omega^2 R}}}}
        \\end{{aligned}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_116(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A train of mass $m=2000$ tons moves in the latitude $\\varphi=$ $=60^{{\\circ}}$ North. Find:
        (a) the magnitude and direction of the lateral force that the train exerts on the rails if it moves along a meridian with a velocity $v=$ $=54 \\mathrm{{~km}}$ per hour;
        (b) in what direction and with what velocity the train should move for the resultant of the inertial forces acting on the train in the reference frame fixed to the Earth to be equal to zero.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        (a) When the train is moving along a meridian only the Coriolis force has a lateral component and its magnitude (see the previous problem) is,
        $\\begin{{aligned}} &2 m \\omega v \\cos \\theta=2 m \\omega \\sin \\lambda \\\\ &(\\text{{Here we have put}} R \\dot{{\\theta}} \\rightarrow v ) \\\\ \\text{{So, }} \\qquad F_{{\\text {{lateral }}}}&=2 \\times 2000 \\times 10^3 \\times \\frac{{2 \\pi}}{{86400}} \\times \\frac{{54000}}{{3600}} \\times \\frac{{\\sqrt{{3}}}}{{2}} \\\\ &=3.77 \\mathrm{{kN}} \\text{{(we write }} \\lambda \\text{{ for the latitude)}}
        \\end{{aligned}}$ 
        (b) The resultant of the inertial forces acting on the train is,
        $\\vec{{F}}_{{i n}}=-2 m \\omega R \\theta \\cos \\theta \\overrightarrow{{e_{{\\varphi}}}}$
        $+\\left(m \\omega^2 R \\sin \\theta \\cos \\theta+2 m \\omega R \\sin \\theta \\cos \\theta \\dot{{\\varphi}}\\right) \\overrightarrow{{e_\\theta}}$
        $+\\left(m \\omega^2 R \\sin ^2 \\theta+2 m \\omega R \\sin ^2 \\theta \\dot{{\\varphi}}\\right) \\overrightarrow{{e_r}}$
        This vanishes if $\\displaystyle \\theta=0, \\dot{{\\varphi}}=-\\frac{{1}}{{2}} \\omega$
        Thus $\\displaystyle \\vec{{v}}=v_{{\\varphi}} \\vec{{e}}_{{\\varphi}}, v_{{\\varphi}}=-\\frac{{1}}{{2}} \\omega R \\sin \\theta=-\\frac{{1}}{{2}} \\omega R \\cos \\lambda$
        (We write $\\dot{{\\lambda}}$ for the latitude here)
        Thus the train must move from the east to west along the $60^{{\\text {{th }}}}$ parallel with a speed, $\\displaystyle \\frac{{1}}{{2}} \\omega R \\cos \\lambda=\\frac{{1}}{{4}} \\times \\frac{{2 \\pi}}{{8 \\cdot 64}} \\times 10^{{-4}} \\times 6.37 \\times 10^6=115.8 \\mathrm{{~m}} / \\mathrm{{s}} \\approx 417 \\mathrm{{~km}} / \\mathrm{{hr}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _problem_1_117(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        At the equator a stationary (relative to the Earth) body falls down from the height $h=500 \\mathrm{{~m}}$. Assuming the air drag to be negligible, find how much off the vertical, and in what direction, the body will deviate when it hits the ground.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        We go to the equation given in 1.111. Here $v_y=0$ so we can take $y=0$, thus we get for the motion in the $x \\dot{{z}}$ plane.
        $\\ddot{{x}}=-2 \\omega v_z \\cos \\theta$
        and $\\ddot{{z}}=-g$
        Integrating, $\\displaystyle z=-\\frac{{1}}{{2}} g t^2$
        $\\displaystyle \\dot{{x}}=\\omega g \\cos \\varphi t^2$
        So $\\displaystyle x=\\frac{{1}}{{3}} \\omega g \\cos \\varphi t^3=\\frac{{1}}{{3}} \\omega g \\cos \\varphi\\left(\\frac{{2 h}}{{g}}\\right)^{{3 / 2}}$
        $\\displaystyle =\\frac{{2 \\omega h}}{{3}} \\cos \\varphi \\sqrt{{\\frac{{2 h}}{{g}}}}$
        There is thus a displacement to the east of
        $\\displaystyle \\frac{{2}}{{3}} \\times \\frac{{2 \\pi}}{{8}} 64 \\times 500 \\times 1 \\times \\sqrt{{\\frac{{400}}{{9.8}}}} \\approx 26 \\mathrm{{~cm}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate
    
    def _init_file_path(self, file_name):
        import pathlib
        from xv.util import locatePath
        rootFolder = pathlib.Path(__file__).parent.absolute()
        file_path = locatePath(file_name, 
                               parent_depth = 6,
                               rootFolder = rootFolder,
                               part_match = False,
                               search_type = 'file',
                               verbose = self.verbose,
                               )
        return os.path.join(rootFolder, file_path)