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Question Number 215608 by ajfour last updated on 11/Jan/25

Commented by ajfour last updated on 11/Jan/25

The charged masses released as shown. Find their speeds when they get in line. I wonder if this gets periodic for certain appropriate ratios!

$${The}\:{charged}\:{masses}\:{released}\:{as}\:{shown}. \\ $$$${Find}\:{their}\:{speeds}\:{when}\:{they}\:{get}\:{in}\:{line}. \\ $$$${I}\:{wonder}\:{if}\:{this}\:{gets}\:{periodic}\:{for}\:{certain} \\ $$$${appropriate}\:{ratios}! \\ $$

Commented by mahdipoor last updated on 11/Jan/25

why negative mass rising up?

$${why}\:{negative}\:{mass}\:{rising}\:{up}? \\ $$

Commented by ajfour last updated on 11/Jan/25

come?on + one pulls it upwards while the negative pair repel..

$${come}?{on}\:+\:{one}\:{pulls}\:{it}\:{upwards}\:{while} \\ $$$${the}\:{negative}\:{pair}\:{repel}.. \\ $$

Answered by mr W last updated on 12/Jan/25

Commented by mr W last updated on 12/Jan/25

say at time t the postion of M is (0, y_1 ) and the position of m is (±x_2 , y_2 ). d_1 ^2 =x_2 ^2 +(y_1 −y_2 )^2 d_2 ^2 =4x_2 ^2 F_1 =((k_e Qq)/d_1 ^2 )=((k_e Qq)/(x_2 ^2 +(y_1 −y_2 )^2 )) F_2 =((k_e q^2 )/d_2 ^2 )=((k_e q^2 )/(4x_2 ^2 )) V=−(dy_1 /dt) (↓) A=(dV/dt)=−(d^2 y_1 /dt^2 ) u=(dx_2 /dt) (→) v=(dy_2 /dt) (↿) a_x =(du/dt)=(d^2 x_2 /dt^2 ) a_y =(dv/dt)=(d^2 y_2 /dt^2 ) MA=Mg+2F_1 sin θ ⇒(d^2 y_1 /dt^2 )=−g−((2k_e Qq(y_1 −y_2 ))/(M[x_2 ^2 +(y_1 −y_2 )^2 ]^(3/2) )) ma_y =F_1 sin θ−mg ⇒(d^2 y_2 /dt^2 )=((k_e Qq(y_1 −y_2 ))/(m[x_2 ^2 +(y_1 −y_2 )^2 ]^(3/2) ))−g ma_x =F_2 −F_1 cos θ ⇒(d^2 x_2 /dt^2 )=((k_e q^2 )/(4mx_2 ^2 ))−((k_e Qqx_2 )/(m[x_2 ^2 +(y_1 −y_2 )^2 ]^(3/2) )) at t=0: y_1 =b x_2 =a, y_2 =0 V=−(dy_1 /dt)=0 u=(dx_2 /dt)=0,v= (dy_2 /dt)=0 ..... ============= i don′t think the problem is solvable.

$${say}\:{at}\:{time}\:{t}\:{the}\:{postion}\:{of}\:{M}\:{is} \\ $$$$\left(\mathrm{0},\:{y}_{\mathrm{1}} \right)\:{and}\:{the}\:{position}\:{of}\:{m}\:{is} \\ $$$$\left(\pm{x}_{\mathrm{2}} ,\:{y}_{\mathrm{2}} \right). \\ $$$${d}_{\mathrm{1}} ^{\mathrm{2}} ={x}_{\mathrm{2}} ^{\mathrm{2}} +\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${d}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{4}{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${F}_{\mathrm{1}} =\frac{{k}_{{e}} {Qq}}{{d}_{\mathrm{1}} ^{\mathrm{2}} }=\frac{{k}_{{e}} {Qq}}{{x}_{\mathrm{2}} ^{\mathrm{2}} +\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${F}_{\mathrm{2}} =\frac{{k}_{{e}} {q}^{\mathrm{2}} }{{d}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{{k}_{{e}} {q}^{\mathrm{2}} }{\mathrm{4}{x}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$${V}=−\frac{{dy}_{\mathrm{1}} }{{dt}}\:\:\:\:\:\left(\downarrow\right) \\ $$$${A}=\frac{{dV}}{{dt}}=−\frac{{d}^{\mathrm{2}} {y}_{\mathrm{1}} }{{dt}^{\mathrm{2}} } \\ $$$${u}=\frac{{dx}_{\mathrm{2}} }{{dt}}\:\:\:\:\left(\rightarrow\right) \\ $$$${v}=\frac{{dy}_{\mathrm{2}} }{{dt}}\:\:\:\:\left(\upharpoonleft\right) \\ $$$${a}_{{x}} =\frac{{du}}{{dt}}=\frac{{d}^{\mathrm{2}} {x}_{\mathrm{2}} }{{dt}^{\mathrm{2}} } \\ $$$${a}_{{y}} =\frac{{dv}}{{dt}}=\frac{{d}^{\mathrm{2}} {y}_{\mathrm{2}} }{{dt}^{\mathrm{2}} } \\ $$$${MA}={Mg}+\mathrm{2}{F}_{\mathrm{1}} \:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {y}_{\mathrm{1}} }{{dt}^{\mathrm{2}} }=−{g}−\frac{\mathrm{2}{k}_{{e}} {Qq}\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)}{{M}\left[{x}_{\mathrm{2}} ^{\mathrm{2}} +\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${ma}_{{y}} ={F}_{\mathrm{1}} \:\mathrm{sin}\:\theta−{mg} \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {y}_{\mathrm{2}} }{{dt}^{\mathrm{2}} }=\frac{{k}_{{e}} {Qq}\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)}{{m}\left[{x}_{\mathrm{2}} ^{\mathrm{2}} +\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} }−{g} \\ $$$${ma}_{{x}} ={F}_{\mathrm{2}} −{F}_{\mathrm{1}} \:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {x}_{\mathrm{2}} }{{dt}^{\mathrm{2}} }=\frac{{k}_{{e}} {q}^{\mathrm{2}} }{\mathrm{4}{mx}_{\mathrm{2}} ^{\mathrm{2}} }−\frac{{k}_{{e}} {Qqx}_{\mathrm{2}} }{{m}\left[{x}_{\mathrm{2}} ^{\mathrm{2}} +\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${at}\:{t}=\mathrm{0}: \\ $$$${y}_{\mathrm{1}} ={b} \\ $$$${x}_{\mathrm{2}} ={a},\:{y}_{\mathrm{2}} =\mathrm{0} \\ $$$${V}=−\frac{{dy}_{\mathrm{1}} }{{dt}}=\mathrm{0} \\ $$$${u}=\frac{{dx}_{\mathrm{2}} }{{dt}}=\mathrm{0},{v}=\:\frac{{dy}_{\mathrm{2}} }{{dt}}=\mathrm{0} \\ $$$$..... \\ $$$$============= \\ $$$${i}\:{don}'{t}\:{think}\:{the}\:{problem}\:{is}\:{solvable}. \\ $$

Commented by ajfour last updated on 12/Jan/25

Gravitation free space or on frictionless horizontal flat plane i meant..sir.

$${Gravitation}\:{free}\:{space}\:{or}\:{on}\:{frictionless}\: \\ $$$${horizontal}\:{flat}\:{plane}\:{i}\:{meant}..{sir}. \\ $$

Commented by mr W last updated on 12/Jan/25

ok. then g=0 in the equstions. it′s still too hard to solve, i think.

$${ok}.\:{then}\:{g}=\mathrm{0}\:{in}\:{the}\:{equstions}. \\ $$$${it}'{s}\:{still}\:{too}\:{hard}\:{to}\:{solve},\:{i}\:{think}. \\ $$

Commented by mr W last updated on 12/Jan/25

is it similar to the three−body problem, that means there is no general closed−form solution?

$${is}\:{it}\:{similar}\:{to}\:{the}\:{three}−{body} \\ $$$${problem},\:{that}\:{means}\:{there}\:{is}\:{no} \\ $$$${general}\:{closed}−{form}\:{solution}? \\ $$

Commented by ajfour last updated on 12/Jan/25

cant say, i have imagined and constructed all of it on my own. Think first we'll determine incline side as a function of angle theta..i have tried but could solve i ll try again.

Commented by ajfour last updated on 13/Jan/25

Say y+z+ssin θ=b (dz/dt)=(((2m)/M))v (dy/dt)=v (dx/dt)=u ⇒ dy(1+((2m)/M))+d(ssin θ)=0 scos θ=x m(((udu)/dx))=((kq^2 )/x^2 )−((kqQ)/s^2 )cos θ ⇒ mu^2 =((kq^2 )/s_0 ^2 )∫((s_0 /(scos θ)))^2 {1−((Q/q))cos^3 θ}d(scos θ) m(((vdv)/dy))=((kqQ)/s^2 )sin θ ((M(2v)d(2v))/dz)=2(((kqQ)/s^2 ))sin θ mv^2 =−((kq^2 )/s_0 ^2 )((M/(M+2m)))∫((Q/q))((s_0 /(ssin θ)))^2 sin^3 θd(ssin θ) ....& even from energy conservation U_0 =((kq^2 )/(4a^2 ))−((2kQq)/((a^2 +b^2 )))=((kq^2 )/s_0 ^2 )(((a^2 +b^2 )/4)−((2Q)/q)) U=((kq^2 )/s^2 )((1/(2cos θ)))^2 −((2kQq)/s^2 )+mu^2 +(m+2M)v^2 or differentiating this ((kq^2 )/(4s^2 ))(2sec^2 θtan θ)(dθ/ds)−((2kq^2 )/s^3 )(((sec^2 θ)/4)−((2Q)/q)) +2mudu+2(m+2M)vdv=0 ....

$${Say}\:\:\:\:\:{y}+{z}+\boldsymbol{{s}}\mathrm{sin}\:\theta={b} \\ $$$$\frac{{dz}}{{dt}}=\left(\frac{\mathrm{2}{m}}{{M}}\right){v}\:\:\:\:\:\frac{{dy}}{{dt}}={v}\:\:\:\:\:\:\frac{{dx}}{{dt}}={u} \\ $$$$\Rightarrow\:\:{dy}\left(\mathrm{1}+\frac{\mathrm{2}{m}}{{M}}\right)+{d}\left({s}\mathrm{sin}\:\theta\right)=\mathrm{0} \\ $$$$\boldsymbol{{s}}\mathrm{cos}\:\theta={x} \\ $$$${m}\left(\frac{{udu}}{{dx}}\right)=\frac{{kq}^{\mathrm{2}} }{{x}^{\mathrm{2}} }−\frac{{kqQ}}{{s}^{\mathrm{2}} }\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:{mu}^{\mathrm{2}} =\frac{{kq}^{\mathrm{2}} }{{s}_{\mathrm{0}} ^{\mathrm{2}} }\int\left(\frac{{s}_{\mathrm{0}} }{{s}\mathrm{cos}\:\theta}\right)^{\mathrm{2}} \left\{\mathrm{1}−\left(\frac{{Q}}{{q}}\right)\mathrm{cos}\:^{\mathrm{3}} \theta\right\}{d}\left({s}\mathrm{cos}\:\theta\right) \\ $$$${m}\left(\frac{{vdv}}{{dy}}\right)=\frac{{kqQ}}{{s}^{\mathrm{2}} }\mathrm{sin}\:\theta\:\: \\ $$$$\frac{{M}\left(\mathrm{2}{v}\right){d}\left(\mathrm{2}{v}\right)}{{dz}}=\mathrm{2}\left(\frac{{kqQ}}{{s}^{\mathrm{2}} }\right)\mathrm{sin}\:\theta \\ $$$$\:{mv}^{\mathrm{2}} =−\frac{{kq}^{\mathrm{2}} }{{s}_{\mathrm{0}} ^{\mathrm{2}} }\left(\frac{{M}}{{M}+\mathrm{2}{m}}\right)\int\left(\frac{{Q}}{{q}}\right)\left(\frac{{s}_{\mathrm{0}} }{{s}\mathrm{sin}\:\theta}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{3}} \theta{d}\left({s}\mathrm{sin}\:\theta\right) \\ $$$$....\&\:{even}\:{from}\:{energy}\:{conservation} \\ $$$${U}_{\mathrm{0}} =\frac{{kq}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }−\frac{\mathrm{2}{kQq}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}=\frac{{kq}^{\mathrm{2}} }{{s}_{\mathrm{0}} ^{\mathrm{2}} }\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{2}{Q}}{{q}}\right) \\ $$$${U}=\frac{{kq}^{\mathrm{2}} }{{s}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2cos}\:\theta}\right)^{\mathrm{2}} −\frac{\mathrm{2}{kQq}}{{s}^{\mathrm{2}} }+{mu}^{\mathrm{2}} +\left({m}+\mathrm{2}{M}\right){v}^{\mathrm{2}} \\ $$$${or}\:{differentiating}\:{this} \\ $$$$\frac{{kq}^{\mathrm{2}} }{\mathrm{4}{s}^{\mathrm{2}} }\left(\mathrm{2sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:\theta\right)\frac{{d}\theta}{{ds}}−\frac{\mathrm{2}{kq}^{\mathrm{2}} }{{s}^{\mathrm{3}} }\left(\frac{\mathrm{sec}\:^{\mathrm{2}} \theta}{\mathrm{4}}−\frac{\mathrm{2}{Q}}{{q}}\right) \\ $$$$\:\:\:+\mathrm{2}{mudu}+\mathrm{2}\left({m}+\mathrm{2}{M}\right){vdv}=\mathrm{0} \\ $$$$.... \\ $$

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