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

Commented by ajfour last updated on 11/Jan/25

$T h e c h a r g e d m a s s e s r e l e a s e d a s s h o w n .$ $F i n d t h e i r s p e e d s w h e n t h e y g e t i n l i n e .$ $I w o n d e r i f t h i s g e t s p e r i o d i c f o r c e r t a i n$ $a p p r o p r i a t e r a t i o s!$
Commented by mahdipoor last updated on 11/Jan/25

$w h y n e g a t i v e m a s s r i s i n g u p ?$
Commented by ajfour last updated on 11/Jan/25

$c o m e ? o n + o n e p u l l s i t u p w a r d s w h i l e$ $t h e n e g a t i v e p a i r r e p e l . .$
	Answered by mr W last updated on 12/Jan/25

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

	$s a y a t t i m e t t h e p o s t i o n o f M i s$ $(0, y_{1}) a n d t h e p o s i t i o n o f m i s$ $(\pm x_{2}, y_{2}) .$ $d_{1}^{2} = x_{2}^{2} + {(y_{1} - y_{2})}^{2}$ $d_{2}^{2} = 4 x_{2}^{2}$ $F_{1} = \frac{k_{e} Q q}{d_{1}^{2}} = \frac{k_{e} Q q}{x_{2}^{2} + {(y_{1} - y_{2})}^{2}}$ $F_{2} = \frac{k_{e} q^{2}}{d_{2}^{2}} = \frac{k_{e} q^{2}}{4 x_{2}^{2}}$ $V = - \frac{{d y}_{1}}{d t} (↓)$ $A = \frac{d V}{d t} = - \frac{d^{2} y_{1}}{{d t}^{2}}$ $u = \frac{{d x}_{2}}{d t} (\to)$ $v = \frac{{d y}_{2}}{d t} (↿)$ $a_{x} = \frac{d u}{d t} = \frac{d^{2} x_{2}}{{d t}^{2}}$ $a_{y} = \frac{d v}{d t} = \frac{d^{2} y_{2}}{{d t}^{2}}$ $M A = M g + 2 F_{1} \sin θ$ $\Rightarrow \frac{d^{2} y_{1}}{{d t}^{2}} = - g - \frac{2 k_{e} Q q (y_{1} - y_{2})}{M {[x_{2}^{2} + {(y_{1} - y_{2})}^{2}]}^{\frac{3}{2}}}$ ${m a}_{y} = F_{1} \sin θ - m g$ $\Rightarrow \frac{d^{2} y_{2}}{{d t}^{2}} = \frac{k_{e} Q q (y_{1} - y_{2})}{m {[x_{2}^{2} + {(y_{1} - y_{2})}^{2}]}^{\frac{3}{2}}} - g$ ${m a}_{x} = F_{2} - F_{1} \cos θ$ $\Rightarrow \frac{d^{2} x_{2}}{{d t}^{2}} = \frac{k_{e} q^{2}}{4 {m x}_{2}^{2}} - \frac{k_{e} {Q q x}_{2}}{m {[x_{2}^{2} + {(y_{1} - y_{2})}^{2}]}^{\frac{3}{2}}}$ $a t t = 0 :$ $y_{1} = b$ $x_{2} = a, y_{2} = 0$ $V = - \frac{{d y}_{1}}{d t} = 0$ $u = \frac{{d x}_{2}}{d t} = 0, v = \frac{{d y}_{2}}{d t} = 0$ $. . . . .$ $=============$ $i {d o n}^{'} t t h i n k t h e p r o b l e m i s s o l v a b l e .$
	Commented by ajfour last updated on 12/Jan/25

	$G r a v i t a t i o n f r e e s p a c e o r o n f r i c t i o n l e s s$ $h o r i z o n t a l f l a t p l a n e i m e a n t . . s i r .$
	Commented by mr W last updated on 12/Jan/25

	$o k . t h e n g = 0 i n t h e e q u s t i o n s .$ ${i t}^{'} s s t i l l t o o h a r d t o s o l v e, i t h i n k .$
	Commented by mr W last updated on 12/Jan/25

	$i s i t s i m i l a r t o t h e t h r e e - b o d y$ $p r o b l e m, t h a t m e a n s t h e r e i s n o$ $g e n e r a l c l o s e d - f o r m s o l u t i o n ?$
	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
	$Say y+z+ssin θ=b (dz/dt)=(((2m)/M))v (dy/dt)=v (dx/dt)=u ⇒ dy(1+((2m)/M))+d(ssin θ)=0 scos θ=x m(((udu)/dx))=((kq^2 )/x^2 )−((kqQ)/s^2 )cos θ ⇒ mu^2 =((kq^2 )/s_0 ^2 )∫((s_0 /(scos θ)))^2 {1−((Q/q))cos^3 θ}d(scos θ) m(((vdv)/dy))=((kqQ)/s^2 )sin θ ((M(2v)d(2v))/dz)=2(((kqQ)/s^2 ))sin θ mv^2 =−((kq^2 )/s_0 ^2 )((M/(M+2m)))∫((Q/q))((s_0 /(ssin θ)))^2 sin^3 θd(ssin θ) ....& even from energy conservation U_0 =((kq^2 )/(4a^2 ))−((2kQq)/((a^2 +b^2 )))=((kq^2 )/s_0 ^2 )(((a^2 +b^2 )/4)−((2Q)/q)) U=((kq^2 )/s^2 )((1/(2cos θ)))^2 −((2kQq)/s^2 )+mu^2 +(m+2M)v^2 or differentiating this ((kq^2 )/(4s^2 ))(2sec^2 θtan θ)(dθ/ds)−((2kq^2 )/s^3 )(((sec^2 θ)/4)−((2Q)/q)) +2mudu+2(m+2M)vdv=0 ....$
	$S a y y + z + s \sin θ = b$ $\frac{d z}{d t} = (\frac{2 m}{M}) v \frac{d y}{d t} = v \frac{d x}{d t} = u$ $\Rightarrow d y (1 + \frac{2 m}{M}) + d (s \sin θ) = 0$ $s \cos θ = x$ $m (\frac{u d u}{d x}) = \frac{{k q}^{2}}{x^{2}} - \frac{k q Q}{s^{2}} \cos θ$ $\Rightarrow {m u}^{2} = \frac{{k q}^{2}}{s_{0}^{2}} \int {(\frac{s_{0}}{s \cos θ})}^{2} {1 - (\frac{Q}{q}) \cos^{3} θ} d (s \cos θ)$ $m (\frac{v d v}{d y}) = \frac{k q Q}{s^{2}} \sin θ$ $\frac{M (2 v) d (2 v)}{d z} = 2 (\frac{k q Q}{s^{2}}) \sin θ$ ${m v}^{2} = - \frac{{k q}^{2}}{s_{0}^{2}} (\frac{M}{M + 2 m}) \int (\frac{Q}{q}) {(\frac{s_{0}}{s \sin θ})}^{2} \sin^{3} θ d (s \sin θ)$ $. . . . & e v e n f r o m e n e r g y c o n s e r v a t i o n$ $U_{0} = \frac{{k q}^{2}}{4 a^{2}} - \frac{2 k Q q}{(a^{2} + b^{2})} = \frac{{k q}^{2}}{s_{0}^{2}} (\frac{a^{2} + b^{2}}{4} - \frac{2 Q}{q})$ $U = \frac{{k q}^{2}}{s^{2}} {(\frac{1}{2 \cos θ})}^{2} - \frac{2 k Q q}{s^{2}} + {m u}^{2} + (m + 2 M) v^{2}$ $o r d i f f e r e n t i a t i n g t h i s$ $\frac{{k q}^{2}}{4 s^{2}} (2 \sec^{2} θ \tan θ) \frac{d θ}{d s} - \frac{2 {k q}^{2}}{s^{3}} (\frac{\sec^{2} θ}{4} - \frac{2 Q}{q})$ $+ 2 m u d u + 2 (m + 2 M) v d v = 0$ $. . . .$
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