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Question Number 215256 by ajfour last updated on 01/Jan/25

Commented by ajfour last updated on 01/Jan/25

$A f t e r t h i s e l a s t i c c o l l i s i o n o n a f l a t$ $f r i c t i o n l e s s s u r f a c e F i n d a f t e r w h a t$ $t i m e a t w h a t y c o o r d i n a t e d o e s p o i n t$ $A c r o s s t h e y - a x i s .$
Commented by ajfour last updated on 02/Jan/25
https://youtu.be/roXqnV-fZ7k?si=JafblceCGUAmx3u- My video lecture discussing a problem of finding spring extension when to each end of it is attached a positive charge q.
	Answered by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 02/Jan/25

	$I_{1} = I_{2} = \frac{{m L}^{2}}{12}$ ${m v}_{1} = J$ $I_{1} ω_{1} = \frac{J L}{4}$ $\Rightarrow ω_{1} = \frac{3 v_{1}}{L}$ ${m u}_{2} = m u \cos 30 °$ $\Rightarrow u_{2} = \frac{\sqrt{3} u}{2}$ ${m v}_{2} = m u \sin 30 ° - J$ $\Rightarrow v_{2} = \frac{u}{2} - v_{1}$ $I_{2} ω_{2} = - J \times \frac{L \cos 30 °}{2}$ $\Rightarrow ω_{2} = - \frac{3 \sqrt{3} v_{1}}{L}$ $\frac{m}{2} (v_{1}^{2} + u_{2}^{2} + v_{2}^{2}) + \frac{1}{2} \times \frac{{m L}^{2}}{12} (ω_{1}^{2} + ω_{2}^{2}) = \frac{{m u}^{2}}{2}$ $v_{1}^{2} + {(\frac{\sqrt{3} u}{2})}^{2} + {(\frac{u}{2} - v_{1})}^{2} + \frac{L^{2}}{12} [{(\frac{3 v_{1}}{L})}^{2} + {(- \frac{3 \sqrt{3} v_{1}}{L})}^{2}] = u^{2}$ $5 v_{1}^{2} - {u v}_{1} = 0$ $\Rightarrow v_{1} = \frac{u}{5}$ $\Rightarrow ω_{1} = \frac{3 u}{5 L}$ $a t t i m e t p o i n t A c r o s s e s t h e y - a x i s :$ $x_{A} = \frac{L}{2} \cos (ω_{1} t) = 0$ $\Rightarrow ω_{1} t = \frac{3 u t}{5 L} = \frac{π}{2}$ $\Rightarrow t = \frac{5 π L}{6 u} ✓$ $y_{A} = v_{1} t + \frac{L}{2} \sin (ω_{1} t) = \frac{L ω_{1} t}{3} + \frac{L}{2} \sin (ω_{1} t)$ $= \frac{L}{2} (\frac{π}{3} + 1) > L$ $==================$ $θ_{1} = ω_{1} t = \frac{3 u t}{5 L}$ $x_{1} = 0$ $y_{1} = v_{1} t = \frac{u t}{5}$ $x_{A} = x_{1} + \frac{L \cos θ_{1}}{2} = \frac{L \cos (\frac{3 u t}{5 L})}{2}$ $y_{A} = y_{1} + \frac{L \sin θ_{1}}{2} = \frac{u t}{5} + \frac{L \sin (\frac{3 u t}{5 L})}{2}$ $θ_{2} = \frac{π}{6} + ω_{2} t = \frac{π}{6} - \frac{3 \sqrt{3} u t}{5 L}$ $x_{2} = \frac{L}{4} - \frac{\sqrt{3} L}{4} + u_{2} t = \frac{(1 - \sqrt{3}) L}{4} + \frac{\sqrt{3} u t}{2}$ $y_{2} = - \frac{L}{4} + v_{2} t = - \frac{L}{4} + \frac{3 u t}{10}$ $x_{D} = x_{2} + \frac{L \cos θ_{2}}{2} = \frac{(1 - \sqrt{3}) L}{4} + \frac{\sqrt{3} u t}{2} + \frac{L \cos (\frac{π}{6} - \frac{3 \sqrt{3} u t}{5 L})}{2}$ $y_{D} = y_{2} + \frac{L \sin θ_{2}}{2} = - \frac{L}{4} + \frac{3 u t}{10} + \frac{L \sin (\frac{π}{6} - \frac{3 \sqrt{3} u t}{5 L})}{2}$ $==================$ $f o l l o w i n g p i c t u r e s a r e s h o w i n g t h e$ $m o v e m e n t o f t h e r o d s a f t e r t h e$ $c o l l i s i o n a s w e l l a s t h e l o c u s o f t h e$ $p o i n t s A a n d D .$
	Commented by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 02/Jan/25

	Commented by ajfour last updated on 02/Jan/25

	$b e a u t i f u l o m g! h o w y o u g e n e r a t e t h i s .$
	Commented by mr W last updated on 02/Jan/25

	$i j u s t h a v e u s e d t h e a p p G r a p h e r$ $l i k e f o l l o w i n g .$ $(w i t h m = u t i n t h e e q u a t i o n s)$
	Commented by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 02/Jan/25

	$w e c a n e v e n g e n e r a t e a n a n i m a t i o n .$ $b u t s i n c e t h i s a p p {d o e s n}^{'} t s u p p o r t$ $a n i m a t e d g i f i m a g e s, s o i j u s t t o o k$ $s o m e s c r e e n s h o t s f o r d i f f e r e n t$ $v a l u e s o f m .$
	Commented by MathematicalUser2357 last updated on 17/Feb/25

	$S o I h a v e a G r a p h e r F r e e :$
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