Two particles A and B of equal masses m are tied with an inextensible string of length 2l. The initial distance between A and B is l. Particle A is given speed v. Find the speed of particle A and B just after the string becomes taut.


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Question Number 25828 by Tinkutara last updated on 15/Dec/17

$T w o p a r t i c l e s A a n d B o f e q u a l m a s s e s$ $m a r e t i e d w i t h a n i n e x t e n s i b l e s t r i n g$ $o f l e n g t h 2 l . T h e i n i t i a l d i s t a n c e$ $b e t w e e n A a n d B i s l . P a r t i c l e A i s$ $g i v e n s p e e d v . F i n d t h e s p e e d o f$ $p a r t i c l e A a n d B j u s t a f t e r t h e s t r i n g$ $b e c o m e s t a u t .$
Commented by Tinkutara last updated on 15/Dec/17

	Answered by ajfour last updated on 16/Dec/17

	$v_{A} = \frac{v}{4}, v_{B} = \frac{3 v}{4} j u s t a f t e r s t r i n g$ $b e c o m e s t a u t t h e s e c o n d t i m e .$
	Answered by mrW1 last updated on 16/Dec/17

	$\Rightarrow v_{A} \cos θ = v_{B} . . . (i)$ $m v \times \sin 30 = {m v}_{A} \sin θ$ $\Rightarrow \frac{v}{2} = v_{A} \sin θ . . . (i i)$ $m v \times \cos 30 = {m v}_{A} \cos θ + {m v}_{B}$ $\Rightarrow \frac{\sqrt{3}}{2} v = v_{A} \cos θ + v_{B} = 2 v_{B} . . . (i i i)$ $\Rightarrow v_{B} = \frac{\sqrt{3}}{4} v$ $(i i) / (i) :$ $\tan θ = \frac{v}{2 v_{B}} = \frac{v \times 4}{2 \times \sqrt{3} v} = \frac{2 \sqrt{3}}{3}$ $\Rightarrow θ = \tan^{- 1} \frac{2 \sqrt{3}}{3} \approx 49.1 °$ $\sin θ = \frac{2}{\sqrt{7}}$ $(i i) :$ $v_{A} = \frac{v}{2 \sin θ} = \frac{\sqrt{7} v}{4}$
	Commented by mrW1 last updated on 16/Dec/17

	$T h a n k s, t h i s i s m y q u e s t i o n . I m a d e$ $f o l l o w i n g i m a g e, b u t c o u l d n o t p o s t .$
	Commented by Tinkutara last updated on 16/Dec/17
	Thank you Sir!
	Commented by mrW1 last updated on 16/Dec/17

	Commented by ajfour last updated on 16/Dec/17

	$d i s a g r e e w i t h l i n e 1 .$
	Commented by mrW1 last updated on 16/Dec/17

	$a f t e r t h e s t r i n g i s g o t t a u t, b o t h a r t i c l e s$ $h a v e t h e s a m e s p e e d i n t h e d i r e c t i o n$ $o f t h e s t r i n g . w h a t i s w r o n g ?$
	Commented by ajfour last updated on 16/Dec/17

	$v_{A} \sin θ = \frac{v}{2} (a l r i g h t)$ $v_{B} + v_{A} \cos θ = \frac{v \sqrt{3}}{2} (f i n e)$ $b u t θ = 90 ° (s h o u l d b e)$ $\Rightarrow v_{A} = \frac{v}{2}, v_{B} = \frac{v \sqrt{3}}{2} .$ $A f t e r t h i s d i s t a n c e d e c r e a s e s$ $a n d i n c r e a s e s a g a i n w h e n$ $t h e r e i s a g a i n a j e r k e x c h a n g e$ $v i a t h e s t r i n g . s t r i n g s h a l l r e m a i n$ $t a u t a f t e r t h i s s e c o n d j e r k a n d$ $v_{A} = \frac{1 v}{4}, v_{B} = \frac{3 v}{4} j u s t a f t e r t h i s$ $s e c o n d j e r k .$
	Commented by ajfour last updated on 16/Dec/17

	Commented by ajfour last updated on 16/Dec/17

	$i t i s a s t r i n g, t h e v e l o c i t y$ $c o m p o n e n t a l o n g t h e s t r i n g$ $l e n g t h w i l l b e e x c h a n g e d .$ $v_{A} (p a r a l l e l t o s t r i n g) = 0$ $v_{B} = \frac{v \sqrt{3}}{2} (a l o n g s t r i n g l e n g t h) .$ $s e p a r a t i o n b e t w e e n A a n d B$ $d e c r e a s e s h e r e a f t e r a n d w h e n$ $a g a i n i n c r e a s e s t o 2 l w h e r e u p o n$ $t h e i r i s a n o t h e r i m p u l s e e x c h a n g e$ $a n d A a n d B w i l l c o m e i n a$ $v e r t i c a l l i n e o f p l a n e (i s h a l l$ $p o s t i m a g e) . A f t e r t h i s$ $t h e i r v e l o c i t i e s b e c o m e p a r a l l e l$ $t o d i r e c t i o n o f o r i g i n a l v e l o c i t y$ $v w i t h v_{A} = \frac{1 v}{4} a n d v_{B} = \frac{3 v}{4} .$
	Commented by ajfour last updated on 16/Dec/17

	Commented by ajfour last updated on 16/Dec/17

	$c o n s i d e r t h e c a s e i n t h i s i m a g e,$ $a n d d o t h i n k a g a i n . .$
	Commented by mrW1 last updated on 16/Dec/17

	Commented by mrW1 last updated on 16/Dec/17

	$M y q u e s t i o n i s i f (1) o r (2) i s c o r r e c t .$ $O r i t i s n o t c e r t a i n w h a t h a p p e n s .$
	Commented by ajfour last updated on 16/Dec/17

	$(2) i s c o r r e c t .$
	Commented by ajfour last updated on 16/Dec/17

	Commented by ajfour last updated on 16/Dec/17

	$I f t i e d w i t h s t r i n g (a n d n o e n e r g y$ $l o s s e s) j u s t t h e f o r c e s a n d$ $v e l o c i t i e s w i l l h a v e o p p o s i t e$ $d i r e c t i o n s .$
	Commented by mrW1 last updated on 16/Dec/17

	$I s e e n o w .$ $T h i s i s b e c a u s e t h e c o l l i s i o n$ $i s p e r f e c t l y e l a s t i c .$ $m_{A} u = m_{A} v_{A} + m_{B} v_{B}$ $\Rightarrow u = v_{A} + v_{B}$ $\Rightarrow u^{2} = v_{A}^{2} + v_{B}^{2} + 2 v_{A} v_{B}$ $\frac{1}{2} m_{A} u^{2} = \frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} v_{B}^{2}$ $\Rightarrow u^{2} = v_{A}^{2} + v_{B}^{2}$ $\Rightarrow v_{A} v_{B} = 0$ $i f v_{B} \neq 0, t h e n v_{A} = 0 .$
	Commented by ajfour last updated on 16/Dec/17

	$t h a n k s f o r t h e n o t i c e S i r!$
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