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Question Number 28585 by Tinkutara last updated on 27/Jan/18

	Answered by ajfour last updated on 27/Jan/18

	$c o l l i s i o n f o r m_{1} i s e l a s t i c .$ $a f t e r c o l l i s i o n, v e l o c i t y o f m_{1}$ $i s r e v e r s e d; k i n e t i c e n e r g y$ $i s r e g a t h e r e d a s s p r i n g c o m e s t o$ $r e l a x e d l e n g t h$ $K_{i} = m_{1} g h$ $m_{1} s t i l l r i s e s f u r t h e r b y^{'} x^{'} t i l l i t s t o p s$ $j u s t w h e n m_{2} b r e a k s o f f t h e$ $s u r f a c e, s o k x = m_{2} g . . . (i)$ $K_{1} = K_{f} + △ U_{g} + △ U_{s p r i n g}$ $m_{1} g h = 0 + m_{1} g x + \frac{1}{2} {k x}^{2}$ $= x (m_{1} g + \frac{k x}{2})$ $u s i n g (i)$ $h = \frac{m_{2} g}{k} (1 + \frac{m_{2}}{2 m_{1}}) .$
	Commented by Tinkutara last updated on 27/Jan/18
	Why collision of m1 is elastic?
	Commented by ajfour last updated on 27/Jan/18

	$o n l y c o n s e r v a t i v e f o r c e s a c t e d$ $o n i t .$
	Commented by Tinkutara last updated on 28/Jan/18
	But what is the datum assumed here?
	Commented by ajfour last updated on 28/Jan/18
	in simpler words please ..
	Commented by Tinkutara last updated on 28/Jan/18
	What is the reference line assumed for calculating potential energy?
	Commented by ajfour last updated on 28/Jan/18

	$w i t h m_{2} a t g r o u n d l e v e l t h e$ $l e v e l o f m_{1} w h e n s p r i n g c o m e s$ $t o n a t u r a l l e n g t h a f t e r c o l l i s i o n$ $i s t a k e n t o b e t h e r e f e r e n c e l e v e l$ $o f g r a v i t a t i o n a l p o t e n t i a l e n e r g y .$
	Commented by Tinkutara last updated on 28/Jan/18

	$B u t t h e n t h e y s h o u l d b e - m_{1} g h a n d$ $- m_{1} g x . O r n o t ?$
	Commented by ajfour last updated on 28/Jan/18

	$N o, s p r i n g i s b e i n g s t r e t c h e d a f t e r$ $t h i s d u e t o m o m e n t u m o f m_{1} .$ $m_{1} i s r i s i n g u p .$
	Commented by Tinkutara last updated on 28/Jan/18
	But potential energy when measured downwards should be taken negative or not?
	Commented by ajfour last updated on 28/Jan/18

	$K_{0} o f m_{1} = - △ U$ $= - (- m_{1} g h) = m_{1} g h$ $j u s t b e f o r e m_{2} h i t s t h e g r o u n d :$ $a f t e r t h i s t h e r e i s m_{2} c o l l i d e s$ $w i t h g r o u n d a n d l o o s e s a l l i t s$ $k i n e t i c e n e r g y . m_{1} c o m p r e s s e s$ $t h e s p r i n g a n d t h e r e a f t e r, b y t h e$ $t i m e s p r i n g s r e g a i n s i t s o r i g i n a l$ $l e n g t h m_{1} r e g a t h e r s i t s K i n e t i c$ $e n e r g y . t h e l e v e l o f m_{1} a t t h i s$ $l e v e l i s n o w t a k e n t o b e i n i t i a l$ $p o s i t i o n t o a p p l y w o r k e n e r g y$ $t h e o r w m . K_{1} o f m_{1} = K_{0} = m_{1} g h .$ $m_{1} r i s e s b e y o n d t h i s p o i n t a n d$ $b y t h e t i m e i t s s p e e d g e t s z e r o,$ $m_{2} i s b r e a k s c o n t a c t .$ $w o r k - e n e r g y e q . f o r m_{1} i s$ $W_{e x t} = K_{2} - K_{1} + △ U_{g} + △ U_{s p r i n g}$ $0 = (0 - m_{1} g h) + (m_{1} g x - 0) + (\frac{1}{2} {k x}^{2} - 0)$ $s p r i n g f o r c e f i n a l l y i s k x = m_{2} g$ $h e n c e$ $m_{1} g h = x (m_{1} g + \frac{k x}{2})$ $h = x (1 + \frac{k x}{2 m_{1} g}) = \frac{m_{2} g}{k} (1 + \frac{m_{2}}{2 m_{1}}) .$
	Commented by Tinkutara last updated on 29/Jan/18
	Thank you very much Sir! I got the answer.
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