Particles of mass m_1 and m_2 (m_2 >m_1 ) are connected by a light inextensible string passing over a smooth fixed pulley. initially both masses hang vertically with mass m_(2 ) at a height X above the floor. if the system is released from rest. with what speed will mass m_2 hit the floor and the mass m_1 will rise a further distance of [(((m_2 −m_1 )x)/(m_1 +m


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Question Number 119648 by aurpeyz last updated on 26/Oct/20

$P a r t i c l e s o f m a s s m_{1} a n d m_{2} (m_{2} > m_{1})$ $a r e c o n n e c t e d b y a l i g h t i n e x t e n s i b l e$ $s t r i n g p a s s i n g o v e r a s m o o t h f i x e d$ $p u l l e y . i n i t i a l l y b o t h m a s s e s h a n g$ $v e r t i c a l l y w i t h m a s s m_{2} a t a h e i g h t$ $X a b o v e t h e f l o o r . i f t h e s y s t e m i s$ $r e l e a s e d f r o m r e s t . w i t h w h a t s p e e d$ $w i l l m a s s m_{2} h i t t h e f l o o r a n d t h e$ $m a s s m_{1} w i l l r i s e a f u r t h e r d i s t a n c e o f$ $[\frac{(m_{2} - m_{1}) x}{m_{1} + m_{2}}] a f t e r t h i s o c c u r .$
Commented byaurpeyz last updated on 26/Oct/20

$p l s h e l p$
	Answered by TANMAY PANACEA last updated on 26/Oct/20

	$m_{2} g - T = m_{2} a$ $T - m_{1} g = m_{1} a$ $a = \frac{g (m_{2} - m_{1})}{m_{1} + m_{2}}$ $v^{2} = 0^{2} + 2 a x$ $v = \sqrt{2 \times \frac{g (m_{2} - m_{1})}{m_{1} + m_{2}} \times x} ⇚ t h i s i s t h e s p e e d$ $b y w h i c h m_{2} s t r i k e g r o u n d$ $p l s c h k$
	Commented byaurpeyz last updated on 28/Oct/20

	$t h a n k s$
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