A block of wood of mass 0.6kg is balanced on top of vertical port 2m high.A 10g bullet is fired horizontally into the block and the embedded bullet land at a 4m from the base of the port.Find the initial velocity of the bullet.


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Question Number 29212 by NECx last updated on 05/Feb/18

$A b l o c k o f w o o d o f m a s s 0.6 k g i s$ $b a l a n c e d o n t o p o f v e r t i c a l p o r t$ $2 m h i g h . A 10 g b u l l e t i s f i r e d$ $h o r i z o n t a l l y i n t o t h e b l o c k a n d$ $t h e e m b e d d e d b u l l e t l a n d a t a 4 m$ $f r o m t h e b a s e o f t h e p o r t . F i n d$ $t h e i n i t i a l v e l o c i t y o f t h e b u l l e t .$
Commented by NECx last updated on 06/Feb/18

$p l e a s e h e l p$
	Answered by mrW2 last updated on 07/Feb/18

	$v_{0} = i n i t i a l v e l o c i t y o f b u l l e t$ $v_{1} = v e l o c i t y o f w o o d w i t h b u l l e t$ $t = \sqrt{\frac{2 h}{g}}$ $L = v_{1} t = v_{1} \sqrt{\frac{2 h}{g}}$ $\Rightarrow v_{1} = L \sqrt{\frac{g}{2 h}}$ $m_{B} v_{0} = (m_{B} + m_{W}) v_{1}$ $\Rightarrow v_{0} = \frac{m_{B} + m_{W}}{m_{B}} \times v_{1}$ $= \frac{m_{B} + m_{W}}{m_{B}} \times L \sqrt{\frac{g}{2 h}}$ $= \frac{10 + 600}{10} \times 4 \sqrt{\frac{10}{2 \times 2}}$ $= 122 \sqrt{10}$ $\approx 386 m / s$
	Commented by NECx last updated on 08/Feb/18

	$n o w i t s u n d e r s t o o d . G r a c i a s!$
	Commented by mrW2 last updated on 08/Feb/18

	$d e n a d a, s e \tilde{n o r}!$
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