A horizontal force of 0.8N is required to pull a 5kg block across a table top at a constant speed. With the block initially at rest,a 20g bullet fired horizontally into the block to slide 1.5m before coming to rest again.Determine the speed v of the bullet,where the bullet is assumed to be embedded in the block.


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Question Number 29216 by NECx last updated on 05/Feb/18

$A h o r i z o n t a l f o r c e o f 0.8 N i s$ $r e q u i r e d t o p u l l a 5 k g b l o c k a c r o s s$ $a t a b l e t o p a t a c o n s t a n t s p e e d .$ $W i t h t h e b l o c k i n i t i a l l y a t r e s t, a$ $20 g b u l l e t f i r e d h o r i z o n t a l l y i n t o$ $t h e b l o c k t o s l i d e 1.5 m b e f o r e$ $c o m i n g t o r e s t a g a i n . D e t e r m i n e$ $t h e s p e e d v o f t h e b u l l e t, w h e r e$ $t h e b u l l e t i s a s s u m e d t o b e$ $e m b e d d e d i n t h e b l o c k .$
	Answered by ajfour last updated on 05/Feb/18

	$f = 0.8 = 50 μ \Rightarrow μ = 0.016$ $m v = (M + m) V$ $μ (M + m) g s = \frac{1}{2} (M + m) V^{2}$ $\Rightarrow V = \sqrt{2 μ g s}$ $v = \frac{(M + m) V}{m} = (\frac{M}{m} + 1) \sqrt{2 μ g s}$ $= (251) \sqrt{2 \times 0.016 \times 10 \times 1.5}$ $\Rightarrow v = 251 \times \sqrt{48} m / s$ $= 251 \times 7 (1 - 0.02)$ $= 1757 - 35 = 1722 m / s .$
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