A body of mass 0.1kg dropped from a height of 8m onto a hard floor bounces back to a height of 2m. Calculate the change of momentum. If the body is in contact with the floor for 0.1s then what is the force exerted on the body? [g=10ms^(−2) ]


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Question Number 48194 by Necxx last updated on 20/Nov/18

$A b o d y o f m a s s 0.1 k g d r o p p e d f r o m$ $a h e i g h t o f 8 m o n t o a h a r d f l o o r$ $b o u n c e s b a c k t o a h e i g h t o f 2 m .$ $C a l c u l a t e t h e c h a n g e o f m o m e n t u m .$ $I f t h e b o d y i s i n c o n t a c t w i t h t h e$ $f l o o r f o r 0.1 s t h e n w h a t i s t h e$ $f o r c e e x e r t e d o n t h e b o d y ?$ $[g = 10 {m s}^{- 2}]$
Commented by Necxx last updated on 20/Nov/18

$p l e a s e h e l p$
	Answered by mr W last updated on 21/Nov/18

	$v_{1} = - \sqrt{2 {g h}_{1}}$ $v_{2} = \sqrt{2 {g h}_{2}}$ $Δ p = m (v_{2} - v_{1}) = m \sqrt{2 g} (\sqrt{h_{2}} + \sqrt{h_{1}})$ $= 0.1 \times \sqrt{20} (\sqrt{2} + \sqrt{8}) = 1.897 N s$ $F = Δ p / Δ t = 1.897 / 0.1 = 18.97 N$
	Commented by Necxx last updated on 20/Nov/18

	$o h . . . . . I t h o u g h t i t s m e a n t t o b e$ $t h e d i f f e r e n c e b e t w e e n t h e f i n a l$ $a n d i n i t i a l m o m e n t u m n o t t h e$ $r e v e r s e . W h y d i d y o u t a k e t h i s$ $s t e p ?$ $T h a n k s$
	Commented by Necxx last updated on 21/Nov/18

	$p l e a s e w h y i s v_{1} n e g a t i v e ?$
	Commented by mr W last updated on 21/Nov/18

	$p a n d v a n d F a r e v e c t o r s . t h e y h a v e$ $d i r e c t i o n . p o s i t i v e d i r e c t i o n i s u p w a r d s .$
	Commented by Necxx last updated on 21/Nov/18

	$o k . I a m r e a l l y g r a t e f u l . I n o w$ $u n d e r s t a n d i t f o r r e a l .$
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