Near the shore a fisherman jumps out of his boat with a velocity of 5.00 ms^(−1) and lands on the shore 0.650 afterwards.The boat moves backwards. The respective masses of the fisherman and the boat are 85.0 kg and 165 kg and the frictional force between the boat and water is negligible. What will be the distance between the fisherman and the both at the instant when he just lands on the shore?


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Question Number 139561 by physicstutes last updated on 28/Apr/21

$Near the shore a fisherman jumps out of his boat with a velocity$ $of 5.00 {ms}^{- 1} and lands on the shore 0.650 afterwards . The boat$ $moves backwards . The respective masses of the fisherman and$ $the boat are 85.0 kg and 165 kg and the frictional force between$ $the boat and water is negligible . What will be the distance between$ $the fisherman and the both at the instant when he just lands on the$ $shore ?$
	Answered by mr W last updated on 28/Apr/21

	$(1 + \frac{85}{165}) \times 0.65 \approx 0.98 m$
	Commented by physicstutes last updated on 28/Apr/21

	$sir how have you accounted for the speed$
	Commented by mr W last updated on 28/Apr/21

	$a s f a r a s n o e x t e r n a l f o r c e a c t s o n$ $a n o b j e c t i n a c e r t a i n d i r e c t i o n, t h e$ $o b j e c t {w o n}^{'} t c h a n g e i t s p o s i t i o n i n$ $t h i s d i r e c t i o n . {t h a t}^{'} s c l e a r .$ $o n t h e o b j e c t ‘ ‘ b o a t + {f i s h e r m a n}^{″} n o$ $e x t e r n a l f o r c e a c t s i n h o r i z o n t a l$ $d i r e c t i o n, t h e r e f o r e i t {d o e s n}^{'} t c h a n g e$ $i t s h o r i z o n t a l p o s i t i o n, i . e . t h e$ $c e n t e r o f m a s s f r o m ‘ ‘ b o a t + {f i s h e r m a n}^{″}$ $k e e p s u n c h a n g e d w h e n t h e m a n$ $j u m p s f r o m b o a t o n t o t h e s h o r e .$ $t h e c e n t e r o f m a s s o f ‘ ‘ b o a t + {f i s h e r m a n}^{″}$ $w a s 0.65 m a p a r t f r o m t h e s h o r e .$ $a t t h e m o m e n t w h e n t h e f i s h e r m a n$ $r e a c h e s t h e s h o r e, t h e f i s h e r m a n$ $i s 0.65 m a w a y f r o m t h e c e n t e r o f$ $m a s s, t h e b o a t m u s t b e \frac{85}{165} \times 0.65 m$ $a w a y f r o m t h e c e n t e r o f m a s s, b u t$ $i n o p p o s i t e d i r e c t i o n, s o t h e d i s t a n c e$ $f r o m f i s h e r m a n t o t h e b o a t i s$ $0.65 + \frac{85}{165} \times 0.65 \approx 0.98 m .$
	Commented by physicstutes last updated on 28/Apr/21

	$sir the boat is said to m o v e b a c k w a r d therefore we have a force$ $involved right ?$ $equal and opposite forces . the man gets a force to move to the shore and$ $the boat gains an opposite force to move backwards . . but accelerations will$ $not be equal since masses are different . right ?$
	Commented by mr W last updated on 29/Apr/21

	Commented by mr W last updated on 29/Apr/21

	$b l a c k = b o a t$ $r e d = f i s h e r m a n$ $C O M = c e n t e r o f m a s s f r o m b o a t & m a n$ $w h e n t h e m a n j u m p s f r o m t h e b o a t,$ $h e a c t s a f o r c e t o t h e b o a t, t h i s f o r c e$ $m a k e s t h e b o a t t o m o v e t o l e f t, b u t$ $t h e b o a t a c t s a l s o t h e s a m e f o r c e i n$ $o p p o s i t e d i r e c t i o n t o t h e m a n a n d$ $m a k e s i h m t o m o v e t o r i g h t . b o t h$ $f o r c e s a r e i n t e r n a l f o r c e s i n s i d e$ $t h e o b j e c t ‘ ‘ b o a t & {f i s h e r m a n}^{″} . o n$ $t h e o b j e c t ‘ ‘ b o a t & {f i s h e r m a n}^{″} i t s e l f$ $n o e x t e r m a l f o r c e a c t s i n t h i s$ $p r o c e s s, t h e r e f o r e t h e C O M o f t h e$ $o b j e c t ‘ ‘ b o a t & {f i s h e r m a n}^{″} r e m a i n s$ $u n c h a n g e d .$
	Commented by mr W last updated on 29/Apr/21

	$i f o n e r e a l l y u n d e r s t a n d s t h e p h y s i c s,$ $h e s e e s t h e b o a t a n d t h e f i s h e r m a n$ $a s a s i n g l e o b j e c t a n d t h a t t h i s o b j e c t$ ${d o e s n}^{'} t m o v e, a l t h o u g h b o t h b o a t a n d$ $f i s h e r m a n a r e i n m o t i o n . w i t h t h i s$ $u n d e r s t a n d i n g h e g e t s t h e c o r r e c t$ $r e s u l t w i t h o u t m u c h c a l c u l a t i o n .$ $i f o n e i s g o o d i n p h y s i c s, h e k n o w s$ $h o w t o c a l c u l a t e t o g e t t h e c o r r e c t$ $r e s u l t . h e r e i s h o w t o c a l c u l a t e :$ $s a y F i s t h e f o r c e b e t w e e n b o a t a n d$ $f i s h e r m a n a s h e j u m p s f r o m b o a t .$ $s a y t h i s f o r c e a c t s i n a s h o r t t i m e Δ t .$ $V = s p e e d o f b o a t (t o l e f t)$ $v = s p e e d o f f i s h e r m a n (t o r i g h t)$ $M V = F Δ t$ $m v = F Δ t$ $M V = m v$ $\Rightarrow \frac{V}{v} = \frac{m}{M}$ $t h e f i s h e r m a n m o v e s d_{1} = 0.65 m i n$ $t i m e t = \frac{d_{1}}{v} . i n t h i s t i m e t h e b o a t m o v e s$ $d_{2} = V t = V \times \frac{d_{1}}{v} = \frac{V}{v} \times d_{1} = \frac{m}{V} \times d_{1}$ $= \frac{85}{165} \times 0.65$ $t h e d i s t a n c e b e t w e e n b o t h i s t h e n$ $d_{1} + d_{2} = 0.65 + \frac{85}{165} \times 0.65 \approx 0.98 m .$
	Commented by peter frank last updated on 29/Apr/21

	$t h a n k y o u$
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