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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-

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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?
$$\mathrm{Near}\:\mathrm{the}\:\mathrm{shore}\:\mathrm{a}\:\mathrm{fisherman}\:\mathrm{jumps}\:\mathrm{out}\:\mathrm{of}\:\mathrm{his}\:\mathrm{boat}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity} \\ $$$$\mathrm{of}\:\mathrm{5}.\mathrm{00}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{and}\:\mathrm{lands}\:\mathrm{on}\:\mathrm{the}\:\mathrm{shore}\:\mathrm{0}.\mathrm{650}\:\mathrm{afterwards}.\mathrm{The}\:\mathrm{boat} \\ $$$$\mathrm{moves}\:\mathrm{backwards}.\:\mathrm{The}\:\mathrm{respective}\:\mathrm{masses}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fisherman}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{boat}\:\mathrm{are}\:\mathrm{85}.\mathrm{0}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{165}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{the}\:\mathrm{frictional}\:\mathrm{force}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{boat}\:\mathrm{and}\:\mathrm{water}\:\mathrm{is}\:\mathrm{negligible}.\:\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{fisherman}\:\mathrm{and}\:\mathrm{the}\:\mathrm{both}\:\mathrm{at}\:\mathrm{the}\:\mathrm{instant}\:\mathrm{when}\:\mathrm{he}\:\mathrm{just}\:\mathrm{lands}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{shore}? \\ $$
Answered by mr W last updated on 28/Apr/21
(1+((85)/(165)))×0.65≈0.98m
$$\left(\mathrm{1}+\frac{\mathrm{85}}{\mathrm{165}}\right)×\mathrm{0}.\mathrm{65}\approx\mathrm{0}.\mathrm{98}{m} \\ $$
Commented by physicstutes last updated on 28/Apr/21
sir how have you accounted for the speed
$$\mathrm{sir}\:\mathrm{how}\:\mathrm{have}\:\mathrm{you}\:\mathrm{accounted}\:\mathrm{for}\:\mathrm{the}\:\mathrm{speed} \\ $$
Commented by mr W last updated on 28/Apr/21
as far as no external force acts on an object in a certain direction, the object won′t change its position in this direction. that′s clear. on the object “boat+fisherman” no external force acts in horizontal direction, therefore it doesn′t change its horizontal position, i.e. the center of mass from “boat+ fisherman” keeps unchanged when the man jumps from boat onto the shore. the center of mass of “boat + fisherman” was 0.65m apart from the shore. at the moment when the fisherman reaches the shore, the fisherman is 0.65m away from the center of mass, the boat must be ((85)/(165))×0.65m away from the center of mass, but in opposite direction, so the distance from fisherman to the boat is 0.65+((85)/(165))×0.65≈0.98m.
$${as}\:{far}\:{as}\:{no}\:{external}\:{force}\:{acts}\:{on} \\ $$$${an}\:{object}\:{in}\:{a}\:{certain}\:{direction},\:{the} \\ $$$${object}\:{won}'{t}\:{change}\:{its}\:{position}\:{in} \\ $$$${this}\:{direction}.\:{that}'{s}\:{clear}. \\ $$$${on}\:{the}\:{object}\:“{boat}+{fisherman}''\:{no} \\ $$$${external}\:{force}\:{acts}\:{in}\:{horizontal} \\ $$$${direction},\:{therefore}\:{it}\:{doesn}'{t}\:{change} \\ $$$${its}\:{horizontal}\:{position},\:{i}.{e}.\:{the}\: \\ $$$${center}\:{of}\:{mass}\:{from}\:“{boat}+\:{fisherman}'' \\ $$$${keeps}\:{unchanged}\:{when}\:{the}\:{man} \\ $$$${jumps}\:{from}\:{boat}\:{onto}\:{the}\:{shore}. \\ $$$${the}\:{center}\:{of}\:{mass}\:{of}\:“{boat}\:+\:{fisherman}'' \\ $$$${was}\:\mathrm{0}.\mathrm{65}{m}\:{apart}\:{from}\:{the}\:{shore}. \\ $$$${at}\:{the}\:{moment}\:{when}\:{the}\:{fisherman} \\ $$$${reaches}\:{the}\:{shore},\:{the}\:{fisherman} \\ $$$${is}\:\mathrm{0}.\mathrm{65}{m}\:{away}\:{from}\:{the}\:{center}\:{of} \\ $$$${mass},\:{the}\:{boat}\:{must}\:{be}\:\frac{\mathrm{85}}{\mathrm{165}}×\mathrm{0}.\mathrm{65}{m} \\ $$$${away}\:{from}\:{the}\:{center}\:{of}\:{mass},\:{but} \\ $$$${in}\:{opposite}\:{direction},\:{so}\:{the}\:{distance} \\ $$$${from}\:{fisherman}\:{to}\:{the}\:{boat}\:{is} \\ $$$$\mathrm{0}.\mathrm{65}+\frac{\mathrm{85}}{\mathrm{165}}×\mathrm{0}.\mathrm{65}\approx\mathrm{0}.\mathrm{98}{m}. \\ $$
Commented by physicstutes last updated on 28/Apr/21
sir the boat is said to move backward 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?
$$\mathrm{sir}\:\mathrm{the}\:\mathrm{boat}\:\mathrm{is}\:\mathrm{said}\:\mathrm{to}\:\boldsymbol{{move}}\:\boldsymbol{{backward}}\:\mathrm{therefore}\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{force} \\ $$$$\mathrm{involved}\:\mathrm{right}? \\ $$$$\:\mathrm{equal}\:\mathrm{and}\:\mathrm{opposite}\:\mathrm{forces}.\:\mathrm{the}\:\mathrm{man}\:\mathrm{gets}\:\mathrm{a}\:\mathrm{force}\:\mathrm{to}\:\mathrm{move}\:\mathrm{to}\:\mathrm{the}\:\mathrm{shore}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{boat}\:\mathrm{gains}\:\mathrm{an}\:\mathrm{opposite}\:\mathrm{force}\:\mathrm{to}\:\mathrm{move}\:\mathrm{backwards}..\mathrm{but}\:\mathrm{accelerations}\:\mathrm{will} \\ $$$$\mathrm{not}\:\mathrm{be}\:\mathrm{equal}\:\mathrm{since}\:\mathrm{masses}\:\mathrm{are}\:\mathrm{different}.\:\mathrm{right}? \\ $$
Commented by mr W last updated on 29/Apr/21
Commented by mr W last updated on 29/Apr/21
black=boat red=fisherman COM=center of mass from boat & man when the man jumps from the boat, he acts a force to the boat, this force makes the boat to move to left, but the boat acts also the same force in opposite direction to the man and makes ihm to move to right. both forces are internal forces inside the object “boat & fisherman”. on the object “boat & fisherman” itself no extermal force acts in this process, therefore the COM of the object “boat & fisherman” remains unchanged.
$${black}={boat} \\ $$$${red}={fisherman} \\ $$$${COM}={center}\:{of}\:{mass}\:{from}\:{boat}\:\&\:{man} \\ $$$${when}\:{the}\:{man}\:{jumps}\:{from}\:{the}\:{boat}, \\ $$$${he}\:{acts}\:{a}\:{force}\:{to}\:{the}\:{boat},\:{this}\:{force} \\ $$$${makes}\:{the}\:{boat}\:{to}\:{move}\:{to}\:{left},\:{but} \\ $$$${the}\:{boat}\:{acts}\:{also}\:{the}\:{same}\:{force}\:{in}\: \\ $$$${opposite}\:{direction}\:{to}\:{the}\:{man}\:{and} \\ $$$${makes}\:{ihm}\:{to}\:{move}\:{to}\:{right}.\:{both} \\ $$$${forces}\:{are}\:{internal}\:{forces}\:{inside} \\ $$$${the}\:{object}\:“{boat}\:\&\:{fisherman}''.\:{on} \\ $$$${the}\:{object}\:“{boat}\:\&\:{fisherman}''\:{itself} \\ $$$${no}\:{extermal}\:{force}\:{acts}\:{in}\:{this}\: \\ $$$${process},\:{therefore}\:{the}\:{COM}\:{of}\:{the} \\ $$$${object}\:“{boat}\:\&\:{fisherman}''\:{remains} \\ $$$${unchanged}. \\ $$
Commented by mr W last updated on 29/Apr/21
if one really understands the physics, he sees the boat and the fisherman as a single object and that this object doesn′t move, although both boat and fisherman are in motion. with this understanding he gets the correct result without much calculation. if one is good in physics, he knows how to calculate to get the correct result. here is how to calculate: say F is the force between boat and fisherman as he jumps from boat. say this force acts in a short time Δt. V=speed of boat (to left) v=speed of fisherman (to right) MV=FΔt mv=FΔt MV=mv ⇒(V/v)=(m/M) the fisherman moves d_1 =0.65m in time t=(d_1 /v). in this time the boat moves d_2 =Vt=V×(d_1 /v)=(V/v)×d_1 =(m/V)×d_1 =((85)/(165))×0.65 the distance between both is then d_1 +d_2 =0.65+((85)/(165))×0.65≈0.98m.
$${if}\:{one}\:{really}\:{understands}\:{the}\:{physics}, \\ $$$${he}\:{sees}\:{the}\:{boat}\:{and}\:{the}\:{fisherman} \\ $$$${as}\:{a}\:{single}\:{object}\:{and}\:{that}\:{this}\:{object} \\ $$$${doesn}'{t}\:{move},\:{although}\:{both}\:{boat}\:{and} \\ $$$${fisherman}\:{are}\:{in}\:{motion}.\:{with}\:{this} \\ $$$${understanding}\:{he}\:{gets}\:{the}\:{correct} \\ $$$${result}\:{without}\:{much}\:{calculation}. \\ $$$$ \\ $$$${if}\:{one}\:{is}\:{good}\:{in}\:{physics},\:{he}\:{knows} \\ $$$${how}\:{to}\:{calculate}\:{to}\:{get}\:{the}\:{correct} \\ $$$${result}.\:{here}\:{is}\:{how}\:{to}\:{calculate}: \\ $$$${say}\:{F}\:{is}\:{the}\:{force}\:{between}\:{boat}\:{and} \\ $$$${fisherman}\:{as}\:{he}\:{jumps}\:{from}\:{boat}. \\ $$$${say}\:{this}\:{force}\:{acts}\:{in}\:{a}\:{short}\:{time}\:\Delta{t}. \\ $$$${V}={speed}\:{of}\:{boat}\:\left({to}\:{left}\right) \\ $$$${v}={speed}\:{of}\:{fisherman}\:\left({to}\:{right}\right) \\ $$$${MV}={F}\Delta{t} \\ $$$${mv}={F}\Delta{t} \\ $$$${MV}={mv} \\ $$$$\Rightarrow\frac{{V}}{{v}}=\frac{{m}}{{M}} \\ $$$${the}\:{fisherman}\:{moves}\:{d}_{\mathrm{1}} =\mathrm{0}.\mathrm{65}{m}\:{in} \\ $$$${time}\:{t}=\frac{{d}_{\mathrm{1}} }{{v}}.\:{in}\:{this}\:{time}\:{the}\:{boat}\:{moves} \\ $$$${d}_{\mathrm{2}} ={Vt}={V}×\frac{{d}_{\mathrm{1}} }{{v}}=\frac{{V}}{{v}}×{d}_{\mathrm{1}} =\frac{{m}}{{V}}×{d}_{\mathrm{1}} \\ $$$$=\frac{\mathrm{85}}{\mathrm{165}}×\mathrm{0}.\mathrm{65} \\ $$$${the}\:{distance}\:{between}\:{both}\:{is}\:{then} \\ $$$${d}_{\mathrm{1}} +{d}_{\mathrm{2}} =\mathrm{0}.\mathrm{65}+\frac{\mathrm{85}}{\mathrm{165}}×\mathrm{0}.\mathrm{65}\approx\mathrm{0}.\mathrm{98}{m}. \\ $$
Commented by peter frank last updated on 29/Apr/21
thank you
$${thank}\:{you} \\ $$

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