Menu Close

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-




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
$${thank}\:{you} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *