two fishermen, of masses 70 and 90 kg stand at opposite ends of their 20 meter...

two fishermen, of masses 70 and 90 kg stand at opposite ends of their 20 meter boat. the boat ( without fisherman) has a mass of 400 kg. There is no wind or current and the boat can move without friction on the water's surface. The 90 kg fisherman walks to the left end of the boat. How far the moved when the fisherman reaches the left end?

Expert Solution

as there are no external forces,centre of mass will remain at rest

centre of mass of the system in initial position

XCM= (M₁X₁ + M₂X₂+M₃X₃) / (M₁+M₂+M₃₎

considering left end of the boat as origin

XCM= (70*0 + 400*10+90*20) / (70+400+90)

XCM=10.3571 m from left end

suppose boat moves k m in the left direction,so centre of mass will remain and rest and will be equal to Xcm

XCM= (M₁X₁ + M₂X₂+M₃X₃) / (M₁+M₂+M₃₎

10.3571=(70+90)(-k)+(400)(10-k)/(70+90+400)

10.3571=(-160k+4000-400k)/560

-560k+4000=5799.976

k=-3.214 m

which means boat will move in right diresction

genius_generous answered 2 years ago

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