Mr. Verma (50 kg) and Mr. Mathur (60 kg) are sitting at the two extremes of a 4 m long boat (40 kg) standing still in the water. To discuss a mechanics problem , they come to the middle of the boat. Neglecting friction with water, how far does the boat move on the water during the progress?
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Thanks for asking the question!
ANSWER::
m₁ = 60 kg m₂ = 40 kg m₃ = 50 kg
Let A be origin of the system .
So , we can say that Mr. Verma and Mr. Mathur are at extreme position of the boat .
Centre of mass = ( 60 x 0 + 40 x 2 + 50 x 4 ) / 150 = 280 /150 = 1.87
Centre of mass will be at a distance of 1.87 from point A which is origin of system .
So , when both comes to the mid - point of boat the centre of mass lies at 2m from A .
Shift in centre of mass = 2 - 1.87 = 0.13 m towards right.
And also as there is no external force in longitudinal direction their centre of mass will not shift .
Therefore , the boat will move 0.13 m towards right.
Hope it helps!
ANSWER::
m₁ = 60 kg m₂ = 40 kg m₃ = 50 kg
Let A be origin of the system .
So , we can say that Mr. Verma and Mr. Mathur are at extreme position of the boat .
Centre of mass = ( 60 x 0 + 40 x 2 + 50 x 4 ) / 150 = 280 /150 = 1.87
Centre of mass will be at a distance of 1.87 from point A which is origin of system .
So , when both comes to the mid - point of boat the centre of mass lies at 2m from A .
Shift in centre of mass = 2 - 1.87 = 0.13 m towards right.
And also as there is no external force in longitudinal direction their centre of mass will not shift .
Therefore , the boat will move 0.13 m towards right.
Hope it helps!
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60 kg man is at origin
50 kg man is 4 m to right of 60 of man
Let us take the distance travelled by the boat w.r.t ground=X
Distance traveled by 60 kg man would be=2+X
Distance traveled by 50 kg man would be=2-X
As center of mass will not move w.r.t ground
60(2+X)-50(2-X)+40X=0
Boat will move
40/3 cm to left
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