Two people, one of mass 90 kg and the other of mass 60 kg , sit in a rowboat of mass 78 kg . With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, 3.5 m apart from each other, now exchange seats. How far will the boat move?

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aristocles aristocles · Answer 1 · 2019-08-04T17:38:10+02:00

Answer:

boat will move by 0.46 m

Explanation:

Here as we know that the two persons are sitting on the boat at rest

And here the boat along with all persons are at rest

There is no other force on this system

So here the center of mass of the system will remains at rest always

so the displacement of the center of the mass will be zero

[tex]\Delta x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}[/tex]

now we can say that boat moves a distance "x" opposite to the motion of 90 kg person

so we will have

[tex]0 = \frac{90(3.5 - x) + 60(-x - 3.5) + 78(-x)}{90 + 60 + 78}[/tex]

[tex]0 = 315 - 90x - 60x - 210 -78x[/tex]

[tex]228 x = 105[/tex]

[tex]x = 0.46 m[/tex]

shubhamchouhanvrVT shubhamchouhanvrVT · Answer 2 · 2022-03-19T12:37:08+01:00

The distance traveled by boat will be X=0.46m

It is known that

Since two persons are sitting on a boat which is at rest

The boat along with all the persons is at rest

There is no other force on this system

So the center of mass of the system will remain constant

so the displacement of the center of the mass will be zero

[tex]\Delta X_{cm}=\dfrac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}[/tex]

now we can say that boat moves a distance "x" opposite to the motion

of 90 kg person

[tex]\Delta X_{cm}=\dfrac{90(3.5-x)+60(-x-3.5)+78(-x)}{90+60+78}[/tex]

[tex]0=315-90x-60x-210-78x[/tex]

[tex]228x=105[/tex]

[tex]x=0.46\ m[/tex]

Hence the distance traveled by boat will be X=0.46m

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