Question

Two people, one of mass \(\large m_1\) kg and the other of mass \(\large m_2\) kg, sit in a row boat of mass \(\large m\) kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, \(\large L\) m apart from each other, exchange seats.

How far will the boat move?

Answer 1

hmmm.. this will look a little bit complex.. i;ll try to make it simple as much as i can ..

we have to solve this in 2 steps..
first we have to find the distance boat moved when the m1 moves to the other end
then we have to find the distance that boat will move in opposite direction when m2 switch his place

for the first part
lets say the velocity of m1 relative to the boat is V
and the velocity of boat and m2 relative to land is U
then the velocity of m1 relative to the land is V - U

so now we have to applied the law of conservation of momentum
m1( V - U ) = U ( m2 + m)
U ( m1 + m2 + m ) = m1(V)
U = [m1(V)] / [ (m1 + m2 + m ) ]
if the time m1 take to travel the distance "L" is "t"
then Vt = L
and the distance that boat moved in opposite direction during that time is
S = ut
= [m1(V)] / [ (m1 + m2 + m ) ] x t
but V = L/t
so...
S1= [m1(L)] / [ (m1 + m2 + m ) ]

now we have to follow the same process again to find the distance that boat moved when m2 switch his place... its the same as above ..only m1 will be m2 and m2 will be m1

from there we will get the distance that boat moved during the motion of m2 is
S2= [m2(L)] / [ (m1 + m2 + m ) ]
now this displacement take in the opposite direction of the first one which means the total distance the boat moved will be the difference of these 2 distances

which means the distance that boat moved is
S2 - S1
= [ (L)( m2 - m1) ] / [ m1 + m2 + m ]

Answer 2

and this answer will be decide on which mass is greater.. m1 or m2 which is not mentioned in the question

Answer 3

Good job Suru :) <3 @***[ISURU]***

There's another way, just find initial center of mass, and final center of mass, then let them equal to each other.

Since it says the boat is rest then center of mass is also at rest in beg and end.

Answer 4

Yeah I think too @study100

Answer 5

\(\large \frac{L-1}{2}\) is the answer?

Answer 6

\[\large X _{i} = \frac{ m_{1}(0) +m(\frac{L}{2}) + m_{2}(L)}{ m_{1}+m_{2}+m }\]

\[\large X _{f} = \frac{m_{2}(x)+ m(\frac{ L }{ 2 } +x)+m_{1}(L+x)}{ m_{1}+m_{2}+m }\]
then set Xi= Xf
\[\large m_{1}(0) + m(\frac{ L }{ 2 }) + m_{2} (L) = m_{2}(x) + m(\frac{ L }{ 2 } + x) + m_{1} (L+x)\]
and then distribute then simplify.

Answer 7

no @JungHyunRan

Answer 8

\[\large 0 + \frac{ L }{ 2 }m + Lm_{2} = m_{2}x + \frac{ L }{ 2 }m + mx + Lm_{1} + m_{1} x\]
then you cancels it
\[\large Lm_{2} - Lm_{1} = m_2 x + mx + m_{1} x\]
\[\large x = \frac{ L (m_{2} - m_{1} ) }{ m_{1} + m_{2} + m}\]
like suru's answer as well :)

Answer 9

This is my way:

Let's suposse the m1 kg mass is FIRST at left. CM position is with respect to the left would be:
\(\huge \frac{(0.m_1)+(L.m_2)+(\frac{L}{2}.m)}{m_1+m_2+m}\) \(\large m\) from the left.

Now let's suposse that the m2 kg mass is FIRST at the left AGAIN: CM position would be:
\(\huge \frac{(0.m2)+(L.m1)+(\frac{L}{2}.m)}{m_1+m_2+m}\) \(\large m\) from the left.

The difference between the two cases is:
\(\Large \frac{(0.m_1)+(L.m_2)+(\frac{L}{2}.m)}{m_1+m_2+m}-\frac{(0.m2)+(L.m1)+(\frac{L}{2}.m)}{m_1+m_2+m}=\frac{L(m_2-m_1)}{m_1+m_2+m}\)

Answer 10

We know No external forces acting and then the center of mass wil not be moved.

Answer 11

Yes, no external force :)
and your way is the least painful one to solve! :)) <3

Answer 12

^_^ thank you

Answer 13

for your help

Answer 14

\(\color{salmon}{\large\heartsuit}\large\text{good job, you're most welcome!}\color{salmon}\heartsuit\)
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