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The problem is as follows:

A person whose mass is $80\,kg$ is standing over a canoe of $6\,m$ in length and whose mass is $400\,kg$. Both are initially at rest as it is shown in the figure from below. Then the person starts walking from one end to the other of the canoe. Find the distance to the dock when the person has reached the other end of the canoe.

Sketch of the problem

The alternatives given are as follows:

$\begin{array}{ll} 1.&5\,m\\ 2.&4\,m\\ 3.&1\,m\\ 4.&3\,m\\ 5.&2\,m\\ \end{array}$

I'm still stuck with this problem. I'm assuming that it has to do with the conservation of the momentum but I don't know how to use it here. I'm assuming when the person walks the boat starts moving to the right as the person walks to the left. But how exactly can I relate it with the lenght?. Can somebody help me here?

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  • $\begingroup$ Postulate: This should be the same for any way the person moves (Slow, fast, bursts, constant accel). Try with constant acceleration, and postulate that it takes one second to reach the end of the canoe. Then postulate that the forces on human and canoe need to be equal (though opposite). Then postulate that the length covered by the canoe and the human need to add up to 6m (canoe goes one way, human the other, so for the human to reach the other end, six meters need to be covered); F=ma; s=1/2*at²; $\endgroup$ – bukwyrm Dec 9 '19 at 11:00
  • $\begingroup$ You just need conserving the centre of mass, you aren't given any velocities, no need for conserving momentum $\endgroup$ – ab123 Dec 9 '19 at 11:03
  • $\begingroup$ This question should be posted in Physics SE. $\endgroup$ – Shubham Johri Dec 9 '19 at 11:04
  • $\begingroup$ @ShubhamJohri Due the nature of Physics SE. they do not help in homework problems as it is stated in this problem and the confusion is more regarding the mathematical aspect of this. $\endgroup$ – Chris Steinbeck Bell Dec 9 '19 at 11:12
  • $\begingroup$ Conservation of momentum/centre of mass are not mathematical laws. $\endgroup$ – Shubham Johri Dec 9 '19 at 11:13
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I believe you have to consider the man and the canoe as a single system, so that any frictional force between the two is an internal force of the system. Assuming resistance due to water is negligible, there is no net external horizontal force on the system, so the $x$ coordinate of its centre of mass is conserved.

Now assume the right edge of the dock to be the origin. Also let the centre of mass of the canoe lie in its middle.

Abscissa of initial centre of mass is$$\frac{m_\text{man}\cdot x_\text{man}+m_\text{canoe}\cdot x_\text{canoe}}{m_\text{man}+m_\text{canoe}}=\frac{80\cdot6+400\cdot3}{80+400}$$Abscissa of final centre of mass, after the man reaches the other end of the canoe, is$$\frac{80\cdot x+400\cdot(x+3)}{80+400}$$Now equate the two and find $x$.


We can solve the above exercise by conservation of momentum too. There are no net external horizontal forces on the system. The initial horizontal momentum is $0$, and is conserved. This means that at any instant,$$m_\text{man}v_\text{man}-m_\text{canoe}v_\text{canoe}=0$$where all velocities are in ground frame. Thus, $v_\text{canoe}=\frac15v_\text{man}$. The leftward distance moved by the man in infinitesimal time in ground frame is $v_\text{man}dt$. The total distance moved to the left is then $\int_0^Tv_\text{man}dt$, where $T$ is the time in which the man reaches the other end of the canoe. Clearly,$$\int_0^T(v_\text{man}+v_\text{canoe})dt=6\implies\int_0^Tv_\text{man}dt=5$$and the distance moved to the left is $5$. Since he was originally $6~m$ away, the man is now $1~m$ away from the dock.

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  • $\begingroup$ Can this problem be solved using the conservation of momentum?. $\endgroup$ – Chris Steinbeck Bell Dec 9 '19 at 11:13
  • $\begingroup$ This problem is a classic exercise in centre of mass concepts. It was designed for this particular topic, and I'm afraid I have never tried solving it using momentum. I can try to solve it using momentum, but it will take a while. @ChrisSteinbeckBell $\endgroup$ – Shubham Johri Dec 9 '19 at 11:17
  • $\begingroup$ @ChrisSteinbeckBell I have added the conservation of momentum approach. $\endgroup$ – Shubham Johri Dec 9 '19 at 11:50
  • $\begingroup$ @ShubhamJori Should'nt be that the man moves to the left then be $-v_{\textrm{man}}$ and that of the boat moves to the right then be $v_{\textrm{canoe}}$, where does it come $v_{\textrm{man}}=\frac{1}{5}=\frac{80}{400}$? The part where I'm stuck at is how you obtained the distance away from the dock?. He walks to the left is only $5\,m$ but doesn't it say that he walks all the length of the canoe which is $6\,m$?. I'm just assuming that the length that I obtained I have to subtract after walking. Can you help me to clear out this doubt please?. $\endgroup$ – Chris Steinbeck Bell Dec 9 '19 at 12:01
  • $\begingroup$ Yes, the canoe moves to the right with $v_\text{canoe}$ and the man to the left with $-v_\text{man}$. But since the total momentum is $0$, I have multiplied the momentum equation by $-1$, which really doesn't matter. You get $v_\text{canoe}=\frac{m_\text{man}}{m_\text{canoe}}v_\text{man}$, i.e. $1/5$ of the man's speed. $\endgroup$ – Shubham Johri Dec 9 '19 at 12:05
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Think of this in a 1-D plane. I assume there's no friction due to water. Now consider the canoe-person system. The center of mass of canoe is at its center, consider this as origin. $3m$ to the right is a particle of mass $80kg$. So the center of mass of this system is at $$\frac{400\times 0+80\times 3}{400+80}=.5m (right of the boat)$$. Now if that particle of mass $80kg$ move to the left by $6m$ the center of mass should remain at the same position (conservation of momentum). For that, the center of mass of canoe-person will shift, which can be calculated by $$\frac{400\times 0+80\times (-3)}{400+80}$$ So the center of mass shifted to left by $.5m$ in the relative frame. Now according to the conservation of momentum, the center of mass should be in the same position. So the canoe should to the right by $1m$ to keep the COM at the same position. So that person is $1m$ away from the center.

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  • $\begingroup$ This is incorrect. The centre of the canoe, the origin of your choice, itself moves to the right. $\endgroup$ – Shubham Johri Dec 9 '19 at 11:09
  • $\begingroup$ Consider the canoe as a frame of reference. Beside our answers yield the same answer. $\endgroup$ – Naman Jain Dec 9 '19 at 11:20
  • $\begingroup$ My earlier comment was posted before you edited your answer. I am removing my downvote as the error has been corrected now. But I also think the last statement needs justification using Physical formulae. $\endgroup$ – Shubham Johri Dec 9 '19 at 11:26
  • $\begingroup$ As regards your frame of reference argument, the centre of mass is not conserved about a translating origin. $\endgroup$ – Shubham Johri Dec 9 '19 at 11:30
  • $\begingroup$ The mathematical calculation would be something like this- Since the COM moved from $.5m$ to $-.5m$ with the canoe as the frame of reference. But the COM should be at rest from the dock as a frame of reference. So the change in COM should be $.5m-(-.5m)=1m$ in order for COM of system to be at rest. $\endgroup$ – Naman Jain Dec 9 '19 at 11:57
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You need to equate the position of the center of mass of the system (man + boat) staying in the same place horizontally as there is no external force on the system. The man's forces are internal to the system.

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