How do I find distance from a dock when a person walks in a canoe?

Question

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.

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?

Answer 1

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 $x$.

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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.

Answer 2

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.

Answer 3

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.