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

Jun 2017
276
6
Lima, Peru
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. Can someone help me here?.
 

topsquark

Math Team
May 2013
2,381
993
The Astral plane
When is momentum conserved (aside from in collisions)? Since the interface with the water and canoe can be considered to be negligible what can you say about the motion of the center of mass of the man-canoe system?

(This also applies to your platform on ice problem.)

-Dan
 
Jun 2017
276
6
Lima, Peru
When is momentum conserved (aside from in collisions)? Since the interface with the water and canoe can be considered to be negligible what can you say about the motion of the center of mass of the man-canoe system?

(This also applies to your platform on ice problem.)

-Dan
The only thing which comes to my mind is to let that the momentum at the beginning is zero and when the person walks to the left the boat moves to the right. Both move at the same speeds but with the direction swapped.

$p_i=p_f$

I'm confused at exactly how should I tackle the center of mass approach that you mention. The source of the confusion is where should I put the center of mass for the person and for the boat?. I can only tell that the center of mass at the beginning will be the same at the end.
 
Jun 2019
493
260
USA
Let $\vec{r}_1$ be the location of the centre of mass of an object of mass $m_1$, and $\vec{r}_2$ the same for mass $m_2$. If you treat $m_1$ and $m_2$ as a single object (or system), the centre of mass of the system will be $\displaystyle \vec{r}_{1+2} = \frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}$.
(Make this a volume integral, and it is how we define centre of mass in the first place.)
 
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Jun 2017
276
6
Lima, Peru
Let $\vec{r}_1$ be the location of the centre of mass of an object of mass $m_1$, and $\vec{r}_2$ the same for mass $m_2$. If you treat $m_1$ and $m_2$ as a single object (or system), the centre of mass of the system will be $\displaystyle \vec{r}_{1+2} = \frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}$.
(Make this a volume integral, and it is how we define centre of mass in the first place.)
Although integrals can be used. Can I have a calc-free version please? :rolleyes:

What would be the way to approach this problem using the conservation of momentum?.
 
Jun 2019
493
260
USA
Conservation of momentum tells you that both objects start stationary, and both end stationary. This isn't enough to solve the problem, so we have to go one step further and say the centre of mass doesn't move. Look at the equation; you can use it to write an equation for how far the boat moves as a function of how far the man moves.
 

skeeter

Math Team
Jul 2011
3,196
1,724
Texas
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. Can someone help me here?.
Center of mass method ...

Let $x=0$ be the position of the dock's end.

center of mass of the person/canoe system is $x_c = \dfrac{400 \cdot 3 + 80 \cdot 6}{400+80} = 3.5 \text{ m}$ from the dock's end

once the person gets to the oppsite end of the canoe closest to the dock, the canoe moves away from the dock a distance x

the center of mass of the system remains stationary for any movement by the person ...

$3.5 = \dfrac{80 \cdot x + 400(x+3)}{80 + 400} \implies x = 1 \text{ m}$


Momentum method ...

$p_0 = 0$ and the person walks 6m in t seconds $\implies v_p = \dfrac{6}{t}$ ... note $v_p$ is velocity relative to the canoe

if $v_c$ is the velocity of the canoe relative to the dock, then $v_c-v_p$ is the person's velocity relative to the dock

$400v_c + 80(v_c-v_p) = 0 \implies v_c = \dfrac{v_p}{6}$

velocity * time = distance $\implies v_c \cdot t = \dfrac{v_p}{6} \cdot t = v_p \cdot \dfrac{t}{6} = \dfrac{6}{t} \cdot \dfrac{t}{6} = 1 \text{ m}$
 
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