How do I find using the conservation of momentum the distance between a dock and a boat?

Jun 2017
337
6
Lima, Peru
The problem is as follows:

A person of mass $m$ is standing in the rear end of a boat of mass $5m$. Assuming the boat does have its engine turned off and put upwards. The person begins to walk over the boat as indicated until reaching the front end of the boat. Using this information. Find the length that the boat has recoiled back which the person should try either to jump or swim to reach the dock.



The alternatives given are as follows:

$\begin{array}{ll}
1.&\textrm{2 m}\\
2.&\textrm{1.2 m}\\
3.&\textrm{1 m}\\
4.&\textrm{3 m}\\
4.&\textrm{2.5 m}\\
\end{array}$

This problem kind of begs for a solution using the conservation of the center of mass. However I'm not sure if does it exist another approach. Let's say, conservation of momentum?. Can this procedure be used?. Can someone help me with this matter as well?. Please!
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
initial momentum of the system is zero

$mv_p + 5mv_b = 0$

$v_b = -\dfrac{1}{5}v_p$

since both move the same amount of time, the boat moves $\dfrac{1}{5}$ the distance the person moves in the opposite direction, $\dfrac{1}{5} \cdot 6 = 1.2$ meters
 

romsek

Math Team
Sep 2015
2,875
1,608
USA
Well first you have to use conservation of momentum to find out how the boat reacts to the person walking on it.
Assume that at $t=0$ the boat and the person's velocity are $0 ~m/s$

$0=m v_p + 5m v_s\\

v_s = -\dfrac 1 5 v_p$

The time it takes for the person to walk to the front of the boat is

\(\displaystyle T = \dfrac{6}{v_p}\)

and in this time the boat moves backwards a distance of

$d=v_s T = 1.5 v_p T = \dfrac{6}{5}~m = 1.2 ~m$

(Skeeter so fast)