1
$\begingroup$

To be clear I have indeed reviewed the question asked by helios321 (Classic man on boat problem). But i have something else to ask related to man on a boat problem.

The man on a boat problem goes like this: A man is standing on one side of a boat and the boat is stationary. We ignore friction between water and boat (and air friction). Thus there are no external forces on the man+boat system. So momentum is conserved, and centre of mass does not move. (Copied from helios321's post)


I know that if the man moves to the other side of the boat the boat moves in the opposite direction. But what i don't understand is :

Let the boat move $x$ m to left and the man $(L-x)$m to right.[L is the length of the boat] then how can we say that
$M_{man}(L-x) = M_{boat}(x)$

$\endgroup$
1
$\begingroup$

As the man begins to move, the boat begins to move in the opposite direction. So when the man has moved, say forward with respect to the boat the boat meanwhile has drifted backwards. If one would calculate their center of mass it would be at the same place as before. And if one would sum up the momentum vectors of the two bodies that is the man and the boat, the resultant would be zero.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Well, before the man started moving, lets find the location of the Centre of mass of our system consisting of the man and the boat.
Let us assume the man to be the origin.
The centre of mass of the boat is at $\frac {L}{2} $ distance from the man.
Hence, the centre of mass is at $\frac {M_{boat} \frac {L}{2}}{M_{boat} + M_{man}} $

Now, when the man moves to the other end of the boat to the right, let us assume that the boat(i.e. its Centre of mass) moves $x$ distance to the left.
Hence, now the man is at $(L-x)$ distance from the previously chosen origin and the boat's centre of mass is at $ \frac {L}{2} - x $ distance from the same origin.

Now, the centre of mass is at $\frac {M_{man}L - M_{man}x + M_{boat}\frac {L}{2} - M_{boat}x}{M_{man} + M_{boat}}$

Now, since the $V_{CM}$ (Velocity of the Centre of mass of the system) is zero. we will equate both the position of centre of mass. And after doing so we get the expression

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

There is no external force on the system right? So shouldn't the centre of mass remain stationary?

Let me give you an example too. A man is standing on the boat and jumps onto the pier. As a result, the boat moves backward and centre of mass of system is still at rest

So back to this problem, the boat also acquires a velocity opposite in direction to the man, though of a lesser magnitude(Considering the boat is heavier than man). Hence the net velocity of centre of mass results to zero

Hope it helped

Edit: Answering the new question Now i hope you believe that the centre of mass is at rest right? So that means its position does not change(As the system is initially at rest). You know that Msystem*Xcm,initial =Mman Xman,initial+MboatXboat,initial Msystem*Xcm,final =Mman Xman,final+MboatXboat,final

As Xcm,initial=Xcm,final, on subtracting we get

Mman*(change in Xman)+Mboat*(change in Xboat)=0 So Mman*(L-x)-Mboat*x=0 (- sign is because boat moves left) I hope you got it now.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ i was able to understand that the center of mass will be at rest. But could you please help me with my new problem regarding the same man on a boat problem. I have edited it now. $\endgroup$ – Koustubh Jain Apr 26 at 11:16
  • $\begingroup$ yes i have added it to the answer. please check and tell me if you got it $\endgroup$ – Vamsi Krishna Apr 26 at 13:21
  • $\begingroup$ thanks but i got the answer myself $\endgroup$ – Koustubh Jain Apr 26 at 19:01
  • $\begingroup$ Great but please upvote as we would like some credit after spending time for this $\endgroup$ – Vamsi Krishna Apr 27 at 4:38
  • $\begingroup$ Well, i tried to but it doesn't count as i have less than 15 reputation $\endgroup$ – Koustubh Jain Apr 28 at 5:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.