# What happens when a man walks on a canoe? what are the forces acting on canoe and man and how to describe it with law of conservation of momentum?

In my physics class our teacher taught us about this canoe-man problem, where a man walks across a canoe and due to the "law of momentum conservation", the canoe attains a constant $$v$$ velocity, and the man attains constant $$u$$ velocity. Is this possible because, each step he exerts a force on the canoe and that gives the canoe and man different velocities, or should I assume that with 2,4,6... steps the canoe stops as $$Mv = mu$$ as in Law of Conservation of Momentum ,and in 1,3,5... steps the canoe starts to move again?

1. The friction of water is considered negligible.

2. If my assumptions are wrong, what kind of motion does the canoe and the man has?

3. Does the canoe has a "go-stop-go-stop" motion?

4. Does it move in constant acceleration?

If you want more clarification on my question please let me know....

• what do you mean by this I assume that with 2,4,6... steps the canoe stops as 'Mv = mu' as in L.O.C.M ,and in 1,3,5... steps the canoe starts to move again ? Commented Oct 22, 2020 at 10:40

the canoe attains a constant 'v' velocity, and the man attains constant 'u' velocity. Is this possible because, each step he exerts a force on the canoe and that gives the canoe and man different velocities,...

If your teacher is suggesting that the velocities of the man and canoe must be constant for momentum to be conserved, I believe your teacher is incorrect. I don't think it matters that the velocities and the momentum of the man and canoe are changing in time. The important thing is that at any instant in time, the instantaneous momentum of the man is his mass times his instantaneous velocity and the momentum of the canoe is its mass times its instantaneous velocity. It is sum of the instantaneous momentum of the man plus canoe that is conserved, or

$$m_{man}v_{man}+m_{canoe}v_{canoe}=constant$$

If the man and canoe velocities are not constant, then the velocities are instantaneous velocities.

Keep in mind that momentum is a vector quantity and that at any instant in time the sign of velocity of the man and thus his momentum, is opposite to the sign of the velocity of the canoe and it momentum, since the man and canoe are moving in opposite directions with respect to the center of mass of the system.

...or should I assume that with 2,4,6... steps the canoe stops as 'Mv = mu' as in L.O.C.M ,and in 1,3,5... steps the canoe starts to move again.

Having trouble following the numbers. For example, if you are referring to the numbered items, there is not 5 and 6. I will just address the significance of 1-4 on the analysis.

1. the friction of water is considered negligible.

This is a necessary condition if the man and canoe are to be considered an isolated system, and in order for momentum to be conserved the system must be appropriately isolated. In the present example we are dealing with horizontal momentum. For conservation of horizontal momentum, there can be no net external horizontal forces acting on the system. Water friction on the canoe would constitute such an external force. (Gravity is not a horizontal force so it doesn't affect the current example).

2. if my assumptions are wrong, what kind of motion does the canoe and the man has.

I don't believe your assumptions are wrong because the details of the motion of the man and canoe don't matter. As long as the system is isolated, at any instant in time the instantaneous momentum of the system is conserved (constant). It equals zero if the center of mass of the system was initially at rest with respect to the water.

3. does the canoe has a "go-stop-go-stop" motion?

Yes, if the man takes a step, stops and waits a while, repeating the sequence until reaching the other end of the canoe. When he stops, his and the canoes momentum with respect to the center of mass is zero. When he takes a step and accelerates, the sum of his and the canoes instantaneous momentum (based on their instantaneous velocities) is zero.

> 4. does it move in constant acceleration?

The COM of the man-canoe system has no acceleration if the system is isolated, that is, if the system is not subjected to a net external horizontal force.

The force that the man exerts on the canoe and the canoe exerts on the man are internal forces having no effect on the motion of the system as a whole. Those internal forces are equal and opposite per Newton's 3rd law.

If it is a step and stop sequence, there is combination of an initial acceleration of the man and canoe with respect to the COM of the system when the man initiates the step and deceleration when the man ends the step (stops).

Hope this helps.

• sorry ,I wasn't refering to the numbered list. I was refering to the steps of the man who walks on the canoe. Commented Oct 23, 2020 at 13:34
• @donthababakka No problem. I'll just edit the statement out Commented Oct 23, 2020 at 13:38

Let's see Newton's second law of motion first :

$$F_{ext} = ma$$

We know how it is applicable for a single object but when we consider two or three bodies at a time, we need to define a common point for all those bodies and apply the above equation for that common point which we call the center of mass.

Now coming to your question :

First of all we all are able to walk on a surface because of friction. What actually happens is that we push the surface backward and in return the surface pushes us forward and hence we move.

In your case , let's say that your mass is $$m_o$$ while the mass of the canoe is $$m_c$$. So when you apply a force on the canoe in the backward direction say $$F_r$$ then the canoe starts accelerating backwards with

$$a_c= \frac{F_r}{m_c}$$

Now you know that canoe also pushes you forward with the same force $$F_r$$ . So your acceleration is

$$a_o = \frac{F_r}{m_o}$$

So you accelerate forward while the canoe accelerates backward.

But what if we consider you as well as the canoe both at the same time and apply Newton's second law for the common point i.e the center of mass (shown as violet point in the picture).

Since there is no external force on that center of mass we conclude that the acceleration of the center of mass is zero.

$$0=m×a_{c.o.m}$$

Also from here , you can conclude that since there is no external force the net change in linear momentum of the system should be zero . This signifies that the linear momentum of you and the canoe cancels each other i.e both are equal and opposite.

Although your and canoe's individual momentum is changing but for the system it is a constant.

And thus the center of mass of your system doesn't move from its initial position.

You are right that the water applies forces (like friction, buoyant) too on the canoe but we generally ignore it in mechanics. However if you will consider that force too then you should count it as the external force and it will act on both the bodies i.e on the center of mass of the system.

NOTE : I have considered the initial state to be at rest If the canoe was initially moving then the common point i.e the center of mass will keep on moving with the constant velocity even if you walk on the canoe .

Hope it helps ☺️.

If the man is walking normally, the canoe will indeed have a stop go stop go motion, in sync with the man's walking.

For simplicity, the teacher has considered the man moving with constant velocity, which isn't a very intuitive example. Instead, imagine a scooter on the plank moving forward with constant velocity. As the scooter starts moving forward because of friction, the plank will start moving backward. Once its velocity is constant, there is no longer a frictional force between the two surfaces, so the plank's velocity is also constant. When the scooter stops, it stops due to backwards friction, and consequently, the plank stops due to forward friction.

Whatever the man does on top of the canoe, the center of mass (COM) moves at a constant velocity. As long as no momentum is transferred due to friction. If you want to find the center of two masses you draw a line between the centers of masses of each object. If the two objects are of the same mass you draw the COM in the middle. If one of the objects is larger you draw it slightly more towards the heavier mass.

Since the man is heavier than the canoe the COM is closer to the man.

Let's start with the scenario where both the man and the canoe are at rest. The COM is also at rest so under these assumptions it will always stay at rest. When the man starts walking he gains some velocity but the canoe gains some velocity in the opposite direction by conservation of momentum. When stops moving the canoe also stops. Since the COM is closer to the man we can reason that the canoe will make bigger movements. Like a seesaw where one of the arms is bigger than the other. The man would be at the short arm (closer, smaller movements) and the canoe at the large arm (further away, larger movements) and the COM in the middle (stationary).

The fact that the canoe makes larger movements can also be seen by looking at momentum. Momentum is defined by $$p=mv$$. When everything is at rest the velocities are all zero so the total momentum is zero: $$p_\text{man}+p_\text{canoe}=m_\text{man}\cdot 0+m_\text{canoe}\cdot 0=0$$ When the man is moving the total momentum is still zero. So we have $$m_\text{man}v_\text{man}+m_\text{canoe}v_\text{canoe}=0$$ We see that a) one of the velocities has to be negative and b) since the mass of the canoe is smaller it has to have a larger speed to cancel the momentum of the man. This larger speed at every time results in bigger movements.

Finally when the COM is moving it will do so at a constant velocity. The same reasoning as above applies but now both objects have some velocity.