# Question about the effects of mass transfer on momentum

Two boats move parallel to and towards each other on a stagnant lake with the same speed. At the moment they are side by side a box is being transferred quickly from the first boat to the other. I am asked the mass of the boat that takes the box with the mass of the first boat, its speed, the box's mass, the other boat's mass and its speed before and after having the box transferred.

I have checked the solution and this is a simple momentum problem and I can understand that there is going to be a change in momentum for each boat since one loses mass and the other gains it but why should the change in momentum be equal for the two objects? I know about the conservation of momentum but the proof given for C.o.M is based on the fact that when two objects interact they exert an equal amount of force on each other for an equal amount of time which results in an equal impulse and thus equal change in momentum. Like when there is a collision or a firing of a cannon ball. There are force pairs that make an equal impulse on both bodies possible. But I can't see that in this example. Where are the force pairs? I mean, yes, we could argue that the box was carried by force but I believe we would still have done the same calculations if the box had just disappeared on one boat and then had emerged on the other. How and why would that work? And how does this compare to an example where there is only one boat moving with constant speed and then a box emerges out of nowhere. How would we calculate the change in velocity?

I hope my understanding of momentum isn't completely flawed.

## 1 Answer

Suppose I'm in the first boat carrying the mass, $m$, and you're in the second boat. And let our closing speed be $v$.

As we pass I drop the mass into your boat. I do this without changing the velocity of the mass, so no force is exerted on me and my boat carries on at an unchanged speed.

But now you find you have a mass in your boat and moving at a speed $v$ relative to you. Unless you want the mass to slide out of your boat and splash into the lake you have to exert a force on the mass to reduce its velocity relative to you to zero. The impulse needed is obviously just the change in momentum of the mass $mv$.

But Newton's third law tells us that while you are exerting a force on the mass it is exerting an equal and opposite force on you, and it exerts the force on you for the same time that you exert a force on it. Therefore the impulse exerted on you is equal and opposite to the impulse you exert on it, and your change of momentum will be $-mv$.

• I had never thought about it like that. Thank you so much! On the other hand, how would the first boat be affected in speed? – Mert Baştuğ Jul 8 '17 at 15:19
• Actually I believe this is not quite true. The boat receiving the box will end up with a different speed, so the change in momentum of the boat is not quite -mv (since the speed of the box will not change by v, but by $(v-\Delta v)$ where $\Delta v$ is the change in speed of the second boat... – Floris Jul 9 '17 at 14:52