A silly question (perhaps) - Conservation of momentum for a cannon firing a cannonball

[Shaye]

Homework Statement:

A canon fires a cannonball of mass 55kg at 35ms^-1. The cannon recoils at 2.5 ms^-1.

What is the mass of the cannon?

Relevant Equations:

p=mv

Maybe a silly question but on the above question using the conservation of momentum:

momentum before firing (0) = momentum after firing (55*35)+(M*2.5)

If I re-range the above it's M = -(55*35)/2.5 = -770kg. I can I reconcile that minus sign (basically get rid of it)?

Thanks

 

[hutchphd]

The momentum and velocity are vector quantities. The velocities are in opposite directions so one of them is negative.

 

[PeroK]

It wasn't the question that was silly: it was having the cannon recoil in the same direction as the cannonball!

 

[Mayhem]

With these kinds of mechanics problems, making a vector drawing is really useful for keeping track of your signs ... also remember units.

 

[Shaye]

Thanks everyone! A DOH! moment for me. Keeping track of the vectors is very useful

 

[kuruman]

I can I reconcile that minus sign (basically get rid of it)?

Actually the question is not silly in my opinion, yes there is a way to reconcile the negative sign without fudging the answer and I think @Shaye will profit from seeing how.

In mechanics problems, one often assigns a direction to an unknown vector and a symbol to its magnitude, goes through the calculations and if the magnitude turns out negative, then the actual direction is opposite to the initially assigned direction. The same method is famously used in circuit analysis when one assigns a direction to the current in a given loop, goes through the calculations and, if said current turns out negative, it actually flows in the opposite direction.

Here we have exactly the same principle at work. @Shaye chose the recoil momentum to be in the same direction as the cannonball, the magnitude of the momentum came out negative after the calculation, therefore $\dots$