Conservation Of Momentum question

[Fladam]

We've been discussing the principle of Conservation of Momentum lately in class and I've been wondering, the principle states that for two bodies:
m1u1+m2u2= m1v1+m2v2

I understand that there are some rules which surround this principle but we haven't been shown them in full and I was thinking in class and came up with the following scenario, so I was wondering what rules would come into play in the following events...

Say a lorry travels at a speed of u=20ms^-1
and has m=2000kg, so p= 40000 kgms^-1

Now if there is a wall made from solid concrete which has u=0ms^-1 but a mass of perhaps 500kg, then it has zero momentum of course (p=0kgms^-1)

My question is, if the scenario arises where the lorry crashes through the wall, the lorry loses some energy from the collision and it has its velocity ( and hence its momentum) reduced.

The wall also breaks into pieces but it gains energy from the collision, now the wall as a whole remains at zero velocity as it is cemented to the ground, however the pieces which fly off all gain momentum. All of the pieces will have variable masses and velocities and so their momentums will be different, would the sum of the momentum of the pieces add up to allow for the change in momentum for the lorry? Or are there some rules which govern the principle of conservation of momentum that this scenario doesn't obey?

Thanks very much for all your answers, this is my first post on here and I look forward to many more :) I apologize if the answer is a blatant break of the rules of the principle also!

 

[McLaren Rulez]

Yes, the pieces flying off carry away the momentum that the lorry lost. Momentum will be conserved in this example since there are no external forces in the direction that the lorry travels in.

However, this is an inelastic collision and therefore you will find that the kinetic energy of the lorry and the flying bits of wall are less than the kinetic energy of the lorry before it crashed. Some energy is spent in breaking the wall.