Momentum vs Kinetic energy

<Suppose a railway ore truck is rolling along its tracks @ velocity v, with mass m. As it passes under a loading station a mass m of ore is dumped in. What is the truck's new velocity. Assuming all the usual simplifications - no friction >

Energy is Continuous.

Truck had 1/mv1 sq. After being dumped with m again, will still have same KE.

So new velocity will be V2 -- valued thus: 1/2 mv1 sq= 1/2.2m v2sq.=mv2 sq

Hence v2=√(0.5 v1sq)

( SORRY Suffix/Superscript not working from CR4)

... and perhaps worth adding that an elastic collision is one in which none of the energy is transformed into internal energy (delfections etc) or into sound, heat, light etc etc.

In the real (macroscopic) world there is no such thing as an elastic collision, but things like billiard balls or air tables are decent approximations. In your example, each piece of ore would tumble, jostle, crack, bounce, settle etc etc, so there would be plenty of energy 'lost' from kinetic energy into other energy forms.

The velocity of the train will not be changed by the offloading of the ore because the ore is offloaded with its kinetic energy and all.

The energy is being absorbed by the bunker so that this energy is not transferred to the train.

The force on the bunker will be equal to the change in momentum of the ore.

Just to put some rough numbers on this. Ignoring the mass of the carriage: suppose that you have 1000 Kg of water travelling at 10 metres per second (that's just above 22 miles per hour). The kinetic energy of the system is 50,000 Joules. Now you drop an extra 1000 Kg of water in the truck, and, it all settles down. The carriage is now travelling at 5 m/s and the new system has 25,000 Joules of kinetic energy. The other 25,000 Joules has been absorbed by the 2000 Kg of water raising the temperature by 0.0125^oC.

(standard temperature and pressure, and, ignoring the potential energy given up by the fall of the second 1000 Kg.)

Momentum is conserved only if the interacting bodies are not influenced by external forces. The external influence of the railroad tracks (assuming the load is dropped on top of the ore car) would qualify as an external influence, therefore momentum is not conserved. Remember that velocity in the momentum equation is a vector having both magnitude and direction.

Energy is always conserved. It may be turned into heat as would be the case for the potential energy of the falling ore (because the tracks constrain the movement of the car no useful work is done). If you assume the ore has zero velocity in the direction of the moving car, work is done to accelerate the ore to terminal velocity of the combined masses. The only place this energy can come from is the initial kinetic energy of the rail car. Rough numbers: the initial kinetic energy of the rail car is equal to the kinetic energy rail car and ore.

After giving this some more thought I'm not sure I agree with my own post. If the ore is viewed as a stationary mass the resulting velocity of the combined mass would still be in the same direction as the ore car. So, momentum is conserved?

That is correct. Any conservation principle only applies to a closed system absent of external influences. So, momentum is always conserved while kinetic energy (KE specifically, total energy is also always conserved) is only conserved in totally elastic collisions. In any inelastic collision, a portion of KE will be converted to another form (heat, sound), but momentum will always remain. Also bear in mind that momentum is reference frame dependant; calculating from one of the colliding bodies' point of view is just as valid as calculating from an external POV (Einstein's elevator), but the POV must be maintained for both the before and after collision calculations.

So, in regards to the OP question, the choice of which method to use is determined by the collision type, selection of a convenient, but consistent, frame of reference, and defining a system of events that is truly closed.