The value of the final speed depends on the mass of the ore.
Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.
Initially, there is only the cart, so the momentum is
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is
. The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
and so the final speed is
Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.
Initially, there is only the cart, so the momentum is
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is
and so the final speed is