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
[tex]p=Mv=(1200 kg)(10.8 m/s)=12960 kg m/s[/tex]
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is [tex]v_f[/tex]. The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
[tex]p=12960 kg m/s =(1200 kg+m)v_f[/tex]
and so the final speed is
[tex]v_f = \frac{12960 kg m/s}{1200 kg +m} [/tex]
