The initial kinetic energy of the cart is
[tex]K= \frac{1}{2}Mv_{i}^2 = \frac{1}{2}\cdot 200 kg \cdot (25 m/s)^2=62500 J [/tex]
After the block is dropped into the cart, if there are no other forces acting on it, the kinetic energy of the new system cart+block must be conserved, so it should be the same as before. But the new mass will be M+m, where m=50kg is the mass of the block. Therefore we can write
[tex]K= \frac{1}{2} (M+m) v_f^2[/tex]
from which we find
[tex]v_f = \sqrt{ \frac{2K}{M+m} }= \sqrt{ \frac{2\cdot 62500 J}{200kg+50kg} }=22.36 m/s [/tex]
