Take into account that the total linear momentum must conserve before and after the collision. Then, you have:
p is the momentum before the collision and p' is the momentum after the collision.
By considering the given information you have:
p = m1*v1 + m2*v2
m1 = 2000 kg
m2 = 2500 kg
v1 = 30 m/s
v2 = 0 m/s
p' = (m1 + m2)v
Replace the previous values into the equation p=p' and solve for v:
![\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=(m_1v_1+m_2v_2)/(m_1+m_2) \\ v=\frac{(2000kg)(30(m)/(s))+(2500kg)(0(m)/(s))}{2000\operatorname{kg}+2500\operatorname{kg}} \\ v=13.33(m)/(s) \end{gathered}]()
Hence, the speed after the collision is 13.33 m/s