Use the Law of Conservation of Linear Momentum to analyze the situation.
The linear momentum p of an object with mass m and speed v is:
According to the law of conservation of linear momentum, the total momentum before the collision must be the same as the total momentum after the collision.
Before the collision, the 1500-kg car is at rest, so its linear momentum is 0.
The total linear momentum before the collision corresponds to the 1200-kg car:
![\begin{gathered} p=(1200\operatorname{kg})(21.1(m)/(s)) \\ =25,320\operatorname{kg}\cdot(m)/(s) \end{gathered}]()
Since the two vehicles lock together after the collision, they can be considered as a single object with the sum of their masses. Let v' be the speed after the collision. Then, the linear momentum of the two cars together, p', is:
![\begin{gathered} p^(\prime)=(m_1+m_2)v \\ =(1200\operatorname{kg}+1500\operatorname{kg})v \\ =(2700\operatorname{kg})v \end{gathered}]()
Since p=p', substitute the value of p and the expression for p' to obtain an equation for v and solve it to find the speed after the collision:
![\begin{gathered} p^(\prime)=p \\ \Rightarrow(2700\operatorname{kg})v=25,320\operatorname{kg}\cdot(m)/(s) \\ \Rightarrow v=\frac{25,320\operatorname{kg}\cdot(m)/(s)}{2700\operatorname{kg}} \\ \Rightarrow v=9.3777\ldots(m)/(s) \\ \therefore v\approx9.38(m)/(s) \end{gathered}]()
Therefore, the speed of the combined cars immediately after the collision is:
