a 1.2 x 10^4 kg railroad car is coasting along a level, frictionless track at a constant speed of 25m/s, when a 3000 kg load is dropped vertically onto the car from above. What will its new speed be assuming the load stays on the car??

Because the track is level and frictionless, the net force on this car-load system will be zero in the horizontal direction. As a result, (by Newton's Second Law of mechanics,) the total momentum of this system in the horizontal direction will stay the same.

Momentum of the car-load system in the horizontal direction, before contact:

Car: [tex]m(\text{car}) \cdot v(\text{car, before}) = 1.2 \times 10^{4}\; \rm kg \times 25\; m \cdot s^{-1} = 3.0 \times 10^{6}\; \rm kg \cdot m \cdot s^{-1}[/tex].
Load: zero (for it is dropped "vertically.")

Combine the two parts to obtain: [tex]p(\text{system, before}) = 3.0 \times 10^{6}\; \rm kg \cdot m \cdot s^{-1}[/tex].

Because the load stays on the car, the car and the load should have the same horizontal velocity after contact. Let [tex]v(\text{system})[/tex] denote that velocity. Momentum of the system after contact:

Car: [tex]m(\text{car}) \cdot v(\text{car, after}) = 1.2 \times 10^{4}\; \rm kg \times (\mathnormal{v}(\text{system}))[/tex].
Load: [tex]m(\text{load}) \cdot v(\text{load, after}) = 3000\; \rm kg \times (\mathnormal{v}(\text{system}))[/tex].

Combine to obtain:

[tex]p(\text{system, after}) =1.5\times 10^{4}\; \rm kg \times (\mathnormal{v}(\text{system}))[/tex].

Because the total momentum of the system will stay the same:

[tex]\begin{aligned}&1.5\times 10^{4}\; \rm kg \times (\mathnormal{v}(\text{system}))\\ &= p(\text{system, after}) \\&= p(\text{system, before}) \\ &= 3.0\times 10^{6}\; \rm kg \cdot m \cdot s^{-1}\end{aligned}[/tex].

Solve for [tex]v(\text{system})[/tex] to obtain:

[tex]v(\text{system}) = 20\; \rm m\cdot s^{-1}[/tex].

In other words, the new velocity of the system would be [tex]20\; \rm m \cdot s^{-1}[/tex].