Two carts, labeled cart 1 and cart 2, collide with one another on a horizontal, essentially frictionless track. How does the momentum of a cart change? What happens to the momentum of the system consisting of the two carts?

Let's use a subscript i for initial and f for final.

The momentum of cart 1 before the collision is p_1i.
The momentum of cart 1 after the collision is p_1f = p_1i +Dp₁.

The momentum of cart 2 before the collision is p_2i.
The momentum of cart 2 after the collision is p_2f = p_2i +Dp₂.

The total momentum of the system beforehand is p_1i + p_2i.

The total momentum of the system afterwards is p_1f + p_2f = p_1i + Dp₁ + p_2i +Dp₂.

Consider Dp₁, the change in momentum experienced by cart 1 in the collision. This comes from the force applied on cart 1 by cart 2 during the collision (it's the area under the force vs. time graph).

Similarly, Dp₂, the change in momentum experienced by cart 2 in the collision, comes from the force applied on cart 2 by cart 1 during the collision (it's the area under that force vs. time graph).

How do the areas under the force vs. time graphs compare?

Newton's Third Law tells us that the areas have equal magnitudes and opposite signs, so Dp₁ = -Dp₂.

The total momentum of the system beforehand is p_1i + p_2i.

The total momentum of the system afterwards is p_1f + p_2f = p_1i + Dp₁ + p_2i + Dp₂ = the total momentum of the system beforehand!

So, one implication of Newton's Third Law is that the momentum of an isolated system is conserved.