Consider the opposite extreme - a maximally inelastic collision of the two identical carts, one initially at rest. That means the carts stick together (perhaps they have sticky material on their ends), and each has the same final momentum $p_{1xf} = p_{2xf}$

Find the final momentum, final speed, and final kinetic energy of the carts in terms of their initial values.

What is the change in internal energy of the two carts?

Initial situation: Just before collision

Final situation: Just after collision

Neglect friction and air resistance

System: Both carts

Surroundings: Earth, track, air (neglect friction and air resistance)

Since the y and z components of momentum don't change, we can work with only x components

From the momentum principle (x components):

$$\vec{p}_{1xf} + \vec{p}_{2xf} = \vec{p}_{1xi}$$

$$2p_{1xf} = p_{1xi}$$

$$p_{1xf} = \dfrac{1}{2}p_{1xi}$$

The final speed of the stuck-together carts its half the initial speed:

$$v_{f} = \dfrac{1}{2}{v_{i}}$$

Final translational kinetic energy:

$$(K_{1f} + K_{2f}) = 2(\dfrac{1}{2}mv^2_{f})$$

$$(K_{1f} + K_{2f}) = 2(\dfrac{1}{2}m(\dfrac{1}{2}v_{i})^2) = \dfrac{1}{4}mv^2_{i}$$

$$E_f = E_i + W + Q$$

$$K_{1f} + K_{2f} + E_{int1f} + E_{int2f} = K_{1i} + K_{2i} + E_{int1i} + E_{int2i}$$

$$K_{1f} + K_{2f} = K_{1i}$$

Combine momentum and energy equations:

$$\dfrac{p^{2}_{1xf}}{2m} + \dfrac{p^{2}_{2xf}}{2m} = \dfrac{(p_{1xf} + p_{2xf})^2}{2m}$$

$$p^{2}_{1xf} + p^{2}_{2xf} = p^{2}_{1xf} + 2{p_{1xf}p_{2xf}} + p^{2}_{2xf}$$

$$2{p_{1xf}p_{2xf}} = 0$$

There are two possible solutions to this equation. The term ${p_{1xf}p_{2xf}}$ can be zero if $p_{1xf} = 0$ or if $p_{2xf} = 0$.