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Example: Maximally Inelastic Collision of Two Identical Carts
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?
Facts
Initial situation: Just before collision
Final situation: Just after collision
Lacking
Approximations & Assumptions
Neglect friction and air resistance
Representations
System: Both carts
Surroundings: Earth, track, air (neglect friction and air resistance)
Solution
Since the y and z components of momentum don't change, we can work with only x components
From the momentum principle (x components):
$${p}_{1xf} + {p}_{2xf} = {p}_{1xi}$$
$$2p_{1xf} = p_{1xi}$$
$$p_{1xf} = \dfrac{1}{2}p_{1xi}$$
The final speed of the stuck-together carts its half the initial speed:
$$v_{f} = \dfrac{1}{2}{v_{i}}$$
Final translational kinetic energy:
$$(K_{1f} + K_{2f}) = 2(\dfrac{1}{2}mv^2_{f})$$
$$(K_{1f} + K_{2f}) = 2(\dfrac{1}{2}m(\dfrac{1}{2}v_{i})^2) = \dfrac{1}{4}mv^2_{i}$$
$$(K_{1f} + K_{2f}) = \dfrac{K_{1i}}{2}$$
From the energy principle:
$$K_{1f} + K_{2f} + E_{int,f} = K_{1i} + E_{int,i}$$
$$E_{int,f} - E_{int,i} = K_{1i} - (K_{1f} + K_{2f})$$
$$\Delta E_{int} = K_{1i} - \dfrac{K_{1i}}{2}$$
$$\Delta E_{int} = \dfrac{K_{1i}}{2}$$
The final kinetic energy of the system is only half of the original kinetic energy, which means that the other half of the original kinetic energy has been dissipated into increased internal energy $\Delta E_{int}$ of the two carts.