Mistake Master
Mistake Master
Two carts collide. With no outside push on the pair, total momentum is the same before and after, every time. Total kinetic energy is not. Elastic collisions keep the full $K_\text{sys}$; perfectly inelastic ones lose as much as possible, with the carts stuck together moving at one shared velocity afterward. Everything in between counts as inelastic. This lesson is about telling the three cases apart, and never mixing up which law applies to which type.
Picture two carts colliding on a frictionless track. Both made of steel. Or both of wet clay. Or one of each. Every version: the collision is over in milliseconds. The track is the same; the carts are different. What happens afterward depends entirely on what the carts are made of, not on the track.
Two questions to ask of any collision:
That asymmetry is the whole topic. Momentum stays conserved here whatever the carts are made of. In the collision models in this lesson, system kinetic energy is conserved only for elastic collisions and decreases for inelastic collisions. The job is to tell which type of collision you have and apply the right rule.
Collisions come in three flavors, sorted by what they do to system kinetic energy:
Two laws are on the table, but they apply differently:
$$\vec p_\text{sys,i} = \vec p_\text{sys,f} \quad \text{always (no outside force).}$$ $$K_\text{sys,i} = K_\text{sys,f} \quad \text{only if the collision is elastic.}$$For a one-dimensional perfectly inelastic collision, the final shared velocity comes straight from momentum conservation:
$$v_f = \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2}.$$Then compute $K_i$ and $K_f$ separately and subtract to find the energy drop.
A 3 kg cart moving at $+4$ m/s strikes a 1 kg cart sitting at rest on a frictionless track. The carts stick. Find the final shared velocity and the change in system kinetic energy.
$$\vec p_\text{sys,i} = (3)(4) + (1)(0) = 12 \text{ kg} \cdot \text{m/s}.$$
After sticking, total mass is $4$ kg, so $v_f = 12 / 4 = +3$ m/s. Both carts move at $+3$ m/s.
$K_\text{sys,i} = \tfrac{1}{2}(3)(4)^2 + 0 = 24$ J.
$K_\text{sys,f} = \tfrac{1}{2}(4)(3)^2 = 18$ J.
Loss: $24 - 18 = 6$ J, transformed into heat, sound, and deformation. Here's the catch: $K_\text{sys,f}$ is not zero, even though the carts stuck. Sticking does not mean stopping.
Momentum was conserved; kinetic energy dropped by 6 J. Both carts continue moving together at $+3$ m/s. The only setup where a perfectly inelastic collision leaves the combined object at rest is when the initial total momentum is already zero, like two equal masses moving toward each other at equal speeds.
The asymmetry between the two laws is the whole point. To make it concrete, run the same initial setup two ways. Two carts, $m_1 = m_2 = 2$ kg, $v_1 = +4$ m/s, $v_2 = 0$.
System momentum afterward, for both cases: $p_\text{sys,f} = (4)(2) = 8$ kg$\cdot$m/s in the elastic case (only cart 2 moves), and $(4)(2) = 8$ kg$\cdot$m/s in the perfectly inelastic case (both carts move together). Same number, exactly. Momentum cannot tell the two cases apart.
Kinetic energy can. $16$ J versus $8$ J. The $8$ J that vanished from the inelastic case went into heat, sound, and deformation. That energy isn't gone in any cosmic sense, just gone from the kinetic-energy ledger. Mechanical $K$ dropped; heat and sound went up to compensate.
The trap. A student looking only at momentum will report "same outcome" for both collisions. They're right about momentum and wrong about the physics. The carts move very differently afterward.
Newton's-cradle special. When two equal-mass carts collide elastically with one at rest, they swap velocities. Cart 1 stops; cart 2 leaves at cart 1's incoming speed. Unequal masses don't swap velocities cleanly, but the system $K$ is still conserved.
This whole topic turns on keeping two ledgers apart, and each trap below smears them together — which quantity is always conserved, how much kinetic energy a perfectly inelastic hit keeps, and when $K_i = K_f$ is actually allowed.
The student has packed two laws together and matched them both to "elastic." But momentum doesn't care whether the collision is elastic. Two cars crashing into each other and crumpling still conserve momentum, the same way two billiard balls do. The deformation does eat kinetic energy, but it does not eat momentum.
Fix. Keep the two ledgers separate. System momentum is conserved in every collision where no external force acts on the pair, every time. System kinetic energy is conserved only in elastic collisions. Two questions, two different yes/no answers, never lumped together.
"Stuck" is not the same as "stopped." The carts share one final velocity, but unless the initial system momentum was already zero, that shared velocity is not zero either. A 3 kg cart at $+4$ m/s hitting a stationary 1 kg cart and sticking still moves at $+3$ m/s afterward. There's $18$ J of kinetic energy in that motion, and only $6$ J was lost out of the original $24$ J.
Fix. Compute the final shared velocity from momentum conservation first: $v_f = (m_1 v_1 + m_2 v_2) / (m_1 + m_2)$. Then compute the final $K$ from that velocity. The only setup that gives $K_\text{sys,f} = 0$ is the one where $m_1 v_1 + m_2 v_2 = 0$ to start with, like two equal masses approaching each other at equal speeds.
Both laws together would lock down both final velocities in a one-cart-strikes-one problem, which is exactly why students reach for them. The catch: writing both assumes the collision is elastic, and writing them for an inelastic case produces a contradiction. Watch what happens with a $2$ kg cart at $+6$ m/s sticking to a stationary $4$ kg cart. Momentum gives $v_f = 12/6 = +2$ m/s. If you also wrote $K_i = K_f$ and solved for the shared $v_f$, you'd get $\sqrt{2 K_i / (m_1 + m_2)} = \sqrt{12} \approx 3.46$ m/s. The two answers disagree because the collision is not elastic. No single shared velocity satisfies both.
Fix. Before writing any equation, decide which type of collision you have. Elastic: both laws apply, use both. Inelastic (including perfectly inelastic): only momentum applies; $K_i = K_f$ is false here, and the missing energy is the answer to a different question (how much went to heat, sound, and deformation).