Collisions and Conservation of Momentum

The question

A 4 kg cart rolls at 5 m/s into a 6 kg cart parked on the track. They lock together and roll off as one. How fast do they go — and where did some of the energy disappear to?

In any collision with no outside push, the total momentum before equals the total after — what one object loses, another gains. That single rule, Σp_before = Σp_after, predicts the outcome of crashes, recoiling rifles, and pool breaks. Kinetic energy is a different story: sometimes it survives the collision (elastic), sometimes it turns to heat and crunch (inelastic).

The key idea

In a closed system (no net external force), total momentum is conserved: Σp_before = Σp_after. Because velocity is a vector, you must add momenta WITH their signs — rightward and leftward partly cancel. Collisions are classified by what happens to kinetic energy: elastic collisions conserve KE (objects bounce apart), perfectly inelastic collisions lose the most KE (objects stick and move as one). Recoil — a gun kicking back, a skater throwing a ball — is the same law run in reverse: starting from rest, the total momentum stays zero, so the two pieces fly apart with equal and opposite momenta.

Momentum is conserved because the forces two colliding objects exert on each other are equal and opposite (Newton's third law) and act for the same time, so the impulses cancel: object 1's momentum drop exactly equals object 2's gain. Kinetic energy is the dividing line. In a perfectly inelastic crash (cars crumpling, clay balls sticking) the most KE is converted to heat, sound, and permanent deformation — yet Σp is untouched. In an elastic collision (pool balls, gas molecules) KE survives too, and the two conservation conditions together fix the outcome. For equal masses hitting head-on into a stationary target, the elastic result is a clean velocity swap. Recoil flips the picture: a 4 kg rifle and a 0.01 kg bullet start at rest with Σp = 0, so after firing m_bullet·v_bullet = −m_rifle·v_rifle — the rifle kicks back just enough to keep the total at zero. **All forms:** $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ always; the restitution $e$ ranges the energy story. **Limiting case:** $e = 1$ is perfectly elastic (KE kept), $e = 0$ perfectly inelastic (maximum KE lost, bodies stick) — every real collision lives between the two; equal masses colliding elastically simply swap velocities. **Connect it:** conservation of momentum IS Newton's third law integrated over the contact time — equal and opposite forces for the same $\Delta t$ give $\Delta p_1 = -\Delta p_2$.

The formula

• ·u = velocities before
• ·v = velocities after (m/s
• ·signed vectors — pick a positive direction). m = masses (kg). General conservation: total momentum before = total after. Perfectly inelastic special case: the objects share one final velocity v = (m₁u₁ + m₂u₂)/(m₁ + m₂). Elastic adds a second condition — KE before = KE after — which pins down both final velocities. Recoil is just the case where total momentum is zero
• ·so the two parts get equal and opposite momenta.

What to remember

• ·Total momentum before = total momentum after (use signed velocities).
• ·Elastic collisions conserve kinetic energy; inelastic ones lose some.
• ·In a perfectly inelastic collision the objects stick and share one velocity.