When two bodies collide and stay together — couplers latching, a bullet in a block, two cars welded by their own bumpers — the collision is perfectly inelastic, and the six-symbol ledger collapses. There is only one velocity afterwards, so $v_1$ and $v_2$ become a single $v$, and the law solves itself: $v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$ — read aloud v equals m-one u-one plus m-two u-two, all over m-one plus m-two. Same cast as before: $m_1, m_2$ the two masses in kilograms, $u_1, u_2$ their velocities before in m/s, and $v$ the one speed they share after.

Read the shape of it: the total momentum on top, the total mass underneath. It is a weighted average of the two starting velocities, weighted by mass — which is why the answer always lands BETWEEN them, and always closer to the heavier body's speed. Check every answer against that bracket before you trust it. And be clear about what was lost: momentum survived in full, while kinetic energy went into heat, sound and crumpled metal. “Inelastic” is the name for that spending. A perfectly inelastic collision spends the most of all — and that is what makes it survivable.