Momentum

Momentum is a vector, so one object gets a positive velocity and the opposite-moving one a negative velocity. For a 2 kg object at +5 m/s and a 3 kg object at −4 m/s the total is $2 \times 5 + 3 \times (−4) = 10 − 12 = −2$ kg m/s, NOT $10 + 12 = 22$. It is fatal in explosions, where the two momenta must sum to zero. Define a positive direction first and sign consistently.

The mark-scheme definition of momentum needs two elements: the product of mass and velocity ($p = mv$) AND the unit kg m/s (or N s). Calling it a force, energy, or (most often) a moment (force × distance, unit N m) scores zero. Momentum is $p = mv$ in kg m/s — never N, N m or J. "Moment" and "momentum" are different quantities.

Impulse is $F \times t = \Delta p$, not $F = \Delta p$. Candidates find $\Delta p$ correctly then report it as a force in N, skipping the ÷ time. If $\Delta p = 30$ kg m/s and $t = 0.5$ s, then $F = 30 / 0.5 = 60$ N, NOT 30 N. Also convert prefixes: 25 ms = 0.025 s before substituting ("ms never stands for minutes").

"Elastic" does NOT mean the objects stretch — it means kinetic energy is conserved. A collision that stores energy in a stretched band is actually inelastic. Exam clue: objects that bounce apart with no KE loss are elastic; objects that stick together ("coalesce") are inelastic. Momentum is conserved in both.

When objects coalesce (inelastic), the momentum after uses the combined mass: $(m_A + m_B)v$. Using only one mass gives the wrong velocity — e.g. a 4 kg + 6 kg pair must use 10 kg, not 4 or 6. Add the masses before dividing the total momentum to find the common velocity.

A conservation-of-momentum statement scores full marks only if it names the condition: total momentum is constant provided no external (net) force acts — i.e. a closed system. Omitting "closed system" / "no external forces" loses a mark, because friction or an external push would change the total momentum.