Conservation of Momentum

Two ice skaters face each other and push off simultaneously. Neither was moving before. After the push, both glide away in opposite directions. No external force acted. Where did their momentum come from? It did not come from anywhere, the system had zero total momentum before, and it still has zero total momentum after: equal magnitudes, opposite directions, cancelling to zero.

This law follows from Newton's Third Law: action and reaction forces are equal and opposite, and act for the same time. The impulses are equal and opposite, so the changes in momentum cancel, total momentum is unchanged.

Astronaut answer: Before: total momentum = 0 (at rest). After: $0 = 2 \times 4 + 80 \times v_{astronaut}$. $v_{astronaut} = -8/80 = -0.1\text{ m/s}$, astronaut moves at 0.1 m/s in the opposite direction to the wrench.

Law of conservation of momentum (closed system): $\Sigma p_{before} = \Sigma p_{after}$, i.e. $m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$; this follows from Newton's Third Law; for a perfectly inelastic collision: $v_f = \Sigma p_{before}/(m_1+m_2)$; for an explosion from rest: $0 = m_1v_1' + m_2v_2'$.

Pause, copy the highlighted law and both special cases into your book before moving on.

We just saw that momentum is conserved in all closed-system collisions. That raises a question: is kinetic energy also conserved in every collision, or only in some? This card answers it → all collisions conserve $\Sigma p$; only elastic collisions also conserve $\Sigma KE$; test by calculating both sides.

All collisions conserve momentum. Only elastic collisions also conserve kinetic energy.

To verify if a collision is elastic: calculate total KE before and after. If $\Sigma KE_{before} = \Sigma KE_{after}$, it is elastic. Even if only 1% of KE is lost to heat, it is inelastic.

All collisions conserve $\Sigma p$; elastic collisions additionally conserve $\Sigma KE$; inelastic and perfectly inelastic collisions lose kinetic energy as heat/sound/deformation; to classify: calculate $\Sigma KE_{before}$ and $\Sigma KE_{after}$, if equal, the collision is elastic.

Pause, copy the highlighted classification rule into your book before moving on.

We just saw that $\Sigma p_{before} = \Sigma p_{after}$ applies cleanly when a collision is collinear (1D). That raises a question: what happens when the collision is glancing, off-centre, so the objects scatter at an angle instead of along one line? This card answers it → momentum is a vector, so it resolves into x and y components exactly like the forces in L02 ($F_x = F\cos\theta$, $F_y = F\sin\theta$), and conservation of momentum applies separately to each component.

In a game of pool, the cue ball almost never strikes another ball dead centre. Instead it clips the target ball off to one side, a glancing (2D) collision, and both balls scatter at an angle rather than travelling along the original line. The law of conservation of momentum still applies, but now you must track it in two directions at once.

NESA syllabus dot point (ACSPH064): "conduct an investigation to describe and analyse one-dimensional (collinear) and two-dimensional interactions of objects in simple closed systems." The method: resolve every velocity into x and y components using the angle from the original line of motion, then apply conservation of momentum to the x-components and the y-components independently, exactly as you resolved forces into $F_x$ and $F_y$ in L02.

2D (glancing) collisions: momentum is a vector, resolve every velocity into components ($p_x = mv\cos\theta$, $p_y = mv\sin\theta$) using the angle from the original line of motion; conservation of momentum applies separately to each axis: $\Sigma p_{x,before} = \Sigma p_{x,after}$ AND $\Sigma p_{y,before} = \Sigma p_{y,after}$; check both to confirm the collision is a valid closed system (NESA ACSPH064).

Pause, copy the highlighted 2D collision method into your book before moving on.

We just saw that momentum is conserved in every collision while kinetic energy is only conserved in elastic ones. That raises a question: if the trolleys visibly slow down when they stick together, where did the momentum go? This card answers it → run the collision yourself and read both rows of the ledger at the same instant.

Try this: answer the Predict first question before you touch anything else, then press Stick together and read the two ledger rows side by side. Now drag the restitution slider slowly from 0 to 1 and watch which row moves and which one does not.