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Collisions & Impulse

Source lecture(s): SC133 Lec 11

Intuition

A collision is physics compressed into milliseconds: forces too large and too brief to track directly. The escape is to integrate over the mess. Impulse — force accumulated over time — equals the momentum change, no matter how wild the force profile. And for the colliding system, total momentum simply passes through unchanged. From car crashes to particle physics, this is the same calculation.

Impulse–momentum theorem

\[\boxed{\,\vec J = \int \vec F\,dt = \Delta\vec p\,} \qquad\text{(} \vec J \approx \vec F_\text{avg}\,\Delta t \text{)}\]

The same \(\Delta p\) can come from a huge force over a short time or a gentle force over a long one. Every safety device exploits this: airbags, crumple zones, bending your knees on landing — all stretch \(\Delta t\) to shrink \(F_\text{avg}\).

The collision taxonomy

Momentum is conserved in all of them (isolated system); kinetic energy distinguishes:

Type Momentum Kinetic energy Signature
Elastic conserved conserved bodies bounce apart cleanly
Inelastic conserved partly lost some heat/deformation
Perfectly inelastic conserved maximum loss bodies stick together

Perfectly inelastic (stick): \(v' = \dfrac{m_1v_1 + m_2v_2}{m_1 + m_2}\) — the center-of-mass velocity; the KE of relative motion is destroyed.

Elastic 1-D (the exam classic):

\[v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1 + \frac{2m_2}{m_1+m_2}v_2 \qquad v_2' = \frac{2m_1}{m_1+m_2}v_1 + \frac{m_2 - m_1}{m_1+m_2}v_2\]

Limiting cases to memorize: equal masses exchange velocities (Newton's cradle); a ball off a wall reverses; a heavy ball barely notices a light one but sends it off at nearly \(2v_1\).

Worked example: ballistic pendulum

A 10 g bullet embeds in a 2 kg block, which swings up 5 cm. Bullet speed?

Two stages, two laws — never mix them:

  1. Collision (momentum only, KE destroyed): \(m_b v = (m_b + M)V\).
  2. Swing (energy): \(V = \sqrt{2gh} = \sqrt{2(9.8)(0.05)} = 0.99\,\text{m/s}\).
\[v = \frac{2.01}{0.01}(0.99) \approx 199\,\text{m/s}\]

Using energy conservation across the embedding would be wrong by a factor of \(\sim 200\) — the classic trap.

Try it live

The collision lab lets you set masses, velocities and elasticity, then watch momentum and kinetic-energy bars during the impact.

Common mistakes

  • Conserving kinetic energy in inelastic collisions. Momentum: always (isolated). KE: only if stated elastic.
  • Dropping signs in 1-D. Velocities are signed; head-on means opposite signs.
  • Two-stage problems solved in one stage (ballistic pendulum above).
  • Forgetting impulse is a vector. A ball bouncing straight back off a wall has \(|\Delta p| = 2mv\), not zero and not \(mv\).

Knowledge graph position

Prerequisites: Linear momentum, Newton's laws. Leads to: Rotation (angular analogues), kinetic theory of gases — pressure is molecular impulse.

Quiz

Q1 (computational). A 1000 kg car at 20 m/s rear-ends a stationary 1500 kg van; they lock together. Final speed, and fraction of KE lost?

Answer

\(v' = 1000(20)/2500 = 8\,\text{m/s}\). \(K_i = 200\,\text{kJ}\), \(K_f = \tfrac12(2500)(64) = 80\,\text{kJ}\) — 60% of the kinetic energy went into crumpled metal. (Loss fraction \(= m_2/(m_1+m_2)\).)

Q2 (conceptual). Why does an egg survive being dropped onto a pillow but not onto concrete, when \(\Delta p\) is identical?

Answer

Impulse \(F\Delta t = \Delta p\) is fixed, but the pillow stretches \(\Delta t\) by ~100×, cutting the peak force by the same factor — below the shell's breaking strength.

Q3 (multiple choice). In an elastic collision between equal masses, one initially at rest, after impact: (a) both move at \(v/2\) (b) they exchange velocities (c) the first bounces back

Answer

(b). The mover stops dead; the target leaves with \(v\) — watch any Newton's cradle or billiards shot.

Q4 (conceptual). A superball and a clay ball of equal mass hit a door at the same speed. Which is more likely to slam it shut?

Answer

The superball: it bounces back, delivering impulse up to \(2mv\) versus the clay's \(mv\). Elastic rebound doubles the momentum transfer.