SG 4 - Impulse and Momentum — Quick Notes

Key Concepts: Momentum and Impulse

• Momentum: $p = m\,v$ (vector; direction = velocity direction); units: kg·m/s.
• Impulse: $J = \Delta p = \sum F\,\Delta t$; units: N·s (which is equivalent to kg·m/s).
• Impulse–momentum theorem: $J = \Delta p$.
• For constant mass: $\Delta p = m\,\Delta v$; thus $J = m\,\Delta v$.
• Relationship to force over time: $J = \int F\,dt\;\approx\;F\,\Delta t\;\text{(for constant }F)$.

Newton’s Second Law in Momentum Form

• Net force relation: $F_{\text{net}} = \frac{dp}{dt}$.
• With constant mass: $F_{\text{net}} = m\,a\; (a = \frac{\Delta v}{\Delta t})$ and $\Delta p = \int F\,dt$.
• Special case: when mass is constant, $\Delta p = m\,\Delta v$ and $J = \Delta p = m\,\Delta v$.

Impulse–Momentum Theorem

• Core statement: a net force applied over a time interval changes an object's momentum.
• Formula: $\sum F\,\Delta t = \Delta p = J$.
• If mass changes, $J = \Delta (m\,v)$ (general form); in these notes, mass is treated as constant.

Force–Time Graphs and Impulse

• Impulse equals area under the force–time graph: $J = \Sigma F \Delta t = \text{Area under }F\text{–}t$.
• Constant net force: area is a rectangle, $J = F\Delta t$.
• Linearly increasing force from 0 to F0 over time t: area is a triangle, $J = \tfrac{1}{2} F_0 t$.

Practical Implications and Intuition

• For the same impulse, a shorter contact time requires a larger average force; a longer stopping time reduces peak force.
• Momentum depends on both mass and velocity; impulse governs how momentum changes.

Quick Problem Highlights (with results)

• Problem: A 3.0 kg object accelerates from 0 to 6.0 m/s.

  • (\Delta v = 6.0\,\text{m/s})
  • (\Delta p = m\Delta v = 3.0\times6.0 = 18\ \text{kg·m/s})
  • Impulse: (J = 18\ \text{N·s})

• Problem: A 1.0 kg object has a 10 N force applied for 0.4 s.

  • (J = F\Delta t = 10\times0.4 = 4\ \text{N·s})
  • (\Delta v = J/m = 4/1 = 4\ \text{m/s})

• Problem: A 0.3 kg ball changes velocity from 5 m/s to -4 m/s.

  • (\Delta v = -9\,\text{m/s})
  • (\Delta p = m\Delta v = 0.3\times(-9) = -2.7\ \text{kg·m/s})
  • Impulse: (-2.7\ \text{N·s})

• Problem: A 5.0 kg object at rest experiences a 20 N force for 2 s.

  • (\Delta v = (F\Delta t)/m = (20\times2)/5 = 8\ \text{m/s})
  • Final velocity: 8 m/s

• Problem: A 1000 kg car stops from 25 m/s in 5 s.

  • (\Delta v = -25\ \text{m/s})
  • (\Delta p = m\Delta v = -25{,}000\ \text{kg·m/s})
  • (F_{avg} = \Delta p/\Delta t = (-25{,}000)/5 = -5000\ \text{N}) (magnitude 5.0 kN)

Worked Graphical Example (FA3)

• If force rises linearly from 0 to 25 N over 5 s: J = area of triangle = (\tfrac{1}{2}\times 5\,\text{s}\times 25\,\text{N} = 62.5\ \text{N·s}).

Summary of Key Formulas and Units

• Momentum: $p = m\,v\,;\quad \text{units: }\ kg\cdot m/s$
• Impulse: J = \Delta p = \sum F\,\Delta t = F\,\Delta t\ (\text{constant }F) \;\; ;\quad \text{units: }\ \text{N·s}
• Change in momentum: $\Delta p = m\,\Delta v$
• Newton’s second law (momentum form): $F_{\text{net}} = \frac{dp}{dt}$
• Impulse–momentum theorem: $J = \Delta p = \Delta(m\,v)$
• Graphical interpretation: impulse equals area under the force–time graph

Closing Takeaways

• Momentum is the product of mass and velocity; direction matches motion.
• Impulse is the force applied over time that changes momentum.
• The impulse–momentum theorem links force, time, and momentum change and is visualized as the area under a force–time curve.