Momentum and Impulse

Momentum

• Definition: The linear momentum (vector) of an object is the product of its mass and its velocity.

  • Symbol: $p$
  • Formula: $p = m v$
  • SI-unit: $\text{kg·m·s}^{-1}$ (same as $\text{N·s}$).
  • Vector nature:
  • Magnitude: $|p| = m |v|$.
  • Direction: always identical to the velocity vector.

• Physical meaning

  • Measures “quantity of motion”.
  • Indicates the difficulty of bringing a moving body to rest or of changing its motion.
  • Conserved in the absence of external (net) forces.

• Common sources of confusion

  • Heavy, slow object vs. light, fast object can possess the same momentum.
  • A body at rest has zero momentum regardless of its mass.

• Everyday relevance: vehicle crashes, sports (kicking a ball, tackling), recoil of firearms, rocket propulsion.

Calculating Momentum – Illustrative Examples

• Example 1: Sedan

  • Data: $m = 2000\,\text{kg}$, $v = +2\,\text{m·s}^{-1}$ (‘‘to the right’’)
  • $p = (2000)(2) = 4000\,\text{kg·m·s}^{-1}$ → rightwards.

• Example 2: Parked truck

  • Data: $m = 5000\,\text{kg}$, $v = 0$
  • $p = 0$. A massive object still has no momentum while stationary.

• Quick mental‐check: doubling either mass or velocity doubles momentum; doubling both quadruples it.

Change in Momentum (Δp)

• Symbol: $\Delta p$, sometimes written $p<em>f - p</em>i$.

• General expression: $\Delta p = m (v<em>f - v</em>i) = m\,\Delta v$.

• If the mass is constant and the object starts from rest ($v<em>i=0$): $\Delta p = m v</em>f \;(= p_f)$.

• Sign matters! Positive change → same direction as chosen positive axis; negative change → opposite.

• Worked micro-example (ball rebounds)

  • Ball mass $m = 2\,\text{kg}$.
  • Thrown towards a wall at $v_i = +5\,\text{m·s}^{-1}$ (right).
  • Rebounds at $v_f = -3\,\text{m·s}^{-1}$ (left).
  • $\Delta p = 2(-3 - 5) = 2(-8) = -16\,\text{kg·m·s}^{-1}$ (16 right→left; i.e.
    momentum has changed direction and magnitude).
  • Interpretation: the wall imparted a leftward impulse of $16\,\text{N·s}$ on the ball (ball gave the wall an equal and opposite impulse).

Newton’s Second Law – Momentum Formulation

• Traditional form: $\Sigma F = m a$.
• Equivalent momentum version (more general):
  $\Sigma F = \frac{\Delta p}{\Delta t} = m \frac{\Delta v}{\Delta t}$.
• Highlights:
  • If $\Delta p = 0$ (isolated body or system), $\Sigma F = 0$.
  • A large force applied briefly can produce the same $\Delta p$ as a small force applied for a long time.

Impulse (J)

• Definition 1: The product of the net force acting on an object and the time interval over which it acts.

• Definition 2: The change in linear momentum of the object.

• Symbol: $J$ (vector).

• Formulae:
  $J = F<em>{\text{net}}\,\Delta t = \Delta p = m (v</em>f - v_i)$.

• SI-unit: $\text{N·s}$ (identical to $\text{kg·m·s}^{-1}$).

• Example – Kicked football

  • Ball mass $m = 1.5\,\text{kg}$.
  • Approaches foot at $v<em>i = -15\,\text{m·s}^{-1}$ (left) and leaves at $v</em>f = +8\,\text{m·s}^{-1}$ (right).
  • $\Delta p = 1.5(8 - (-15)) = 1.5(23) = 34.5\,\text{N·s}$ to the right.
  • If contact time $\Delta t = 0.20\,\text{s}$, the average force magnitude:
    $F = \frac{\Delta p}{\Delta t} = \frac{34.5}{0.20} = 172.5\,\text{N}$ (directed rightwards on the ball; leftwards on the foot).

Conservation of Linear Momentum

• Statement: In an isolated (closed + no external net force) system, the vector sum of the linear momenta of all bodies remains constant.
  $\sum p<em>{\text{before}} = \sum p</em>{\text{after}}$

• Isolation criteria

  • Negligible external forces (friction, air resistance, support forces cancel, etc.).
  • Internal forces (e.g.
    collision forces) always occur in equal and opposite pairs → do not alter the system’s total momentum.

Two-Body System Master Equation

$m<em>1 v</em>{1i} + m<em>2 v</em>{2i} = m<em>1 v</em>{1f} + m<em>2 v</em>{2f}$

Collision / Interaction Scenarios

• Elastic collision (ideal)

  • Both momentum and kinetic energy are conserved.
  • Example: billiard balls, perfectly hard spheres.

• Inelastic collision (real-world majority)

  • Momentum conserved, kinetic energy not (transformed to sound, heat, deformation).
  • Basketball in the transcript: ball loses speed after bounce – height after rebound lower than drop height.

• Perfectly inelastic collision (‘‘sticky’’)

  • Objects stick together; final velocities equal.
  • Equation: $m<em>1 v</em>{1i}+m<em>2 v</em>{2i}=(m<em>1+m</em>2) v_f$.

• Explosion / Separation (reverse problem)

  • Objects initially as one body ($m<em>1+m</em>2$) split into parts; often internal chemical or mechanical energy supplied.
  • Pre-event momentum $=(m<em>1+m</em>2) v_i$; post-event described by the two-body equation above.

Worked Examples of Momentum Conservation

Example A – Two carts stick together (perfectly inelastic)

• Cart 1: $m<em>1 = 10\,\text{kg}$, $v</em>{1i} = -10\,\text{m·s}^{-1}$ (left).
• Cart 2: $m<em>2 = 5\,\text{kg}$, $v</em>{2i} = +25\,\text{m·s}^{-1}$ (right).
• Combined final velocity $v<em>f$: $10(-10)+5(25)=(10+5)v</em>f \Rightarrow -100+125=15 v<em>f \Rightarrow 25=15 v</em>f$
  $v_f = +1.67\,\text{m·s}^{-1}$ (right).

Example B – Rocket fuel-tank separation (explosion type)

• Before explosion: combined mass $m=2+3 = 5\,\text{kg}$, upward velocity $v_i = +10\,\text{m·s}^{-1}$.

• After explosion: part A $m<em>A = 2\,\text{kg}$, $v</em>{Af} = +14\,\text{m·s}^{-1}$; part B $m<em>B = 3\,\text{kg}$, $v</em>{Bf}= ?$.

• Conservation:
  $(5)(10)=2(14)+3v<em>{Bf} \Rightarrow 50=28+3v</em>{Bf} \Rightarrow v_{Bf}=\frac{22}{3}=+7.33\,\text{m·s}^{-1}$ (upwards).

• Maximum height of part A (after separation)

  • Kinematics: $v<em>f^2=v</em>i^2+2a\,\Delta y$ with $v<em>f=0$ at top, $v</em>i=14$, $a=-9.8$.
  • $0=14^2+2(-9.8)\,\Delta y \Rightarrow \Delta y=\frac{14^2}{19.6}=10\,\text{m}$.
  • Initial height $50\,\text{m}$ → $h_{\max}=50+10=60\,\text{m}$.

Example C – Basketball dropped & bounced (inelastic)

• Mass $m=0.63\,\text{kg}$. Contact time with ground $\Delta t = 0.016\,\text{s}$.
• Velocity on impact (from graph) $v_i = -6.9\,\text{m·s}^{-1}$.
• Velocity leaving ground (from energy to 1.3 m rebound) shown to be $v_f = +5.05\,\text{m·s}^{-1}$.
• Impulse:
  $J = m(v<em>f - v</em>i) = 0.63(5.05 - (-6.9)) = 0.63(11.95)=7.5\,\text{N·s}$ upward.
• Net average force:
  $F_{\text{net}} = \frac{J}{\Delta t}=\frac{7.5}{0.016}\approx 4.7\times10^2\,\text{N}$ upward.
• Ground’s actual contact force (normal) is larger because gravity acts downward concurrently:
  $F<em>{\text{normal}}=F</em>{\text{net}} + mg \approx 470 + (0.63)(9.8) \approx 476\,\text{N}$.

Graphical Interpretation (Basketball Question)

• Velocity-time area = displacement (geometric link to prior kinematics lecture).
• Straight-line sections in graph suggest constant acceleration ≈ $g$ on both ascent and descent; hence air resistance negligible (slope magnitude unchanged).
• Position-time sketch would resemble an inverted “V” for fall, followed by a smaller “∧” for rebound.

Ethical / Practical Notes

• Automotive safety design increases collision time (airbags, crumple zones) → reduces force by spreading impulse.
• Sports technique: “follow-through” when hitting or kicking a ball extends $\Delta t$, lowering peak forces on athlete’s body.
• Rocket stage jettison conserves momentum; engineers use explosive bolts and differential masses to control final velocities.

Formula & Unit Summary (Quick Reference)

• Momentum: $p = m v$ (\text{kg·m·s}^{-1})
• Change in momentum: $\Delta p = m (v<em>f - v</em>i)$
• Newton II: $\Sigma F = \dfrac{\Delta p}{\Delta t} = m a$
• Impulse: $J = F_{\text{net}}\,\Delta t = \Delta p$ (\text{N·s})
• Conservation (two-body): $m<em>1 v</em>{1i} + m<em>2 v</em>{2i} = m<em>1 v</em>{1f} + m<em>2 v</em>{2f}$.
• Perfectly inelastic (stick together): $m<em>1 v</em>{1i} + m<em>2 v</em>{2i} = (m<em>1 + m</em>2)v_f$.
• Explosion / recoil: $(m<em>1+m</em>2)v<em>i = m</em>1 v<em>{1f} + m</em>2 v_{2f}$.
• Maximum height (projectile/rocket): $v<em>f^2 = v</em>i^2 + 2 a \Delta y$ (with $a=-g$).

Conceptual Connections to Prior Physics Topics

• Momentum is a direct extension of Newton’s laws (linking force and time to motion change).
• Energy conservation complements, but does not replace, momentum conservation—both must be examined in collisions.
• Graphical kinematics (area–displacement, slope–acceleration) enables qualitative assessments without formulas, building on earlier graph skills.

Common Mistakes & Tips

• Forgetting that momentum and impulse are vectors; always assign a positive direction before calculation.
• Mixing units (e.g.
  using grams, km·h⁻¹) → always convert to SI.
• Impulse ≠ force; impulse depends on both force and interaction time.
• In explosions, external forces (e.g.
  gravity) can often be neglected during the very short separation interval but must be included for subsequent motion analysis.
• When two bodies stick, kinetic energy is lost even though momentum is conserved.