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 pf - pi.

  • General expression: \Delta p = m (vf - vi) = m\,\Delta v.

  • If the mass is constant and the object starts from rest (vi=0): \Delta p = m vf \;(= 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{\text{net}}\,\Delta t = \Delta p = m (vf - 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 vi = -15\,\text{m·s}^{-1} (left) and leaves at vf = +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{\text{before}} = \sum p{\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

m1 v{1i} + m2 v{2i} = m1 v{1f} + m2 v{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: m1 v{1i}+m2 v{2i}=(m1+m2) v_f.

  • Explosion / Separation (reverse problem)

    • Objects initially as one body (m1+m2) split into parts; often internal chemical or mechanical energy supplied.
    • Pre-event momentum =(m1+m2) 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: m1 = 10\,\text{kg}, v{1i} = -10\,\text{m·s}^{-1} (left).
  • Cart 2: m2 = 5\,\text{kg}, v{2i} = +25\,\text{m·s}^{-1} (right).
  • Combined final velocity vf: 10(-10)+5(25)=(10+5)vf \Rightarrow -100+125=15 vf \Rightarrow 25=15 vf
    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 mA = 2\,\text{kg}, v{Af} = +14\,\text{m·s}^{-1}; part B mB = 3\,\text{kg}, v{Bf}= ?.

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

  • Maximum height of part A (after separation)

    • Kinematics: vf^2=vi^2+2a\,\Delta y with vf=0 at top, vi=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(vf - vi) = 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{\text{normal}}=F{\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 (vf - vi)
  • 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): m1 v{1i} + m2 v{2i} = m1 v{1f} + m2 v{2f}.
  • Perfectly inelastic (stick together): m1 v{1i} + m2 v{2i} = (m1 + m2)v_f.
  • Explosion / recoil: (m1+m2)vi = m1 v{1f} + m2 v_{2f}.
  • Maximum height (projectile/rocket): vf^2 = vi^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.