Lecture 6 Vectors

Newton’s Third Law

  • Concept: Newton's third law states that for every action, there is an equal and opposite reaction.

  • Interaction of masses:

    • Let m1 and m2 be two interacting masses.

    • Force on m1 due to m2: F12

    • Force on m2 due to m1: F21

    • Relation: F12 = -F21

      • Opposite directions

      • Equal magnitudes

Conservation of Momentum in 2D and 3D

  • Definition: Momentum vector defined as p = mv.

  • Derivation of Force from Momentum:

    • F = m dv/dt = dp/dt

  • For two masses m1 and m2:

    • Forces of interaction: F12 = -F21

    • Momentum of m1: p1

    • Momentum of m2: p2

    • Then: dp1/dt = F12 and dp2/dt = F21

    • Adding these: d/dt(p1 + p2) = F12 + F21 = 0

      • Therefore, p1 + p2 = constant (Conservation of Momentum)

Example of Momentum Conservation

  • Scenario: Collision of two particles with initial velocities u1, u2 and final velocities v1, v2.

  • In the absence of external forces, condition:

    • m1u1 + m2u2 = m1v1 + m2v2

  • Example: A snooker ball with an initial velocity of 2 m/s collides with an identical resting ball.

    • After collision:

      • Ball 2 moves at 1 m/s at a 60° angle to the original direction of Ball 1.

    • Calculate final velocity of Ball 1:

      • Define directions: u1 = 2i (initial) and u2 = 0

      • Calculate v2:

        • v2 = 1(cos 60i + sin 60j) = (1/2)i + (√3/2)j

      • Conservation of momentum gives:

        • v1 = u1 + u2 - v2 = 2i - (1/2)i - (√3/2)j

        • Results in v1 = (3/2)i - (√3/2)j

      • Final angle analysis:

        • Dot product v1 · v2 = (1/2)(3/2) + (-√3/2)(√3/2) = 0, indicating a 90° separation after collision.

        • Final speed of Ball 1:

          • |v1| = √((3/2)² + (-√3/2)²) = √(9/4 + 3/4) = √(3) m/s

Newton’s Universal Law of Gravitation

  • Concept: General law of attraction proposed by Newton between all matter.

  • Influences:

    • Explains motions of celestial bodies (e.g., moons and planets).

  • Definitions:

    • Let r be the position vector of m2 relative to m1.

    • Define r̂ as unit vector in direction of r: r̂ = r / |r|

    • Gravitational force: F21 = -G(m1 m2) / r² r̂ = -G(m1 m2) / r³ r

    • Where G = 6.67 × 10⁻¹¹ m³/s²/kg.

Remarks on Newton’s Universal Law of Gravity

  1. Direction: F21 is directed towards -r, indicating attraction of m2 to m1.

  2. Magnitude: |F21| = G(m1 m2) / r², representing an inverse square law.

  3. Position Vector Relationship:

    • Position vector of m1 relative to m2 is -r.

    • Thus, F12 = -G(m2 m1) / r³ (-r), aligning with Newton’s third law.

  4. Equivalence of Masses:

    • m1, m2 are gravitational masses; experimentally equal to inertial masses in Newton’s second law (Principle of Equivalence).

    • Example: In a lift, one cannot distinguish between gravitational acceleration and acceleration in free space.

  5. Example: Two 1 kg masses at distance 1 m apart experience a very small force of attraction: G(1)(1)/1² = 6.67 × 10⁻¹¹ N (negligible unless m1, m2 are large).

Gravity Near the Earth’s Surface

  • For a mass m near Earth’s surface (mass M, radius R):

    • Gravitational force: F = GMm/R² = mg

    • Results in g = GM/R², a constant acceleration towards Earth’s center.

  • Numerical Calculation:

    • g = (6.67 × 10⁻¹¹)(5.98 × 10²⁴)/(6.37 × 10⁶)² ≈ 9.81 m/s².