Newton's Third Law of Motion

The Third Law describes the nature of forces as interactions between bodies. It states: "To every action, there is always an equal and opposite reaction."

If a body A exerts a force on body B (the 'action', F⃗AB\vec{F}_{AB}), then body B simultaneously exerts a force on body A (the 'reaction', F⃗BA\vec{F}_{BA}). These two forces are equal in magnitude and opposite in direction.

F⃗AB=−F⃗BA\vec{F}_{AB} = -\vec{F}_{BA}

Key Points:

  • Action and reaction forces act on different bodies. This is a critical point. Because they act on different bodies, they do not cancel each other out.
  • They are always equal in magnitude and opposite in direction.
  • The forces are of the same nature. For example, the gravitational pull of the Earth on the Moon is paired with the gravitational pull of the Moon on the Earth.
    Action-reaction pair example

Misconception Clarification:

  • Action and reaction do not cancel out because they act on different objects.
  • For motion analysis, apply Newton’s second law separately to each object.

Conservation of Linear Momentum

The Third Law leads directly to one of the most fundamental principles in physics: the law of conservation of linear momentum. It states that if the net external force on a system of particles is zero, the total momentum of the system remains constant.

Derivation: Consider two colliding bodies, A and B. The only forces are the internal forces they exert on each other, F⃗AB\vec{F}_{AB} and F⃗BA\vec{F}_{BA}.

From the Second Law:

F⃗AB=dp⃗Bdt\vec{F}_{AB} = \frac{d\vec{p}_B}{dt} and F⃗BA=dp⃗Adt\vec{F}_{BA} = \frac{d\vec{p}_A}{dt}.

From the Third Law: F⃗AB=−F⃗BA\vec{F}_{AB} = -\vec{F}_{BA}.

Therefore, dp⃗Bdt=−dp⃗Adt  ⟹  dp⃗Adt+dp⃗Bdt=0\frac{d\vec{p}_B}{dt} = -\frac{d\vec{p}_A}{dt} \implies \frac{d\vec{p}_A}{dt} + \frac{d\vec{p}_B}{dt} = 0.

ddt(p⃗A+p⃗B)=0\frac{d}{dt}(\vec{p}_A + \vec{p}_B) = 0

This means the total momentum of the system, P⃗total=p⃗A+p⃗B\vec{P}_{total} = \vec{p}_A + \vec{p}_B, does not change with time. It is conserved.

Example:

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s80\ m/s, what is the recoil speed of the gun?

Solution:

  1. Identify the System: Consider the gun and the shell as a single isolated system.

  2. Initial State: Before firing, both the gun and the shell are at rest. The total initial momentum of the system is zero. P⃗initial=0\vec{P}_{initial} = 0

  3. Final State: After firing, let the velocity of the shell be v⃗s\vec{v}_s and the recoil velocity of the gun be v⃗g\vec{v}_g. Let the direction of the shell be the positive direction. So, vs=+80 m/sv_s = +80\ m/s. The total final momentum is P⃗final=msv⃗s+mgv⃗g\vec{P}_{final} = m_s \vec{v}_s + m_g \vec{v}_g.

  4. Apply Conservation of Momentum: Since the forces involved in firing are internal to the system, the total momentum is conserved. P⃗initial=P⃗final\vec{P}_{initial} = \vec{P}_{final}

    0=msvs+mgvg0 = m_s v_s + m_g v_g

  5. Solve for Recoil Velocity:

vg=−msvsmgv_g = - \frac{m_s v_s}{m_g}

vg=−(0.020 kg)(80 m/s)100 kg=−0.016 m/sv_g = - \frac{(0.020\ kg)(80\ m/s)}{100\ kg} = -0.016\ m/s

The recoil speed (the magnitude of the velocity) is **0.016 m/s**. The negative sign indicates that the gun moves in the direction opposite to the shell.