Motion of Centre of Mass (COM)

Imagine n tiny chunks of matter, each with mass mi and position ri. Their common balance point – the centre of mass – sits at R and obeys

$$M\mathbf R \;=\; m_1\mathbf r_1 + m_2\mathbf r_2 + \dots + m_n\mathbf r_n \tag{6.7}$$ :contentReference[oaicite:0]{index=0}

  • Speed of the COM: Differentiate once to get $$M\mathbf V = m_1\mathbf v_1 + m_2\mathbf v_2 + \dots + m_n\mathbf v_n \tag{6.8}$$ Here V is the COM’s velocity and vi the velocity of each particle. :contentReference[oaicite:1]{index=1}
  • Acceleration of the COM: Differentiate again to reach $$M\mathbf A = m_1\mathbf a_1 + m_2\mathbf a_2 + \dots + m_n\mathbf a_n \tag{6.9}$$ :contentReference[oaicite:2]{index=2}
  • Link with forces: Newton’s second law for every particle turns (6.9) into $$M\mathbf A = \mathbf F_1 + \mathbf F_2 + \dots + \mathbf F_n \tag{6.10}$$ Internal forces cancel out in pairs, leaving only the external push or pull $$M\mathbf A = \mathbf F_{\text{ext}} \tag{6.11}$$

So the COM moves exactly as if all the mass sat at that single point and every external force acted there. Internal fireworks can never deflect the COM by themselves. A classic illustration: a shell flying on a parabolic path can explode mid-air, but its COM still glides along the same old parabola under gravity alone. :contentReference[oaicite:4]{index=4}

Linear Momentum of a System

The total momentum P is the vector sum of each particle’s momentum:

$$\mathbf P \;=\; \mathbf p_1 + \mathbf p_2 + \dots + \mathbf p_n = m_1\mathbf v_1 + m_2\mathbf v_2 + \dots + m_n\mathbf v_n \tag{6.14}$$ :contentReference[oaicite:5]{index=5}

Insert (6.8) and you get a neat relation with the COM: $$\mathbf P = M\mathbf V \tag{6.15}$$ :contentReference[oaicite:6]{index=6}

Differentiate once more: $$\frac{d\mathbf P}{dt} = M\mathbf A \tag{6.16}$$ Combine this with (6.11) to extend Newton’s second law to the whole system: $$\frac{d\mathbf P}{dt} = \mathbf F_{\text{ext}} \tag{6.17}$$ :contentReference[oaicite:7]{index=7}

  • Momentum conservation: If all external forces vanish, then $$\frac{d\mathbf P}{dt}=0 \;\Rightarrow\; \mathbf P=\text{constant}$$ Equivalently, the COM cruises along at constant speed and in a straight line. :contentReference[oaicite:8]{index=8}

Key Take-aways

  1. The COM’s motion depends only on external forces, never on internal ones.
  2. $$\mathbf R,\,\mathbf V,\,\mathbf A$$ of the COM come from mass-weighted sums of positions, speeds, and accelerations of individual particles.
  3. $$M\mathbf A = \mathbf F_{\text{ext}}$$ packs the entire system into a single “effective particle.”
  4. Total momentum links directly to the COM: $$\mathbf P = M\mathbf V$$.
  5. No external force ⇒ total momentum stays fixed and the COM coasts at constant speed.

High-Yield Ideas for NEET

  • Explosion/fragment problems: Internal forces cancel, so the COM keeps its original projectile path. (Great for multi-shrapnel questions.)
  • System as a single particle: Replace a complicated body by its COM to apply $$\mathbf F = M\mathbf A$$ quickly.
  • Momentum conservation: Zero external force means $$\mathbf P$$ stays constant – vital in collision and recoil questions.
  • COM velocity formula: $$\mathbf V = \dfrac{m_1\mathbf v_1 + m_2\mathbf v_2 + \dots}{M}$$ – a favorite in two-body motion and rocket-ejection problems.

Keep practicing: treat every cluster of masses as a single superhero mass at its COM, and tough mechanics questions turn into plain sailing!