Linear Momentum of a System of Particles

Think of momentum as the “oomph” a moving object carries. For one particle we write \( \mathbf{p}=m\mathbf{v} \), where m is the mass and \( \mathbf{v} \) is the speed-and-direction (velocity) vector. :contentReference[oaicite:0]{index=0}

1. Quick Refresher: One Particle

• Momentum formula: \( \mathbf{p}=m\mathbf{v} \).
• Newton’s second law in its “momentum outfit”: \( \mathbf{F}= \dfrac{d\mathbf{p}}{dt} \). Here \( \mathbf{F} \) is the net force. :contentReference[oaicite:1]{index=1}

2. Gather the Gang: Many Particles

For n particles we simply add their momenta: \[ \mathbf{P}= \mathbf{p}_1+\mathbf{p}_2+\dots+\mathbf{p}_n = m_1\mathbf{v}_1+m_2\mathbf{v}_2+\dots+m_n\mathbf{v}_n . \] :contentReference[oaicite:2]{index=2}

3. Momentum Meets the Center of Mass

The total mass \( M=m_1+m_2+\dots+m_n \). We call the speed of the center of mass (CM) \( \mathbf{V} \). A neat result pops out: \[ \mathbf{P}=M\mathbf{V}. \] So the system’s whole momentum equals the total mass times the CM speed-and-direction. :contentReference[oaicite:3]{index=3}

4. Pushing the Whole Crowd

Differentiate the last equation with time: \[ \dfrac{d\mathbf{P}}{dt}=M\mathbf{A}, \] where \( \mathbf{A}=\dfrac{d\mathbf{V}}{dt} \) is the CM acceleration. But Newton already tells us \( \dfrac{d\mathbf{P}}{dt}= \mathbf{F}_{\text{ext}} \), the sum of all outside forces. Therefore \[ \mathbf{F}_{\text{ext}} = M\mathbf{A}. \] :contentReference[oaicite:4]{index=4}

5. When No Outside Force Acts

Set \( \mathbf{F}_{\text{ext}}=0 \) and you immediately get \[ \dfrac{d\mathbf{P}}{dt}=0 \quad\Longrightarrow\quad \mathbf{P} = \text{constant}. \] That’s the conservation of linear momentum. Because \( \mathbf{P}=M\mathbf{V} \), the CM cruises at constant speed in a straight line whenever outsiders leave the system alone. Component-wise: \( P_x=c_1,\; P_y=c_2,\; P_z=c_3 \) (three separate constants). :contentReference[oaicite:5]{index=5}

6. Everyday-Style Examples

• Mid-air firework: A shell follows its usual arc, then bursts into many bits. Those internal blasts cancel each other, so the CM keeps gliding along the same parabola as if nothing happened. :contentReference[oaicite:6]{index=6}
• Radioactive split: A moving radium nucleus breaks into a radon nucleus plus an alpha particle. They zip off in different directions, yet their combined CM keeps the original straight-line path. In the CM frame the two fragments zoom back-to-back. :contentReference[oaicite:7]{index=7}
• Binary stars: Two equal-mass stars orbit each other. Viewed from far away, their CM drifts smoothly through space while each star traces a wiggly path. In the CM frame each star simply moves in a neat circle, exactly opposite its partner. :contentReference[oaicite:8]{index=8}

7. Why Work in the CM Frame?

Jump into the CM frame and life simplifies: the CM stays put, and only the internal shuffle remains. This trick turns complex tracks (explosions, decays, star dances) into clear, often symmetric motion, making calculations friendlier. :contentReference[oaicite:9]{index=9}

High-Yield NEET Nuggets

1. Conservation of linear momentum: \( \mathbf{P}= \text{constant} \) when \( \mathbf{F}_{\text{ext}}=0 \).
2. Link between momentum and CM: \( \mathbf{P}=M\mathbf{V}. \)
3. Internal forces never change \( \mathbf{P} \); only external forces can.
4. Component-wise conservation (\( P_x, P_y, P_z \)) helps solve “back-to-back” emission problems.
5. Switching to the CM frame often turns messy multi-particle motion into clean, exam-friendly diagrams.

Keep practicing sample problems—every time you check momentum before and after an event, you build rock-solid intuition that pays off on test day!