System of Particles & Center of Mass: Master Notes

Chapter Overview

Center of Mass (CM) allows us to treat any complex distributed system as a single point particle. In IIT-JEE (Advanced), master problems center on continuous mass CM integrals, Zero External Force shift problems ($m_1\mathrm{\Delta }{x}_1+m_2\mathrm{\Delta }{x}_2=0$), oblique elastic/inelastic collisions, and variable mass rocket dynamics.

---

1. Center of Mass Coordinates & Standard Geometries

$${\overrightarrow{R}}_{\text{cm}}=\frac{\sum m_i{\overrightarrow{r}}_i}{\sum m_i}=\frac{1}{M}\int \overrightarrow{r}dm$$

1.1 High-Yield Standard Bodies CM Cheat Sheet (From Base/Center)

Geometry	Mass Distribution	Position of Center of Mass $y_{\text{cm}}$	Memory Trick
Semicircular Wire Ring	Linear $\lambda$	$\frac{2R}{\pi }\approx 0.637R$	Ring is outer perimeter
Semicircular Flat Plate (Disk)	Surface $\sigma$	$\frac{4R}{3\pi }\approx 0.424R$	Disk mass pulled toward center
Hollow Hemispherical Shell	Surface $\sigma$	$\frac{R}{2}=0.5R$	Exactly halfway up
Solid Hemisphere	Volume $\rho$	$\frac{3R}{8}=0.375R$	Heavier at base
Hollow Cone	Surface $\sigma$	$\frac{h}{3}$ (from base)	$1/3$ from base
Solid Cone	Volume $\rho$	$\frac{h}{4}$ (from base)	$1/4$ from base (tapering)

---

2. Motion of Center of Mass & Conservation of Momentum

$${\overrightarrow{P}}_{\text{total}}=M{\overrightarrow{v}}_{\text{cm}},\sum {\overrightarrow{F}}_{\text{ext}}=M{\overrightarrow{a}}_{\text{cm}}=\frac{d{\overrightarrow{P}}_{\text{total}}}{dt}$$

2.1 The "Man on Boat / Plank" Shift Rule ($\sum F_{\text{ext},x}=0$)

If no external horizontal force acts on the system:

$$a_{\text{cm},x}=0⟹v_{\text{cm},x}=0⟹\mathrm{\Delta }{X}_{\text{cm}}=0$$$$\boxed{{\displaystyle m_1\mathrm{\Delta }{x}_1+m_2\mathrm{\Delta }{x}_2+\cdots +m_n\mathrm{\Delta }{x}_n=0}}$$

Universal Shift Formula: When person $m$ walks distance $L$ relative to a movable plank of mass $M$:

$$\mathrm{\Delta }{x}_{\text{plank}}=-\left(\frac{m}{m+M}\right)L$$

---

3. Collisions & Coefficient of Restitution ($e$)

$$e=\frac{\text{Velocity of Separation along Line of Impact (LOI)}}{\text{Velocity of Approach along Line of Impact (LOI)}}=\frac{v_{2\parallel }-v_{1\parallel }}{u_{1\parallel }-u_{2\parallel }}$$

Multi-Mode DiagramCenter of Mass & 1D/2D Collision Dynamics

Option 1: Publication-Grade Scientific Vector SVG

Collision along the Line of Impact (LOI) with coefficient of restitution e and impulse integration.

Momentum Conservation: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Restitution Coefficient: $e = \frac{v_2 - v_1}{u_1 - u_2}$

Option 2: Interactive 60 FPS Canvas Physics Engine

Simulate 2-body collision with variable restitution coefficient e.

Restitution (e): 0.8

Restitution ($e$):0.8

Option 3: Interactive 3D Spatial Viewport

3D collision trajectory.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Momentum Invariant:$P_{\text{total}} = \text{const}$

Option 5: Reactive Parametric Physics with Live Calculus

Final velocity calculator for collisions.

Mass m₁: 2 kg

Restitution e: 1

v1 final ($v_1$):$0.0 m/s$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

3.1 Post-Collision Velocities in 1D

$$v_1=\left(\frac{m_1-e{m}_2}{m_1+m_2}\right)u_1+\frac{m_2(1+e)}{m_1+m_2}{u}_2$$$$v_2=\frac{m_1(1+e)}{m_1+m_2}{u}_1+\left(\frac{m_2-e{m}_1}{m_1+m_2}\right)u_2$$

3.2 Kinetic Energy Loss in Inelastic Collisions

$$\boxed{{\displaystyle \mathrm{\Delta }{K}_{\text{loss}}=\frac{1}{2}\mu (u_1-u_2{)}^2(1-e^2)}}$$

Where $\mu =\frac{m_1{m}_2}{m_1+m_2}$ is the reduced mass of the two-body system!

---

4. Variable Mass Dynamics (Rocket Propulsion)

When mass enters or leaves a system at rate $\frac{dM}{dt}$ with velocity ${\overrightarrow{u}}_{\text{rel}}$ relative to the body:

$$\boxed{{\displaystyle M\frac{d\overrightarrow{v}}{dt}={\overrightarrow{F}}_{\text{ext}}+{\overrightarrow{u}}_{\text{rel}}\frac{dM}{dt}}}$$

• Thrust Force: ${\overrightarrow{F}}_{\text{thrust}}={\overrightarrow{u}}_{\text{rel}}\frac{dM}{dt}$ (forward when exhaust is expelled backward).
• Tsiolkovsky Rocket Equation (Gravity-free space, exhaust speed $u$):$$v(t)=v_0+u\ln \left(\frac{M_0}{M(t)}\right)$$

---

5. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 — Wedge Recoil & Spring Collision

Question:
A block of mass $m$ slides down a smooth curved track of mass $M$ initially at rest on a frictionless floor. The track has a smooth horizontal exit. If the block is released from height $h$, find:

1. The speed $V$ of the track when the block leaves horizontally.
2. The speed $v$ of the block at the moment of exit.

Step-by-Step Solution:

• 1. Conservation of Horizontal Momentum ($\sum F_{\text{ext},x}=0$):$$mv-MV=0⟹V=\frac{m}{M}v$$
• 2. Conservation of Mechanical Energy:$$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}M{V}^2=\frac{1}{2}m{v}^2+\frac{1}{2}M{\left(\frac{m}{M}v\right)}^2=\frac{1}{2}m{v}^2(1+\frac{m}{M})$$$$v^2=\frac{2gh}{1+\frac{m}{M}}=\frac{2Mgh}{M+m}$$$$\boxed{{\displaystyle v=\sqrt{\frac{2Mgh}{M+m}},V=\sqrt{\frac{2m^{2}gh}{M(M+m)}}}}$$

---

PYQ 2: JEE Advanced 2019 — Oblique Collision of Elastic Spheres

Question:
A moving ball of mass $m$ collides elastically ($e=1$) with an identical stationary ball of mass $m$. The collision is oblique (non-head-on). Prove that the two balls move at right angles (${90}^{\circ }$) to each other after the collision.

Step-by-Step Solution:

1. Conservation of Linear Momentum:

   $$m\overrightarrow{u}=m{\overrightarrow{v}}_1+m{\overrightarrow{v}}_2⟹\overrightarrow{u}={\overrightarrow{v}}_1+{\overrightarrow{v}}_2$$

   Squaring both sides:

   $$u^2=({\overrightarrow{v}}_1+{\overrightarrow{v}}_2)\cdot ({\overrightarrow{v}}_1+{\overrightarrow{v}}_2)=v_1^2+v_2^2+2({\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_2)\text{--- (1)}$$

2. Conservation of Kinetic Energy ($e=1$):

   $$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2⟹u^2=v_1^2+v_2^2\text{--- (2)}$$

3. Comparing (1) and (2):

   $$v_1^2+v_2^2=v_1^2+v_2^2+2({\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_2)⟹2({\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_2)=0⟹\boxed{{\displaystyle {\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_2=0}}$$

   (The post-collision trajectories are strictly perpendicular ($\theta ={90}^{\circ }$)).

---

6. High-Yield Formula Sheet

System	Formula	Notes
Shift of Plank	$\mathrm{\Delta }x=-\frac{m}{m+M}L$	No external horizontal force
Energy Loss in Collision	$\mathrm{\Delta }K=\frac{1}{2}\mu v_{\text{rel}}^2(1-e^2)$	Inelastic collisions
Thrust Force	$F_{\text{thrust}}=u_{\text{rel}}\frac{dm}{dt}$	Rocket / Conveyor belt
Solid Hemisphere CM	$y_{\text{cm}}=\frac{3R}{8}$	From flat base
Hollow Hemisphere CM	$y_{\text{cm}}=\frac{R}{2}$	From flat base