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APEX CLASS ESTD 2018

IITJEE: 1D Collisions [Restitution (e)] & Center of Mass Tracker

By

Ashish Kumar Bahekar

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Apex Class • IITJEE Momentum & COM Lab

1D Collisions (Restitution e) & Center of Mass Tracker

Explore how Momentum is ALWAYS conserved while Kinetic Energy depends on Restitution (e)!

🎱 Initial: m₁ = 2kg (u₁ = 10 m/s →) & m₂ = 2kg (u₂ = 4 m/s →). Pick Collision Type:
BEFORE IMPACT (u₁ – u₂ = 6 m/s Approach): 2kg u₁ = 10 m/s 2kg u₂ = 4 m/s AFTER ELASTIC IMPACT (e = 1 → Identical Masses SWAP Velocities!): 2kg v₁ = 4 m/s! 2kg v₂ = 10 m/s! Kinetic Energy Audit K_i = 116J K_f = 116J Loss = 0J 🏆 Elastic Equal-Mass Shortcut: Whenever m₁ = m₂ and e = 1, velocities simply exchange (v₁ = u₂, v₂ = u₁)! BEFORE IMPACT (u₁ = 10 m/s, u₂ = 4 m/s → Approach = 6 m/s): 2kg u₁ = 10 m/s 2kg u₂ = 4 m/s AFTER INELASTIC IMPACT (e = 0.5 → Separation v₂ – v₁ = 0.5 × 6 = 3 m/s): 2kg v₁ = 5.5 m/s 2kg v₂ = 8.5 m/s Kinetic Energy Audit K_i = 116J K_f = 102.5J Loss = 13.5J ⚡ Heat Loss Formula: ΔK_loss = ½ [m₁m₂/(m₁+m₂)] (1 – e²)(u₁ – u₂)² = ½ (1)(1 – 0.25)(36) = 13.5 J! BEFORE IMPACT (Total Momentum P_i = 2(10) + 2(4) = 28 kg·m/s): 2kg u₁ = 10 m/s 2kg u₂ = 4 m/s AFTER PERFECTLY INELASTIC IMPACT (e = 0 → Bodies Stick & Move at v_com = 7 m/s!): 2+2=4kg Common v = 28 / 4 = 7.0 m/s Kinetic Energy Audit K_i = 116J K_f = 98J Max Loss 18J 🧱 Perfectly Inelastic Rule: Momentum is STILL 100% conserved (P_f = 4 × 7 = 28), while KE loss is maximum (18 J)! Drop H₀ = 16 m 1st Rebound H₁ = e² H₀ 2nd Rebound H₂ = e⁴ H₀ 🏀 Bouncing Ball Formulas: Speed after nᵗʰ bounce vₙ = eⁿ u₀ | Height after nᵗʰ bounce Hₙ = e²ⁿ H₀!
Restitution (e = Sep / App)
e = 1.0 (Elastic) e = 0.5 (Inelastic) e = 0.0 (Sticks!) e = √(H₁ / H₀)
e = (v₂ – v₁) / (u₁ – u₂)
Linear Momentum (P)
28 kg·m/s (Conserved!) 28 kg·m/s (Conserved!) 28 kg·m/s (Conserved!) Impulse ΔP = m(1+e)u₀
Conserved in ALL Collisions
Final Velocities (v₁, v₂)
v₁ = 4, v₂ = 10 m/s v₁ = 5.5, v₂ = 8.5 m/s v₁ = v₂ = 7.0 m/s vₙ = eⁿ √(2gH₀)
Notice v_com = 7 m/s in all 3!
Kinetic Energy Loss (ΔK)
0.0 J (100% KE Kept) 13.5 J Lost as Heat 18.0 J (Max Loss!) ΔK = (1 – e²) mgH₀
ΔK = ½ μ (1 – e²)(u₁ – u₂)²
🎯 Select a High-Frequency JEE Center of Mass Problem Type:
m₁ = 2kg m₂ = 6kg ★ COM Fulcrum r₁ = [m₂ / (m₁+m₂)] d = (6/8)×12 = 9.0 m r₂ = 3.0 m ⚖️ Golden Moment Rule: About the Center of Mass, m₁r₁ = m₂r₂ (2 × 9m = 6 × 3m = 18)! COM is always closer to the heavier mass. 💥 Shell Explodes at Peak (x = R/2) Frag 1 (m) drops at x₁ = R/2 ★ COM Lands at R! Frag 2 (m) at x₂ = 3R/2! 🎆 5-Second Explosion Trick: Internal explosion cannot change COM path! M·R = m(R/2) + m(x₂) → x₂ = 3R/2! Removed Cavity (r = R/2, Area = A/4) O (0,0) ★ New COM = -R / 6 🧀 Cavity Subtraction Formula: X_com = (A₁x₁ – A₂x₂) / (A₁ – A₂) = [πR²(0) – π(R/2)²(R/2)] / (¾ πR²) = -R / 6!
Master COM Equation
r₁ = m₂ d / (m₁ + m₂) M X_com = m₁x₁ + m₂x₂ X = (A₁x₁ – A₂x₂)/(A₁ – A₂)
Vector Position Formula
COM Shift Direction
Toward Heavier Mass Zero Shift (Follows Parabola) Away from Cut-out Cavity
Physical Intuition Check
Standard JEE Result
m₁ r₁ = m₂ r₂ R = (m₁x₁ + m₂x₂) / M X_com = -R / 6
Direct 10-Second Substitution
🔥 3 Must-Memorize Center of Mass Positions for JEE Main:
• Semicircular Ring: 2R / π  |  Semicircular Disc: 4R / 3π  |  Hollow Hemisphere: R / 2  |  Solid Hemisphere: 3R / 8  |  Solid Cone: H / 4 (from base).

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