Collision Theory

Mass of the first body kg and for elastic collision coefficient of restitution,

Let initially body moves with velocity v after collision velocity becomes .
Let after collision velocity of M block becomes .
By conservation of momentum

or
or ....(i)
Since ,
or or
or ......(ii)
Substituting value of in E (i) from E (ii)

or

Step 1: In elastic collision of equal masses: vecv₁ ⊥ vecv₂
Step 2: Hence: theta₁+theta₂=90^∘

Step 1: Initial kinetic energy Kᵢ = \tfrac12 m(2v)² + \tfrac12 (2m)v² = 3mv²
Step 2: Final velocity after perfectly inelastic collision vf = (2mv + 2mv)/(3m) = (4v)/(3)
Step 3: Final kinetic energy Kf = \tfrac12 (3m)((4v)/(3))² = (8)/(3)mv²
Step 4: Fractional loss (Kᵢ-Kf)/(Kᵢ) ≈ 0.5

Step 1: Momentum is conserved: m₁ v = (m₁ + m₂)V
Step 2: V = (m₁ v)/(m₁ + m₂) < v

Step 1: In an elastic collision of two identical masses where one is initially at rest, both momentum and kinetic energy are conserved.
Step 2: It can be shown from vector momentum conservation that the velocities after collision are perpendicular.
Step 3: Hence, theta₁+theta₂ = 90^∘

Step 1: Initial kinetic energy: Kᵢ = (1)/(2)m(2v)² + (1)/(2)(2m)v² Kᵢ = 2mv² + mv² = 3mv² Step 2: Total momentum before collision: vecp = (2mv)hati + (2mv)hatj Step 3: After perfectly inelastic collision, combined mass = 3m Velocity after collision: vf = frac√((2mv)² + (2mv)²)3m = \frac2√(2)3v Step 4: Final kinetic energy: Kf = (1)/(2)(3m)(\frac2√(2)3v)² = (4)/(3)mv² Step 5: Percentage loss in energy: % loss = (Kᵢ - Kf)/(Kᵢ) × 100 = (3 - (4)/(3))/(3) × 100 ≈ 62%