Question

A block of mass 2kg moving with 2m/s collides head do with another block of equal kept at rest
find : -
(a) maximum possible loss of Kinetic energy
(b) if the actual loss of kinetic energy 1/2 the maximum then find
Coefficient of restitution?​

Answer 1

Mass of first block , m₁ = 2 kg

Speed of block , v₁ = 2 m/s

Mass of second block , m₂ = 2 kg

Let final velocity of second block = v₂

Using law of conservation of momentum ,

2 x 2 = (2+2) v

v = 1 m/s

Therefore ,

(a) Loss in Kinetic Energy in elastic collision = (1/2) x 2 x 2² v - (1/2)(2+2) x 1²

= 4 - 2

= 2 J

(b) Actual loss = Maximum Loss / 2 = 2/2 = 1 J

(1/2) x 2 x 2² - (1/2) x 2 x v₁² + (1/2) x 2 x v₂² = 1

4 - (v₁² + v₂²) = 1

4 - [(1 + e²) x 4] / 2 = 1

2(1 + e²) = 3

1 + e² = 3/2

e² = 1/2

e = 1 / √2

Hope it helps!