Question

A body of mass m/2 moving with velocity v_0 collides elastically with another mass of m/3. find percentage change in KE of first body?​

Answer 1

It has given that, mass of body = m/2 is moving with velocity v₀ collides elastically with another mass of m/3 which is at rest.

We have to find the percentage change in kinetic energy of first body.

solution : we have to find the velocity of first particle using formula,

v₁ = (2m₂u₂)/(m₁ + m₂) + (m₁ - m₂)/(m₁ + m₂)u₁

here m₁ = m/2 , m₂ = m/3 , u₁ = v₀ , u₂ = 0

so, v₁ = 2(m/3 × 0)/(m/2 + m/3) + (m/2 - m/3)(v₀)/(m/2 + m/3)

= 0 + (m/6)v₀/(5m/6)

= v₀/5

now change in kinetic energy , ∆K = K_i - K_f

= 1/2 (m/2)v₀² - 1/2 (m/2) (v₀/5)²

= 1/2 (m/2) v₀² [ 1 - 1/25]

= 1/2 (m/2) v₀² × (24/25)

percentage change in kinetic energy of first body = ∆K/Ki × 100

= 1/2 (m/2) v₀² × (24/25)/ 1/2 (m/2) v₀² × 100

= 24/25 × 100

= 96 %

Therefore the change in kinetic energy of first body is 96 %