Question

A body of mass ‘M’ at rest is struck by a body of mass ‘m’. Show that the fraction of KE of mass m transferred to the struck particle is \(\frac {4mM}{(m+M)^2}\).

Answer 1

m₁ = m, u₁ = u

m₂ = M, u₂ = 0 V₂ = ?

By conservation of momentum

m₁ u₁ = (m₁ + m₂) v₂

∴ v₂ = \(\frac {m_1u_1}{m_1+m_2}\)

k E of body struck after collision

E₂ = \(\frac{1}{2}\) m₂V₂² = \(\frac{1}{2}\)m \((\frac{2mu}{m+M})^2\)

= \(\frac{2Mm^2U^2}{(M+m)^2}\)

Initial K E E₁ = \(\frac{1}{2}\) m₁ u₁² = \(\frac{1}{2}\) mu²

∴ Fraction of initial K E transferred is