Question

A body of mass m₁ strikes a stationary body of mass m₂. If the collision is elastic and head- on, the fraction of kinetic energy transmitted by the first body to the second body is -

• A. $\frac{m_{1}m_{2}}{m_{1} + m_{2}}$
• B. $\frac{2m_{1}m_{2}}{m_{1} + m_{2}}$
• C. $\frac{4m_{1}m_{2}}{(m_{1} + m_{2})^{2}}$
• D. $\frac{2m_{1}m_{2}}{(m_{1} + m_{2})^{2}}$

Answer 1

Answer: (C)

$\frac{4m_{1}m_{2}}{(m_{1} + m_{2})^{2}}$

Answer 2

$\frac{1}{2}$m₁u²

=$\frac{1}{2}m_{1}v_{1}^{2}$+$\frac{1}{2}m_{2}v_{2}^{2}$

e = $\frac{v_{2} - v_{1}}{u}$ Ž u = v₂

– v₁ m₁u = m₁v₁ + m₂v₂

m₁u = m₁v₁ + m₂(u + v₁)

(m₁ – m₂)u = (m₁ + m₂)v₁ Ž v₁ = $\frac{\left( m_{1} - m_{2} \right)u}{m_{1} + m_{2}}$

v₂ = u + $\frac{\left( m_{1} - m_{2} \right)u}{m_{1} + m_{2}}$ = $\frac{2m_{1}u}{m_{1} + m_{2}}$ Ž KE_i = $\frac{1}{2}$m₁u²

KE_transfer = $\frac{1}{2}m_{2}v_{2}^{2}$ =

$\frac{1}{2}m_{2}\frac{4m_{1}^{2}u^{2}}{(m_{1} + m_{2})^{2}}$

Fraction = $\frac{KE_{trasnsfer}}{KE_{i}}$ = $\frac{4m_{2}m_{1}}{(m_{1} + m_{2})^{2}}$