Question

What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of $5$ times its mass? (Assume the collision to be head-on elastic collision)

• A. $50.0\%$
• B. $66.6\%$
• C. $55.5\%$
• D. $33.3\%$

Answer 1

Answer: (C)

Velocity after collision
$V_2=\frac{\left(m_2-m_1\right)u_2+2m_1{u}_1}{m_1+m_2}$
$V_2=\frac{(5m-m)0+2m\cdot u_0}{m+5m}=\frac{u_0}{3}$
$\%\Delta KE=\frac{\frac{1}{2}5m{\left(\frac{u_0}{3}\right)}^2-0}{\frac{1}{2}m{u}_0^2}\times 100$
$=\frac{5u_0^2}{9u_0^2}\times 100=\frac{500}{9}=55.6\%$