Question

When a moving body collides elastically with a stationary body, of n times its mass, then the fraction of kinetic energy transferred to the stationary body is _______.

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

Answer 1

Answer: (A)

A. $\frac{4n}{(1+n)^2}$

Answer 2

Let $m_1 = m$ and $m_2 = nm$. Let the initial velocity of the moving body be $u$. Initial kinetic energy $KE_i = \frac{1}{2}mu^2$. Final velocity of the stationary body $V_2 = \frac{2m_1}{m_1+m_2}u = \frac{2m}{m+nm}u = \frac{2u}{1+n}$. Kinetic energy transferred to the stationary body $KE_{transfer} = \frac{1}{2}m_2V_2^2 = \frac{1}{2}(nm)(\frac{2u}{1+n})^2 = \frac{2nm u^2}{(1+n)^2}$. Fraction of kinetic energy transferred = $\frac{KE_{transfer}}{KE_i} = \frac{\frac{2nm u^2}{(1+n)^2}}{\frac{1}{2}mu^2} = \frac{4n}{(1+n)^2}$.