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)

Answer 1

For elastic collision
$u=v_2-v_1...\left(1\right)$
From momentum conservation:
$mu=m{v}_1+5m{v}_2$
$\Rightarrow u=v_1+5v_2....\left(2\right)$
From (1) and (2)
$v_2=\frac{u_1}{3}$

Alternatively:
For a head on elastic collision
$v_2=\frac{m{u}_1}{m+5m}+\frac{m{u}_1}{m+5m}$

$=\frac{2u_1}{6}\text{or}\frac{u_1}{3}$

Initial kinetic energy of first mass: $K=\frac{1}{2}m{u}_1^2$
Final kinetic energy of second mass
$K^{\prime }=\frac{1}{2}\times 5m{\left(\frac{u_1}{3}\right)}^2$
$K^{\prime }=\frac{5}{9}\left(\frac{1}{2}m{u}_1^2\right)=\frac{5}{9}K$
% energy transferred $=\frac{\frac{5}{9}K}{K}\times 100=55.6$%
$\Rightarrow$ kinetic energy transferred $=55\%$ of initial kinetic energy of first colliding mass