Answer
The fraction of kinetic energy that is lost in this collision is 0.955
Work Step by Step
(a) We can find the initial kinetic energy $KE_1$.
$KE_1 = \frac{1}{2}mv^2$
Note that $v' = \frac{m}{m+M}~v$
We can find the kinetic energy $KE_2$ after the collision.
$KE_2 = \frac{1}{2}(m+M)(v')^2= \frac{1}{2}(m+M)(\frac{m}{m+M})^2~v^2$
$KE_2 = \frac{1}{2}(\frac{m^2}{m+M})~v^2$
We can find an expression for the fraction of kinetic energy that is lost in the collision.
$\frac{\Delta KE}{KE} = \frac{KE_1-KE_2}{KE_1}$
$\frac{\Delta KE}{KE} = \frac{\frac{1}{2}mv^2 - \frac{1}{2}(\frac{m^2}{m+M})~v^2}{\frac{1}{2}mv^2}$
$\frac{\Delta KE}{KE} = 1 - \frac{m}{(m+M)}$
$\frac{\Delta KE}{KE} = \frac{(m+M) - m}{(m+M)}$
$\frac{\Delta KE}{KE} = \frac{M}{(m+M)}$
(b) $\frac{\Delta KE}{KE} = \frac{M}{(m+M)}$
$\frac{\Delta KE}{KE} = \frac{0.380~kg}{(0.018~kg+0.380~kg)}$
$\frac{\Delta KE}{KE} = 0.955$
The fraction of kinetic energy that is lost in this collision is 0.955
