Answer
(a) $\frac{4}{3}v$
(b) elastic
Work Step by Step
(a) We can find the final speed of the second car as follows:
According to the law of conservation of momentum
$mv+\frac{1}{2}m(0)=m(\frac{v}{3}+\frac{1}{2}mv_2)$
$\implies \frac{2}{3}mv=\frac{1}{2}mv_2$
$\implies v_2=\frac{4}{3}v$
(b) We know that
$K.E_i=\frac{1}{2}mv^2$
and $K.E_f=\frac{1}{18}mv^2+\frac{16}{36}mv^2=\frac{1}{2}mv^2$
since $K.E_i=K.E_f$, thus the collision is elastic.
