Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball...

The correct answer is B Elastic
Let mass of ball 1=2m
Mass of ball 2=m
Velocity of ball 1 before collision =u
Velocity of ball 1 after collision $=\frac{u}{3}$
Velocity of ball 2 before collision =0
Let final velocity of ball 2$=v_2$
Momentum will be conserved because there is no outside force acting on the system.
Prior to and following a collision, using conservation of linear momentum.
$(2m)u+m(0)=(m)v_2+(2m)(\frac{u}{3})$
or $v_2=\frac{4u}{3}$
Now,
$$\begin{gathered} e=\frac{\text{velocity of separation}}{\text{velocity of approach}} \\ e=\frac{v_2-v_1}{u_1-u_2} \\ e=\frac{\frac{4u}{3}-\frac{u}{3}}{u-0}=1 \end{gathered}$$
Thus, collision is elastic.