Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in contact (figure). One of the bobs is released after being displaced by $10^{\circ}$ so that it collides elastically head-on with the other bob.
$(a)$ Describe the motion of two bobs.
$(b)$ Draw a graph showing variation in energy of either pendulum with time,for $0 \leqslant t \leqslant 2T$,where $T$ is the period of each pendulum.

(N/A) Consider the diagram where bob $B$ is displaced through an angle $\theta$ and released.
At $t=0$,suppose bob $B$ is displaced by $\theta=10^{\circ}$ to the right. It is given potential energy $E_{1}=E$. Energy of $A$,$E_{2}=0$.
When $B$ is released,it strikes $A$ at $t=T/4$. In the head-on elastic collision between identical masses,they exchange velocities. Thus,$B$ comes to rest and $A$ gets the velocity of $B$. Therefore,$E_{1}=0$ and $E_{2}=E$.
At $t=2T/4$,$B$ reaches its extreme right position when $KE$ of $A$ is converted into $PE=E_{2}=E$. Energy of $B$,$E_{1}=0$.
At $t=3T/4$,$A$ reaches its mean position,when its $PE$ is converted into $KE=E_{2}=E$. It collides elastically with $B$ and transfers its entire energy to $B$. Thus,$E_{2}=0$ and $E_{1}=E$. The entire process is repeated.
$(b)$ The values of energies of $B$ and $A$ at different time intervals are tabulated below:

Time $(t)$	Energy of $B$ $(E_{1})$	Energy of $A$ $(E_{2})$
$0$	$E$	$0$
$T/4$	$0$	$E$
$2T/4$	$0$	$E$
$3T/4$	$E$	$0$
$4T/4$	$E$	$0$
$5T/4$	$0$	$E$
$6T/4$	$0$	$E$
$7T/4$	$E$	$0$
$8T/4$	$E$	$0$