The bob $A$ of a pendulum,released from a horizontal position to the vertical,hits another bob $B$ of the same mass at rest on a table as shown in the figure. If the length of the pendulum is $1\,m$,calculate:
$(a)$ The height to which bob $A$ will rise after the collision.
$(b)$ The speed with which bob $B$ starts moving.
Neglect the size of the bobs and assume the collision to be elastic.

(N/A) When ball $A$ reaches the bottom point,its velocity is horizontal. Since the collision is elastic and the masses are equal,the velocities are exchanged.
$(a)$ In an elastic collision between two bodies of equal mass where one is at rest,the moving body comes to rest and the stationary body acquires the velocity of the moving body. Therefore,bob $A$ comes to rest at the bottom point after the collision. The height to which it will rise is $0\,m$.
$(b)$ The speed with which bob $B$ starts moving is equal to the speed with which bob $A$ hits bob $B$. Using the law of conservation of energy for bob $A$ as it falls from height $h = 1\,m$:
$v = \sqrt{2gh}$
$v = \sqrt{2 \times 9.8 \times 1}$
$v = \sqrt{19.6} \approx 4.43\,m/s$.