A point mass of 1 kg moving horizontally with speed u hits a...

First, find the velocity of the $2$ kg bob just after the collision using conservation of mechanical energy during its upward swing:

$\dfrac{1}{2} M v_M^2 = M g h$

$v_M = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.8} = \sqrt{16} = 4$ ms$^{-1}$.

For a 1D elastic collision, the velocity of a target mass $M$ initially at rest, struck by a mass $m$ moving with initial velocity $u$, is given by:

$v_M = \dfrac{2m}{m + M} u$

Substituting the given values ($m = 1$ kg, $M = 2$ kg, $v_M = 4$ ms$^{-1}$):

$4 = \dfrac{2 \times 1}{1 + 2} u$

$4 = \dfrac{2}{3} u$

$u = 6$ ms$^{-1}$.

Answer: $6$ ms$^{-1}$