A very heavy block of mass M moving horizontally with a speed of 2...

In a 1D elastic collision where a very heavy object ($M$) collides with a light stationary object ($m$), the light object acquires approximately twice the initial velocity of the heavy object.

$v_{\text{bob}} = 2v = 2 \times 2 = 4$ ms$^{-1}$.

By conservation of mechanical energy for the pendulum swing, the kinetic energy of the bob just after the collision is converted entirely into gravitational potential energy at its maximum height:

$\dfrac{1}{2} m v_{\text{bob}}^2 = mgh$

$h = \dfrac{v_{\text{bob}}^2}{2g} = \dfrac{4^2}{2 \times 10} = \dfrac{16}{20} = 0.8$ m.

Answer: $0.8$ m