Answer
a) The velocity of the ball just before the collision has a magnitude of $4.85m/s$ and is directed along the $+x$ direction.
b) The velocity of the ball just after the collision has a magnitude of $0.97m/s$ and is directed along the $-x$ direction.
Work Step by Step
a) Since air resistance is negligible, we can apply the principle of conservation of mechanical energy:
$$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=0$$ $$\frac{1}{2}(v_f^2-v_0^2)+g(h_f-h_0)=0$$
We take the starting point to be before the ball was released, where $v_0=0$, and the end point to be just before it hit the block, where it has velocity $v_f$ and $h_f-h_0=-d_{wire\_length}=-1.2m$. Therefore, $$\frac{1}{2}v_f^2+g(h_f-h_0)=0$$ $$v_f=\sqrt{-2g(h_f-h_0)}=4.85m/s$$
The velocity $\vec{v_f}$ is directed in the $+x$ direction.
b) The ball, whose mass is $m=1.6kg$ and whose initial velocity is $\vec{v}=+4.85m/s$, hits the block, whose mass is $M=2.4kg$ and whose initial velocity is $\vec{V}=0$. We assume the total linear momentum is conserved, so $$M\vec{V}+m\vec{v}=M\vec{V}_f+m\vec{v}_f$$ $$2.4\vec{V}_f+1.6\vec{v}_f=7.76 (1)$$
Since the collision is elastic, the kinetic energies are conserved, too. $$\frac{1}{2}(MV^2+mv^2)=\frac{1}{2}(MV_f^2+mv_f^2)$$ $$2.4V_f^2+1.6v_f^2=37.64 (2)$$
Solving (1) and (2), we find the velocity of the ball to be $\vec{v}_f=-0.97m/s$, which has a magnitude of $0.97m/s$ and is directed in the $-x$ direction.
