Answer
a) The velocity of the 2nd block is $+0.51m/s$
b) The ratio is $1.48\times10^{-3}$
Work Step by Step
a) First, let's look at the collision with the 1st block.
The bullet, whose mass $m=4g=4\times10^{-3}kg$ and initial velocity $\vec{v}=+355m/s$, hits the block, whose mass $M=1.15kg$ and initial velocity $\vec{V}=0$. After the collision, the block moves at $\vec{V_f}=+0.55m/s$. We assume the total linear momentum is conserved, so the velocity of the bullet after the collision is $$M\vec{V}+m\vec{v}=M\vec{V}_f+m\vec{v}_f$$ $$\vec{v_f}=\frac{0+m\vec{v}-M\vec{V}_f}{m}=+196.9m/s$$
Then, we take a look at the collision with the 2nd block.
The bullet now has initial velocity $\vec{v}=+196.9m$ and hits the 2nd block, whose mass $M=1.53kg$ and initial velocity $\vec{V}=0$. After the collision, we have a block-bullet system, whose mass $M+m=1.534kg$ and whose velocity is $\vec{V_f}$, which we need to find.
Again, we have $$M\vec{V}+m\vec{v}=(M+m)\vec{V}_f$$ $$\vec{V}_f=\frac{0+m\vec{v}}{M+m}=+0.51m/s$$
b) The total KE before the collision, which comprises of KE of only the bullet, is $$KE_b=\frac{1}{2}m_{bullet}v_{bullet}^2=252.05J$$
The total KE after the collision, which comprises of KE of block 1 and KE of the bullet-block 2 system, is $$KE_b=\frac{1}{2}m_{block1}v_{block1}^2+\frac{1}{2}m_{system}v_{system}^2=0.373J$$
The ratio is $\frac{0.373}{252.05}=1.48\times10^{-3}$
