#### Answer

$.26v_0;.31v_0$

#### Work Step by Step

a) We find that the blocks, after the collision, have a velocity of:
$v_{1f}=\frac{M-nM}{m+nM}v_{1i}+\frac{2nM}{M+nM}v_{2i}$
$v_{1f}=\frac{M-nM}{m+nM}v_{0}$
$v_{2f}=\frac{2M}{m_1+m_2}v_{1i}+\frac{M-nM}{M+nM}v_{2i}$
$v_{2f}=\frac{2M}{m+nM}v_{0}$
If n is less than three, we see than 2M will be greater than $|M-nM|$. This would make the second block faster, meaning the blocks will never again collide.
b) If n is equal to 4, we see than 2M will be less than $|M-nM|$. This would make the first block faster, meaning the blocks will collide a second time.
c)
We find:
$v_{1f}=\frac{-9M}{11M}v_{0}$
$v_{2f}=\frac{2M}{11M}v_{0}$
They will collide again. Continuing to use the equations $v_{1f}=\frac{M-nM}{m+nM}v_{1i}+\frac{2nM}{M+nM}v_{2i}$ and $v_{2f}=\frac{2M}{m_1+m_2}v_{1i}+\frac{M-nM}{M+nM}v_{2i}$ until $|v_{1f}|$