Answer
a) The velocity of the first log is $-1.53m/s$
b) The velocity of the second log is $+1.08m/s$
Work Step by Step
a) At the start, both the lumberjack and the log are at rest. $\vec{v}_{0j}=\vec{v}_{0l}=0$, so the initial total momentum $\sum\vec{p}_0=0$. After that, the lumberjack runs to the other end of the log with $\vec{v}_{fj}=+3.6m/s$ . We need to find the velocity of the log and then $\vec{v_{fl}}$.
The principle of conservation of linear momentum states that $$m_j\vec{v_{fj}}+m_l\vec{v_{fl}}=\sum\vec{p}_0=0$$ $$\vec{v_{fl}}=-\frac{m_j\vec{v_{fj}}}{m_l}$$
We have $\vec{v_{fj}}=+3.6m/s$, $m_j=98kg$, and $m_l=230kg$. Therefore, $$\vec{v_{fl}}=-1.53m/s$$
b) When the lumberjack jumps on the second log, he maintains his speed of $+3.6m/s$, which we would consider his initial speed on the 2nd log: $\vec{v}_{0j}=+3.6m/s$, while the 2nd log is at rest $\vec{v}_{0l}=0$. Until the lumberjack comes to rest on the 2nd log, the velocity of the lumberjack and the log are the same $\vec{v}_{fj}=\vec{v}_{fl}=\vec{v}_{f}$.$$m_j\vec{v_{0j}}+m_l\vec{v_{0l}}=(m_l+m_j)\vec{v}_f$$ $$\vec{v_{f}}=\frac{(m_j\vec{v_{0j}}+m_l\vec{v_{0l}})}{m_l+m_j}=\frac{m_j\vec{v_{0j}}}{m_l+m_j}$$
We have $\vec{v_{0j}}=+3.6m/s$, $m_j=98kg$, and $m_l=230kg$.
$$\vec{v_f}=+1.08m/s$$
