A lumberjack (mass = 95.7 kg) is standing at rest on one end of a floating log (mass = 218 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +2.54 m/s relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water. (a) What is the velocity of the first log (again relative to the shore) just before the lumberjack jumps off? (b) Determine the velocity of the second log (again relative to the shore) if the lumberjack comes to rest relative to the second log.

To solve this problem, we can apply the principle of conservation of momentum.

(a) Let's denote the velocity of the first log just before the lumberjack jumps off as v1, and the velocity of the lumberjack as v_lumberjack relative to the shore. Since the lumberjack is initially at rest and then attains a velocity of +2.54 m/s, we have:

v_lumberjack = +2.54 m/s

According to the law of conservation of momentum, the total momentum before the lumberjack jumps off is equal to the total momentum after the lumberjack jumps off. The total momentum considers both the lumberjack and the log.

Before the lumberjack jumps off:
Total momentum = Momentum of log + Momentum of lumberjack
Total momentum before = (mass of log * velocity of log) + (mass of lumberjack * velocity of lumberjack)

Since the lumberjack is initially at rest, the momentum of the lumberjack is zero. Therefore, the total momentum before the lumberjack jumps off is given by:

Total momentum before = (mass of log * velocity of log)

After the lumberjack jumps off, the total momentum is only due to the log because the lumberjack is no longer on it. We have:

Total momentum after = (mass of log * velocity of log) + 0
= (mass of log * velocity of log)

Since the momentum is conserved, we can equate the two expressions for total momentum to get:

(mass of log * velocity of log) = (mass of log * velocity of log)

Therefore, the velocity of the first log just before the lumberjack jumps off is equal to the velocity of the lumberjack, which is:

v1 = v_lumberjack
= +2.54 m/s

So, the velocity of the first log just before the lumberjack jumps off is +2.54 m/s relative to the shore.

(b) To determine the velocity of the second log, we need to apply the conservation of momentum again.

Again, consider the total momentum before and after the interaction between the lumberjack and the second log.

Total momentum before = (mass of log * velocity of log)
= (218 kg * +2.54 m/s)
= 554.92 kg·m/s (taking two decimal places)

Total momentum after = (mass of log * velocity of log) + (mass of lumberjack * velocity of lumberjack)

In this case, we want to find the velocity of the second log when the lumberjack comes to rest relative to it. When the lumberjack comes to rest relative to the second log, the velocity of the lumberjack is zero (v_lumberjack = 0). Therefore, we have:

Total momentum after = (mass of log * velocity of log) + (0)

Since momentum is conserved, the total momentum before and after must be the same. Hence, we have:

(mass of log * velocity of log) = (mass of log * velocity of log) + 0
554.92 kg·m/s = (218 kg * velocity of second log)

To find the velocity of the second log, rearrange the equation and solve for it:

velocity of second log = (554.92 kg·m/s) / (218 kg)
≈ 2.54 m/s (taking two decimal places)

Therefore, the velocity of the second log, relative to the shore, is approximately +2.54 m/s when the lumberjack comes to rest relative to it.