A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 245 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +2.7 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 just before the lumberjack jumps off? (Indicate the direction of the velocity by the sign of your answers.)
m/s
(b) Determine the velocity of the second log if the lumberjack comes to rest on it.
m/s

(a) Well, the lumberjack is running to the other end of the log, which means he's pushing off the log with his feet. It's like a really extreme game of hopscotch. Since there's no friction to slow him down, the momentum he gains from running is transferred to the log. So, just before the lumberjack jumps off, the log will have a velocity of -2.7 m/s in the opposite direction.

(b) Now, when the lumberjack hops onto the second identical log, we need to consider conservation of momentum. Since there's no external force acting on the system, the total momentum before and after is the same. So, if the first log has a velocity of -2.7 m/s, the second log will have a velocity of +2.7 m/s to balance things out. Otherwise, we'd have some pretty wobbly lumberjacks and logs floating around.

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the lumberjack jumps off the log should be equal to the total momentum after he jumps.

Momentum (p) is equal to mass (m) multiplied by velocity (v).

1. Before the lumberjack jumps:
- The mass of the lumberjack is 98 kg.
- The mass of the log is 245 kg.
- The velocity of the lumberjack is 0 m/s (standing at rest).
- The velocity of the log is also 0 m/s.

To find the total momentum before the jump, we can add up the momenta of the lumberjack and the log:

Total momentum before the jump = (mass of the lumberjack x velocity of the lumberjack) + (mass of the log x velocity of the log)

Total momentum before the jump = (98 kg x 0 m/s) + (245 kg x 0 m/s) = 0 kg·m/s

2. After the lumberjack jumps:
- The lumberjack's mass remains the same at 98 kg.
- The mass of the log remains the same at 245 kg.
- The velocity of the lumberjack is +2.7 m/s relative to the shore.
- The velocity of the log is unknown.

To find the velocity of the log just before the lumberjack jumps off, we can set up the equation:

Total momentum after the jump = (mass of the lumberjack x velocity of the lumberjack) + (mass of the log x velocity of the log)

Total momentum after the jump = (98 kg x 2.7 m/s) + (245 kg x velocity of the log)

Since momentum is conserved, the total momentum before and after the jump should be the same:

0 kg·m/s = (98 kg x 2.7 m/s) + (245 kg x velocity of the log)

To find the velocity of the log, we rearrange the equation:

(245 kg x velocity of the log) = -(98 kg x 2.7 m/s)

velocity of the log = -(98 kg x 2.7 m/s) / 245 kg

Simplifying this calculation:

velocity of the log = -27.54 m/s

Therefore, the velocity of the first log just before the lumberjack jumps off is -27.54 m/s. The negative sign indicates that the log is moving in the opposite direction to the lumberjack's velocity.

To find the velocity of the second log if the lumberjack comes to rest on it, we can use the same principle of conservation of momentum:

Total momentum before the jump = Total momentum after the jump

(98 kg x 0 m/s) + (245 kg x -27.54 m/s) = (98 kg x velocity of the lumberjack on the second log) + (245 kg x velocity of the second log)

Since the lumberjack comes to rest on the second log, the velocity of the lumberjack on the second log is 0 m/s:

(245 kg x -27.54 m/s) = (0 kg·m/s) + (245 kg x velocity of the second log)

Solving for the velocity of the second log:

velocity of the second log = (245 kg x -27.54 m/s) / 245 kg

Simplifying this calculation:

velocity of the second log = -27.54 m/s

Therefore, the velocity of the second log, if the lumberjack comes to rest on it, is -27.54 m/s. Again, the negative sign indicates movement in the opposite direction to the lumberjack's velocity.

To answer these questions, we can apply the principle of conservation of momentum.

(a) Before the lumberjack jumps off the first log, the system consisting of the lumberjack and the log has zero net external force acting on it, so the total momentum of the system is conserved. We can use the equation:

m1v1 + m2v2 = m1v1' + m2v2'

where
m1 = mass of the lumberjack (98 kg)
v1 = initial velocity of the lumberjack (0 m/s since they are at rest)
m2 = mass of the log (245 kg)
v2 = initial velocity of the log (0 m/s since it is at rest)
v1' = final velocity of the lumberjack (2.7 m/s)
v2' = final velocity of the log (unknown)

Plugging in the values, the equation becomes:

(98 kg)(0 m/s) + (245 kg)(0 m/s) = (98 kg)(2.7 m/s) + (245 kg)(v2')

This simplifies to:

0 = 264.6 kg·m/s + (245 kg)(v2')

To find v2', isolate v2':

(245 kg)(v2') = -264.6 kg·m/s
v2' = (-264.6 kg·m/s) / (245 kg)

Calculating this gives us:

v2' ≈ -1.08 m/s

So the velocity of the first log just before the lumberjack jumps off is approximately -1.08 m/s, indicating that it is moving in the opposite direction relative to the shore.

(b) Now, let's apply the conservation of momentum principle again when the lumberjack jumps onto the second log. This time, the lumberjack comes to rest on the second log, so the final velocity of the lumberjack, v1', is 0 m/s.

Using the same equation as before:

m1v1 + m2v2 = m1v1' + m2v2'

we can plug in the known values:

(98 kg)(2.7 m/s) + (245 kg)(-1.08 m/s) = (98 kg)(0 m/s) + (245 kg)(v2')

Simplifying:

(264.6 kg·m/s) + (-264.6 kg·m/s) = (245 kg)(v2')
0 = (245 kg)(v2')

Therefore, the velocity of the second log when the lumberjack comes to rest on it is 0 m/s. The second log is initially at rest and remains at rest when the lumberjack jumps onto it.