A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 270 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +2.6 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.)
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, as long as no external forces are acting.
The momentum of an object is given by the formula:
momentum = mass × velocity
Let's denote the velocity of the first log just before the lumberjack jumps off as V1 and the velocity of the second log (which is initially at rest) as V2.
Before the lumberjack jumps off, the total momentum of the system is zero since both the lumberjack and the first log are initially at rest. Therefore:
Total initial momentum = 0
After the lumberjack jumps off, the total momentum of the system is still zero because there are no external forces acting on the system. Therefore:
Total final momentum = 0
The momentum of the lumberjack can be calculated using the formula:
momentum of lumberjack = mass of lumberjack × velocity of lumberjack
The momentum of the first log can be calculated using the formula:
momentum of first log = mass of first log × velocity of first log
The momentum of the second log can be calculated using the formula:
momentum of second log = mass of second log × velocity of second log
Since the total initial and final momentum is zero and the mass of the lumberjack and the second log are the same, we can write the following equation:
0 = momentum of lumberjack + momentum of first log + momentum of second log
Substituting the formula for momentum into the equation, we get:
0 = mass of lumberjack × velocity of lumberjack + mass of first log × velocity of first log + mass of second log × velocity of second log
Plugging in the given values:
0 = (98 kg) × (+2.6 m/s) + (270 kg) × V1 + (270 kg) × V2
Since the second log is initially at rest (V2 = 0 m/s), we can simplify the equation to:
0 = (98 kg) × (+2.6 m/s) + (270 kg) × V1
Now, we can solve for V1:
(-98 kg × 2.6 m/s) = (270 kg × V1)
-254.8 kg*m/s = 270 kg × V1
V1 = (-254.8 kg*m/s) / (270 kg)
V1 = -0.944 m/s
Therefore, the velocity of the first log just before the lumberjack jumps off is -0.944 m/s (indicating that it is moving in the opposite direction of the lumberjack's jump), or you can simply say the direction is towards the shore.
To find the velocity of the first log just before the lumberjack jumps off, we can apply the law of conservation of momentum. The law of conservation of momentum states that the total momentum of a closed system remains constant before and after an event.
In this case, the system consists of the lumberjack and the log. Initially, the momentum of the system is zero since both the lumberjack and the log are at rest. After the lumberjack runs to the other end of the log, the momentum of the system will still be zero since the lumberjack and the log have equal and opposite momenta.
Let's assign positive velocities to the right and negative velocities to the left. Using the principle of conservation of momentum, we can write:
(m1)(v1) + (m2)(v2) = 0
Where:
m1 = mass of the lumberjack
v1 = velocity of the lumberjack
m2 = mass of the log
v2 = velocity of the log
Substituting the given values:
(m1)(2.6 m/s) + (270 kg)(v2) = 0
Solving for v2:
(98 kg)(2.6 m/s) = - (270 kg)(v2)
v2 ≈ -0.946 m/s
Therefore, the velocity of the first log just before the lumberjack jumps off is approximately -0.946 m/s (in the opposite direction of the lumberjack's velocity).