A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 200 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +4.0 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.
IF friction is neglected :), then the center of mass remains at the same position. This leads to
98*4+200V=0
solve for V. Negative number means in the opposite direction.
for the second log,
98*4=(200+98)V
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the lumberjack jumps must be equal to the total momentum after the lumberjack jumps.
(a) Before the lumberjack jumps, the total momentum is zero since both the lumberjack and the log are at rest. After the lumberjack jumps and runs to the other end of the log, he attains a velocity of +4.0 m/s relative to the shore. As a result, the total momentum after the jump must also be zero (since the lumberjack does not have any momentum relative to the shore).
Let's assume the velocity of the log just before the lumberjack jumps off is "v". Since the total momentum before the jump is zero, we can write:
(98 kg)(0 m/s) + (200 kg)(v) = 0
Simplifying this equation, we get:
200v = 0
v = 0 m/s
Therefore, the velocity of the first log just before the lumberjack jumps off is 0 m/s.
(b) To find the velocity of the second log if the lumberjack comes to rest on it, we can again apply the conservation of momentum. Since the total momentum must be conserved, we have:
(98 kg)(+4.0 m/s) + (200 kg)(0 m/s) = (98 kg)(v2) + (200 kg)(0 m/s)
Simplifying this equation, we get:
392 kg*m/s = 98v2
Dividing both sides by 98 kg, we find:
4.0 m/s = v2
Therefore, the velocity of the second log if the lumberjack comes to rest on it is 4.0 m/s in the positive direction (relative to the shore).