A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 280 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +2.8 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.)
in m/s
(b) Determine the velocity of the second log if the lumberjack comes to rest on it.
in m/s
2006-12-06 07:04:09 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
a) Momentum before = momentum after
0 = m1*v1 + m2*v2
where m1 and v1 is the data on the lumberjack, m2 & v2: log.
b) Momentum before = momentum after
m1*v1 + 0 = (m1+m2)*v3
2006-12-06 07:24:46 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
This is two simple conservation of momentum problems. Use m1v1 = m2v2 to solve each.
2006-12-06 07:24:06 · answer #2 · answered by Anonymous · 0⤊ 0⤋