physics question?

Question

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

Answer 1

I answered this question 6 hours ago. Maybe you need me to give more explanation.

a) Momentum before = momentum after

Momentum before = 0 because nobody had jumped yet - no velocity. So the equation is
0 = m1*v1 + m2*v2
where m1 and v1 is the data on the lumberjack, m2 & v2: log. Solve for v2. The sign will be negative indicating the log goes in the direction opposite the run. (I see the question says "velocity of the first log just before the lumberjack jumps off?" The log velocity is the same just before as just after.)

b) Momentum before = momentum after
The lumberjack has velocity v1 when he jumps for the 2nd log. The 2nd log has 0 velocity before the jump. After the jump, the mass of the lumbverjack and the log are essentially one body of mass m1+m2 with a new velocity v3
m1*v1 + 0 = (m1+m2)*v3

I hope this helps better.

Answer 2

REASONING The system consists of the lumberjack and the log. For this system, the sum of the external forces is zero. This is because the weight of the system is balanced by the corresponding normal force (provided by the buoyant force of the water) and the water is assumed to be frictionless. The lumberjack and the log, then, constitute an isolated system, and the principle of conservation of linear momentum holds.


Look at #7 on the link... you'll have to change the numbers and do the math but it will explain it to you

Answer 3

I think this problem should be solved by conservation of momentum.

In the first case, the initial momentum of the system is zero, and it ends up with a lumberjack momentum of 98kg * 2.8m/s = 274.4 kg m/s
The 1st log must have a momentum opposite to it (-274.4), so its speed should be -0.98m/s (-274.4/280)

Then we have a lumberjack with a momentum of 274.4 landing on a 280kg log. The total momentum must be the same, but on a system with a higher mass. The speed of this system is

v= 274.4 / (280 + 98) = +0.726m/s

Answer 4

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm thats a hard 1

Answer 5

im sorry. wish i could help but that stuff just doesnt click in my head.

luv
cara