Frictionless problem including momentum

[Erenjaeger]

Homework Statement

A lumberjack (mass = 111 kg) is standing at rest on one end of a floating log (mass = 259 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +2.93 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 (again relative to the shore) just before the lumberjack jumps off? (b) Determine the velocity of the second log (again relative to the shore) if the lumberjack comes to rest relative to the second log.

(not sure if the title is fitting, because not 100% this is actually a momentum question)[/B]

Homework Equations

p=mv[/B]

The Attempt at a Solution

so i found the momentum of the lumerjack by mv which was 111x2.93=325.23 kg m/s which has to equal the momentum of the log and we know the mass of the log is greater than that of the lumberjack it is 259kg so obviously for the momentum to be conserved the velocity has to be a lot slower. so i went 325.23kg m/s = 259kg⋅Vm/s and solved for velocity by dividing 325.23 by 259 which was 1.256 but the velocity as a vector is going to be in the other direction so the answer is -1.256m/s.
Im struggling on finding the velocity of the second log after the lumberjack jumps on it relative to the shore. [/B]

 

[Doc Al]

Im struggling on finding the velocity of the second log after the lumberjack jumps on it relative to the shore.

Hint: During the interaction of the lumberjack and the second log, total momentum is conserved. So what's the total momentum?
Hint 2: The lumberjack and log end up moving with the same velocity, which is what you need to solve for.

 

[Erenjaeger]

Hint: During the interaction of the lumberjack and the second log, total momentum is conserved. So what's the total momentum?
Hint 2: The lumberjack and log end up moving with the same velocity, which is what you need to solve for.

if total momentum is conserved then the final momentum is going to be the same as the initial momentum which was 325.23 kg m/s (this is just the lumberjack) is that going to = the final momentum ? or do i have to also consider the momentum of the log even though its moving in the opposite direction?
So I am assuming that the second log is going to slow the momentum down because its a lot heaver than the lumberjack so how could i solve for the combined velocity of the two??

 

[Doc Al]

if total momentum is conserved then the final momentum is going to be the same as the initial momentum which was 325.23 kg m/s (this is just the lumberjack) is that going to = the final momentum ? or do i have to also consider the momentum of the log even though its moving in the opposite direction?

Initial momentum of the system (lumberjack + second log) = final momentum of the system. Initially, only the lumberjack is moving so only he has any momentum. So that momentum you calculated is the total momentum.

So I am assuming that the second log is going to slow the momentum down because its a lot heaver than the lumberjack so how could i solve for the combined velocity of the two??

When the lumberjack jumps on the second log, his speed will slow down (and the log will speed up!). They end up with the same speed V. So set up an expression for the total momentum of lumberjack and log in terms of V. (Then you can set it equal to the value of total momentum you calculated earlier and solve for V.)

(Note: Realize that when working out what happens with the second log you can forget all about the first log. You're done with that part of the problem.)

 

[Erenjaeger]

Initial momentum of the system (lumberjack + second log) = final momentum of the system. Initially, only the lumberjack is moving so only he has any momentum. So that momentum you calculated is the total momentum.When the lumberjack jumps on the second log, his speed will slow down (and the log will speed up!). They end up with the same velocity V. So set up an expression for the total momentum of lumberjack and log in terms of V. (Then you can set it equal to the value of total momentum you calculated earlier and solve for V.)

P_f=P_o so, 325.23 kg m/s = m_{lumberjack and second log} ⋅ V
So, 325.23 kg m/s = 370kg ⋅Vm/s
V= 325.23/370=0.879m/s
is this correct??

 

[Doc Al]

P_f=P_o so, 325.23 kg m/s = m_{lumberjack and second log} ⋅ V
So, 325.23 kg m/s = 370kg ⋅Vm/s
V= 325.23/370=0.879m/s
is this correct??

Looks good to me!