# Frictionless problem including momentum

• Erenjaeger
In summary, a heavy lumberjack standing at rest on one end of a floating log runs to the other end, attaining a velocity of +2.93 m/s relative to the shore. He then jumps onto an identical floating log that is initially at rest, neglecting any friction and resistance between the logs and the water. The velocity of the first log just before the lumberjack jumps off is calculated using conservation of momentum. The velocity of the second log after the lumberjack jumps on it is determined by setting the initial and final momentum equal and solving for the combined velocity of the lumberjack and log, which is found to be 0.879 m/s.
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]

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]

Erenjaeger said:
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.

Doc Al said:
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??

Erenjaeger said:
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.

Erenjaeger said:
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.)

Doc Al said:
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.)
Pf=Po so, 325.23 kg m/s = mlumberjack and second log ⋅ V
So, 325.23 kg m/s = 370kg ⋅Vm/s
V= 325.23/370=0.879m/s
is this correct??

Erenjaeger said:
Pf=Po so, 325.23 kg m/s = mlumberjack 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!

Erenjaeger

## 1. What is frictionless problem including momentum?

Frictionless problem including momentum is a theoretical concept in physics where the effects of friction are ignored in order to simplify the analysis of a system's motion. This allows for a clearer understanding of the relationship between an object's mass, velocity, and momentum.

## 2. How is friction related to momentum?

In the context of a frictionless problem, friction is not related to momentum as it is not considered in the analysis. However, in real-world scenarios, friction can affect an object's momentum by slowing it down or changing its direction.

## 3. What are some examples of frictionless problems including momentum?

Some examples of frictionless problems including momentum include a hockey puck sliding on ice, a ball rolling down a smooth ramp, and a satellite orbiting the Earth.

## 4. How is momentum conserved in a frictionless system?

In a frictionless system, momentum is conserved due to the principle of conservation of momentum, which states that the total momentum of a closed system remains constant. This means that in the absence of external forces, the initial momentum of a system will be equal to the final momentum.

## 5. How does friction affect the kinetic energy of a system?

In a frictionless system, friction does not affect the kinetic energy as it is not present. However, in a real-world scenario, friction can convert some of the kinetic energy of a system into other forms of energy, such as heat or sound.

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