A Lumberjack Mass 98 Kg
Frictionless problem including momentum
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Homework Statement
A lumberjack (mass = 111 kg) is standing at rest on one finish 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 +two.93 m/s relative to the shore, and and so hops onto an identical floating log that is initially at residual. Fail any friction and resistance between the logs and the h2o. (a) What is the velocity of the first log (again relative to the shore) just before the lumberjack jumps off? (b) Decide the velocity of the second log (once more relative to the shore) if the lumberjack comes to rest relative to the second log.
(non sure if the title is plumbing equipment, because not 100% this is actually a momentum question)[/B]
Homework Equations
p=mv[/B]
The Endeavour at a Solution
so i plant 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 information technology is 259kg so obviously for the momentum to be conserved the velocity has to be a lot slower. and then i went 325.23kg m/s = 259kg⋅Vm/s and solved for velocity by dividing 325.23 by 259 which was one.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]
Answers and Replies
Hint: During the interaction of the lumberjack and the second log, total momentum is conserved. And then what'south the total momentum?Im struggling on finding the velocity of the second log after the lumberjack jumps on it relative to the shore.
Hint 2: The lumberjack and log cease upwardly moving with the same velocity, which is what yous need to solve for.
if total momentum is conserved so the final momentum is going to be the same every bit the initial momentum which was 325.23 kg m/s (this is just the lumberjack) is that going to = the terminal momentum ? or do i accept to also consider the momentum of the log fifty-fifty though its moving in the opposite direction?Hint: During the interaction of the lumberjack and the 2nd log, full momentum is conserved. So what'due south the total momentum?
Hint ii: The lumberjack and log cease up moving with the same velocity, which is what you need to solve for.
So im assuming that the second log is going to wearisome the momentum down because its a lot heaver than the lumberjack so how could i solve for the combined velocity of the two??
Initial momentum of the system (lumberjack + 2nd log) = final momentum of the system. Initially, only the lumberjack is moving and then only he has whatever momentum. So that momentum yous calculated is the total momentum.if full momentum is conserved then the concluding momentum is going to be the same as the initial momentum which was 325.23 kg m/southward (this is merely the lumberjack) is that going to = the final momentum ? or exercise i have to also consider the momentum of the log even though its moving in the reverse direction?
When the lumberjack jumps on the second log, his speed will slow down (and the log will speed up!). They end up with the aforementioned speed V. Then gear up up an expression for the total momentum of lumberjack and log in terms of V. (Then you tin can set information technology equal to the value of total momentum you calculated earlier and solve for V.)So im assuming that the second log is going to slow the momentum downwards because its a lot heaver than the lumberjack and then how could i solve for the combined velocity of the 2??
(Annotation: Realize that when working out what happens with the 2nd log you can forget all about the first log. Yous're done with that part of the trouble.)
Pf=Po then, 325.23 kg grand/s = chiliadlumberjack and second log ⋅ VInitial momentum of the system (lumberjack + 2d log) = last momentum of the arrangement. Initially, only the lumberjack is moving so but he has any momentum. So that momentum you calculated is the total momentum.When the lumberjack jumps on the second log, his speed volition tedious down (and the log will speed upwardly!). They terminate up with the same velocity V. So fix an expression for the total momentum of lumberjack and log in terms of V. (Then you can set up information technology equal to the value of full momentum you calculated earlier and solve for 5.)
So, 325.23 kg g/s = 370kg ⋅Vm/s
V= 325.23/370=0.879m/due south
is this correct??
Looks good to me!Pf=Po so, 325.23 kg 1000/s = mlumberjack and 2d log ⋅ 5
So, 325.23 kg m/s = 370kg ⋅Vm/s
Five= 325.23/370=0.879m/southward
is this correct??
Suggested for: Frictionless trouble including momentum
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A Lumberjack Mass 98 Kg,
Source: https://www.physicsforums.com/threads/frictionless-problem-including-momentum.880730/
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