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ssm A lumberjack (mass $=98 \mathrm{kg} )$ is standing at rest on one end of a floating log (mass $=230 \mathrm{kg}$ ) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of $+3.6 \mathrm{m} / \mathrm{s}$ 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? (b) Determine the velocity of the second log if the lumberjack comes to rest on it.

$a) -1.5 \mathrm{m} / \mathrm{s}$, $b) +1.1 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 7

Impulse and Momentum

Moment, Impulse, and Collisions

University of Washington

Simon Fraser University

University of Sheffield

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this problem. A lumberjack is standing on log and everything is at rest in the beginning of the problem. So we know that the initial momentum off the system p total and I can call that p total initial is zero. Nothing's moving. Then the lumberjack begins to run at a speed of 3.6 meters per second. So there's two parts of the question one. How fast is the first log move when the lumberjack is moving and then secondly, what happens when he jumps onto the second log? So let's take the first park. If our system here is the jogger, that sorry, the log and the logger Um then we know the initial momentum of the system is zero. But after some kind of action takes place, which in this case is the running. The total momentum would be that of the lumberjack and the log. So the Lumberjacks mo mentum is going to be 98 kilograms times his velocity, which is 3.6 meters per second. The logs mo mentum is the logs mass 230 kilograms. But we don't know the velocity of the log, so but we do know that whole quantity is equal to zero because initial momentum and final momentum have to be the same if there's no external forces. So doing the math here, we can say that this quantity here is 352.8 kilograms meters per second and I can put the to 30 times the velocity of the law and on the other side, giving it a negative sign to 30 times the velocity off the log and then dividing both sides by 230 kilograms. The velocity of the law comes out to negative 1.5 leaders per second. So as the person the logger, the lumberjack moves to the right, the log moves to the left at a speed of 1.5 meters per second. So that's per day of the problem. Part B is what happens to the or what is the velocity of the second log when the lumberjack lands on it. So we have a new system to look at before a scroll down. We're looking at what happens to the lump the this system here when the lumberjack lands on it. So basically we have an initial momentum. The system consists of the lumberjack and the second Law. We're looking at this part of the problem here, and three initial momentum in this case is not zero because the person was running. So let's set it up. The initial momentum total momentum initial is going to be the moment, um, off the lumberjack. Okay, The momentum of the log is zero, cause it's sitting there. So the momentum of the lumberjack is all the only thing that's moving. And then when they combined, when the lumberjack lands on the log, the total momentum is going to be the momentum of the I guess I can write lumberjack plus log because they are one system. Okay, So to figure out the quantities here we have the mass of the lumber, Jack, I'll just do it in symbols first. And then we can plug numbers in times the initial velocity of the lumberjack. And then for this one, we have the combined mass of the lumberjack, plus log times, the velocity of the lumberjack times, the log I'll just call that be final. It's the velocity of the system after they combine and we know that these quantities air also equal to each other. So plugging my numbers in for the mass of the lumberjack I have 98 kilograms. The velocity of the person before landing on the second log was 3.6 meters per second. Then we have the combined mass of 98 kilograms and 230 kilograms times the velocity which we're calling the final the velocity of the system after the man lands on the lock. So again, this quantity on the left is 3 52.8 kilogram meters per second and our mass combined is 328. So to solve for our velocity final, we just divide those sides by 328 and our velocity final is 1.1 leaders per second. So after the impart packs the log and the lumberjack moved to the right at 1.1 meters per second

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