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Problem 109 Hard Difficulty

A wagon with two boxes of gold, having total mass 300 $\mathrm{kg}$ . is cut loose from the horses by an outlaw when the wagon is at rest 50 $\mathrm{m}$ up a $6.0^{\circ}$ slope (Fig. 8.50$) .$ The outlaw plans to have the wagon roll down the slope and across the level ground. and then fall into a canyon where his confederates wait. But in a tree 40 $\mathrm{m}$ from the canyon edge wait the Lone Ranger (mass 75.0 $\mathrm{kg}$ ) and Tonto (mass 60.0 $\mathrm{kg}$ ). They drop vertically into the wagon as it passes beneath them. (a) If they require 5.0 $\mathrm{s}$ to grab the gold and jump out, will they make it before the wagon goes over the edge? The wagon rolls with negligible friction. (b) When the two heroes drop into the wagon, is the kinetic energy of the system of the heroes plus the wagon conserved? If not, does it increase or decrease, and by how much?


(a) $16159J$ $\\$
(b) $\Delta K=-5562J$


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Video Transcript

mhm. We know that in collision off all kinds. Since there is no external Nate, focus on the system the total mo mentum before the collision. Equality total mo mentum after the collision. So in that case, we can write Haven is called toe feature. If the collision is between two particles A and B, then we get the situation where the moment amount for particle is given by pay is a culture every so we can write the collision equation in the following form, which is like this equation number two. We also know that the kinetic energy is given by he is equal to one half MV Square. Let's say equation number three and the gravitational potential energy is given by you is a call to M. G Y. Let's say a question Number four. Now we have three stages. The first stage is the designing off the wagon along the inclined path. The second stage is the impact off the Long Ranger and Toronto with the wagon. And the third stage is the motion off the wagon from the trade. Who did please now the first stage since there is no other force than gravity on the wagon. The energy is concerned. So the gravitational potential energy off the wagon at the starting point is totally converted into kinetic energy at the bottom off the inclined path. Then from Equation three and four, we can get we even is a call to square root off two g buy. But why is it called toe LCN Tita that we know So our equation, we even will be square root off do G l science teacher. Now we plug our value for Ellen Theta, so we get to be able. So it's a two g is 9.8 and it's 50 and take a is 60 degree. Yeah, so answer ever be 10.1 major per second. This is the velocity off Hagen just before the impact of the two heroes with it. Now in the second case, since there is no external horizontal net force on the system the wagon Long Ranger and Toronto, the total moment, um, off the system is conserved in the horizontal direction. And since the system, most one object after the impact, so they have the same same velocity after the impact. So in that case, we can write mhm. We ate two is equal to the B two is equal to be to. Now we plug our values for M E M B and we even into equation to. So we will get 300 in two 10.1 into sorry, plus 1 35 in 20 Is it well too 300 blows 1 35 Vito. Okay, so from this, we will get V two will be equivalent toe 6.98 m per second. Now in the third stage, the calculated time it takes for the Wigan to reach the cliff as form so the time can be returned. Else he is equal to distance upon velocity we to. So it's a 40 upon 6.298 and the time will be 5.73 seconds. Since the time it takes for the wagon to reach the place is more than five against. Then the two heroes will be able to jump out safely from the wagon before it falls off the place. Now assume that the two heroes jumped from the height off 20.6 m, which is the safe height for the approximate Mars. So using a proper kinda Matic equation. We get the speed off the two heroes just before the impact with the Reagan as followed. So it's a we square F y is equal to the I've I square bless to a why they love I where I represent initial stage in every present final stage. But we know this square me. I is equal to zero plus two into minus 9.8 into minus 0.6, so we will have B B. I is equal to three point 43 m per second. So the total kinetic energy off the system just before the impact is given is a call to K. Avon Bliss. Give me one. So instead, off K, even I can write one half m a. We even square. So one half Emmy is 300. We even square. We even is 10.1 square plus. Instead, off gave me when I can write one half m. B, which is 1 35 and be we even square, which is 3.43 square, gives the answer. 16,159 June and the kinetic energy of the system after the impact can be calculated as K two is equal to one half m e plus M B, which is 300 plus 1. 35 we two square, which is 6.98 square. So we will help the answer. 10,597 June. Since Kevin is greater than Cato, the kinetic energy is not consult during the impact, and it decreases by given Minus. K two is a call to 16,159 minus 10,597. So we can write. Tell Turkey is a call to minus 5562. To who?