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Jack and Sill are standing on a crate at rest on the frictionless, horizontal surface of a frozen pond. Jack has mass 75.0 $\mathrm{kg}$ , Jill has mass 45.0 $\mathrm{kg}$ and the crate has mass 15.0 $\mathrm{kg}$ . They remember that they must fetch a pail of water, so each jumps horizontally from

the top of the crate. Just after each jumps, that person is moving away from the crate with a speed of 4.00 $\mathrm{m} / \mathrm{s}$ relative to the crate. (a) What is the final speed of the crate if both Jack and sill jump simultancously und in the same direction? (Hint: Use an inertial coordinate system attached to the ground, (b) What is the final speed of the crate if Jack jumps first and then a few seconds later Jill jumps in the same direction? (c) What is the final speed of the crate if fill jumps first and then Jack, again in the same direction?

(a) $3.56 \mathrm{m} / \mathrm{s}$

(b) $5.22 \mathrm{m} / \mathrm{s}$

(c) $4.66 \mathrm{m} / \mathrm{s}$

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Numerade Educator

University of Washington

Hope College

McMaster University

{'transcript': "So when this exercise have two characters character, a off mass equal to 75 kg and character be off Mexico to 45 kg, these two characters air standing in a great and it's great. It's floating in a frictionless surface and it has mass equal to 15 kg. So one thing we know about the system is that when one of the characters jump out of the crate, its velocity is always four minus the velocity off the create. Okay, so in question A, we want to find what will be the velocity off Great off the crate, this velocity here in red when both characters A and B jump out of the great at the same time so we can use a conservation off the new momentum because there is no external force. So we have that, um, so in the beginning, the system stationary. So we have that zero has to be equal to the momento of character. A plus the mo mentum off character B minus the mo mentum off the crate, and it's minus because of a lot. So the crates will be pointing to the negative direction. Okay, so we have that this is equal to M a times v Sorry times for my mis v blocks M B times four minus V on which V is the velocity we want to find and this is equal to M C times We okay now we just try just to be to the values and isolated the quantities we have. So we have here and may plus and be times for minus we. So we have the mass off character, a 75 and the mass of Peace 45. So we have, uh 100 on 20 times for minus V is equal to 15 time magazine so we can isolate V and we have that, uh, 135 V is equal to 480. And by this we find that when the two characters jump at the same time, the velocity off the crate will be equal to 3.55 meters per second. Okay, No question B is the same scenario two characters in a crate. But now we have to consider when the character A jumps first and after a while, character be jumps. OK, so let's write it here. What will be the velocity of character. A when character A jobs first. So we have that the velocity of character A that'll call v a prime will be equal to the velocity. Off will be two for minus the velocity off the boat. Okay, this is the the condition that the exercise gave us for the velocity off. The character is always four minus the velocity off the boat. Okay, this is from rest frame. Okay? So we can find what the prime is by using conservation off off. Lena. Momentum from this stage to the stage on which everything is at rest. So we have that zero is going to be equal to P a minus. What? I'll call B B C, which is the linear momentum off the create plans the character be so this is equal to and eight times V a, which we know it is able to form in minus were crying. Um, and this is equal thio BBC, which is m b plus m c times we crime. So we can just be suit the values and find that with prime will be equal to um So I'm sorry. So the prime will be equal to 2.22 m per second. Okay, so this will be developed of the great in this case. But we want to find that the loss of the great Well, both characters jump so right after character be jumps. Okay, so we can use conservation off linear momentum again. But this time we use the conservation from this scenario one to this scenario too. So we have that, um, the B c has to be equal to, uh, minus. BBC has to be equal to, uh, Phoebe minus pc. Okay, so we know off minus BBC meant to B c S, m b plus M c times t prime. And this has to be equal to and be times for minus v. No, this time ISS four minus the final velocity because the character be will jump after character a minus the mo mentum off the great. So m c times we okay, so we can try to isolate all of this. So we have that, uh, V is going to be equal to and be times four plus and be sorry, plus m c times the prime over and be plus m c. So we just be sued The values and B is 45 kg and since 15 and we find that the velocity in this case when character A jumps is going to be equal to 5.22 meters per second. Oh, and also obviously we should be prime, which is what we found here. Okay, now in question. See, we want instead of character a jumping first we want that character be jumps and then character A jumps right after. Okay, eso In this case, the velocity if you be prime will be equal to four minus we crime. So again, it's the same method that we used so question be so, uh, pick the Mo mentum address has to be equal to the mo mentum off this first scenario. So zero has should be equal to t B mine's B A C, which is the mo mentum off character A and the crate. Okay, so this is m B times will be prime, which is four minus b prime. This has to be equal to I am a plus m c times with Brian. Okay. So we can isolate to be crying, and in the end, we have, uh, we can find something like this. So if we should be stood some things you have that $19 for 45 Prime is going to be equal to 45 times for Okay, So I'm just putting multiplying this with this and and be by minus four prime and then passing and be minus my on B minus and be the prime to the other side of the equation and summing everything up so his So he finds that the prime So the velocity off the create after the character be jumps is going to be able to wind one points 33 meters per second. And now, as we did in question be, we can use conservation of linear momentum between the scenario one on which only character B has jumped and the scenario tube. So in this case, we have that the mo mentum off the system crate Poehler's character a hash be conserved in scenario 12 scenario too. So we have that minus p c A has to be equal to p a minus specie. Okay, So, to be student values you have that this is M A plus EMC times we prime This has to be able to I m a times four minus v uh, minus M C V. Okay, so we just isolate the V factory we have. And in the end, we have something that looks like this. So we have nine. TV is going to be equal to 90 times one point 33 which is this term, um, miners plus 300 which is going to be eight times four. Okay, so we find that in this case, the final velocity after the character, A jumps will be able to 4.67 m per second and this concludes the exercise."}