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Welcome to this numerary tutorial.
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My name is mark s.
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The date today is july 20th, 2020.
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These sets of questions are a bit tricky in the wording.
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So the questions were reposted with all the accurate units, punctuation grammar, and format corrected.
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So if we organize the data properly, then we don't really have to struggle on solving the question.
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As much as we have to go in a sequential format one by one until we solve the data in its entirety.
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So the first question, we have a 300 gram cart moving at 1 .6 meters per second on a theoretically frictionless track.
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When it collides with a larger one kilogram car at rest after the collision small car recoils the opposite direction at 0 .830 meters per second.
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What is the speed of the large car after the collision? express your answer with the appropriate unit.
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So what we need to do is go down to solution number one here that i've outlined.
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Obviously the collision is an elastic collision because assume an object hits another object and its kinetic energy is not conserved and it just stops and there's a combined velocity of both objects that would be an elastic collision so whenever you have a system try to find its opposite or its antonym to understand the relationship with respect to the difference or the differential.
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So we have to exclude what does not pertain to the problem.
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So it's obviously not an inelastic collision.
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It is then, therefore, an elastic collision.
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So access the formula for elastic collision.
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Which is posted here.
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And then we can go back up and look at the data.
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And for m1, it's 0 .3 kilograms or 300 grams.
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So another importance is to keep the units in kilograms for mass and meters per second for velocity.
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Very important.
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So 300 grams is 0 .3 kilograms.
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Mass of second body, 1 kilogram.
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Initial velocity of first body, that's u1, 1 .6 meters per second.
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Initial velocity of second body, it's at rest, so 0 .00 meters per second.
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Final velocity of first body, 0 .830 meters per second, and then the final velocity of second body, b2, is not known or unknown.
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So all we have to do is plug the data into the components of the equation, and we can then solve for v2 algebraically, which is 0 .231 meters per second.
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The second question here, it states that a 3 -kilogram block of wood sits on a frictionless table, a 3 gram bullet fired horizontally at a speed of 380 meters per second, goes completely through the block emerging at a speed of 180 meters per second.
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So the bullet goes through the block, and then when it exits and penetrates the entire block, exits are through and through gunshot, through the wood block, it's leaving at 180 meters per second.
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So what is the speed of the block immediately after the bullet exits the wood block? express your answers with the appropriate units.
04:22
Okay, so we will go down to solution number two here.
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And it might be a good practice if you're not experience with collision physics problems to state both the inelastic collision and elastic collision formulas and determine their answers appropriately.
04:45
So if we start with the inelastic collision, which means that let's say the bullet hit and did not exit the block and both the bullet and the wood block had a final combined speed or velocity.
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That would be a purely inelastic collision.
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However, that only occurs as the bullet is penetrating through the block material.
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But after it exits, then it becomes an elastic collision because they're separating and some of the kinetic energy is conserved.
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So it doesn't hurt to apply both equations so that an understanding is occurring over the function of time.
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In this case, the time it takes to pass through the wooden block, which is not relevant to the question.
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But it gets an understanding, a conceptual understanding of what is actually going on.
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The question can be understood.
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It can be interpreted more than just words in the problem.
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But actual understanding the totality of what's going on.
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So if we just apply the inelastic collision, this is actually a partially inelastic collision problem.
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So we can just start with the inelastic collision formula stated right up here.
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And so we put in all the units.
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So the mass of the first body, three kilograms.
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If we go back up here, we have three kilograms.
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So the first body is the woodland.
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Block and the second body is the bullet.
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So we enter in all the data, the mass of first body, three kilograms.
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The initial velocity of the first body, 0 .00 meters per second.
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Mass of second body, m2, three grams, but it's got to be converted to kilograms, so 0 .003 kilograms.
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The initial velocity of second body, 380 meters per second.
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That's the bullet traveling.
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To impacting the wood block.
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So that's the initial velocity of the second body, the bullet.
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And then the final velocity of combined body...