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A 4800 -kg open railroad car coasts along with a constant speed of 8.60 $\mathrm{m} / \mathrm{s}$ on a level track. Snow begins to fall vertically and fills the car at a rate of 3.80 $\mathrm{kg} / \mathrm{min}$ . Ignoring friction with the tracks, what is the speed of the car after 60.0 $\mathrm{min}$ ? (See Section 2 of "Linear Momentum.")

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8.2 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 9

Linear Momentum

Motion Along a Straight Line

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

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momentum conservation problem. Ah, your initial momentum movie called to your final momentum. Okay? And it's all in one de eso initially you have when you called, I'm gonna note the car by sea end snow by esa Initially just have the cars mass times, cards, velocity, Um, after snow falls, uh, in for a certain period of time, you have the cars mass plus the snow's mass times the car's new mass velocity. Okay, the car's new velocity B C prime will just be m cbbc over m c plus m s. Now we know I'm seeing do CV, so that's all great. But M s is actually a rate of change. The rate in which um, snow is added to the car times the total time during which snow it's added, uh, so emcee as 4800 kilograms, VC is a 0.6 meters per second and ah ah, that is divided by 4800 kilograms. Uh, plus 3.8 kilograms per minute that you multiply by delta T, which is 60 minutes. So the minutes go away, you're left with kilograms and meters seconds. Ah, and so the answer you get is 8.21 meters per second and this is the final speed of the car

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