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This is College Physics Answers with Shaun Dychko We imagine that a .1 milligram piece of paint is orbiting around the earth and it’s a piece of space debris, and it’s going to hit the window of a spaceship. And what force will it apply on that window? When we’re writing down the information that we know, we’ll take this as our opportunity to convert units into m.k.s. units; meters, kilograms and seconds. And so this is .1 times ten to the minus three kilograms. The initial speed of this paint chip is four times ten to the three meters per seconds, and it’s in contact with the window for 6.00 times ten to the minus eight seconds. Now what the question does not tell us is what the final speed of the paint chip is. And we have to assume that it’s going to be zero. Because any other assumption would prevent us from being able to answer the question. Because we have to know what the final speed is, in order to calculate the force it applies. Because the force is going to be it’s change in momentum divided by time. And we need to know what the final momentum is of the paint chip. So we need some final speed, and so it sticks to the window and does not puncture it. Although, when we get our final answer you have to wonder if maybe it would puncture it. But anyway, we have the change in momentum; mass times final speed, minus mass times initial speed all divided by time. So factor out the <i>m</i>, and we have <i>m</i> times <i>v f</i> minus <i>v i</i> over <i>t</i> So, that’s the mass of .1 times ten to the minus three kilograms, multiplied by a final speed of zero, minus an initial speed of four times ten to the three meters per second. Divided by six times ten to the minus eight seconds. Which works out to negative 6.67 times ten to the six newtons. That’s about seven million newtons. And the negative sign just indicates that the force applied on the paint chip is in the opposite direction to it’s initial velocity.

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Submitted by bac28192 on Sun, 10/13/2019 - 11:38

Submitted by bac28192 on Sun, 10/13/2019 - 11:39

In reply to You state that the 0.100 mg… by bac28192