(II) A meteorite has a speed of 90.0 m/s when 750 km above th...

Question

(II) A meteorite has a speed of 90.0 m/s when 750 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.25 m.

(b) How much work does the sand do to stop the meteorite (mass = 575 kg)?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A 632 kg satellite at a height of 325 kilometers above the surface moves at 85.6 m per second vertically down towards Mars. After some time, it hits the surface and rests after penetrating 5.65 m beneath the surface, ignoring air resistance, calculate the work done by the Martian soil that made the satellite stop. H the radius of MARS is 3390 kilometers. The mass of Mars is 6.39 multiplied by 10 to the power of 23 kg. And the gravitational acceleration on Mars is 3.72 m per second squared. So our end goal, the final answer that we're ultimately trying to solve for is we're trying to figure out what the work done by the Martian soil that made the satellite stop because the satellite is flying from space down to the surface of Mars and the soil basically acts like a, as a slow, it's basically slowing down this and it's causing the satellite to come to a stop. So it's slowing down this speed since it's speeding towards the surface of Mars and to help the soil will help the satellite come to a stop. And we're trying to figure out what our final answer that we're trying to find is we're trying to figure out the work done by the soil that come, that helps the satellite come to a stop. And that's the final answer that we're trying to solve for. We're also given some multiple choice answers. Let's read them off to see what our final answer might be. And let's also note that they're all in units of jewels A is negative 1.00 multiplied by 10 to the power of nine B is negative 6.98 multiplied by 10 to the power of eight C is negative 3.83 multiplied by 10 to the power of eight D is negative 4.82 multiplied by 10 to the power of five. OK. So first off, let us write down all of our known and unknown variables. Let's note that the initial speed of the satellite is equal to 85.6 m per second. Let us also note that the speed of the satellite just before it hits the surface or strikes the surface. Let's call this vs, we do not know what this value is. We do know. However, the mass of the satellite, which we're gonna say is M which is equal to 625 kg. And the initial height above the surface, let's call this H is equal to 325 kilometers. So 325 kilometers which is equal to let me convert to meters. 3.25 multiplied by 10 to the power of 5 m. Awesome. And moving right along the distance penetrated by the satellite on Mars, let's call this D and D is equal to 5.6 5 m. Awesome. Moving right along. Let's also note that the radius of Mars, which we're gonna call RM is equal to 3390 kilometers, which is equal to 3.39 multiplied by 10 to the power of 6 m. And we also know that the mass of Mars, let's call this capital M is equal to 6.39 multiplied by 10 to the power of 23 kilograms. And the universal gravitational constant. Let's call this G, this is an outright given to us in the prom itself. But we should recall that it's 6.67 multiplied by 10 to the power of negative 11 in its units are newtons multiplied by meters squared per kilograms squared. Awesome. And since now that we wrote down or wait, we forgot one more. Don't forget that we're told that the gravitational acceleration on Mars, it's call as G capital M is equal to 3.72 so 3.72 m per second squared. OK. Now, we wrote down everything that we know and don't know. So with that in mind, moving right along, let us also note that since there is no friction, the mechanical energy would be conserved. Therefore, we could write that the initial energy is equal to the final energy. And we can also write that one half multiplied by lower case M multiplied by V I squared plus negative G multiplied by capital M multiplied by lower case M all divided by RM plus H is equal to one half multiplied by M multiplied by V S squared plus negative G multiplied by capital M multiplied by lower case M all divided by RM. So now we need to take this equation and we need to rearrange it to isolate and sulfur vs. So when we do that, we will get that, VS is equal to the square root of V I squared plus two. So V I squared plus two multiplied by G multiplied by capital M multiplied by one divided by RM minus one divided by RM plus H. OK. So now at this point, we need to plug in all of our known variables and solve for the numerical value of VS. So let's do that. So vs is equal to the square root of And we know that V I is equal to 85.6 m per second. And this value is squared plus two multiplied by 6.67 multiplied by 10 to the power of negative 11 newtons multiplied by meters squared per kilogram squared multiplied by 6.39 multiplied by 10 to the power of 23 kg, which is multiplied by one divided by 3.39 multiplied by 10 to the power of 6 m minus one, divided by 3.39. So 3.39 multiplied by 10 to the power of six L. The value of 3.25 multiplied by 10 to the power of 5 m. Awesome. So with that in mind, when we take that monster of an equation and plug it into our calculator, we will find that it's equal to 1485.638 m per second when we round to three decimal places. So now we need to start solving for our final answer. But to do that, let's figure out what we're gonna denote our different variables. So let us call the work done by the Martian soil to be WS and that the work done by the gravity as the satellite penetrates the surface to be WG and the network to be W And let us also note that the final speed of the satellite will be VF equals 0 m per second. Because it's not gonna be moving once it hits the ground, the gravitational acceleration on Mars is also GM equals 3.72 m per second squared. So now at this point, we can recall and use the equation for the change in kinetic energy of the satellite as it penetrates the surface, which we can write as delta ke is equal to one half multiplied by M multiplied by VF squared minus. So we can't forget our extra parenthesis here. So it's VF squared parenthesis parenthesis. Awesome minus one half multiplied by M multiplied by vs square, which is equal to one half multiplied by N multiplied by zero squared minus one half multiplied by M multiplied by VS squared. Awesome. So now we can simplify and write that delta ke is equal to one half multiplied by M multiplied by VS squared. Therefore, we could write that W equals WG plus WS. So now, at this point, we must recall that the work energy principle states that the work done is equal to the change in kinetic energy. Therefore, we could write that delta ke is equal to WG plus WS. And now we need to solve for WS which gives us. So this is rearranging this equation to isolate and solve for WS which will give us that WS is equal to delta ke, which is eke minus WG, which we can write as WS is equal to negative one half multiplied by M multiplied by VS squared minus M multiplied by GM multiplied by D. So now we need to plug in on all of our known variables to solve for WS. So when we do that, we will get that WS is equal to negative one half multiplied by 632 kilograms, multiplied by 1485.6 38 meters per second squared minus 632 kilograms. multiplied by 3.72 meters per second squared, multiplied by our penetration distance which is 5.65 m. Awesome. So when we plug that into our calculator, we will find that the work done by the soil or the ground is equal to negative 6.98. When me round to two decimal places multiplied by 10 to the power of eight jewels and this is rounded to two decal places and more specifically, it's rounded to three significant figures. And that's our final answer. Fre we did it negative 6.98 multiplied by 10 to the power of eight jewels. So looking at our multiple choice answers, the correct answer has to be the letter B negative 6.98 multiplied by 10 to the power of eight jewels. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.

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(II) A meteorite has a speed of 90.0 m/s when 750 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.25 m.
(b) How much work does the sand do to stop the meteorite (mass = 575 kg)?

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