Like

Report

A $0.150-\mathrm{kg}$ frame, when suspended from a coil spring, stretches the spring 0.050 $\mathrm{m}$ . A $0.200-\mathrm{kg}$ lump of putty is dropped from rest onto the frame from a height of 30.0 $\mathrm{cm}$ (Fig. 8.42$)$ . Find the maximum distance the frame moves downward from its initial position.

$0.232 \mathrm{m}$

You must be signed in to discuss.

{'transcript': "Okay, So in this exercise, I have a frame off mass m f. And this frame is being suspended by a spring. And we know that when the spring carries the frame alone, it stretches by a distance. Y f Okay, We also have a mass that is being held 30 centimeters from above the the frame. And we released this mass, and this mass starts to fall. And we know that when the mass hits the frame, the linear momentum off the mass will pushes all the system downwards. And in the end, we want to know what will be the maximum distance that the system will go after the ball hits the frame. Okay, so here I wrote some the given information. So we have so we know the mass of the frame MF We know the mass off the ball em. We know that the initial velocity off the ball zero because it was at rest in the beginning, we know the distance between the initial distance between the ball and the frame, 30 centimeters. And we also know how much the spring stretches when it's hanging the frame alone. Y f. And this is equal to seven centimeters. Okay, so, uh, to solve this exercise, we can consider conservations off some quantity. So first, let's see where we have conservation off linear momentum. Okay, So since we have the force off the spring and the force off the gravitational force, we can only apply conservation of linear momentum at the moment off the collision between the ball m and the frame and f So we have that. Take this as the ball just before it hits the frame. It will have a final velocity that I'll call M v f. Okay. And this is the system just after the ball hits the frame. So just after the ball hits, the frame will treat ball and frame as a single body off the Thomas M s plus m and this body will be will have a initial velocity downwards off zero. So we have the conservation off linear momentum between these two cases. So we have that the linear momentum off the ball with its final velocity, V f has to be equal to m f plus m times V zero. Okay, so we have that we can isolate with zero. So v zero the velocity the initial velocity off the body ball plus frame is going to be equal to, um over m f plus m times the final velocity off the ball. Besides, all right it in red. So we know that this is the final velocity off the ball. Okay, so how do we find the velocity off the ball? Because we know em. We know MF such calculate zero. We have to know the earth so we can use kind of medics to calculate the f. So we have that the ball. It's moving due to the acceleration off the gravitational force downwards. So we have that the Newton's second law for the ball is going to be equal to m times A equals two m times G. The acceleration off the gravitational force. So we have that acceleration off the ball is going to be equal to gravity. And we know that the kind of medical expression for the for the displacement of a body that moves in a constant acceleration is going to be equal. So, delta why is going to be equal to the initial velocity in the Y direction time? See bus one half off a sorry g t squared. Okay, we know that Delta Y is citizen 2 m, so we can find the time it takes for the ball to hit the frame to be equal to the square. Root off two times 0.3 over 9.8 and this is equal to zero point 25 seconds. Okay? And we also know that the velocity of a body that is moving, uh, with an external acceleration is so the final velocity minus the initial velocity off the body, is going to be equal to the acceleration, which in our case, is g times t. So we want to the final velocity off the ball. So we take the time it takes to reach the the frame. And we have that V F is going to be equal to 9.8 times 0.25 which is 2.42 5 meters per second. Now we can to be stood to this into our expression for conservation of linear momentum and will find that the initial velocity off this body here off mass MF plus M will be, um, equal to select me. Write it here. So you have that V zero will be equal to just be seen that the values 1.28 6 m spur second. So let's keep this result here. So this is something that we have to use. Um, okay, so we use the conservation of linear momentum, and from this we found to the initial velocity off the system right after the ball hits the frame. And there is also one quantity that is conserved and the scenario that is the mechanical energy, so mechanical energy. So let's look again at our scenarios and let's see, where do we have a conservation off mechanical energy? So notice that from scenario that I'll call one two scenario, too. We do not have conservation off mechanical energy, and that is because the collision between the frame between the ball and the frame sorry is going to be elastic in elastic collision. And we know that the collision is in elastic because the ball is going to be fluid to the frame and both will be treated as a single object. Okay, so from two. So for one from 1 to 2, we don't have conservation of mechanical energy. So now let's look so let's look to the final scenario. That is where the whole system, it's down a distance. D. So in this case, no mechanical energy was lost, so we can consider the conservation off mechanical energy in here. Okay, so we have that. So first noticed that in the scenario three, the velocity off the whole system is going to be zero. Okay, so let me call this V Capital F zero in this case. Okay, so we have that, uh, the mechanical energy off scenario, too. So the gravitational energy off scenario two plus three kinetic energy off scenario two plus, uh, plus a spring potential energy off scenario to which I'll call, which I'll just call s for spring. So as to this has to be equal to the gravitational potential off scenario three plus the, uh, kinetic energy of scenario three plus the spring potential off scenario story. Okay, so let's just be stirred the values. And we have that that the gravitational potential off scenario two will be just a m f. Sorry. MF plus M g times D. Because I chose my coordinate system to have the zero value at the bottom. So in the beginning will have that uh, we have the resistance d and in the end, the gravitational potential Zero. So let me put it here also zero for the final gravitational potential we have that the kinetic energy in the beginning is going to be equal to I m f plus m times V zero that we calculated from the conservation of linear momentum squared over two. And in scenario three, the kinetic energy is going to be zero because the whole system will be at rest, plus the spring potential energy. So we have that this is going to be equal to the spring constant times the amount that the spring, uh, is stretchered. So I call it Y f for the beginning over two. So y f squared. And in the end, this is going to be equal to K over to, uh, y f plus d squared. OK, so why I f. Because this is among the spring is already stretched when it is hanging the the frame. Okay, Okay. So for all the values from the conservation off energy, the only value that we don't know besides the quantity that we want to find, that is the It is the spring constant cave. Okay, so how do we find this spring? Constant? So we can find the spring constant from, uh, again, Newton's second law in the first scenario. So pick the Newton's second law for the first scenario on the on the system spring plus frame. Okay, so in this case, we have the gravitational force pulling the spring and the frame downwards, and we have the spring force pulling the system upwards. Okay, So from Newton's second law, we have that, um Okay, X Sorry. Kyff is going to be equal to M f times G, so we can isolate K. And we find that is going to be equal to M s G over Y f. And this is a call to to NT one Newton spare meter. Okay, so now that we have K, we know v zero, we can just be stood all the values that we have and try to isolate D in this expression, so in the left hand side will have. So we have, uh, So for the spring energy 20. Sorry. 21. Time 0.7 squared over two, plus the gravitational potential energy, which is going to be equal Thio 0.25 times 9.8 times D plus the kinetic energy, which is 0.25 times the initial velocity squared, 1.38 6 squared over to and this is going to be equal to the final energy, which is of just the spring energy. So 21/2 time 0.7 plus D squared. Okay, So we can just calculate the values. And we have that, um, this the spring energy in the top is just zero point 0514 Okay. What for five. Sorry, Thisted. I'm here off. The gravity is going to be 3.43 D. And the term from the kinetic energy is 0.33 62 and determine the bottom. We can expand it. So we have that. This is going to be equal to, uh, 1.47 d A. Plus 0.5145 plus standpoint. Five d squared. Okay. Eso we have ah, prosthetic expression. The prosthetic formula. So let's we can see also that we can cancel out disturb With this term, we can subtract these two terms and we are left with our final quadratic formula, which is 10.5 d squared is going to be equal to sorry. Uh, plus 0.33 62 minus 1.96. Sorry, there is a d here. Uh, this has to be equal to zero. So we just use vascular a formula to serve this. And we have that there will be two possible values for the distance. The the displacement off the system and he's going to be equal. Thio 1.96 plus minus the square root off 1.96 squared plus four times 10.5 times 0.33. 62 and this all over two times 10.5. So so solving for the plus and minus from the plus, we'll find a positive value and from the miners will find a negative value for D, where, since displacement is a positive quantity will only take the positive value. So from this will find that this is going to be 0.20 nine 5 m. So this will be the displaced, the maximum displacement off the system"}