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Requlred Infonmation Jatn Arocicing Lenno stroke by hitting balls Jqainst giound an anglc 800" respcct ta Ine vertcaiPart ? 07Tne balli Daves hl; racquclhcight...

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Requlred Infonmation Jatn Arocicing Lenno stroke by hitting balls Jqainst giound an anglc 800" respcct ta Ine vertcaiPart ? 07Tne balli Daves hl; racquclhcightaf 74.0 cm above666Uevercedcortbobol % # lewvcs the racquct Is 270 Ms and Ihc verc l conpcneni WmSI vclocity? Emttt & pogitve value KO Doll / upwalid savd a dstace % 172 m beforo It teaches the wall; what l5 and 0 ncgatva value / the Doll t downtaidDultruntes

Requlred Infonmation Jatn Arocicing Lenno stroke by hitting balls Jqainst giound an anglc 800" respcct ta Ine vertcai Part ? 07 Tne balli Daves hl; racqucl hcightaf 74.0 cm above 666 Uevercedcortbobol % # lewvcs the racquct Is 270 Ms and Ihc verc l conpcneni WmSI vclocity? Emttt & pogitve value KO Doll / upwalid savd a dstace % 172 m beforo It teaches the wall; what l5 and 0 ncgatva value / the Doll t downtaid Dultruntes



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Figure 16.46 This child's toy relies on springs to keep infants entertained. (credit: By Humboldthead, Flickr) The device pictured in Figure 16.46 entertains infants while keeping them from wandering. The child bounces in a harness suspended from a door frame by a spring constant. (a) If the spring stretches 0.250 $\mathrm{m}$ while supporting an 8.0 -kg child, what is its spring constant? (b) What is the time for one complete bounce of this child? (c) What is the child's maximum velocity if the amplitude of her bounce is 0.200 $\mathrm{m} ?$

For this problem on the topic of momentum and collisions, We are given a list of objects and we are told that each object is dropped from a height of 85 centimeters. And we have also were also given the values of the height that they reach after their bounds. We want to use this information to determine the coefficient of restitution for each of these objects. Now we know that we can use the equation for the coefficient of restitution, or heights, which is absalon, and absalon is equal to the square root of the height from which the object bounces, or to which the object bounces each one divided by capital H, which is the height from which the object is initially dropped. Now, if we do example calculation for the golf ball, we have this to be the square root from our table. This is 62.6 centimeters divided by capital H. The height from which the golf ball is dropped 85 centimeters. This gives us a coefficient of restitution of zero point 85 818 If we apply the same calculation to all of these objects, we can get the coefficient of restitution for each one. So for the golf ball, we show that that 0.858 Similarly for the tennis ball at 0.712 for the billiard ball, zero 0.804 with a handball, zero point 752 with a wooden ball, zero point 603 for the steel ball, the steel ball bearing zero point 597 for the glass, marble, zero point 658 for the pole of rubber bands, zero point a 28 And lastly, the coefficient of restitution for the hollow hard plastic balls, 0.6 88

Hi, friends. This is the problem based on concept of conservation, of energy, a wall of moss and moving with velocity 10 m per second. So its initial kind of technology is how can be funny squared. I'm into tiny squares, so it will be 15 final kind of technology is 70% off and he shall kind of technology half. Um, the F Square is called 2.7 and two 50 m. So from here, final velocity, you will get eight point 367 m per second. Hence answer is correct. Mhm.

Everyone here it is. Given first of all hits with initial velocity of 7 24 7 m per second. It gets compressed by Yeah, six millimeters. I mean yeah, yeah. Oh needed before coming to rest. We have to find the acceleration. So we finally square is scared to be initially square Last two years 07507 inscribed Let's do into here. We are finding that is 0.6 acceleration. You will get 41 double 6.67 meters per second squared. Handsome D option is correct.

Okay, guys. Now let's do problem 36 from chapter eight. And this is once again a combination of problems. There is kin o matics involved, and there is a conservation of energy. So it says that two children are playing a game in which they try to hit a small box on the floor with a marble fire from a spring loaded gun that is mounting on the table. The target boxes horizontally distance all the target boxes horizontal distances to 0.2 meters from the edge of the table. And so in figure for eight. Okay, Bobbie compresses this gun spring 1.1 centimeters from the center of the marble. Well, he compresses the spring 1.1 centimeters, but the marble falls 27 centimeters short of the box. And they want to know how far this girl Rhoda, should compress the spring to basically hit the box, which is 2.2 meters away. Well, this is obviously a Kinnah matics problem here because they have this this box thing. Okay, And there's a spring inside this box and there is a box. There is another box that is a certain distance away. Okay, search and distance D And we know ideas because D is equal to 2.20 meters. Now, this entire problem is very interesting. All we need to do is simply work backwards and understand the principles involved here. All right, so let's start in order to solve this. What do we know? Well, we know that. You know, if you stick a ball in here and it fires out right, there's obviously conservation of energy occurring, okay? And we know that we know that there is obviously potential energy of this spring, right, Because this spring is being compressed north firing out, and when it fires out, there is kinetic energy. And so we want to know basically what this, like velocity is what is the velocity that this kid is using? You know what velocity due this fall fly out. And with that velocity we can solve for the time we consult for the time required. And when we saw for the time require we confined the velocity needed and that we can solve for how far this spring has to compress in order for it to reach that velocity. So a lot of working backwards here. We're missing a lot of information. Like, what is K right? And what is K? We know what X is. We don't know what is at the moment, but we'll we'll get to this. But first, let's recognize that you can cancel some stuff out so we can cancel out the 1/2. But 1/2. We know what Delta X is in this problem. Because when when this boy, when this boy Bobby first compresses it and he stretches it. Ah, 1.1 centimeters, which is 0.11 meters. Okay, so let's keep that in mind. And 0.11 meters, 0.11 meters square is 1.21 times 10 to the negative four. Oh, man, that's a lot of right out. All right. 1.21 times 10 to the negative. Four meters. Times X is equal to Ah, whatever the mass is, we don't even know what the massive this thing is. Mass times the final square. What is the mass? Why does the knocking of the mass in this problem? Well, anyway, well, we know that we can find the velocity somehow, so I don't even think we need the mass. Okay, so let's find the velocity. How do we do that? Well, we know that there is a kid, a matics equation, right? But first, let's explain this. When the ball shoots out, it's at a search and horizontal velocity, and it travels certain distance D and in a certain Ahmet off time travels that distance deep. So we know that when it reaches the ground there is a final velocity of zero, but we don't know what the initial velocity is. This is very interesting because there's a kid, a Matics equation that deals with this right and isn't that Kenna Max equation? The final square is equal to be initial square plus two a Delta D, where we recognize that the final zero and we recognize that A is actually G because its gravity acting downwards. So it's two g, Delta D is equal to the initial square, and so RV initial is equal to the square root of two G Delta D, and all we have to do simply substitute in what we know. We know that Bobby Waas 27 centimeters short off the 2.20 meters, so we know that we know that he was 2.20 meters minus 0.27 meters is equal to what 2.20 minus 0.27 is equal. Oh, hold up a 2nd 2.20 minus 0.27 is equal to 1.931 point 9 30 meters. So he was at 1.93 meters. Okay, So v initial is equal to the square to 9.8 meters per second square times 1.93 meters, 1.93 meters and that will know what his initial offer and then So let's actually do that. Okay, So two times 9.8 times 1.93 What is that? That's 37 something. We square root it. 6.15 So his velocity was 6.15 Okay, 6.1 five Ah, 641645 meters per second. And we know he traveled 1.93 meters. And so isn't this more like Isn't this less complicated now? Because we know he starts out with this lost e. He travels this distance. So isn't the in the ex direction equal to D in the ex direction over Delta T. Because Delta t remains the same right, So 6.15 meters per second is equal to 1.93 meters over some amount of seconds. And we just saw for that. So 1.93 divided a 6.15 Well, that's 0.31 That's so Delta T is equal to 0.3138 seconds. 0.3138 seconds. Okay, so now all we have to do is recognize that we have to work backwards again. So if the ball is going to fall in this amount of time, because gravity acting on it is the same right acting on it is the same. So we know that it wants to travel this amount to 0.20 meters and we have this much time. So let's erase all the other information here. So we have more space, right? And since we have this time amount, why don't you go ahead and and work on it and try to figure out what else do you think we'll need to do before I actually finished the rest of the problem knowing that we have time and knowing that Ah, your spring constant is Oh, that's actually very interesting. I forgot to mention one thing since we know what are the lossy initial velocity is, we can actually go back and solve for the spring Constant. Right? And let's go ahead and do that before we go on to the stuff that would, like, close off everything. So we know that's 1/2 K Delta X square, 1/2 and the square. I'm assuming that mast is actually negligible or something. Uh, actually, yeah, I think so. I think mass is actually not going to be a factor in this problem. So we have 1/2 and then what is is your point 011 meters square. Times K is equal to 1/2 and 6.15 meters per second square. And does it even give you a mass for the ball? How far short of press this ball it does not. Does this say that mask is negligible? Ah, no, it does not. So you know what I actually think we mask off the ball is negligible in this case because ah, the only thing acting when is gravity right? Mg So net force is equal to net force is equal to m A And what's the force? It's mg, right? So mass cancels out and your acceleration is gravity. And if that's the case, then yeah, masses literally negligible in this in this scenario. So all you have to do solve for your K because we're ready. We already have RV and stuff, right, Sol, for your k So you can reuse this K in the problem when we work backwards. Right? And now let's start from that point where we left off, we have We know that the ball is going to take 0.3138 seconds toe fall from this point to this point and we know that we want to travel to 0.0.0.20 meters. Right? So we have 2.20 meters over 0.3138 seconds. And so what will What is the necessary velocity? Well, let's do that. 0.3138 Well, I need to have a velocity of about seven. So equals to be and V is equal to about 7.1 meters per second, right? And now if we know it needs this velocity Well, let's go back over here. Let's go back over here and conservation of energy once again, once you have your K that you solved down here, okay? And you know, the, you know, the you don't know the compression yet because that's what you want to solve for, and that is equal to 1/2. Mm. The square where M is negligible. Okay, so let's solve. Let's solve for Delta X. So the one halfs cancel out too. So Delta X is equal to the square root of K. Oh, shoot, Shoot, Man, this is not enough space. So Delta X is equal to the square root of these square over K. And let me make this square root sign bigger. Okay? And you know, obvious these over here 7.0 meters per seconds all over. Kay, the value you found here and that is the separation you need in order for for the girl to shoot it from this this spring gun to hit this box. Okay. And that's all for this problem.


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