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1,Ina5 projectile motion , oaded spring = experimenl Eun spring horizontal then placed cn & compressed range ofthe horizontal able 25 crn by a ball ball is 1,6...

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1,Ina5 projectile motion , oaded spring = experimenl Eun spring horizontal then placed cn & compressed range ofthe horizontal able 25 crn by a ball ball is 1,67 wilh # ball at the after neight 0f 1.0 m moment it left the 'release the trigger ol tne above &round The spting gun and the 'spring gun; {b) the force constant sprirg gun; Firid {a) the speed effect ofau, of trie spting of the enore Kfrction inside

1,Ina5 projectile motion , oaded spring = experimenl Eun spring horizontal then placed cn & compressed range ofthe horizontal able 25 crn by a ball ball is 1,67 wilh # ball at the after neight 0f 1.0 m moment it left the 'release the trigger ol tne above &round The spting gun and the 'spring gun; {b) the force constant sprirg gun; Firid {a) the speed effect ofau, of trie spting of the enore Kfrction inside



Answers

The launching mechanism of a toy gun consists of a spring of unknown spring constant, as shown in Figure P5.39a. If the spring is compressed a distance of 0.120 $\mathrm{m}$ and the the gun fired vertically as shown, the gurcan launch a 20.0 -g projectile from rest to a maximum height of 20.0 $\mathrm{m}$ above the starting point of the projectile. Neglecting all resistive forces, (a) describe the mechanical energy transformations that occur from the time the gun is fired until the projectile reaches its maximum height, (b) determine the spring constant, and (c) find the speed of the projectile as it moves through the equilibrium position of the spring $(\text { where } x=0),$ as shown in Figure P5. 39 $\mathrm{b}$ .

So in this case, we should first find the work done by friction. This would be, of course, equaling the frictional force multiplied by the length or the the length of the barrel of the gun. You can say L so this would be equaling negative 0.3 to Newton's multiplied by uh so 15 centimeters, or 150.15 meters. And this is equaling negative 4.8 times 10 to the negative third, Jules. And so we can then, ah, say that the work is equaling the change in energy or the change in mechanical energy, so this would be equal in 1/2 and he squared minus 1/2 K X square. So all of the potential energy the elastic potential energy of the spring inst in the toy gun is going to be converted into the kinetic energy of the bullet or the rubber ball. Uh, and so we can then say that essentially negative 4.8 times 10 to the negative third jewels will be equaling 1/2 times the mass of the rubber ball, which we know to be 5.3 grams, or 0.53 kilograms times the velocity squared as the velocity as it as it exits the barrel minus 1/2 times the spring constant of the spring. 8.0 Newtons per meter multiplied by the compression of the spring. And this would be equaling five centimeters or 50.5 meters quantity squared and essentially recon user ta 4 85 or 89 in order to solve for V And so he is gonna be equaling 1.4 meters per second Approximately. This would be the velocity of the rubber ball as it leaves the barrel of the toy gun. That is the end of the solution. Thank you for watching.

In the institution, it is beyond that if the origin of the coordinate system lies at the center of mass, then we have to tell that some of moments of the masses of the system about the center of mass. So as we can say that for this occasion, if I assume some masses that is much and want them to. So we know that the center of mass is given by the R. Is equal to the M. One R one plus M two or two Plus M three. R. 3. Upon this is one plus M two plus M three. Now it is given did origin B center off moss of body. So we can say that. So are vectors should equals to zero. So from here we can say that. And when our one vector plus M two or two vector plus N three. R. Three vector is equal to this is zero. So these are the moment of masses. So we can say that some of moment of moth about center of mass is zero. So this is the answer of our given fusion. And for that we can say option, this is the correct choice. Thank you.

So here, for part A. Were describing the energy transformations that take place. So first we have the mechanical energy. So the you can say that the mechanical energy of the system transforming into the potential energy of the spring. Or we can say elastic potential energy. This elastic potential energy then converts itself into kinetic energy and then the kinetic energy of the projectile then transforms into gravitational potential energy wants to projectile reaches its maximum height. So for part B we can relate, we can use the conservation of mechanical energy and say that then the potential energy of the spring would be equaling to the gravitational potential energy. Or we can say 1/2 times the spring constant. K. Multiplied by the compression squared equaling M G H. And so the spring constants K would be equaling 22 M G H divided by the compression squared. So this would be to Multiplied by .020 kg or 20g, Multiplied by 9.8 meters per second squared, multiplied by 20 0.0 m. And this is going to be divided by then .1- 0m quantity squared. The spring constant is then found to be 544 newtons per meter. And then for part C, we can use conservation of energy to find the initial speed. So we can say that then the kinetic energy would be equal to the gravitational potential energy. And we are we have essentially one half times the mass times the velocity squared equaling m G H masses cancel out. And we are left with the velocity equaling the square root of two. G. H. This is the square root of then two multiplied by 9.8 m per second squared, Multiplied by 20 m. And we find that then the initial velocity is going to be equal to 19 0.7 meters per second. That is the end of the solution. Thank you for watching

Question number 43 when the spring is released than its initial spring potential energy. Half gay XY Squire is converted toe the gravitational potential energy off the given mass, which is M G. That's so no substituting the values how into his spring constant which is 1200 newtons per meter, into the formation in the spring. This is equal to the mosque. 0.15 kg into 9.8 m per second in Squire and hide this 5 m. Solving this for X, we will get zero point 035 meter.


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