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Both (hem edenence [Kc #Jmt non Ieru omentum Cnjnd Both of them experience zero momentum change_Questlon 14 (4 polnts) A20-kg mass moving at 5.0 m/s suddenly collid...

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Both (hem edenence [Kc #Jmt non Ieru omentum Cnjnd Both of them experience zero momentum change_Questlon 14 (4 polnts) A20-kg mass moving at 5.0 m/s suddenly collides head-on with 3.0-KB mass at rest If the collision Is pertectly Inelastic; what Is the speed of the masses just after the collision?25 m/s10 mls2.0 m/s0 m/s, since the collision is inelasticQuestion 15 (4 points) 340-g air track cart travelingat 1 .25 mls suddenly collides elastically with a

Both (hem edenence [Kc #Jmt non Ieru omentum Cnjnd Both of them experience zero momentum change_ Questlon 14 (4 polnts) A20-kg mass moving at 5.0 m/s suddenly collides head-on with 3.0-KB mass at rest If the collision Is pertectly Inelastic; what Is the speed of the masses just after the collision? 25 m/s 10 mls 2.0 m/s 0 m/s, since the collision is inelastic Question 15 (4 points) 340-g air track cart travelingat 1 .25 mls suddenly collides elastically with a



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Two objects slide over a frictionless horizontal surface. The first object, mass $m_{1}=5 \mathrm{~kg}$, is propelled with speed $v_{i 1}=4.5 \mathrm{~m} / \mathrm{s}$ towards the second object, mass $m_{2}=2.5 \mathrm{~kg}$, which is initially at rest. After the collision, both objects have velocities which are directed $\theta=30^{\circ}$ on either side of the original line of motion of the first object. What are the final speeds of the two objects? Is the collision elastic or inelastic? $$ \begin{array}{llll}
& \begin{array}{l} \text { Speed of first } \\ \text { body } \end{array} & \begin{array}{l} \text { Speed of sec- } \\ \text { ond body } \end{array} & \text { Collision } \\ \hline \text { (a) } & 5.1962 \mathrm{~m} / \mathrm{s} & 2.5981 \mathrm{~m} / \mathrm{s} & \text { elastic } \\ \hline \text { (b) } & 2.5981 \mathrm{~m} / \mathrm{s} & 5.1962 \mathrm{~m} / \mathrm{s} & \text { inelastic } \\ \hline \text { (c) } & 2.5981 \mathrm{~m} / \mathrm{s} & 5.1962 \mathrm{~m} / \mathrm{s} & \text { elastic } \\ \hline \text { (d) } & 5.1962 \mathrm{~m} / \mathrm{s} & 2.5981 \mathrm{~m} / \mathrm{s} & \begin{array}{l} \text { Cannot } \\ \text { determine } \end{array} \end{array} $$

Hi. In the given problem, there are two balls Mass of the first ball is M one is equal to two kg and mass of the second one is M two which is one below ground. The initially speed means speed before collision. Initially speed off the first wall Is M one i. is equal to 25 mita for second. And initially speed off the second ball is and that is we not. We won. I and we do. I. You just give an ass 20 meter for second. Now, after the collision speed of the second, ball speed of the first ball means we one F. That is equal to minus 2.5 m/s. Here, negative sign represents that the direction of motion of the first of all is reversed. Similarly, the direction of motion of the second ball is also reversed so its velocity not the speed, its velocity will be minus 35 meter per second. Now we will find the kinetic energies of the system comprising of the two balls before collision and after collisions. So first of all, total kinetic energy of the system before collision is E. G I. And that will be given us half into em burn into even I square glass, half Into empty into V two I square. So plugging in or non values E k I is equal to half into two kg into square of 25 m/s plus half into one program into 20 meter per second to the whole square. So here it comes out to be 625 glass, 200 Jews. Or we can say the kinetic energy of the system before collision is 825 jewels. Now in the similar method car net energy of the system after collision and that is E G A F. Which has given us half and one we even F square plus half M two V two F sq Plugging in non values. Again, this is half into two kg into for everyone. F this is minus 2.5 to the whole square plus half, into one kg Into -35 to the whole script. So here it comes out to be 6.25 plus six, 112.5 jules. When adding it, it comes out to be 618 .75 Jews means it is clear that the kinetic energy of the system is decreasing after the collision. It is less than that before the collision, so as kinetic energy of the system is decreasing, so it must be and in elastic collision. And moreover, as the balls are not sticking together of part a coalition, so it is not are perfectly last, perfectly inelastic collision hands here, our option is correct. According to which the coalition is simply In the last eight, not the perfectly in the last day. Thank you.

If this problem were asked to determine whether this is a an elastic, perfectly in elastic or elastic collision. So we have a ball of two kilograms traveling at 25 meters per second, which collides head on with a one kilogram bowl traveling at 20 meters per second and then they reverse direction. So the kind of bounce off of each other now this to killing ground ball is now moving at 2.5 meters per second and the one kilogram bull is now moving at 35 meters per second. And so, if you look at her definition for an elastic collision and an elastic collision, a elastic collision has conservation of kinetic energy, which means our kinetic energy at the beginning is equal to our kinetic energy total at the end. In a in elastic collision, there is no conservation of kinetic energy in a perfectly in elastic collision. The kinetic energy is still do not equal, which makes it an elastic collision. But they also have the same final velocity, which means they stick together at some point or they started off together, so we can immediately roll out that it is not a perfectly in elastic collision. So it has to either be an elastic elastic or cannot be determined so we can determine our total Connecticut ready at the beginning by calculating the individual kinetic energies and then adding them together. So I get 1/2 2 times 25 squared, plus 1/2 one 20 squared. Now, even though this is going left, kinetic energy is a scaler, so it won't have a direction. Even if it did, it would get squared out and become positive. So here you get a kinetic energy of 825 jewels, and then we can apply the same concept to the final. So we'll have 1/2 to 2.5 squared plus 1/2 one 35 squared, and this gives me a kinetic energy of 618 0.75 jewels. Now, because these two values are different, it must therefore be on in elastic collision

In this problem, we need to define what is an in elastic collision, an elastic collision and a perfectly in elastic collision. So in the elastic collision are mo mentum is conserved. So our momentum at the beginning is equal to our momentum at the end. And our kinetic energy total at the beginning is also equal to my kinetic energy at the end. So that was for an e elastic collision for an an elastic collision. My momentum is still constant. So my momentum at the beginning is equal to my moment at the end again totals. But my kinetic energy at the beginning does not equal my kinetic energy at the end. So there was usually a loss of kinetic energy there, a perfectly in Alaska collision is a special case in usually which they stick together. So something like an object coming towards another object and they collide and moved together as one mass. So in this case, my momentum is still evil before it after. But my kinetic energy is not equal, which tells us that this isn't in elastic collision. A perfectly in elastic collision also loses the most amount of energy. So if we look at our situation. We basically just need to calculate our kinetic energy at the beginning, our kinetic energy at the end and then compare them to each other. If they're equal, then it is elastic. If they're not equal than it's in elastic and then because he do not stick together, I can already rule out that they're perfectly in elastic. So my kinetic energy at the beginning it is no kinetic energy as 1/2 M V squared. So when half two times 25 squared plus 1/2 one 20 squared now you can include direction here, but because it's getting squared, it's going to cancel it anyway, so it doesn't really matter, either whip. So I get about 825 jewels for this one and this my kinetic energy before they've collided. And then in the after I have 1/2 two 2.5 squared plus 1/2 one 35 squared, and this gives me 618.75 jewels again because they're different values. This tells us that it isn't in elastic collision

Problem # 196. In fact, there is a body of mass five Kg moving with a velocity velocity. This is mass, this is you won okay along a straight line collided with the body. Two kg moving along the same line with the velocity. Okay, so another masses ah of two Kg. It is moving in the same line moving in the same line. So either in this direction or in this direction with velocity. You want. Okay, the collagen perfectly inelastic magnitude of the velocity of composed mass after the collision. So we'll be considering two cases first. We should write whatever it is being given five kg MAs four m per second. Five kg mass four m/s and then I am too two kg MAs An eight m/s. U. N. Is either eight m/s or uh minus eight m per second. If it's going in this direction, if we support these two positive. So there will be two cases. Case one Case one is when you equal to this get me the president. So momentum will be conserved from the previous question. They will be getting be equal to and when you win plus him to you too. Buy in one place. I am too. So the same one and one is Uh huh five kg five and two four plus 2 into eight, Divide by five Plus 2. Doing deep Bliss 16 37 36 x seven meter per second. So when answer should be this and second the equal to and when you on formula will be the same. But only thing is here it will be negative in place of positive. So 20-16 x seven pulled by seven meter per second. Okay so option numbers B and uh the korean. Okay option maybe. Andy are the correct answers. Thank you.


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