two-objects-move-toward-each-other-collide-and-separate-if-there-was-no-net-external-force-acting-on

Question

Two objects move toward each other, collide, and separate. If there was no net external force acting on the objects, but some kinetic energy was lost, then
(A) the collision was elastic and total linear momentum was conserved
(B) the collision was elastic and total linear momentum was not conserved
(C) the collision was not elastic and total linear momentum was conserved
(D) the collision was not elastic and total linear momentum was not conserved

Vipender Yadav · Accepted Answer

so her in this problem were given that two objects move towards each other, the collide and then separate on were also given that the net external force acting on the object is zero on the kind of energies also lost in the collision. So this is lost energy is lost Now there are four options given and we&#x27;re total that which option is correct now since the Net force acting on the object is zero. So the total leaner moment of will be conserved so we can contribute by this that linear momentum will be conserved next. They are saying that kinetic energy is lost during the collision. So the collision cannot be elastic collision. It has to be non elastic, so collision is not elastic. Option C matches with the given statement. Therefore we can conclude that options C is correct. I hope you have understood problem. Thank you

Ze-Han Lee · Answer

So in this problem, we&#x27;re considering until ah, pure any elastic process. So the pureeing any last year process means that Suppose we have to two blocks and won them too. And, ah, someone moves with moves with the velocity v and M to stand at rest. So after a few seconds, you see that they will stick together. All right, So this is them. Why, I&#x27;m too. And they move along with the velocity. Let&#x27;s say be prime. All right. So for this process Ah, the moment, um, is always considered. However, the energy is not conserved, right? Because the part of the energy part of the kind of energy is transferred in thio some other form, a form of energy, like the internal energy or even transferring to heat. Okay, so the energy is not considered the only consider momentum. Here is the moment of

Abhishek Jana · Answer

so linear momentum is conserved during boot, elastic and any lasting coalitions. And for perfectly elastic collisions. Kinetic energy is conserved as well, because no energy is lost during hit. So for inelastic collision, let&#x27;s look at an example of two blocks. So if they&#x27;re on track and if there is some friction in the system, then after they hate each other or during their travel, there will be some form off heat due to the friction between ground and the blocks. So kind of the energy wouldn&#x27;t be concerned, but for elastic collision, since there is no other form of energy present there, so it will become served us well, so e option will be the correct.

Maria Gabriela Cota Moreira · Answer

So it&#x27;s exercise. We have a perfect elastic collusion between two particles. So particle A and particle be so In the beginning, Particle A is moving in the horizontal access with initial velocity V zero and the park will be is stead still on? After that, the particle a collides head on with particle B and both particles changes Velocities of particle begin a velocity V b Q and particle way gained a A velocity of V h. Okay, so in exercise A let me write it here. We want to find what fraction off the initial kinetic energy that I&#x27;ll call. Okay, zero goes for particle A and what fraction goes to particle be okay, So we have We want to find expressions for this and this. It was so starting with the fraction with the expression from the fraction off cannot energy that goes to park away. So we have that the initial kinetic energy is just a massive particle. A times the initial velocity squared over two and then the final kinetic energy of particle A is it with you the massive Parco a V H two squared or Richard? Um so this fraction, uh, begin becomes V H two squared over 30 squared. Okay, for the more we also have that leszek collusions have, uh, a relation between the initial and final velocity off this first particle here. So I can write that they the final velocity V two is equal to, um, a minus M b over a blues M b these times be sorry, v zero. Then you should also forgo it. Okay, This is a equation. Uh, that works for elastic collisions on Lee on, and it is written in Equation 8.24 in the test book. Okay, so we just use this and to be astute in our expression, so we have that. Okay, He&#x27;s zero. So, So big V A squared. We have I am a minus m b over. I am a plus m b squared. Okay. The V zero squared cancels out with zero from the dominator. So this is the first expression we wanted to find. The second expression is for the amount off the fraction of energy that goes to particle Be What? So, um, we can argue that since we&#x27;re working with a plastic collusion, we have conservation off kinetic energy such that uh um que a plus k b has to be equal to zero, which also implies that if I divide the conservation off kinetic energy by K zero in both sides, we have that the some off the fraction that goes to particle A with the some off the fraction, with the fraction that goes to particle be has to be equal to one. Okay, this is go thesis school from this come from conservation of energy. So conservation off kinetic energy. Okay, okay. So in this case, I can isolate the fraction k B zero to be one minus k over K zero. So I only pick one minus the expression I found previously. So one miners m a minus m b over, um a plus m b Everything squared. So performing this objection, we find that the fraction that goes to, uh, K B is equal to and May waas MBI squared. So this is going to be a really big term. So here, this one becomes m a plus m b squared. So I&#x27;m going to expand this here inside off this parenting, so am re squared. Plus m b squared was too, um, may and be miners and may minus them B squared. I&#x27;ll also expand this so we have and may squared less M b squared minus shoe and may M b Okay, so because of this minus sign here and may squared cancels with Amy squared and B squared also cancels with and B squared. And we can sum up these two terms such that we have that the fraction of energy the fraction of the initial energy that goes to particle B is going to be equal to 4 a.m. a. M B over m A plus m b squared. So this is the final answer. All questions. A now question. Be asked us to evaluate the americal value off this fractions. Okay, que zero and K B zero for some cases. So the first case is when the mass off particle way is equal to the mass off particle. Be so we can go back to the case expression. And we have that when m A is equal to m B, this term become zero. OK, zero goes to zero, so que que zero is equal to zero. Okay. Which means that the the particle a well, we lose all its initial velocity will be standing still after the collision and from the conservation off kinetic energy we have that baby zero will be obviously equal to one. And the second case is when, um sorry. Um, A is equal to five times and be okay. So this time I&#x27;ll rewrite the expression again. Here. Eso we have that This is going to be I am a, uh sorry. Um, so this would be six and sorry, uh, for M b over six. M b squared. Okay, because I said be stewed this by five MB. So five and B minus and be becomes for M B and five and B plus and B six and b. Okay, so from this, we can cancel all today&#x27;s with this, and we have to over three. So two squared over three squared is equal to 4/9. So the first answer and now sorry, this is not over. KB is over zero. So now we can once again use conservation off, uh, kinetic energy and but one minus for over nine. And find that the amount of energy that will go to particles to park O B is going to be 5/9. Okay, Uh, now, questions see asked us to find the values off the mass ratio and may end over and be such that the kinetic energy will be equal, equally shared between the two particles. So questions see Okay, we have We want to find this fraction in the case where K B over zero is equal to K A over zero. And we know from conservation off kind of energy that this fraction will be will have to be equals. You wanna have? Okay, so it just pick the expression from K P over K zero that we found on question A and, uh, an equal to the expression we found for K A over K zero. So we pick this expression here, they&#x27;ll call one and put this equals to this expression to So we have expression. One equals true expression to this leads to, um m m a sorry for em, eh? And be over. Um um, A plus and B squared has to be equal to m A minus m b over a may plus, um, be both terms squared. So notice that we can cancel out the denominators and expanding this term here on the right hand side, we have four m A and be equal to M a squared plus M b squared minus two. Um, a m B. Okay, so if we divide, uh, the both sides of the equation by m. B squared, we&#x27;ll have that. So remember that we want to find this fraction here. I&#x27;m sorry. So we want to find this fraction here. So dividing both sides of the equation by and B squared we have that, uh, So I m a over m b squared plus one minus. So passing for, um a m b to the other side of the equation minus six. Uh um a over m B. This is And this is it for 20 So notice that this is a credit credit IQ formula on which my variable is m A over M b. So solving the quadratic formula, we have that the truth possible? Uh, answers are six plus minus the square root off 32 over to, and this leads to the two possible fractions. We can have so five point 83 and the second one to be 0.172 on this concludes the exercise

Joseph Petrullo · Answer

for this problem, you have to look at two things. The first thing is that the total Momenta is zero initially and after some collision, we have that m one V one plus m two V two is equal to zero. Now, if the collision is elastic, then we know that 1/2 m one V one squared. We call this view and I squared initially plus 1/2 m two v two I squared is equal to 1/2 m one B one squared plus 1/2 m two V two squared. Now what? We can notice that the only way to solve this equation for any arbitrary valued of the masses and velocities is that the initial and final speeds of each object must be the same. To see this, we can take a look at this second equation here. So in the second equation, obviously, if V one I is equal to be one and if v two I is equal to be to then this equation will be satisfied. Furthermore, when we look at Equation one, we can consider the fact that say, we have originally as a stated in the question that the two of the two masses approach each other such that the total momentum is zero. Then what will have here is since we know that from equation one m one V one plus m two V two is equal to zero where this is the final condition. Then initially we know that we&#x27;re going tohave some m i or some m one V one I plus m two v two I Now what we notice is that if we keep it, if we want it to be true, that this m one V one I plus them to veto is also equal to zero. So we&#x27;ll have both equations satisfied. We can notice that if V one I is equal to minus V one and V two I is equal to minus V two. Then what will have is that the actual value for the speed which is the absolute value of this number, will be the same before and after. And just by negating both of the velocities if they if they add to zero at the end, then by negating both of the velocities, we will just get that It also adds to zero in the beginning. Initially, therefore, in order to satisfy both these two equations for any arbiter cherry value of masses or velocities. It must be true that V one I is equal to V one and V two I is equal to the two. But we can&#x27;t forget that we&#x27;re talking about speeds here and so we want the absolute values here.

Kristela Garcia · Answer

so we have two main types of collisions. We have an elastic and elastic in an inelastic collision. The total kinetic energy at the beginning does not equal are total kinetic energy at the inn in an eat elastic collision are kinetic energy at the beginning. Does equal are kinetic energy at the end, and so we have the conservation of kinetic energy conservation of kinetic energy in a perfectly in elastic collision. Because it is an end elastic, collision C is automatically both because kinetic energy is not concert more specifically in a perfectly in elastic collision, two objects will stick together and travel with the same. Find the velocity now in both situations in both in elastic and elastic, Mo mentum is conserved. Our total momentum at the beginning is equal to our total mental at the end. So be is correct and true. Aye is true. And in this situation you do have a maximum amount of kinetic energy loss. O. D is also so. See, it&#x27;s false

Problem

Two frictionless carts (mass $=500$ g each) are s…

Problem 7 Medium Difficulty

Answer

C

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