INT Consider a partially elastic collision in which ball A of mass m with initial velocity (vix)A collides with stationary ball B, also of mass m, and in which 1/4 of the mechanical energy is dissipated as thermal energy. Find expressions for the final velocities of each ball. Hint: Mathematically there are two solutions; however, one of them is physically impossible.

Video transcript

Hey, everyone in this problem, we're asked to consider a collision between two cars. Car one and car two of equal mass two F in which car one is initially moving with velocity V and car two is initially stationary at a red traffic light. And the collision is partially elastic with 2/5 of the mechanical energy dissipated as heat and sound. And we're asked to find the final velocities of the cars after the collision. We're given four answer choices A through D, each of them containing different final velocities for car one anchor two in terms of that initial velocity V. All right. So let's get started and let's think about this information we're giving. So we have our initial case before the collision and then we have our final case after the collision. Now, initially both carves M one, sorry, car one and car two both have a mass of two F. So M one is gonna be equal to M two and that's gonna be equal to two F and that's gonna be the same in the final case. And the mass of the card does not change during the collision. So we have M one equals M two equals two M or both. Now, initially, we're told that car one has a velocity of V. So we're gonna say V one knot is just equal to V and with the one subscript indicates what the car we're talking about. OK. So car one and then the not subscript indicates that we're in our initial time point. Now, the initial velocity of car two V two knot. Well, we're told that it's initially stationary. So its initial velocity is just going to be OK. And in the final case, we have V one F and we have V two F and that is what we were looking for those two values. OK. The final velocity for both cars. All right. So we have a collision. We're told some information about mechanical energy. Let's start with our conservation of momentum. OK. We know we have this collision, we know we have conservation of momentum. So let's look at the conservation of momentum first and see where that gets. Now, we're called at the conservation of momentum tells us that the initial momentum peanut, it's gonna be equal to the final momentum pf And the momentum in this case is gonna be made up of two things because we have two objects in our system. OK. So we have the initial momentum of car. What was the initial momentum of car two is equal to the final momentum of car one who lost the final momentum of cartoon, we call that momentum is given by a mass multiplied by velocity. And so our equation is gonna become M one V one dot plus M two, V +29 is equal to M one V one F plus M two V two F. Substituting in what we know. The second term on the left-hand side is gonna go to zero because V two knot is zero. Substituting in our mass, we have two M multiplied by the initial velocity of current one, which is V that's gonna equal to two M multiplied by V one F plus two M multiplied by B two. And you'll see that this two M mass can be divided. So it's in every single term, so we can divide that out and we're left with the initial velocity V gonna be equal to the sum of the two final velocities. OK. So we have an equation here. We have two unknowns, V one F and V two F that we're trying to solve. So with the one equation, it's not going to be enough to solve. We're given information about the mechanical energy. So we can switch over to the mechanical energy as well and see what that tells us for. Now, we're gonna try to write one of these unknown values in terms of the other. So let's go ahead and write V two F in terms of V one F. So we can write V two F is equal to V minus V one F and we're gonna call this equation one. And we're gonna use that later with our mechanical energy information. All right. So let's give ourselves some more space and start working with the mechanical energy. Now, we're told that 2/5 of the mechanical energy is dissipated. What that means is that the final mechanical energy is 3/ the initial mechanical energy. OK. So 3/5 of our initial mechanical energy. And in this case, our me, our mechanical energy is just kinetic energy. OK? We have no springs, we have no um gravitation to worry about here. These cars are not on a cliff or anything like that, they're on the ground, they're at the same height. So we don't have to worry about the gravitational potential energy spring potential energy. So we have just our kinetic energy for our mechanical energy. We have 3/5 that initial kinetic energy is going to be equal to the final kinetic energy. OK? Because 2/5 is lost and dissipated as heat. OK? Now, our kinetic energy just like our momentum is made up of two components, one for each car. So we have 3/5 K, one knot plus K, two knot is equal to K one F plus K two F. And recall that kinetic energy is given by one half MV squared. So we have 3/5 multiplied right? Um One, the one knot squared and we missed the one half. There we go. One half, M one V one knot squared plus one half and two V two knot squared. This is gonna be equaled to one half M one V one X squared plus one FM two V two F squared. And what you can see if you look at this equation, we were given information about the masses M one and M two. We know both initial philosophies on the left hand side. So we're gonna be left with just V one F and V two F as unknowns. OK? So now we have the same two unknowns. We have two unknowns with two equations, which means we're gonna be able to solve. OK. So we do not, we know that that's zero. So the second term in our brackets on the left hand side is gonna go to zero. We're gonna be left with 3/ multiplied by one half, multiplied by two M multiplied by V squared. This is gonna be equal to one half multiplied by two M multiplied by V one F squared. What one? Huh? Multiplied by two M multiplied by V two F squared. Well, we actually know V two F square V two F, we can write as V minus V one F so we can write this as V minus V one F all squared. And now our equation has only V one F as the unknown. And we've eliminated this V two F by using our substitution method. OK. Let's go ahead and try to simplify. So every single term here has the one half multiplied by two F. So we can divide it by one half and by two F, those were all divided. We're gonna be left with 3/5. V square is equal to V one F squared plus V minus V one F word. We're gonna go ahead and expand on the right hand side again, we want to solve for V one F. So we have three bits V squared is equal to V one F squared plus V squared minus two V multiplied by V one F plus V one F. Now this looks really messy but this is gonna be a quadratic equation from V one F. OK. We have a B one F squared term, we have a V one F term and we have just a constant term with this V OK. Remember V is our initial velocity. So that is just a number. So if we move our 3/5 V squared to the right hand side, we can write this as zero is equal to two V one F squared minus two V multiplied by V one F plus 2/5 V squared. OK? And now we can use our quadratic equation to solve for V one F A R A value is gonna be the coefficient corresponding to V one F squared which is two B is gonna be the coefficient corresponding with V one F which is negative to B and see who's gonna be the coefficient corresponding with their constant term, which is 2/5 V squared. Hm. So our quadratic formula V one F is gonna be equal to negative B which is two B B plus or minus the square root of B squared negative to V squared minus four multiplied by A which is two multiplied by C just 2/5 V squared. And all of this is divided by two A. So two multiplied by two. Alrighty, we need some more space here and we're gonna simplify. So we can simplify this in the numerator, we're gonna have four V squared. We're gonna have minus if we multiply this out 16 5th B squared. OK? So we can write this as two V plus or minus the square root 4/5 V square divided by four. Now four, we know the square of forest too V square, we take the square root we get V so we can write this as to be plus or minus two V multiplied by the square root of 1/5 divided by four, simplifying V divided by two plus or minus the divided by two root five. And then to rationalize the denominator. On the second term, we have V divided by two less or minus route five B divided by 10. OK? So we've just found B one at. So this is the value for V one F OK? We have this plus or minus. We're gonna have to decide which of those makes sense. Let's look at what this means for V two F first. OK. So if we go back up to where we have this equation for V two F and we substitute in what we just found, we get that V two F is gonna be equal to V minus V divided by two plus or minus root five B divided by 10. Now, the V minus V half, we're gonna be left with V divided by two. Then we're saying subtract this plus or minus root five B divided by 10. Now, because we're subtracting that. What's gonna happen is our plus or minus is gonna turn into a minus plus. OK? So when V one F uses the plus sign, V two F will happen minus sign and vice versa. OK? So we have V divided by two minus plus root five B divided by 10 and that's gonna be our V two F. Now we're left again to figure out what sign makes sense. No, if car one were to move faster than car two in the original direction after the collision, it would have to pass through it or jump over it. Ok? We know that that can't happen. So that means that the final speed of car two, if they're in the same direction must be greater. Now, route five divided by 10 is going to be smaller than a half. So these values are gonna be positive they're definitely gonna be moving in the same direction. So what we need to do is take the minus for V one F, the plus for V two back. Ok. So we're gonna be taking the minus. Yes, the plus for B two F and the minus or V one F. Ok. Otherwise again, B one F would have to car one would have to move through car two, which is not possible. So if we go to our answer choices and we can compare what we have. Ok. Now, option C and D were very simple with a quarter V or half V. Ok. So those ones we can eliminate, we're left with A or B which have these very similar results. OK? And the difference is whether we choose the positive or negative. So we said for V one F, we needed to choose the negative so that it was smaller than V two F in the same direction. And so the answer choice here is going to be b thanks everyone for watching. I hope this video helped see you in the next one.