this is the first section in the chapter on elastic collisions in one dimension chapter four and this is about direct impact and newton's law of restitution so by direct impact what we mean is two particles things that models of particles um directly impact each other like this so they move towards each other like this they impact and then they move off in their their own direction so they it may be that they move in opposite directions after the impact it may be that actually the way that they impact is that one of them is at rest um and then after the impact so this is at the top here this is before the impact this is what's going on and after the impact down here and it may be that after the impact they both move in the same direction but the one uh that wasn't moving this one here moves off at a faster speed than the one behind it so if this was u um and here starting off u was zero for example and this is a v1 and v2 um v2 would be have to be greater than v1 because it's moving off ahead of the one behind it yeah it can't be less than the one behind it otherwise they well it wouldn't work so we do have a coefficient linked to the the way or the speed at which these two objects collide and the speed at which they separate and it's called the coefficient of restitution coefficient of restitution restitution and the letter we use for his letter e its value is between 0 and 1. it tells us about the elasticity of the collision okay so elasticity of collision so how much they sort of bounce after the collision well if the value of this coefficient of restitution e equals one is what we call their perfectly elastic perfectly elastic in other words you know if they were um you know going together at 10 meters per second then when they separate the speed of separation would be 10 meters per second we'll talk a bit more about the um these speeds of approach and speed and separation in a minute if the value of e is zero then we have things which are inelastic or perfectly in elastic which basically means when they collide they stop moving completely so imagine plasticine where you got two bits of plasticine thrown towards each other like this and then what happens is that they just become one like blob like this and they stop moving that would be an example where the coefficient of restitution is zero whereas things that are almost perfectly elastic might be things like uh snooker balls or pool balls where you know that they got a great deal of bouncing power so how is this coefficient of restitution worked out so e is this coefficient it changes depending on the type of particles that are involved in the collision but the way it's measured is it's the speed of separation speed of separation divided by the speed of approach speed of approach so in other words what's the speed that they're moving away from each other so speed moving away from each other from each other after collision or after the collision whereas the speed of approach is the speed moving toward each other toward each other before the collision before the collision and it all depends where whether the objects are moving in the same direction or opposite directions um i don't want to sort of talk about rules like you know if they're moving towards each other always do this or if they're moving away from each other always do this if they're moving in the same direction always do this rather than usual this is a bit of common sense let's look at the question and we can quite easily work out what the speed of separation is and what the speed of approach is whether we need to add the velocities together or subtract them all depends on the question okay in each part of this question two diagrams so the speed and directions of motion of the two particles a and b just before that's here and just after the impact our collision the particles move in a straight horizontal plane find the coefficient of restitution in use case all right so let's just remind ourselves of some things coefficient restitution should always be between zero and one um the way that we work it out is it's the speed of uh separation speed of i'll just put scp for separation divided by the speed of approach speed of approach now although i don't like formulae let's see if we can see some like general rules so let's look at the separation so we've got two particles now if these two particles were moving away from each other and v1 and v2 the speed of separation would be v1 plus v2 because you've got the combined effect of them moving in opposite directions if after the collision v1 and v2 they're moving in the same direction now in this case v2 would be greater than v1 for it to be moving away and the speed of separation would be v2 minus v1 yeah because if v2 was moving off at 10 and uh v1 was moving towards it at nine then actually um the speed of separation would only be one so we subtract in that case okay um so this is for separation my let's look at approach so i think when things approach each other it could be that they're approaching each other and they're moving in this same direction so we've got u1 and u2 now for this to happen u1 would have to be greater than u2 otherwise it would never catch up u1 will be greater than u2 and esprit of approach would be um [Music] u2 minus u1 yes i have a think about that the spread of approach would be u 2 minus u 1 more talk a bit but more about it we go through the examples and if they were moving in opposite directions before the collision then the speed of approach would be u1 plus u2 okay so the reason i don't like rules for this is because uh when they're approaching you add when they go in the same direction you add when again in the opposite direction when they um when they're separating when you approach you do and they're going in the same direction you subtract i suppose when they go in separate and again in the same direction you also subtract but it it's easier just to work it out using a little bit of common sense rather than rules but that's just to give you an idea what's going on so a so at what speed does a separate well one's at rest uh the other one is moving away at two so they separate at two meters per second and they approach at eight because the other one's at rest so that becomes a quarter that's our coefficient restitution for a b okay right after the impact let's have a look well b is moving away um at five and a is behind that four so actually it only looks like um b is moving away at one meter per second so the speed of separation would be one and i suppose if we look at the rules that we had down here then it's like v2 minus v1 that makes sense and then what's the speed of approach well um that's going to be three yeah a would be appearing to move towards a at three meters per second because b is already moving so the speed of um approach is three so that's just a third there so again we could look at our speed of approach where we would do u2 minus u1 and 4c again let's just use a bit common sense to work this out speed of separation well if they're moving away we add them together so that's nine and before the impact they're moving towards each other so we add so that's going to be 18 which is a half okay two particles are traveling in the same direction on a smooth surface with speeds four and three respectively let's start to do a diagram a b and we're told they're moving in the same directions like this and when it says respectively it means the same order so a is four b is three four and three they collide directly and immediately articulation continue to travel in the same directions okay so they continue going like this with speeds 2 and v respectively to nv respectively given that the coefficient of restitution between a and b is a third find v okay so e is the speed of separation over speed of approach now that's e is a third now the speed of separation since they're moving in the same direction is going to be um [Music] v is going to be bigger than 2 by the way says v minus 2. and the speed of approach is going to be four minus three which is one right so this should be easy enough to solve a third equals v minus two over one we can cross multiply so we get one equals three v minus two so that means that one third equals v minus two which means that v equals a third plus two so we could say two in the third meters per second and it has to be bigger than the two otherwise we know that we've done something wrong something's not quite worked out okay so we've got these two masses and we need to change the masses into kilograms so we have a and b so a is going to be 0.2 kilograms and b is going to be 0.4 kg and we're told that they're traveling in opposite directions towards each other so let's put these in here and it says that their velocities are five and four so five four it says they collide and after collision we just told their velocity so we don't know which direction so i'm going to assume they actually change direction if i am wrong then when i work out my answer for v1 or v2 i will get a negative answer which means it's actually going in the other direction and i'm going to choose this direction as my positive direction yep you could use the other direction you will still get the same answer and we're also told that the coefficient restitution is half now why have we been given the masses this time well because we need a second equation because you've got two unknowns and that equation is going to be the conservation of linear momentum which we've done already in chapter one along with things like impulse okay so conservation in the momentum you you're allowed just to write clm the examiner will know what that means so before the collision what's the total momentum well that's going to be 0.2 times 5 plus 0.4 times by negative 4. look at the direction of the four and then after the collision it's going to be i'm gonna have to write it down here because i'm running out of space um naught point to v1 now let's have a look oh my v1 is going against my opposite so it's not 0.2 times negative v1 plus 0.4 v2 right let's tidy this up so naught point two times five is one um 0.4 times by 4 is going to be 1.6 that's negative 1.6 equals and it's going to be negative 0.2 v1 plus 0.4 v2 okay so i have negative 0.6 equals this that side hasn't changed i probably want to write this in a form that's going to be a little bit nicer to use as an equation so i'm going to times everything by negative 10 so that will give me 6 equals 2 v1 minus four v two i could divide that by two i'll leave it like that for the time being that's fine now i'm gonna wanna get an equation from my um coefficient of restitution now what's the speed of separation well because of the directions i've chosen it will be v1 plus v2 and before the collision they're moving towards each other that will be five plus four right that will give me two by cross multiplying two v one plus v two equals nine so if expanded brackets i will get two v one plus two v 2 equals 9. so here's my second equation here so i want to solve simultaneously now if i actually make 2v1 the subject here i can nicely substitute that into the other equation yeah so all i've done there is taken away 2v2 from both sides i'm now going to substitute it into this so that'll give me 6 equals so where i've got 2 v1 i'm going to put 9 minus 2v2 minus 4v2 so that gives me minus 3 if i take away 9 from both sides minus 6 v2 right so what does that give me v2 equals negative 3 divided by negative 6 which is a half so my direction is correct let me put this on the diagram this is going to be a half and it's going to be moving in that direction so next step is to work out a v1 so i can now substitute that half let me substitute that half into here yeah back into that same equation again so if i do that i could substitute into the first one again doesn't matter too much which one so i'm going to have 6 equals 2 v1 minus 4 v2 now v2 is a half so 6 equals 2 v1 minus 2 says add 2 to both sides so 8 equals 2 v1 which gives me v1 equals 8 over 2 which is 4. so there we go so that also means that i got the direction for v1 correct so let's just make that really clear on this diagram here like this so um we said that v2 is a half so this is a half here meters per second so that's the direction it's going in and v is four so that's going in this direction so my assumptions about the directions work great right first we need to do is to draw a diagram so we've got these two balls p and q so i will attempt to fit the diagram up here so p q and their masses are 3m and 4m whenever you get this thing respectively written in a question it just means basically um the direct the order of the things written is the same as the order of the first things that we're given so uh p goes with the free m q goes with the four m i'm gonna put kg here so i don't i know that these are masses because there's other stuff as well i don't wanna miss these out um they're moving it says they're moving towards each other like this in a horizontal direction and the speed of p is 3u and the speed of q is 2u and in part a i want to show what the speed is of q after the collision now that means that i need to make some assumptions about what v1 and v2 are and i'm gonna assume that after the collision they actually both change directions okay and i need to keep that in mind as i work through the question and i also need to just decide in a direction which is positive and i'm going to choose this direction as positive so part a show that the speed of q after the collision is u over seven brackets 15 e plus one so we basically want to show that this is what v two equals so we're gonna need two equations we'll have one equation for the coefficient of restitution e we're told it's e and that is the speed of separation that's the v1 plus v2 on my diagram because they're moving in opposite direction and a speed of approach which is the 3u plus 2u because of my diagram and as we're told in the question they're moving towards each other so that would be so those two together make five u so that would be five e u equals v one plus v two okay so quick equation number one i'm gonna keep that one to one side and my other equation is going to be from the conservation of linear momentum so i want the total momentum before multi uh and that equals the momentum after so before the collision the momentum of p is 3m times by 3u the momentum of q is 4m times by 2u um ah now that's going to be negative 2u because of the direction just spotted that so times by negative 2u almost missed that out then after the collision uh the momentum of p is going to be 3m times by negative v1 because again that's going against my opposite direction plus 4m times by v2 so this can all be tidied up to give me 9 mu minus 8 mu equals negative 3 m v1 plus four m v two now every term has got m in it so they can all cancel out so i'll cancel this out there and then i'll have nine u minus 8u which is just u so u equals negative 3v1 plus 4 v2 right so here's my second equation now what do i need to do i need to put these two equations together simultaneously that is this equation and this one and since i need to get rid of um v1 and i just want an equation for v2 i'm going to make v1 the subject and i'm going to make v1 the subject of this one because it's just a bit easier so if i do that v1 equals 5 eu minus v2 and now what i'm going to do is substitute that equation into here and i'll write it down on the left hand side so once i do that i will have u equals uh negative 3 times by 3 v1 in other words negative 3 times by 5 eu minus v2 plus 4 v2 let's expand the bracket so i'll have u equals negative 15 eu plus three v two plus four v two now remember i want to make v the subject so i can put seven v two i'll put those together equals um u plus 15 eu right we're almost there now um i can factorize so i've got 7v2 equals u and then in brackets 15 e plus 1 and then the last step is v 2 equals u over 7 15 e plus 1 as required okay so now we can move on to part b let's just separate this out there's a lot of working here don't want to mix stuff up so part b uh given that the direction of p is unchanged find the possible values of e now we need to look back at our question i assumed and my equations are based on v1 changing directions now if v1 its direction is unchanged then my value of v1 will be negative it will be less than zero so you've gotta watch out for that so i'll just write that down um if the direction of p is unchanged okay then my value of v1 will be negative which means that it will be less than zero so in other words v1 is less than zero so we're going to do is want to find out an equation for v1 or an expression for v1 and make that less than zero and then follow through and then make e the subject right so can i work out v1 uh yes i can i've got an expression for v2 which is here yeah and what i'm going to do i'm going to plug that value of v2 into now where did i have an expression for v1 here we go let's put it in here yeah it says we're missing about so that will be v1 equals 5 eu minus v2 now v2 is this you just go down 15 e plus one now we've said well that needs to be less than zero so 5 5 or 15 5 eu five e u minus u over seven fifteen e plus one it needs to be less than zero that means that it's changed actually opposite to the direction i've got so what we're basically saying is that it's going this way yeah so let's tidy this up which means expanding the brackets then refactorizing it to get the e out so if i expand a brackets higher 5 eu minus 15 over 7 eu minus um u over seven is less than zero right um now the five eu minus the 15 over 7 so 5 minus 15 over 7 is 20 over 7. so basically what i've got is 20 over 7 eu and then i'm going to move to u over seven across so i end up with less than u over seven next step um i'm going to divide both sides by u actually let's do that here so i've got u on both sides which i can divide by so that's gone so what i basically have now is 20 over 7 e is less than 1 over 7. so now e is less than so if i divide one over seven and divide that by twenty over seven i get one over twenty so that would be my range of values that the coefficient friction needs to be sorry not the coefficient friction coefficient restitution needs to be less than 1 over 20 for that to take place okay part c part c so in part c it says given that the magnitude of the impulse of p on q is 80 m u over 9. let's get rid of this we can see atm u over 9 find the value of e right in part c we're given the magnitude of the impulse of p on q which is 80 mu over 9 find the value of e now the magnitude of the impulse um on of p on q um or q on p it doesn't matter which one we choose if i draw what those impulses look like one here and one here it doesn't actually matter which one we use to work out the value of e i'm going to use q simply because i i know its initial velocity to you and the final velocity of uh v2 was given in the question now if i use p particle p or ball p to work out uh in order in my formula the issue is if v1 is wrong then um that actually means that the rest of my working is going to be wrong so i'm going to say here consider q so i'm going to draw a little diagram for q like this and i'm going to put on its velocity before which is 2 and that way it's velocity after now i worked that out didn't know v2 and it was given in the question u over 7 15 e plus one and the impulse is going to look like this that's the actual impulse of p on q or i could you work out the impulsive q on p get the same thing and that's given as 80 m u over 9. so i now use my uh impulse equal to change in momentum to let this out so impulse is change in momentum so when i do substitute that now in the impulse is atmu over nine so let's write that down at m u over nine that equals the massive kiosk as i should have written that down that was four m wasn't it it's mass for him so four m times by v so v is going to be um [Music] u over seven times by 15 e plus one okay now i'm taking um this direction as positive here the same direction as the impulse and then i subtract the momentum before so that's going to be 4m times by now it's going to be negative to u because of the direction it's going the 2u is going against my positive direction so times by negative 2u negative 2 u okay so the rest of this is just algebraic it's just actually um making e the subject working out what e is now um [Music] let's tidy this up so i've got 80 mu over 9 equals so i'm going to expand the brackets here so i'm going to get 4 times 15 which is 60 over 7 so i'll have 60 over 7 uh m u e new and then i will have plus then it's going to be four over seven m u okay that's expanding fully the first brackets then the second bracket is going to be plus 8 m u so the first thing i'll notice that every term has got mu in it so let's get rid of those like that i'm going to continue my working over here due to lack of space just about hopefully squeezing in here right so let's write down what we have now now that we've crossed all those mu's out so 80 over 9 equals 60 over 7 e plus 4 over 7 plus eight okay right i can put everything called in one size i'd have 60 over seven e okay that's going to be equal to 80 over nine minus eight minus four over seven so let's work that out so that's going to be um eighty over nine minus four over seven minus eight okay that gives me 20 over 63 so i should be able easily to work out e e is going to be equal to 20 over 63 20 over 63 divided by 60 over 7 and that leaves me with e equal to 1 over 27 so that's the value of e there so there's our final answer and it it it fits with what we we did before because we said that e needs to be less than one over twenty and it is if i got an answer of e greater than one over twenty twenty it's probably telling me i've i've not done something quite right i should now be able to do exercise four a on pages 73 to 76 of the textbook and i suppose the important thing here is that e equals the speed of separation over speed of approach and i suppose what we could say is that if we have two particles that are separating like this okay we've got v1 v2 so in this case v2 would be greater than v1 then to work out the speed of separation we do v2 minus v1 if our two particles were separating like this going in opposite directions then the speed of separation [Music] would be v1 plus v2 this and if we now look at our speed of approach if two things are approaching um like this one of them could be at rest by the way okay and we've got v1 and v2 in this case v1 would have to be greater than v2 to get closer to it then the speed of approach is going to be v1 minus v2 and if two things are approaching and they are approaching going in opposite directions like this then the speed of approach is going to be v1 plus vt