An atomic nucleus of mass $m$ traveling with speed $v$ collides elastically with a target particle of mass $2m$ (initially at rest) and is scattered at $90 ^\circ$.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Particle one with mass m is approaching the second particle with mass 2m which is initially at rest and particle one is approaching with a speed of v 1 and we'll say that the x-axis goes along the original velocity of particle one and the y-axis is straight up. So momentum is conserved in the x-direction and it's also conserved in the y-direction which means that the total momentum after the collision in the x-direction equals the total momentum of the system before the collision in the x-direction and then the same is true for the y-direction. So in the x-direction, we have the total momentum before is mv 1 and then afterwards, the total x-momentum is a component of particle two in the x-direction because particle one has no momentum in the x-direction after the collision since it goes straight up. So we'll take the x-component of this momentum here which is gonna be cos Θ 2 since it's the adjacent leg we are interested in times the momentum of particle two which is its mass 2m times its speed v 2 prime. And then in the y-direction, there should be no momentum in the y-direction after the collision since there's no momentum in the y-direction before the collision since this particle is moving entirely on the x-direction. Now to say that there's no y-direction momentum doesn't mean things don't move in the y-direction, in fact they are, but the thing is that they move such that the momenta is equal in the opposite directions along the y-axis. So we have mv 1 prime since we are told that this thing goes straight up; its momentum after collision is entirely in the y-direction, upwards, and that's gonna equal the momentum of particle two y-component downwards. So that is mass 2m times its speed v 2 prime times sin of Θ 2— sin gives us the opposite leg of this momentum triangle. And then since the m's cancel, we can reduce the equation 1 to this point here we'll get v 1 equals 2 times v 2 prime times cos Θ 2 and then in the y-direction, we can cancel out the m's and get the v 1 prime is 2 times v 2 prime sin Θ 2. And we have a third equation we can work with too which we need because we have three unknown's; we don't know v 1 prime, we don't know v 2 prime and we don't know Θ 2 so we need a third equation to help deal with three unknown's and that's gonna come from energy. We can say that kinetic energy of the system beforehand which is just the kinetic energy of particle one since particle two is not moving before the collision; we are gonna say mv 1 squared and never mind the one-half because it's gonna cancel everywhere so I didn't bother writing the one-half. So it's mv 1 squared equals m times v 1 prime squared that's the kinetic energy of particle one after the collision, you know, without writing the one-half and then plus the mass of particle two which is 2m times its speed squared after the collision and then the m's also cancel—I guess I didn't need to write that either but. So we have equation 3 is v 1 squared equals v 1 prime squared plus 2v 2 prime squared. So now we have an algebra issue where we need to eliminate some of the unknown's in order to solve for one of them. Let's eliminate this Θ 2; we are gonna use some trigonometry trickery by saying that cos squared of any angle plus sin squared of that same angle equals the number 1. And so we are gonna take equation 1, square it and then add it to equation 2 squared and we'll end up from equation 1, we'll have 4 times v 2 prime squared times cos squared Θ 2. And then from equation 2, we'll have 4 times v 2 prime squared times sin squared Θ 2 and on the left-hand side, we have v 1 squared and then we have v 1 prime squared. And these two terms have a common factor, 4v 2 prime squared, and we can factor that out which gives us cos squared plus sin squared in brackets which is this trigonometry identity, the number 1. So this whole right hand side now is just 4v 2 prime squared. And then we can subtract the v 1 squared from both sides or move to the right hand side, whichever way you like to say it, and then that gives us v 1 prime squared is 4v 2 prime squared minus v 1 squared. Of course that doesn't quite answer what is v 1 prime but it does achieve an elimination of this Θ 2 variable and so that means that we can turn our attention to equation 3 and get rid of this v 2 prime and then we'll have an equation containing only v 1 prime and v 1, which we know. So equation 3 rewritten slightly to solve for v 2 prime, we can move this to the left hand side which makes v 1 squared minus v 1 prime squared then divide both sides by 2 giving us v 2 prime squared equals v 1 squared minus v 1 prime squared all over 2 and then we substitute that into this part here so I have rewritten this line here but instead of v 2 prime, I have written or I should say, instead of v 2 prime squared, I have substituted for v 2 prime squared by writing v 1 squared minus v 1 prime squared all over 2 in red there. And we'll reduce the 4 over 2 to 2 and then distribute that into the brackets so that gives us 2 times v 1 squared minus 2 times v 1 prime squared minus v 1 squared and this 2v 1 squared minus v 1 squared makes just v 1 squared and then move this to the left hand side by adding it to both sides gives us 3v 1 prime squared and then divide both sides by 3, take the square root of both sides and we get that v 1 prime is the speed of the particle one before the collision divided by square root 3, there, and then v 2 prime squared, we can look back over here and now substitute our answer for v 1 prime squared that we just figured out, v 1 squared over square root 3 but since we are squaring this, we'll use this one so that's v 1 squared over 3 plugged in for v 1 prime squared and you know we divide both of them by 2 and then that's v 1 squared over 2 minus v 1 squared over 6 give it a common denominator by multiplying this by 3 over 3, you get 3v 1 squared minus v 1 squared all over 6 which is 2 over 6 and then take the square root of both sides after reducing that 2 over 6 to 1 over 3 and you get v 2 prime is v 1 over root 3; that happens to be the same thing as v 1 prime. And I ended up answering part (b) before part (a) but that's fine so now we go look at part (a) and take one of these equations, whichever one you like, for the x-direction or the y-direction, doesn't matter which you choose. And I have rewritten it here; v 1 is 2 times v 2 prime times cos Θ 2 and solve for cos Θ 2 by dividing both sides by 2v 2 prime. And so that's v 1 divided by 2 over v 2 prime which is the same as v 1 over 2 multiplied by the reciprocal of v 2 prime— the reciprocal of it is root 3 over v 1— and then the v 1's cancel giving us that cos Θ 2 is square root 3 over 2. So that means Θ 2 is the inverse cosine of square root 3 over 2 and that happens to be from one of our special triangle's where this is 90, 30, 60, 1, 2, root 3 so inverse cos of root 3 over 2 is this adjacent over hypotenuse which is 30. And next question is what fraction of the initial kinetic energy is transferred to the target particle which is the one that was initially at rest which we have called particle two here. So that's gonna be the kinetic energy of particle two after the collision because it has no kinetic energy to begin with so its change in kinetic energy is just gonna be whatever its final kinetic energy is after the collision. So that's one-half times its mass of 2m times its speed v 2 prime squared divided by one-half mass of the incoming particle times its speed before collision squared and the one-halves cancel, the m's cancel and we are left with 2 times v 2 prime which is v 1 over root 3— v 2 prime is v 1 over root 3— and we have to square that and then times by 1 over v 1 squared; instead of having fractions on top of fractions, I'd like to keep it all in one line and so we have only one numerator and one denominator to work with. So instead of dividing by v 1 squared, I'm multiplying by 1 over v 1 squared. So we can see that they cancel here and this is 2 over 3 because the square root 3 is gonna get squared so we have the fraction of kinetic energy transferred to particle two is two-third's of the initial kinetic energy.