Paper III number 13. (first part) Would someone like to finish it please?
I'm only going to do a sketch, because I don't think this is very difficult and you should be able to manage it.
If the particle hits completely inelastically, then the vertical velocity immediately after the collision is 0. So the time taken to reach the ground is g2h. The horizontal velocity is unchanged at 8ghcosα and so the distance travelled after the collision with the ceiling is 4hcosα. Obviously the distance travelled before the collision is the same as in the elastic collision case.
After a bit of algebra grinding, you end up with D=4hcosα[2sinα−4sin2α−1−1]
From here, just differentiate in the obvious way and group terms and you should be able to get something manageable. You end up with a term involving 4sin2α−1 and one involving 4sin2α−11 and I found it helpful to rewrite the 1st term as 4sin2α−14sin2α−1 so those terms could be grouped.
In my working, I did actually get to where you can see that π/4 would be a root for the derivative, but you can always substitute sinα=cosα=π/4 to show you get 0.
I also got to the point where you could demonstrate algebraically that D′(π/4−ϵ)>0,D′(π/4+ϵ)<0 and so we definitely have a maximum. If you had to, you could always demonstrate this using a calculator, as they were allowed back then.
I'm only going to do a sketch, because I don't think this is very difficult and you should be able to manage it.
If the particle hits completely inelastically, then the vertical velocity immediately after the collision is 0. So the time taken to reach the ground is g2h. The horizontal velocity is unchanged at 8ghcosα and so the distance travelled after the collision with the ceiling is 4hcosα. Obviously the distance travelled before the collision is the same as in the elastic collision case.
After a bit of algebra grinding, you end up with D=4hcosα[2sinα−4sin2α−1−1]
From here, just differentiate in the obvious way and group terms and you should be able to get something manageable. You end up with a term involving 4sin2α−1 and one involving 4sin2α−11 and I found it helpful to rewrite the 1st term as 4sin2α−14sin2α−1 so those terms could be grouped.
In my working, I did actually get to where you can see that π/4 would be a root for the derivative, but you can always substitute sinα=cosα=π/4 to show you get 0.
I also got to the point where you could demonstrate algebraically that D′(π/4−ϵ)>0,D′(π/4+ϵ)<0 and so we definitely have a maximum. If you had to, you could always demonstrate this using a calculator, as they were allowed back then.
Would it be sufficient to argue that the difference in ranges is zero when particle just touches ceiling and also when α=π/2 so α=π/4 must be a maximum.
OK, I wasn't sure if that was what was stopping you.
From there, what I'd do is resolve horizontally and vertically for the forces at A, splitting the rod reaction into Rx and Ry. Then you know 13Rx=21Ry, which is enough to let you solve for T, and from that, solve for k.
Edit: crossing of posts doesn't make it terribly obvious, but this is in response to Q14,
Would it be sufficient to argue that the difference in ranges is zero when particle just touches ceiling and also when α=π/2 so α=π/4 must be a maximum.
Not to my eyes, no - there's no reason to expect the different to behave 'symmetrically'.
Not to my eyes, no - there's no reason to expect the different to behave 'symmetrically'.
But there must be a maximum in between the two zeros and as this is the only time the derivative is zero surely we can conclude that it is a maximum. It does not require any assumption of symmetry.
But there must be a maximum in between the two zeros and as this is the only time the derivative is zero surely we can conclude that it is a maximum.Sorry, I misunderstood you. I thought you were claiming the derivative must be zero at pi/4 simply because the difference was zero at the point where it touches the ceiling and when theta = pi/2.
What you are saying is fine if you can show pi/4 is the only place where the derivative is zero. At the point I got to, it was equally easy just to show the derivative had the "/-\" behaviour near pi/4 (which I tend to prefer as being a more "direct", though it doesn't really matter).
OK, I wasn't sure if that was what was stopping you.
From there, what I'd do is resolve horizontally and vertically for the forces at A, splitting the rod reaction into Rx and Ry. Then you know 13Rx=21Ry, which is enough to let you solve for T, and from that, solve for k.
Edit: crossing of posts doesn't make it terribly obvious, but this is in response to Q14,
I don't think anyone has posted a solution to this question yet, I had a quick look and couldn't find anything, but sorry I someone has already done it .
Ahh, I see now - on some paper sites the older STEP papers are mislabeled. I think I'm correct in thinking that you have just done STEP I question 8 .
aahh ok sorry about that i thought it seemed like an easier question, just imagined it would be III because of the t substitution.. oh well, now i know at the very least i got the question right !
I've just looked at this really quickly (I'm off to college soon), but I think there is something wrong with your first bit: I don't think that the volume halves when the snowman is halve its initial height (which from your working it seems to). I think you need to find differential equations for the radii of the head and the body, to find the time at which the snowman is half its height. Ive got to go now, but my answer for the ratio is:
I don't think it's asking for the volume to halve (at least not on the version of the paper that I have). It says "When Frosty is half his initial height, find the ratio of his volume to his initial volume". And half height should mean half radius if the snowballs are right on top of each other. Or am I being stupid? (pretty likely!)
I don't know if it's relevant but somewhere in my working I got t=81(4π3VU)32+c (after integrating the DE for the volume of the upper sphere). This could possibly help to determine at which time the upper snowball melts completely - but the problem is I don't see how to find the constant, hence this bit is completely useless atm.
I think the problem is, the head could melt slower or faster than the body, which means you could reach half the initial height without the head or body necessarily being exactly half the respective radius.
When i attempted this question, this is what plagued me... I don't think you can assume that the two spheres are going to amount to half the initial height and be half the initial radii.