Applied Mathematics: Linear Motion Problem

alan4cult

I've a problem with the 1997 Q1(b)(iii) in Applied Maths
Here's the question:


I can do the first two parts but the last part is proving difficult. What causes the particle to return to q? It's not a collision so I'm led to believe it decelerates until the final velocity is 0 and then accelerates back towards q but I'm unsure what I have to do to answer the last part. Can somebody post a solution?

Thanks,
Alan

alan4cult

It's ok I worked it out. I'm still wondering about other ways to do the last part?
I may as well post a solution for anyone who wants it.
.
(b)(i)
This part is fairly easy and there are many ways to do it. Let the distance from p to the point of collision be S1 and from Q to the same point S2. Therefore S1+S2=d

Particle P
u=2u
a=f
s=S1
t=T(same for both)

s=ut+[1/2]at^2
S1=2uT+[f/2]T^2

Particle Q
u=3u
a=-f
s=S2
t=T

s=ut+[1/2]at^2
S2=3uT-[f/2]T^2

d=S1+S2=2ut+[f/2]T^2+3uT-[f/2]T^2
d=5uT
T=d/5u

(b)(ii)
Before the second particle comes to rest i.e. before v=0 i.e. v>0

v=u+at
v>0

u+at>0
3u-[fT]>0
3u-f(d/5u)>0
Multiply by 5u
15u^2-fd>0
fd<15u^2

(b)(iii)
This is the one that caught me out. Firstly the second particle goes towards p, then comes to a stop and heads back towards q. The collision will occur at q if fd=30u^2 so we need to prove this. The best way is to forget about the second particle and concentrate on what the first particle is doing.

FOR FIRST PARTICLE
u=2u
a=f
t=d/5u
s=d(as it travels the complete distance from p to q)

s=ut+[1/2]at^2
d=2u(d/5u)+(f/2)(d^2/25u^2)
d=d[2/5+fd/50u^2) --- Factor out d
1=2/5+fd/50u^2
50u^2=20u^2+fd
fd=30u^2

That seems enough to get the full marks but I'm still open to suggestions on other ways to this part (b)(iii).

NoQuarter

http://boards.ie/vbulletin/forumdisplay.php?f=380
youll have better luck here mate!

ZorbaTehZ

You could just argue that since it's constant acceleration and therefore symetric, it would take twice the time it would take Q to return to q. Therefore:

d/5u = 2(3u/f)
d/5u = 6u/f
fd = 30u^2

alan4cult

Yep, that's pretty much how I was first thinking but, I wasn't sure what was causing the particle to return. When I think about it you could just say that s=0 at q

s=ut+[1/2]at^2

0=3ut-f/2t^2
0=t(3u-f/2t)
t=0
or
3u=f/2t
6u=ft
When t = d/5u
6u=f(d/5u)
30u^2=fd

Sorry, a bit long winded with the equations but since yesterday that came to me and I hadn't written it down so had to work it here to make sure it was right.