[BL1993]

Times in case one: t1, t, t2
Times in case 2: T1, T, T2

I probably shouldn't have sued capitals and should have called them x, y and z sorry about that.

[japester]

@BL1993, I've just been tinkering around with Q1 (b) (iii) again (a sucker for punishment!) and my intuition tells me that the answer you're getting of 2t cannot be right because the time for the original journey is actually t+t1+t2 = 2t and the time for the journey with the speed limit must intuitively be longer than 2t. I am taking the distance to be covered as 3vt/2 (based on the original journey) and then I note that the time for the acceleration part of the second journey takes (2/3)t1 (this is the same t1 as in the first journey) and the time for the deceleration for the second journey takes (2/3)t2 (again, this t2 refers to the time for deceleration in the first journey - intuitively, these times are lower than the original times, as expected, because it will take less time for the body to get to the lower speed given the same acceleration and likewise for the deceleration.

So then I form an equation for the distances of the 2 journeys to get the time T, where the car is traveling with constant velocity in the 2nd journey. So, 3vt/2 = 2vt1/9+2vt2/9+(2v/3)T, giving me a value for T of (23/12)t (I am leaving t1+t2 = t here). Now, I can find the total time for the second journey by adding this to the accceleration and deceleration times to get (23/12)t + (2/3)t = (31/12)t. So the time for the second journey is a little over 2 and a half times the time taken for the original journey.

I could still be wrong about this final answer but I've a very strong feeling that the time must be greater than 2t, based on intuition alone. If I'm wrong, I'd be very interested to see anyone elses solution.



@ everyone

I've just done Q2 and the answers I'm getting match with those of a few people. They are:

(a) -i - 8j , shortest distance = 49.61m (b) 28.96deg<=theta<=60deg

[MedMan101]

Originally Posted by japester

@BL1993, I've just been tinkering around with Q1 (b) (iii) again (a sucker for punishment!) and my intuition tells me that the answer you're getting of 2t cannot be right because the time for the original journey is actually t+t1+t2 = 2t and the time for the journey with the speed limit must intuitively be longer than 2t. I am taking the distance to be covered as 3vt/2 (based on the original journey) and then I note that the time for the acceleration part of the second journey takes (2/3)t1 (this is the same t1 as in the first journey) and the time for the deceleration for the second journey takes (2/3)t2 (again, this t2 refers to the time for deceleration in the first journey - intuitively, these times are lower than the original times, as expected, because it will take less time for the body to get to the lower speed given the same acceleration and likewise for the deceleration.

So then I form an equation for the distances of the 2 journeys to get the time T, where the car is traveling with constant velocity in the 2nd journey. So, 3vt/2 = 2vt1/9+2vt2/9+(2v/3)T, giving me a value for T of (23/12)t (I am leaving t1+t2 = t here). Now, I can find the total time for the second journey by adding this to the accceleration and deceleration times to get (23/12)t + (2/3)t = (31/12)t. So the time for the second journey is a little over 2 and a half times the time taken for the original journey.

but you got t1+t+t2 wrong! its equal to t not 2t! and then you simply sub in 2v/3 for v in the original equation to get 2t in part (iii)

[BL1993]

Oh damn, 2t is the time i get for the time it travels at constant velocity the second time >.<. Let me do it out again. :L

[BL1993]

I did the question out again and I get the same result as you japster. 31t/12 is correct.

[japester]

@medman101, I'm taking it that from (ii) t1+t2 = t. Therefore, for the first journey, the total time spent traveling is t1 + t + t2 = t + (t1+t2) = t + t = 2t. Hope this helps to see where I'm coming from

[japester]

I've just done Q3 and Q4. The answers I'm getting seem to be consistent with a number of people. They are

Q3(a) (i) 50 (ii) 71.56 deg (b) (i) 56.31 deg (ii) 0.7071 (that was a long question to do 3 hours would be much better for this exam - one slip and you're done for!!)

Q4(a) 2.1528m/s (b) 6 simultaneous equations the tensions are 73N, 24N and 21.9N

[Geog ariphic]

Originally Posted by japester

@ everyone

I've just done Q2 and the answers I'm getting match with those of a few people. They are:

(a) -i - 8j , shortest distance = 49.61m (b) 28.96deg<=theta<=60deg

I remember getting everything there except 60 degrees, and if you've confirmed this with others then i assumed i rushed it an got it wrong.
Thanks

[randylonghorn]

Lord's sake, lads, would ye forget about it and go do something fun!

[MedMan101]

Originally Posted by randylonghorn

Lord's sake, lads, would ye forget about it and go do something fun!

Randylonghorn: Posts: 15,269.....says you..!

[randylonghorn]

Over 6 years!

And none of them are dissecting an applied maths exam the Saturday night after my LC finished, I'll guarantee you that!

[Geog ariphic]

Now Randy, calm yourself.
Noone forced you to come here. Some of us are interested in doing mathemtical and/or engineering or just science courses next year, and some of us just can't leave a problem alone until we've solved it satisfactorily.
Let us have our little chat.

[japester]

I've just done Q5 and I got the following answers:

(a)

(i) To prove this, I used the coefficient of restitution formula (NEL) and the PCM formula to get 2 equations. Using the NEL formula I end up with Vq = 2ue + Vp and using the PCM formula I end up with Vq = -(1/3)u - (2/3)Vp. I equate these to end up with Vp = (-u/5)(1+6e). We can be certain that the sphere P will rebound provided that the value of Vp is negative (it was going in the positive direction to begin with along +ve x axis). Vp can only be negative if (1+6e) > 0. As the range of allowable values of e are 0<e<1 then it must be the case that 1+6e is always positive and so we can conclude that the sphere P rebounds for all value of e.

(ii) To prove this I follow a similar path to get equations involving Vp using NEL and PCM. Then I end up equating these to get an expression for Vq in terms of e. This time it is Vq = (-4/5)(1-4e). For sphere Q to rebound this must turn out to be positive (as it was travelling in the -x direction before impact). So (1-4e) must turn out to be negative. This can only happen if e>0.25. So the range of values of e for which Q will rebound is 0.25<e<1

(b)

(i) I am ending up with k = (1-e)/2 here

(ii) e = 0.75

[Geog ariphic]

i had the same answer for k.

For the first part, Vp, i had that or something quite similar, but it was positive! I just lied and said it was the opposite direction, but obviously it affected everything else and i couldnt get a real value of e. (i.e. I believe i obtained 14.43... As ONE of its possible value)

[godtabh]

Originally Posted by randylonghorn

Over 6 years!

And none of them are dissecting an applied maths exam the Saturday night after my LC finished, I'll guarantee you that!

Boards wasnt invented the saturday the day after my applied marys exam. Great subject. Still have murphy's book and look over it from time to time. Long living cert though