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STEP 1999 Paper 2 Question 11
By Neil Morrison (P1462) on Thursday, August 3, 2000 - 07:27 pm:

An automated mobile dummy target for gunnery practice is moving anti-clockwise around the circumference of a large circle of radius R in a horizontal plane at a constant angular speed w. A shell is fired from O, the centre of the circle, with initial speed V and angle of elevation a. Show that if V2 < gR, the no matter what the value of a, or what vertical plane the shell is fired in, the shell cannot hit the target.

[because, of course, the shell cannot travel R, but I'm giving the whole question;]

Assume now that V2 > gR and that the shell hits the target and let b be the angle through which the target rotates between the time at which the shell if fired and the time of impact. Show that b satisfies the equation:

g2b4 - 4w2V2b2 + 4R2w4 = 0

[I got this just by adding the bits up]

Deduce that there are exactly two possible values of b.

[This is the bit I don't get. Why are there two values? The quadric equation would have to have its middle SP below the b axis or have the other two SPs both sitting on the b axis. And the former is not possible as the constant is < 0.]

Let b1 and b2 be the possible values of b and let P1 and P2 be the corresponding points of impact. By considering the quantities (b12 + b22) and b12b22 or otherwise show that the linear distance between P1 and P2 is 2Rsin{[w/g](V2 - Rg)1/2}

Neil M

By David Loeffler (P865) on Thursday, August 3, 2000 - 10:58 pm:

Well, I suppose that the quartic can be regarded as a quadratic in b2. This will have either two, or zero, positive roots since the constant term is positive. In fact, it has exactly two, because V2>gR, so the minimum is less than the minimum of g2b4 - 4gb2rw2 + 4r2w4 = (gb2 - 2rw2)2 = 0.
Thus there are two distinct positive values of b2, giving rise to two distinct positive (and two negative) roots of the quartic.

Hope this helps.

David

By Neil Morrison (P1462) on Friday, August 4, 2000 - 10:18 am:

I think I had got to that stage, but I was trying to win the logic battle. But the problem is it says exactly two possible values. And in your solution, we have 2 positive and 2 negative, the negative ones (say A and B) can be regarded as 2p + A and 2p + B. The only way I could resolve this is to show that the "2p +" values end up being equal to the original positive values, so in effect there are only two distinct solutions.

Thanks anyway

Neil M

By David Loeffler (P865) on Friday, August 4, 2000 - 01:55 pm:

That would actually be impossible, surely, since p is transcendental? But the physical situation implies that we're not looking for "position" angles which are mod 2p, but for the angle the target has actually moved through. This will not be mod 2p as, for example, 3p would correspond to one and a half circuits, not to half a circuit. The target won't turn around and go backwards, so perhaps that is why we can discard the negative solutions.

By Neil Morrison (P1462) on Friday, August 4, 2000 - 04:43 pm:

But the net position of the target will be the same after 3p as if it had moved p backwards.
However, I agree that this will not work. It was the only way I could think of to justify the question. I think its badly worded.

Neil M