# Oxford PAT 2013, Question 21

## 12 thoughts on “Oxford PAT 2013, Question 21”

1. Beatrice says:

Hello.
Would it have been correct to use v² = u² +2as (one of the kinematic equations) and assume that ma = f thus a=f/m? The answer is the same, but I’d like to know if it would have been considered as correct or if there are similar situations for which this wouldn’t work..

Thank you very much for this extremely useful site!

1. Hi Beatrice — yes it is absolutely fine to use the kinematic equation v² = u² +2as here. The only real condition for using this equation is that a is a constant, which it clearly is here.

2. Lucy smithy says:

Hi,

Where did the value of -2 come from?

Thanks

1. You mean where did the -2 come from in part b: y ordinate of detection = -2R? The particle starts at y = 0 and moves in a circle of radius R with the detector as a diameter , therefore the particle hits the detector again at the ‘other side’ of the diameter, which is a distance of twice the radius away. Therefore the particle hits the detector at y = -2R.

Is that what you meant?

3. Hi Joel — yes, sorry I must have had a bit of a brain fade here because this answer is just full of major stupid errors. I will redo it.

4. Rachida says:

Hi, for part (b) why there is a minus 2r ( y=-2r)? The y axe is directed downwards, So the coordinate y will be positive. Can you explain that for me?

1. Hi Rachida, thanks for your comment.

Yes well spotted, there is a little arrow on the bottom of the y axis, which presumably means that the y axis is supposed to point in the opposite direction to the standard direction. I didn’t notice this so I wrote the answer as a negative displacement along the (standard) y axis. If the y axis is pointing down, the displacement should be a positive quantity.

Thanks for pointing this out.

5. Retnemmoc says:

For part (a) you have the increase in ‘v’as ft/m, yet you’ve equated it to as sqrt(2dm/f), which is just ‘t’ according to your working, unless I’m missing something.

1. Hi, thanks very much for the correction — it’s the same error as described immediately above. I’ve fixed it now.

6. ghghuy says:

Hello! My answer for (a) is slightly different. The velocity increase term I got was sqrt(2fd/m). I think that if you take a look at the way you set your term, you’ll see that you made a slight algebraic error. Please do correct me if I misunderstood, however. I used the relationship “work done = change in kinetic energy”, so I may have made a mistake.

1. Hi, thanks for the correction. You are absolutely right of course: the last step was a stupid mistake. I have fixed the image so it’s right now.

7. The answer to part (c) is a bit messy so here is a more detailed explanation. The pixels on the screen can be considered to be bins that are little squares of side Δy. Two beams will only be distinguishable if they hit different pixels. To ensure that two beams will definitely always hit different pixels they need to be at least Δy apart when they hit the screen. Since we derived a formula for y in terms of v the previous part, we can work out Δy in terms of Δv and then solve for Δv.