# Oxford PAT 2009, Question 26

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## 3 thoughts on “Oxford PAT 2009, Question 26”

1. Could an alternative solution be to use the horizontal acceleration a=F/m where F = eV/d so a = eV/md and t= d sqrt(2m/eV) (since d=½ a t^2) and then substitute it in the equation s = ½ g t^2 obtaining that s=½ g d^2 2m/eV ? The final solution differs from yours but I can’t see where the mistake is

1. I think the difference between our answers comes from a different reading of the question.

I think the question is saying that the electron is accelerated by the electron gun and then, after it comes out of the electron gun, it travels a distance d from the gun to the screen.

I think your answer is interpreting the question as saying that the electron travels a distance d through the electron gun, and hits the screen at the end of the gun. Given that interpretation I think your reasoning is correct. But I am pretty sure that this is not the intended interpretation.

Does that make sense to you?

1. Yes I understand your interpretation and after reading the question another time I think you are right, thank you.