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AS Physics Question: Projectile Motion + Motion questions in General

I know this should be in the study help physics section but there are less people browsing that forum and so less possibility of getting help!

1. Protons in a particle accelerator drop 1.2µm over the 1700m length of the accelerator. How fast are they travelling?

Help!

I'm struggling with these sort of questions a lot ... is there any trick or something to remember how to do these questions... i have never understood to this day how to do questions to do with constant acceleration and SUVAT equations etc.

But lets just start by seeing this question and if i understand how to do it!

Help!
Thanks!

Reply 1

spex

Well they only give you useful information, so start with that and work up.

I see it fell a bit due to gravity while it was going around the accelerator, so you could find the unknown in s = ut + .5at^2 (time) and then you have a quadratic in terms of t (you know u = 0 and a = 9.81).

Then you have the time taken to fall that small amount, which can then be used to find the speed it travelled with horizontally, ie s= ut+.5at^2 again. (this time a = 0, because we're considering lateral movement)

And finally, because they want the resultant velocity (taking account of vertical and horizontal speed), it's a matter of finding the vertical speed with a suitable equation. You already know the horizontal speed, so now you can find the resultant with some pythagoras, and if you were so inclined, the direction it travels in.

As a side note, the resultant velocity will probably be almost identical to the horizontal velocity, as the vertical component is so small relative to the horizontal. It'll vary of course, but only after a lot of decimal places.

Reply 2

neomilan

spex

I see it fell a bit due to gravity while it was going around the accelerator, so you could find the unknown in s = ut + .5at^2 (time) and then you have a quadratic in terms of t (you know u = 0 and a = 9.81).

is s = 1.2µm or 1700m

Reply 3

Zygroth

Work with the horizontal and vertical directions separately. Vertically, you know s = 1.2 x 10^-6, a = 9.8 and u = 0. You can find t, which must be the same for both vertical and horizontal motion. Then you have a = 0, s = 1700 and t horizontally, so you can work out the speed.