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# Can you check my answers (Circular motion, particle accelerator) watch

1. I think I have got the wrong answers for this question because they don't seem right.

You are told that:
• Protons are injected into a proton synchrotron at a speed of 8.0 x 106 ms-1.
• Protons have mass of 1.7 x 10 -27 kg
• Diameter of path taken by protons is 400m (so radius is 200m)

Part (a) : Calculate the force that has to be provided to produce the circular path.

I used F= mv2/r and got 5.44 x 10-16 N as my answer but idk if it's right.

Part (b) : Before reaching their final energy the protons travel around the accelerator 420,000 times in 2.0 s. Calculate the total distance travelled by a proton in the 2.0s time interval.

For this, I used v = 2πr/T where T = (2.0/420000) , and rearranged to make r the subject. I got 6.06 m as my answer which doesn't seem right.

If I just use T = 2.0 s then the answer I get is 2.5 x 106 m which sounds about right but then isn't T supposed to be the time period for just one orbit?

Part (c): Unless a vertical force is applied the protons would fall as they move through the horizontal channel. Calculate the distance a proton would fall in two seconds.

For this, I used suvat and ended up getting 1.6 x 107 m as my answer which definitely doesn't sound right considering the diameter of the circular synchrotron is only 400m...
I got this answer for whichever suvat equation I used.
For the second I get 5.28x10^8 m, to 3sf. Circumference of the circle multiplied by number of passes
For the third I get 19.6m, using s=ut+0.5at^2, where a=g, t=2,u=0
3. (Original post by jessyjellytot14)
Part (b) : Before reaching their final energy the protons travel around the accelerator 420,000 times in 2.0 s. Calculate the total distance travelled by a proton in the 2.0s time interval.

For this, I used v = 2πr/T where T = (2.0/420000) , and rearranged to make r the subject. I got 6.06 m as my answer which doesn't seem right.

If I just use T = 2.0 s then the answer I get is 2.5 x 106 m which sounds about right but then isn't T supposed to be the time period for just one orbit?
2.0 s is not the period. 2.0 s is the time that the proton travel around the accelerator for 420 000 times. Using the method proposed by the an_atheist, you would arrive the correct answer.

(Original post by jessyjellytot14)
Part (c): Unless a vertical force is applied the protons would fall as they move through the horizontal channel. Calculate the distance a proton would fall in two seconds.

For this, I used suvat and ended up getting 1.6 x 107 m as my answer which definitely doesn't sound right considering the diameter of the circular synchrotron is only 400m...
I got this answer for whichever suvat equation I used.
Good to know that you are exercising your "common sense"...
4. (Original post by Eimmanuel)

Good to know that you are exercising your "common sense"...
I do try
5. (Original post by jessyjellytot14)
I do try
Don't be surprised that many A students do not use it in doing physics problems and then claim that they understand what they are calculating.

All the best for your revision and coming exam!

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