Projectile Motion bombarding cancer tumors

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SUMMARY

The discussion centers on calculating the required acceleration for high-energy ions to deflect sideways while bombarding cancer tumors. The ions travel at a speed of 5.0 x 10^7 m/s and must move 2.0 cm sideways after passing through a 5.0 cm long acceleration region. The initial calculations yielded an acceleration of 4.4 x 10^-11 m/s², but the expected answer is 6.52 x 10^12 m/s². Participants suggest reviewing the application of kinematic equations, particularly yf = yi + vi t + 1/2 a t², to clarify the discrepancy.

PREREQUISITES
  • Understanding of kinematic equations, specifically yf = yi + vi t + 1/2 a t²
  • Basic knowledge of projectile motion principles
  • Familiarity with units of measurement in physics, particularly meters and seconds
  • Experience with problem-solving in physics contexts
NEXT STEPS
  • Review the application of kinematic equations in projectile motion scenarios
  • Study the effects of acceleration on particle trajectories in physics
  • Learn about high-energy ion bombardment techniques in cancer treatment
  • Explore error analysis methods in physics calculations
USEFUL FOR

Physics students, medical physicists, and researchers involved in cancer treatment methodologies will benefit from this discussion.

Grapz
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Homework Statement


You are asked to consult for the city's research hospital, where a group of doctors is investigating the bombardment of cancer tumors with high energy ions. The ions are fired directly toward the center of the tumor at speeds of 5.0 x 10 ^ 7 m/s. To cover the entire tumor area, the ions are deflected side-ways by passing them between two charged metal platest hat accelerate the ions perpendicular to the direction of their initial motion. The acceleration region is 5.0 cm long, and the ends of the acceleration plates are 1.5 m from the patient. What acceleration is required to move an ion 2.0 cm to one side?


Homework Equations


d = vt (1)
yf = yi + vi t + 1/2 a t^2 (2)


The Attempt at a Solution



I find the time it takes for it to hit the target
d / v = t
1.5 / 5.0 x 10 ^ 6 = 3.1 x 10 ^ - 7 s

Next i do use equation 2.

0.02 = 1/2 a ( 3.1 x 10 ^ -7) ^ 2
a = 4.4 ^ 10 - 11 m/s^2

The answer in the book says the answer is 6.52 x 10 ^ 12

Can someone help me out and straighten things out. The question i find is a bit ambigious, not exactly sure if i interpreted it right. thanks
 
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maybe you could try working backwards to see what you've done wrong.
i think that you've used some values in the wrong places
 
I have tried to work backwards.

When i do that, i get that the y value is 0.3 :/
 

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