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 5.6 ×106m/s. To cover the entire tumor area, the ions are deflected sideways by passing them between two charged metal plates that 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.
I used the equation s= vt + .5at^2
where s is the change in position, v is initial velocity, t is change in time and a is acceleration. I did it twice, once for the x direction and once for the y direction. Using the x one, i found time to be 2.679x10^-7, and with the y one a=4.126x10^13 but i guess that's not right so i don't know where i went wrong.
2 answers
10
Realize that the acceleration only occurs while between the 2 plates so you must calculate the time in the plates which is 8.9 x 10^-9 sec. Also, you must determine what vertical velocity is needed to reach total distance of 2 cm during the time it takes to travel 1.5 m. Then determine what accel is needed during the time ion is between the plates to produce that velocity. Remember also that the ion will have already moved vertically a small distance during the time it is accelerating, so the total distance it has to travel is less than 2 cm, ie., (2 - x)cm.
Desired velocity = (2 -x cm)/2.68x10^-7)
accel needed=desired velocity/time between plates (this will be in terms of x)
Then use the equation s=(1/2)at^2, remembering that s = x
Desired velocity = (2 -x cm)/2.68x10^-7)
accel needed=desired velocity/time between plates (this will be in terms of x)
Then use the equation s=(1/2)at^2, remembering that s = x
1 answer