Asked by VJC
An electron starting from rest acquires 4.20 keV of kinetic energy in moving from point A to point B. How much kinetic energy would a proton acquire, starting from rest at B and moving to point A?
Determine the ratio of their speeds at the end of their respective trajectories.
Determine the ratio of their speeds at the end of their respective trajectories.
Answers
Answered by
Damon
I assume that the energy comes from asn E field between A and B.
The force is proportional to charge and E so the force magnitude is equal (opposite sign though )
The distance is the same so the work done is the same.
So the kinetic energy is the same (1/2) m v^2
so
(1/2) Mproton Vproton^2 = (1/2) Melectron Velectron^2
so
Vproton^2/Velectron^2 = Melectron/Mproton
The force is proportional to charge and E so the force magnitude is equal (opposite sign though )
The distance is the same so the work done is the same.
So the kinetic energy is the same (1/2) m v^2
so
(1/2) Mproton Vproton^2 = (1/2) Melectron Velectron^2
so
Vproton^2/Velectron^2 = Melectron/Mproton
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