A proton follows a path with initial speed v0 = 2.4×10^6 m/s and passes through a point P with speed wanted by energy conservation; what is the speed at P?

• A. 0.6×10^6 m/s
• B. 1.2×10^6 m/s
• C. 2.4×10^6 m/s
• D. 3.6×10^6 m/s

Answer: (B)

Energy conservation for a charged particle moving in an electric field says that the total mechanical energy—the sum of kinetic energy and electric potential energy—stays the same as it travels. The proton starts with kinetic energy K_i = (1/2) m v0^2. If, at point P, its speed is v, then its kinetic energy is K_f = (1/2) m v^2. Since energy is conserved, any decrease in kinetic energy must be exactly balanced by an increase in electric potential energy.

If the speed at P is half of the initial speed, v = v0/2, then the final kinetic energy is K_f = (1/2) m (v0/2)^2 = (1/4) K_i. That means the kinetic energy has dropped to one quarter of its initial value, and the remaining energy has become electric potential energy. Plugging in v0 = 2.4×10^6 m/s, halving gives v = 1.2×10^6 m/s, which is the speed at P.