# [SOLVED] A proton follows the path shown in (Figure 1) . Its initial speed is v0 = 1.6×106 m/s .?

A proton follows the path shown in (Figure 1) . Its initial speed is v0 = 1.6×106 m/s .?

What is the proton’s speed as it passes through point P?

For neatness and convenience, I’ll use the notation 1.6E-19 (for example) instead of 1.6×10⁻¹⁹.

________________________________

Proton’s initial kinetic energy K1 = ½mv₀²

= ½ x 1.67E-27 x (1.6E6)²

= 2.138E-15J

The potential energy at the start point P1 = kqQ/r

= 9E9 x 1.6E-19 x -10E-9/0.003

= -4.8E-15 J

The potential energy at the end point P2 = kqQ/r

= 9E9 x 1.6E-19 x -10E-9/0.004

= -3.6E-15 J

Since energy is conserved:

K1 P1 = K2 P2

K2 = K1 P1 – P2

. . .= 2.138E-15 (-4.8E-15) – (-3.6E-15)

. . .= 9.38E-16 J

½mv² = 9.38E-16

v = √[2 x 9.38E-16 / 1.67E-27]

. .= 1.06E6 m/s

__________________________________

Of course the solution can be written much quicker than that. I would usually just write:

½mv₀² kqQ/r1 = ½mv² kqQ/r2

½mv² = ½mv₀² kqQ(1/r1 – 1/r2)

and then substitute values.