We have enabled User registrations.
0 votes

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?

I've gotten 2.56E6 and 2.0E6 as answers and mastering physics will not accept either one as correct. Please help!

1 Answer

0 votes
selected by
Best answer

For neatness and convenience, I'll use the notation 1.6E-19 (for example) instead of 1.6x10⁻¹⁹.


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.

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.