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## Homework Statement

A proton and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, each with an initial speed of 0.010c. what is their distance of closes approach, as measured between their centers?

## Homework Equations

u = proton mass = 1.67262158 × 10-27 kilograms

e = elementary charge = 1.60217646 × 10-19 coulombs

Kinetic energy Ke= 1/2 mv^2

U = Kq1q2/r

K = 9.0exp9

deltaKe + deltaU = 0

## The Attempt at a Solution

(deltaKe1 + deltaU1) + (deltaKe2 + deltaU2) = 0 where 1 represents the u mass proton and 2 represents the alpha particle.

((0-.5u(.01c)^2) + (Ke*2e/r)) + ((0-2u(0.01c)^2) + (Ke^2e/r)) = 0

solve for r =2.45 *10^-14

the answer in the textbook is 1.93 * 10 ^-14

Iv done this over and over and over and am getting this 2.45 answer. please help!!!!! it is very frustrating! thanks!!!