Consider the \"collision\" between two helium nuclei (alpha particles\"). These
ID: 1773814 • Letter: C
Question
Consider the "collision" between two helium nuclei (alpha particles"). These particles have a mass equal to four protons and a charge equal to two protons. Two helium nuclei are heading directly towards each other with the same speed, so v1 =- v2 where v1 and v2 are the velocities of the particles. A detector determines that the speed of the particles is v 2.90x106 m/s when they are a distance d, 1.59x1012 m apart If de is the closest distance the particles approach each other, what is the ratio of the initial to final distances, dy/d? Give your answer to at least three significant digits. Remember: Momentum conservation requires that v1 =-v2 during the entire "collision". Total momentum is equal to zero.)Explanation / Answer
increase in potential energy = decrease in KE
kq^2*[1/df - 1/ di] = 0.5*mv^2 - 0 = 0.5*(4*1.6726e-27)*2.9e6^2 = 2.8133*10^-14 J
[1/df - 1/ di] = 2.8133e-14/[9e9*(2*1.6e-19)^2] = 3.05*10^13
multiplying by di,
di/df - 1 = 3.05e13*1.59e-12 = 48.5
di/df = 49.5 answer