Consider the \"collision\" between two helium nuclei (\"alpha particles\"). Thes
ID: 1772064 • 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 vi = 4.62x106 m/s when they are a distance di = 2.92x10-12 m apart. If df is the closest distance the particles approach each other, what is the ratio of the initial to final distances, d/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
mass of alpha particle m1 = m2 = 4*1.67*10^-27 kg
charge q1 = q2 = 2*1.6*10^-19 C
speed of the particle v1 = v2 = 4.62*10^6 m/s
at distance di = 2.92*10^-12 m
potential energy of the system Ui = k*q1*q2/di
kinetic energy of the system Ki = (1/2)*m1*v1^2 + (1/2)m2*v2^2
at distance df
potential energy of the system Uf = k*q1*q2/df
kinetic energy of the system Kf = 0
Uf + Kf = Ui + Ki
k*q1*q2/df = k*q1*q2/di + (1/2)*m1*v1^2 + (1/2)m2*v2^2
9*10^9*3.1*10^-19*3.2*10^-19/df = 9*10^9*3.1*10^-19*3.2*10^-19/(2.92*10^-12) + (1/2)*4*1.67*10^-27*(4.62*10^6)^2 + (1/2)*4*1.67*10^-27*(4.62*10^6)^2
df = 6.25*10^-15 m
ratio di/df = 467 <<<----------ANSWER