Consider the \"collision\" between two helium nuclei (\"alpha particles\"). Thes
ID: 1487278 • 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 = 1.5x106 m/s when they are a distance di = 3.4x10-12 m apart.
If df is the closest distance the particles approach each other, what is the ratio of the initial to final distances, di/df?
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
at a distance di
KEi = 0.5*m1*vi^2 + 0.5*m2*vi^2
PEi = -k*q1*q2/di
at a distance = d
KEf = 0
PEf = k*q1*q2/d
from energy conservation
KEf + PEf = KEi + PEi
0 + k*q1*q2/d = 0.5*m1*vi^2 + 0.5*m2*vi^2 + k*q1*q2/di
q1 = q2 = 2*1.6*10^-19
m1 = m2 = 4*1.67*10^-27 kg
9*10^9*4*(1.6*10^-19)^2/d = (0.5*4*1.67*10^-27*(1.5*10^6)^2) + (0.5*4*1.67*10^-27*(1.5*10^6)^2) + (9*10^9*4*(1.6*10^-19)^2/(3.4*10^-12))
d = 6.023*10^-14 m <<-----answer